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0,
U(xo +
and consider the open of
x0 + yo
in
yo'E'pl'p2'.. "pn)
If
TP.
0
denotes the open neighborhood U(x0;c/2;p1,P2,...$Pn) X U(ye;c/2;P1,p2,...,pn) in the product topology on
then clearly
V x V,
(x,y) E U
implies
that
pk[x + Y - (xo + Yo)] < Pk(x- xo) + Pk(Y - Yo) (k = 1,2,...,n),
< E
from which we conclude that addition is continuous.
Similar arguments,
whose details are omitted, establish the continuity of inversion and
Q
scalar multiplication.
All the spaces described in Section 1.2 are topological linear spaces.
Now let us state some further results concerning topological linear spaces.
The proofs are reasonably straightforward and are
left to the reader.
Theorem 2.1.2.
Then for each defined by
y E V
Let
be a topological linear space over
(V,T)
and each
a E I, a # 0,
py(x) = x + y, x E V,
Ta(x) = ax, x E V,
and
:
the mappings V -' V,
ya are surjective homeomorphisms.
f.
pY: V - V,
defined by
Moreover, the
2. Topological Linear Spaces
34
r or Va
image under p the image under
Va
of a convex set is again a convex set, and
of a linear subspace is again a linear subspace.
The fact that translation in a topological linear space
(V,T)
is a homeomorphism is very useful, as it often allows us to reduce a "global" question to a "local" one.
For example, suppose that
is some collection of open sets in translation; show that
U E U
that is,
(V,T). that is invariant under
implies that
is a base for the topology
U
U
T
To
U + x E U, x E V.
it is then sufficient
to demonstrate that
U
base at the origin.
Similarly we shall see in Chapter 3 that a
contains some subset that is a neighborhood
linear transformation from
(V,T)
to
(V,T)
is continuous on
V
if and only if it is continuous at the origin. It is also worth remarking that the continuity of scalar multiplication in a topological linear space implies that every open neighborhood of the origin in Theorem 2.1.3. (i)
Let
W C V
If
V
is an absorbing set.
be a topological linear space over #.
(V,T)
is a linear subspace, then the closure of
W
is
a linear subspace.
W C V
If
(ii)
topology on
W
is a linear subspace and
induced by
then
T,
(W,T')
T'
is the relative
is a topological linear
space. (iii)
If
W C V
is a closed linear subspace, then
a topological linear space if
T'
open sets are sets of the form
is the topology on
{x + W I x E U), U E T,
(V/W,T')
V/W
is
whose.
that is,
T',
is the usual quotient topology. Theorem 2.1.4. over
#
a E A,
and let
V
Let
I.
be topological linear spaces
denote the product of the topological spaces
with the product topology
defined componentwise. over
(Va,Ta), a E A,
Then
(V,T)
T
V V.
and with linear space operations is a topological linear space
35
2.2. Finite-Dimensional Topological Linear Spaces
From
2.2. Finite-Dimensional Topological Linear Spaces.
Theorem 2.1.1 we see that the spaces &, II 1 < p
0
such that
U
will bq denoted by
E
The set
F
is said to be
of the origin in
there
V
E C aU.
It is easily seen that the definition of boundedness in a topolo-
gical linear space reduces in the case of seminormed linear spaces to that given in Definition 1.3.2. Theorem 2.4.1.
,
the gauge of
Let
(V,T)
be a topological linear space over
be a convex balanced absorbing set, and let
B C V
let
B.
q
be
Then
int(B) C Bo C B C B1 C cl(B).
(i)
i
B = B1
(ii)
(iii) (iv)
if
B
is open.
B = B
if
B
is closed.
If
V -SIR
q
is continuous, then
Bo = int(B)
and
B1 = cl(B). q
(v)
If
(vi)
Proof.
x E int(B). U
of
x
:
V - IR B
is continuous if and only if
is bounded, then
q
is a norm.
From Theorem 1.4.1 we know that Since
such that
int(B)
0 E int(B).
Bi C B C B1.
Let
is open there exists an open neighborhood
U C int(B).
Moreover, by the continuity of
2. Topological Linear Spaces
.42
la - 11
0
scalar multiplication, there exists an
an =
But let
1
- 1/n,
Then
.
1)q(x)
q(anx)
=
(anx) C :B1 C B, n = 2,3,...
and we conclude that
(n = 2,3,...),
- n < I
1
However, since
.
scalar multiplication is continuous, we deduce that Therefore
limnanx = x E cl(B).
and part (i) is proved.
B1 C cl(B),
Parts (ii) and (iii) follow immediately from part (i). q
If
:
V -+IR
is continuous, then
is open, as it is the
Bo i
that
int(B) C Bi C B
We conclude at once from
inverse image of Bo = int(B).
BI = cl(B).
A similar argument shows that
i
If as
Conversely, suppose
q(0) = 0 < 1.
an open neighborhood of we have
0
q(x) < 1, x E U.
neighborhood of that is,
Bo = int(B) i
is continuous, then
q
q
0
such that
such that, if
is continuous at
0.
and let
0 E int(B),
U c int(B).
But then, for any
0 E int(B),
entails that
e > 0, eU
x = eu E eU,
be
U
int(B) C BI,
Since
is an open q(x) = eq(u) < c;
then
The inequality
lq(x) - q(y)l < q(x - y)
(x,y E V),
which is valid for any seminorm by Proposition 1.1.1, then shows that q
is continuous, indeed even uniformly continuous, on
V.
This proves
part (v) of the theorem. Finally, suppose
B
is bounded.
If
then
x # 0,
Hausdorff, there exists an open neighborhood
U
of
0
since
T
such that
is
43
2.4. Seminorms and Convex Balanced Absorbing Sets
By Proposition 2.3.2 we may assume that
x f U.
Because
is bounded, there exists some
B
and since
if
ab < 1,
then
ab(x/ab) = x
such that
b
E abU d U,
Hence
U.
as
and this holds for
b > I/a,
By the definition of
x E bB.
ab > 1,
is balanced,
U
as
B C all,
such that
b > 0
But this implies that
x/ab E U.
contradicting the choice of any
for which
a > 0
is absorbing, there also exists some
B
Clearly then
x E bB.
is also balanced.
U
it then follows
q
q(x) > 1/a > 0.
that
Therefore it
then
x f 0,
q(x) 7 0,
and so
is a norm.
q
This completes the proof of part (vi) and the theorem.
The last portion of Theorem 2.4.1 allows us to characterize those topological linear spaces that are normed linear spaces. Theorem 2.4.2. @.
be a topological linear space over
(V,T)
Then the following are equivalent: (i)
T
Let
There exists a norm on
V
such that the norm topology and
coincide. (ii)
T
Proof.
contains a bounded convex open neighborhood of the origin.
Clearly, if there exists a norm
the norm topology coincides with
r,
then
(x
bounded convex open neighborhood of the origin.
(jx+j
such that
V
on
< 11
is a
Hence part (i)
implies part (ii).
Conversely, suppose of the origin.
B E T is a bounded convex open neighborhood
By repeating the argument used in proving Proposition
2.3.3, we may assume without loss of generality that balanced. B
B
From Theorem 2.4.1 (vi) we conclude that the gauge
is a norm on
V.
T.
First, from Theorems 2.4.1 (iv) and (v) we note that I
q(x) < 1),
q
of
It remains to prove that the topology generated
by this norm coincides with
B = (x
is also
and hence for each aB = (x
I
a > 0
q(x) < a).
we have
2. Topological Linear Spaces
44
To show that the norm topology coincides with T {aB
in view of Theorem 2.1.2, that the family
of open
a > 0)
forms a neighborhood
To accomplish this it suffices to prove that, if
base at the origin.
is any open neighborhood of the origin, then there is some
U E T a > 0
for which
since
B
aB CU.
But if
is such a neighborhood, then,
U
is bounded, there exists some where
aB c U,
Thus
I
T
neighborhoods of the origin in the topology
it suffices to show,
b > 0
such that
B C bU.
a = 1/b > 0.
Therefore part (ii)
of the theorem implies part (i).
O
Theorem 2.4.2 clearly provides us with a means of proving that certain topological linear spaces are not normed linear spaces. example, consider the seminormed linear space
(Cm([0,2n]),P),
For
where
n
P = {pn
Pn(f) =
I
E IIf(k)IL
n
k=0 To show that this is not. a normed linear space it is sufficient to
prove that there exists no norm on TP
topology and the topology
Cm([0,2n])
coincide.
for which the norm
In view of Theorem 2.4.2,
this assertion can evidently be established by showing that rP contains no bounded open sets.
Thus suppose that is, for each
E C Cm([0,2n])
n = 0,1,2,...,
is bounded and open -- that
and
E E
h E E
be fixed.
Then by the definition of
c > 0
and
i,
Pk 'pk
.
.
.
in
, p
Mn > 0
there exists an
pn(f) < Mn, f E Ef n = 0,1,2,...,
P
TP.
TP
such that
Furthermore, let there exists some
such that
n
1
U(h;c;Pk .Pk ,...,py ) C
Set
r = supj=1,2,,nkj, r > 0.
that
and so
Clearly
n
2
1
and assume, without loss of generality,
Pk (f) < pr(f), f E C '([0,2n]), j = 1,2,...,n,
U(h,e,pr) C E.
Now for each 0 < t < 2n.
b > 1
Obviously
define
hb(t) = h(t) + (c sin bt)/2rbr,
hb E Cm([0,2n]), b > 1.
Indeed, elementary
45
2.4. Seminorms and Convex Balanced Absorbing Sets
computations reveal that the kth derivative of value of
k,
depending on the
hb,
is one of the following four functions: c sin bt
h
(k)
(t) + 2rbr -k ,
h(k)(t) + c cos bt 2rbr-k
Consequently,
r pr(h - hb) =
E 11h(k)
k=0 e
2r
r k=0
-
1
br -
k
c(r + 1) 2r
< c, as
b > 1.
In particular, we can then conclude that 1hbn)11m < pn(hb.1 < Mn, n = 0,1,2,...,
for all
clearly leads to a contradiction since, when and
lI(e sin bt)/2rbr-n11.
E
b > 1.
and so
But this
n > r,
11(e cos bt)/2rbr-n1j.
as one chooses by a suitable choice of Therefore
hb E E, b > 1,
can be made as large
b > 1.
cannot be both bounded and open, and so
Tp
is
not a norm topology.
It is, however, the case that the metric topology on C ([0,2n)) determined by the metric P(f,g) =
does coincide with
E
n
pn (f - g)
n=0 2 [1 + pn(f - g)]
Tp.
(f,g E C '([0,2n]))
The details are left to the reader.
2. Topological Linear Spaces
46
We begin this section with two definitions.
2.5. Frechet SSpaces.
Definition 2.5.1. §.
Then
V
a
lim x
aor
Definition 2.5.2. linear space over then
be a topological linear space over
is a Cauchy net (sequence) in
such that
x E V
(V,T)
is said to be complete (sequentially complete) if, when-
(x ) c V
ever
Let
Let
If T
6.
there exists some
T,
= x.
(V,T)
be a locally convex topological
is metrizable and
(V,T)
is complete,
is'called a Frechet space.
V
It is obvious that every Banach space isa Frechet space. also be apparent from the following discussion that where
It will
(C"0([a,b}),[pn)),
is a Frechet space, but,
pn(f) = EZOljf(k),jm, n
as we have seen in the preceding section, it is not a Banach space.
As we know from Theorem 2.3.1, if
(V,T)
is a locally convex
topological linear space, then there exists a family of seminorms on
V
such that
(V,P)
is a seminormed linear space and
P
T = TP.
We shall see that the Frechet spaces are precisely those complete locally convex topological linear spaces such that
is countable.
P
Before we establish this we wish to prove a lemma, which is of some independent interest. Lemma 2.5.1.
such that
Let
P = (p
(V,P)
be a seminormed linear space over
is countable.
family of seminorms
Q = (qm)
such that
(i)
(V,Q)
is a seminormed linear space over
(ii)
(V,P)
and
(iii)
Proof.
(V,Q)
m = 1,2,3,...,
qm+1(x) 0
p1(x) < e/2,.. ,p m(x; < e%') c- (x
c (x and so the identity mapping
V
-n
qm (x) < s)
pI(x) < c,...,pm(x) < e),
is obviously a topological isomor-
phism.
O
We can now characterize the metrizable locally convex topological linear spaces.
Theorem 2.5.1. linear space over (i)
(ii)
V
T
is metrizable.
(V,P)
If
T
at the origin.
Let
pn
T =
TP.
(Un)
that forms a neighborhood base at
Clearly we may assume, without loss of generality,
balanced absorbing set. 2.3.3.
of seminorms on
is metrizable, then there exists a countable
Un+l, n = 1,2,3,...,
Un
P = (pn)
is a seminormed linear space and
family of convex open sets
that
be a locally convex topological
Then the following are equivalent:
There exists a countable family
such that Proof.
(V,T)
Let f.
and that each
Un
is a convex
The latter is possible because of Proposition
be the gauge of
Un, n = 1,2,3,...
.
Then from the
argument used in the proof of Theorem 2.3.1 we see that seminormed linear space and
T =
TP,
Thus part (i) implies part (ii).
where
P = (pn).
(V,P)
is a
2. Topological Linear Spaces
48
Because of Lemma 2.5.1,
Conversely, suppose that part (ii) holds.
we may assume that
Evidently the sets
n = 1,2,3,..., x E V. n,k = 1,2,3,...,
pn(x) < pn+l(x),
is such that
P = (pn)
Un k
(x
+
pn(x) < 1/k),
form at the origin, a neighborhood base for
T
consisting of convex balanced absorbing sets.
Define Pn(x - Y) P(x,Y) =
E
(x,y E V).
2n [1+pn(x-Y))
n= 1
defines a metric on
that is trans-
It is easily verified that
p
lation invariant, that is,
p(x - y,0) = p(x,y), x,y E V.
V
the metric topology and T agree it suffices, since
p
To see that is translation
invariant, to examine what happens at the origin. But let
and suppose
n > 1
pn+l(x) < 1/2n+1
pl(x) < p2(x) :ists some
B
is balanced in the
need not be balanced.
be a locally convex topological linear space and
1,2,3,...
If
.
(xn)
converges to
converges to
(1j
((i,'n)
rv,i)
int(B)
then
i
(1',Ti
dcfine-l a subset
be a seminormed linear space over
17.
letting
Letting
We have
of
0
such that p(x) < Mp, x E E (Definition 1.3.2).
E C aU (V,T)
B,B1,...,Bn
following:
I.
p E P
E
to be bounded if for each
V
fcr every open neighborhood such that
prove that
0,
0.
crave that this is equivalent to the definition that
a > 0
prove
C,
s also haIanL.ed.
U
of the origin in
V
E
there
is bounded if
there exists some
(Definition 2.4.1).
be a topological linear space over
be bounded subsets of
V,
4
and
prove each of the
2.6. Problems
53
is bounded.
(a)
Uk_1Bk
(b)
BI
(c)
aB
(d)
x + B
is bounded for each
(e)
cl(B)
is bounded.
Let
18.
is bounded.
Bn
is bounded for each
a E 4.
x E V.
be a topological linear space over
(V,T)
is a compact subset of
V,
prove that
V
Let
19.
of
E
V
i
any sequence of elements of
E, the sequence
Let
20.
21.
that converges to
V
V
is normable -- that is, there exists a norm on topology and the product topology number of spaces
V
Letting
*22.
(xn)
is
converges to
0.
Prove that every
be a family of normable topological linear
Let
letting
and
(an)
is bounded.
Prove that the topological product
spaces.
0
(anxn)
be a topological linear space.
(V,T)
Prove
4.
is bounded if and only if, whenever
is a sequence of scalars in
Cauchy sequence in
is bounded.
be a topological linear space over
(V,T)
that a subset
B
In particular,
is bounded.
B
conclude that every convergent sequence in
4.. If
a
of the spaces
V
a
V
such that the norm
T coincide --
if and only if the
that are non-zero is finite.
(V,T)
be a topological linear space over
and
W CV be a closed linear subspace, prove each of the
following:
(a)
If
V
is locally convex, then
If the topology on
V
V/W
is locally convex.
is given by the family of seminorms
(p J,
a
then the quotient topology on (Pa),
is given by the family- of scrin-mr',
where 'pa(x + W) =-infyEWpa(x + y), x E V. (b)
If
Note that, if then
V/W
V/W
is a Frechet space, then
V V
V/i'
is a Freshet sr,_
is an arbitrary, complete, locally convex spa c,
need not be complete.
2. Topological Linear Spaces
54
23.
Let
(Va,T a)
be a family of locally convex metrizable
topological linear spaces over duct
V
of the spaces
V
a
countably many of the spaces
Prove that the topological pro-
f.
is metrizable if and only if at most V
are nonzero. of
24.
A topological linear space
(V,T)
is said to be locally
bounded if there exists a bounded open neighborhood of the origin in V.
Prove that every locally bounded topological linear space is
metrizable. *25.
A locally convex topological linear space
to be bornological if, whenever V
that absorbs every bounded set in
B C V
there exists some
a 0
E f
is said
(V,T)
is a balanced convex subset of
A
V
(that is, for every bounded
such that
B C aA
whenever
jal > ja01), the origin is contained in the interior of
A.
Prove
that every metrizable locally convex topological linear space is bornological.
CHAPTER 3
LINEAR TRANSFORMATIONS AND LINEAR FUNCTIONALS
3.0.
Linear transformations and linear functionals
Introduction.
play a central role in functional analysis, and after defining these concepts, we shall examine a number of concrete examples.
Then we
shall discuss some basic results concerning linear transformations, the most important being the equivalence between th^ notions of continuity and boundedness, and the fact that the space of all contin-
uous linear transformations from a normed linear space to a Banach space can be made into a Banach space in a natural manner.
The last
section of the chapter contains a few fundamental results pertaining to linear functionals.
A characterization of continuity of linear
functionals is established, and the question of the existence of sufficiently many nonzero continuous linear functionals on separate the points of
V
V
to
This will set the scene
is considered.
for the Hahn-Banach Theorem to be discussed in the next chapter. 3.1.
Linear Transformations.
Let us begin by defining linear
transformations and linear functionals. Definition 3.1.1. A mapping
T
:
VI
V2
Let
VI
and
V2
be linear spaces over
is said to be a linear transformation
T(ax + by) = aT(x) + bT(y)
L'(V1,V2). If
logical linear spaces, those be denoted by
(V1,TI)
T E L'(V1,V2)
L(V1,V2).
55
if
(x,y E V; a,b E $).
The collection of all linear transformations from will be denoted by
b.
and
(V2,T2)
V1
to
V2
are topo-
that are continuous will
3. Linear Transformations and Functionals
56
Some authcrs use the term "linear operator' instead of "linear transformation". L'(V1,V2)
It is readily seen that space over
$
can be made into a linear
by defining
(T + S)(x) = T(x) + S(x)
(aT) (x) = aT(x)
(T,S E L'(V1)V2); x E V1; a E $).
Clearly, when
We shall always assume that this has been done. and
are topological linear spaces,
V2
V1
is a linear sub-
L(V1,V2)
L'(V1,V2).
space of
defined on all of
V1.
is always
T E L'(V1,V2)
Also it should be expressly noted that
It is, however, possible to discuss linear but this
V1,
transformations that are only defined on subsets of
introduces additional difficulties that we prefer to avoid -- for example, the definition of linear space operations in
L'(V1,V2).
It
is evident though that the domain of a linear transformation must be a linear space, in any case. V1 = V2 = V,
When for
and
L'(V,V)
V2 = $,
we shall generally write
we give the elements of
Definition 3.1.2.
Let
linear transformations in V.
If
(V,T)
L'(V,$)
be a linear space over
V
absolute-value topology, then the elements of
element in
V'
by
x';
conjugate,
Then the
$
L(V,$)
has the usual are called
V.
similarly
or dual space of
$.
will be denoted by V and 3 generic
L'(V,$)
and its generic elements by
and
are called linear functionals on
L'(V,$)
continuous linear functionals on
L(V)
a special name.
is a topological linear space and
The linear space
V1 = V
respectively; and when
L(V,V),
and
L'(V)
x*; V.
L(V,$) V*
will be denoted by
V*
will also be referred to as the
57
3.1. Linear Transformations
Obviously, since linear functionals are special instances of linear transformations, afy results that we establish for the latter objects will automatically hold for the former. Let us look at a few examples of linear transformations and As with the examples in Chapter 1, the verifications
functionals.
t
of the following assertions are left to the reader. Let
Example 3.1.1.
m x n
V1 = UP
A = (aij).
real matrix
and
V2 = U
,
and consider the
Then
(x « (xl,x2,...,xn) E IK )
T(x) = A )Cn
defines a linear transformation in
If we consider On
L'(42n,JF ).
and fl with any topologies under which they are topological linear spaces (e.g., any of the norms If
m
then
1,
T
p
,
1 < p < W)
then T E L(1
(C(f0,1]),Ij IIm).
Let
defined for each f E C([0,1])
Let
(V,P) = (C (f0,l]),(pn)),
pn(f) = IIflIn = =ollf(k)IIW (V,P)
(0 < t < 1).
where = 0,1,2,...).
is a seminormed linear space, and the transformation
defined for each
f E C "Q0,11)
1(V).
T
by
(Tf)(t) = fl(t)
belongs to
Then the trans-
L(V).
Example 3.1.3.
Then
).
by
(Tf)(t) = fp f(s) ds
belongs to
,11
T E
Example 3.1.2. formation
11-11
(0 < t < 1)
58
Linear Transformations and Functionals
3.
The elements of
Cm([0,1])*
are called distributions.
Discus-
sion of this important class of continuous linear functionals can be found, for example, in [E1, pp. 297-418; E3, pp. 46-132; Sh,
Y,
We, however, shall content ourselves with this
pp. 28-30, 46-52].
passing mention. Example 3.1.4.
Let
gical space and let For each
t
0
be a locally compact Hausdorff topolo-
X
(V,11.11.)-
E X
define
(f E C0 (X)) .
x*(f) = f(t0) Then
x* E C(X)*. More generally, if
µ E M(X),
then
x*(f) = SX f(t) dµ(t) defines an element
x* E C(X)*.
(f E C0(X))
The converse cf this last assertion
is also true, but we shall not prove it in its full generality.
We
shall, however, establish it for a spacial case in the next chapter.
let
Example 3.1.5.
Let
(V'11-11) = (M(X) ,
11 11) ,
total variation norm.
X
be a locally compact Hausdorff space and
where as usual the norm in
For each
f E C(X)
x* E M(X)*.
elements of
M(X)*
is the
define
fX f(t) dµ (t) Then
M(X)
E M(X)).
However, it is not generally the case that all are obtained in this manner.
For example, if
X
is not compact, then the-preceding formula defines an element of M(X)*
for any
f E C(X).
Example 3.1.6.
Let
(V,1I.Il) _
) CO
and let (an) E Q1.
Then
x*((bn}) =
E b
n=1
a n n
((bn3 E c0)
S9
3.1. Linear Transformations
defines an element of
Moreover all of
co.
is obtained in this
co
This is actually just a special case of Example 3.1.4 when
way.
X = (n
since in this case
n = 1,2,3,...)
I
Example 3.1.7.
(L1([-n,n],dt/2n),II.111),
Let
dt
denotes Lebesgue measure on
(C
where
o
and let
For each integer
fnnf(t)e-int
f(n) = zn
_
belongs to
we define
n
(f E L1([-n,n],dt/2Tr)).
dt
and the transformation
f E L1([-n,n],dt/2n),
(V2,II.1I2)
where
denotes the locally compact space of the
Z
integers with the discrete topology.
Then f E C0(Z),
M(X) = QI.
T,
T(f) = f,
defined by
That is, the Fourier
L(V1,V2).
transformation is a continuous linear transformation from L1((-n,n],dt/2n) into
Co(l)
to
The fact that
Co(Z).
T
L1([-n,n],dt/2n)
maps
is nontrivial and is known as the Riemann-Lebesgue
Lemma [E2, p. 36]. Example 3.1.8.
Let
and let if
p =
q, 1
1 < q < cD,
and
q =
be such that if
1
p =
(I < P < m),
(Lptµ),II Ilp
(v,11-11) = (Lp(X,S,µ),il Jlp)
where
1/p + 1/q = 1,
Then, if
g E Lq(p),
x*(f) = fX f(t)g(t) defines an element
x* E Lp(&)*.
If
If
p = 1,
if
(X,S,µ)
(Xa) C S
Lq(µ).
the formula
(f E Lp(µ))
1 < p < cc,
assertion is also valid; that is, every element of mined by an element of
q = m
then the converse Lp(N)*
is deter-
We shall prove this in Section 8.3.
then the converse assertion need not be valid.
However,
is a measure space such that there exists a family of disjoint subsets of
a-finite when restricted to
Xa,
X
with the properties that
and if
E E S
and
µ(E) < m,
µ
is
then
60
Linear Transformations and Functionals
3.
for at most a countable number of
E fl xa # 0
uous linear functional on for some
then every contin-
a,
is given by the preceding formula
L1(µ)
In particular this is true if
g E L.(µ).
(see, for example, [DS1, pp. 289-290]).
topological space and
If
X
is a-finite
µ
is a locally compact
is a regular Borel measure, then it is
µ
always the case that every continuous linear functional on obtained from some When
L1(µ)
is
[E1, pp. 215-220,239-240].
g E L.(&)
then, except in trivial cases, there always exist
p = m,
continuous linear functionals on preceding formula for, some
that are not given by the
Lm(µ)
g E L1(µ) (see [DS1, p.,296]).
3.2. Some Basic Results Concerning Linear Transformations.
The
linearity property of linear trapsformations is of fundamental importance, and its consequences will be used repeatedly in the remainder of this volume.
The reader should pay special note to the role lin-
earity plays in the proofs of the results in this section, as they are typical of the importance of linearity.
Subsequently the role of
linearity may not always be as clearly displayed as it is here. Proposition 3.2.1. and let
Let
T E L'(V1,V2).
V1
and
V2
be linear spaces over
t
Then
(i) T(0) = 0. The range of
(ii)
R(T) = {y
+
is a linear subspace of T
(iii)
Proof.
T
sinee
and hence
whenever
x E V1),
V2.
is injective, then
Clearly, if
T
Conversely, suppose that T
for some
T(x) = 0
T-1
implies
exists and
T-l
x = 0.
E L'(R(T),V1).
Since the proofs are quite elementary, we give only that
for part (iii). x = 0.
y E V2, y = T(x)
is injective if and only if
If
(iv)
T,
is injective, then x,y E V1
is a linear mapping, we have x - q
T(x) = 0
0,
that is,
implies
x = y.
x = 0.
and
T(x) = 0
T(x) = T(y).
implies Then,
0 = T(x) - T(y) = T(x - y), Consequently
T
is injective
0
61
3.2. Basic Results on Linear Transformations
Proposition 3.2.2.
linear spaces over
4
(V1,TI)
Let
(V2,T2) be topological
and
Then the following
T E L'(V1,V2).
and let
are equivalent: (i)
at
x
0 (ii)
(iii) (iv)
There exists some
T
x
E VI
such that
is continuous at the origin in
T
is uniformly continuous on It is obvious that
Proof.
(iv) implies (iii) implies (ii)
then
But then
T(x) E W2.
neighborhood of the origin in y - z E'U1,
Consequently
U2 E T2
If
V1,
W1 E TI
of
and hence, if
T(x0).
such that,
x0
is an open
U1 = W1 - x0
T(y) - T(z) = T(y -
then
is
then from Theorem 2.1.2
V2,
Thus there exists some open neighborhood x E W11
x0.
is an open enighborhood of
W2 = U2 + T(x0)
we see that
V1.
is continuous at
T
Suppose that
an open neighborhood of the origin in
such that
V1.
T E L(V1)V2).
implies (i).
if
is continuous
T
y,z E V1
are
z) E W2 - T(x0) = U2.
T, is uniformly continuous.
Therefore (i) implies (iv), and the proof is complete.
0
One of the must useful results concerning the continuity of linear transformations between normed linear spaces is the relationship between 'boundedness' and continuity given in the next theorem. Definition 3.2.1. linear spaces over
i
and
Let
and let
be bounded if there exists some
T E L'(V1,V2).
M > 0
be normed Then
T E L'(V1,V2)
is bounded, then we define
IIT1I = inf(M
I
is said to
such that
IIT(x)112 < MIIx111 If
T
IIT (x)112 < MllxjI1, x E V1)
(x E V1).
62
Linear Transformations and Functionals
3.
Let
Theorem 3.2.1.
linear spaces over
(V2,I1-i12)
T E L'(Vl,V2).
and let
i
and
(V1,11.111)
be normed
Then the following
are equivalent: T E L(Vl,V2).
(i)
T
(ii)
Moreover, if
is bounded. T
is bounded, then
IIT(x)112 IITI1
xEV
1
1
x#0 IiT(x)112
sup xEV
1
Ilx111 0
and so
such that
'
0,
liT(z)112 < 1.
IIT(y)112 < 2llylil/e.
y = 0,
T
is
is continuous at the origin in
But then for any y E V1, y
then reveal that
Conversely, suppose
is continuous.
Then, in particular,
and so there exists some
11T(x)112 < 1.
that
T
(x,y E V1)
- T(y) 112 < Mlix - ylil
1lxlll < s,
implies
set z = cy/21lylil. Simple computations
Obviously this estimate is
and therefore, taking
M = 2/c, we conclude
is bounded.
The remainder of the proof is left to the reader.
0
63
3.2. Basic Results on Linear Transformations
The notation
L(V1,V2).
a norm on
Theorem 3.2.2.
spaces over
t.
is not haphazard, as
IITII
actually defines
11.11
Indeed, we have the following result: Let
and
(V1,11.111)
be normed linear
(V2,II.112)
Then
is a normed linear'space over
(i)
(T E L(V1,V2)).
xEpV IIT(x)IIZ
IITII
where
t
1
IIXIII =1 is a Banach space over
(ii)
is a Banach space over
(V2,11.112)
(V2,11.112)
is a Banach space, let
Cauchy sequence; that is, given integer
I.
The proof of part (i) is routine and is omitted.
Proof.
ing that
such that
N
n,m > N
s > 0,
we see at once that
x E V1
Thus, since
to
V2
- T(x)112 = 0.
V2
implies that
a,b E §
IITn
- Tall < e.
Since
(x E V1),
IIx1l1
is a Cauchy sequence in
T(x),
in
V2
V2
for each
such that
We claim that the mapping
defines a linear transformation
To prove the linearity of and
Assumbe a
is a Banach space, there exists for each
some element, call it
limnlITn(x) V1
(Tn(x))
(T.) C L(V1,V2)
there exists some positive
IITn(x) - T(x)112 0
n,m > N,
Therefore
IIT(x)1I2 < MIIxIIl.
It remains only to show that
(n = 1,2,3,...),
converges to
there exists
for
n
T
in
N
such that,
then
1f n(x) - Tm(x)112 N,
IITn (x) from. which it follows by the definition of for
n > N.
That is,
and so
(T n)
converges to
that T
IITn -
TII < c,
in
is a Banach space.
Since the scalar field
f
0
can always be considered to be a
Banach space, we have the following corollary: Corollary 3.2.1. i.
Then
Let
be a normed linear space over
is a Banach space over
f.
b5
3.2. Basic Results on Linear Transformations
Note that the norm in
here is, of course, that in
V*
and not the norm in
V* = L(V,4),
despite the notation.
V,
On the
surface this may appear confusing, but it will not be so in practice, Moreover, the
since the context will make clear which norm is meant. method adopted helps to simplify notation.
It is perhaps worthwhile to write out explicitly the various ex-
pressions for x* E V*
indicated in Theorem 3.2.1.
IIx*II, x* E V*,
Thus for
we have
We close this section with two further general results about linear transformations and,normed linear spaces. Proposition 3.2.3.
linear spaces over
6
Let
(V1,I1.111)
and let
and
be normed
(V2,11.112)
T E L(V1,V2).
Then the following
E L(R(T),V1),
where
are equivalent: T-1
(i)
T-1
exists and
sidered to be a linear subspace of There exists some
(ii)
mIIxI Proof.
m > 0
l
such that
< IIT(x)I12
If part (ii) holds, then clearly
L'(R(T),V1).
(x E V1).
T(x) = 0
T-1
implies
T-1
x = 0,
exists and
Inequality (ii), however, says precisely that
IlT-1(y) II1 < (1/m) IIY112
that is,
is con-
(V2,11.112).
and from Proposition 3.2.1(iii) and (iv) we see that belongs to
R(T)
(y E R(T)),
E L(R(T),V1).
The converse implication
s apparent:
0
Linear Trar.sfor,!-atiorr
3.
66
it < I /In.
Clearly, we see from the preceding, that Proposition 3.2.4.
linear spaces over
Let
(Vilii
he nnrmed
(V,
is finite dimensiona,t, then
If
t.
and
{1)
any;
L'(V1,V2) = L(V1,V2).
Some Basic Results Concerning Linear Functionals.
3.3.
is a linear space over
then it is easily seen that
C,
be considered as a linear space over
R.
V
If
V
can also
Our first concern in this
section is to investigate the connection between linear functionals on
V
and linear functionals on
C
as a linear space over
linear space over
as a
V
This connection will be useful in discussing
R.
Second, we show that a linear functional on
the Hahn-Banach Theorem.
a topological linear space is continuous if and only if its kernel, that is,
(x
(
x E V, x'(x) = 0),
is a closed linear subspace of
V.
This is also a particularly useful result, as will be seen in Chapter 5.
Finally, we shall give an example of a topological linear space
that has no nonzero continuous linear functionals. Definition 3.3.1.
then
x'
V
If
xi = Re(x')
Let
and
V
x' = Im(x'),
are real linear functionals on
x', V.
that is,
C xi
Let
a,b E R and
x'
x,y E V.
R,
and let and
respectively, then
xi
x2
and
Moreover,
x'(x) = xi(x) - ix'(ix) = xZ(ix) + ix(x) Proof.
If
V.
be a linear space over
are the real and imaginary parts of x;
C.
considered as a linear space over
is said to be a real linear functional on
Proposition 3.3.1. x' E V'.
be a linear space over
V
Let
is a linear functional on
(x E V)
Then on the one hand,
x'(ax + by) = axi(x) + aix2(x) + bxi(Y) + bix2(y) = axi(x) + bx;(Y) + i[ax2(x) + bx2(Y)], while on the other hand, x'(ax + by) = xi(ax + by) + ix;(ax + by).
67
8.3.Basic Results on Linear Functionals
Equating real and imaginary parts, we conclude at once that are linear functionals over
x2
xi
and
IF_
For the second part of the proposition we note first that for each
x E V
we have x'(ix) = xi(ix) + ix2(ix).
But also, since
x' E V',
x'(ix) = ix'(x) = i[xi(x) + ixZ(x)] ix
- x'(x).
'(x)
Again equating real and imaginary parts, we deduce that and so
xi(ix) = -x2(x), x2(ix) = xi(x), x E V, x'(x) = xj(x)
ix '(X)
= XI (x) - ixi (ix) = x2(ix) + ix2(x).
U
The last portion of this proposition shows us how to express a linear functional in terms of either its real or imaginary part alone.
Conversely, the next proposition shows us how to define linear func-
tionals over C
by using real linear functionals.
The details are
straightforward and are left to the reader. Proposition 3.3.2. x'
Let
be a linear space over C
V
be a real linear functional on
for each
x
in
V,
then
x'
V.
is in
If
and let
x'(x) = xi(x) - ixi(ix)
V'.
The next theorem can be easily derived from the preceding two propositions, we omit the proof. Theorem 3.3.1.
C
and let
x' E V'.
Let
(V,T)
be a topological linear space over
Then the following are equivalent:
68
Linear Transformations and Functionals
3.
x' E V*.
(i)
(ii)
Re(x')
is a continuous real linear functional on
V.
(iii)
Im(x')
is a continuous real linear functional on
V.
First we
Now we turn to the second concern of this section. need a few definitions. Definition 3.3.2. W
V
Let
be a linear space over
be a proper linear subspace of
Then
V.
W
W e Wl,
maximal linear subspace if, whenever
and let
i
is said to be a either W = WI
or
W1 = V.
W c V
It is easily seen that a proper linear subspace
is
maximal if and only if it is of codimension one, that is, for any xD E V - W
the linear subspace spanned by
Equivalently, W
is maximal if and only if
Definition 3.3.3. x' E V'. of
N(x') _ (x
Then
Another common name for
N(x')
Proposition 3.3.3.
V
Proof.
N(;,')
E V - N(x'),
But if
Let
is of dimension one. and let
§
is called the kernel
N(x')
x E V,
is the null space of
be a linear space over
N(x')
then, since
x'(xo) # 0,
N(x') U (x0) we see that
[x'(x)/x'(xo)}xo) = 0.
Hence
x - [x'(x)/x'(xo)]xo E N(x'), and so
and let V.
is maximal it clearly suffices to show that, if
then the linear span of
x''(x -
I
is a proper linear subspace.
x - [x'(x)/x'(xo)]x0 E V is such that
x'.
is a maximal linear subspace of
It is evident that
To prove that 0
x E V, x'(x) = 0)
V.
x'.
x' E V', x' 4 0.- Then
x
I
V/W
is all of
be a linear space over
V
Let
[xo) U W
x E N(x') + [x'(x)/x'(x0)]xo.
Therefore the linear span of
N(x') U (xo)
is
V.
is all of
V.
69
3.3. Basic Results on Linear Functionals
be a linear space over
and let
Corollary 3.3.1.
Let
V
x',xi,x,,,...,xn be in
V'.
Then the following are equivalent:
(i)
is linearly dependent on
x'
4
xi,x...,x'.
(ii) N(x')' flk_1N(xk). The proof is left to the reader.
Next we establish the indicated characterization of continuous linear functionals. Theorem 3.3.2.
Let
x' E V'.
and let
Then the following are equivalent:
x' E V*.
(i)
is closed.
N(x')
(ii)
Proof.
Clearly (1) implies (ii), and if
is closed, and that
tion 3.2.2; c > 0
some
x'
x'
is not continuous.
W = (x
then obviously
x' + 0,
that
Then, by Proposi-
is not continuous at the origin, and so there exists
such that every open neighborhood
contains some point Let
x'.= 0,
On the other hand, suppose that
(ii)'implies (i). N(x')
be a topological linear space over
(V,T)
(
for which
x
U
of the origin
Ix'(x)I > c.
x E V, x'(x) = e/2j.
We claim that
W = N(x') + y 0
for some
yo E V.
Indeed, if
y E V - N(x'),
x'(x + [c/2x'(y)jy) = s/2, x E N(x'),
that
then we see immediately and so
N(x') + [s/2x'(y)jy C W.
Conversely, since there exist some and so that
N(x')
a E I,
is of codimension one, if
z E W,
and
z = x + ay,
x'(z) = ax'(y) = s/2. W = N(x') + yo,
where
x E N(x') Thus
such that
a = s/2x'(y),
then
and we conclude
yo = [s/2x'(y)jy.
Consequently, by Theorem 2.1.2,
W
is a closed subset of
as it is the translate of the closed subspace Thus there exists an open neighborhood
U0
of
N(x'), 0
and
such that
V,
0 f W.
70
3.
U0 n W = 0.
x0 E Uo
Now there exists some Then clearly
a = s/2x'(x0).
x'(axo) = x'([s/2x'(xo))xo) = s/2
Therefore
+al < 1,
such that
Ix'(x0)I > s.
contradict-
axo E W,
implies that
But
axo E Uo.
and so
U0 tl W = .
ing the fact that
If
U0
Moreover, by Proposition 2.3.2, we may assume that
is balanced. Let
Linear Transformations and Functionals
is continuous.
x'
x' E V'
and
x' # 0,
then Theorem 3.3.2 gives a simple and
useful characterization of the continuity of
However, there is
x'.
a more fundamental question concerning linear functionals that we have Do there exist any nonzero (continuous) linear func-
not yet faced:
.shortly that it is always the case, when
while it may very well be the case that is whether V'
and
x,y E V. x # y,
or x*(x) # x*(y)?
x'(x) # x1(y) V # (0),
does there exist an
then
V'
(0),
V
(0),
that
V*
(0).
A related question
separate the points of
V*
We shall see
V?
tionals on a given (topological) linear space
V;
T
that is, given x* E V*
or
x' E V'
V'
such that
Again we shall see that, when
separates the points of
V,
but this need not
be the case for V*.
In the next chapter we shall establish general conditions on
V
that ensure the existence of sufficiently many nonzero continuous linear functionals on
to separate the points of
V
shall see that this is always the case whenever
V
V.
Indeed, we
is a locally
convex topological linear space -- equivalently, a seminormed linear space.
Somewhat different proofs of these facts will also be given In both cases the Hahn-Banach Theorem, in either its
in Chapter 5.
analytic or geometric form, will be instrumental in establishing the desired result.
that V'
The remainder of this section is devoted to showing
always contains sufficiently many nonzero elements to
separate the points of linear space
V
V
for which
and to giving an example of a particular V*
fails to have this property.
71
3.3 Basic Results on Linear Functionals
For the sake of completeness we make the following Cefinition: Definition 3.3.4.
be a linear space over
V
Let
F C V is said to separate points if,
family of linear functionals whenever
x' E F
there exists some
x,y E V, x # y,
Then a
E..
such that
x'(x) } x'(y). V # (0)
Note, in particular, that, if points, then
only if
V'
and that
(0)
'
x'(x) = 0, x' E F,
Theorem 3.3.3.
x = 0.
be a linear space over
V
Let
Proof.
xo E V,
Let
W0 C V
linear subspace
xo
Then
x0
W.
ordering into W1,W2 E W.
We claim that there exists a
0,
'
of codimension one such that
Indeed, consider the family that
Clearly
V,
that is,
Ua
E AWa
E W
and
Moreover,
W0
V/W0
Thus for each such that
dent that
W
Consequently,
has a maximal element,
will be greater than one and so
x1 E V - W0
and for which
that properly contains
y E W0
is an upper bound for
Wa < a E AWa, a E A.
such that
Wo,
x E V
x = axo + y.
x' E V', N(x') = W01
xI
will be linearly
x0 - ax, f Wo, a E f.
easily verified that the linear span of
W0.
E AWa
is of codimension one, since if this is not the
there will exist some x0
and
Wo.
case, then the dimension of
independent of
such
is a linearly ordered subset
Ua
by Zorn's Lemma [DS1, p. 6], the family which we denote by
WI C W2,
whenever
V = (Wa)a E A
W C V
We introduce a partial
(0) E W.
as
WI < W2
Then, as is easily verified,
W.
x0 f W0.
W of linear subspaces
01
W by setting
Suppose that
of
of
{,V # (0).
separates points.
V'
W
separates
F c V'
separates points if and
F C V'
implies
and
W0 U (x1)
It is then
is an element of
thereby contradicting the maximality there exists a unique Define and
x'(x) = a.
x'(xo) = 1.
a E 4
and
It is then evi-
Linear Transformations and Functionals
3.
72
Consequently, if construction, if x' E V'
x'(y
0
0
xo = yo
such that
Therefore
E V, y
y0,z
)
0
we see that there exists some
z0 # 0,
x'(x0) =
x'(yo) - x'(z0) = 1.
that is,
1;
# x'(z
then, by the previous
# z0,
and
),
separates points.
V'
0
As we have indicated, if then it may be the case that
LJ
V
to)
V*
(0),
is a topological linear space, and so a fortiori
does
V*
A concrete example of this is provided by the
not separate points.
topological linear spaces
([0,1],dt), 0 < p < 1,
L
where
dt denotes
P
Lebesgue measure on Theorem 3.3.4.
(0,1].
L([0,1],(jt)* _ [0j, 0 < p < 1.
Suppose that
Proof.
x* E L([0,1],dt)*
f E L([0,1],dt) p
exists some s, 0 < s < 1,
set
such that
fs = k[0"] f,
and
Ix*(f)I
where
teristic function of the interval
X[0 s] and set [O,s],
x* # 0. 1.
denotes the characf` = f - fl. S
S
Clearly
fs,fs
E Lp([0,1],dt), 0 < s < 1,
and
p)p = f'l f (t) `p dt reveals that of
s
(j1fIIIp)p
(0 < s < 1)
is a continuous monotone increasing function
and that
((IfIIIp)p = 0 and
p )p
Consequently there exists some 1
('Ifs( 1i1,)p =
(1
Then there
For each
= (1Ifllp)P.
so, 0 < so < 1,
For this
so
such that
we also have
73
3.3. Basic Results on Linear Functionals
(TIES 1IP)P =
j'o .f(t) - fs (t)Ip
dt
0
0 =
If(t)Ip dt
fsl 0
If(t)Ip dt
= f1
JSo If(t)Ip dt
-
(IIfIIp)P - (Ilfs
II
p)P
0
- (Ilfllp)P
Ix*(f)I > 1,
Now since
x*
and the linearity of
we deduce via the triangle inequality
that either
Ix*(fI )I > 1/2 0
or Ix*(fs
0
Define
fl = 2f5
where
,
i =
1
or
is such that
2
0
Ix*(fs )I > 1/2. 0
It is then evident that
fl E Lp([O,1],dt), Ix*(f1)I > 1,
IIfIIIp =
2(1 - 1/P)
Repeating the argument with f2 E Lp([0,1],dt)
such that
IIf2I1 p
=
f1
Ix*(f2)I
and"
IIfIIp
in place of > 1
f,
we obtain some
and
2(1-1/p)I(flop = 22(1-1/0IIfIIP.
Continuing in this manner, we construct a sequence
(fn} CL ([O,I],dt) p
such that
Ix*(fn)I > 1 and Ilfn(Ip = 2n(1-1/P)IIf1Ip, n = 1,2,3,...
.
Linear Transformations and Functionals
3.
74
we have
0 < p < 1,
But since
1
thereby contradicting the contin-
Ix*(fn)I > 1, n = 1,2,3,...,
while
limnllfnll p = 0,
and so
- 1/p < 0
uity of- x*.
L([0,1],dt)* _ (0).
Therefore
3_4. Problems.
1.
i
(Proposition 3.2.1)
(a)
T(0) = 0.
(b)
The range of
(c)' If
T
T
be linear spaces over
V2
and
V1
Prove each of the following:
T E L'(V1,V2).
and let
Let
is a linear subspace of
is injective, then
T-1
V2.
exists and
T-I E L'(R(T),VI). 2.
(Theorem 3.2.1)
linear spaces over
IITII
3.
and let
space over
1,
Prove that
= sup{jjT(x)112
1
= sup(IIT(x)I12
1
I.
'
0)
x E V1, j1x1l1 < 1) x E V1, 11x111 = 13. and
Let
Prove that
(V2,II.112)
(L(V1,V2),II.II)
be normed
is a normed linear
where
IITII
= sup(IIT(x)112
(Proposition 3.2.4)
linear spaces over sional, then
T E L(V1,V2).
= sup(IIT(x)jj2/1{x111 I'x E V1, x
(Theorem 3.2.2)
linear spaces over
4.
4
be normed
and
Let
6.
1
Let
)'
E V1, 11x111 = 1).
(V1,11-111)
Prove that, if
L'(V1,V2) - L(V1,V2).
and
(V1,11.111)
be normed is finite dimen-
75
3.4 Problems
S.
Let
VI
and
be linear spaces over
V2
E CV
be a linear transformation, and let
let
#,
T
:
be any subset of
V2
V1 VI.
I
is symmetric.
(a)
If
E
is symmetric, prove that
(b)
If
E
is balanced, prove that
(c)
Tf
E
is convex, prove that
(d)
Give an example to show that a nonconvex set may have a
T(E)
is balanced.
T(E) T(E)
is convex.
convex image.
6.
Let
and let
T
maps bounded sets in 7.
V1
Let
T E 1'(V1,V2). that
T
y E V2.
8.
and
(V1,11-111)
T E L'(VI,V2).
E
and Let
be linear spaces over
T
A,
that is,
and
#
V1
T(x) = y, x E A,
let
and suppose for some
is identically zero.
Let
and
(V2,11.112)
spaces over
I.
composition
TS E L(VI,V3)
If
V2.
be an absorbing subset of
A
is constant on Prove that
into bounded sets in
V1
V2
be nonmed linear spaces over (V2111'112 ) Prove that T E L(VI,V2) if and only if
S E L(VI,V2)
and
(V3,11.113)
be normed linear
T E L(V2,V3),
prove'that the
and that
IITSII < IITIIIISII
I.
9. Let
Let (V1,II'111) and (V2,11.II2) lje normed linear spaces over T E L(VI,V2) be an isomorphism onto V2 and suppose
T-1 E L(V2,V1).
(a)
Prove that
(b)
If
V2
IIT-'II > IITII-I
is a Banach space, prove that
V1
is also a Banach
space.
10. with
Let
(V1,II.111)
V1 } (0).
is complete.
and
(V2,11'II2)
Prove that, if
be normed linear spaces over
L(V1,V2)
is complete, then
V2
3. Linear Transformations and Functionals
76
over
11.
Let
§,
let
be normed linear spaces
and
T,Tn E L(V1,V2), n = 1,2,3,...,
n = 1,2,3,...
If
and let
converges to (L(V1,V2),I1-II) and [xn) converges to x in iTn (xn)J converges to T (x) in (V II II2)
x,xn E 4'1,
Let
12. i
and let
(V1,P)
and
positive number
Let
c
T
:
in
T
prove that
be seminormed linear spaces over T E L(V1,V2)
Prove that
if and only
for all
q[T(x)] < c p(x)
such that
and a
p E P
there exists a seminorm
q E Q
x E V1.
he a mapping that is additive (that is,
U,.- 1k
T(x - y) = T(x) + T(y),
where
Prove
and continuous.
x,y E IR)
is linear; that is, there exists some, c EIR such that
T
that
[T n)
(V2,Q)
T E L'(V1,V2).
if for every seminorm
13.
.
T(x) = cx.
(Corollary 3.3.1)
*14.
x',xi,...,x'
let
on
xi..... x'
x E V,
16.
over
Q:
prove that
(Theorem 3.3.1) and let
x' E V'.
is linearly dependent
x'
Let
V
be a linear space over C V.
If
x'(x) = xl'(x)
Let
(V,T)
be a topological linear space
Prove that the following are equivalent:
x' E V.
(b)
Re(x')
is a continuous real linear functional on
V.
(c)
Im(x')
is a continuous real linear functional on
V.
17.
Let
V
.
be a linear space over C
If
x' E V'
and
Ix'(x)I < 1
for all
x E E.
V.
and
- ix(ix)
x' 6 V'.
(a)
subset of that
if and only if
Prove that
and
i
N(x') D 11 =1N(xk).
be a real linear functional on
xi
where
V.
(Proposition 3.3.2)
15.
let
be in
be a linear space over
V
Let
and let
Re[x'(x)] < I
E
for all
be a balanced x E E,
prove
77
3.4 Problems
Letting
18.
C
be a linear space over
V
x',y' E V',
and
prove each of the following: (a)
If
Re[x'(x)] < Re[y'(x)]
(b)
If
IRe(x'(x)]I < IRe[y'(x)]I
x' = ry' for some
20.
x' E V',
for all
= y'.
then
x E V,
be a normed linear space over (lx*II
Let
be a topological linear space over
(V,T)
If
C.
prove that
x'
x'
r E IR.
Let
19.
x* E V*,
then
x E V,
for all
= IIRe(x*)II. and let
§
Prove that the following are equivalent:
T 0.
(a)
x' E V*.
(b)
N(x')
(c)
x'
is not dense in
V.
is bounded on some subset
U c V
such that
int(U)
contains the origin.
subset
is a proper subset of
x'(U)
(d)
for some nonempty open
f
U C V.
be a Banach space over
Let
21.
be closed linear subspaces of
V.
unique representation in the form
and let
Suppose that each x = y + z,
Prove that there exists a constant
z E N.
f
K
M
x E V
with
and
N
has a and
y E M,
such that
Ilyll < Kllxll, 'x E V. Let
22.
W C V is,
V
be a linear space over
is a linear subspace of x'(W)
is a bounded subset of Let
23.
V
x' £ V'
24. (V1.I1.II1)
Let
and
V.
If
such that
x E V.-. W,
W = N(x')
f
and let
that
W
be a maximal
prove that there exists a and
x'(x) = 1.
be a normed linear space over
be its completion (see Theorem 1.1.1).
isometrically isomorphic to V.
W,
If
W C N(x').
prove that
f,
x' E V'.
and let
is bounded on
x'
be a linear space over
linear subspace of unique
V
f
f
and'let
Prove that
V*
is
78
Linear Transformations and Functionals
3.
25.
over
I.
Let
(V,7)
be an n-dimensional topological linear space is also an n-dimensional topological
Prove that
linear space.
26.
Give an example to show that
Let
V* # Vt 27.
Let
(V,II'11) = (C([0,1]),l1'II )
(a)
Let
T E L(V)
be defined by (t E [0,1); f E C([0,l])).
T(f)(t) = tf(t)/(l + t2)
Find
I`TII. (b)
Let
be defined by
S E L(V)
(t E (0,1]),
S(f)(t) = fI K(t,s)f(s) ds where
Find
K(t,s)
[0,1) x [0,1)
is continuous on
f E C([0,1]).
and
llS,I. (c)
Let
x* E V*
be defined by x*(f) = f(to)
for fixed (d)
to E [0,1) Let
y* E V*
and all
y*(f)
for fixed
*28. of
V
y E C([0,l])
and all
f E C([0,1]).
consisting of those functions
derivative of
Find
and let
T E L'(W,V) f.
(lx*It.
fp f(t)y(t) dt
Let
and second derivatives on Let
Find
f E C([0,1)).
be defined by
be defined by Prove that
f
T(f) = f", T-1
be the subspace
W
that have continuous first
and are such that
[a,b)
IJy*11.
where
exists and that
f(a) - f(b) = 0. f"
is the second
R(T) = V.
79
3.4 Problems
29.
and let
Let
family of linear functionals F = (p 9t (f) = f(t), f E C([0,1]). *30.
c* ..f1
I
Prove that
In Example 3.1.6 we saw that
tE F
F C V* [0,1]], where
separates points.
c* = 11.
and exhibit the general form of an
be the
Prove that
x* E c*.
CHAPTER 4
THE HAHN-BANACH THEOREM: ANALYTIC FORM
4.0. Introduction.
The Hahn-Banach Theorem is, together with
the Uniform Boundedness and Open-Mapping Theorems, one of the most important theorems of functional analysis.
In its analytic form,
discussed in this chapter, the theorem assures us that a linear functional on a linear subspace of a linear space that is bounded
by a seminorm can always be extended to the entire space in such a way that the seminorm boundedness is retained.
The proof of the
theorem and a number of its consequences will be given in Sections 4.1 and 4.2.
Among these consequences is the fact that there exist
sufficiently many continuous linear functionals on a nontrivial The
locally convex topological linear space to separate points.
remaining sections of this chapter are devoted to a sampling of various applications of the Hahn-Banach Theorem and its consequences, and to a proof, in the last section, of Helly's Theorem.
Although
the proof of the latter theorem makes no direct use of the HahnBanach Theorem as developed in this chapter, it does depend on a simple case of the geometric form of the theorem, and hence the proof provides a motivation for the exposition of the succeeding chapter.
4.1. The Hahn-Banach Theorem: Analytic Form. linear space over
*
and
W c V
also that
p
is a seminorm on
tional on
W
such that
Suppose
is a linear subspace.
V
and that
y'
V
is a
Suppose
is a linear func-
It
Iy'(x)I < p(x) Then can
y'
be extended to a linear functional on all of
80
(x E W). V
that
81
4.1. Hahn-Banach Theorem: Analytic Form
is also bounded by
That is, does there exist some
p?
such
x' E V'
that
(i)
(ii)
x'(x) = Y, (x)
(x E W).
Ix'(x)I < p(x)
(x E V).
The Hahn-Banach Theorem tells us that such an extension is always possible, although it is, in general, not unique. First, we
To prove this result we proceed in several stages. show how to make such an extension for linear spaces over the codimension of
IR
when
is one and then apply Zorn's Lemma (DS1, p.'61
W
to obtain the general result for arbitrary linear spaces over
II:.
t = C will then be established with the help of
The theorem when
some of the results in Section 3.3.
one.
exists some (i)
(ii)
be a linear space over W C V
and let
V,
is such that
y' E W'
If
V
Let
Lemma 4.1.1'.
seminorm on
x' E V'
p
y'(x) < p(x), x E W,
(x E V).
x0 E V - W.
Let
V
is spanned by
as
Then, since W U {xoj,
x = ax0 + y
x'(x) = ay + y1(y)
Clearly such an
x'
agrees with
W
is of codimension one,
and thus every
for some
xl,x2 E W.
and
a E IR
is a linear functional on y'.
x E V
can
y E W.
for some suitable choice of
To show that
x'
we need to look more closely at how one chooses Let
then there
x'(x) < p(x)
x'
W
be a linear subspace of codimension
such that
be uniquely expressed as
tion to
be a
p
(x E W).
we see that
y.
let
x'(x) = y'(x)
Proof.
We define
IR,
V
whose restric-
is also bounded by y.
Then
Y'(xl) - Y'(x2) = Y'(xl - x2) < p(xi - x2)
=P(xl+x0-x2-xo) P(xl + x0) + P(-x2 - xo),
4. Hahn-Banach Theorem: Analytic Form
82
and so we have -P(-x2 - x0) - y'(x2) < p(x1 + xo) x1 E W
Thus for fixed
we deduce that
(-p(-x2 - x0) - y'(x2)
is bounded above
Clearly
bl < b2.
x2 E W) C !R
and so has a least upper bound, call it
(p(xl + xo) - y'(xl)
larly
+
Let
y
Note that by the choice of
I
xl E W)
be any real number such that y
we have
To complete the proof we need to show that the (
chosen here is such that
x = axo + Y.
b2.
bl < y < b2.
(x E W).
-P(-x - x0) - Y'(x) < y < P(x + xo) - Y, (x)
the
Simi-
b1.
has a greatest lower bound
x'
x'(x) < p(x), x E V.
defined for Suppose
We consider three cases.
Case 1.
a = 0.
Then
x'(x) = y1(y) < p(y) - p(x).
Case 2.
a > 0.
Then from
y < p(y/a + xo) - y'(y/a)
we deduce
that
ay + y1(y) < ap(y/a + xo) = P(ax0 + Y),
and so
x'(x) < p(x).
Case 3.
a < 0.
Then from
-p(-y/a - x0) - y'(y/a) < y
we deduce
that
ay + y'(y) < -ap(-y/a - xo)
= P (axo + Y), and again we have
x'(x) < p(x).
This completes the proof.
0
4.1. Hahn-Banach Theorem: Analytic Form
83
A few remarks are in order before proceeding.
First we note that
the result is not more general than the one alluded to in our introIndeed, since
ductory remarks. that
is a seminorm, it is easily seen
p
if and only if
x'(x) < p(x)
The lemma was
Ix'(x)I < p(x).
stated in this seemingly stronger form since it remains true, as given, and with the same proof, if we replace the assumption that
p
is a seminorm with a slightly weaker assumption; namely, we need only
assume that
p
:
is such that
V -- IR
(1)
p(x) > 0,
(2) (3)
p(x + y) < p(x) + p(y). p(ax) = ap(x)
(a E I42;
a > 0; x,y E V) .
These remarks apply also to the next theorem and will be used in proving a geometric form of the Hahn-Banach Theorem (Theorem 5.2.1) However, when
in the next chapter. that
4 = C, we need the hypothesis
is a seminorm.
p
The ' functional given by Lemma 4.1.1 is clearly not generally unique.
Indeed, it is apparent that it is unique for a fixed
xo E V - W
if and only if
inf (p(x + x
- y'(x)) =
°)
xEW
sup { -p(- x - x) - y'(x)?.
xEW
°
Theorem 4.1.1 (Real Hahn-Banach Theorem). Let space over
let
142,
linear subspace.
be a seminorm on
p
If
y' E W'
then there exists some
V,
is such that
x' E V'
V
and let
be a linear W C V
y'(x) < p(x), x E W,
such that
x'(x) = y'(x)
(i)
(x E W).
(ii) x' (x) < p (x) Proof.
[DS1, p. 6].
be a
(x E V).
The proof is a standard application of Zorn's Lemma Let us call a pair
(u',U)
an extension of
y' provided
(1)
U C V
(2) (3)
u' (x) = y' (x)
(x E W).
u'(x) < p(x)
(x E U).
is a linear subspace such that
U D W.
4. Hahn-Banach Theorem: Analytic Form
84
Let
be the family of all extensions of
U
(y',W) E U.
(ui,U1) < Z,"2)
if
U1 C U2
is a partial ordering on
ordered subset of
U.
a
imal element, call it We claim that
by
Let
U = Ua Ua
and
V0 = V.
Clearly
V0
and let
Since
V0 # V1,
(x',V0),
V0 = V,
V1
of codimen-
z' E Vi
such
that is,(z',V1) > (x',Vo).
this contradicts the maximality of
Therefore
has a max-
be the linear space spanned
V1
is a linear subspace of
is an extension of
(z',VI)
U
If this were not true, we would get the
sion one, and so, by Lemma 4.1.1, there exists some that
is an
(u',U)
(x',Vo).
x0 E V -. V0
V0 U (x0).
where
Then
0
jy*(x)l < 1
E
then there exists
x*(x) = y*(x), x E W.
at the origin, and so there exists an in,
If
it is continuous
and seminorms whenever
4. Hahn-Banach Theorem: Analytic Form
86
Define
pk(x) < c, k = 1,2,...,n, x E W.
sup{pl(x),...,pn(x))
P(x) =
(x E V) .
c
Then, by essentially the same argument as that used in the proof of Lemma 2.5.1, we see that
Moreover, given we deduce that that.is,
x E W,
and so
p(x/a) < 41
ly*(x/a)l < 1;
However, since this holds for each
ly*(x)l < p(x), x E W.
x' E V'
x'(x) = y*(X), x E W,
and
lx'(x)l < p(x), x E V.
6 > 0
and
x E V
Furthermore, if k = 1,2,...,n,
a > p(x),
Consequently by the
there exists some
Hahn-Banach Theorem (Theorem 4.1.2) that
V.
Then from
a > p(x).
let
pk(x/a) < c, k = 1,2,...,n,
ly*(x)l < a.
we conclude that
is a seminorm on
p
is such that
sucl.
pk(x) < 6c,
then
sup(p1(x),...,Pn(x))
p(x) = and so
lx'(x)l < 6.
by Proposition 3.2.2,
< 6,
c
Thus
x'
is continuous at the origin, and so,
x' = x* E V*. Cl
Corollary 4.2.1. and let -Proof.
W
(V,P)
Let
V # [0).i Then Suppose
V*
be a seminormed linear space over
separates points.
x,y E V, x # y.
Then
xo = x - y # 0.
be the one-dimensional linear subspace of
define
y*(axo) = a, ax0 E W.
to verify that x* E V*
y* E W*.
points.
spanned by
x0
and
Using Theorem 2.2.1 it is not difficult
Hence, by Theorem 4.2.1, there exists some
such that x*(x) = y*(x),
In particular,
V
Let
x E W.
x*(x0) = x*(x - y) -
I
# 0,
and
V* separates
0
4.2. Consequences of the Hahn-Banach Theorem
over
V
and let
f
'
[0).
f.
T
(0).
be a seminormed linear space over
(V,P)
x* E V*
then there exists some
x E V, x # 0,
If
V*
Then
Let
Corollary 4.2;3.
be a seminormed linear space
(V,P)
Let
Corollary 4.2.2.
such that
x*(x) = 1.
Since, by Theorem 2.3.1, the concepts of seminormed linear spaces and locally convex topological linear spaces are equivalent, we conclude at once that Theorem 4.2.1 as well as Corollaries 4.2.1, 4.2.2, and 4.2.3 are all valid when
such a space and
V # (0).
separates points whenever
V*
In particular,
linear space.
is a locally convex topological
(V,T)
V
is
Thus we see that the topological linear where
is Lebesgue measure on
spaces
Lp([O,1],dt), 0 < p < 1,
[0,1],
are not locally convex since, by Theorem 3.3.4,
dt
Lp([0,1],dt)* = (0), 0 < p < 1.
These results are, of course, also valid in the special case in which
V
Moreover, in this case several
is a nonmed linear space.
of these results can be sharpened. Theorem 4.2.2. and let some
W C V
x* E V*
be a normed linear space over
Let
be a linear subspace.
If
y* E W*,
such that
(i) x*(x) = Y*(x)
(ii)
t
then there exists
(x E W) .
Ux*II = IIY*II
Proof.
Let
p(x) = IIY*II IIxII, x E V, IIY*II
II
p
where as usual
IY*(x)I.
X11 =1 xEW
Then
p
is evidently a seminorm on
V
and
Iy*(x)I < p(x), x E W.
Hence by the Hahn-Banach Theorem (Theorem 4.1.2) there exists some x' E V'
such that
x'(x) = y*(x), x E W,
and
I x' (x) 1 < p (x) = IIY*III1xII
(x E V) .
4. Hahn-Banach Theorem: Analytic Form
88
This, by Theorem 3.2.1, x' = x* E V* and
I1x*11 < Ily*II
Moreover, from
sup
IIY*II
114 -1
IY*(x)I °
II X11p= 1
Ix*(x)I
sup
Ix*(x)I
xEW
xEW
0,
x0 E V - W.
then there exists some
If
such that
x* E V*
x*(x) = 0
(i)
f,
(x E W).
x*(xo) = d.
(ii)
(iii)
llx*11 = 1.
Proof. Let
W0 CV be the linear subspace spanned by
W U (x 0)
and define
y* (axo + y) = ad Clearly
y*
is a linear functional on
y*(y) - 0, y E W.
W0, y*(xo) = d,
Moreover, we claim that
Indeed, note-'first that if
IIxII -
I1ax0
(a E f ; y E W).
IlY*ll
and
I.
x = axo + y, a f 0,.y E W,
then
+ YII - Ila (xo - (-Y/a) l II > laid.
Hence
Iy*(x)I = laid < Ilxil;
by the definition of
d,
that is,
given
s > 0
iiy*II < 1.' On the other hand,
there exists some
y E W
such
4.2.
Consequences of the Hahn-Banach Theorem
that
llx0 -
Set
It < d + a.
z E WO,llzll = 1.
89
z = (x0 - y)/11x0 - yll.
Then
and
Y11
d > -(-d--+--CT Since
c > 0
is arbitrary, it follows from the definition of
that lly*ll > 1,
ilr*lI
and thus llr*ll = I.
An application of Theorem 4.2.2 completes the proof.
0
Note, in particular, that Theorem 4.2.3 applies whenever a closed linear subspace and
W
is
x0 E V.- W.
A number of corollaries can be obtainel from the preceding results. We reave the proofs to the reader. Corollary 4.2.4. let
W C V
Let
d - infy E Wllxo - yll > 0, (i)
x0 E V - W.
x*(x0) * 1.
(iii)
llx*ll = 1/d.
x0 E V, x0 + 0,
(i) x*(x0)
If
then there exists some x* E V* such that (x E W) .
(ii)
(ii)
be a normed linear space over
x*(x) = 0
Corollary 4.2.S. If
(V,It.l1)
be a linear subspace, and let
be a norued linear space over
Let
then there exists some
x* E V*
1lx0II.
Ilx*II - 1.
Moreover,
sup l[x Oil - llx*ll
_l x* E V*
l x* (X ) I 0
.
such that
I.
4.
90
Corollary 4.2.6. and let
x0 E V.
Let
be a normed linear space over
(V,1j.I1)
x*(xo) = 0, x* E V*,
If
Corollary 4.2.7.
Let
let
W C V
and
x*(x) = 0, x E W,
then
W C V
imply that Let
x*(x0) = 0,
(V,1+.!)
be a linear subspace.
4,
x0 = 0.
be a normed linear space over
(V,11-11)
be a closed linear subspace, and let
Corollary 4.2.8. and let
Hahn-Banach Theorem: Analytic Form
x0 E V. then
If
4,
x* E V*
x0 E W.
be a normed linear space over Then the following are equiva-
lent:
(i)
(ii)
cl(W) - V. If
x* E V*
is such that
then
x*(x) = 0, x E W,
x* = 0.
As the reader may suspect, a number of these results for normed linear spaces have valid analogs in the context of seminormed linear spaces, (i.e., locally convex topological linear spaces).
We shall
return to this in Section 5.3 after we obtain a geometric version of the Hahn-Banach Theorem.
In the remaining sections of this chapter we shall examine various applications of the Hahn-Banach Theorem and its consequences. Other uses of this important theorem will occur in subsequent chapters. 4.3. The Hahn-Banach Theorem and Abelian Semigroups of Transformations.
We wish to establish an extension of the Hahn-Banach
Theorem asserting that linear functionals on subspaces bounded by a' seminorm can be extended to the entire space in such a way as to
preserve not only this boundedness but also the action of the linear functionals with regard to certain families of linear transformations. The statement of the next theorem will make this rather vague assertion precise.
First, however, we need a definition.
Definition 4.3.1. that
G C LT(V)
Let
V
be a linear space over
i.
Suppose
is a family of linear transformations such that
(i)
T,S E G
(ii)
TS = ST
implies
TS E G. (S,T E 'v).
4.3. Abelian Semigroups of Transformations
Then on
G
91
is said to be an Abelian semigroup of linear transformations
V.
Theorem 4.3.1. Let seminorm on
formations on Suppose
let
f,
p
be a
G be an Abelian semigroup of linear transp[T(x)] < p(x), x E V
such that
V
W C V
be a linear space over
V
and let
V,
y' E W'
is a linear subspace and
and
T E G.
is such that (x E W).
(a)
ly'(x)l < p(x)
(b)
T(x) E W
(x E W; T E G).
(c)
y'[T(x)J = y'(x)
(x E W; T E G).
Then there exists some (i)
(ii)
(iii)
x' E V'
such that
x'(x) = y1(x)
(x E W).
IX'(x)) < p(x)
(x E V).
x'[T(x)] = x'(x)
Proof.
(x E V; T E G).
As might be expected, the idea of the proof is to arrange
things so that the Hahn-Banach Theorem (Theorem 4.1.2) can be applied.
To this end we begin by defining a new seminorm on x E V
for each
V:
we define
P[TI(x)+...+Tn(x)j po(x) - inf(
)
n
where the infimua is taken over all possible finite subsets (T1,T2,...,Tn) c G.
Since
we see at once that
0 < po(x) < p(x), x E V.
verify that
p
is a seminorm and
p0(ax) - ja`p0(x), x E f,
Furthermore, let
definition of
po
x,y E V
there exist
as
and suppose
p
has this property. s > 0.
T1,...,Tn; S1,...,S®
p[T1(x)+...+Tn(x)J < po(x) + n P[S1(Y)+...+s (y)] m
p[T(x)] 0
is
Po(x + Y) < Po(x),+ P0(Y), x,y E V.
is a seminorm on
V
such that
p(x), x E V.
93
4.3. Abelian Semigroups of Transformations
Moreover, if
x E W,
T1,T2,...,Tn
then for any
G
in
we have
jy[T1(x)]+. ..+y'[Tn(x)]I
ty`()I =
n
n =
Iy'[ E T (x) k=1k n
0
and
The reader should compare the remarks following Lemma 4.1.1. G
Clearly, if V,
V
p(x + y) < p(x) + p(y),
p(x) > 0,
consists only of the identity transformation on
then Theorem 4.3.1 reduces to the Hahn-Banach Theorem. Let us look at an application of Theorem 4.3.1 to the existence
of so called Banach limits.
Consider the faiAily
sequences of complex numbers.
c
of all convergent
Then from well-known properties of
convergent sequenceg we know that to every such sequence corresponds a unique complex number
limkak
there
(akj
that has the following
properties: If
(I)
ck - aak + bbk, k = 1,2,3,..., then
[ak),(bk) E c, If
(ii)
ek = 1, k
For each
(iii)
where
(ck) E c
(ak) E c,
and
1,2,3,...,
n = 1,2,3,...,
then
(ck) E c
where
a,b E C
and
limkck = a limkak + b limkbk. then if
and
limkek = 1.
ck = ak+n, k = 1,2,3,..., limkck = limkak.
Is it possible to define a notion of "limit" for all bounded sequences of complex numbers that will satisfy all of these properties and will reduce to the ordinary limit if the sequence under consideration is convergent?
A rephrasing of this question in func-
tional analytic terms will indicate how we might proceed to answer it.
Note first that the collection of all bounded sequences of complex numbers is a Banach space
le with the norm
((akj E
11(a011- - sup (ak+ k
and that
c
is a closed linear subspace of
in Example 1.2.4. that seen,
limkak
f .
This was mentioned
The preceding discussion says, among other things,
defines a linear functional on
lli,mka.j < II(akIll., (ak) E c.
c,
and, as is easily
Thus the question we have posed is
4.3. Abelian Semigroups of Transformations
95
equivalent to asking whether there exists a (continuous) linear functional on
f
tional on
c
that satisfies the three properties of the limit funcindicated above and agrees with the ordinary limit
functional on
c.
Obviously a straightforward application of the'}lahn-Banach
Theorem (Theorem 4.1.2) with the seminorm
p(ta k)) = U(ak)l1.
(or of Theorem 4.2.2) yields an extension to e m of the limit functional on c that satisfies properties (i) and (ii). In order to
ensure that property (iii) will also hold for the extension obtained we appeal to Theorem 4.3.1.
To this end we define ck = ak+l, k Clearly Moreover,
G
T E L(Q
by
m)
T((ak)) = {ckwhere
(ak) E em,. and set G - (Tn In = 1,2,3,...). is an Abelian semigroup of linear transformations on
if
p((ak)) = jh(ax)JIm, (ak) E em,
pCT((ak))
I
t
m.
then
= JJT((uk))11m
sup Iak
k>2
sup iak
k>1 - 11(a OIL = p((ak)),
and, from property (iii) of the limit functional on then
limkTn((ak)) = limkak, n = 1,2,3,...
.
c,
if
(ak) E c,
Thus all the hypotheses
of Theorem 4.3.1 are fulfilled, and an application of the theorem immediately gives the next result. Theorem 4.3.2. (i)
(ii)
If
(ak) E c,
Given
k = 1,2,3,...,
There exists some thin
n = 1,2,3,..., then
x* E * such that
x*((ak1) = limkak. if
(ak) E Im and
x*((ck)) = x*((ak)).
ck = ak+n,
96
Hahn-Banach Theorem: Analytic Form
4.
The continuous linear functional
obtained in Theorem
t
on
x*
m
We shall take a look at a
4.3.2 is generally called a Banach limit.
different method of obtaining such functionals in Section 9.6. 4.4. Adjoint Transformations.
Suppose
are nontrivial seminormed linear spaces over 4.2.2 we know that
(V2,P2)
Then from Corollary
4.
There exists a simple, and
Vk # (0), k = 1,2.
quite useful, relationship between transformations certain elements in ' L'(V2,Vi).
and
(V1,P1)
T E L(V1,V2)
and
To be precise we make the following
definition:
Definition 4.4.1.
linear spaces over
4,
Let
(V1,P1)
and
let
then we define a mapping
and
Vk f (0), k = 1,2.
T* E L'(V*,V*)
T E L(V1,1'2),
(x1 E V1; X2 E V*),
2
2
being called the adjoint of
T.
It is easy to verify that the definition of
an element of
If
by setting
T*(x* )(xl) = x2[T(xl)]
T*
be seminormed
(V2,P2)
T*
indeed defines
L'(V*,V-).
With the aid of the Hahn-Banach Theorem we can easily establish several results about the adjoint when the spaces
V1
and
V2
are
normed linear spaces. Theorem 4.4.1. spaces over
Let
and suppose
4
(i) T* E L(V*,V*) (ii)
T*
and
T E L(V1,V2).
and
(V2,11.112)
Then
IIT*I1 = IIT11
is injective if and only if
R(T)
(V2.II.112). Proof.
Let
x1 E V1
and
be normed linear
x*2 E V.
Then
IT*(x2)(x1)I = Ix*2[T(xl)II
IIx*llliTUIIxllll,
is dense in
97
4.4. Adjoint Transformations
Thus from Theorem 3.2.1 we see that T* E I,(VZ,V,*) and IIT*II < IITII. Moreover, suppose c > 0 and let xl E VI be such that IIx1II = 1 and IIT(x1)II2 > IITII - a, which is possible by the definition of IITII. Then by Corollary 4.2.5 there exists some x2 E V2* such that 11x211 = 1 and x2[T(x1)] = IIT(xl)II2' But then we have and 30
IIT* (x2) II < IITII IIx2I1.
IT*(x2)(x1)I = Ix2[T(x1)]I
(IT(x1)II
-a
> IITII from which we conclude at once that since
IIx2II = 1,
IIT*(x2)II > IITII
this entails also that
s > 0 is arbitrary, we obtain
IIT*II > IITII
IIT*II > IITII.
However,
- a.
- a,
and since
Thus Theorem 4.4.1(i)
is proved.
In order to establish the equivalence in Theorem 4.4.1(ii) us first suppose that T* it suffices, by Corollary that
we see that
6ince
T*(x2) = 0.
be such that
xl E V1,
shows that
Therefore
T*
is dense in
T*(x2) = 0.
Then
vanishes on
x2
via the continuity of
However, given such an
then x2 = 0.
R(T)
and so
(V 2'11-112 )
x2 = 0..
and let
x*[T(xl)] = T*(x2)(xl) = 0,
R(T),
from which we deduce,
and the denseness of
x
is such
x2 E V*
is injective, it follows that
T*
x2 E VZ
is dense
R(T)
T*(x2)(x1 ) = x2[T(xl)] = 0, xl E V1,
Conversely, suppose
x2 = 0.
4.2.8, to prove that, if
0, x2 E R(T),
x2(x2)
x2 E V*,
To show that
is injective.
let
R(T),
that
is injective, and this completes the proof.
Theorem 4.4.1 easily yields the following corollary.
The
details are left to the reader. Corollary 4.4.1.
Let
a normed linear space over that
(V1,II'II1)
I,
IIT(x)112 > =11x111, x E V1,
are equivalent: (i) 'T (ii)
T*.
is surjective. is injective.
be a Banach space and
and suppose for some
T E L(V1,V2)
m > 0.
is such
Then the following
4.
98
Hahn-Banach Theorem: Analytic Form
We shall return to the notion of the adjoint in subsequent secA
tions, particularly when we discuss Hilbert spaces in Chapter 13.
similar idea will also appear in the investigation of reflexivity of normed linear spaces in Section 8.1. 4.5. Separability of bility of
V*
V
at least in the case that
is a
We recall the following definition:
Definition 4.5.1.
Then
V,
implies that of
normed linear space.
i.
Here we wish to show that the repara-
V*.
(V,T)
Let
be a topological linear space over
is said to be separable if it contains a countable dense
V
subset.
Theorem 4.5.1. If
is separable, then
V*
Proof.
(xn) C V and
be a normed linear space over
Let
Let
is separable.
V
be a countable dense subset of
(x*J
4.
be so chosen, using the definition of
Ixn(xn)I ? IIxnII/2, n = 1,2,3,...
D C V
Let
.
IIxnII,
V*
and let
that
1
IIxnII
denote the family
of all finite linear combinations with (complex) rational coefficients of (xn).
of
V
Clearly
is countable.
D
spanned by
(xn).
Thus it is apparent that dense in
Obviously
is dense in
D
W denote the linear subspace
D C W
particular,
x* E V*
D
is dense in
if and only if
is such that
x*(xn) = 0, n = 1,2,3,...
.
W
W.
is
- x*II.= 0.
limkllxn
x*(x) a 0, x E W.
Since
there exists some sequence from the set
such that
and
V
V.
Suppose then that
V*,
Let
(xn)
(xn),
However,
In
is dense in call it
(x* k
k
Ilxnk - x*II
=
(xnk - x*) (x) { IIxII p
1 I
I (x*nk - x*) (xnk ) I
= Ixn (xnJI k
> Ilxn II/2 k
k
(k - 1,2,3,...),
99
4.6 Annihilators
from which it follows that
= 0.
linkllxn k
implies that
limkllxn
11
=
But
k
and so llx*ll = 0,
lIx*II,
Consequently from Corollary 4.2.8 we conclude that is dense in
Therefore
V.
- X*" = 0
limkllxn
11
that is, x* = 0. W,
and hence
D,
is separable.
V
It should be noted that the converse of this result need not he For example,
valid.
li = .f
is not.
CD
is a separable Banach space, whereas
fI
In this section we wish to introduce the
4.6. Annihilators.
notion of annihilator and to use it for describing the dual spaces of subspaces and quotient spaces of normed linear spaces.
It will be
apparent when we discuss Hilbert spaces in Chapter 13 that the concept of the annihilator is a natural generalization to normed linear spaces of the notion of an orthogonal complement in Euclidean spaces and in inner-product spaces in general. Let
Eppfinition 4.6.1.
over
f.
If
E C V,
E1
then
(x*
x* E V*, x*(x) = 0, x E E)
I
is called the annihilator of (E*)j W (x
I
on
V
EL C V*
E.
If
E* C V*,
then
x E V, x*(x) - 0, x* E E*)
is called the annihilator of Thus
be a topological linear space
(V,T)
E*.
is the set of all continuous linear functionals
that vanish identically on
E,
and
(E*)1 C V
common zeros of the continuous linear functionals on. V to
E*.
It is evident that
E'
and
(E*)1
is the set of that belong
may be rather trivia],
for example, if "V* _ (0).
The proof of the next proposition is straightforward and is left to the reader.
Hahn-Banach Theorem: Analytic Form
4.
100
Proposition 4.6.1. over
f,
and
E C V,
El C V*
is a normed linear space, then
V
If
E* c :V*.
Then
is a linear subspace.
F.l C V*
(i)
(ii)
be a topological linear space
(V,T)
Let
is a closed
linear subspace. (iii)
is a closed linear subspace.
c V
(E*) A.
(iv) (v)
E C (E)t.
1
V T (0),
is a normed linear space,
V
If
E CV
and
El # (0).
is a proper linear subspace, then
Among other things, the precedirg proposition says that E C (El)l
whenever
The next result gives us a condition
E c :V.
under which the containment is equality. Theorem 4.6.1.
W C V
If
be a normed linear space over
Let
is a linear subspace, then Since
Proof.
c,t!W` C (W,)l.
W C (WL)t
But if
and
cl(W) = (WL)1 is closed, we see that
(Wl)1
X0 E (W')1 - cl(W),
then,
since
cl(W)
i- a closed lirear subspace, we see from Theorem 4.2.3 that there x' E V*
exists some x t cl(W;. that
such that
In particular,
x*(xo) = 0, T'aerefore
x*(xo) # 0
x* E W.
x*(x) - 0,
and
However,
x
0
E (Wl)1
thereby contradicting the choice of
implies
x*.
cl(W) _ (W). C)
Corollir n
and let
42.6.1.
V
Let
(V,11.11)
be a norm-ad linear space over
be a linear subspace.
Then the following are
equivalent: (i)
(ii)
W
is a closed linear subspace.
W - (W1)1
Next we shall use the concept of an annihilator to describe the spaces of continuous linear functionals on the subspaces and quotient spaces of normed linear spaces.
101
4.6. Annihilators
First we make a general definition. Definition 4.6.2. linear spaces over isometry if
Let
(Vl,ll-ji
y
A mapping
1.
be normed
and
)
V1 - V2
:
is said to be an
Such a mapping is also said to
IIcP(x)ji2 = Wxljl, x E V1.
be isometric.
Theorem 4.6.2. and let
W C V
be a normed linear space over
Let
be a closed linear subspace.
jective isometric isomorphisms between (V/W)*
and
Proof.
If
cp
cp
V*/Wl - W*,
Moreover, suppose
is an isomorphism.
x*(x)
and between
y*(x), x E W.
defined by It is easily seen
is well defined.
Theorem 4.2.2, there exists some and
W*,
defines a continuous linear functional :
cp(x* + WL)(x) = x*(x), x E W, that
and
then clearly the equation
x* + Wl E V*/WL,
Thus the mapping
W.
V*/Wi
Wi.
(x* + Wl)(x) = x*(x), x E W, on
I
Then there exist sur-
x* E V*
y* E W*.
such that
IIx*ll = IIY*II
0
there exists k
in
I
we have
113
4.10. Helly's Theorem
n E 1akxkllllxs11
11
n
< (M + e) c > 0
Since
!J
E akxk+) k=1
is arbitrary, we conclude at once that part (i) implies
part (ii).
Conversely, suppose part (ii) holds.
ck = 0, k = 1,2,...,n,
If
then part (i) is trivial on taking
xg = 0.
Thus without loss of
generality we may assume that some
ck
Moreover, we claim that
.we may assume that xi,x*,...,x*
be a maximal linearly independent subset of
xi,x2,...,x*, m < n,
where we have for convenience possibly renumbered the
xi,x2,.... xn,
and
ck.
lix11 < M + c
are linearly independent on the foi-
Suppose they are linearly dependent and let
lowing grounds:
xk
0.
Then, given and
c > 0,
suppose
xa E V
xk(x`) = ck, k = 1,2,...,m.
then there exist some
bkj E I
is such that
Now, if
m < k < n,
such that m
xk =
E
j=1 bkjxj
and hence m
xk(xel =
j=1 j]
E bk x*(xQ) =
m
E bk c j=1 j
.. 7
However, from part (ii) of the theorem we see that
Ick
-
m
m
E bkjcj1 < M I+xk -
E -1
j-1 and so
0,
xk(xe) = ck, k = m + 1,...,n.
This shows that we may assume, without loss of generality, that xi,x*,...,xn
are linearly independent.
114
Hahn-Banach Theorem: Analytic Form
4.
Finally, we assume for the purposes of the proof that
E = gt.
This is purely a matter of convenience, and the same arguments mutatis
mutandis as given below apply in the case T
We define a mapping
by setting
V - JE
:
4 = C.
(x E V).
T(x) _ (x1*(x),x2(x),...,xn(x))
Clearly
is linear.
T
Moreover, we claim that
is surjective by
T
being a proper
the following,argument:
If
linear subspace of 1,
is contained in some linear subspace of It
of codimension one.
Thus there exist
T(V),
then
T(V) # n p,
a1,a2,...,a
in
akxk(x) = 0, x E V.
of which are zero, such that
not all
IP.,
But this
E'
clearly contradicts the linear independence of hence
T
xi,x2,...,xn
Denote the standard basis vectors in a by ek = k
(ekl,ek2,...,ekn),
Let
xk E V
e > 0,
set
K. = {x
and so T(K9)
I
x E V, (jxjj < M + cj.
Obviously
K,,
Furthermore, we claim that
has a nonempty interior that contains the origin in Indeed, let
be = (M + c)/(n supk=l,2,...,nhIxk(I).
< be, k = 1,2,...,n,
n
E akxkI1 < E Iak{ k=1 k=1 Thus, if
Then, if
we have
n
IIxk11 < M +
6-
thenEnk=1 a x
Iakl < be, k = 1,2,...,n,
k k
n=1
akxk) = q=1 akek E T(K,).
Consequently
open ball about the origin of radius T(K,)
that is,
¢ 1, k, ekj = 0 if j ekk be such that T(xk) = ek, k = 1,2,...,n.
is convex and balanced.
T(K9),
ek;
where
1,2,..,n.
Given
and
is surjective.
be
in
E Ke,
T(K,)
(IFS, f
IIm) ,
and so
contains the
and so
has a nonempty interior that contains the origin. Parenthetically we remark that
T(K,)
having a nonempty interior
follows at once from the Open-Mapping Theorem (Theorem 7.2.1) on noting that
T
is continuous.
115
4.10. Helly's Theorem
xa E Ks
Recall now that we wish to prove the existence of some
Suppose no such
(cl,c2,...,cn). T(K9)
that is,
xk(xs) = ck, k = 1,2,...,n;
such that
exists.
xa
T(x6) = c =
Then
T(K6).
c
Since
is a convex balanced set with a nonempty interior that contains
the origin and
there exist
T(K9),
c
bl,b2,...,bn
in
IR
such that
n (a)
(b)
E b c k=1 k k
I
> 0.
n E bkxk(x)I
0,
x0 E V -. W.
then there exists some
Prove
x* E V*
such that
(b)
x* (x) = 0 x*(xo) = 1.
(c)
IIx*I)
(a)
1/d.
(x E W) .
Let
(Corollary 4.2.S)
3.
over
117
Problems
4.11.
Prove that, if
f.
be a normed linear space
(V,11-11)
x0 # 0,
x0 E V,
then there exists some
such that
x* E V*
(a)
x* (x0) = llx0II
(b)
llx*Il -
Moreover, prove that
lixll= o
sup
lIx*lE
x VI
0
*
such that for no
xo # 0,
X0 *(X) = ilxoll
x E V,
= 1,
IIx(I
and an
V
is, give an example of a normed linear space
xo i V*,
is it true that
.
be a normed linear space over f, be a linear subspace, and let x0 E V. Prove that
let
Let
5.
inf yEW
llx o -
Let
6.
YII = sup(I x* (x0) I
(V,7)
and let
x0 E V,
exists on x E V.
that
Prove that the dual result to Problem 3 is not valid;
*4.
p
x* E V*
I
W C N (x*) )
x* E V*, lix*Il = 1,
be a topological linear space over be a continuous seminorm on such
hat
x*(x0) s p(x0)
let
f,
Prove that there
V.
and
WCV
Ix*(x)I < p(x),
Use this to give another proof of Corollary 4.2.1.
7.
Let
(V,P)
be a seminormed linear space over
be a closed convex balanced subset of there exists an
x* E V*
such that
V.
4
and let
Prove that for any
x*(x0) > I
x0
E E
Ix*(x)1 < 1,
and
xEE. 8.
Let
(V,P)
be a seminormed linear space over
be a convex balanced neighborhood of the origin in 0 E int(U).
such that
Prove that for any
x0 f U
V,
there exists an
f
and let
that is,
x* E V*
U
4. Hahn-Banach Theorem: Analytic Form
118
x*(xo) > sup lx*(x)I.
xEU (Corollary 4.2.6)
9.
over
(Corollary 4.2.7)
W C V
@, let
Let
then
and
W C V
and let
4
be a normed linear space
(V,11-11)
x*(x) = 0, x E W,
xo E V.
implies that
xo E W.
(Corollary 4.2.8)
11.
over
for all
be a closed linear subspace, ahd let
Prove that, if 'x* E V* x*(x0) = 0,
x*(xo) = 0
xo = 0.
then
10.
over
Prove that, if
xo E V.
and let
!,
x* E V*,
be a normed linear space
Let
be a normed linear space
Let
Prove that the follow-
be a linear subspace.
ing are equivalent: (a)
cl(W) = V.
(b)
If
Define
x* E W*
of
to all of
into
IR
so that
*l4
then
x* = 0.
Give two different extensions
with the same norm as
I xlJ < 1).
such that
ax + by
and
V
x*[(x,y,z)] = x.
II23
x*.
be a normed linear space over
B1 = {x i x E V, B1
by
Let
13.
x*(x) = 0, x E W,
and let W= ((x,y,z) I x+ 2y= z= 0).
Let
12. x*
is such that
x* E V*
are in
f(ax + by) = af(x) + bf(y) B1.
Prove that
4
and let
let f be a mapping from
Furthermore, f
whenever
x, y,
can be extended to all of
f E V'.
Let
V
be a linear space over
IR
and let
P C V
be such
that
Then
(a)
x,y E P
(b)
x E P
P
mean that
and
and
a,b > 0 -x E P
imply
is called a convex cone. y - x E P.
imply
Prove that
ax + by E P.
x = 0.
For
0.
be a normed linear space over
Let
*15.
using the embedding of such that
V
in
1.2.2).
prove that there is a set
V**,
the bounded functions on
X
.(see Example
X
is a'separable normed linear
Conclude that, if
space, then
Without
I.
isisometrically isomorphic with a subspace of the
V
Banach space
is isometrically isomorphic to a subspace of
V
III) .
(tm. II
16.
from C
be a Banach space over
Let
into- V,
then
if
If
f.
is a mapping
x
is said to be analytic on a domain
x
Cl C C
iim iix(z + h) - x(z)iI
h--0 exists for every on
imply
x > 0
and
can be extended to
y'
imply
fl P # 0 if and only if (-x + W) fl P # ¢. x E W
we
x E V,
such that for all
V
be a linear subspace of
W
Let
Q,
then
z E 0, z + h E 0.
the mapping
is analytic on Q is entire if for all
h
x*(x)
and some
bounded entire function, then *17.
Let
and
C,
M > 0.
U = int(D).
define' IIfIIE = sup(jf(z)l
I
Let
f(z) = 1
(b)
Let
W
for each
z E U
IIx(z)II < M
V
: C
is a
C,
let
r denote
If
f E C(D)
and
E C D,
z E E). =0
akzk,
a0,...,an are in
where
be a polynomial. Prove that IIf1}U = IIfIIr. for any f E C(D).J
[Note that
be any linear subspace of
polynomials and is such that
x
x
is a constant.
x
the unit circle, and let
(a)
The mapping
is bounded if
x
Prove that, if
denote the closed unit disk in
0
is analytic
x
[x*(x)](z) = x*[x(z)]
in the usual complex-variable sense.
is analytic on
x
z E C
and
x* E V*
If
: C -- C defined by
IIfIIU - IIfIIr
there exists some
C(D)
for all
µz E M(r)
f (z) - f r f (C)
(C)
C,
IIfIIU = IIftlD
that contains all f E W.
Prove that
such that
(f E W).
120
Wr of
consider the subspace
[Hint:
on
Hahn-Banach Theorem: Analytic Form
4.
r
that are restrictions to
tional
y' E WI.
defined by
consisting of all functions
C(I')
of members of
I'
y1(f) = f(z)
and the func-
W,
for fixed
z E U.
Use
Theorems 4.1.2 and 4.8.2.1 (c)
For each
n E Z
un(C) = Cn, C E D.
define
Prove that
rInjein6I
Jr un(C) where
z = reie E U.
(d)
For
(Note that
0 < r < 1
un E W, n = 0,1,2,... .)
define
Pr(s) _
rlnleins
}+
n =
Compute ETTn
Pr (0
2n
-
t)eint dt
and compare the answer with
fl- un(C) where
z = rele E U. (e)
Since, as seen in (d), we have
fr f(C) dµz(C) = 2L f"n f(elt)Pr($ - t) dt whenever
f = un, n E Z,
and
z = reie E U,
and since every f E C(F)
may be approximated uniformly by trigonometric polynomials, it can be shown that
f(z) = Jr f(C) de _
where
z = reie E U.
(C)
fnn f(elt)Pr(e - t) dt
2n
Prove that
Pr(8 - t) =
1 1
- r2
- 2r cos(0 - t) + r2
(f E W),
Problems
4.11.
P
121
To summarize, we have proved that,
is called the Poisson kernel.
r
if
W
is a linear subspace of
and is such that
that contains all polynomials
C(D)
(1filU = (IflLr., f E W,
then
for each z = reie E U
the Poisson integral representation
I -
f(z) - 2n fnn
18.
Let
f E W.
denote all the bounded subsets of
B(IR)
4.3.1 to prove the existence of a set function E E B(1R)
f(eit) dt
- 2r cos (e - t) + r
1
is valid for each
r2
p
IR.
Use Theorem
defined for all
with the following properties:
If E,F E B(IR) "and E A F = 0, then p(E U F) = p(E) + p(P). (b) If E E BOR) , then p (E + t) = p (E) , t E IR. (c) If E,F E 8(IR) and E C F, then p(E) < p(F); (a)
(d)
If
E E 8(a)
Lebesgue measure of 19. (V 2'11.112)
is Lebesgue measurable, then
is the
E.
(Corollary 4.4.1)
Let
(V1,j.111)
a normed linear space over
is such that
p(E)
JjT(x)112 > mllxl,l, x E VI,
be a Banach space and
and suppose
f
for some
m > 0
L(V1,V2) .
Prove that
the following are equivalents: (a)
T
(b)
T*
20.
Let
is surjective. is injective.
(V1,11.111)
linear space over
f
be a Banach space and
and suppose
JIT(x) 112 > m1Jx111, x E VI,
T E L(V1,V2)
for some m > 0.
be a nonmed is such that
Prove that
(R(T) '11-112)
is a Banach space. Letting (V1,11.111) a:.d (V2,l1.j12) be Banach spaces over and S,T E L(V1,V2), prove each of the following: 21.
(a)
(aS)* = aS*, a E I.
(b)
(S + T)* = S* +
(c)
If
T-1
If
(V3,
(d)
(e)
q(T) = T*,
defined by 22. f
then
the family
T*(V2) c V*
L(V*,V*),
be normed linear spaces over is injective if and only if
T
separates the points of
V1.
be a normed linear space over
Let
23.
and
#
is an isometric isomorphism.
cp
Prove that
T E L(V1,V2).
and let
to
L(V1,V2)
and
(V1,11.l1)
Let
then
(AT)* = T*A*.
is the mapping from
T(T)
If
is dense in
R(T)
is also a Banach space over
then
A E L(V2,V3),
T*.
exists and
(T 1), _ (T*) 1.
Analytic Form
Hahn-Banach Thecrem:
4.
122
T E L(V).
Suppose that for some
T(x) = ax
and
x E V
where
T*(x*) = bx*,
x* E V*
and
Prove that
a # b.
a,b E #,
and let
#
we have
x* (x) = 0. VI . iF,
Let
24.
be represented by an Prove that of
V2 = e,
T E L(V1,V2).
and
real matrix
m x n
T
can
(aij) (see Example 3.1.1).
is represented by the matrix
T*
Then
the transpose
(aji),
(aij).
Let
25. defined by
(V.11-11) = (co, I1
)
and let T E L(V) be the mapping
T((an)) _ (an/n), (an) E co.
and find an expression for 26.
f
E C V,
let
#,
R(T)
is dense
T*.
(Proposition 4.6.1)
space over
Prove that
Let
and let
(V,T)
be a topological linear
E* CV*.
Prove each of the
following:
(a)
El C V*
(b)
If
V
is a linear subspace.
is a normed linear space, then
E-L
c V*
is a closed
linear subspace.
C V
(c)
(E*)
(d)
E C (E')i.
(e)
If
1
V
is a closed linear subspace.
is a normed linear space,
i proper linear subspace, then
E1 # (I
.
V T (0),
and
E C V
is
123
4.11. Problems
(Theorem 4.6.2)
*27. !
be a nonmed linear space over
Let
W CV be a closed linear subspace.
and let
Prove that there
exists a surjective isometric isomorphism between
be a topological linear space over
*28.
Let
(V,T)
E,F C V
and
E*,F* C V*.
(a)
Prove that, if
E C :F,
(b)
Prove that, if
E* C F*,
(c)
If
prove that
t.
and
F
I.
Let
then
(F*)1 L- (E*) 1.
are closed linear subspaces and
E = F.
Fl
El
I
and a subset
V
for
E* C V*
[(E*)1J1 # E*.
29.
Let
Let
T E L(V1,V2), let' T* E 1(V2,V*I)
N(T) = (x
be normed linear spaces over
(V1,Il-ll1)- and
x E V1, T(x) = 0)
I
be the adjoint of
N(T*) _ (x*
and
and
N(T*)
T,
let
x* E V2, 1*(x*) = 0).
I
N(T)
W.
and
Fl C EL
then
Give an example of a space
(d)
which
E
(V/W)*
are called the kernel or null space of
T
and
T*.
Prove each of the following:
(a) (cl[R(T)])1 = N(T*) (b) cl[R(T)] - [N(T*)]1. (c)
{cl[R(T*)])1 . N(T).
(d)
cl[R(T*)] C [N(T)]L,
spaces, then
*30.
Let
and
R(T)
f,g E L1(IR,dt)
are Banach
and
h ( Lp(IFt,dt),
Prove each of the following:
(b)
f * h E L (IR,dt). llf * hilp < IIfIII Ilhlip.
(c)
f * (g*h)= (f *B) *h.
(e)
a (f * h)
(f)
£ * h = -h * f.
(a)
V1
cl[R(T*)] _ [N(T)]'.
(Proposition 4.7.1)
1 < p < *°.
and if
(d) .(f+g) *h =f *h+gh. (af) * h = f * (ah) , a E C.
Moreover, prove that, if
f E Lp(IR,dt)
and
1/p + l/q = 1, 1 < p < m, then f * h E C(IR)
h E Lq(Il3,dt),
and
Ilf * hllm < Ilfllpllhllq
124
Finally, if
f E L1OR,dt)
f E Lp(1R,dt)
Analytic Form
Hahn-Banach Theorem:
4.
h E Lp(IR,dt), I < p < m,
and
or
h E Lq(IR,dt), 1/p + 1/q - 1,- prove that
and
Ts(f) * h = f * Ts(h) = Ts(f * h), s E IR,
where
Ts(g)(t) = g(t - s),
t E IR.
Let
31.
defined on
be a complex-valued function of bounded variation
g
[0,1)
and define x*(f) = fl f(t) dg(t)
Prove that
x* E (C([0,1]))*. (Theorem 4.8.2)
*32.
(f E C([O11])).
Let
be a loca))y compact Hausdorff
X
Prove that the following are equivalent:
space.
(a)
x* E C0(X)*.
(b)
There exists a unique
µ E M(X)
such that
x*(f) = fX f(t) dµ(t) Prove that the correspondence between tive isometric isomorphism between
be a sequence in
(xk)
C0(X)*
V
such that
and
defines a surjecM(X).
limkllxkll - 0.
sequence in
i
any
there exists some finite set
M > 0,
p
and
be a normed linear space over
Let
33.
x*
(f E C0 (x)) .
that does not converge to zero.
and let
!
(ak)
Let
be a
Prove that, given
(b1,...,bn)
of scalars
such that
n
n
E bleak I > rl{I E bkxkll k=1 k=1 *34. E
(cf
Let
(V,T)
be a topological linear space over
be a convex balanced compact subset of :
x0 E E
f E F)
be a family of scalars.
such that
f(xo) = cf
for all
V.
Let
i
and let
F C V*
and let
Prove that there is some f E F
if and only if
4.11.
Problems
125
n
n
E a c
k=1 k k for all choices of where
ck - cf k
.
sup(I E a f (x)
f1,...,fn
k=1 k k in
F
x ( E)
ard scalars
a1,...,an
in
I
CHAPTER S THE HAHN-BANACH THEOREM:
Introduction.
5.0.
GEOMETRIC FORM
In the preceding chapter we discussed at
some length the Hahn-Banach Theorem, as well as its consequences and The form of the theorem we proved was an analytic one,
applications.
involving the possibility of extending linear functional;-.
However,
the Hahn-Banach Theorem also enables us to obtain a considerable amount of geometric information about topological linear spaces, much of which is generalizations of well-known theorems concerning Euclidean spaces.
We now wish to turn to an exposition of this
geometric aspect of the Hahn-Banach Theorem.
Our treatment here
will not be as long nor as detailed as the preceding one. After introducing the concepts of linear variety and hyperplane, and establishing some basic results concerning such objects, we shall prove a geometric form of the Hahn-Banach Theorem.
assert that, if and
Lo
K
This theorem will
is a convex absorbing set in a linear space
is a linear variety disjoint from
K,
V
then there exists a
real hyperplane
L
one side of
We shall use the Real Hahn-Banach Theorem (Theorem
L.
that contains
L
0
and is such that
K
lies on
4.1.1) to prove this theorem and shall show that the two theorems are equivalent.
Subsequently we shall use the geometric form of the Hahn-Banach
Theorem to reestablish some of the consequences of the Hahn-Banach Theorem discussed in Section 4.2, and to extend some of those results, which we have proved only for normed linear spaces, to locally. convex topological linear spaces.
We conclude the chapter in Section
5.4 with the statement, without proof, of several additional results of a geometric nature.
126
127
5.1. Linear Varieties and Hyperplanes
To begin with we wish
S.I. Linear Varieties and Hyperplanes.
to introduce the notions of linear variety and hyperplane in a linear space, and to give some elementary results concerning such objects. I
can always be considered
Throughout this chapter, whenever we speak
IR
a linear space over
over
V
We recall that a linear space
V
of, for example, a real linear subspace of tional on
V,
we mean a linear subspace of
linear space over linear space over
IR
V1 considered as a
or a linear functional on
IF..
Thus, for instance,
as a linear space over The linear space
or a real linear func-
V = C
considered as a
can be thought of
and it is then the same space as
IR,
W = IFi
V
over,
IR
IR2.
is clearly a real linear subspace
of V = C. Definition 5.1.1. L C V
Let
be a linear space over
V
I.
A ,set
is said to be a (real) linear variety if there exists a (real)
linear subspace
W C V
(real) linear variety
and some L
x0 E V
such that
L = x0 + W.
A
is said to be a (real) hyperplane if W
is a maximal (real) linear subspace.
Thus a (real) linear variety is a translate of a (real) linear subspace, and a (real) hyperplane is a translate of a maximal (real) linear subspace.
Clearly in the case that
# = lR
hyperplane and real hyperplane are identical.
the notions of
Recall also that a
maximal linear subspace is by definition (Definition 3.3.2) a proper linear subspace.
It is easily seen that every linear variety is a real linear variety, but not conversely.
For example,
IR
subspace, and hence a real linear variety, in linear variety in
C.
Also,
IR C C
plane that is not a hyperplane.
is a real linear C,
but
IR
is not a
is an example of a real hyper-
An example of a hyperplane that is
not a real hyperplane is provided, for instance, by C C C2. There is an intimate connection between hyperplanes and linear functionals, as shown by the following proposition:
S. Hahn-Banach Theorem: Geometric Form
128
Proposition 5.1.1.
be a linear space over
V
Let
0
Then the following are equivalent:
L C V.
is a (real) hyperplane.
L
(i)
Therexists a (real) linear functional
(ii)
and some (real) number
such that
a E f
Moreover, the (real) hyperplane only if
L = (x
on
x'
Suppose
is a hyperplane, say
L
is a linear subspace of codimension one. (x1) U W
L =.x
+ W,
spans
Let
x1 E V -r W.
defines a linear functional on then obviously
L = (x
that is,
(
other hand, if
L = x
0
L = (x
x1 E V
the kernel of
Clearly, if
x'(x) = ax'(x1) = a,
shows that
and
I
and
and so then
x E L,
and
L
The arguments when
a = 0
x'(x) = a),
such that x'.
a = x'(x0),
If
if and only if
where
x E L.
x' E V', x' # 0.
x'(xI) = 1.
Let
xo = ax1
From Proposition 3.3.3 we know We claim that
x = xo + y, y E N(x'), Thus
then
x0 + N(x') C L.
On the
x'(-xo + x) = -x'(xo) + x'(x) = -a + a
-xo + x E N(x'); that is,
+ N(x'),
Then
(c E I; y E h')
x' f 0.
is a maximal linear subspace.
L = x0 + N(x').
W
contains the origin.
Conversely, suppose
N(x')
V
x'(x) = a),
W = L
Then there exists some W = N(x'),
where
and the formula
V,
x' (cxl + Y) = c
that
0,
contains the zero vector if and
L
0
x0 E W,
V,x'
x ( V, x'(x) = a).
I
a = 0.
Proof.
and
and
x E x0 + N(x').
Hence
is a hyperplane.
L
is a real hyperplane or
x'
is a real
linear functional are essentially the same and are left to the reader.0
The description of hyperplanes in terms of linear functionals will be quite useful in the succeeding development.
129
5.1. Linear Varieties and Hyperplanes
Some other elementary Eroperties of linear varieties and of convex sets are collected in the next two propositions.
Their proofs
are left to the reader. Proposition'S.1.2. over
t.
Let
be a topological linear space
(V,T)
Then
(i)
If
is a (real) linear variety, then
L C V
is a
cl(L)
(real) linear variety. (ii)
closed or
If
L C V
is a (real) hyperplane, then either
L
is
cl(L) a V.
Clearly, in the case of real linear varieties and real hyperplanes we are considering
Proposition 5.1.3. (i)
If
K C V
as a topological linear space over
(V,T)
Let
be a linear space over
V
(ii)
Then
is convex, then the following sets are convex:
aK, a E #; x + K, x E V; T(K), {x .
#.
If
K1,K2 C V
If
{Kct J
(
T(x),E K), T E L (V).
are convex, then
K1 + K2
and
KI
- K2
are convex. (iii)
E A
is a family of convex sets in
then
V,
aor is convex.
(1 K
(iv)
If
E CV,
then
(1
K
is convex, and if
KI C V
is
KDE
K convex
convex and
then
K1 D E,
KI D
(1
K.
KDE K convex
The last portion of the proposition says that the intersection of all the convex sets that contain a set set that contains
E
is the smallest convex
E.
Note also that a (real) linear variety is a convex set.
S. Hahn-Banach Theorem: Geometric Form
130
Before continuing, another definition is necessary. Definition 5.1.2.
he a real linear functional on
x'
the sets (x
be a linear space over
V
Let
[x
x'(x) > a], (x
I
x'(x) < a)
I
plane
L = (x
I
A set
x'(x) = a).
E C V
if
F
half-spaces determined by
L,
and
if
is said to lie strictly on
E
lies on one side of
E
is said to lie on one side
L = (x
x*(x) = a)
I
and
L
Obviously half-spaces are convex sets. linear space and
x'(x) > a),
is contained in any one of the
L
L
I
a EIR and
are called half-spaces determined by the real hyper-
of the real hyperplane
one side of
Then for each
V, x' # 0.
x'(x) < a), (x
I
and let
0
V
If
E n L = 0.
is a topological
is a real hyperplane,
x*
being a real continuous linear functional, then the first two halfspaces indicated in the preceding paragraph are open sets and the latter two are closed.
The following proposition is evident from the definition: Proposition 5.1.4.
Let
for some (i)
E c (x I
(ii)
E C (x I
a E IR
E
and let
V, x' # 0,
or
F C (x
I
L
or
E C (x
L = (x
I
let
x'
x'(x) = a)
if and only if either
x'(x) > a).
lies strictly on one side of
x'(x). < a)
let
t,
Then
E C V.
lies on one side of
x'(x) < a) E
be a linear space over
V
be a real linear functional on
I
L
if and only if either
x'(x) > a).
A necessary and sufficient condition for a convex set to lie strictly on one side of a real hyperplane is provided by the next theorem.
Theorem S.I.I.
Let
V
be a real hyperplane, and let
be a linear space over K C V
be convex.
are equivalent: (i)
(ii)
K
lies strictly on one side of
K n L = 0.
L.
f,
let
L c V
Then the following
Linear Varieties and Hyperplanes
5.1.
The validity of part (i) implying part (ii) is established
Proof.
linear functional on
V,
let
L = (x
If
K
x'(x) = a).
However, since
continuous function of
b
b - 1
b
x'(x2)
and
bo, 0!< bo < box I
at
on
bxI + (1 - b)x2 E K, 0 < b < 1, (1
- b)x'(x2)
is clearly a x'(xl)
taking the values
[0,1],
at
Consequently there exists some
0.
x'[b0xI + (1 - bo)x2] = a.
such that
1,
L,
we see that there exist
K fl L = D
is convex,
K
be such that
x'(x1) < a < x'(x2).
x'[bx1 + (1 - b)x2] = bx'(x1) +
and so
be a real
x'
a E IR
does not lie strictly on one side of
for which
xl,x2 E K
let
and let
x' # 0,
then from Proposition 5.1.4 and some
K n L a 0,
Conversely, suppose
by definition.
!
131
Thus
contrary to assumption.
+ (1 - b0)x2 E K fl L,
Therefore part (ii) of the theorem implies part (i), and the proof is complete.
We conclude this section with the following proposition: Proposition 5.1.5.
over
and let
i
one side of
int(E)
(ii)
cl(E)
(iii)
E C V
lies on
then
lies strictly on one side of lies on one side of
V,
let
x'
T 0,
L.
L.
Using Proposition 5.1.1, let
Proof.
tional on
int(E) # p,
If
is closed.
L
(i)
be a topological linear space
(V,T)
be a real hyperplane.
L C V and
L
Let
and let
be a real linear func-
x'
a E IR
be such that
L - Lx
x'(x) = a).
Without loss of generality, we may suppose that
E c (x
x'(x) < a),
as otherwise we merely need replace
and
a by If L
Ll
{x
I
x'
by
-x'
-a.
is not closed, then by Theorem 2.1.2 the hyperplane x'(x) = a + 1)
S.1.Z we see that at once that
L1
is also not closed, and from Proposition
is dense in
int(E) n L1 # p,
int(E) c (x.( x'(x) < a).
Hence
V.
Since
int(E) # 0,
contradicting the fact that L
is closed.
it follows
S.
132
Hahn-Banach Theorem: Geometric Form
Moreover, from the proof of Proposition 5.1.1 we see that the real hyperplane choice of
L
is of the form
x0 + N(x')
continuous by Theorem 3.3.2. int(E) C (x
one side of
I
Thus
is closed, and so
{x
is
x'
is an open set,
x'(x) < a)
I
shows that
x(x) < a)
is a homeo-
V
Consequently, since translation in
xo.
morphism, we see that N(x') = -xo + L
and
for some suitable
lies strictly on
int(E)
L.
(x
Part (iii) of this proposition is now apparent, as
I
x'(x) < a)
is closed.
5.2. The Hahn-Banach Theorem:
Geometric Form.
The geometric
content of the Hahn-Banach Theorem is that certain kinds of sets The precise
can be separated from one another by real hyperplanes.
meaning of this will become clear in the statement of the theorem and in some further geometric consequences to be mentioned in Section 5.4.
There are various other equivalent geometric formu-
lations of the theorem in addition to the one presented here (see, for example, [DS1, pp. 412, 417, and 418; El, pp. 116-118; KeNa,
pp. 22 and 23; T, pp. 142 and 151; W1, pp. 46-51, 219
and 2203).
Before we continue to the statement and proof of the Hahn-Banach
Theorem in its geometric form we wish to introduce what could be called the gauge of a convex absorbing set.
Since, however, we have
already used the term "gauge" in another context (Section 1.4) in
dealing with convex balanced absorbing term again here.
Definition 5.2.1. K C V
Let
V
be a linear space over
be a convex absorbing set. rK(x) = inf(a If
K
I
Then for each
K,
x E V
we set
a > 0, x E aK).
and it would be a seminorm on
In general, though,
and let
$
were a.convex balanced absorbing set, then
just the gauge of tion 1.4.2.
sets, we shall not use the
We content ourselves with the following definition:
rK
V
rK
would be
by Prafosi-
need not lle a seminorm.
5.2.
133
Geometric Form
Hahn-Banach Theorem:
However, it is easily seep by the same arguments as those used in the proof of Proposition 1.4.2 that
has the following properties.:
rK
(x E V).
rK(x) > 0 (2) rK(0) - 07.
(1)
(x,Y E V).
rK(x + y) < rK(x) + rK(Y)
(3)
(a E 0; a > 0).
rK(ax) - arK(x)
(4)
We shall use the next lemma in the proof of the Hahn-Banach Theorem. Let
Lemma 5.2.1.
be a linear space over
V
be a convex absorbing set. that
V
If
and let
§
is a real hyperplane such
L c V
then there exists a real linear functional
K f1 L =
K C :V
on
x'
such that L = (x
(i)
'
x'(x) = 1). (x E V).
-rK(-x) < x'(x) < rK(x)
(ii)
Since
Proof.
is a real hyperplane, there eld sts, by
L
Proposition'S.1.1, a real linear functional
an
a E tR such that
0 E K,
as
K
I.
= (x
'
Since
Then
Let
x/b E K,
L.
Moreover,
x E V
0.
I
,
b > 0
x'(x/b) < 1; b,
Consequently for each
x'(x) < 0,
we conclude that
x E V
x'(x) = 1). K
as
is such that
that is,
we also have
and
x' = y'/a
Hence
L = (x
such that
K C (x
and suppose
and hence
this holds for any such
and thus
'
K fl L.= J, from Theorem 5.1.1 we see that
on one side of x'(0) - 0.
V
a
and
0,
y' (x) = a).. Since K fl L =
is absorbing, we see that
is a real linear functional on
V, y'
on
y'
lies strictly 0 E K
and
x E bK.
x1(x) < b.
Since
x'(x) < rK(x), x E V. x'(-x) < rK(-x),
-rK(-x) < -x'(-x) = x'(x), x E V.
We can now state and prove a geometric form of the Hahn-Banach Theorem.
0
be a linear space over
V
Let
Theorem 5.2.1.
such that L = (x
I
and let
is a linear variety
then there exists a real hyperplane
K fl Lo = 0,
where
x'(x) = 11,
j
Lo C V
If
be a convex absorbing set.
K C V
Geometric Form
Hahn-Banach Theorem:
5.
134
is a real linear functional on
x'
V,
such that
(i) L0 C L. (x E
x'(x) < rK(x)
(ii)
K C'(x
(iii)
x'(x) < 1).
I
Since
Proof.
is a linear variety, it is a real linear
Lo
W0 CV and
variety, and,so there exists a real linear subspace x0 E V
some if
xo E W0, Let
Lo C W
x0 f W0,
since
contrary to hypo-
K n Lo T 4>,
(x0) U W0 is a maximal real linear subspace of W.
W0
is a real hyperplane in. W.
Lo
W,
Moreover,
be the real linear subspace spanned by
and
Now consider of
and so
Lo = W0
then
W C V
thesis.
Thus
L0 = x0 + W0.
such that
Clearly
V).
It is easily seen that
K fl W.
is a convex absorbing set.
computed on the linear space
as a subset
rKflW, restricted to W.
rK
is precisely
W,
K A W,
Furthermore, we claim that
This is evident from
rKnW(x) = inf(a
(
a > 0, x E a(K fl W))
= inf(a
a > 0, x E aK f1 W)
= inf(a
a > 0, x E aK)
(x E W) .
= rK(x) The second of these equalities
subspace.
Finally, we note that
is valid since
W
(K fl w) fl L. = 4>,
is a real linear
as
K Cl Lo = 0
Hence, by Lemma 5.2.1, there exists a real linear functional on
W
such that
Lo = (x
I
x E W, y'(x) = 1)
and
.
y'
y'(x) < rK(x),
x E W.
Next, appealing to the Real Hahn-Banach Theorem (Theorem 4.1.1) and the remarks following Lemma 4.1.1, we deduce the existence of"a
Hahn-Banach Theorem:
5.2.
real.-linear functional
x'
135
Geometric Form
on
V
such that
x'(x) = y'(x), x E W,
x'(x) < rK(x), x E V.
and
.Let
L = (x
that is,
x'(x) = 1).
Moreover, if
Lo C L.
and
1
K C (x
Thus, when
'
Then
is a convex absorbing set and
is a linear
Lo
there exists a real hyperplane
variety such that
K fl L. = 0,
that contains
and is such that
0
rK(x) < 1,
then
x E K,
V,
x'(x) < 1;
and so
x'(x) < 1).
K
L
is a real hyperplane in
L
K
lies on one side of
L.
L
The
improvement on this provided by the next corollary is precisely what
was used in the last portion of the proof of Helly's Theorem (Theorem 4.10.1).
Corollary 5.2.1. I
K C V
and let
Let
(V,T)
be a topological linear space over
be a convex absorbing set for which
then there exists a closed real hyperplane where
x*
(i) L0 C L. (ii)
(iii)
I
x*(x) = 1), V,
such that
.
int(K) C (x cl(K) C {x
x*(x) < 1). x*(x) < 1).
Immediate from Proposition 5.1.5 and Theorem 5.2.1.
Proof.
Note, in particular, that, if set, then
L = (x
is a continuous real linear functional on
int(K)
K
K
0
is an open convex absorbing
lies strictly on one side of the real hyperplane
L.
Incidentally, it is not necessary to use the analytic form of the Hahn-Banach Theorem to prove the geometric form, as we just did.
But we prefer doing it this way, rather than take the time to develop the machinery necessary for an independent proof.
A proof without
the use of the analytic form of the theorem can be found, for instance, in [WI, pp. 46-51, 219, and 220].
We have alluded at several points to the fact that the geometric form of the Hahn-Banach Theorem (Theorem 5.2.1) is equivalent to the
136
Geometric Form
Hahn-Banach Theorem:
S.
We shall now show that this is indeed the case when
analytic form.
§ = a The equivalence of the analytic form of the Hahn-Banach Theorem over an arbitrary field
§
(Theorem 4.1.2)
and Theorem
5.2.1 is an easy consequence of the theorem we present next. Since we shall only consider
we shall drop the adjective
* = 1R,
"real" for hyperplanes in the statement and proof of the theorem. Theorem 5.2.2.
be a linear space over
V
Let
Then the
1R.
following are equivalent: Suppose
(i)
(k E V) ,
and
K c V
Suppose
a linear variety such that
If
is such that
y' E W'
x' E V'
such that
x'(x) < p(x), x E V. is a convex absorbing set.
If
Lo C V
is
then there exists a hyper-
K fl Lo
L=(xJ x'(x)=1), x'EV', such that L0CL, KG(xI x'(x) 0),
then there exists some
x'(x) = y'(x), x E W,
y'
is such that
be a linear subspace.
y'(x) < p(x), x E W,
and
IR
p(x + y) < P(x) + p(y) p(ax) = ap(x)
W C V
(ii)
V
:
P (x) > 0
(a) (b) (c) and let
p
Lo = (x
I
is a
K = (x
+
x E W, y'(x) = 11. W,
x E V, p(x) < 11,
Then, by Proposition 5.1.1,
and so a linear variety in
used in proving Proposition 1.4.1, we see that absorbing set in
V.
Since
V.
If
then, by the same arguments as-those K
y'(x) < p(x), x E W,
is a convex.
we have
K f) L
o
= .
Hence, applying part (ii), we deduce the existence of a hyperplane L = (x x E V, x'(x) = 1) in V, where x' is a linear functional I
on
V,
such that
137
Geometric Form
Hahn-Banach Theoref:
5.2.
Lo C L, K C (x
x'(x) < rK(x),
and
x.' (x) < 1),
x E V. Now, if
x E V,
then from the definition of
we see that
K
a > 0, x E aK)
x'(x) < rK(x) = inf(a = inf(a
a > 0, x/a E K)
= inf(a
a > 0, p(x/a) < 1)
= inf(a
a > 0, p(x) < ai
= P (x) and, moreover, this also shows that
x'(x) < p(x), x E V,
Thus
Kf1L=f,. x'(x) = y'(x),
The proof will be completed once we show that x E W.
Recall that
in
as is
W,
Lo = (x
L fl W = (x
x E W, y'(x) = 1)
I
x E W, x'(x) = 11.
I
x'
Clearly Lo C L fl W,
Lo C :L.
Indeed, since
as
to
W.
We claim that Lo = L 4) W.
L n W
and
Lo
The latter is obvious
is a linear functional on
W
since the restriction of
is a hyperplhne
are hyperplanes, then, as can be
seen, for example, from the proof of Proposition 5.1.1, there exist xl,x2 E W
Lo = x1 + N(y')*
such that
where, as before, we consider and compute shews that since
N(x') N(y')
and
N(y')
and so
L (1 W.
from which we deduce at once,
are linear subspaces, that But then
-x1 + x2 E N(x'),
N(y') = N(x')
both spaces are maximal linear subspaces of W.
Hence
since W,
and
Lo = xl + N(y') C x2 + N(y') _
Thus to show the opposite containment it clearly suffices to y'(x2) = 1.
Suppose
y'(x2) = a.
If
a = 0,
xl + N(y') C x2 + N(y') = N(y'),
xl E N(y').
a ' 0.
W
x1 + N(y') = Lo C L (1 W = x2 + N(x')
C -x1 + x2 + N(x'), N(x')
L () W = x2 + N(x'),
to de a linear functional on
x'
Now
is not identically zero on
prove that
so
W.
-x1 + x2 + N(x') = N(x').
N(y') C N(x'), x'
in
and
then clearly
x2 E N(y')
and
from which we conclude that
But this contradicts the fact that
y'(xl) = 1.
Thus
Hahn-Banach Theorem:
5.
138
But then
x2/a E Lo c L fl W.
Consequently
y'(x2) a 1
Therefore
and
Lo = L fl W.
z' = x'
Lo
that
- y'
on
Since
W.
we see at once that z' (x) = x' (x) - y' (x) -l- 1 = 0, x E Lo;
Lo c N(z').
that is,
a = 1.
that is,
x'(x2/a) = 1/a = 1;
Next consider the linear functional
Lo = L fl w,
implies that
y'(x2/a) = I
and
x2/a E W,
Geometric Form
' N(z').
Moreover, since
and
Lo
0
Furthermore, we claim that
we see
0 E N(z'),
Indeed; on
N(z') = W.
the one hand we have Lo - Lo = x1 + N(y') - [xi + N(y')]
= N(Y') - N(y') = N(y'),
and on the other hand,
Lo - Lo C N(z') - N(z') = N(z').
Hence as
N(y') c N(z').
y'(x1) = 1,
However, since
N(y') # N(z').
we see that
maximal linear subspaces of
xI E Lo C N(z')
W,
x1 f N(y'),
and
Since the kernels are
we must then conclude that
But this last assertion says precisely that
N(z') = W.
x'(x) = y'(x), x E W,
This completes the proof.
so part (ii) implies part (i).
0 x'
and
is worthwhile noticing for itself.
Its
The argument used in the preceding proof to show that y'
agreed on the space
W
content is precisely the next proposition. Proposition 5.2.1. L C V
Let
V
be a linear space over
I
and let
be a hyperplane that does not contain the zero vector.
x',y' E V1
are such that
x'(x) = y'(x), x E L,
then
x'
If
- y'.
5.3. Some Consequences of the Hahn-Banach Theorem Revisited.
In
section 4.2 we discussed many consequences of the Hahn-Banach Theorem and used the analytic form of the theorem
(Theorem 4.1.2) to prove
5.3. Consequences of the Hahn-Banach Theorem
139
Now we wish to reconsider some of these results and show how
them.
the geometric form of the Hahn-Banach Theorem (Theorem 5.2.1) can be used in their proofs. Theorem 5.3.1.
linear space over
and
x0 E V, x0 T 0,
But, if
x*(xo) # 0.
that
4
be a locally. convex topological
(V,r)
V T (0).
Then
separates points.
V*
Because of linearity, it is clearly sufficient to-show
Proof.
that, if
Let
then there exists some x0 T 0,
then, since
convex, there exists a convex open neighborhood x0
x* E V*
such
is locally
$
of
K
0
such that
As always, such a neighborhood is absorbing since scalar
K.
multiplication is continuous. such that
int(K) = K # 0,
Thus
and
K
K
is a convex absorbing set Lo
11, Lo = Q, - %here
= (xoi.
Therefore, by Corollary 5.2.1, there exists a closed real hyperplane L = (x
I
y*(x) = 1) = x0 + N(y*),
real linear functional on
V,
where
y*
is some continuous
K fl L = . However, by
such that
Proposition 3.3.2 and Theorem 3.3.1, if x*(x) = y*(x) - iy*(ix) then
x* E V*
and
x*(xo) = Y*(xo) - iy*(ixo)
Corollary 5.3.1.
linear space over
f.
Corollary 5.3.2.
linear space over then
x
0
4,
Let If
Let
(V,T)
V # (0), (V,T)
and let
0. \;
O
be a locally convex topological then
V* # (0).
be a locally convex topological
x0 E V.
If
x*(xo) = 0, x* E V*,
- 0.
Corollary 5.3.3.
linear space over let
(x E.\V ),
x0 E V.
x*(xo) = 0,
If
then
Let
(V,T)
be a locally convex topological
let
W C V
be a closed linear subspace, and
x* E V* x0 E W.
and
x*(x) = 0, x E W,
imply that
exists a convex open neighborhood K = -x
Let
0
absorbing set, and
of
U
is closed, there
is an open convex
K
K fl Lo = 0.
is a linear variety such that
Lo
..
U (1 W
such that
x0
Then
L0 = -x0 + W.
and
+ U
W
Then, since
x0 f W.
Suppose that
Proof.
Geometric Form
Hahn-Banach Theorem:
S.
140
Consequently, by Corollary 5.2.1, there exists a closed real hyperL = (x
plane
functional on y*(-x
and so
y*(x
But, if
x E W,
then
)
0
is a continuous real linear
y*
L0 CL and
such that
V,
= 1,
0
where
y*(x) = 1),
I
as
-1
)
-x
-x0 + x E L0
In particular,
K fl L = p.
0
E L
0
= -x
0
+ W G L.
and
Y*(x) = Y*(-x0 + x + x0) = Y*(-xo + x) + y*(xo) = 1 that is,
W,
x0
Consequently, if
W C N(y*).
structed a real continuous linear functional y*
then we have conon
V
such that
This, however, contradicts the
y*(x0) # 0.
and
y*(x) = 0, x E W,
1 = 0;
-
hypotheses of the corollary.
Since
W
by
x* E V*
Indeed, define
x*(x) = y*(x) - iy*(ix), x E V.
is a linear subspace'of
V
x*(x) = 0, x E W.
see at once that
and But
y*
vanishes on
we
W,
x*(x0) = y*(x0) - iy*(ix0) # 0,
contrary to assumption.
Therefore
x E W. 0
Corollary 5.3.4.
linear space over
i
Let
(V,T)
and let
be a locally convex topological
W C V
be a linear subspace.
Then the
following are equivalent:
(i) cl(W) = V. (ii)
If
x* E V*
is such that
x*(x) = 0, x E W,
then
x* = 0.
5.4. Some.Further Geometric Consequences of the.Hahn-Banach Theorem.
As already mentioned, the geometric form of thb Hahn-Banach
Theorem also provides us with some information about when various sorts of sets in a topological linear space can be separated from one another by a real hyperplane.
We would now, like to state a
141
5.4. Geometric Consequences of the Hahn-Banach Theorem
The proofs, which can be based on our
sampling of such results.
Details and
previous work in this chapter, are left to the reader.
further results can also be found in [DS1, pp. 409-418; E1, pp. 116144; K, pp. 186-188, and 243-245; KeNa, pp. 22 and 23; W1, pp. 46-51, 219 and 220].
E1,E2 C V. E2
if
L C V
real hyperplane
A
and
E1
be a linear space over
V
Let
Definition 5.4.1.
E2
is said to separate
EI.
and
lie on different sides of
L,
and it is said
and
E2
lie strictly on
to strictly separate different sides of
and let
0
and
E1
if
E2
EI
L.
Thus, in view of Propositions 5.1.1 and 5.1.4, we see that and
L = (x
are separated by
E2
real linear functional on
E1
and
E1 C (x
if
E2
x'(x) = a),
EI C (x
or vice versa.
x(x) > a),
E2 C (x
if
V,
I
x'(x) < a)
i
x'
strictly separates
L
E2 C :(x
I
x'(x) > a),
5.2.1) says that, if K is a convex absorbing set and linear variety disjoint from that separates
and
and
K
Lo
is a
then there exists a real hyperplane
K,
Corollary 5.2.1 says that, if
Lo.
then there exists a closed real hyperplane that separates
int(K) # 0, K
Lo
One can also separate pairs of disjoint convex sets, provided they meet certain requirements. Theorem 5.4.1. Let §
and let (i)
plane
L (ii)
or
The geometric form of the Hahn-Banach Theorem (Theorem
vice versa.
L
is some
and
x1(x) < a)
Similarly and
where
E1
K1,K2 C V If
(V,T)
be nonempty disjoint convex sets.
int(K1) # 1,
that separates If
and
KI
real hyperplane
L
K1
K2
then there exists a closed real hyperand
K2.
are open, then there exists a closed
that strictly separates
Corollary 5.4.1. linear space over
be a topological linear space over
§,
Let, (V,T) let
K C V
K1
and
K2.
be a locally convex topological be a nonempty closed convex set,
142
x0 E V.
and let plane
x0 f K,
If
linear space over Then
K
that contain
Let
(V,T)
and let
§
Geometric Form
then there exists a closed real hyper-
that strictly separates
L
Corollary 5.4.2.
set.
Hahn-Banach Theorem:
S.
and
K
[x
0).
be a locally convex topological
K CV be a nonempty closed convex
is the intersection of all the closed half-spaces K.
A final result, analogous to Proposition 5.1.5, is the next proposition.
Eroposition 5.4.1. oveil
that
and
§
K1,K2 C V
and let
int(K1) # . then
K2,
Let
L
If
L
be a topological linear space
(V,T)
be nonempty disjoint convex sets such is a real hyperplane that separates
K1
is closed.
We shall use some of the results from this section when we discuss the Krein-Mil'man Theorem in Section 11.2. 5_5. Problems.
I.'
(Proposition 5.1.2)
space over
§,
If
(a)
Letting
(V,T)
be a topological linear
prove each of the following:
is a (real) linear variety, then
L C V
cl(L)
is a
(real) linear variety. If
(b)
closed or
2.
is a (real) hyperplane, then either
L C V
L
is
cl(L) = V.
(Proposition 5.1.3)
Letting
V
be a linear space over
$,
prove each of the following: (a)
If
K C V
If
(Ka)
is convex and
T E L'(V),
then
(x
I
T(x) E K)
is convex. (b)
is convex.
is a family of'convex sets in
V,
then
fl
K
a s
143
Problems
S.S.
(c)
E e V,
If
co(E) =
let
Prove that
K.
fl
co(E)
KDE K convex K1 e V
is convex, and if
(V,T
Let
3.
I
x*(x) < a)
(a)
cl(U) = H.
(b)
int(H) = U.
4.
Let
and let
If
Let
d
let is the disd = 1/ilx*U.
be a topological linear space
(V,T)
Lo CV is a linear variety such that
there exists a closed real hyperplane x*
If
from the origin, prove that
L
x*
and prove that
x*(x) = 1).
I
If
let
a E IFS,
K CV a convex absorbing set for which
and
#
and
x*(x) < a)
I
L = (x
(Corollary 5.2.1)
5.
V
be a normed linear space over §,
tance of the hyperplane
over
U = (x
and
(V,11-11)
x* E V*, x* T 0,
KI D co(E).
be a topological linear space over .
is a continuous real linear functional on H = (x
then
KI a E,
is convex and
K f1 Lo = 0,
L = (x
is a continuous.real linear functional on
I
int(K) # 0.
prove that
x*(x) = 1), V,
where
such that
Lo C L
(a) (b)
int(K) C ix x*(x) < 1) (c) cl(K) C (x I x*(x) < 1). (Proposition 5.2.1)
6.
L C V
let
V
be a linear space over
and
be a hyperplane that does not contain the zero vector.
Prove that, if then
Let
x',y' E V'
are such that
x'(x) = y'-(x), x E L,
x' = y'.
(Theorem 5.4.1)
*7.
space over
#,
and
Letting
K1,K2 C V
(V,T)
be a topological linear
be nonempty disjoint convex sets
prove each of the following: (a)
plane
L
If
int(K1) # 0,
that separates
then there exists a closed real hyperK1
and
K2.
If
(b)
and
KI
real hyperplane
(Corollary 5.4.1)
8.
gical linear space over and let
Let
K2.
then there exists a closed
x0 f K,
that strictly separates
L
and
be a nonempty closed convex set,
K C V
t,
K1
be a locally convex topolo-
(V,T)
Prove that, if
x0 E V.
real hyperplane 9.
are open, then there exists a closed
K2
that strictly separates
L
Geometric Form
Hahn-Banach Theorem:
S.
144
K
and
(x ). 0
Give an example in the plane of two nonempty disjoint closed
convex sets that'can be separated, but not strictly separated. 10.
(Corollary 5.4.2)
linear space over Prove that contain 11.
be a locally convex topological
(V,T)
K C V
be a nonempty closed convex set.
is the intersection of all the closed half-spaces that
K
K.
(Proposition 5.4.1)
space over
int(K1) f p.
separates
K1
and
Prove that
and only if
then
is a real hyperplane that
L
is closed.
L
and
(0)
convex subsets of
V.
and let
E1,E2 C V.
can be separated (strictly separated).
be a linear space over
V
f
can be separated (strictly separated) if
E2
E1 - E2
Let
Prove that if
K2,
and
E1
be a topologdcal linear
(V,T)
be a linear space over
V
Let
12.
Let
K1,K2 CV be nonempty disjoint convex sets
and let
4
such that
*13.
Let
and let
f
Prove that
E1
I
and
and let E2
E1,E2 C V
can bg strictly
separated if and only if there exists a convex absorbing set such that 14.
over
U
(E1 + U) fl E2 = 0.
Let
i,
(V,T)
let
E C V
be a locally convex topological linear space be a closed convex subset, and let
a compact convex subset.
Prove that, if
exists a closed real hyperplane K.
be
(Hint:
L
E fl K - 0,
K C V
then there
that strictly separates
see Chapter 2, Problem 10(c).]
be
E
and
Problems
S.S.
1S.
Let
(V,T)
145
be a topological linear space over
i.
Prove
that the following are equivalent: (a)
Every subspace of finite codimension is dense in
(b)
There exist no closed hyperplanes in
V.
V.
CHAPTER 6
THE UNIFORM BOUNDEDNESS THEOREM
We turn our attention in this chapter to
6.0. Introduction.
another fundamental theorem of functional analysis, the Uniform Boundedness Theorem.
It asserts, for example, that a family of
continuous linear transformations between two Banach spaces that is pointwise bounded is actually uniformly bounded.
The proof of this
theorem, which is an application of the Baire Category Theorem, together with the proof of the related Banach-Steinhaus Theorem, will be given in Section 6.2.
In Section 6.1 we recall the notion
of category and the Baire Category Theorem, and use the latter to This result will, in turn, provide us with
prove Osgood's Theorem.
The
the key step in the proof of the Uniform Boundedness Theorem.
remaining sections of the chapter are devoted to some applications of the Uniform Boundedness and Banach-Steinhaus Theorems.
We shall
see additional applications in subsequent chapters. 6.1. The Baire Category Theorem and Osgood's Theorem.
The
purpose of this section is to recall, without proof, some facts about the concept of category in metric spaces and to prove Osgood's Theorem.
An application of the latter in the next section will give
us the Uniform Boundedness Theorem. For the sake of completeness we make the following definitions: Definition 6.1.1. Then more,
E E
Let
be a topological space and
is said to be nowhere dense if is said to be of category I if
of nowhere dense subsets of E
X
is not of category
X;
E
I.
146
int[cl(E)] = 0. E
E C X.
Further-
is the countable union
is said to be of category II if
147
6.1. Baire Category Theorem and Osgood's Theorem
If
then,, for example, the Cantor ternary set in
X = IR,
[0,1]
is nowhere dense, the rationals are of category I, and the irrationals are of category II.
The details of the next proposition are left to the reader. Let
Proposition 6.1.1.
be a topological space and
X
E C X.
Then the following are equivalent: (i)
(ii)
such that (iii)
E
is nowhere dense. 0 C X
If
is open, then there exists some open set
U C 0
u n E m a.
cl[X ~ cl(E)] = X.
Moreover. if
is a closed set of category II, then
E
int(E) # P.
We are primarily interested in the notion of category in the context of metric spaces.
The most important result in this connec-
tion is the next theorem. Theorem 6.1.1
(Baire Category Theorem).
Every complete metric
space is of category II in itself.
Discussions and proofs of this theorem can be found, for example, in[BaNr, pp. 76-80; DS1, p. 20; Ry, pp. 139 and 140; W2,pp. 178-180]. Osgood's Theorem, to be proved here, is a classic example of the application of the Baire Category Theorem.
Before we state and
prove the theorem we require one further definition. Definition 6.1.2.
Thp function
(t
I
f
Let
X
be a topological space and
:
X - IR.
is said to be lower semicontinuous if
t E IR, f(t) < a) is a closed subset of Clearly, if
f
f
X
for each a E IR.
is continuous, it is lower semicontinuous.
converse need not be the case.
The
148
6.
(Osgood's Theorem).
Theorem 6.1.2 space and suppose
Uniform Boundedness Theorem
{fa)a
be a topological
X
is a family of real-valued lower semi-
E A
continuous functions defined on
X.
E C X
If
is a set of category
there exists some
t E E
II such that for each
Let
Mt > 0
for which
M > 0
such that
sup fa(t)'< Mt,
aEA then there exists some open set
0 C :X
and some
sup fa(t) < M.
aEA tEC Proof.
For each positive integer
En = [t Since the functions
Moreover, evidently Un=1 En,
define
f(t) < n, a E A).
are lower semicontinuous, it is apparent
a
is closed, being the intersection of closed sets.
En
that each
f
I
n
E C Un-1 E.
Thus, since
E,
and hence no
is of category 11, we conclude that there exists some
for which
is not nowhere dense, that is,
En
0 Cltarly, if
0
int(En ) } . 0
int(Ep ), then 0
sup fa(t) < no = M.
aEA
0
t E 0
Another phrasing of Osgood's Theorem is worthwhile noting: a family of real-valued lower semicontinuous functions on
X
bounded above at each point of some set of category II in
X,
the family is uniformly bounded above on some open subset of
if
is
then X.
When we apply this in the particular context of normed linear spaces and linear transformations, we shall see that the open set can be replaced by the whole space.
This is essentially the point of the
Uniform Boundedness Theorem.
6.2. The Uniform Boundedness Theorem and the Banach-Steinhaus Theorem. theorem.
We begin at once with the statement and proof of the main
149
6.2. Uniform-Boundedness and Banach-Steinhaus Theorems
Theorem 6.2.1 (Uniform Boundedness Theorem). and
be normed linear spaces over
(V2,11.112)
(Ta)ct E A
for each
x E E
and suppose
is a set of category II such that
E C V1
If
L(V1,V2).
C
6
Let
M > 0
there exists some
for which
x
IITa(x)I12 < Mx.
sup
aEA then there exists some
M > 0
such that
sup IITa11 < M.
a It is evident that for each
Proof.
continuous, real-valued function on
sups
E Afa(x)
exists some
or E A
the function
is a continuous, and hence lower semi-
fa(x) = (ITa(x)II2, x E V1,
V1.
For each
x E E,
Thus by Osgood's Theorem (Theorem 6.1.2) there
< Mx.
M' > 0
and an open set sup
0 C V1
such that
IITor (x)112 < M'.
aEA x E 0
In particular, there exist some
B(xo,6) _ (x
I
IIx -
x01f1 0
such that
and
IITa (x)11< M' . 2
of E A
x E B(x0,6) Now, if
y E V1, IIyI11 < 6,
then clearly
y + x0 E B(x0,6),
and so
[IT (y)112 0
there exists some
for which
sup IITnil < M. n Then the estimates
IIT(x)112 = lim IITn(x)112 n
< lim inf IIT n II n
IIxil,
(xEV1)
<MIIxjIl reveal that
T E L(V1,V2)
and
IITII < lim in nlITnI1.
By employing Theorems 6.2.1 next corollary.
(i)
I.
6.2.2 it is easy to prove the
The details are left to the reader.
Corollary 6.2.2. spaces over
and
O
Let
Suppose
For each
(V1,11.111)
and
(Tn) c L(V1,V2)
x E V1
be Banach
(V2'11-112)
is a sequence such that
there exists some
Mx > 0
such that
supra IITn(x)112 < M. (ii)
There exists some
E C B(x0,6) _ (x
I
IIx - x0111 < 6)
and such that for each which
xO E V1,
x E E
some
such that
6 > 0, E
there exists some
and a set
is dense in B(xo,6) zx E V2
for
limn Tn(x) = zx.
Then there exists a unique element (a)
T(x) - limn Tn(x)
(b)
11T11
lim infra IITn11.
T E L(V1,V2)
such that (x E V1).
6. Uniform Boundedness Theorem
152
In particular, the conclusions of the corollary hold if the (Tn(x))
sequence
converges for each
in some dense subset of
x
and hypothesis (i) is satisfied.
V1
The.Banach-Steinhaus Theorem is not generally valid if the (Tn)
sequence
(T ) C L(V1,V2).
is replaced by a net
This is the
of
Corollary 6.2.2
case since a Cauchy net need not be norm bounded.
does, however, remain valid if one replaces the sequence by a net. The details are left to the reader.
It is also worthwhile remarking that some sort of category assumption is required in order to ensure the validity of the Uniform Boundedness Theorem. space
As an example of this, consider the linear
of all complex-valued polynomials defined on
V
We note that
usual pointwise operations. x(t)
Ek=O aktk, ak E C,
for which
x(t) E V
IR
if and only if
where there exists some integer
ak = 0, k > N(x).
with the
N(x) > 0
If we set
(x(t) E V),
jlx(t)Ij = sup Iakl then it is easily verified that
is a normed linear space.
It is not, however, a Banach space because, for example, it is apparent that V
Ek=1 k-2tk
V,
but it is approximated in the norm of
by polynomials.
For each positive integer
n
m x*[x(t)] = x*( E
define
n-1 aktk
)
k=0
Clearly
xn
n-1
,shows that
xn E V*
and
E
ak
k=0 11x*jj < n.
E
=
ak
(x(t) E V).
k=0
is a linear functional on
I
by
x* E V* n
V,
and
n sup Iakl = nhIx(t)jj k
Actually
I1xn+1 = n
since, if
1S3
6.2. Uniform-Boundedness and Banach-Steinhaus Theorems
then IIyII - 1 and xn[y(t)] {x;) E V and (IIx;,II) is unbounded.
y(t) - I to tk,
then there does exist some
x(t) E V,
However, if
n. Thus we see that
Mx > 0
for
which
sup lxn[x(t)]I <Mx in
Indeed, if
aktk
x(t) =
n < m,
and
then
n-1 Ixn[x(t)]I =
E
I
akI < n sup Iakl
k=0
k
= n Ox(t)II
< (m + 1)Ilx(t)II, whereas if
in > in,
then
m
Ixntx(t)]I = IkE o Hence, if
Mx . (m + 1)Ilx(t)ll,
akI
(m + 1)llx(t)II
we see that
sup lx[x(t)]I < Mx. n Consequently we conclude that the Uniform Boundedness Theorem fails to hold in this case.
lies in the fact that
V
The reason for the failure, of course,
is not of category 11 in itself.
We have not given the most general form of the Uniform Boundedness and Banach-Steinhaus Theorems, preferring instead to concentrate only on these results in the context of normed linear spaces. will he sufficient for our needs.
This
These theorems and their corol-
laries do, however, have valid analogs in a more general context.
We state two of them here without proof and refer the reader for further discussion to [DS1, pp. 51-55; El, pp. 462-466, 476-477 and 480-483; 'K, pp. 168-170; Wl, pp. 116-118 and 223-225].
6. Uniform Boundedness Theorem
1S4
over
and suppose
#
(V2,T2)
Mx > 0
be Frechet spaces that defines the
Vk
(T 0)a E A C L(V1,V2)
If
there exists some
x E V1
and
is the metric on
pk
Tk, k = 1,2.
topology
(V1,T1)
Let
Theorem 6.2.3.
and if for each
for which
sup p2(Ta(x),0) < Mx
aEA 6 > 0
there exists a
e > 0,
then, given
such that
(a E A) .
p2(Ta(x),0) < c
sup
xEV
pl(x,0) 0,
and
we set
n
U(T;a;x1,x2,.... xn) = IS I S E L(V1,V2), IIS(xk) - T(xk)1120, n EZ, n>0, xl,x2,...,xn
in V1)
and
U Uso (T).
Uso
TE L(V1,V2) The proof of the next proposition is left to the reader. Proposition 6.3.1. linear spaces over (i)
L(V1,V2)
Uso
topology
A net so(T)
be normed
Then
is a base for a Hausdorff topology
such that
linear space over (ii)
f.
and
Let'
(L(V1,V2),so(T))
so(T)
on
is a locally convex topological
I.
(Ta) e L(V1,V2)
if and only if
converges to
T E L(V1,V2)
limof IITa(x) - T(x)II2 = 0
in the
for each
x E V1.
As indicated, the topology generated by strong operator topology for If both
V1
and
V2
L(V1,V2)
Uso
will be called the
and denoted by
so(T).
are Banach spaces, then we can obtain a
completeness result with the strong operator topology.
6. Uniform Boundedness Theorem
156
over
and
Let
Theorem 6.3.1.
is sequentially complete.
Then. (L(V1,V2),so(T))
I.
is a Cauchy sequence in
[Tn) ' L(V1,V2)
Suppose
Proof.
(Tn(x))
Then from Proposition 6.3.1(ii) we see that for each
sequence in such that
zx E V2
some
Since
be Banach spaces
(V2,II.II2)
x E VI,
so(T).
is a Cauchy
and hence there exists
limnlITn(x) - zXII2 = 0, x E VI.
is a Banach space, we may apply the Banach-
(VI,II:II1)
Steinhaus Theorem (Theorem 6.2.2) to conclude that the formula T(x) = zx = lim Tn(x)
(x E V1)
n
defines an element that
[Tn)
T E L(VI,V2).
converges to
Therefore
in
T
It is apparent from the definition so(T).
is sequentially complete.
(L(V1,V2),so(T))
0
An application of the Uniform Boundedness Theorem (Theorem 6.2.1) proves the next result. Theorem 6,3.2.
be a Banach space and
Let
a normed linear space over sequence in
so(T),
If
f.
(Tn) c L(VI,V2)
then there exists some
M > 0
is a Cauchy such that
sup IITnII < M. n Proof.
[Tn(x))
Since
(T
n
is a Cauchy sequence in
hence norm bounded. for which
Theorem
Thus for each
supnIITn(x)II2 < Mx.
so(T),
for each x E VI
we see that
x E VI
there exists some
series E
0 (1 - a)k.
a
Mx > 0
O is such that
a E IR
and the inverse of
and
An appeal to the Uniform Boundedness
completes the proof.
Recall that, if a # 0.
is a Cauchy sequence in
)
11
- at < 1,
then clearly
can be computed by the geometric
We now wish to apply the sequential complete-
157
6.3. Strong Operator Topology
ness of
when
(L(V),so(T)),
analog of this fact in
Theorem 6.3.3. suppose
T E 1(V). T
(i)
V
is A Banach spat., to obtain the
L(V).
Let If
be a Banach space over
(V, II II)
I(x) = x, x E V,
where
- TII < 1,
III
and
$
then
is bijective.
(ii) T-1 E L(V). Proof.
Then, since
s > 0.
Let
N
T-I
converges to
(I - T)k)
(iii)
III
in
that there exists an TUk Hence, if n > m > N, < s. .m III -
II
(I -
E
k=m
T)k(x)II
m > N,
such that, if
so(T).
thf-i
E 11(1 - T)k(x)II
k=m E
k=m
III
-
TIIkIIXII
(x E V)
< s1IxII Consequently it follows easily that Cauchy sequence in S E L(V)
in
so(T)
(Ek=O (I
T[S(x)] = T[lim n
T)k) C L(V)
E
k'=0
Moreover, if
x E V,
(I - T)k(x)]
n - lim (T[ E (I - T)k1(k)) n
k=0 n
= lim ([I - (I - T)] [ E (I - T)kI(x)). n k-0
= lim ([ E- (I n
is a
and hence converges to some element
by Theorem 6.3.1.
so(T)
-
k=0
nEl (I - T)k)(x)) k=1
then
158
6.
= lim ([I - (I - T) n +
Uniform Boundedness Theorem
1]
(x))
n
x - lim (I - T) n +
I (x)
n = X.
The last equality is valid since and
- TII < I.
III
tion we obtain T
that
Hence
T
II(I - T)n+ lcx)II < III. - TIIn+ llixII
is surjective.
S[T(x)] = x, x E V,
By a similar computa-
from which it follows at once
is injective.
Therefore
T
is bijective, and
The geometric series
E
_
,(I
S = T-1.
- T) k
is called the Neumann series.
If we set m = 1 -
III - TII > 0, then from IIx - T(x)II < III - TII IIxII, we immediately deduce that JIT(x)II > IIxII - III TII IIxII = mIIxII.
x E V,
Proposition 3.2.3 would then ensure that of
L(R(T),V).
T-1
exists as an element
As we have seen, however, a good deal more can be
deduced.
An examination of the proof also reveals that the sequence (F. ;O(I - T)k) L(V),
is even a Cauchy sequence in the norm topology on
and so, by the same arguments as those used in the proof,
T-1 - I =OCI - T)k.
This form of the result will be used in Chapter
13 when we discuss the spectral theory of bounded linear transformations on Hilbert spaces. The theorem was stated in this particular form merely to indicate an application of the sequential completeness of the strong operator topology.
6.4. Local Membership in
Lq(IR,dt).
First we make precise what
we mean by local membership. Definition 6.4.1.
function on R and Lq(ff dt)
Let
g
1 < q < 4a.
if for every
be a Lebesgue measurable complex-valued Then
g
is said to belong locally to
a,b E 1R, -- < a < b < a.,
the function
6.4. Local Membership in
X(a,b]g E Lq(IRdt), tion of
1S9.
Lq(1R,dt)
where
denotes the characteristic func-
X(a,b]
(a,b].
Obviously, if
g E Lq(R,dt),
then
g
belongs locally to
but the converse need not be valid.
Lq(R,dt),
The next result will
give sufficient conditions under which local membership implies membership. Theorem 6.4.1.
Let
1 < q < co
and
1/p + l/q =
is a Lebesgue measurable complex-valued function on (i)
g belongs locally to
IR
g
such that
Lq(]Rdt).
(ii). SIR Jf(t)g(t)j dt < m
Then
Suppose
I.
(f E Lp(R,dt)).
g E Lq (lR dt) . Proof.
is trivial.
If
g(t) = 0
for almost all
t,
positive integer
n
It is apparent that
For each
define
xn(f) = fn f(t)g(t) dt Lp(R,dt),
then the conclusion
So we may assume that this is not the case.
xn
(f E Lp(R,dt).
is a well-defined linear functional on
and an application of Holder's inequality reveals that
I xn(f)I 0
limnk=lank
a'
and write
m
E ank(xk - a) =
k=1
E ank(xk - a) +
m
E
ank(xk - a).
k=m+1
k=1
We choose
so large that for
k > m
Ixk - aI < ZM where
M = supnE. = llankl. Such a choice of m is possible, as limkxk = a. Keeping m fixed, we next choose N such that, if
n > ti,
then
m IankI
k=1
= 1
Then the following are equivalent:
(i)
(a)
A = Cank)n,k
Ixk - al < 2.
163
6.5.A Result in the Theory of Summability
This choice of
limn ank = 0, k = 1,2,3,...
is possible since
N
n > N
Consequently, combining these estimates, we see that for is
CO
ank(xk - a)l < E (ankllxk - al
E
+
lankllxk - al
k=m+1
k=1
k=1
.
e
Me
1
3
Next define a sequence
(xk)
of complex numbe rs by
xk=0 for 1 m.
x((yk)) _ - l l ankl , It follows at once = 1 Iankl, n = 1,2,3,... .
is a Toeplitz matrix, we have
165
6.6.Divergent Fourier Series
lim xn((xk n
lim
ankxk
z n k=1 = lim xk
((xk} E c),
k
and from the Uniform Boundedness Theorem (Theorem 6.2.1) we conclude that
0 sup 11x*11 = sup E Ia n n n k=1 nk
2(2n+ 1) 122n
k0(k+1)n
n
n
2
2n
E(k+l) f0 0
k
2n
2
Isin t( dt
2
(k + 1)n 'kn 2
k=0
Isinj2n+ I)tj dt
Z 2n+1 (k +2
2n E
_ 2
1
kn
sin t dt
2
1
= n kE0 (k+l) where again an elementary change of variable has been effected. Hence
4
Zn
1
LIDnLII > 2 E
k=0
n
(n
k+1
from which it follows at once that . sup IIDnIII =
O
n Not only do there exist diverge at
t = 0,
f E C([-n,n])
but the set of all such
whose Fourier series f
is of category II in
Even more can be said, as is shown by the following
C([-n,-n].
corollary:
Corollary.6.6.1.
Let
denote the set of
E
that n
sup
k=-n
n Then
E
is of category
E f(k)I
0
there exists some
such that
suPnlhTn(X)112 < (b)
Mx'. There exists some
x0 E V1,
E C B(xo,6) _ (xl EIx - x0I1l < 6)
and such that for each
x f E
some
such that
6 > 0, E
there exists some
and a set
is dense in zx E V2
B(xo,6)
for whion
limnT,(x) - zx.
Prove that there exists a unique element (i)
(ii)
T(x) = limnTn(x), x E V1.
IJT!J < lim infnJITnil.
T E L(V1,V2)
such that
6. Uniform Boundedness Theorem
172
Let
(Theorem 6.2.3)
*10.
spaces over
the topology
0
and suppose
(Taja
Vk
that defines and sup-
C L(V19V2)
E A
for which
Mx > 0
there a;ists some
x E VI
be Frechet
(V2,T2)
and.
is the metric on
pk
Suppose
Tk, k = 1,2.
pose that for each
(V1,T1)
sup p2(Ta(x),0) < Mx
aEA 6 > 0
there exists some
e > 0,
Prove that, given
such that (a E A).
p2(Ta(x),0) < a
sup
xEV 1 p1(x,0) 0,
then there exists a sequence
be Frechet spaces
is such that
T E L'(Vl,V2)
and suppose
f
category II in
(V2,(g})
and
(Vl,(pk))
such that
(k n}
limnan = 0.
(i)
k
(ii)
(iii)
> n, n = 1,2,3,...
n
cl[T(Un)] D (y
I
.
(n = 1,2,3,...).
y E V2, qk (Y) < an n
Proof.
and so
n
Let
T(V1) - Un
=
Since
it follows that there exists some We claim, moreover, that of the origin, Since set.
pn
Clearly
be a positive integer. 1mT(Un).
m0
for which
Suppose
int(cl[T(Un)]} # o.
y E int(cl[T(Un)]}.
is a seminorm, it is evident that
Un
is a convex balanced
Thus from Propositions 2.3.2(iii) and.(iv), since
we see that
int(cl(T(Un)]}
-y E int(cl[T(U )J}, n
V2,
contains an open neighborhood
cl[T(Un)]
arguing as follows:
Um = 1mUn,
V1
is of category II in
T(V1)
is convex and balanced.
is linear,
T
Therefore
0 = y/2 +1(-y)/2 E int(cl[T(Un)]),
and so
from which it is apparent that
cl[T(Un)I
contains an open neighbor-
hood of the origin.
Consequently, by the definition of the topology in exist some
an > 0
and some positive integer
cl[T(Un)] D (Y
I
kn
V2, there
such that
Y E V2, ql(Y) < an,...,gk (Y) < n} n
y E V2, qk (Y) < an),
(y
n
as
qm(y) < qm
+
1(y), y E V2, m = 1,2,3,...
dent that we can choose so that
limnan = 0
Lemma 7.2.2.
over
I
and
Let
and suppose
category II in
V2.
aA
and
kn
.
It is, moreover, evi-
successively in this fashion
kn > n.
-
(V1,(pk))
T E L'(V1,V2)
ID
and
(V2;(gm))
is such that
For each positive integer
n
be Frechet spaces T(V1)
choose
is of an > 0
7. Open Mapping and Closed Graph Theorems
184
limn cn = 0.
(i)
If
T
[y
cl[T(Un))
(ii)
Un = (x
such that
kn > n
and a positive integer
I
(
y E V2, gkn(Y) < en) = Wn,
x E V1, pn(x) < 1/n2), n = 1,2,3,...
The existence of the sequences
Let
(kn)
is
The other cases are proved in a similar manner.
n = 1.
for which
and
(6n)
.
We give the argument only for the
of course ensured by Lemma 7.2.1. case
.
T(2Un) D Wn, n = 1,2,3,...
is a closed mapping, then Proof..
where
y E W1.
We need to show the existence of some Now from Lemma 7.2.1 we know that
T(x) = y.
and hence there exists some
x1 E U1
such that
we see that
Prom the definition of
W2
since
there exists some
cl[T(U2))
W2,
gk3[y - T(xI) - T(x2)J < e3.
x E 2U1
cl[T(U1)}
W1,
gk2[y - T(x1)] < c2.
y - T(x1) E W2. x2 E U2
Hence,
such that
Continuing in this fashion, we obtain,
by repeated applications of Lemma 7.2.1, a sequence
(xn) C V1
such
that
(a)
xn E Un.
(b)
qk
(n = 1,2,3,...).
[y - T(Ek = Ixk)) < en + 1
n+l We claim that linear space
L2 n) _ (`k = lxk)
is a Cauchy sequence in the seminormed
(V1,(pk)).
Indeed, giver any seminorm
p.,
n > m > j,
if n
pj (zn - zm) = pj (k - E + lxk) n E
p.(xk)
k=m+ I
J
n
E pk(xk) k=m+I
it,
(zn)
con-
(T(zn)) c V2
Moreover, the sequence
qm
is any semi-
then
[y - T(zn)] < to + 1' + 1
Since to
y
limnen = 0,
we deduce immediately that
in the seminormed linear space However, since
we see that
T
T(x) = y.
(T(zn))
converges
(V2,(gm)).
is a closed mapping, from Proposition 7.1.1 Furthermore, we observe that n
pl(x) = pl(lim
E xk)
n k=1 n
< lim sup
n < lim sup n
E p (x ) 1 k
k=1
n £ pk(xk)
k=1
n2 6
< 2.
Therefore
x E 2U1,
and the proof is complete.
We can now state and prove the Open Mapping Theorem.
0
7. Open Mapping and Closed Graph Theorems .
186
be Frechet spaces over
(V2,(gm))
T
If
V2.
and
T E L'(V1,V2)
and suppose
I
is of category II in
T(V1)
is such that
(V1,{pk))
Let
en Mapping Theorem).
Theorem 7.2.1'(
is a closed
mapping, then
Proof.
kn > n, n = 1,2,3,...,
and positive integers
en > 0
Let
T(2Un) Z) Wn = (y
be so chosen that i
is open.
U C V1
is surjective.
T
(ii)
Un = (x
is open whenever
T(U) C V2
(i)
where
y E V2, qkn(y) < cn),
x E V1, pn(x) < 1/n2), n = 1,2,3,...
.
Such choices are
possible by Lemmas 7.2.1 and 7.2.2. is linear, to extablish part (i) it clearly suffices
T
Since
then
is an open neighborhood of the origin in
U
to show that, if
is an open set in
T(U)
some positive integer as
such that
x + cUn C: UP
Consequently
is open.
U
e > 0
and some
n
then there exists
x E U,
But if
V2.
VI,
T(x + cUn) = T(x) + (;)T(2Un) C T(U). However,
T(2Un) J Wn,
and so
T(x) + (e/2)Wn.
T(U)
is an open neighborhood of the origin in
Since
(e/2)Wn
we conclude that
V21
T(U)
is open.
Next suppose y # 0,
y E V2.
If
then there exists some
kn
for which
latter were not the case, that is, if then we would have qm(y) < (,,m, 1(y),
z = (en/2gkn(y)Jy
and so
Then
0. '
as
kn > n
contrary to assumption.
y = 0,
gkn(z) - cn/2 < en, x E 2Un
gkn(y)
and if
such that
T((2gk(y)/cnlx) . y. n
If the
gkn(y) = 0, n = 1,2,3,...,
qm(y) = 0, m = 1,2,3,...,
Thus there exists some deduce that
T(0) = y,
y - 0, then
and so
T(x) - z.
Therefore
T
and Set
z E Wn C T(2Un).
From this we
is surjective.
D
It is instructive at this point for the reader to refer back to the proof of Helly's Theorem (Theorem 4.10.1) and see how the Open Mapping Theorem could be used to shorten it.
187
7.3. Closed Graph Theorem
Let us give two rather easy corollaries of the theorem. Let
Corollary 7.2.1. spaces over
and suppose
I
and
(Vl,(pk))
be Frechet
(V2, (gm))
is a closed mapping.
T E L'(Vl,V2)
Then the following are equivalent: The range of
(i)
V2.
is surjective.
T
(ii)
is of category II in
T
In particular, if
is a closed linear transformation between
T
is surjective or its range is of
T
Frechet spaces, then either
T
Moreover, the Open Mapping Theorem says that, if
category I.
is
This immediately gives us
surjective, then it is an open mapping. the next corollary. Corollary 7.2.2.
spaces over then
f.
Let
be Frechet
(V2,(gm))
is a closed bijective mapping,
T E L'(V1,V2)
If
and
(Vl,(pk))
T-1 E L(V2,V1).
As a corollary to the corollary we have the following useful result:
Corollary 7.2.3. Tl
and
(V,T2)
from
V
be a linear space over
are Hausdorff topologies on
T,
are Frechet spaces over
f.
If
Clearly
I(x) = x, x E V. (V,T1)
to
(V,T2).
such that
Tl D T2,
I
:
and'
(V,T1)
Tl = T2.
then
I
and suppose
i
V
defined
V
is linear, bijective, and continuous
Hence, by Corollary 7.2.2,
linear, bijective, and continuous from
so
V
Consider the identity transformation
Proof.
by
Let
(V,T2)
to
I-1 = I
(V,T1),
is
and
T1 C T2.
'7.3:
The Closed Graph Theorem.
If
T
is a closed bijec6tive
linear transformation between two Frechet spaces, then Corollary 7.2.3 asserts that
T
is continuous.
In particular,
T
is also
a closed bijective linear transformation between Freohet spaces, and hence it must have a continuous inverse -- that is,
T
is continuous.
7. Open Mapping and Closed Graph Theorems
188
Thus we see that every closed bijective linear transformation between Frechet spaces is continuous.
Actually such more than this is true:
every closed linear transformation between Frechet spaces is continuous.
This is the content of the Closed Graph Theorem, one of the
most important consequences of the Open Mapping Theorem. Theorem 7.3.1 (Closed Graph Theorem). be Frechet spaces over
(V2,{qm))
closed mapping, then Proof.
Since
t.
(Vl,{pk))
Let
T E L'(V1,V2)
If
and is a
T E L(V1,V2).
T
is a closed mapping,
G(T),
the graph of
is a closed linear subspace in the topological product space
T,
V1 x V2.
It is not difficult to verify that this product space is obtained from the seminormed linear space pair
(x,y) E VI x V2
k,m a 1,2,3,...
.
we have
Moreover,
where for each
(V1 x V2,(rkm)),
rI.((x,y)] a pk(x) 4 qm(y), (VI x V2,{rk.)), is complete -- that
is, it is a Frechet space, and so
is a Frechet space.
G(T)
The
details are left to the reader.
Define the mapping Clearly
S
S
:
G(T)
is linear and surjective.
is a sequence that converges to k,m a 1,2,3,...,
by
VI
S[(x,T(x))] a x, x E V1.
{(xnT(xn))) e G(T)
Suppose that
(x,T(x))
G(T).
in
Then for each
we would have
lim (Pk(xn - x) + gm[T(xn) - T(x)]) a 0, n from which it is apparent that lim Pk(S[(xn,T(xn))] - S[(x,T(x))]) = lim pk(xn - x) a 0
n for each
n k = 1,2,3,...
S E L(G(T),VI).
then reveals that
.
Hence from Theorem 1.5.1 we conclude that
An appeal to the Open Mapping Theorem (Theorem 7.2.1), S(U) C VI
is open whenever
if C G(T)
is open.
Note that in the preceding argument it suffices to consider sequences, rather than nets,"since the spaces and hence complete metric spaces.
G(T)
and
VI
are Frechet spaces
189
7.3. Closed Graph Theorem
is any open neighborhood of the origin in
W
Now suppose
and consider the open neighborhood by
U = ((x,T(x))
I
borhood of the origin in exists some
of the origin in
U
Then
x E V1, T(x) E W].
(x,T(x)) E U
z E S(U),
S[(x,T(xD] i z,
such that
given
G(T)
is an open neigh-
S(U)
Moreover, if
V1.
V2
then there and so
T(z) = T(x) E W.
Therefore
and
T[S(U)] C :W,
is continuous.
T
The example of a closed discontinuous mapping in Section 7.1 demonstrates that the assumption of linearity is necessary for the validity of the Closed Graph Theorem.
Our most frequent use of the Closed Graph Theorem will be in the context of Banach spaces.
Because of this, and because the
proof is also instructive, we wish to restate and reprove the theorem in this setting.
Theorem 7.3.2 (Closed Graph Theorem). be Banach spaces over
mapping, then Proof.
x E V1,
If
f.
Let
(V1,11'1[1)
T E L'(V1,V2)
and
is a closed
T E L(V1,V2).
As before, define
where
G(T),
S
G(T)
the graph of
linear subspace of the Banach space
11(x,T(x))11 - 11x111 ` IIT(x)112' x E V1.
i,
V1
by
S[(x,T(x))] - x,
is considered to be a closed
V1 x V2
Clearly
with the norm S
is linear and bi-
jective, and the estimate
IIS[(x,T(x))]111 - 11x111 shows that
S E L(G(T),V1).
IIxIII * IIT(x)112 = II(x,T(x))II
Consequently, by Corollary 7.2.2,
S_1
E L(V1,G(T))
Thus, in particular, we see that
11x111
(x E v,)
hIT(x)112 - 11(x,T(x))1I
7. Open Mapping and Closed Graph Theorems
190
from which we deduce that (1ls-11l
11T(x)112
0
such that
11x111 < M11x112, x E V,
some
m > 0
such that
11x112 < mllxlll, x E 'V'
I
Consider the identity mapping
is a linear mapping from
show that
I
be a linear space
are norms on
some
Proof.
V
Let
is continuous.
(V,11.111)
!.
such that
V
If there exists
then there exists I(x) = x, x E V.
onto
(V,11.112).
Clearly
We need to
We shall accomplish this by using the
Closed Graph Theorem.
So suppose that limn1lxn - xlll = 0
[x n) C V, x,y E V,
and
Jimnl11(xn)
are such that
.
Then for each
- y112 = 0.
n
we
have
111(x)
- Al = l1x -
Y111
1lx-xn111+lixn-Y111 lix - xnlll + MlII (xn) from which it follows that Thus we see that
I
- Y112,
I(x) = y.
is a closed mapping and from the Closed
Graph Theorem (Theorem 7.3.2) conclude that
I
is continuous..
O
191
7.4. A Uniform Boundedness Theorem
In particular, if
jjxn1 < MllxIt2, x E V,
are equivalent norms on
then
11-J!,
and
1H12
V.
As in our treatment of the Uniform Boundedness Theorem we have not discussed the most general forms of the Open Mapping and Closed Graph Theorems.
For other developments of these two fundamental
results the reader is referred to [DS1, pp. 55-58; E1, pp. 419-458; K, pp. 166-168; KeNa, pp. 97-100].
Also it should be remarked that,
although we employed the Open Mapping Theorem to obtain the Closed Graph Theorem, it is possible to do the opposite: one can first prove the Closed Graph Theorem and then use it to deduce the Open Mapping Theorem.
This is done, for example, in [KeNa, pp. 97 -100).
A Uniform Boundedness Theorem for Continuous Linear
7.4.
Functionals.
As a first application of the Closed Graph Theorem we
prove a version of the Uniform Boundedness Theorem '(Theorem 6.2.1)
for continuous linear functionals on Frechet spaces. Theorem 7.4.1. suppose
(xo,)a E A C
V.
be a Frechet space over
(V,(pk))
Let
If for each
x E V
and
t
there exists some
AIxa(x)l < Mx, then for each c > 0 there E exists some open neighborhood U of the origin in V such that sups
'
sup.
such that
Mx > 0
e, x E U.
E Proof.
tions on
Consider the space
of all bounded f-valued func-
A with the norm Oi. = sups E AIf(a)t, f E B(A).
cated in Example 1.2.2, each
B(A)
x E V
define
(B(A).11.11,)
on A
T(x)
by
As indi-
is a Banach space over T(x)(a) = x*(x), a E It.
I.
For
Then
01
T(x) E B(A)
since
sup JT(x)(afl = sup Ix*(x)l < Mx
aEA
aEA
Clearly, the correspondence Moreover, we claim that
T
x
T(x)
defines an element
is continuous.
to the Closed Graph Theorem (Theorem 7.3.1).
T E L'(V,B(A)).
To prove this we appeal
7. Open Mapping and Closed Graph Theorems
192
(xn) C V, x E V,
Indeed, suppose (xn)
converges to
V
in
x
and
(T(an))
and
are such that
f E B(A)
converges to
in
f
B(A).
The latter assumption asserts that
l in IIT (xn) - fllm = l im sup I T (xn) (a) - f (a) I
n aEA
n
lim sup (x*(xn) - f(a)I
n aEA
= 0. Thus, in particular, for each x* E V*,
a
and so
nan
lim x*(x )
T(x)(a) - f(a), a E A,
a E A,
x(x)
a
limnxa(xn) = f(a).
Consequently
for each a E A.
T(x) = f,
that is,
is a closed
T
and so
But
mapping.
Hence the Closed Graph Theorem (Theorem 7.3.1) leads us to conclude that 6 > 0
some
and some seminorm
IIT(x)IIm < c.
sumption that U = (x
I
T E L(V,B(A)).
Therefore, given t > 0, pn
such that, if
there exists
pn(x) < 6,
then
We need only one seminorm because of our standing aspk(x) < pk
pn(x) < b),
+
1(x), x E V, k = 1,2,3,...
.
Setting
we see that sup Ix*(1)I - sup IT(x)(a)I aEA aEA o'
= IIT (x) II.
<e
(x E U),
which completes the proof.
7.5.
1.2 that
C]
Sow Results on Norms in C([0,1])
C([0,11).
We know from Section
is a Banach space under the usual supremum norm
IIfIL =
sup
t E [0,1]
If(t)l
(f E C([0,11))
In this section we wish to give some results concerning other possible norms on
C([0,1]).
limnfn(t) = f(t), 0 < t < 1.
it is the case that
are equivalent norms on
Clearly
to
and
{fn) C C([0,1])
and
limnllfn - f11. - 0
limnIlfn - f1l = 0,
Then
11.11
and
C([0,1]).
Consider the identity mapping
Proof.
if
are such that
f E C([0,1])
and
{fn} C C([O,1])
11.11
and suppose that, whenever
is a Banach space over t
(C([0,1]),11.1I)
such that
C([0,l])
11.1 be a norm on
Let
Theorem 7.5.1.
193
C([0,1])
Some Results on Norms in
7.5.
I
from
I
(C([0,1]),11-11.)
is linear and bijective.
Moreover,
are such that
f,g E C([0,1])
then for each
limnIll(fn) - g11 a 0,
t,
we have
0 < t < 1,
If(t) - g(t)I < If(t) - fn(t)1 + Ifn(t) - g(t)I < IIf - fnll. + Ifn(t) - g(t)I.
f - g,
I (f)
Hence
as
limnlll (fn) - g1I = 0 implies that g(t), 0 < t < 1.
lianI(fn)(t) - limnfn(t)
Thus
is a closed
I
mapping and hence continuous by the Closed Graph Theorem (Theorem, 7.3.2).
Consequently there exists some f E C([0',1]),
M > 0
such that
11f11
MIIfIL,
from which it follows at once, by the Two Norm Theorem
(Theorem 7.3.3),
that
0.11
and
11.11.
are equivalent.
o
The last step of the argument could have appealed to the Open Mapping Theorem, rather than the Two Norm Theorem. since
I
is continuous, linear, and bijective, we see from Corol-
lary 7.2.2 that
that is,
To be precise,
11.11
I-1
and
is continuous, and hence 11.11.
I
is a homebmorphism;
are equivalent.
There are, of course, many norms that one can. introduce into C((0,1]).
Theorem 7.5.1 says.that any such complete norm whose
notion of convergence is at least as strong as pointwise convergence must be equivalent to the supremi
norm.
7. Open Mapping and Closed Graph Theorems
194
is
C([O,l])
under which
C([0,1])
An example of a norm on
not a Banach space is given by the L1-norm:
(f E C([0,1])).
nfill = f0 (f(t)I dt It is easy to verify that
11.111
ljfjll < jjfljm, f E C([0,1]). If
is a norm on
and that
C([0,1])
(C([0,1]),11-111)
were a Banach space,
then by the Two Norm Theorem (Theorem 7.3.3) there would exist some However, consider
jf)Im < m1If1j1, f E C([0,1]).
such that
m > 0
-
for each positive integer
n
the function _
fn(t) _
(
2
(0 < t < n),
)t + n
n < t < 1).
fn(t) = 0 Then simple computations reveal that
and
Ilfn1I. = n
Ilf nil, = 1,
in contradiction to the previous estimates.
n = 1,2,3,...,
is not a Banach space.
Consequently
A Criterion for the Continuity of Linear Transformations
7.6.
on
defined by
fn E C([0,l]),
As should be evident by now,'one of the chief uses of the
Q
Closed Graph Theorem is to establish the continuity of a given linear transformation.
In the preceding examples the proof of continuity
was auxiliary to some other purpose, but now we wish to discuss a situation in which it is the central concern. tions on
T E L'(f.2)
Namely, we seek condi-
that will ensure the continuity of
shall see that this is the case whenever shift transformation on
£2.
T
T.
We
commutes with a certain
To be precise, we. make the following
definition: Definition 7 . 6 . 1 .
mation defined by k = 2,3,4,..., Thus
S
Let
S E L'(12)
S({ak)) = [bk),
for each
where
denote the linear transforbl = 0
and
bk
= ak
- 1
(ak) E f2.
shifts all the components of a sequence
{ak) E 'e2
one position to the right and puts a zero in the first position.
7.6. Continuity of Linear Transformations on
195
£2
We have the following criterion for the continuity of Theorem 7.6.1.
T E L'(£Z)
If
em = 1.
em = (ek) C £2
that is,
then we set
(x) m
m = 1,2,3,...
=
ek = 0, k # m, ((ak))m
am'
I(x)m1 < 11x112, x E £2, for each
.
Consequently, for each Sn,m
belongs to
(x)m
we
is the mth component of the sequence
It is immediately evident that
mapping
T E L(L2).
m = 1,2,3,...,
For each
the sequence such that
x = (ak) E £2,
If
m = 1,2,3,...; x.
then
To begin with we wish to establish some notational con-
Proof.
ventions that will simplify the proof. denote by
TS = ST,
and
T E L'(.f2):
defined by
n,m = 1,2,3,...,
Sn m(x) = (anT[en])m,
Clearly each
£Z.
we claim that the
x = [ak) E -2' is linear and
Sn m
ISn,m(x)I = I(anT[e'J)ml = Ian(T(enJ)mI
= I(T[enJ)mliani
(x E
(T[enJ)mINxil2 shows that
Sn m
is continuous from £2
Now we claim that, if a closed mapping.
show that, if that is,
To see this, since
(xj)
limjllx3Il2
T E L'(£2)
0
is a sequence in and
limjliT(x3)
(y)m = 0, m = 1,2,3,...
to
and
T £2
P.1
C.
TS - ST,
then
T
is
is linear, it suffices to and
- yII2 = 0,
y E £2 then
are such y - 0;
that
.
But it is easily seen that for each
m = 1,2,3,...,
m X]
(ak)
E akek + ST(zm)
(j
l,2,3,...
(z3)k = am + k' k = 1,2,3,... Since TS = ST, TSII1 = SmT, m = 1,2,3,..., and so
it follows
_
k= where
at once that
.
7. Open Mapping and Closed Graph Theorems
196
m
E (akT[ek])m + (SmT[zm])m
(T(xI])m =
k=1 m E 1Sk,m(xl k
The equations are valid since zeros in the first
m
Thus, since each
because
(S°Tfzlm])a = 0
Sm
introduces
positions when applied to any sequence. Sk
m'
k = 1,2,...,m; m = 1,2,3,...,' is a
continuous linear functional on
we conclude that for each
t2,
m = 1,2,3,..., m lim (T(xI))
j
m
E S
= lim
(xI
k= 1 k'm
J
m
E Sk,m((0))
=
k=1 = 0.
However,
m
limi IIT(x3)
- y1I2 = 0
1,2,3,...,' and hence
Therefore
T
limjl(T[x3] - y),I = 0,
y = 0.
is a closed mapping, and so, by the Closed Graph
Theorem (Theorem 7.3.1),
7.7.
implies that
T
is continuous.
Separable Banach Spaces.
In the preceding three sections These
we have given some applications of the Closed Graph Theorem.
are, of course, indirectly also applications of the Open Mapping Theorem, since the former was derived from the latter.
In this and
the next section we shall give two instances of direct applications of the Open Mapping Theorem. Theorem 7.7.1. C,
If
(V,11.11)
is a separable Banach space over
then there exists a closed linear subspace
such that
V
is topologically isomorphic to
W
of
LI/W,
([1,11.111)
where the
quotient space is taken with the usual quotient norm
III x + will = inrcfwllx + ylll
(x E L1)
197
7.7. Separable Banach Spaces
Proof.
(zk) C BI - (z
is separable, there exists a sequence
V
Since
TI(x) - Ek - lakzk
x - (ak) E LI we define in
(v,11-11)
that is dense in
z E V, lizll < 1)
l
since
T1 E L(LI,V).
Let
verified that
w
k - 1llakzkll W - (x
k = I l ak 1
The series converges = Ilxll1 . Clearly then
x E ll, TI(x)
I
For each
B1.
Then it is readily
01.
is a closed linear subspace of
and so
li,
ll/w
is a Bane:ch space with the quotient norm (Theorem
Now define T T
:
by setting
V
LI/W
T(x + W) = TI(x),
is evidently a well-defined injective continuous linear transfor-
mation from
that
T1
W - (x
as
V,
is surjective.
x E L1, T1(x) - 0). Moreover,
l
To see this it suffices to show
is surjective. z E V
Suppose integer
to
lI/w
we claim that T
kI
and
such that
Then there exists some positive
llzll < 1. llz
-
zk111 < 1/2.
so there exists some positive integer 112(z - zkl) - zk211 < 1/2;
that is,
Furthermore,
k2 > kI llz - zkl
Izkn/2n
-
111 < 1/2a, a = 1,2,3,...
- zk2/2ll < 1/22.
(zk ) C BI
Since
.
112(z - zkl)ll < 1,
such that
tinuing in this fashion, we obtain a sequence llz
LI.
V
Con-
for which
is a Banach
-
space, we conclude that where
zz - (aa)
and hence
is such that am a 1/2n -
am - 0, m } kn, n - 1,2,3,... then
lzkn/2n - 1,
z - Ln
w - z/11z11 E BI,
If
if a - kn,
z ' 0
Hence
T1
T E L(LI/W,V)
and
is any element of
so there exists some
that is, TI(llzllx) - z. Consequently
.
z = TI(xz),
xw
for which
V,
TI(xw)
w,
is surjective.
is bijective, and from the Open
Mapping Theorem (Theorem 7.2.1) or Corollary 7.2.2 it follows that T-1
is continuous.
Therefore
t1/W
and
V
are topologically iso-
morphic.
o
It should be remembered, and is not difficult to prove, that, if
V
is a separable Banach space and
subspace, then
V/W
able Banach space.
is separable.
W C V
is a closed linear
In particular,
l1/W
is a separ-
7. Open Mapping and Closed Graph Theorems
198
from Example 3.1.7 that, if
f(k)
in Cow.
L1([-n,n],dt/2n)
The Category of
7.8.
f C L1([-n,n],dt/2n) _
2n
We recall
and
`.nn f(t)e-ikt dt -
(k E Z),
w
then
Z
where, as before,
f E Co(1.),
denotes the locally compact
space of the integers with the discrete topology.
It can also be
w
f - f
shown that the linear transformation to
from
L1([-n,n],dt/2n)
is injective, as will be done in Corollary 13.7.1.
C0(Z)
over, it is evident that
More-
f E L1([-n,n],dt/2n).
I1fIIm < IIf1I1,
Combining these observations we see that the Fourier transformation
f
w f
is an injective continuous linear transformation from
to
However, the Fourier
transformation is not surjective; indeed, the range of the Fourier transformation is a set of category 1. L1([-n,n],dt/2n)w = (f
We set
Theorem 7.8.1.
I
We shall now prove this. f E L1([-n,n],dt/2n)].
L1([-n,n],dt/2n)
is a set of category I in
(C0 (SE.) , II' Ilm) -
Proof.
Define
T(f)
f, f E L1([-n,n],dt/2n).
Then
T
is an
injective continuous linear mapping from (Co(a),II III).
Suppose the range of
is of category II in T
C0(Z).
T,
to
that is,
L1([-n,n],dt/2n)w
Then from Corollary 7.2.1 we see that
is surjective, and so from Corollary 7.2.2 it follows that
T 1 E L(C0(Z),L1([-n,n],dt/2n)). Consequently for each f E L1([-n,n],dt/2n). However, consider the functions -n < t < n, n = 0,1,2,..., in Section 6.6.
!1f1I1 < IIT-IIIIIfllm
Dn(t) = [sin(n + 1/2)t]/[sin t/2],
that is, the Dirichlet kernels introduced
We proved there that
supnJIDnil,
On the other hand, since we know that for each Dn(t) = F
_
-ne
ikt ,
-n < t < n,
using the_well-known facts that
n = 0,1,2,...,
some straightforward computations .
199
7.9. Problems
fnn eikt
1
for
k = 0,
dt = 0
for
k f 0
dt =
I
pnn e ikt
2n
(k E 7d),
n = 0,1,2,...,
reveal that for each w
D(k) =
1
Dn(k) = 0
for
-n < k < n,
for
Ikl> n
and so the inequality
Thus On 11. = 1, n = 0,1,2,...,
IIT-IIIIIDnIIm
lIDnlll < cannot hold for all
(k E Z).
n.
But this contradicts the previous estimate, and therefore LI([-n,n],dt/2n)
must be ofrategory I in
It should,nevertheless, be,noted that
norm dense subalgebra in
(Corollary 7.1.1)
1.
spaces such that :
X
U
Let
is a
(See, for example, [Rul, p.9].)
(X,T)
and
is a Hausdorff topology.
is continuous, then
Y
2.
L1([-n,n],dt/2n)
Problems.
7.9.
f
II W).
C0(Z).
f
be topological
(Y,U)
Prove that, if
is closed.
be it normed linear space over
Let
0.
If
are both closed, does it necessarily follow that S + T
S,T E L'(V) is closed? 3.
be a nonmed linear space over .
Let
T E L'(V)
is a closed linear transformation and
subset of
V,
is
T(E)
E C V
necessarily a closed subset of
If
is a closed V?
7, open Mapping and Closed Graph Theorems
200
Suppose (a)
f
:
If
and
(X1,11), (X2,12),
Let
4.
X1
X2
f
and
and
g
X2
:
(X3,T3)
be topological spaces.
X3.
are closed, prove that
g
g o f
X1 - X3
:
is`not necessarily closed. (b)
g.* f
:
t.
and
over
f
Let
graph of
be a normed linear space'over
and
(V1,T1)
V1 x V2
T E L'(V)
If
I.
T
is closed.
be topological linear spaces
be neighborhood bases at the ,origins
respectively.
V2,
in
'(V2,T2)
U2
and
U1
is closed whenever
T - XI
prove that
and let
T
is closed, prove that
g
is closed.
(V;11'fl)
Let
and
V1
is continuous and
f
X1 - X3
71.E f,
6.
of
If
Let
Prove that the
T E L'(V1,V2).
is closed if and only if
n{T(U1) + U2
U1 E U1, U2 E U2) - o). 1
7.
(Corollary 7.2.1)
chet spaces over
Let
(V 1,{pk))
and suppose
f
and
T E L'(V1,V2)
(V2,{qm)) be
Fre-
is a closed mapping.
Prove that the following are equivalent: (a)
R(T)
(b)
T
8.
is of category II in
is surjective.
(Corollary 7.2.2)
chet spaces over
9. I.
(V1,{pk))
and
(V2,{qm))
T E L'(V1,V2)
be
Fre-
is a closed
T-1 E L(V1,V2). and
Let
(V2,1I.((2)
Suppose that for some injective
T-1 E L(V1,V2). OV2
Let
Prove that if
f.
bijective mapping, then
over
V2.
Prove that
V1
be normed linear spaces T E L(V1,V2),
we have
is a Banach space if and only if
is a Banach space. 10.
be a normed linear space and
Let
a Banach space over a linear subspace
If
f.
DT
of
T V1
(V2'11-112)
is a linear transformation mapping into
V2,
that is,
T E L'(DT,V2),
be
7.9. Problems
T
then
201
(xn)
is said to be closed if whenever and
x E V1.
y E V2
are such that
Prove that, if exists some
T E L'(DT,V2)
M > 0
for which
T E L(V1,V2).
if and only if
and
V1
R(T)
and
(V1,ll'll1)
is closed in
T E L(DT,V2).
is isomorphic to
0
V1/N(T)
be Banach spaces over
be injective and closed.
if and only if there exists some
V2
DT
be Banach spaces over
Prove that
T E L'(V1,V2)
and let
then
is closed.
R(T)
Let
12.
T(x) = y.
11T(x)112 < Mllx111, x E DT
and
Let
11.
and
is closed and bounded, that is, there
is a closed linear subspace of
and let
x E DT
DT,
and
limnllxn - x111 = 0
it is the case that
limn11T(xn) - Y112 - 0,
is a sequence in
Prove that c > 0
i
R(T)
such that
11x111 < clIT(x)112, x E V1.
f :.Xl
X2
and
(X2,T2)
be topological spaces and let
be a mapping.
Suppose
Ti
T, e T2
such that that
(X1,T1)
Let
13.
prove that *14.
f
Let
(that is,
is Hausdorff.
(X2,T2) :
(V,l1'l1)
cp(x) - x + W, x E V:
Let
Let
cp
Prove that
and such
T2)
(X1,T1) - (X2,TZ)
:
:
cp
V - V/W
is continuous,
I
and let
W C V
V
V/W.
be a .separable Banach space over V.
Prove that
i
V/W
and let is separ-
able.
and let that
Let
f
:
and V1
V2
f E L(V1,V2).
be
be the natural mapping
maps the open unit ball of
be a closed linear subspace of
16.
X
is closed.
be a Banach space over
onto the open unit ball of
W C V
f
(X1,T1) -. (X2,T2)
a closed linear subspace.
1S.
is weaker than
T2 If
is another topology for
be such that
be Banach spaces over
y* o f E Vi, y* E V.
Prove
f
7. Open Mapping and Closed Graph Theorems
202
Suppose
is a norm on
j1j-1(1
(i)
define
f E L1([-n,n],dt/2n)
For
17.
L1((-n,n],dt/2n)
such that
is a Banach space.
(L1([-n,n],dt/2n)lIJ
is a continuous linear functional on
x;
(ii)
xk(f) = f(k), k E Z.
(L1([-n,n],dt/2u),III -ICI)
for each
the Banach space
k E Z.
are equivalent norms on the space L1.([-TT,TI] ,dt/2n), where as usual Prove that
and
jI1
II
111-111
jf(t)I dt.
jjfjjl = 2n 18.
E C Z and define
Let
CE _ {f
(a)
Prove that
(b)
ECZ M>0
constant that
*19.
CE
is a closed linear subspace of
is said to be a Sidon set if there exists Some such that
Ek = -mIf (k)
x E V,
Suppose
jjxjj < 1).
there exists a number
k > 1
corresponds an
such that
x E B1
20.
x = En
21.
Let
(V,T1)
-
lix
=0
§
and let
has the following property:
all < 1/k.
(an)n = 0
x E B1
there
Prove that to each
of points of
A
an/An
V
to a Banach space
T1
and
and
T2
(V,T2)
J,
T1 } T2.
W
is continuous.
be topologies for a linear space
V
are both Frechet spaces over
Prove
that the following are equivalent: (a)
Prove
Use Problem 19 to prove that a closed linear map from a
Banach space
that
a E A
A C V
such that to each
there corresponds a sequence
such that
< Mjjfjj.,, f E CE.
be a normed linear space over
Let I
f c C([-n,n]), f(k) = 0, k f E).
is a Sidon set if and only if Ek = -mlf(k)l < m, f E CE.
E e 7Z
B1 a (x
I
§.
such
7.9. Problems
203
(b)
The topology
T with subbase
(c)
The topology
TI n T2
(d)
There is an
U2
is not Hausdorff.
x C V, x
each TI-neighborhood
is not complete.
TI U T2
such that
0,
'
x E UI + U2
for
of the origin and each T2-neighborhood
UI
of the origin.
There is a sequence
(e)
to zero relative to 22.
Un T
tion
ITn(Vn),
23.
be Frechet spaces over
onto any Frechet space
whenever. U
F
converges
Ix n)
T2.
6.
let n E L(VW).
n
prove that every continuous linear transforma-
is open in
Let,(V1111-11, )(V2"1.112)
,
T C L(VI,V3)
and
x E VI
the equation
T(x) = S(y)
Prove that the mapping
F
is open,
from
be Banach spaces
(V3"1.113 )
S E L(V2,V3).
A(x) = y
T(U)-
that is,
W.
and
Let
f.
every
W
from
is open in
over
such that
relative to
0
be a topological linear space and for each
W
If
y
(V n, n), n = 1,2,3,...,
Let
W
Let
V
in
(xn)
and to
T1
Suppose that for
has a unique solution V1
to
V2
y.
so determined is
linear and bounded.
24. over all
Let
(V1,1H11) , (V2'11.112) ,
Suppose
f.
x E VI,
T E L(VI,V3)
where
25.
and let
Let
X
If
where
Suppose
f
:
X - C
sequence
defined by
y
(yk
If
for
A
is
p
for every
g E Lp(X,p),
fg
is
prove that
1/p + 1/q = 1, 1 < p < m.
I < p,q
0.
Define
is the first deri-
is closed.
T(f)(t) = f(t) + xfl K(t,s)f(s) ds where
f'
If
is continuous.
T-l
there exists a measure
t E [0,1J
such that f(t) = fI g(s) dµt(s).
30.
Let
W
those functions
be the linear subspace of f
in
that have continuous first and
C([0,1])
second derivatives and satisfy linear differential operator
f(O) = f(1) = 0. T
consisting of
C([0,1])
:
W
C([0,1]),
Let
a0,a11
and
Assuming that the differential equation solution
f E W
for each choice of
exists and is continuous.
be the
defined by
T(f)(t) = a0(t)f"(t) + al(t)f'(t) + a2(t)f(t) where the fixed coefficients
T
(t E (0,1]),
aZ are in
T(f) = g
g E C([0,1]),
C([0,lJ).
has a unique prove that
T
205
7.9. Problems
M
subspaces
and
x = y + z,
where
V = M c8 N.
If
V
is a direct sum of two of its linear'
V
A linear space
31.
is s Banach space, them a projection
and
N(P)
R(P)
P2 = P,
that is,
is an
P
and a direct
such that for some
M
V = M E ,N.
N,
is a Banach space
V
If
prove that
L(V),
is a closed linear subspace
V
other closed linear subspace (a)
has a unique decomposition In this case we write
z E N.
and
y E M
idempotent transformation in sunimand of
x E V
if every
N
and
P E L(V)
is a projection,
are direct summands of
V
and
R(P) E) N(P).
V
of
V.
be A Banach space and let
V
Let
(b)
Suppose that
V = M
N
be a direct summand
M
and define
P
:
whenever x= y± z E V, y E M, z E N.
P(x) = y
M
V
by setting
Prove that
P
is
a projection.
be an infinite-dimensional separable Banach
Let
32.
A Schauder basis in
space over f.
such that for every
,such that ak
:
V
*33.
T
is a set
(xkI C V, sup x*(xn) n n sup
E
k-E l
1
1
n k=1 `O
=
,
However, there exists no x E c0, IIxIIW < 1 , for which
I x* (x) l _ IIx*Ils
as can be seen from the following argument: Suppose 'x = (ak) E c0, IIxIIm < 1.
Since
lanl < 1,
and hence
linkak = 0,
Ix*(x)I =
there exists some
#
s k=1 CD
k =l k#n
I
a
Ln n.
n
such that
8. Reflexivity
214
< k
1
1
l
k'
*
n'
k'n
Ilx*Il,
=
Therefore
c
is not reflexive.
0
D
Another interesting question is the following one: V
is a nonempty closed convex set in a Banach space
Suppose
K
and let
d = infx E Kllxll. Clearly, if 0 E K, then d = 0, and there exists some point
x0 E K
On the other hand, if
0
for which then
K,
lix0ll = d,
d > 0,
clear under what conditions there exists an llx0II = d.
Indeed, there may exist no such
reflexive, then such an
x0 E K
namely,
xo
0.
but it is not entirely x
0 x0.
E K
for which
However, if
V
is
with minimal norm always exists.
The proof of this requires some knowledge of weak topologies and will be delayed until Chapter 9 (Corollary 9.9.2). tion we shall prove that such a unique
x0 E K
In the next sec-
exists whenever
V
is uniformly convex.
The notion of reflexivity also plays an important role in the study of'weak topologies on Banach spaces, a subject we shall pursue in some detail in Chapters 9 and 10.
8.2.
Uniform Convexity and Mil'man's Theorem.
As indicated in
the preceding section, it is of some interest to develop means for determining whether a given Banach space is reflexive. mination is not always easy.
Such a deter-
There is, however, one property of a
Banach space that is amenable to verification and ensures that the space is reflexive. norm.
This property is the uniform convexity of the
The main purpose of this section is to. establish this result.
215
8.2. Uniform Convexity and Mil'man's Theorem
Definition 8.2.1.
is said to be uniformly convex if for each
Then
t.
there exists some
IlyII < 1,
be a normed linear space over
Let
and
such that, whenever
6 > 0
yll > e,
llx -
x,y E V,
e > 0
11x11 < 1,
Il(x + y)/211
then
Strictly speaking it is not a linear space that is uniformly convex, but rather the norm on the space, and if one takes the same linear space with two different norms, then one space may be uniformly convex and the other space may fail to be so.
For instance, consider
Then we claim that
is uniformly convex, whereas first that, if
(IIx-
and
x - (al,a2)
12)2+ (11-
is not.
(
are in
y = (bl,b2)
(IR2'11.112)
Indeed, note 1R2,
2
2
k=1
k=1
then
E (ak+bk)2+a E (ak-bk)2
-YA1
2
=
4 E (2ak2
2 + 2bk)
k=1 2
=
1
2
E ak) + Z( E bk)
2( k=1
k - i
(Ilxl12)2
+
2(l1Yll2)2.
2
But then, if e > 0 is given, let 6 > 0 be such that (1 -
and
6)2
1
-
c2/4.
lix - Y112 > s (lilx
Then for any x,y E 1R2,
(1x112 < 1, IIY112
0).
1>1 -6 f
2
Thus
is not uniformly convex.
(IR
Note also that the norass
on
11.112
are equivalent norms
and
V. Another important class of uniformly convex spaces is the inner Indeed, suppose
product spaces introduced in Example 1.2.10. is a linear space over
and
§
$
:
V x V
§
if {jxjj = ($(x,x))112, x E V.
formly convex.
is an inner product.
is a normed linear space
Then, as indicated previously,
over
V
V
is uni-
The space
We shall prove this later when we discuss inner
product spaces in detail in Chapter 13.
In particular, every Hilbert
space is uniformly convex.
Geometrically the notion of uniform convexity is an indication of the "roundness" of the closed unit ball about the origin in the space.
6 > 0
It says, in effect, that, given such that, if
least a distance
segment between
c
x
s > 0,
there exists a
are in the closed unit ball and are at
x,y
from one another, then the midpoint of the line and
y
is a least a distance
face of the closed unit ball.
6
from the sur-
Note that uniform convexity thus as-
serts more than that the midpoint of the line segment between two points in the closed unit ball lies in the interior of the ball.
It
actually gives some measure of how far in the interior the midpoint lies.
The weaker notion that
jjxjj < 1, jjyjj < 1
implies jj(x + y)/211 < 1,
is generally called strict convexity, or rotundity.
We do not pur-
sue this notion further here, but instead refer the interested reader to (Da, pp. 111-115; K, pp. 342-347; Wi, pp. 107 and 108).
217
8.2. Uniform Convexity and Mil'man's Theorem
We now prove that every uniformly convex Banach space is reflexThe proof is due to Kakutani [Kt] and makes crucial use of the
ive.
corollary to Helly's Theorem (Theorem 4.10.1). Theorem 8.2.1 (Mil'man's Theorem). space over
0.
If
(V,l.II)
be a Banach
Let
is uniformly convex, then
is reflex-
V
ive.
Given
Proof.
x0 E V if
such that
x** E V**,
we must show that there is some
1(x0) = x**.
Since this can obviously be done
we may without loss of generality assume that
x0* - 0,
Ilxo*II - 1. From the definition of integer
there exists some
k
Ix;*(yk)I > 1 - 1/k.
k = 1,2,3,..., and
Ilxkli = 1
IIxO**II
we see that for each positive
On setting
xk - exp[-i arg x**(yk)]yk
we obtain a sequence
(x;} C V*
x*0*(xk) > 1 - 1/k, k = 1,2,3,...
Now for each positive integer s = 1/n
to Helly's Theorem with
for which
y* E V*, IIyk11 = I,
n
such that .
we may apply Corollary 4.10.1
to deduce the existence of some
such that
xn E V
(i)
llxnllIlx*II+n=I+I x**(xk0
(ii)
x*
k
(xn) -
)
Repeating this for each (xn) c V
n = 1,2,3,...,
(k - 1,2,...,n).
we obtain a sequence
whose elements satisfy conditions (i) and (ii) and which
is such that for each
n
(m>n).
xn(xm) - xn(xn) = x0*(xn) We claim that
(xn) CV is a Cauchy sequence.
Indeed, suppose this is not the case. s > 0
and a strictly increasing sequence
such that
- x
Ilx
nk + 1
Then there exist some (nk)
11 > s, k - 1,2,3,...
nk
.
of positive integers For this
s > 0
8. Reflexivity
218
some
there exists, by the uniform convexity of
such that, whenever have
and
IlylI < 1,
IIxII < I.
6 > 0
Ilx - yp > c/2,
we
Thus, on the one hand, since
ll(x + y)/2I1 < 1 - 6.
Ilxnkll < 1 + 1/nk we see that xnk 111+1nkl1,
II1+n1n1 k
I
s
-l+nk-2'
l+lnk +
and hence
+ 1 + xnk I
xnk Il2(1+lnk Consequdhtly
IIxnk + 1 +
2(1 - n)
(n ' 1,2,3,...),
8. Reflexivity
220
and so
Hence
contradicting the previous estimate.
J;x0 + yoll > 2,
x0 = y0. x* E V*.
Finally, let
x* = 0,
If
we may assume without loss of generality that (xk) C V*
T(x0)(x0) - 0,
then
IIx*1I = 1.
be as before and for each nonnegative integer
and
Let
apply
n
Corollary 4.10.1 to Helly's Theorem to obtain a sequence x,,xixZ,... such that
V
of elements of
(a)
jjx;j+
(b)
J{x'
(c)
xk(xn) = x0*(xk)
3. 1
I < lix**Il +
=
1
(n
+n
1,2,3,...).
(k = 0,1,2,...,n; n
Note that we now have applied Corollary 4.10.1 to the sequence whereas before we applied it to
x0,x*,x2,...,
x1* x2,x3,...
Repeating our previous arguments, we deduce that Cauchy sequence in such that of
x0*(xn7 = xn(y0), n = 1,2,3,...
.
x0*(x0) = x0(y0)
Therefore, since V
(xn)
From the uniqueness
shows that
x0
YO'
x**(x*O).
is arbitrary, we see that
x* E V*
is a
y0 E V. IIYOf)
which we have established, we conclude that
x0,
whereas
and so
and hence there exists some
V,
.
x0*,
'r(x0)
is reflexive. 11
The converse of Mil'man's Theorem may fail. reflexive Banach spaces is uniformly convex.
(V,11.fl)
Indeed, there exist
such that no norm equivalent to
For the details the reader is referred
to [K, p. 361].
We shall obtain some necessary and sufficient conditions for a space to be reflexive in Chapters 9 and 10 (Theorem 9.9.1, Corollary 9.9.1, and Corollary 10.3.1) when we discuss weak topologies. We note also that all finite-dimensional Banach spaces are reflexive.
This follows at once from Mil'man's Theorem, the uniform
.convexity of
(l
and
(In,II.JI2), n = 1,2,3,..., and the
221
8.2. Uniform Convexity and Mil'man's Theorem
fact that all such spaces are topologically isomorphic to either
(Cn,11.112) (Theorem 1.3.1).
or
(eI1i-1I2)
One further comment about Mil'man's Theorem is necessary in reference to the proof: the argument used in the proof to show that the sequence
constructed by means of Corollary 4.10.1 to
(xn),
Helly's Theorem, is a Cauchy sequence, can readily be modified to obtain the following proposition.
The details are left to the reader
We shall use the proposition in the proof of Theorem 8.2.2. Proposition 8.2.1. linear space over
If
f.
be a uniformly convex normed
Let
(xn) C V
is a sequence such that
(i) limnllx11 = 1, limn,kl{xn t xkll = 2,
(ii)
then
(xn)
is a Cauchy sequence.
Let us now see how the two questions explicitly raised at the and of the preceding section can be answered in uniformly convex spaces.
First we address ourselves to the existence of an element
of minimal norm in a closed convex set. Theorem 8.2.2. space over
i.
If
there exists a unique Proof. set
d
If
x
0 E K,
infx E KIIxtj > 0.
limnllxn11 = d.
limnllxn/dII = 1.
be a uniformly convex Banach
Let
K C V
Clearly,
0
is a nonempty closed convex set, then
such that
EK
then clearly
xo = 0.
(xn) C K
Let
11x011 = infx E KIIxII
Moreover, for each
{I-- - 1
2
the estimates
and
8. Reflexivity
222
because
The latter estimate is.valid
limn,k1l(xn + xk)/dll = 2.
show that K
is convex.
Thus from Proposition 8.2.1 we conclude that
is a (xn/d) x0 E V be such that
Cauchy sequence in
V,
limnl1xn - x011 = 0.
Obviously, by the choice of
that
K
and so is
is closed, we have
(xn).
x0 E K
Suppose that there exist'some 11x0
that there exists some .However, since
and
(xn)
6 > 0
y0 E K, 11yO11 - d,
such that
and the fact
lix011 = d.
Then by the uniform convexity of
y011 = c > 0.
-
Let
and (V,11-11)
we see
(1(x0'+ y0)/211 < (1 - 6)d < d.
is convex
K
Ilxo + yo 1 > inf Ilxii = 2 xEK
,
which contradicts the previous estimate. Therefore
xo = yo,
and
x0
is unique.
The completeness assumption cannot be dropped. Corollary 8.2.1. space over
I.
Let
K C V
be a 'uniformly convex Banach
is a nonempty closed convex set and
then there exists a unique
K,
y0
If
x0 E K
such that
UY.-x011-if lly0-xll. A second corollary to Theorem 8.2.2 provides us with the promised improvement of Theorem 8.1.4. Corollary 8.2.2. space over
4.
If
r.0 E V, 11x611 = 1,
Let
(V,1l.l1). be a uniformly convex Banach
x* E V*, x* 4 0,
then there exists a unique
such that x*(xo) = l1x*ll.
1 Proof.
5.1.1
origin. 0
L,,
L
Let
L = (x
I
x E V, x*(x) = llx*lI).
is a closed hyperplane in In particular,
L
V
Then by Proposition
that does not contain the
is a nonempty closed convex set such that
and hence by Theorem 8.2.2
there exists a unique x0 E L
223
1 < p
0. Clearly x*(xo) = iix*ii, from which it follows that lixo(i > 1. is uniformly convex and
On the other hand, since
hence reflexive by Mil'man's Theorem, we deduce from Theorem 8.1.4
that there exists some Thus
yo E L
and
0
E V, i1yo1i = 1,
such that x*(yo)
Consequently
11yoii < 11xo11.
the unique element in Therefore
y
as
yo = xo,
lixoii = 1,
is
and the proof is complete.
Mil'man's Theorem to prove that the spaces
0
We now wish to apply
Lp(X,S,o), 1 < p < As a corol-
introduced in Example 1.2.4 are reflexive Banach spaces. lary we shall see that
x0
of minimal norm.
L
Reflexivity of L (X,S,µ), 1
0,
and
a,b E IR, a > 0
then
aP + bP < (a2 + b2)P/2.
(i)
(a2 + b2)"'2
(ii)
Proof.
If either
< a
2(p- 2)/2(aP
or
b
+ bP).
is zero, then the inequalities are
trivial, and we may assume without loss of generality that neither a
nor
b
Obviously,
is zero.
a2/(a2
+
b2) < 1
b2/(a2
and
and so aP (a2
bP
a
p/2
2
b
b2)
[a2
(a
a2 a2
= 1,
p/2
2 ii
a2
b2)p 2 +
+
b2 b2
+ a2
+
b2
b21
+
b2) < 1,
8. Reflexivity
224
as
ap + by < (a2 + b2)p/2
Thus
p/2 > 1.
To prove part (ii) of the lemma we note that the inequality is trivial if
p = 2.
If
p > 2,
then we set
1/p' + 1/q' = 1,
Then
q' = p'/(p' - 1) = p/(p - 2).
p' = p/2 > 1,
and
and so by
Holder's Inequality for finite sums we see that a2 + b2 < [(a2)p'
[ap +
+
(b2)p']1/P'[lq'
bp]2/P[2(P - 2)/p]
(a2 + b2)p/2 < 2(p - 2)/2(ap + bp).
It is thus apparent that
We shall use this lemma to show that convex for
Lp(X,S,µ)
0
is uniformly
p > 2.
Theorem 8.3.1. 2 < p < m,
lq']1/q'
+
then
(X,S,p)
Let
be a positive measure space.
If
is a uniformly convex Banach
(L
space.
Proof.
Let
f,g E L (X,S,p).
We then apply Lemma 8.3.1 twice:
P
once with a = If(t)I
a = If(t) + g(t)I b = Ig(t)I,
and
and
b = If(t) - g(t)I;
to deduce for almost all
and once with t E X
that
If(t) + g(t)Ip + If(t) - g(t)IP < [If(t) + g(t)I2 + If(t) - g(t)I2]p/2 and
[If(t)I2 +
Ig(t)I2)P/2 < 2(P - 2)/2
[If(t).Ip + 1g(t)1p1-
Furthermore, using the definition of the absolute value of a complex number, we see that for almost all
If(t) +
g(t)JP
t E X
+ If(t) - g(t)IP < [If(t) + g(t)I2 + If(t)
[21f(t) 12
0
let
e,0 < e < 2,
in
g
(IIBIIp)p]
[ (IIf Ip)p + (Ilip)p] - ({If
Ilp)p
- (2)p 6)p
Therefore
Lp(X,S,µ)
Corollary 8.3.1.
Let
is uniformly convex.
(X,S,µ)
If
Apply Mil'man's Theorem (Theorem 8.2.1).
Proof.
It is, furthermore, true that p,
be a positive measure space. is a reflexive Banach space.
then
2 < p < co,
0
1 < p < m,
is reflexive for all
Lp(X,S,µ)
and we shall establish this fact in a moment.
that, however, we wish to give another lemma.
Before
In the interest of
completeness we make the foliowing definition: Definition 8.3.1.
Then a mapping
p
:
V
Let
W
V
and
W
be linear spaces over
is said to be antilinear if (x,y E V; a,b E f).
cp(ax + by) = aT(x) + sp(y) If
then clearly "linear"
I = IR,
t.
and
"antilinear"
are identi-
cal.
Lemma 8.3.2. 1 < p < m.
formula
If
Let
(X,S,µ)
1/p + 1/q = 1,
be a positive measure space and let then for each
g E Lq(X,S,µ)
the
8. Reflexivity
'226
xg(f) = fX f(t)g(t) dj,(t) xg E Lp(X,S,&)*.
defines an element g
xg
(f E Lp(X,S,4))
Moreover, the correspondence
defines an antilinear isometry from
to
(Lp(X,S,µ)*,II II) .Proof.
Obviously
is linear, and by HUlder's Inequality
xg
it is clear that
Ixg(f)I < I11IgIIfllp IlgilgllfII
xg E Lp(X,S,P)*
Thus
and
Furthermore, let r ,
since
(q
E
TgT
, .,0-1
-
- -
--.,,.,
I
]q/P L1
(q
1) p du(t)/P
g
q/p
Mq 1,
and so f E Lp (X, S,µ)
and
IIf Iip = 1.
Moreover,
xg(f) = [ 1T, ]q/PfX Ig(t)Iq dµ (t)
[fX Ig(t) Iq dµ(t)]1 - 1/P = IIgIIq. Thus
,
we'see that
- 1)p = q,
IIfIIp =
IIxgII < IIgIIq be defined almost everywhere on
f -
(f E Lp(X,S,µ)).
p
IIxgII = IIgIIq The last assertion of the lemma is now apparent.
q/p q)
X .
by the Then,
8.3. Reflexivity of
Theorem $82. then 1 2, the assertion is established by Corollary
If
Therefore suppose
Lp(X,S,µ)
Lemma 8.3.2 asserts that
1 < p < 2.
can be identified via the antilinear isometry pace of
with a closed linear sU But
be a positive measure space. If is a reflexive Banach space.
(X,S,µ)
Let
227
as
2 < q < m,
< 2,
1
Lq(X,S,µ)*,
and so
where
g - xg
1/p + 1/q
Lq(X,S,µ)*
1.
is reflexive
by Theorem 8.1.2(ii), since it is the dual space of the reflexive Banach space
Lq(X,S,p).
Therefore, by Theorem 8.1.2(iii),
Lp(X,S,µ)
is reflexive.
Corollary 8.3.2. lei
If
I < p < m.
(X,S,I) \e ; positive measure space and
Let
then the mapping
1/p + l/q = 1,
g - x9*,
defined by xg(f) = fX f(t)g(t) dµ (t)
(f E Lp(X,S,µ)),
is a sur}ective antilinear isometry from
(Lq(X,S,µ),[[Alq"", to
(Lp(X,S,µ)*,11.11) Proof.
isometry from is such that
We know by Lemma 8.3.2 that Lq(X,S,µ)
g
Lp(X,S,p)*.
to
x**(xg) - 0, g E Lq(X,S,µ).
reflexive, there.exists some
f E Lp(X,S,µ)
xg
Suppose Since
is an antilinear x** E Lp(X,S,µ)** Lp(X,S,N)
such that
is
T(f) = x**.
But then
fX f(t)g(t) do (t) = x'f w T(f)(xg) x**(x*) g
= 0
from which it folle*s that
f = 0.
.(g E Lq(X,S,i+)),
8. Reflexivity
228
Therefore from Corollary 4.2.8 to the Hahn-Banach Theorem we is surjective.
g - xg
conclude that'the mapping
0 L1(X,S,µ).
We close this chapter with'a'few remarks about
is not uniformly
Except in trivial situations, convex.
E1,E2 E S
For suppose
finite measure and let
are two disjoint sets with positive
f - XEl/µ(E1)
and
denotes the characteristic function of XEk
is easy to verify that
g - XE2/µ(E2),
where
Ek, k - 1,2.
Theh it
IIf - gill - 2,
and
11f111 = lIBlil = 1
whereas
Il(f + g)/2111 It also happens that
fails to be reflexive inless
L1(X,S,µ)
it is finite dimensional (see Section 11.3). However, it can be shown that L.(X,S,µ),
at least when
Example 3.1.8.
is a-finite.
can be identified with
This was alluded to in
We state this result precisely in the next theorem.
Theorem 8.3.3. space.
µ
L1(X,S,µ)*
Let
Then the mapping
(X,S,µ) g
xg,
be a a-finite positive measure defined by
xg(f) = fX f(t)gTtT dµ(t)
(f E L,i(X,S,µ)),
is a surjective antilinear isometry from
11m)
to
A proof of this result, which we do not give, can be found in [DS1, pp. 289 and 290; Ry, pp. 246-248].
It is perhaps worthwhile mentioning that the mappings in Corollary 8.3.2 and Theorem 8.3.3 can be replaced by surjective linear isometries if we define
x*
by
x*(f) = fX f(t)g(t) dµ(t) where
g E Lq(X,S,µ), 3 < p < m,
and
(f E L(X, pS,µ)),
1/p + l/q = 1.
We have chosen the antilinear isometry so that the description
229
8.4. Problems
of
will coincide with the usual characterization of the
L2(X,S,p)*
continuous linear functionals on the Hilbert space terms of inner products.
L2(X,S,µ)
in
This latter result will be discussed in
Section 13.4.
Problems.
8.44.
be a Banach space over
Let
1.
the canonical embedding of reflexive, then
T(V)
into
V
V**.
T
denote
V
is not
is of category I in
Use the canonical embedding
2.
Tnd let
Prove that, if
T
:
4
V**.
V - V**
to prove that
every normed linear space
V
subset of a Banach space.
(Compare this with Problem 15 of Chapter 4.)
be a Banach space over
Let
3.
V
is isometrically isomorphic to a dense
t.
Prove that, if
is not reflexive, then its successive second duals
V**, V****,...,
are all-distinct.
4.
Let
and let V**
T1
(V1,11-111). and
V2
and
T2
and
into
denote the canonical embeddings of V2*,
respectively.
denote the adjoint of the adjoint of
on
S.
Prove that
6.
Recall that
V.
V'
Prove that
T.
T**
T** o T1 = T2 o T.
denotes the space of all linear functionals V
in
(V')' = V"
by
T'
Prove that
T'
is surjective
V
is more useful than
is finite dimensional.
Conclude that
V -. V",
T
:
V
V**'
V.
(Proposition 8.2.1)
nonmed linear space over such that
t E L(V1,V2),- let
T'(x)(x') = x'(x),x' E V'.
if and only if
7.
If
into
V1
is not reflexive.
(c,11-11.)
It is possible to embed
defined as
be Banach spaces over I
(V2'11.112 )
I.
Let
Prove that, if
be a uniformly convex (xn} C V
is a sequence
8. Reflexivity
230
(i) limnllxnll = (ii)
limn kiixn + xkll = 2,
is a Caucliy sequence.
(xn)
then
8.
convex set and
.llyo 9.
and let
Prove that, if
t.
yo
K c V
is a nonempty closed
then there exists a unique
K,
x0ll = infx E Kllyo -
PK
:
V
while if
the unique point of
E K
such
xll .
as follows:
K
yo f K.
0
PK(yo) = xo,
let
such that
K
"nearest point mapping"
PK
E K,
y
If
llyo
let
where
x
is 0
- xoli = infx E Kllyo - xll Prove that this
exists and is unique by Corollary 8.2.1).
10.
0
be a nonempty closed convex locally compact set.
'K C V
Define a mapping
(xo
x
be a uniformly convex Banach space over
Let
PK(yo) = yo,
be a 'uniformly convex
Let
(Corollary 8.2.1)
Banach space over
that
1
is continuous.
be a uniformly convex Banach space over
Let
then there exists a unique
x* E V*, x* # 0,
By Corollary, 8.2.2, if
4.
such that x*(xo) Let yo = llx*llx0. V. such that is the unique point of Uyoll = llx*ll and 0 x* (yo) = Ilx*llllyoll. Define a mapping T : V*- V by T(x*) = yo, x
0
E V, llxoll = 1,
Then
llx*ll.
y
x* 4 0,
and
T(0) = 0.
Prove that
T
is norm-preserving and sur-
jective, but need not be injective (and hence need not be linear). 11.
§.
Let'
(V,i.ll)
be a.uniformly convex normed linear space over
If x,y E V are such that lix -,y]] = llxll
x = ay *12,
for some
be a Banach space over
Let
x0 E V,
if
limnllxnll = Nxoll, (a)
(*):
(*).
prove that.
then
limnllxn - xoll
(For the case
L. :Consider the
given a sequence
limnx*(xn) *x*(x0), x' E V*,
Prove that, if
the property
llyll,
a E 1.
following convergence projrty and
+
fxn) c V
and if
0.
is uniformly convex, then
V
V = Lp(X,S,µ), 1 < p c m,
this
has
231
8.4. Problems
is called the Radon-Ripsz Theorem.)
Property (*) does not characterize uniform convexity.
(b)
Prove this by giving an example of a nonuniformly convex Banach space that has the property (*).
13.
Let
(V,II.11)
t.
be a uniformly convex Banach space over
If c > 0 is given, let 6(e) be such that if IIxu1 < 1, IIyII < 1, and Iix - yll > c, then II(x + y)/211 < 1 - b(e). If x* E V*, x E V, Ix*(x) - 11 < 6(c)/2). Prove that, 11x*II - 1, let E = (x if x,y E E for some x* E V* and if IIxII < 1, IIyII < 1, then I
Ilx-y11<e. 14.
V
Let
numbers such that
and give
V
the
be the space of all sequences xn j 0
£2
of complex
for at most finitely many values of
norm, so that
(a)
Prove that
(b)
Give an example of an
II(xn)II2 = (en= 1
n I,xn12)112.
is uniformly convex.
is it true that
1I(xn)112 = 1,
(xn)
x* E V*
such that for no
x*((xn)) = Ijx*Ij.
(xn) E V,
Does this contradict
Corollary 8.2.2?
be a normed linear space over t. V is said to be rotund if IIxUI < 1 and IIyII < 1, x # y, imply II(x + y)/21I < 1. 15.
Let
(V,11-11)
Prove that every uniformly convex normed linear space is
(a)
rotund.
Prove that every finite-dimensional rotund space is uni-
(b)
formly convex.
Give an example to show that the assumption of finite dimen-
(c)
sionality cannot be dropped in problem (b). 16.
over
I
Letting
be a uniformly convex normed linear space
and letting
W C V
be a linear subspace, prove each of the
following;
(a)
W
(b)
If
convex.
is uniformly convex.
W
is a closed linear subspace, then
V/W
is uniformly
8. Reflexivity
232
be a uniformly convex normed linear space over f and let 1 < p < m. Prove that, given c > 0. there exists some 6p > 0 such that ilxii < 1, ilyli < 1, and fix - r11 > c imply 17.
Let
(V, ii ii)
II (x + y)/211P < (1 - 6P) [ (IIx1IP + IIyIIP)/2). *18. If
be Banach spaces over
(Vn,Ii'IIn), n = 1,2,3,...,
Let
let
1 < p < m,
((Vn))
.f
denote the space of all sequences
P
x = (xn), xn E Vn,
§.
with componentwise
such that I= 1(1ixn11n)P < °D, Define a norm
addition and scalar multiplication.
11.11p
QP(IVn))
on
by
II(xn)Ilp = [
(11xnlin)P)1/P.
E n = I
(tP (IV
(a)
Prove that
(b)
If
1 < p < W and
t ((Vn))
is
1q((Vn)}.
and only if each
1 < p < m. V
6 > 0
P((V
lp(V)
prove that the dual of
P(IVn))
))
n
is reflexive if
when
Vn = V, n = 1,2,3,...,
is uniformly convex if and only if
(It can be shown that
convex if and only if each
Vn
is a Banach space.
1/p +'l/q = 1,
denote
is uniformly convex.
for all
)
Conclude that
Prove that
there exists a
p
is reflexive.
Vn
t (V)
Let
(c)
n)),11-11
Vn
ep(IVn))
is uniformly
is uniformly convex and, given
c > 0,
that satisfies the uniform convexity condition
simultaneously.
See, for instance, the article M.M. Day,
"Some more uniformly convex spaces", Bulletin of the American Mathematical Society, 47, 504-507 (1941).) *19.
space.
(Theorem 8.3.3)
Let
Prove that the mapping
(X,S,µ) g
xg,
he a a-finite positive measure defined by
x*(f) = fX f.(t)g(tT (Wt) is a surjective. antilinear isometry from
(f E L1(X,S,µ)), (L.(X,S,µ),iI.Iim)
(LI ((,S,µ)*, 11'11) 20.
Give an example of an
x* f L1([0,1))
x* E Lm([0,1])*
such that
to
233
8.4. Problems
M > 0,
be a reflexive Banach.space over
Let
21.
and suppose
and
(ca }a E A C I
CV*.
(xa}a E A
§,
let
Prove that
the following are equivalent:
There exists some
(a)
x* (x) = C a
(b)
a
l
a
such that
jjxjj < M
and
a E A.
IEaaacaI < MIIEaaaxa"JI,
possible subsets a
x E V
are not zero.
(aaJa CA
C :t
where the sums are taken over all where at most a finite number of the
CHAPTER 9
WEAK TOPOLOGIES
9.0.
Introduction.
In this and the following chapter we shall
discuss topologies on linear spaces that are generated by families of linear functionals.
This topic is a fundamental and substantial
one in functional analysis, and our treatment is neither definitive After a discussion of some of the general properties
nor exhaustive.
of such topologies, we shall specialize to two particular ones -the weak and weak* topologies -- and restrict our attention primarily to normed linear spaces.
These two topologies, the first on
the second on the dual space
functionals on is, by
V*
are generated respectively by the
V*,
continuous linear functionals on
and
V
V,
and by the continuous linear
that are determined by the elements of
V, that
i(V).
The general properties of the weak and weak* topologies will be developed in Sections 9.2 and 9.3, and in Section 9.4 we shall prove one of the most important theorems of the subject, the Banach-Alaoglu Theorem.
This theorem, in the form we shall prove, says that a set
in the dual space of a Banach space is compact in the weak* topology if and only if it is closed in that topology and norm bounded.
This
result provides us with an appropriate replacement in dual spaces of the standard Euclidean characterization of compact sets as the closed and bounded sets.
Some consequences of the Banach-Alaoglu Theorem
concerning metrizability and sequential compactness in the weak and weak* topologies will also be-discussed.
Sections 9.5 through 9.8
contain various applications of the Banach-Alaoglu Theorem. In Section 9.9 we shall give a characterization of reflexive Banach spaces in terms of the compactness of the norm closed unit
234
235
9.1. F-Topologies
ball.in the weak topology and an application of this result to show that in a reflexive Banach space every nonempty norm closed convex The final section is devoted
set contains a vector of minimum norm.
to a'theorem cpncerning adjoint transformations. The general subject will be pursued further in Chapter 10, in which we shall prove the Krein-9mulian and Eberlein-Amulian Theorems.
F-topologies.
9.1.
Our goal in this section is to describe
how one can introduce a topology into a linear space of families of linear functionals on
V
by means
and to-examine some of the
V
basic properties of such topologies.
To be more precise, suppose let F c V'
V
is a linear space over
be a family of linear functionals on
define a topology
TF
on
such that
V
x E V
in the topology
x' E F,
TF
functionals on
(V,TF).
We wish to
V.
(x ) C V
converges to
a
if and only if
and for which the functionals in
and
is a locally convex
(V,TF)
topological linear space for which a net
f
limox'(xa) = x'(x), F
Defining a topology
are continuous linear TF
on
V
that satis-
fies the last two requirements is not difficult. Indeed, recalling our earlier discussion of topologies in seminormed linear spaces (Section 1.5), we define a family P = (px, x E V
I
and
x' E F)
of seminorms on
x' E F.
Then the topology
TP
generated by
px,,...,Px,) = [y 1
where
c > 0, n E Z, n > 0,
is arbitrary, is such that only if
x E V
I
F
is a seminorm.
P, that is, the topology whose consists of sets of the form
k
and the choice of (xa)
converges to
limax'(xo) = x'(x), x' E F.
the elements of
px,(x) = Ix'(x)i, px,
Y E V, px,(x - y) < s, k = 1,2,...,n),
n
2
by setting
It is easily seen that each
neighborhood base at a point .U(x;a;px
V
xi,x2l,...,x' x
in
Tp
in
F
if and
Moreover, it is apparent that
are continuous with respect to
Ti,.
9. Weak Topologies
236
However, the pair
may not be a seminormed linear space
(V,TP)
-- equivalently, dot a locally convex topological linear space -since we have made no requirement that should imply
Ix'(x)1 = px,(x) = 0, x' E F,
If this were done, then from Proposition 1.5.1
x = 0.
and Theorems 1.5.1 and 2.3.1 we could immediately deduce that
(V,TP}
was a locally convex topological linear space with the desired proIt is easily seen that
perties.
if and only if
has the indicated property
F c V'
separates points.
F
We summarize this discussion in the next theorem, leaving the details of the proof to the reader. Let
1!'he orem 9.1.1.
separates points.
F C V'
be a linear space over
V
If
P = (px,
(x E V, x' E- F)
px, cx) = I x' (x) l P
then
is a family of seminorms on
seminormed linear space over logy on
(i)
(ii)
if
(V,TF)
A net
(V,P)
TF = TP
is a
is the topo-
P, then
is a locally convex topological linear space over (xa) c V
converges to
x E V
in
TF
f.
if and only
If
x' E F,
then
x'
is a continuous linear functional
(V, TF) . (iv)
of
Moreover, if
determingd by the family
V
such that
,
limof x'(xa) = x'(x), x' E F. (iii)
on
I.
V
and suppose
where
x' E F)
I
4
F
TF
is the weakest topology on
for which the elements
V
are continuous.
We shall often refer to the topology
TF
constructed in this
way as the F -topology.
Before discussing some basic results about F-topologies, we wish to present a useful alternative way of considering such topologies. Suppose points.
V
For each
is a linear space over x' E F
we set
f
ix, _ I
and
F c V'
separates
and define the papping
237
9.1. F-Topologies
T
:
nx'E Fix' by setting T(x) = (x'(x)); x E V. is that element of the product space nx'E
V
T(x)
That is, whose
F tx,
x'th coordinate
Let us consider
x'(x).
is
and the product space
TF
nx'
with the F-topology
V
as a linear space with coor-
E F ix,
dinatewise addition and scalar multiplication, and the usual product topology
Then
P.
T
linear spaces, and The mapping
V - nx'
:
T
T-I
hence
V
T(V)
:
E F ix,
is linear and continuous.
and
x = y,
so
T(x) = T(y),
separates points.
F
as
Considering
exists.
as a topological
T(V)
space with the relative topology inherited from nx,E T-1
at once that
phism between
then
is a bijective continuous linear mapping, and
V - T(V)
:
are topological
(nx' E F ix "
is also injective since, if
T
x'(x) = x'(y), x' E F, Thus
and
(V,TF)
is continuous.
(V,TF)
Hence
(T(V),PT(V))'
and
F
fx
we see
is a linear homeomor-
T
where
denotes PT(V)
the relative topology on F-topology
TF
on
and thus one can consider the
T(V),
as just the relative topology-on
V
sidered as a subspace of
T(V)
con-
11
x' E F ix'"
The usefulness of this alternative description of the F-topology is exemplified by the following theorem, whose proof is immediate on recalling that a countable product of metric spaces is again a metric space: Theorem 9.1.2. F t V'
Let
separates points.
V
be a linear space over If
F
*
and suppose
is countable, then the F-topology'
is metrizable.
TF
The next two results are straightforward, and their proofs are left to the reader. Proposition 9.1.1. suppose
c TF
TF
Fk C V'
Let
V
be a linear space over
separates points, k = 1,2.
If
i
F1 C f2,
and theri
2.
1
Proposition 9.1.2. Let (V,T) be a loc.I1ly convex topol linear space over I. If F c V* separates poi11Cs, then Tie t,1
9. Weak Topologies
238
x' E F
We noted in Theorem 9.1.1 that each linear functional on
Actually
(V,TF).
continuous linear functionals on
is a continuous
is precisely the set of
F
provided
(V,TF),
F
is a linear
space.
Theorem 9.1.3.
he a linear space over
V
Let
'
Then the follow-
is a linear subspace that separates points.
F e V'
and suppose
ing are equivalent:
(i) x* E F. (ii)
is a continuous linear functional on
x*
As already indicated,part (i) implies part (ii).
Proof.
versely, suppose
x*
is a continuous linear functional on
and, without loss of generality, assume that continuous at F
0,
such that, if
We claim that
x*
x " ...,px,),
x E U(0;e;px
x E 1>k= 1N(xk).
..Ipx ),
" px ".
is a linear subspace of
subspace.
Hence, if x E 1
m = 1,2,3,...,
and so
=
V,
Then
in
then
lx*(x)l < 1.
x 'x
...,x
xk(x) = 0, k = 1,2,...,n,
and so
and so
1N(xk),
that is,
Corollary 3.3.1 we see that
But each
lx*(x)l < 1.
fik
=
1N(xk)
x*
is a linear
then mx E fly= 1N(xk), flk
=
.
Consequently
1N(xk) C N(x*).
Thus from
is linearly dependent on
and hence can be expressed as a linear combination of
these elements of Therefore
n
Ix*(mx)l < 1, m = 1,2,3,...
x*(x) = 0. x E flk = 1N(xk); x',xl,...,x'
x1l,x2....,x'
n
2
1
is
x*
n
2
is a linear combinatior of
Indeed, suppose
N(xk)
and
Con-
(V,TF)
Since
x* # 0.
e > 0
there exists some
x E U(O;e;p'"p 1
implies
(V,TF).
F.
x* E F
as
F
is a linear space, and the proof is
complete.
This result, combined with Corollary 5.3.3 to the Hahn-Banach Theorem in its geometric form, immediately yields the following corollary:
U
239
9.2. Weak and Weak* Topologies
Corollary 9.1.1.
he a linear space over
V
Let
is a linear subspace that separates points.
F c :V1
and suppose
4
W c V
If
is a
linear subspace, then the following are equivalent:
The Weak and Weak* Topologies.
9.2.
such that
x'(x) = 0, x E W.
and
1
(V,TF).
x' E F
W, then there exists some
xo
If
(ii)
x'(xo) =
is a proper closed linear subspace of
W
(i)
For the remainder of this
chapter we concentrate our attention on two particular F-topologies, known generally as the weak and weak* topologies.
In this section
we define these topologies and discuss some elementary results. Definition 9.2.1. linear space over
F = V* C V set
'.
Let
(V,T)
he a locally convex topological
Then the F-topology on
is called the weak topology on
V
corresponding to In this case we
V.
TF = Tw.
By Corollary 4.2.1 to the liahn-Banach Theorem
V* C V'
separates
points.
In view of the discussion in the preceding section we see that the basic neighborhoods in {Y
and that a net
{x ) c V
are sets of the form
f%
Y' E V, lxk(x)
converges to
1,2,...,n}
- xk(Y)l < e, k x E V
in
that is,
Tw;
(Y
{xa)
converges weakly to
x,
if and only if
lim x*(x
a
)
= x*(x),
of
x* E V*.
Furthermore, from Theorems 9.1.1 and 9.1.3 we see that is a locally convex topological linear space over x' E V' if
is a continuous linear functional on
x' E V*;
that is, a linear functional on
4
(V,Tw)
V
(V,Tw)
and that
if and only
is weakly continuous
if and only if it is continuous.
We deduce the next proposition at once from Proposition 9.1.2.
9. Weak Topologies
240
Proposition 9.2.1.
linear space over
Let
(V,T)
Then
I.
be a locally convex topological
fw C T.
The weak* topology is an F-topology, not on
V,
but on
V*.
Before we can define this, we need to make a few preliminary observations.
Suppose
over J.
is a locally convex topological linear space
(V,T)
We claim, in analogy with our discussion of reflexive
normed linear spaces in Section 8.1, that every element of
(V*)'.
x E V
defines an
Indeed, as before, we define for each
x E V
T(X)(X*) = X*(X)
Then it is readily verified that T
:
V
T(x) E (V*)', x E V,
is an injective linear mapping.
(V*)'
identified with
Hence
as a linear subspace of
T(V)
separates points, since
T(V)
(X* E V*).
(V*)'.
and that V
can be
Moreover,
T(x)(x*) = x*(x) = 0, x E V,
implies
X* = 0.
We can now define the weak* topology. Definition 9.2.2.
linear space over to
TF = Tw
(V,T)
be a locally convex topological
Then the F-topology on
t.
F= T(V) c (V*)'
case we set
Let
V*
corresponding
is called the weak* topology on
V*.
In this
.
Again, in view of Section 9.1, we observe the following facts:
The basic neighborhoods in
Tw*
are of the form
U(X';C;T(X1),T(x2),...,T(Xn))
_
(y«
- (Y' =
Y* E V*, IT(xk)(x*) - T(xk)(Y*)< < c, k = 1,2,...,n) Y* E V*, lx*(Xk) - y*(xk)I < c, k = 1,2,...,n)
U (x' ;c;xl,x2,...,xn)
241
9.2. Weak and Weak* Topologies
and
converges to
(x*)
weak* to
Tw*;
in
x*
that is,
if and only if
x*,
converges
(x*)
x*(x), x E V.
Furthermore,
is a locally convex topological linear space over
(V*,Tw*)
is continuous in the topology
x" E (V*)'
weak* continuous if and only if it belongs to
Note that, if
V
if and only if
That is, a linear functional on
x E V.
for some
x" - T(x)
Tw*
and
4,
V*
ig
T(V) C (V*)'.
is a nonmed linear space, then
V*
is a
Banach space, and it can be equipped with both a weak and weak* topology by taking
F c (V*)'
V**
to be
and
T(V), respectively.
Propositions 9.1.1 and 9.1.2 then imeediately show that the following proposition is valid: Proposition 9.2.2. 4
be a nonmed linear space over
Let
and denote the norm topology on
V*
by
T.
Then
Tw* c T'
c: T.
We shall investigate the weak and weak* topologies on Banach spaces in considerable detail in this and the following chapter. As we have indicated, the notions of continuity and weak continuity for linear functionals on locally convex topological linear spaces are equivalent.
The same is true for linear transformations,
in general, between Frechet spaces. Theorem 9.2.1.
over
I
and suppose
Let
and
(V1,T1)
T E L'(V1,V2).
(V2,T2)
be Frechet spaces
Then the following are equi-
valent:
(i)
(ii)
T E,L(V1)V2). T :
Proof. a
T1
(V1,Ti) - (V2,T?)
Suppose
then
T E L(V1,V2)
neighborhood of
k - 1,2,...,n.
Then
is continuous.
0.
Let
xk E Vi,
and let
U(O;c;y1*,y2,....yn)
xk(x) - yk[T(x)), x E V,
and if
x E U(0;s;x*,x2,...,xn),
Iyk[T(x)]l - Ixk(x)I < c, k - 1,2,...,n;
T(x) E U(0;s;yi,y2,...,y*).
Thus
T
that is,
is weakly continuous at
0,
be
9. Weak Topologies
242
and hence
T
is weakly continuous by Proposition 3.2.2.
Hence
part (i) implies part (ii).
that
is weakly continuous.
T
Conversely, suppose
is a closed mapping and
T
by proving that
T E L(V1,V2)
applying the Closed Graph Theorem (Theorem 7.3.1).
Then clearly
we again define
y* E V*
mind, for each
fices, by Theorem 9.1.3, to show that
tinuous.
(xn) c V1, x E V1,
converges to
Then for each
is assumed to be con-
x* E Vi.
Hence
New suppose {xn)
is continuous, and so
T: (VI,T') -+ (V2,T2)
weakly continuous, and
To see this it suf-
is weakly continuous.
x*
y* E V2
But this follows at once since
With this in
x*(x) = y*[T(x)], x E V1.
x* E V.
We claim that
x* f Vi.
We shall show
T1
in
x
(T(xn))
and
are such that
y E V2
and
converges to
y
in
T2.
we have, on the one hand,
y* E V*
lim y*[T(xn)J= n
and, on the other hand, lim y*[T(xn)]
n
lim x*(xn
n
= x* (x) = y*[T(x)]. Thus
y*(y) = y*[T(x)], y* E V2,
and so from Corollary 4.2.1 to
the Hahn-Banach Theorem we conclude that Therefore
T
y = T(x).
is a closed mapping and so continuous by the
Closed Graph Theorem.
0
Thus a linear transformation between two Frechet spaces ig continuous if and only if it is weakly continuous.
Note that the only
if portion of this equivalence, that is, part (i) of Theorem 9.2.1 implying part (ii), linear space.
remains valid fos any locally convex topological
243
9.2. Weak and Weak* Topologies
As indicated in Proposition 9.2.1, the weak topology on a locally convex topological linear space
is weaker than the original
(V,T)
topology, so that, in general, a subset Tw
must be closed. in
E C V
that is closed in
but the converse need not always hold.
T,
Nevertheless, if we restrict our attention to certain algebraic types of sets in
V,
then the two notions of being closed are equivalent.
Theorem 9.2.2.
linear space over
Let
(V,T)
be a locally convex topological is a nonempty convex set.
K C V
and suppose
I
Then the following are equivalent: (i)
K
is closed in
Tw.
(ii)
K
is closed in
T.
Proof.
Clearly part (i) implies part (ii), as
assume that
K
is closed in
Tw C T, so we
From Corollary 5.4.2 to the geo-
T.
metric form of the Hahn-Banach Theorem we know that 'K
is the inter-
section of all the closed half-spaces that contain
Since each
of these half-spaces is of the form some
a E 2
(x
!
K.
x E V, x*(x) < a)
for
and for some real continuous linear functional on
V,
we see that each such half-space is weakly closed. Hence
K
is closed in
and part (ii) of the theorem implies
Tw,
part (i).
U W
In particular, we note that a linear subspace convex topological linear space
V
of a locally
is closed if and only if it is
weakly closed.
As an easy corollary of the theorem, whose proof is left to the reader, we have the next result: Corollary 9.2.1. linear space over to
x E V,
t.
If
(V,T) (xo,) c V
then there exists a net
finite sum of the form (y,)
Let
converges in
T
be a locally convex topological is a net that converges in
(y) c V,
ECr aOaxo, aOt > 0,
to
x.
where each 1,
y,
such that
Tw
is a
9. Weak Topologies
244
Thus, if a net {y,)
net
then there is a
x,
whose elements are convex linear combinations of the eleand which converges to
(xa)
ments of
converges weakly to
(x.)
X.
T and
Concrete examples of sets whose closure in
do not
V = £2 -with the (E;.
For example, let
coincide are easy to come by.
Tw
IIakI2)112.
usual norm topology given by the norm
= 1,2,3,...
ek = 0, k # m, em = 1, Item
that (em)
in
denote the sequence defined
em = (em) E £2
Let, as in Section 7.6, by
II(ak)II2 =
Then it is easily seen
.
from which it follows that the sequence
- en 112 = ,/2, m # n,
In particular,
does not converge in the norm topology.
£2
the origin does not belong to the norm closure of the set consisting em, m
of the sequences
However, from Corollary 8.3.2 we know that fied with
Indeed, if
£2.
x* E £2
and
then
x*(x) _ E Z. IakFk' x - (ak) E £2.
each
x* E £Z
x*
£2
can be identi-
corresponds to
(bkI E £2,
But then we see that for
lim x*(em) = lim bm = 0, m
as
,
F;k
m
l(bk(2 < . Thus the origin belongs to the closure of the
set consisting of the sequences
em, m = 1,2,3,...,
The analog of Theorem 9.2.2 with to be valid.
l
in
replaced by
T'w.
Tw*
fails
For instance, it is not difficult to give examples of
proper closed linear subspaces that are weak* dense.
This will
follow easily from the next theorem, which is of independent interest. Theorem 9.2.3.
be a normed linear space over
Let
f
and set
B1 = (x
B** = (x** 1
If
T
T(BI)
:
V
V**
I
(
x E V, IIxfl < 1),
X** E V**, IIx**II < 1).
is the canonical embedding of
is dense in
Bi*
in
(V**,TW*).
V
into
V**,
then
245.
9.3. Completeness in the Weak and Weak* Topologies
Proof.
and suppose
xo* E BI*
Let
open neighborhood of
in
x
Tw*.
such that
x0 E V, jjxo,! < 1,
exists some
is an
U(xo*;c;x*,X*,...,xn)
It suffices to show that there
T(xp) E U(xp*;e;x*,x*,...,xn). b = e/2 maxk
To this end let
It is readily seen that
`lyo*11 < I
yo* _ (1 - 6)xo**.
and set
,nI1xkjj
= 1,2,
and
y** E U(x**;c;x*,x*,...,xn).
From Corollary 4.10.1 to Helly's Theorem we deduce the existence of
x0 E V
some
such that
Ijx011
5
xk(xo) = yo*(xk), k = 1,2,...,n. Ixo*(xk)
-
- lly**ji)/2 < 1 and
Ily0*11 + (1
T(xo)(xk)l
Evidently
and
xo E B1,
Ixo*(xk) - xk(xo)l
Ixo*(x) - y**(x*)I (k = 1,2,...,n)
< c
shows that
T(x0) E U(xo*;a;x*,x*,...,xn).
Therefore
is weak* dense in
T(R
Corollary 9.2.2. If
T(V)
T
: V
V**
is dense in
T(V)
is weak* dense in of
T(V)
4.3.
0
be a nonmed linear space over .
Let
is the canonical embedding of
V
into
V**,
then
(V**,Tw*).
In particular, if c0), then
B**.
is a nonreflexive Banach space (for example,
V
is a proper norm closed linear subspace of V**,
V**
that
so that the norm closure and weak* closure
do not coincide.
Completeness in the Weak and Weak* Topologies.
From this
point through Chapter 10 we shall concentrate our attention solely on normed linear spaces.
We do this primarily to give the develop-
9. Weak Topologies
246
ment of the succeeding sections a somewhat more concrete flavor than would be the case if we tried to treat the various matters discussed in their fullest generality.
Some partial references to more general
treatments will be indicated at appropriate points.
Our concern in this section will be the completeness properties of the weak and weak* topologies. sequentially complete when quentially complete provided topology is complete if Theorem 9.3.1. (i)
(ii)
M > 0
sequence in
be a normed linear space over is a Cauchy sequence in Supkllxkll
0
such that
supkllx*Il < M. (iii)
is a Banach space, then
If
(V*,Tw*)
is sequenti-
ally complete. (iv)
is a reflexive Banach space, then
If
(V,Tw)
is
sequentially complete. Proof.
The proofs are all essentially applications of the Uni-
form Boundedness Theorem (Theorem 6.2.1). (xk) C V
is a Cauchy sequence in
Cauchy sequence for each
x* E V*,
(V,Tw).
For instance, suppose Then
(x*(xk))
and so there exists some
is a
Mx* > 0
such that
SuPIT(xk)(x*)I = SuPIx*(xk)I < Mx* Hence by the Uniform Boundedness Theorem and Theorem 8.1.1 there exists some
M > 0
for which
Sup Ilxkll = SUP IIT(xk)II < M. This proves part (i) of the theorem.
247
9.3. Completeness in the Weak and Weak* Topologies
A similar argument establishes part (ii), and part (iii) is proved through an application of the Banach-Steinhaus Theorem (Theorem 6.2.2).
The details are left to the reader.
If
sequence in
(T(xk)]
the weak* topology on (T(xk))
in
(V,Tw),
as
Therefore
is a Cauchy
then it is readily verified from the basic
(V,TW),
definitions that
once that
(xk}
is a reflexive Banach space and
(V,11-11)
is a Cauchy sequence in
Since
(V**,7w*).
is sequentially complete, it follows at
V**
converges in
(V**,7w*),
and so
(xk)
converges
T(V) = V**.
is sequentially complete, and part (iv) of
(V,TW)
the theorem is proved. 11
The argument used to prove the last portion of Theorem 9.3.1 is worth noting, as it is a special case of a general principle.
Namely,
if some property has been shown to hold for the weak4 topology, then the same property must be valid for the weak topology on any reflexive Banach space
V.
This follows immediately from the definition of the
weak and weak* topologies in fact that
T(V) = Vj*.
V
and
V**,
respectively, and the
We shall make repeated use of this observa-
tion in the sequel.
Among other things, Theorem 9.3.1 shows that
is sequentially complete in both the weak and weak* topo-
1 < p < -=,
logies.
Lp(X,S,µ),
It is well to note that a nonreflexive Banach space may or
may not be weakly sequentially complete, depending on the space in question.
For example,
weak topology when
µ
L1(X,S,µ)
is sequentially complete in the
is a-finite, whereas
c 0
Theorem 9.3.2. space.
Then
Proof.
Then, since
Let
(X,S,p)
(LI(X,S,µ),TW)
Suppose that
is not.
be a a-finite positive measure
is sequentially complete.
(fk) c LI(X,S,N)
is a weak Cauchy sequence.
L1(X,S,p)* = L.(X,S,p) (Theorem 8.3.3), we see that
(fE fk(t) d"(t)J
is a Cauchy sequence of complex numbers for each
9. Weak Topologies
248
E E*S,
as the characteristic function
L.(X,S,µ).
of
XE
belongs to
E
In particular, each such sequence converges, and so we
can'define a set function
a
on
S by
a(E) = lim rF fk(t) dµ (t)
(E E S).
k It can be shown that
a
is a bounded complex-valued measure on
,that is absolutely continuous with respect to portion of this assertion is the fact that is,
a
is countably additive.
a
X
The nontrivial
µ.
is a measure -- that
We do not 'give the details of this
here but instead refer the reader to the discussion of the VitaliHahn-Saks Theorem (see, for example, ;DS1, pp. 158-160; Y, pp. 70-72]), of which the preceding assertion is a'special case. Now, by the Lebesgue-Radon-Nikodym Theorem (see, for example, DS1, p. 176]), we deduce the existence of some
f E L1(X,S,µ)
such
that
a(E) = lim fE fk(t) dµ(t) k
s fE f(t) di+.(t) Furthermore, if
(E E S) .
is any finite linear combination of characteristic
g
functions of sets in
S,
then the preceding identity shows that
lim fX fk(t)g(t) do(t) = fX f(t)g(t) dµ(t) k
But such functions are norm dense in
LW(X,S,µ),
from which we deduce
by a simple triangle inequality argument that
lim fX fk(t) t tT dp(t) = fX f(t)F(t dp(t)
(h E Lm(X,S,N)).
k
Therefore and su
(fk)
L1(X,S,µ)
To see that
converges weakly to
f,
as
L1(X,S,p)* = L.(X,S,W)
is weakly sequentially complete.
c
0
O
is not weakly sequentially complete we need
only exhibit a0 weak Cauchy sequence in
In-order to do this we need to know that
c0
that does not converge. c*
can be identified with
9.3. Completeness in the Weak and Weak* Topologies
Ll,
The identification
a fact whose proof we leave to the reader.
corresponds
(ak) E tl
is obtained in the expected manner, namely, to
249
defined by
x* E co,
0 x*(y) =
(y - (bk) E co).
E lbkak k
define
n
Now for each positive integer
and
by
Then we see that for any
bk = 1, k = 1,2,...,n, bk = 0, k > n.
x = (ak) E .Cl
yn = (bk) E co
n>m m
n
(x*(Yn)
x*(Ym)I
-
E ak -
(
k=1
E akl
k=1
n
E
k=m+ 1 n E
9
C V
x* E V*
net.
as here constructed is a Cauchy and
y E A
a,6 E A, where a > y
x*(xa) - x*(x6) = x"(x*) - x"(x*) = 0,
a > 6,
if and only if a D 9.
and so
is such that
and (x
9 > y,
we obtain
is a weak Cauchy
9. Weak Topologies
252
if
is such that
x E V
each
however, contradicts the fact that Therefore
.(xa)
then for
(V,TW),
x* E V*
Furthermore, if
then for any a > y
from which we conclude that
x* E V*,
in
x
because,
V,
we have
x"(x*) = limax*(xa) = x*(x) = T(x)(x*) for
Hence
x*(xa) = x"(x*).
converges to
x* E y,
is such that
and 'y E A
each
(x a)
to any element of
limax*(xa) = x*(x).
we have
x* E V*
Tw
cannbt converge in
(xa)
But
x"
x" = T(x) E V**.
is discontinuous on
does not converge in
(V,TW),
This, V*.
and so
(V,TW)
is not complete.
0
It is easily seen that-, if we begin with a nonreflexive normed
then the proof that
linear space when
V
is not complete
(V,TW)
Indeed,
is infinite dimensional can be somewhat shortened.
in this case we need only take
The existence of If, however,
V
x"
and argue as before.
x" E V** -, r(V)
is now assured by the definition of reflexivity.
is reflexive, then we must use the discontinuous
linear functional
x"
as'done in the proof of the theorem.
Next let us give some necessary and sufficient conditions for the convergence of a sequence in the weak or weak* topology.
The
proofs are straightforward and are left to the reader. Theorem 9.3.4.
be a normed linear space over
Let
I.
Then (i).
only if
A sequence supkIIxkI
0
V*.
Clearly, since
such that
E C aBi.
is weak* compact, since scalar multiplication is a homeo-
morphism in
(V*,7w*)
(Theorem 2.1.2), and so
E
is weak'.compact,
as it is a weak' closed subset of a weak* compact set. assume that
Thus we may
E - Bi.
Now for each x E V
let KT(x) -
(p
a E i,. !al < J'T(x)1J - JjxjJ).
256
9. Weak Topologies
Clearly each
then T(x*) - {x*(x)) and
claim that topology on suppose and
IIx
I x*(x)I < 1`x*IUUjxtI < I1x1j.
is a compact subset of
T(B!)
T(B!).
is a net in
[x*(x))
Since
We
in the relative
T(V*)
considered as a subspace of
T(V*)
and
The latter assertion is valid since, if
T(B1) C IIx E V KT(x).
x* E B!,
0T(x) _ f,
is a compact subset of
KT(x)
nx
Indeed,
E V 1T(x)
T(B*) C IIx
E V KT(x)
is compact in the product topology, by Tikhonov's'
E V KT(x)
Theorem {Ry, pp. 166 and 167], we see that there exists a subnet (x*(x))
of, (x*(x))
(x;(x))
converges to
and some ip = (cpx) E II
X E V KT(x)
in the product topology.
cp
such that
Moreover,
cp E T(B*).
To see this we first show that an element
x*
and suppose
of
B!.
s > 0.
x*(x) = cpx, x E V,
With this in mind, let
x,y E V
defines
be given
Let
Ux* [a I aEt, (a - cpxl a
in
TT(F) C
If
and
ao
c > 0,
let
is such that
y E F
be such that
jx*a(y)
- x*(y)l < e/3
then we also have
lxa(x) - x*(x)l :SIX*
- xa(Y)I ' Ixa(Y)
- x*(Y)I + ix*(Y) - x*(X)!
9. Weak Topologies
260
IIx
I
I
* Ilx*II lly - xll
<e whenever
a > ao.
Thus
converges to
(xQ,)
Note that, if
F C V
separates the points of F-topology
TT(F)
on
Theorem 9.4.2. B* = (x*
and let
I
x*
in
(V*, Tw*) .
is norm dense in
V*,
and
V,
D then s(F) C V**
does indeed define an
T(F)
V*.
Let
(V,11-II)
be a.normed linear space over
4
Then the following are
x* E y*, IIx*Il < 1).
equivalent:
(i)
is separable.
(V, II Il)
The.weak* topology
(ii)
T'w*
V*
on
restricted to
is
B1*
metrizable. If
Proof.
dense subset of V*
V V.
F C V
is separable, let
be a countable norm
Then by Theorem 9.1.2 the topology ,Tr(F)
on
is metrizable, and hence by Lemma 9.4.1 the weak* topology re-
stricted to
B*
is metrizable.
Thus part (i) implies part (ii).
Conversely, suppose the weak* topology on
Bi
is metrizable.
Then, in particular, there exists a sequence of open neighborhoods Un of the origin in
T'w*
such that
Cn°. lUn a (0). /Without loss
of generality we may assume that Un = Un(0,an,Fn) _ (x*
where
Fn C V
countable. x E F. that
x* E I1
x* E V*, Ix*(x)I < an. x E n
is a finite set.
Moreover, suppose
Then
I
Let
F = (.r
x* E V*
1Fn.
n is such that
Clearly
F
Ix*(x)I < an, x E Fn, n - 1,2,3,...,..which shows . lUn
- (0).
Hence
x* = 0.
is
x*(x) a 0,
261
9.4. Banach-Alloglu Theorem
denotes the collection of all finite linear combinations
F
If
of elements of
with rational coefficients, then F is obviously
F
a countable subset of
V
V by the previous
to the Hahn-Banach Theorem.
4.2.8
Corollary
observation and Therefore
that is norm dense in
V
is separable, and part (ii) of the theorem implies
0
part (i).
It is evident that
in.the theorem could be replaced by ar;
Bi
norm closed bounded ball (x*
I
x* E V*, IIx*li < a).
Similarly, the
proof of the first portion of the theorem shows that the weak* topology on any norm bounded subset of
is metrizable, provided V
V*
is separable.
These observations, combined with the fact that compactness and sequential compactness are equivalent in metric spaces, give us the next two corollaries. Corollary 9.4.4.
space over E
f.
If
be a separable normed linear
Let
is weak* closed and norm bounded, then
E C V*
is weak* sequentially compact. Let
Corollary 9.4.5 (Helly's Selection Theorem .
a separable normed linear space over such that
supnuxn11 < M,
then
f.
(xn) c V*
(V,1111)
be
is a sequence
has a subsequence that is
(xn)
weak* convergent to some point in
If
[x*
I
x* E V*, 1(x*iI < M).
For reflexive spaces the preceding results combined with Theorem 4.5.1 give us the following corollaries.
The details are, left to
the reader.
Corollary 9.4.6.
over
f
and let
be a reflexive Banach space
Let
B1 = (x
I
x E V, (ixil < 1).
Then the following are
equivalent:
(i) (ii)
is separable. The weak topology
Tw
restricted td
B,
is metrizable.
9. Weak Topologies
262
Let
Corollary 9.4.7.
space over E
t.
is weakly closed and norm bounded, then
E C V
If
be a separable reflexive Banach
(V,JH-Jj)
is weakly-sequentially compact. Furthermore, by either arguing as in the proof of Theorem 9.4.2
mutatis mutandis, or by applying that result to the dual and bidual
canonical embedding of
and using some of the properties of the
V
of i normed linear space
into
V
we can prove the next result.
V**,
Again the details are left to the reader. Theorem 9.4.3. let
and
BI = {x
I
Let
be a normed linear space over
(V,11I)
x E V,
Ilxjj < lj.
t
Then the following are equi-
valent:
(i) (ii)
(V*, 11.11)
is separable.
The weak topology
Tw
on
V
restricted to
BI
is metri-
zable.
As seen by the example of the closed unit ball in compact set need not be weak* sequentially compact.
£*,
a weak*
Corollary 9.4.4
asserts that weak* compactness implies weak* sequential compactness if
V
is a separable Banach space, while Corollaries 9.4.2 and 9.4.1
together assert that weak compactness implies weak sequential compactness in separable reflexive Banachispaces.
It is an extremely
profound, and perhaps surprising, fact that weak compactness and weak sequential compactness are equivalent in any Banach space.
This
is the content of the Eberlein-9mulian Theorem (Theorem 10.3.1), to be proved in the next chapter.
No such result is available for the
weak* topology.
The Banach-Alaoglu Theorem has a valid analog in the context of locally convex topological linear spaces. nor use this generalization.
We neither discuss
The interested reader is referred to
[K, pp. 245-249; KeNa, p. 155; W1, pp. 236- 241].
9.5. Banach Spaces as Spaces of Continuous Functions
265
In this
Banach ch Spaces as Spaces of Continuous Functions.
9.5.
short section we shall see how every complex Banach space can be represented as a space of continuous functions on a compact Hausdorff topological space. Theorem 9.5.1.
be a Banach space over
Let
there exists a compact Hausdorff topological Space
X
Then
C.
such that
is isometrically isomorphic to a closed linear subspace
(V,il.ll)
of (C(X),I1'll.). Proof.
Let
X = B* = (x*
Banach-Alaoglu Theorem (Theorem 9.4.1),
is a compact Hausdorff
X
topological space in the relative weak* topology. define
fx(x*) = x*(x), x* E X.
function on
and since
X,
topology if and only if x E V,
$
V
:
C(X)
Clearly
fx
For each
(x*)
converges to
x*
converges to
x*(x)
a
fx E C(X).
defined by
x E V
is a complex-valued
(x*(x))
we know at once that
mapping
Then, by the
x* E V*, IIx*II < 1).
I
in the weak* for each
It is easily seen that the
$(x) = fx, x E V,
is linear.
Moreover,
Ut (x) Iim = IlfxJ. sup Ifx(x*)1
X* EX sup
x*EB*
IT(x)(x*)1
= IIT(x)II
shows that
$
is an isometry.
The conclusion of the theorem is now evident.
The function
fx
is, of course, just
T(x)
0 restricted to
8
9. Weak Topologies
264
Obviously a similar result for Banach spaces over
!H
and spaces
Moreover, there
of real-valued continuous functions is also valid.
is an analog for arbitrary locally :onvex topological linear spaces:.
,every such space is topologically isomorphic to a linear subspace of
where
C'(X),
is a suitable locally compact Hausdorff topo-
X
Refer to Example 1.2.1 for the definition of
logical space.
C'(X).
The details are available, for example, in [K, pp. 250 and 251].
9.6.
We discussed Banach limits in
Banach Limits Revisited.
Section 4.3'as an application of the Hahn-Banach Theorem.
Recall
that the aim was to extend the notion of the limit of a convergent sequence in a reasonable way to all bounded sequences, "reasonable" meaning that the extension should be linear, invariant under shifts, and in agreement with the usual notion of limit for convergent sequences.
limits on
We now want to again prove the existence of such Banach but this time we use the Banach-Alaoglu Theorem.
t.,
Actually this time we only consider real sequences -- that is,
the Banach space of some (i)
(ii)
jakj E tRR
if
00
For each
converges, then x*((ak].) = limkak.
n = 1,2,3,...,
if
(ak) E LRR
and
ck F ak + n,
x*((ak)) - x*(jck)).
then
(ak) E
If
We wish to prove the existence
lR.
such that
x* E (£.R)*
k = 1,2,3,..., (iii)
over
LWR
and
ak > 0, k - 1,2,3,...,
then
x*((ak)) > 0.
Condition (iii)
was not imposed on a Banach limit before.
The
virtue of this condition is elucidated by the following lemma: Lemma 9.6.1. (i)
(ii)
For each
if
that
x* E
ek = 1, k = 1,2,3,...,
If
k = 1,2,3,..., (iii)
Suppose
then
n = 1,2,3,...,
is such that
then if
x*((ek)) - 1.
(ak) E
and
Ck
ak t n'
x*((ak))
= x*((ck)), (ak) E 1R, ak > 0, k - 1,2,3,..., 40
-
then
x*((ak))
0.
265,
9.6. Banach Limits Revisited
{ak) E Lm
Then for each
lie inf : k < x*((ak)) < lie sup ak.
k
k Proof.
(ak) E LR and
Let
Consider a sequence
be given.
Then there exists
ak < lim supmam + e.
such that
N
some positive integer
e > 0
(bk) E Lm defined by bk = min(ak, lim sup am + c)
Clearly
and
(ak)
(k = 1,2,3,...).
differ in at most a finite number of com-
{bkj
ponents, and hence there exists some positive integer
n
such that
(ck) be the sequence defined Let bk+ n' k = 1,2,3,... Then x*((ak)) = x*((ck)) k = 1,2,3,... ck = ak + n s b k+ n,
ak + n
by
for k > N.
.
.
x*((bkj)
part (ii) of the hypotheses.
by
But it is evident that
bk < lie supmam + c, k = 1,2,3,...,
and so from parts (i) and (ii) of the hypotheses we conclude that x*((ak)) = x*({bkj)
< (lim sup am + c)x*({ek)) m Its sup as + a.
m Since
c > 0
The inequality
is arbitrary, we see that
lim infmam < x*((akj)
x*({ak)) < lie supmam.
is proved in a similar
manner.
In particular, if of the lemma, then and so
x*
x* E (LR)*
satisfies all three conditions
x*((ak)) = limkak
whenever
(akj
is convergent,
is a Banach limit.
The proof of the next theorem then reduces to showing the existence of some Lemma 9.6.1.
x* E (LRR)*
that satisfies all three conditions of
9. Weak Topologies
266
(i)
(ii)
(ak} E e is convergent, then
If
For each(n = 1,2,3,...,
(iv)
x*({ak)) = limkak.
(ak) E Lm
if
If .(ak) E ,, ak > 0, k = 1,2,3,..., If
such that
and
ck = ak
+ n
x*((ak}) = x*({ck)).
then
k = 1,2,3,..., (iii)
x* E ( R)*
There exists some
Theorem 9.6.1.
x*((ak)) > 0.
then
then
(ak) E
limkinf ak < x*((ak)) < limksup ak. As indicated, wee need only prove the existence of some
Proof.
With
that satisfies the three conditions of Lemma 9.6.1.
x* E (.em)*
m
this in mind, for each positive integer
we define
E ak
xm({ak})
((ak)'E tR).
k - l
xm E (IR)*
It is easily seen that
Il(ek)lim =
1
ahd
an = 1,2,3,...
Let j
F.
= 1,2,3,...
xm({ek}) = 1, we conclude that
denote the closure of
(x*
and that the family
(F.]
rw* ),
Since
B*
x* E r1
1,2,3,...),
has the finite intersection property; Fj
have a nonempty inter-
is weak* compact, from the Banach-Alaoglu
Theorem (Theorem 9.4.1), we see that 1Fj.
r r_ 1Fj } (.
We claim that
Indeed, it is evident that
11x*jj < 1.
sider the-open weak* neighborhood
that
R
((L)*,
in
x* E (gym)*, I1x*II < 1)
that is, any finite number of the sets
x* E r r. 1F.,
m > j)
(xm
It is apparent that
.
I
Let
Iixm11 = 1,
.
Fj c gi
section.
But, since
and 'jIxmjE < I.
x*
has the desired properties.
Moreover, given
U(x*,c,(ek))
of
e > 0,
x*.
we see that there exists a positive integer
xo E U(x*,e,{ek]).
Hence
Ix*((ek)) - x*({ek))I = Ix*((ek1)
< e,
-
11
con-
Since
m
such
9.7. Fourier Series in
267
Lp((-n,n],dt/2n), I < p < -
and so x*((ek)) = 1 and
is arbitrary.
as a > 0
11x*11 = 1,
__RRak
(ak) E
A similar argument shows that, if k
x*((ak1) > 0,
then
1,2,3,...,
0,
x*((ak)) > 0,
rsince
m
To prove that
x*
satisfies part (ii) of the theorem it clearly
suffices to show that, if then
x*((ak)) - x*((ck)).
ck = ak + 1, k = 1,2,3,..., We note first that
(ak) E fm
and
m+l xm{ak)) - xm((ck)) - n
k=1
ak -
E ak)
k=2
al - as + 1)
(m = 1,2,3,...).
Ixm((ak)) - xm((ck))l < 211(ak)'L m
(m = 1,2,3,...).
s
a Consequently
Now let
a > 0
be given and consider the weak* neighborhood
U(x*,a/3,(ak),(ck)). a > 61+(ak)'fm/a
Then, since
such that
1F,
x* E (1:
there exists some and so
xm E
lx*((ak)) - x*((ck))I < lx*((ak)) - xm((ak))I
+ Ixm((ak)) - xm((ckN + Lx*((ck)) - x*((ck})j
2{I(ak)I1m
24
0
is arbitrary, and
the proof is complete.
9.7.
Fourier Series of Functions in
0 Lp([-n,n),dt/2n),
1
j,
we see that if
dt
2n f_n
m
m
However, given
dt =
n dt =
({ck))(t)e-ijt
an
n
k= -n m
m
an eikte-llt
k
-
Em (1
m
)c k(2n
dt)
J
=(1- Ani ) c, since
1fnn ei(k - J)t dt = 1 for k = j,
(k,j E$).
2n fnn ei(k - j)t dt = 0 for k # j The last assertion is easily verified.
Hence we find that
10) = lim _ fnn an ({ck)) m
(t)e-ijt
dt
m
= lmm (1 - n_lZLi)cj (j E 7L).
= C. Therefore part (ii) of the theorem implies part (i).
E)
The second implication in Theorem 9.7.1 could also have been proved by arguing directly from the Banach-Alaoglu Theorem (Theorem 9.4.1) since
{x*
I
x* E Lq([-n,n],dt/2n)*, jjx*j( < M)
closed norm bounded set and hence weak* compact. left to the reader.
is a weak"
The details are
9. Weak Topologies
272
In the case
p - 2
ck = f(k), k E Z, !.k =
-Jck12
one can show that
for some
{ck)
is such that
f E L2([-n,n),dt/2n)
if and only if
We shall return to this when we discuss Hilbert
spaces in Chapter 13.
Multipliers.
9.8.
An argument like the one utilized in the
preceding section can be used to characterize those to
L(LI(gidt),LpMdt))
T
belonging
that commute with convolution.
We make
the following definition: Definition 9.8.1.
1 < p < m and
If
T E L(LI(gt,dt),Lp(R,dt)),
then. T is said to be a multiplier from L1(JR,dt) to Lp(,R,dt) T(f) * g - T(f
if
g), f,g E L1OR,dt).
Before we characterize multipliers, at least in the case we note the following lemma:
1 < p < m,
Lama 9.8.1.
There exists a sequence
[uk) C LI(ki,dt)
such
that Jjukul - 1, k - 1,2,3,...
(i)
(ii) If f E LI OR,dt), then '
Proof.
X[-l/k,1/kl. [-1/k,l/k).
Take
.
limklluk
f - "I - 0.
uk - (k/2)X[-l/k,l/k)' k
1,2,3,...,
where
is the characteristic function of the closed interval
0
The sequence
{uk)
is called in approximate identity for
LIOR,dt).
Theorem 9.8.1.
Let
1 < p.< o and T E L'(LIQR,dt),Lp(4i,dt)).
Then the following are equivalent: (i)
There exists some
h E Lp QR,dt)
such that
T(f) a f * h,
f E LI ( ,dt) . (ii)
T
is a multiplier from
LlOR,dt) to
L OR,dt).
P
273
9.8. Multipliers
Proof.
That part (i) implies part (ii) is immediate from Pro-
If
is a
is a sequence that satisfies the
(un) c LIOR,dt)
conclusions of Lemma 9.8.1, then for each
IIT(f)
T
Conversely; suppose that
positions 4.7.1(1) through (iv). multiplier.
f E L1( t,dt)
we have
- T(un) * flip = IIT(f) - T(un * f)Iip 1, x* E W, as if x* E W, then there exists some n for which x* E K n , and so x* 4 F°0(1 F°1 1 (1 F° n-1 l1n =
hence there exists some
j,
0 < i < n
Ix*(x) - xo*(x)I > 1, x E F3.
that is,
-
1,
such that
x* f FOtt
The indicated estimate is
now apparent.
Now consider the mapping
Clearly S E L(JI,co)
V* -- co
S
IISII < supk1lxklll,
and
defined by
as
S(x*) = (x*(xk)).
limkllxklll = 0.
Moreover, inf
IIS(x*)
- S(x*)II
x* E W*
0
=
[sup Ix*(xk) - xo(xk) inf x* E W* k
> 1.
Thus from Corollary 4.2.4 to the Hahn-Banach Theorem and the fact that
c*
can be identified with
existence of some sequence
(a)
(b)
£
[ak) E .21
(Example 3.1.6) we deduce the such that
l°= lakx*(xk) = 1,
k=1
akx* (xk) = 0
(X* E W).
9. Weak Topologies
282
But, since
is obviously an element of
x0 - I;. lakxk
preceding equations say precisely that x* E W*.
y*[T(xo)] = 0, y* E V2,
means that
xo(xo) = then
W = T*(VZ),
However, since
since
x0 = 0,
x*(xo) = 0, x* E W,
and hence that
T(xo) = 0,
This conclusion, however, contra-
is injective.
T
x*(xo) = 0,
and
1
the
and so from Corollary 4.2.6 to
the Hahn-Banach Theorem we conclude that
dicts the fact that
V1,
xo(xo) = 1.
W = Vi,
Consequently
and part
(iii) of the theorem implies part (i).
Therefore all three parts of the theorem are equivalent.
A construction similar to that used in the last portion of the proof will be used again during the investigation of neighborhood bases for the bounded weak* topology to be discussed in Section 10.1.
Problems.
9.11.
(Theorem 9.1.1)
1.
suppose
F C V'
Let
separates points.
P = (px,
where
px,(x) = tx'(x)I, x E V
of seminorms on Furthermore, let family
V
such that
TF = TP
I
and
Prove that, if x' E F),
I
and (V,P)
x' E F,
then
P
is a family
is a seminormed linear space.
be the topology on
V
determined by the
and prove the following:
P
(a)
be a linear space over
V
(V,TF)
is a locally convex topological linear space over
4. (b)
if
(xo) CV converges to
limox'(xa) = x1(x) (c)
on
A net
If
x' E F,
for each then
x'
x E V
in
TF
if and only
x' E F.
is a continuous linear functional
(V , TF) . (d) F
TF
is the weakest topology on
are continuous.
V
for which the elements
283
9.11. Problems
2.
suppose
F c V separates points.
then the F-topology 3.
be a linear space over
V
Let
(Theorem 9.1.2)
TF
FI c F2 ,
TF c TF
be a linear Apace over
V
Let
I
Prove that, if
separates points..
Fk c V', k = 1,2,
then
and
I
is countable,
F
is metrizable.
(Proposition 9.1.1)
and suppose
Prove that, if
2.
I
4.
(Proposition 9.1.2)
logical linear space over
suppose
be a locally convex topo-
(V,T)
Prove that, if
separates
F C V*
TF c T.
points, then 5.
Let
I.
(Corollary 9.1.1) F c V'
that, if Wc V
Let
V
be a linear space over
6
is a linear subspace'that separates points.
and
Prove
is a linear subspace, then the following are equi-
valent:
and
V
Let
F2
are linear subspaces of C TF ,
TF
I
V'
and suppose
f
both
that separate points.
FI
Prove
FI = F2.
then
2
1
over
be a linear space over
6.
7.
for which
and x'(x) = 0, x E W.
I
that, if
(V,TF),
then there exists some x" E F
xo f W,
If
(ii)
x'(xo) =
is a proper closed linear subspace of
W
(i)
(Proposition 9.2.2)
be a normed linear space
Let
and denote the norm topology on
I1*
by
T.
Prove that
T`''*cTwcT. 8.
(Corollary 9.2.1)
gical linear space over converges in each
9. f.
to
(V,T)
be a locally convex topolo-
Prove that, if
f.
(xa} C V
then there exists a net
x E V,
being a finite sum of the form Eaa0axa,
y,
I'aaOa = 1,
over
Tw
Let
such that
(y$}
(Corollary 9.2.2)
Prove that, if
converges in Let
r
:
V -+ V**
T
to
aOa
is a net that (y0} C :V'
> 0,
x.
be a normed linear space is the canonical embedding
9. Weak Topologies
284
of
V
into
then
V**,
(V**,TW' ).
is dense in
T(V)
J
10.
be a Banach space over .
Let
(Theorem 9.3.1)
and prove each of the following: (a)
If
(xk) C V*
M > 0
there exists some (b)
*11.
over
I
such that
over
I
(a)
only if
(Theorem 9.3.3)
supklIxkII < M.
and suppose
be a normed linear space
Let
is a complete locally convex topo-
(V*,Tw*)
(Theorem 9:3.4)
is finite-dimensional.
V
Prove that
f.
be a normed linear space
Let
and prove the following:
A sequence
and only if
in
x
limkx*(xk) = x*(x)
supkI`xkII < W and
A sequence
converges to
(xk) c V
some norm dense subset of (b)
then
is sequentially complete.
(V*,Tw*)
logical linear space over 12.
(V*,Tw*),
is a Cauchy sequence in
(V,Tw)
for each
if and is
x*
(V*, (I.1I) .
(xk) C V*
x*
limkxk(x) = x*(x)
and
supkIIxkil < cD
converges to
in
if
(V*,Tw*)
for each
in
x
some norm dense subset of (V, II 1) .
r 13.
(Corollary 9.3.1)
Let
be a a-finite positive
(X,S,p)
Prove that the follow-
(fk) C LI(X,S,µ).
measure space and suppose ing are equivalent: (i)
(ii)
supkllfkul < m,
(fE fk(t) dµ(t)) *14.
vectors
f E L1(X,S,µ)
There exist, some
converges weakly to
(fk)
Let
E
the
and for each
nth
E E S
the sequence
converges. be a 'subset of
1 < m < n < m),
(xmn
such that the sequence
f.
coordinate is
is
1,
0.
Prove that the origin of
P,
where the m,
example is due to von Neumann.)
mth_ coordinate of
E
xmn
and all other coordinates are
is in the weak closure of
[.p
that no sequence of elements of
1 < p < , consisting of the
converges weakly to zero.
E,
but (This
285
9.11. Problems I .S. Prove Let
16.
over
that
Co (lit)
is the union of countably many sets that
V
are closed and nowhere dense in Recall that both
17.
c*
(V,Tw).
and
can be identified with
co
has two weak* topologies
Thus
TF1
18.
Let
TF2
and
T F1 c T F2
we have neither
(V,11.11)
on
nor
is a sequence in
(xk)
for r2
.21.
F1 = r1(c)
co .. ca*. Prove that
:
are incomparable -- that is,
£1
FF2 C TF1.
be a Banach space over f.
is weak* sequentially closed in that, if
TFl
F2 = r2(co)
c -- c** and
:
the topologies
where
TF2'
and
fl
for r1
L,,(IR) .
be an infinite-dimensional normed linear space
(V,11.11)
Prove that
t.
is weak* dense in
V'.
V*
Prove that
V*
(Sequentially closed means limkxk = x*,
and
then
x* E V*.) 19.
yet
(V,1I.11)
Ba - (x*
let
be a Banach space over
x* E V*, IIx*II < a).
I
i
Prove that
and for
a > 0
is closed in
Ba
(V*,TW+).
*20.
Give an example of a Banach space
V
for which (V*,T*
(x*
I
x* E V*,
21.
over
i
(Corollary 9.4.1)
and let
norm bounded, then 22.
is'not compact in
UIx*11 = 1)
E c :V*. E
§
and let
(V,11.1I)
Prove that, if
be a normed linear space E
is weak* closed and
is weak* compact.
(Corollary 9.4.2)
space over
Let
W ).
Let
E c V.
(V,11.11)
be a reflexive Banach
Prove that the following are equi-
valent:
(i)
E
is a compact set in
(ii)
E
is a closed set in
23.
over I.
(Corollary 9.4.3)
Let
(V,Tw). (V,Tw)
(V,11.11)
and a bounded set in
(V,11.11).
be a-normed linear space
Prove that., if (x*) a C V* is a net such that
M,
9. Weak Topologies
286
(x*)
then
(x*
x* E V*,
I
24.
be a separable normed
(Corollary 9.4.5)
Prove that, if
f.
then
is a sequence such
(xn) C V*
(xn)
has a subsequence
(x*) (x*
(Corollary 9.4.6)
Let
space over f
is weak* closed and
be a separable normed
Let
converges to some point of 26.
E C V*
is weak* sequentially compact.
E
supnIIxnjl < M,
(V*,
Prove that, if
I.
linear space over
that converges to some point of
Let
(Corollary 9.4.4)
norm bounded, then
that
in
UUx*jj < M)
linear space over
25.
(x*)
has a subnet
x* E V*,
I
(V,11.11)
and let B1 = (x
(V*,Tw*).
in
jjx*ij < M)
be a reflexive 9anach
x E V, lixjI < 1).
I
that
Prove that the
following are equivalent:
is separable.
(i)
The weak topology
(ii)
27.
(Corollary 9.4.7)
Banach space over
I.
norm bounded, then 28.
over
f
E
restricted to
B1
is metritable.
be a separable reflexive
Let
Prove that, if
E C V
is weakly closed and
is weakly sequentially compact.
(Theorem 9.4.3)
and let
Tw
Let
B1 = (x
I
(V,li'i:)
x E V,
be a normed linear space
jxjj < 1).
Prove that the follow-
ing are equivalent:
is separable.
(i)
The weak topology
(ii)
Tw
V
on
restricted to
is metri-
B1
zable.
29. F. C :V,,'
in
V*
Let
be a separable Banach space over
be a convex subset of if and only if
imply that 30.
(V,11 -:j)
Let
V*.
(xn) C E
Prove that and
E
f
and let
is weak* closed
limnx*(x) = x*(x), x E V,'
x* E E.
be a Banach space over
f
and let
287
9.11. Problems
BI = (x
restricted to
V
family
over
Let
and let
t
Let
32.
over
be a linear subspace of V.
F c V*
if and only if
V*
(V2,T2)
and
(VI,TI)
and let
4
contains a countable
V*
be a locally convex topological linear space
(V,T)
is weak* dense in
F
is metri:.able, then
be topological linear spaces
be any mapping from
T
(VI,T1)
V1
is continuous for every
:
then
x* E
Ijx*ji, *
semicontinuous on Let
subspace
(V,1H1)
W c V*
W e V*
over in
x E WI
such that
prove that
37.
Let
x*
in
Defining
4.
is lower
(V,11.I)
[1(xn)]
W
A linear
4.
x* E V* - W
Prove that a sub-
is weak* closed.
be nonmed linear spaces If
(x n)
converges weakly to
converges weakly to
T(x)
in
V2.
be a reflexive Banach space over
4.
If
is a closed linear subspace of
flexive.
x*(x) f 0.
and
T E L(V1,V2).
V1,
W C V
p
be a normed linear space over
is saturated if and cnly if
and let
4
x* E V*, prove that
is said to be saturated if for each
Let
36.
n
(V*,Tw ).
there exists some space
and let
4
converges to
(x*)
be a nonmed linear space over
V* -E by p(x*) = 35.
(VI,TI) -* i;
i`x*ll < lim infnl1xnl1
Let
34.
p
:
Prove that, if
*
(V*,Tw ),
V2; that is,
be a normed linear space over
Let
(xn) e V*, x* E V*.
Prove that
V2.
to
is continuous, if and only if
(V2,T2)
x* o T
33.
Prove that
separates points.
F
is continuous relative to the weak topology on T :
on
that separates points.
F
31.
BI
Tw
Prove that, if the weak topology
x E V, jjxh < 1).
I
V,
prove that
V/W
x
is re-
T
9. Weak Topologies
288
We have seen that it is possible for a subset
38.
to have
that, if then
that converges weakly to
E
x E Z2
(Proposition 9.7.1)
39.
f E LI([-n,n],dt/2n)
Let
g E Lp([-n,n),dt/2n), 1 < p
0
there exist some
a > 0
if and only if for each
U E
(i)
and
x1,x2,...,xn
V
in
and
x* E UflaBi
such that
U(x*;c;xl,x2,...,xn)flaBi,
{y*
I
Y* E V*, IIy*ll < a, Ix*(xk) - y*(xk)I < c, k = 1,2,...,n)
is contained in Tbw*
(ii)
U.
is a topotogy cn
V*
locally convex topological linear space over
is a
f.
Tw* C Tbw*
(iii)
We call
ent that a set EflaB*
(V*,Tbw*)
such that
w
the bounded weak* topology on V*. It is apparw ) if and only if E C V* is closed in (V*,
is closed in
once since each
aBi
(V*,Tw«)
for each
a > 0.
This follows at
is weak* closed.
Next we wish to give three descriptions of a neighborhood base at the origin for the bounded weak* topology.
The first description
involves a fairly long proof, but the other two follow easily from the first.
10. Krein-9mulian and Eberlein-gmulian Theorems
292
Theorem 10.1.1.
be a normed linear space over 0.
Let
Then the sets U(0,l,{xk)) = {x*
I
X. E V*, Ix*(xk)I < 1, k = 1,2,3,...),
(xk) is any sequence in- V such that neighborhood base at the origin for Tbw* where
Proof.
So let
limkllxkll = 0,
form a
tbw*
First we must show that each such set belongs to be such that
(xk) C V
a positive integer
limk IIxkII = 0
such that
n
and let
Choose
a > 0.
We claim that
IIxkII < 1/a, k > n.
U(0,1,[xk))flaBi - U(0;1;x1,x2,.... xn)f)aBi.
Clearly
U(0,1,(xk))flaB C U(0;l;xl,x2,...,xn)()aBI. Conversely, suppose
x* E U(0;l;xl,x2,...,xn)f aB*.
Ix*(xk)I < 1, k = 1,2,...,n,
whereas if
k > n,
Then
then
Ix*(xk)i < Ilx*IIIIxkII < a(a) = 1, and we see that
x* E U(0,1,[x
)flaB1*.
k
Thus
U(0,1,(xk))(1aBI = U(0;l;xl,x2,...,zn)flaBi, and the latter set is open in the relative weak* topology on Since the result holds trivially when U(0,1,(xk)) E Tbw*
a = 0,
aB
we conclude that
It is evident that the intersection of any two sets of the form U(0,1,(xk))
contains a third, so to prove that these sets are a
neighborhood base at the origin it remains to show only that every U E Tbw* given
U E
such the 0 E U contains one of these sets; that is, w such that 0 E U, it suffices to prove the existence
of a sequence
(xk) C V
such that
limkllxkll - 0
The proof of this will require a bit of labor.
and
U(0,1,(xk)) C U.
293
10.1. The Bounded Weak* Topology
F° _ [x*
w
is open in
e > 0
x* E V*, Ix*(x)l < 1, x E F).
I
is a weak* closed subset of
F°
Note that each such U
Now, since
V*.
there exist, by Proposition 10.1.1(i), some
,
xi,x2,...,xn
and
is a finite set, then we write
F C V
In general, if
in
V
such that
U(0;1;xi,x2,...,xn)f1B1* C U.
Obviously, on setting where
F0f1B* C U,
yk = 2xk/c, k = 1,2,...,n,
F1 = [yl'y2''"'yn)
we wish to define a particular sequence V.
we see that
Using a similar argument, [Fn)
of finite subsets of
These sets will be obtained as follows: Beginning with
choose a finite set
F1,
EI C V
such that
Having I;yll < 1, y E E1, and (F1 U E1)° fl 2Bi C U. Set defined Fl,F2,.... Fn, choose a finite set En C V such that {jyj: < 1/n, y E En, and (Fn U En)° fl (n + 1) Bl* C U. Set Fn + 1 = Fn U En.
F2 = F1 a E1.
Of course, it is not entirely clear that such choices of the sets can be made.
show that, if
En
To see that this can be. done ii clearly suffices to
F1,F2,...,Fn
are given, then
En
with the appropriate
properties can be found.
Suppose this is not the case. such that
Then for any finite set
E C V
we have
jjyjj < 1/n, y E E,
(FUE)°fl (n + 1)BI-.Uc where
Uc
denotes the complement of
U
in
V*,
whereas
F°flnB,* C U.
Consider the family of sets
S- ((Fn U E)° fl (n + 1) B* fl UC Clearly all the sets in
S
I
E C V, E finite, jlylj < 1/n, y E E) .
are weak* closed subsets of
Ch + 1)Bi,
which, by the Banach-Alaoglu Theorem (Theorem 9.4.1), is weak* compact. Moreover,
S
E1,E2,...,Ea
has the finite intersection property since, if are finite subsets of
V
such that
11r1`
y E Ek,
10. Krein-Smulian and Eberlein-$mulian Theorems
294
then
k = 1,2,...,m,
UTr,
nI[(v UEll0f, (n 4 1)B1f, Uc] _ [rnL( Ek)]°1i (n -, 1)Bif, Uc k=1 k=1 . 11
Consequently we see that f1S[(nUF.)°( (n + 1)B* flUC] # . Suppose, however, that
x* E flc[ (Fn U E)° fl (n +. 1) B* fi Uc] . Then, in particular, we see that lx*(x)j < 1 for each x E V such that Ilxll < 1/n, from which we deduce at once that j1x*Ij < n. Thus
fl[(FnUE)°fl (n + 1)Bi FIUc] C [Fnfl (n + 1)B*fiUc] AnB= S
FnfinB*f1U°, I
and so that
F0flnBillUc # . This, however, contradicts the hypothesis Therefore there exists some finite set
F0finB* C U. n
such that
ilylj < 1/n, y E E11,
(Fn U En) °
and
We have now shown that there exists a sequence subsets of F
V
such that
F0flnB* Z U.
is a countable subset of
sequence, say
and let
shows at once that
n
of finite
Clearly
This is apparent
all but a finite number
have norm less than or equal to
F
We claim, furthermore, that x* E U(0,1,{xk))
F = Un= 1Fn.
limkIjxkjj = 0.
on noting that for each pcsitive integer of the members of
(Fn)
V, and so we may enumerate it as a
Moreover,
{xk).
Let
En e V
fl (n + I) BI c U.
U(0,1,(xk)) C U. Then
n > jx*jj.
x* E Fn0 i;nB* C U.
Thus
1/n.
Indeed, suppose
jx*(xk)I < 1, k = 1,2,3,..., U(0,1,{xk)) c U,
and
the proof its complete.
G A couple of observations are in order. for which
0 E U,
it is clear that, if
constructed in the proof such that happens that
jx*
I
First, given
(xk) c V
w
is the sequence
U(0,1,{xk)) CU,
jx*(xk)I < 1, k = 1,2,3,...) c U.
ent from the definition of F. °
U E
then it also
This is appar-
Moreover, it is also evident that
295
10.1. The Bounded Weak* Topology
we have
c, 0 < e < 1,
for any
jx*
U(0,c,{xk})
(
where, of course,
U(0,c,{xk}) c_ U,
x* E V*, jx*(xk)I < e, k = 1,2,3,...}.
The latter
observation immediately yields the following corollary: Corollary 10.1.1.
Let
U(O,c,{xk}) _ {x* [xk}
where
I
be a normed linear space over
(V,II'1!)
c, 0 < c < 1,
Then for each
6.
the sets
x* E V*, lx*(xk)I < c, k = 1,2,3,...}, V
is any sequence in
such that Tbw*
limk;JxkU = 0,
form
a neighborhood base at the origin for
The final description of a neighborhood base for
w
is con-
tained in the next corollary. Corollary 10.1.2.
be a normed linear space over
Let
Then the sets
K° _ {x* where
E V*, `x*(x) < 1, x E 1(},
is any compact subset of 7bw* base at the origin for K
Suppose
Proof. a > 0.
K C V
(V,11.11),
form a neighborhood
is compact in the norm topology and
Then, since
is a metric space, we see that
K
is
totally rounded, and so there exist xl,x2,...,xn in K such that K C Uk- 1{x x E V, - xkII < 1/2a). Let x* belong to the set I
IIx
U(0;1/2;x1,x2,...,xn)(laB*. xk, k = 1,2,...,n,
Then, if
such that
Ix*(x)I
- xkll < 1/2a,
IIx
lx*(x
-
xkIl + 1
a(2a) + 2 = 1.
there exists some and so
xk)I + jx*(xk)1
lIx*IIIIx 1
x E K,
2
0.
10. Krein-gmulian and Eberlein-Amulian Theorems
296
il(0;1/2;x1,x2,...,xn)ElaB* C K°,
Thus
10.1.1(i) we conclude that
K0 E Tbw*.
are compact, then
K1,K2 C V
and so from Proposition Clearly
and if
0 E K°,
is compact and
K1 U K2
(K. UK 2)° C Ki fl K2.
which
for
But from Theorem 10.1.1 we know that there exists a
K° C U.
(xkJ C V
sequence
is
K C V
then there exists some norm compact
0 E U,
such that
w*
U E
Consequently it remains only to show that, if
such that
and
iimk(lxkI( - 0
U(0,1,(xk)) C U.
Moreover, as remarked after Theorem 10.1.1, we may even assume that (x*
X. E V*, Ix*(xk)I < 1, k = 1,2,3,...J C U.
I
K = (xkJ U(c)
Clearly then
is a norm compact subset of V
V
a nd
We saw in Proposition 10.1.1(iii) that the bounded weak* topology We now wish to show that the
is stronger than the weak* topology.
continuous linear functionals on provided
(V*,T4"*),
(i)
x"
(ii)
x"
Proof.
and
is continuous.
) - 4
is continuous.
T"`* C Tbw*
it is evident that part (i) implies x":
(V*,
w* )
-- §
is contin-
Then from Theorem 10.1.1 we see that there exists some sequence
(xkJ C V
for which
x* E U(0,1,(xkJ).
limkllxkll = 0
then
Ix"(x*/c)I < 1. ,Now define T(V*)
and
Moreover, for each
x* E
Then
-+ I
On the other hand, suppose
part (ii). uous.
(V*,
Since
§
Then the following are equivalent:
(V*,Tw ) w* :
are precisely those on
)
be a Banach space over
Let
x" E (V*)'.
suppose
w*
V*
is a Banach space.
V
Theorem 10.1.2.
(
Thus T
:
Ix"(x*)I < 1
c, 0 < c < 1,
whenever if
((x*/c)(xk)l < 1, k = 1,2,3,..., Ix"(x*)I < e
V* -- c0
whenever
by setting
is a linear subspace of
cop
and so
x* E U,(O,c,(xkj).
T(x*) _ (x*(xk)J, x* E V*.
and we define a linear
297
10.1. The Bounded Weak* Topology
functional
on
y*
We need to note first that
is well-defined; that is, if
y*
and so x* - x2 E U(0,e,(xk))
xi(xk) = x2(xk), k = 1,2,3,..., all
Hence
e, 0 < c < 1.
and so
x"(xi) = x"(x2).
for
c, 0 < e < 1,
for all
Ix"(xi - xz)I < s Thus
then
But if, T(x*) = T(x2),
x"(x*) = x"(x2).
then
T(x*) = T(x2),
y*[T(x*)] = x'(x*), x* E V*.
by setting
T(V*)
is well-defined and clearly
y*
linear.
Furthermore, suppose limnII(xn(xk))`Im = 0;
that is,
the zero sequence in
co.
positive integer k a 1,2,3,...
Consequently
.
y*
N
is a sequence such that
[xn] C V*
e > 0,
Then, given
n > N,
such that, if
there exists some
then
Ixn(xk)I < e,
Iy*[T(xn)]I = Ix"(x*)I < c
Hence
co,
noted in Example 3.1.6, (ak) E t1
which we again denote by
y*
to a continuous y*.
Since, as
we deduce that there exists a
co = ti,
such that
is a con-
y*
by Proposition 3.2.2.
T(V*),
From Theorem 4.2.2 we see that we can extend linear functional on
n > N.
for
is continuous at the origin, and so
tinuous linear functional on
sequence
converges to
((x-(x,,))) = (T(x*))
y*[T(x*)] = Fk = lx*(xk)ak, x* E V*.
But then
x"(x*) = y*[T(x*)] =
E x*(xk)ak
k=1 n
- lim n
E x*(xk)ak
k=1 n
= lim x*( E akxk)
k=1
n OD
= x*( E akxk)
k=1 = T( E akxk)(x*)
k=1
(x* E V*).
10. Krein-.A`mulian and Eberlein-Amulian Theorems
298
x* E V*
The penultimate equality is valid since obviously an element of Therefore
V,
-
lakxk
and so, by Theorem 9.1.3,
x" = i(F.k= lakxk),
is a weak* continuous linear functional on
is
x"
V*.
We are now almost in a posi-
The Krein-9mulian Theorem.
10.2.
F
limkiixk" = 0.
and
(ak) E t1
as
and
However, one additional
tion to prove the Krein-9mulian Theorem. preliminary result is necessary. Lemma 10.2.1.
tional on
uous on
is continuous on
V
K C V
If
(V,T2).
and
(V,T1)
Let
topological linear spaces over
he locally convex
(V,T2)
and suppose that a linear func-
I
if and only if it is contin-
(V,T1)
is convex, then the following are
equivalent: (i)
K
is closed in
(V,T1).
(ii)
K
is closed in
(V,T2).
Proof.
If
Therefore suppose empty.
If
is empty or all of
K
is closed in
K
xo f K,
(V,T1), K j V,
and
K;
on
and some
tional
x*
Let c= a - x*(xo) > 0
(V,T1)
U.= (y (V,T1)
I
and
x0 E U.
(V,T2)
Moreover,
L
that strictly separates
(V,T2).
such that
a E Ik
x*(xo) < a < x*(x),
and set
have the same continuous linear is an open set in
U
Uf1K = q,
x*(xo) - x*(z)l < e = a - x*(xo) contradicts the fact that closed in
is non-
Y E V, lx*(x0) - x*(Y)I < e).
functionals, it is apparent that clearly
K
that is, there exists some continuous real linear func-
x E K.
Since
and
then by Corollary 5.4.1 and Proposition 5.1.1
there exists some closed real hyperplane xo
then the result is trivial.
V,
because if
shows that
x*(z) > a,
as
Thus
then
which
x*(z) < a,
z E K.
and
(V,T2),
z E U A K,
K
is
299
10.2. Krein-mulian Theorem
Therefore part (i) of the lemma implies part (ii), and the same argument mutatis mutandis establishes the reverse implication. Theorem 10.2.1 (Krein-gmulian Theorem). Banach space over
K C V*
and suppose
I
be a
Let
Then the fol-
is convex.
lowing are equivalent: (£)
(ii)
K
is closed in
Proof.
spaces over
is closed in
(V*,Tw*),
From Proposition 10.1.1(ii) and Theorem 10.1.2 we see (V*,Tbw*)
and
are locally convex topological linear
with the same continuous linear functionals.
I
Lemma 10.2.1, therefore, K
KflaB*
x* E V*, ('x*jj < 1).
+
(V*,Tw*)
that
the set
a > 0,
For each
Bi = (x*
where
(V*,Tw*).
is closed in
(V*,
K
T-bw * ).
is closed in
(V*,Tw*)
By
if and only if
However, from the remark following
Proposition 10.1.1 we know that
K
is closed in
w (V*,
precisely
)
when part (ii) of the present theorem holds, which completes the proof. 11
Before utilizing this result in the proof of the Eberlein-gmulian Theorem we examine a few corollaries.
The first two have simple
proofs, which are left to the reader, Corollary 10.2.1. over
K CV is convex.
and suppose
I
be a reflexive Banach space
Let
Then the following are
equivalent: (i)
(ii)
where
K
is closed in
For each
B1 = (x
(
a > 0,
valent:
W C V*
the set
KflaBI
is closed in
(V,Tw),
x E V, jixil < 1).
Corollary 10.2.2. suppose
(V,1 ).
Let
(V,
is a linear subspace.
be a Banach space over
4
and
Then the following are equi-
10. Krein-Amulian and Eberlein-mulian Theorems
300
is closed in
W
(i)
is closed in
WflB*
(ii)
(V*,TW*).
Bi = (x*
Corollary 10.2.3.
x* E V*, llx*11 < 1).
be a Banach space over
(V, I1 !l)
4
and
Then the following are equi-
is a linear subspace.
W c V*
suppose
I
Let
where
(V*,TW*),
valent:
(i)
is closed in
W
(ii)
(V*,TW*),
Proof.
W
If
B* = [x*
1
I
such that
x* E W, j`x* - x*11 < e).
is closed in
E D [x*
implies part (ii) on taking
`
Clearly
xo = 0
and
it is immediately obvious that for each
a = 0.
Thus part (i)
Since scalar multiplica(V*,TW*)
a > 0
(Theorem 2.1.2),
we have
(2a/e)(E - x*)
while the assertion is trivially true if
Obviously part (ii) entails that
so for each
is weak* closed
e = 1.
tion and translation are homeomorphisms on
(V*,TW*),
E
x* E W, 11x*11 < 1).
Conversely, suppose part (ii) holds.
is closed in
then set 'E = Wf1B
(V*,TW*),
x* E V*, jjx*jI < 1).
and norm bounded, and
that is closed in
E C W
e > 0
and some
x* E W,
some
E D [x*
where
(V*,TW*).
There exist a norm hounded set
and
Wfl(s/2)B* C E - x0*,
a > 0
WflaB*
= Wn(es)B1
= a[wn (2)B1] c La (E - xo) Is
W
is a linear subspace.
WflaB* = (2a7c)(E - xo)flaB*,
Hence, for each
a > 0,
we deduce that
Therefore, by the Krein-9mulian Theorem, we see that closed in
(V*,TW*),
(V*,T*).
and the latter set is closed in W
is
and part (ii) of the corollary implies part (i).0
301
10.2. Krein-gmulian Theorem
In other words, Corollary 10.2.3 asserts that, if a Banach space, then a linear subspace
W c V*
(V,11-11)
is
is weak* closed if
and oily if it contains some weak* closed norm bounded set
E
that
in turn contains some open ball in the normed linear space Finally, we wish to apply Corollary 10.2.3 to derive a result for the weak* topology analogous to Theorem 9.2.2, which asserts that a nonempty convex set is closed if and only if it is weakly closed.
A lemma is necessary, which we state in a fr-,n suitable to
It is valid, however, in a more general context, as can
our needs.
be seen in [DS1, pp. 415 and 416]. Definition 10.2.1. suppose
Then the convex hull of
E C V.
n E akxk
co(E) = (
ak > 0,
k=I If
be a linear space over
V
Let
The proof is left to the reader.
E
is defined as
n E ak = 1, xk E E, n = 1,2,3,...).
k=1
I
is a topological linear space over
(V,T)
convex hull of
E
then the closed
I,
is defined as the closure in
It will be denoted by
and
§
(V,T)
of
co(E).
co(E).
It is not difficult to see that
co(E)
and co(E)
are, respec-
tively, the smallest convex and the smallest closed convex sets that contain
E.
Indeed, it can be shown that co(E) = li(K
I
K
E, K convex)
and
co(E) = (1(K
I
K a E, K closed and convex).
The details are left to the reader. We shall discuss an important theorem about convex hulls in the next chapter.
Now, however, we need only the next result.
Lemma 10.2.2.
suppose
Then
K1,K2 c V*
Let
be a Banach space over
are convex sets that are compact in
co (K1 U K2) = co (K1 U K2) .
I' and (V*,Tw*).
10. Krein-Julian and Eberlein-gmulian Theorems
302
Theorem 10.2.2. K CV*
suppose
Let
be a Banach space over
(V,11.11)
is a convex set that is closed in
W denotes the linear subspacq of V* spanned by
$
and
(V*,Tw*).
If
K,- then the
following are equivalent: (i)
W
is closed in
(ii)
W
is closed in For simplicity we assume that
Proof.
f a C
(V*,Tw*).
Since
is left to the reader.
topology on
Kn
is a wbak* compact convex set, and so from Lemma 10.2.2
we see that
is weak* closed.
Kn + co(Kn U -Kn)
tion 9.2.2 the we&k* topology on logy on Kn
x* E V', 11x*11 < l).
1
Bana;h-Alaoglu Theorem (Theorem 9.4.1) it is apparent that
From the each
is norm closed and for each positive
W
set Kn - Kf1nBi, where Bi - (x*
n
The case where
is weaker than the norm
it follows at once that part (i) implies part (ii).
V*
Conversely, suppose
integer
f - 1R.
Tw*
and hence each
V*,
Kn
However, by Proposi-
is weaker than the weak topo-
V*
is closed in
(VIII)
closed in
Now set
Kn
is
.
Irt is evident that each x E W can
K' = co (K U -K) .
x =k = lakxk
be written as
Since each
(V*,TW).
is convex, it follows from Theorem 9.2.2 that each
ak > 0, k = 1,2,...,n.
-
where
= m + 1akxk'
Thus, on setting
xk E K and
a = 5.;- lak > 0,
we see
that x E aK', as Ii. l (ak/a) xk - "'k = m + l (ak/a) xk E K1. Moreover, from which it follows at once that
0 E K',
Consequently, given such that
x E
there exists some positive integer
x E W,
n > nx.
We claim this implies that
Indeed, it is evident that x E W,
then
exists some x = n y x
x E nK', n > ax. jx
for some
for which
x=
3x
C W. Un = lK' n
Clearly
x E nxKjx.
y E Kjx. [j n
aK' C bK', 0 < a < b.
I < jx < nx,
Consequently, since
]y
n
-
x jj) j(j
n +
11
nx = lK'.
On the other hand, if
K' = U% lK!, If
W = Un
- jz
o).
and so there then
303
10.3. Eberlein- mulian Theorem
x E jxK3x
we conclude that E Un a
K
y E
and so W = Un = 1Kn'
1nKn,
But
since
being a norm closed linear subspace of
W,
Banach space over
#,
is a
and so, by the Baire Category Theorem (Theorem
6.1.1), we conclude that there exists some
also weak* closed and norm bounded.
no
such that the norm
Kn is no Thus, appealing to Corollary
closed set K' has a nonempty interior in
W must be closed.in
10.2.3, we see that
V*,
W.
However,
(V*,Tw*).
Therefore part (ii) of the theorem implies part (i).
We have now reached the
The Eberlein-gmulian Theorem.
10.3.
0
point where we can prove the Eberlein-gmulian Theorem, which says that a weakly closed set in a Banach space is weakly compact if and only if it is weakly sequentially compact.
The proof is quite long
The key results used in prpving the "only if" portion
and involved.
of the theorem are the Hahn-Banach, Uniform Boundedness, and BanachAlaoglu Theorems, together with the fact that the weak and,ngrm closures of a convex set in a Banach space coincide; the proof of the "if" portion of the theorem employs the Uniform Boundedness 14nd
Banach-Alaoglu Theorems, the coincidence of the weak and norm closures of a convex set in a Banach space, and the Krein-9mulian Theoxefi.
Theorem 10:3.1 (Eberlein-gmulian Theorem). Banach space over
#
and suppose
Let
(V,11.11)' be a
E CV is closed in
(V,TW).
Then
the following are equivalent: (i)
E
is comp ct in
(ii)
E
is sequentially compact in
Proof.
Suppose that
be a sequence.
(V,T*).
E
(V,T*).
is weakly compact and let
We must show that
(xk]
(xkJ+
E
has a weakly coevergeAt
subsequence. Let
W denote the norm closure of the linear space spanned by
304
10. Krein-3mulian and Eberlein-gmulian Theorems
is a separable Banach space over
Then
(xk).
Bi = (x*
I
From Theorem 9.4.2 we see that the
x* E W*, 1Ix*I1 < 11.
weak* topology on
restricted to
W*
Set
t.
B*
W
is metrizable,
being
separable, and from the Banach-Alaoglu Theorem (Theorem 9.4.1) we see that
B*
Thus
is weak* compact.
is a compact metric space
B*
A similar
in the weak* topology and hence a separable metric space. argument reveals that for each positive integer nBZ
is a separable metric space in the weak* topology on
so the weak* topology on Let
the space
n,
F0 C W*
W* = to
then, since
once that
F0
W,
as if-
x*(x) = 0, x* E W*.
Furthermore, extending each
Let
and
t*(x) = 0,
it follows at
W*,
x = 0,
and so
x* E F0 V,
F
0
separates
by Theorem 4.2.1 to
we obtain a countable set
F = (x*),
Since* E is weakly compact, it follows that Indeed, consider §x* _ I.
W
Hence, by Corollary 4.2.6 to the
some continuous lineP functional on F C V*.
x E W
is weak* dense in
E[ahn-Banach Theorem, we conclude that points.
W*.
is a family of linear functionals on
F 0
that separates the points of x* E F0,
must also be separable.
1nBi
be a countable set that is weak* dense in
First we note that
and
W*,
T
:
V -. Ox*
E V* §x*,
where
E
is norm bounded.
T(x) = (x*(x))
and
Then from the discussion of F-topologies following Theo-
rem 9.1.1 we know that a can be considered as the relative topology on
T(V)
viewel as a subspace of the topological product space
Furthermore, for each fixed y* E V* x* E V* }x*' seen that, if PK* denotes the projection of IIx* iY*,
*y*
then
Py* o T(E) = (y*(x)
I
and so there exists some
sup I T (x) (Y*) I
xEE
=
x E E)
E V* §x*
onto
is a compact subset of
M. y* > 0
sup I Y* (x)
xEE
it is easily
such that
< My*,
Consequently Corollary 6.2.1 to the Uniform Boundedness Theorem
leads us to conclude that there exists some M > 0
such that
305
10.3. Eberlein-9mulian Theorem
sup 11 T(x)(1 = sup JJxJJ < M, xEE xEE that is,
is norm bounded.
E
Hence the sequence
is a bounded sequence of numbers
(xi(xk))
Consequently there exists a
and thus has a convergent subsequence. {xk(1)}
(xk)
such that
(x1*(xk(1))}
converges.
The same argument applied to the sequence
(x2(xk(2))}
assures us
subsequence
of
(xk(2))
of the existence of a subsequence of
(xk),
such that
is also convergent.
and
.
(xn(yk))
Now
Then evidently
(yk)
converges for each
(yk C E,
yo E E
some
converges, n
to be the diagonal sequence; that is, set
(yk)
k = 1,2,3,...
and
E
n = 1,2,3,...
is weakly compact.
and
Furthermore, we claim that n
W
which
(yk),
in
0
Tw
Moreover,
is weakly closed,.and so
is given and let
Given
limkxn(yk) = a.
A
such that
y
it
y
E W..
0
I n(yk)
-
e > 0,
a) < c/2,
contains infinitely
we see that there exists some
yko E U(yo,c/2,xn).
V,
0
Since each weak neighborhood of
many terms of
y
(yk3.
limkx(yk) = xn(yo), n
there exists some positive integer k > N.
Hence there exists'
W. is a norm closed linear subspace of
follows from Theorem 9.2.2 that
Indeed, suppose
(xk),
.
such that every open neighborhood of
(yk) c W
yk - xk(k),
is a subsequence-of
contains infinitely many terms of the sequence since
(x*l(xk(2)))
each of which is a subsequence of
(xn(xk(.))}
and is such that
Now take
Clearly
converges.
and hence
Continuing in this fashion, we obtain the se-
(xk(j)), j m 1,2,3,...,
quentbs (xk}
(x2(xk(2))}
(xk(1)),
of
k
0
N
for
Thus
jxn(yo) -a1 _1xn(yo) -x*Yk)I +[xn(yk) -a[ 0 c
c
2
2
- f, from which we conclude that
xn(yo)
a,
since
s
is arbitrary.
10. Krein-9mulian and Eberlein-9mulian Theorems
306
To prove that
(yk)
actually converges weakly to
this were not the case.
xo E V*,
Then there would exist some tyk }
and a subsequence
e > 0,
(yk}
of
lxo(yk
Suppose
x* E V*.
for each
limkx*(yk) - x*(yo)
only to show that
it remains
yo
some
such that
- Yo)i > 6
= 1,2,3,...).
(j
I
However, since some
z
0
and
{yk],),(z E
is weakly compact, there exists
E
such that every open neighborhood of' z0
E E,
contains infinitely many terms of that
is a subsequence of
(Yk1}
since
(yk},
all of
V
.
But the sequence
xn(yo - z0) = 0,
consists of extensions to
(xn)
F0
of the family of linear functionals
separates the points of
W,
Moreover,
.
we must alsp have
and so
lim.x*(yk.) - x*(Yo), n = 1,2,3,..., n = 1,2,3,...
As before, we deduce
(yk ).
xn(zoi, n = 1,2,3,....
limjx*(Yk )
and
zo E W
Tw
in
W
on
and so we conclude that
that
yo = z0.
Thus, on the one hand, we see that every open neighborhood of yo
in
Tw
other hand, the inequalities show that shows that
(yk
contains infinitely many terms of
converges to
(yk)
Therefore
(xo(Yk - yo)I > c,
U(yo,s,x), j
Yk
while on the
= 1,2,3,....
yo
in
.
This contradiction
(V,Tw).
is weakly sequentially compact, and part (i) of
E
the theorem implies part (ii). Conversely, suppose if
T
E
is weakly sequentially compact.
denotes the canonical- embedding of
V
into
Now,
then, as
V**,
noted previously (e.g., in the proof of Theorem 9.9.1), T
:
(V,Tw) - (T(V),Tw*)
compact in
(V,TW)
is a homeomorhpism.
if and only if
T(E)
Thus
E
will be
is compact in
(T(V),Tw*),
and the latter will clearly be the case (Theorem 9.4.1) if we can show that
T(E)
is norm bounded and closed in
Since
(V**,Tw*).
compactness and sequential compactness are equivalent in
§,
it
follows, by essentially the same argument as that given'in the first half of the proof, that
T(E)
is norm bounded.
It remains then
307
10.3. Eberlein-Smulian Theorem
of
T(E)
is contained in
(V**,Tw*)
in
T(V),
since
the closure of
is weakly closed.
E
T(E)
T(E)
(V**,Tw*)
in
Thus, if
xp*
lies
belongs to
we must prove that
(V**,Tw*),
in
To accomplish this
T(E).
prove that the closure of
if in
is weak* closed -- that is, that the closure
r(E)
only to show that
x*o* E T(V).
This will require considerable work. First, suppose exists some
y E E
x*,x2 ,...,x* such that
are in
V*.
We claim that there
lary 4.10.1 to Helly's Theorem, of course, shows that such a always exists.
We require, however, that
is in the weak* closure of there exists some
yk E E
T(E),
as
{yk))
{yk)
E
converges weakly to
But since
y.
y E V x*0*
k
T(yk) E U(xp*;1/k;xi,xl,.... xn),
such that
the existence of a subsequence
y E E.
for each positive integer
and the weak sequential compactness of
that
Corol-
x*'(xk) = xk(y), k = 1,2,...,n.
E
of
then allows us to deduce (yk)
The element
and a y
y E .E
such
belongs to
E,
is weakly closed. - In particular then
IxQ*(xm) - x (Y)I < Ix0*(xm) - T(yk)(xm)I
+ Ixm(Yk) - x*(Y)I
< k + Ixm(yk )
(m = 1,2,...,n).
- x* (Y)
)
from which we conclude that
x**(x*)
x*(y), a = 1,2,...,n.
With this preliminary observation made, we now note that to prove
x0* E T(V)
it suffices, by Theorem 9.1.3, to show that
is a weak* continuous linear functional on
V*,
need only show, in view of Theorem 3.3.2, that
N(xp*)
is closed in
(V*,Tw*).
(x*
I
X* E V*, x**(x*) = 0)
Finally, to estkblish this we need only
prove, by Corollary 10.2.2 to the Krein-Smulian Theorem, that N(x**)f1Bi
is closed in
(V*,Tw*),
x0*
and to. do this we
where, as usual
10. Krein-mulian and EberleinAmulian Theorems
.308
(x* To this end, let
exists some
belong to the weak* closure of
x*
and let c> 0
V*
in
such that
yl E E
we have chosen
The fact that
x**(xO*) = x;(y1).
N(x**)flBt
x* E N(x*o*)f1Bi_ such that in
Y1'y2' ..''yn
N(x0*)(1B1
By our previous observation, there
be given.
is in the weak* closure of some
x* E V*, lix*il < jJ.
I
Suppose now
x* E U(x*,c/2,y1). and
E
x*
implies the existence of
x*,xZ...,X*
in
N(x***)f B*
such that x0*(xa*) * ;*(yk)
and
0,1,2,...,k - 1; k = 1,2,3,...,n)
(m
x* E U(x*;e/2;ylty2'. ''ym)' m = 1,2,3,...,n.
exists some
yn
E E
such that
x**(x*) = x*(yn + 1)' m = 1,2,...,n,
1
by our previous observation. xn
+ 1
Choose
E U(x6;e/2;y1,'2'"
(yk) c E
and
''yn + 1) (x,*) C :V* such that
xn +
E N(x**)f B*
so that
In this way we obtain sequences
(a)
x* E N(x**)f1B*
(b)
x**(xm) = xm(Yk)
(c)
xm E U(x*;c/2;y1,y2,...,ym)
Moreover, since
Then there
(m = 1,2,3,...).
(m = 0,1,...,k - 1; k
(m = 1,2,3,...).
xm* E N(x**)n Bt, m = 1,2,3,...,
(d)
x**(x**) = xm(Yk) = 0
(e)
llxmll < 1
1,2,3,...).
we see that
(m - 1,2,...,k - 1; k = 1,2,3,...). (m
1,2,3,...).
Furthermore, statements (b) and (c) show that (f)
lxp*(xo*) - x*(Yk)1 < 2
Now, since
(yk) c E
and
E
(k = 1,2,...,m; a - 1,2,3,...).
is weakly sequentially compact
and weakly closed, there exists a subsequence of weakly to some
yo E E.
we may assume without loss of generality that weakly to
yo.
by statement (d)
Then
(yk)
that converges
In order to avoid some notat.ional awkwardness, (yk3
itself converges
x*(yo) = limkxm(yk) -.0, a = 1,2,3,...,
x*(yk) = 0, k > m -.1.
since
309
10.3. Eberlein- mu1ian Theorem
(yk)
Finally, since al,a29...,an then
IIz
in
converges weakly to
ak > 0,
i,
lak = 1
-
there exist
yo,
such that,"if
z s 1nk
a lakYk'
- yoll < e/2. This follows at once from Theorem 9.2.2 and m = n,
Thus, for
Corollary 9.2.1.
we see by statement (f) that
n
n xn(z)I
Ix0*(
E akx0) - xn( E akyk)I
k1
k=1
n
0
such that
Our next task is to describe
I`T*(y*)I[ > ml(y*11, y* E C(E)*.
more precisely.
T*
(L1(Z)/i(E))*
From Theorems 4.6.2 and 8.3.3 we see that
be identified with
I(E)1 a LI(Z)* = L.(f),
M(E),
can
while the Riesz Repre-
sentation Theorem (Theorem 4.8.2) shows that with
is surjective
T*
can be identified
C(E)*
the space of bounded, regular, complex-valued 8orel
measures on
E.
elements of
M(E)
Since
E C r
is closed, it is easily seen that the
can be identified with those measures in
that have support contained in following definition of
µ
E.
Thus for each
µ E M(E)"
M(I')
the
is meaningful:
(k) = fE a
tiT
dµ (eit)
(k E Z)
Obviously 11;11- = supk 6ZIµ(k) I < JJµJJ, µ E M(E), The complex conjugation in the definition of the manner in which we have identified
µ
L1(b)*
and so
.
µ E LCD (Z) .
is necessitated by with
L(g).
You
will recall (Theorem 8.3.3) that this identification involved an antilinear, rather than a linear, mapping. Given
µ E M(E),
we see that
E
k = -w
T*(µ) _ {bk} E I(E)1 c Lm(Z)
akbk a T* (1+) (a)
= µ [T (a) ] = fE a(eit) dµ(eit)
a
E k = -wak
fE eikt
dµ
(eit)
and
337
11.6. Banach-Stone Theorem
CO
=
k=E
(a = (ak) E L1(b)).
akl+(k)
The interchange of integration and summation is valid since the series expression for
We then see at once
converges uniformly.
a
E M(E).
that
An application of Theorem 9.10.1 immediately yields the following theorem:
Theorem 11.5.1.
E C f= (elt
Let
I
be a closed
t E(-n,n])
Then the following are equivalent:
set.
is a Nelson set.
E
(1)
for each is such that a kbk = 0 I;akezkt = 0, eit E E, then there Ek= -
(bk) E Lm(L)
If
(1i)
for which
(ak) E L1(Z)
exists some
µ E M(E)
such that
There exists an
(iii)
M > 0
111iji < MjjIjj
µ(k) = bk, k E Y. such that
= M sup I µ (k) I
(µ E M(E) ) 11
kEZ (iv) µ E M(E), defines a norm on equivalent to lIµII _ f µl (E), µ E M(E).
that, is
M(E)
Analogs of this result that are valid for arbitrary locally compact Abelian topological groups, as well as some specific examples of Helson sets, can be found in [E3, pp. 233-237; HR2, pp. 563-565; Rul, pp. 114-120].
The Banach-Stone Theorem.
11.6.
Suppose that
and
XI
X2
are compact Hausdorff topological spaces that are homeomorphic. T
:
X1
is a homeomorphism, then for each
X2
T(f)(tI) = f[T(t1)], ti E X1. and that T
T
:
C(X2) - C(XI)
is surjective, because if
by setting
f(t2)
f E C(X2)
It is easily verified that is an isometric isomorphism. g E C(X1)
g[T-1(t2)], t2 E X2,
Thus we see that a homeomorphism between
T(f) E C(X1) Moreover,
and we define
f E C(X2)
then clearly
T(f) = g.
X1
and
X2
If
define
gives rise
11. Krein-Mil'man Theorem
338
to an isometric isomorphism of
onto
C(X2)
The content of
C(XI).
the Banach-Stone Theorem is that the converse assertion is valid.
In order to prove this result we need to know the extreme points of the closed unit ball it Lemma 11.6.1. and let
Bl- = (µ
I
Let
M(X).
be a compact Hausdorff topological space
X
ext(B*) _ (a6t where
6t E M(X)
Proof.
µo
B*
co(B).
IaI
= 1, t E X)
,
is defined by
1
for
t E E,
6t(E) = 0
for
t
E,
E C: X. B = (abt
I
a E U,
and the weak* closure
B c B*,
since
Let
a E Q:,
I
6t(E) =
for any Borel set
Then
µ E M(X), 11µA < 1).
IaI = 1, t E X}.
of
co(B)
is weak* closed and convex.
co(B)
Suppose
Clearly
is contained in µo E M(X)
Bi
and
By Corollary 5.4.1 there exists a closed real hyperplane
that strictly separates
µo
and
co(B),
and Proposition 5.1.1 there exists some
b-e=
s1E Re[ E co(B)
and so by Theorem 9.1.3 f E C(X)
such that
fX f(s) dµ(s)]
< Re( .rX f(s) dµo(s)I = b
for some suitable
e > 0.
In particular,
Re[ afX f(s) d6t(s)) = Re[af(t)] I. Bi = co(B),
Consequently
ext(Rf) c B.
we conclude that
On the other hand, let E Bi
and so from Theorems 11.2.2 and 9.4.1
and suppose
t E X, a E C, Ial = 1,
aot = bµ2 + (1 - b)µ2, 0 < b < 1.
are such that
Obviously
1'l,112
HµkII ' 1, k = 1,2.
We wish to show that
plish this we shall show that, whenever
abt = µl = µ2 To accomis such that
f E C(X)
f(t) = fX f(s) dot(s) = 0, it happens that
,fX f(t) dl`k(t) = 0
(k = 1,2)
and then shall apply the liesz Representation Theorem (Theorem 4.8.2) and Corollary 3.3.1 to conclude that there exists some which
µk = ak6t, k = 1,2.
ak E C
for
The desired conclusion will then be an
easy consequence of this fact.
However, in order to prove the first
assertion we shall have to work a little. Suppose such that
U
is any open neighborhood of
(Ih0II < 1
and
ho (s) = 0,
s E U.
t,
and
Since
X
ho.E C(X) is a normal
topological space, being compact and Hausdorff, there exists, by Urysbhn's Lemma [W2, p. 551, some h(t) = 1,
and
h(s) = 0, s f U.
bfX h(s) AL (S) + (1
-
h E C(X)
such that
IIhilm < 1,
Then
b)fX h(s) dµ2(s) = of h(s) d6t(s)
ah(t) = a.
is
11. Krein-Mil'man Theorem
340
But
a E C, tat = 1,
a
implies that
closed unit ball about the origin in
is an extreme point of the and since
C,
IfX h(s) dok(s)I <JIhil-IIµkII < 1
(k = 1,2),
a = fX h(s) dµk(s)
(k = 1,2).
we deduce that
Similarly, since IIh + holly, < 1, we see that bfX [h(s) + ho(s)) dµ1(s) +
(1
- b)fX [h(s) + ho(s)) d1+2(s)
= a[h(t) + ho(t)] a, and so
(k - 1,2),
fX [h(s) + ho(s)) d1+k(s) = a from which it follows that
fX ho(s)
on an open neighborhood of
Now suppose
(k - 1,2).
is any function that vanishes
ho E C(X)
Thus we have shown that, if
.
0
then
t,
fX ho (s) dlAk(s) = 0 and
f E C(X)
f(t) = 0.
(k = 1,2).
We wish to show that
fX f(s) dpk(s) - 0
(k = 1,2).
Clearly we may assume without loss of generality that
< 1.
IIfII
00
For each positive integer such that of
t
normal.
n
let
If(s)I < 1/n, s E Un,
such that
cl(Wn) C Un.
Un
be an open neighborhood of
and let
Wn
be an open neighborhood
The latter is possible because
Appealing to Urysohn's Lemma, let
hn(s) = 1, s E cl(wn); hn(s) = 0, s f U
;
n
t
hn E C(X)
X
be such that
and 'IIhnII. < 1.
Set
is
341
11.6. Banach-Stone Theorem
{gn) C C(X); gn(s) = f(s), s E Wn
Then
gn = hnf, n gn(s) - 0, s
U
and
;
is evident that
fX[f(s) - gn(s)] Wn
(f - gn)(s) = 0
and
I(f - gn((m < 1
s E Wn,
for
whence by our previous observations we see that
n = 1,2,3,...,
as each
Moreover, it
llgnllm < 1/n, n
n
(k = 1,2; n = 1,2,3,...),
0
is an open neighborhood of
t.
But
limnl(f - gn - film = 0,
and so
fX f(s) dµk(s) = lim fX [f(s) - gn(s)] dµk(s) n
= 0
(k = 1,2),
which is what we set out to prove. Hence, considering see that
k - 1,2.
for which
as elements of
we
C(X)*,
and so by Corollary 3.3.1 there
N(6 t) C N(µk), k = 1,2,
ak E C
exist
µ2
and
6t,µ1,
µk = ak6t,
where obviously
(ak( = 1,
Consequently a&t = bµ1 + _ [ba
(1
+
- b)µ2
- b)a2)6t,
(1
1
and so
a - bal +
- b)a2.
(1
closed unit ball in whence
a6t
C,
Since
is an extreme point of the a
al - a2,
= µl = µ2'
Therefore each measure in so
a
we conclude at once that
B
is an extreme point of
B*,
and
B - ext(Bt).
A second lemma will also be useful. Lemma 11.6.2. and let by
p(t)
Bi = {µ
(
Let
X
be a compact Hausdorff topological space
µ E M(X), ((µ1l < 1).
6t, t E X,
then
cp
If p : X - ext(Bp)
is a homeomorphism from
X
is defined onto
when the latter set is taken wi $ the relative weak* topology.
cp(X)
11. Krein-Mil'man Theorem
342
Clearly
Proof.
by Lemma 11.6.,1.
is an injective mapping of
cp
into
X
ext(Bi)
Consider the weak* neighborhood U(6 t;c;fl,f2,...,fn)flq(X)
of
6t
cp(X).
in
k = 1,2,...,n,
Than for each Wk
Is
is an open neighborhood of
in
t
X,
But, if
also sucha neighborhood.
- fk(t)i < r
Ifk(s)
I
and hence
=
lWk
is
then
s E W,
IfX tk(u) dbs(u) - fX fk(u) dbt(u)J = Ifk(s)
-
fk(t)J
(k = 1,2,...,n),
< e
that is,
nn
W =
cp(W) C U(6t;c;fl,f2,...,fn)(lp(X).
Thus
cp
is continuous and hence a homeomorphism, as
X
is
compact.
0
Theorem 11.6.1 (Banach-Stone Theorem). compact Hausdorff topological spaces.
Let
XI
and
X2
be
Then the following are equi-
valent:
(i)
X1
and
X2
are homeomorphic.
(ii)
are isometrically and
and
linearly isomorphic. Proof.
The proof that part (i) implies part (ii) is contained
in the introductory remarks to this section. T
:
C(X2) ^ C(XI)
Conversely, suppose
is an isometric surjective linear isomorphism.
Then the adjoint mapping
T*
:
M(X1) -- M(X2),
defined by
T*(µ1)(f) = fX f(t2) dT*(1+1)(t2) 2
= fX T(f)(t1) dv1(t1) 1
= µ1[T(f))
(f E C(X2)),
11.6. Banach-Stone Theorem
343
is an isometric linear isomorphism from 4.4.1).
Moreover,
M(X2) (Theorem
to
µ2 E M(X2)
and
is defined by
E M(X1)
fxIf(t1) dµ1(t1) = fx2T T*(µ1) = µ2.
then evidently Theorem 9.10.1. T*
M(X1)
is surjective since if
T*
Let
.(f)(t2)
dµ2(t2)
(f E C(X1)),
Alternatively, one could just apply
Bik = {µ
I
µ E M(Xk), llµl( < 1), k- 1,2.
Since
is linear and isometric, a simple argument shows that T*[ext(Bil)] = ext(Bi2).
Moreover, it is apparent from the definition of the weak* topology that
T*
(M(XO),TW*) -- (M(X2),Tw
:
a homeomorphism from Now let
cpk
:
tk E Xk, k = 1,2.
onto
Bll
is continuous. since
Bit,
ext(Bik), k = 1,2,
Xk
By Lemma 11.6.2 each
Bll
for which ext(Bil)
Ia(tI)J = 1,
Indeed, if
Clearly
If
t1 E X1,
T(tl) E X2
T* c cpl
then
and some
:
X1
'(t1) E V T*
T* o p1(X1)
We claim, moreover, that the mapping
T* o yl(tl) = a(t1)bT(tl)' t1 E X1,
f0 E C(X2)
is
= btk is a homeomorphism onto
yk
by Lemma 11.6.1 and the fact that
ext(B12).
onto
a homeomorphism.
determined by
for some
T*
be defined by 9 (tk)
its range with the relative weak* topology. T* o cy1(t1) = a(tl)6T(tl),
Hence
is weak* compact.
maps is
a : X1
X2,
is continuous.
is such that f0(s2) = 1, s2 E X2,
then
a(t1) - a(ti) fx2f0(s2) doT(t1)(s2)
f
T(f0)(S1 ) d6t (s1)
X 1
1
(tI E XI),
= T(f0)(t1)
from which the continuity of a Furthermore, we claim that +(t 1) = 6T(tl), t1 E X1,
is evident. $
:
X1
- Y2(X2) ,
defined by
is a bijective continuous mapping.
The
bijectivity is easily established, and its proof is left to the
11. krcin-Mil'man Theorem
344
t,
To establish the continuity of
reader.
consider the open neighborhood
of
U
6t2 = t(t1)
let
and
in the relative weak*
6t 2
topology given by U = U(6t ;c;fl,f2,...,fn) nc92(X2) 2
= 16s
I
s2 E
X2,
Itk(52) - fk(t2)I < s, k = 1,2,...,n},
2
where, of course,
fl,f2'...,fn
belong to
C(X2).
Since a is of
W1 C X1
continuous, there exists some open neighborhood
t1
such that a
Ia(t1) - a(s1)I < [2
sup
k= 1,2,...,n
W2 C X1
existence of an open neighborhood
Ifk(t2) - fk[T(sl)]I = IfX2fk(u2
-
sl E W
then for each
W = WI n w2,
of
t1
such that
(s1 E W2; k = 1,2,...,n).
IT(fk)(tl) - T(fk)(sl)I
0
such that
Prove that for
I1µI1 = 1I(X) < 1.
there exist Ek
tl,t2,...,tn
= II zkI
n
Let
of an
K.
IfX f(t) 16.
T
be topological linear spaces
Prove that, if
is an extremal subset of T(K),
µ E M(X)
f E C0(X)
(V2,T2)
T E L(V1,V2).
an extremal subset of
and let
is the image under
T(K)
K.
E zkf(tk k=I
in
X
and
and
< c.
be a locally convex topological linear space
K c V
be a compact convex set.
If
prove that the following are equivalent:
K
is any
11. Krein-Mil'man Theorem
348
(i) co(E) = K. (ii)
(iii)
ext(K) c cl(E). For. every real continuous linear functional supx
it is the case that
and let
over
X*(X) = supx
assumes its maximum at an
K
we saw that if
is a closed subset of
E
co(E)
a locally convex topological linear space and
co(E)
of compactness for
may not be dropped -- that is, give an
example of a locally convex topological linear space
in
such that some extreme point of
E c V
closed subset
is compact,
Give an example to show that the assumption
ext[co(E)] C E.
then
Prove that every
K.
In Theorem 11.2.2
*18.
Kx*(x).
be a compact convex set.
K C V
real continuous linear functional on extreme point of
E
V
be a locally convex topological linear space
(V,T)
Let
17.
EE
x* on
and a
V
is not
co(E)
E.
Give an example of a locally convex topological linear
19.
and a compact convex subset
V
space
(Hint: an example can be found in
not closed. 20.
If
(a)
space over
K CV such that
and
§
ext(K)
is
IR3.)
is a finite-dimensional topological linear
V
is a compact convex set, prove that
K C :V
K = co[ext(K)]. (b)
subset 21.
Give an example of a Banach space K C V Let
such that Bp = (f
I
V
and a compact convex
K t co[ext(K)]. t E L ORdt), 1Ifil < 11, 1 < p < m. P P
Using
the uniform convexity of the norm
I
f E Lp(1R,dt), IIfIIp = 1).
(This provides another proof of a portion of Proposition 11.1.1.)
CHAPTER 12
FIXED POINT THEOREMS
Introduction.
12.0.
Fixed point theorems come in various forms,
of which we shall discuss two. mappings
p
One form says that all continuous
from certain kinds of sets
K
fixed points -- that is, there exists some The point in
K
into themselves have to E K
such that
that is fixed will, in general, depend
cp(to)
to.
on
We shall discuss, without proofs, some theorems of this form
p.
in Section 12:1.
The second £orm of fixed point theorem we shall consider is one which says that certain kinds of continuous mappings from certain kinds of sets into themselves have fixed points.
We shall prove two
such theorems in Sections 12.2 and 12.3, one dealing with contraction
mappings on complete metric spaces, and one, the Markov-Kakutani Fixed Point Theorem, dealing with commuting families of continuous affine mappings on compact convex sets. In the final two sections of the chapter we shall give applications of these results to the proofs of the existence and .unique-
ness of solutions to certain ordinary differential. equations, and
to the existence of a Haar measure -- that is, a nonzero, positive, translation invariant, regular dorel measure -- on a compact Abelian topological group.
12.1.
The Fixed Point Property.
Before discussing contractions
and the Markov-Kakutani Fixed Point Theorem in the next two sections we wish to make some general remarks about fixed points and various fixed point theorems.
349
12. Fixed Point Theorems
350
Definition 12.1.1.
Let
The topological space
to.
If p
is said to be a fixed paint for
is continuous, then to E X cp(to)
be a topological space.
X
X
X
if
if said to have the fixed point
property if every continuous mapping y : X - X
has a fixed point.
with the usual topology, has
X = [-1,1],
.The topological space
:
p
the fixed point property, as we can see from the following argument: Suppose cp $(t) =
from
:
[yp(t)
[-1,1]
- t]/2
Then the mapping
is continuous.
[-1,1] - [-1,1]
t E [-1,1], is also a continuous mapping
where
to itself; moreover,
$(-1) > 0
and
$(1) < 0.
If
then clearly -1 or 1 is a fixed point for
t(-1) = 0 or $(1) = U,
If this is not the case, then
#(1) < 0 < $(-1),
p.
and so by the
intermediate value theorem of calculus we deduce the existence of some to,
such that
-1 < to < 1,
Thus, in any case,
y
9(t0) = 0,
that is,
has a fixed point, and thereby
the fixed point property.
p(to) = to. [-1,1]
has
The analogous result is valid in any space
Theorem 12.1,1 (Brouwer Fixed Point Theorem).
8i : {x I
x E 1, 11x1f
I
2
The sets (n - 1,2,3,...)
have the fixed point property.
We do not prove this result, since it would involve a rather lengthy digression from our main concerns.
A proof that involves
a minimum of topological machinery is available in [DS1, pp. 468-4701, and a more topological proof can be found in [Du, pp. 340 and 341].
The space I itself does not have the fixed point property.
For example, if x0 E J', xo # 0, then the mapping T defined by y (x) = x + x
0
for
x E UP,
IFf -- II ,
is clearly continuous, but
has no fixed point.
Furthermore, even though the sets
8i
in U'
are closed
bounded convex sets that have the fixed point property, it is not the
X
351
12.1. The Fixed Point Property
case that such sets in an arbitrary Banach space have the fixed point
)
R and let
over
K = if
I
f E CR(IZ),
It is easily seen that
but we claim that
K
11f 11.
K
< 1,
lim f(t) = 1, lim f(t) _ -1).
is a closed bounded convex subset of CR(R),
does not have the fixed point property.
To see this, consider the mapping by
(CR (R),,
For instance, consider the Banach space
property.
T(f)(t) = f(t - 1), t E R.
fEK,
T
:
CR(R)
--
CR(It),
defined
Clearly T is continuous, and if
then sup If(t - 1)1
UT(f)II. -
tEIR sup I f(t)
t E IR
and
lim T(f)(t) =
t-.+a, lim T(f)(t) =
t-_m Thus
T
restricted to
lim f(t - 1) a 1, t -++W
lim f(t - 1) = -1.
t -' -ao K
is'i continuous mapping from
K
to
K.
However, T has no fixed point, as can be seen from the following argument: Suppose
f0 E K
f0(t - 1) = f0(t), t E R. fo(0).
were such that
T(fo) = fo;
that is,
Then, in particular, we would have
f0(k), k - tl,t2,t3,.:., which contradicts the facts that
f0 E K. Consequently fo(k)k = -1, as $o(k) - 1 and link +. does not have the fixed point property. Note also that T is
limk K
even linear.
preceding example
K
fails to have the fixed point
property because it is not a compact set, as compact convex sets in
12. Fixed Point Theorems
352
Banach spaces all have the fixed point property..
This is an immedi-
ate consequence of the next theorem.
Theorem 12.1.2 (Schauder-Tikhonov Theorem). locally convex topological linear space over compact convex set, then
1.
Let If
be a
(V,T)
K c V
is a
has the fixed point property.
K
The proof of this result is also omitted, since it depends on Proofs are available in
the Brouwer Fixed Point Theorem.
[DS1, pp. 453-456; El, pp. 161-163].
Our approach to fixed points will be rather the reverse of that indicated in Theorems 12.1.1 and 12.1.2. for conditions on sets
K
Instead of looking
ensuring that every continuous mapping
into itself has a fixed point -- that is, ensuring that
p
of
K
has the fixed point property -- we shall seek conditions on the
K
y
mappings
12.2.
that will ensure their having fixed points.
Contraction Mappings.
The theorem to be proved in this
section is really a topological result and does not use any of the techniques of. functional analysis.
However, its applications
(e.g. to proofs of the existence and uniqueness of solutions to differential equations) are so typical of fixed point theorems that inclusion of the theorem seems desirable.
The Schauder-Tikhonov
Theorem also has ianportant applications in the theory of differential
equations, but, as indicated, its proof is not so accessible (see, for example, [E1, pp. 164-166]). Definition 12.2.1. T
:
some
V
V.
Then
T
b, 0 < b < 1,
For example, if x,y E f22'
then
Let
(V,p)
be a metric space and suppose
is said to be a contraction if there exists such that
p(T(x),T(y)) < bp(x,y)
(V,p) _ (e 2'1
12),
T(x) = T({ak)) _ (ak/2},
is clearly a contraction.
where
where
for x,y E V.
p(x,y) =
llx
- y112,
x = (ak),E Q2,
353
12.2. Contraction Mappings
Theorem 12.2.1. T
:
be a complete metric space.
(V,p)
Let
T
is a contraction, then
V -. V
Proof.
b, 0
0
V, we see that there exist some
V.
=
and so there exists
lkW,
is compact.
K
1/n < e,
If n is any positive integer such that
k/n < ek = bk/m < 6, k = 1,2,3,...,m, and so
then
m
(n)K c U
n)W c U.
k=1 K c nU,
Thus
and
is bounded.
K
u
Theorem 12.3.1 (Markov-Kakutani Fixed Point Theorem). (V,T)
be a topological linear space over
affine mappings from (i)
(ii)
V
(T,S E G).
TS = ST
(T E G) .
T(K) C K
Then there exists some Proof.
For each positive integer
n
=
i k=0 n
n E
Tn(K) c K, n = 1,2,3,..., T E G. Tn(K)
and each
T E G we define
1
K and the assumption that
is continuous, it follows that
T(xo) = xO, T E G.
n
Tk
TO = I, the identity mapping from
vexity of that
for which
xo E K
T
where
such that
V
to
K c V be a
is a family of continuous
G
Suppose
nonempty compact convex set.
and let
§
Let
V
to
T(K) C K
V.
From the con-
it follows at once
Moreover, since each is a compact, and hence
T E G
12. Fixed Point Theorems
356
closed, subset of integers and
Furthermore, if
K.
T,S E G,
TS(x) = Tn[Sm(x)] E Tn(K), as
then
This last observation allows us to
TnSm(K) C Tn(K)(1Sm(K).
and so
TnSm(x) - ST(X) E Sm(K),
we also have
and since TS = ST,
Sm(x) E K,
are positive
m
and
n
E =`{Tn(K)
conclude that the family of sets
+
n = 1,2,3,..., T E G)
is a family of closed subsets of the compact set finite intersection property.
Suppose for each case.
x0 E (1!.
Hence
(K) E
'IT (K) E ETn(K)
We claim
ETn(K).
T E G by tRe following argument:
Then there exists some
the topology
T
of the origin in all the sets in
that has the
K
0 Suppose this is not the T(x0) # xo.
such that
T E G
is a fixed point
x
Since
is Hausdorff there exists an open neighborhood V E,
Now
T(xo) - xo f U.
such that
there exists some
n
belongs to
xo E Tn(K), n - 1,2,3,...
and so, in particular,
Thus for each positive integer
x0
U
yn E K such
that
n-1 xo = n E Tk(yn)
k=0 and hegce, T
being affine
nE T(xo) - xo = T(n nE 1Tk(yn),
k=0
a nl
n
n k E T (yn)
k=0
I n- 1 k n
k=1 T'(yn)
1Tk(yn)
-
E T (yn)
k0
- yn
n
Consequently
T(x0) - xo E (n)(K - K),
as
TT(yn) E K
and
yn E K,
n This, however, implies that
K - K 4 nU, n = 1,2,3,..., because
if there-were some positive integer
n
for which
K - K C nU, then
.
357
12.4. Picard Existence Theorem
(n)(K
K
But
contrary to the choice of
- xo E U,
U.
is compact, and so by the continuity of addition in
the set
(V,T)
T(xo)
K) C U, and so
-
K - K
must also be compact.
there exists some positive integer
n
Thus by Lemma 12.3.1 This
K - K C nU.
such that
contradiction of the previous conclusion allows us to conclude that is indeed a fixed point for each
x0
T F. G.
0
The result indicated at the beginning of the section is now an immediate corollary. Corollary 12.3.1. over T
:
and let
4
V - V
Equations.
(V,T)
be a topological linear space If
be a nonempty compact convex set.
is a continuous affine mapping such that
there exists some
12.4.
K C V
Let
xo E K
for which
T(K) C K,
then
T(xo) = xo.
The Picard Existence.Theorem for Ordinary Differential
We now wish to give a typical application of fixed point
theorems to the question of the existence and uniqueness of solutions to ordinary differential equations.
The fixed point theorem we use
is the one concerning contraction mappings on complete metric spaces.
Consider then the initial value problem Y'(t) = f[t,Y(t)] Y(to) - Yo, where
y'
denotes the derivative of
y
with respect to
t.
solution to the initial value problem we mean some function that is defined on some closed interval which
By a x - x(t)
[to - 6,to + 6], 6 > 0, for
x'(t) = f[t,x(t)], to - 6 < t < to + 6,
and
x(t0) = y4.
The idea behind the use of the fixed point theorem to prove the existence of a solution to the initial value problem is to convert the latter into an equivalent problem whose solution is a fixed point of some contraction mapping.
The means for doing this is provided
by the next lemma, whose elementary proof is left to the reader.
12. Fixed Point Theorems
358
Suppose
Lemma 12.4.1.
is a continuous real-valued function
f
defined on a suitable domain in
and
IR2
Then the following
6 > 0.
are equivalent:
is a differentiable function defined on
(i) y
[to - 6,t
+ 6]
such that (t0 - 6 < t < to 3 6)
y'(t) = f[t,y(t)) Y(t0) = Y0.
is a continuous function defined on
(ii) y
[t0 - 6,to +,63
such that
(t0 - 6 < t < t0 + 6
Y(t) = yo + ,rt f[s,Y(s)I ds 0
The precise domain of definition of vague.
f
has purposely been left
This defect will be remedied in the next theorem.
The lemma tells us that problem if and only if
x
is a solution to the initial value where
F(x) = x,
T(y)(t) = y0 + ft f[s,Y(s)) ds. 0
Thus to apply our theorem about contraction mappings we must be able, to introduce an appropriate complete metric space such that the indicated here is a contraction on that space.
T
This necessitates
some further restrictions on 'f.
Theorem 12.4.1 (Picard Existence Theorem).' Let and
(to,y0) E
and suppose (i)
(ii)
f
u 2. :
D = ((t,s)
Set
D - IR
I
is such that
f is continuous.
There exists some
M > 0
such that
If(t,sl) - f(t,s2)I < MIs1 - s2I
whenever
It - t0+ < a
and
a > O,b > 0,
It - toI < a, Is - y0I < bj
Isk - Yol < b,k = 1,2.
359
12.4. Picard Existence Theorem
If
M0
sup(t,s)
E
DJf(t,s)l
6 = min(a,b/M0,1/2M), then there
and
exists a unique differentiable function
defined on
x
[t0 - 6,to + 6]
such that x'(t) = f[t,x(t)]
(t0 - 6 < t < to + 6)
x(t0) = yo.
Consider the subset
Proof.
defined by:
(CR([t0 - 6,to + V = (y
yE
I
CR((t0-6,t0+6J), +Y(t)-y01<M06, to-6 0
(b)
x*(fo) = 1,
(c)
Ilx*II = 1.
Evidently
K
f E C(G),
belong to
seen that
K
whenever where
f E C(G)
and
f(t) i 0, t E G.
fo(t) = 1, t E G.
as the continuous linear functionals K
for each
t E G.
Moreover, it is easily
is norm bounded, weak* closed, and convex.
by the Banach-Alaoglu Theorem (Theorem 9.4.1) compact convex subset of
w*
(C(G)*,T
).
xt(f) = f(t),
K
Consequently
is a nonempty
363
12.5. Haar Measure on Compact Abelian Topological Groups
For each
define
s E G
It is then evident that
Tsf(t) = f(t,- s), t E G and f E C(G).
TS E L(C(G)), s E G.
Ts
Let
denote the adjoint transformation of
(f E C(G)).
T*(x*)(f) = x*(TSf) Thus
T*
C(G)"
C(G)*
:
Ts:
is linear, and from the definition of the weak* topology,
s
f E C(G) that
and
t+ 5(f) = TtTS(f)
TSTt(f) = TS+ t(f) =
Moreover, since
the group
sot E C,
f(t) > 0, t E G,
and
x* E K
we have
s E G,
for
being Abelian, we conclude
G
TS(K) c K, s-E G.
Finally, we claim that
TST* = T*T*, sot E G. Indeed, if
is weak" continuous.
TS, s E G,
it is immediately apparent that each
then for
and
f E C(G)
Tsf(t) _
as
T*(x*)(f) = x*(TSf) > 0,
Furthermore,
f(t - s) > 0, t E G.
T*(x*)(fo) = x*(TSfo) = x*(fo)
1
and ITS(x*)(f)(
IX*(TSf){
< llx*IH sf1lg,
=114. which shows that Consequently
IIT*(x*)Il = 1.
= (TS
G
TS(x*) E K
Hence
s E G)
(f E C(G)), whenever
x* E K.
satisfies all the hypotheses of
the Markov-Kakutani Fixed Point Theorem (Theorem 12.3.1), and so there exists some
x* E K o
such that
T*(x*) = x*s E G. s
o
01
From the Riesz Representation Theorem (Theorem 4.8.2) we then deduce the existence of a unique
M(G)
such that
xa(f) = fG f(t) dµ(t)
(f E C(G)).
12. Fixed Point Theorems
364
Combining this last observation with a standard approximation argument we obtain for each Borel set
E C G
and each
s E G
1+(E + s) = fG XE+ $(t) dlL(t)
= fG TsXE(t) dµ(t)
= fG XE(t) do(t) = P(E). Thus
µ E M(G)
satisfies properties (i) through (iii).
To see that
is unique, suppose that
µ
properties (i) through (iii). we find that'for each
fGSG f(s
a E M(G)
also satisfies
Then by Fubini's Theorem (Ry, p.269]
f E C(G),
t) do(s)de(t) = fG(fG T-tf(s) dµ(s)]da(t)
= fG(SG
f(s) dp(s))do(t)
= fG f(s) dµ(s) fC do(t)
= fG f(s) dµ(s) and similarly that
fGfG f(s + t) d4(s)do(t) = fGfG f(s + t) da(t)dµ(s)
= fG f(t) da(t). Hence
f,'; f(t) d(N - a)(t) = 0
(f E C(G)), ,
365
12.6. Problems
and so by the Riesz Representation Theorem (Theorem 4.8.2) we conclude N = a.
that
is the unique measure that satisfies properties
µ
Therefore
(i) through (iii).
The measure
u whose existence is assured by Theorem 12.5.1,
µ,
is called the Haar measure on the compact Abelian group G.
The
uniqueness of the Haar measure, as defined here, is due to the requirement that
Without this restriction the measure
µ(G) = 1.
is only unique up to a nonnegative multiplicative constant.
µ
General
discussions of Haar measure on locally compact topological groups are available, for example, in [Ba, pp.125-187; El, pp.247-254; HR1, pp.184-230; Lo, pp.112-120; Nb, pp.49-119; We, pp.33-42].
The Markov-Kakutani Fixed Point Theorem can also be used to reprove some of our previous results (for example, Theorem 4.3.1 and the existence of Banach limits).
The details are left to the reader.
The latter application is also an example of the use of the MarkovKakutani Fixed Point Theorem in the study of so called invariant
means on topological groups, a subject that we do not pursue here: The interested reader is referred to [E1, pp.159-161; Gn, pp.230-261].
Problems.
12.6.
In
*1. for which
let
(f2,11-16)
H
be the set of all sequences
JXnj. 0.
cannot contain an uncountable number of
pairwise disjoint open subsets. (c)
If
U C G
tinct points in
G,
is open and
(tk]
is a countable set of dis-
then there exists some
that (t. + U) fl (tk + U) # 0.
j
and
k,
j # k,
such
CHAPTER 13
HILBERT SPACES
13.6.
Introduction.
Hilbert spaces.
Our concern in this last chapter is
The theory of Hilbert spaces is an extremely rich
and well-developed area of mathematics, and we shall only be able to give a very modest introduction to the subject, but nevertheless an introduction that we hope will provide the reader with a knowledge
of some of the fundamental results of the theory and convince him of the utility and attractiveness of further study. We shall begin with several sections devoted to some basic facts about Hilbert spaces, part of which can be deduced immediately from the development of the preceding chapters.
Included in these
sections are such results as the Cauchy-Schwarz Inequality and the Parallelogram and Polarization Identities.
Next we shall turn
our attention to several of the theorems that make Hilbert spaces rather unique: the Orthogonal Decomposition Theorem, which says that every closed linear subspace of a Hilbert space is an orthogonal direct summand; the Riesz Representation Theorem for the continuous linear functionals on a Hilbert space, which says that all such functionals are obtained from elements of the original space by means of the inner product; and the existence of orthonormal bases in a Hilbert space.
These results, together with some elementary facts about selfadjoint and unitary continuous linear transformations on Hilbert spaces, will then be used to study Fourier analysis in L2([-n,nj,dt/2n) to obtain some results from the theory of probability and random'
series, to prove some theorems in ergodic theory, and to generalize a classic theorem in the theory of analytic functions to a certain
371
372
13. Hilbert Spaces
L2([-n,nj,dt/2n).
subspace of
The final three sections of the chapter deal with the spectral theory of continuous linear transformations on Hilbert spaces, cul-
minating in a spectral decomposition theorem for compact self-adjoint transformations that generalizes a standard result of linear algebra concerning the diagonalization of matrices.
We shall develop only
enough general spectral theory to enable us to establish the spectral decomposition theorem.
Basic Definitions and Results.
13.1.
We begin with the defini-
tion of a bilinear form on a linear space. Definition 13.1.1.
mapping
(i)
t
:
V x V - §
Let
'
be a linear space over
A
t.
is said to be a bilinear form on
V
if
t(x,Y + z) = t(x,Y) + t(x,z),
(ii) $(x + Y,z) = *(x,z) + t(Y,z), (iii) (iv)
t(ax,y) = at(x,y), t(x,ay) = at(x,y)
A bilinear form x E V;
implies
t
V
on
(x,y,z E V; a E t).
is said to be nonnegative if
t(x,x) > 0,
to be positive definite if it is nonnegative and x = 0;
and to be symmetric if
A bilinear form
t
t(x,y) = t y-,x
is thus a mapping from
V x V
t(x,x) = 0 x,y E V.
,
to
that
*
is linear in its first argument, linear in its second argument if t = R,
and antilinear in its second argument if
note that for any bilinear form
t
we have
i = C.
t(x,x) = 0
Also we when
x = 0.
We can now give a definition of an inner product space that is shorter than that given in Example 1.2.10. Definition 13.1.2. is a
Let
V
be a linear space over
definite, bilinear form on
said to be an inner product on be an inner product space.
V,
and the pair
V,
(V,$)
f.
then
If
t
t
is
is saidlo
373
13.1. Basic Definitions and Results
In general, given an inner product space
(V,#),
we write
and denote the space itself by the pair
*(x,y) = (x,y), x,y E V,
Besides the examples of inner product spaces cited in Example V = £2
1.2,10, we mention here the spaces
E
([ak),(bk)) =
with the inner product
akbk
((ak),(bk) E t2)
k=1 and
V = L2(X,S,µ)
with the inner product (f,g E L2(X,S,µ)).
(f,g) = fX f(t)g(t) dµ(t)
The verifications are left to the reader. As indicated previously, an inner product space be made into a normed linear space on setting
can
114 = (x,x)112, x E V.
In order to prove this, however, we need a preliminary result of considerable usefulness.
Theorem 13.1.1 (Cauchy-Schwarz Inequality). an inner product space over
t.
Let
I(x,Y>l < (x,x)112(Y,Y>112
Moreover, given
be
Then
(x,y E V).
x,y E V,
I(x,Y)J = (x,x)112(Y,Y)112
if and only if Proof.
x
Given
and
y
are linearly dependent.
x,y E V,
we claim that there exists some
such that I(x,Y)1 2 = ;x,x>(Y,Y)
- (y - ax,y - ax)(x,x).
a E 4
13. Hilbert Spaces
374
Indeed, if
x = 0,
then the equation is easily seen to hold with
If
x } 0,
then
a = 0.
(x,x) # 0
and we set
a = (y,x)/(x,x).
Then (y - ax,y - ax) _ (y,y) - a(x,y) - a(Y,x) + aa(x,x) 2
2
x
_ (Y,Y) -
(x, ) x,x
)
(x,x)
2
+
x, ) x,x
2
(Y,Y) -
(X,X)
which is equivalent to the desired equation. The conclusions of the theorem follow immediately from the validity of the preceding equation.
a
It should be noted that the inequality portion of,the theorem remains valid, although with a different proof, if we assume only is a symmetric nonnegative bilinear form.
that
The positive
definiteness of the inner product is, however, necessary for the second assertion of the theorem.
We shall make use of this obser-
vation in Section 13.14.
As a corollary we obtain the fact that inner product spaces can be normed.
Corollary 13.1.1.
over
Let
be an inner product space
i and set 114 = (x, x) 1/2, x E V.
linear space over
Proof. lixIl - 0
(V,11.11)
is a normed
4.
It is evident that
implies
Then
x - 0.
I,axj( = jai jlxii
Moreover, if
,
x,y E V,
x E V, a E 4, and then, appealing to
the Cauchy-Schwarz Inequality, we see that
IIx+Y1(2= (x+Y,x+Y) _ (x,x) + (Y,x) + (x,Y) + (Y,Y)
375
13.1. Basic Definitions and Results
llxll2 + 2Re(x,y) + IIYI12 < IIxII2 + 2I (x,Y) I IIxII2 + 21Ix11IIYII
IIYII2
+
+
IIYlI2
(114 + IIxII) 2 Thus
llx + Yll < IIxII + Ilyli, x,y E V, and so
11-11
is a norm.
0
We repeat the next definition from Section 1.2 for the sake of completeness.
Definition 13.1.3. over
0.
If
(V,II.Il)
Let
(V,(.,.))
be an inner product space
is a Banach space over
f,
then
V
is said
to be a Hilbert space.
Of course, the norm here, and in all analogous situations, is 114 = (x,x)1"2, x E V.
Among the examples of inner product spaces
we have mentioned previously,
ffn,Cn,f2,
and
L2(X,S,p)
are all
Hilbert spaces.
One further corollary of the Cauchy-Schwarz Inequality will be useful.
Corollary 13.1.2. over on
f.
be an inner product space
Let
Then the inner product is a continuous f-valued function
V x V,
where
V
is taken with its norm topology determined by
the inner product. Proof.
If
xo,yo E V,
then
I(x,Y) - (xo,yo)I < [(x,Y) - (x.Y0)I + I(x,Y0) - (xo,Y0)I
I(x,y - yo)) + I(x - x0,0)1
IIxIIIIY - y011 + lix - xolll1Yoll
0
13. Hilbert Spaces
376
The Parallelogram and Polarization Identities.
13.2.
In this
section we examine two fundamental identities in inner product spaces: the Parallelogram and Polarization Identities.
We shall use the
former identity to give a characterization of those normed linear spaces that are inner product spaces.
Geometrically the Parallelogram Identity says that the sum of the squares of the lengths of a parallelogram's diagonals is equal to the sum
of the squares of the lengths of its sides.
Precisely,
we have the following theorem: Theorem 13.2.1 (Parallelogram Identity).
an inner product space over V
Let
be
Then
IIx+yk2+llx-YII2=2IIx112+211Y112 Proof.
For any
x,y E V
(x,y E V).
we have
Ilx+Yll2+llx-Y112=(x+Y,x+y)+(x-Y,x-Y) = (x,x) + (Y,x) + (x,Y) + (Y,Y) + (x,x) - (Y,x) - (x,Y) + (y,y) 211x12 + 211Y112.
A similar direct computation also establishes the Polarization Identity, which allows one to express the inner product in terms of the norm.
Theorem 13.2.2 (Polarization Identity). inner product space over
t.
If
4 = 1F2,
(x,Y) = li2 I2 whereas if
(x,Y) =
Let
be an
then
(x,y E V),
I2
li
0 = C, then IIx
2
11x--2--112 + illx
112'
-
.11x
2
(x,Y E V).
13.2. Parallelogram and Polarization Identities
377
The form of the Polarization Identity suggests a means of ir.'roducing an inner product into a normed linear space.
The next result
asserts that such an attempt will succeed provided the norm satisfies the Parallelogram Identity. Theorem 13.2.3.
be a normed linear space over
Let
If
Ilx+YII2+Ilx-Y112=211x11`'+21,yd2 then there exists an inner product
(x,y E V),
on
V
such, that
11x11 = (x,x>1/2, x E V. Proof.
ately.
We shall consider the cases
Suppose first that
= JR
t
(x,Y) = lix We claim that
2
t = JR
and
and define
is an inner product on
V
separ-
V x V
by
(x,YEV')-
11" X12
-
12
t _ 1 on
such that
Ilxll
=
(x,x)1/2
The latter assertion is evident, since x E V.
It thus follows that
and only if
x = 0.
more, given
x,y E V,
(x,x) > 0, x E V,
Hence
and
(x,x) = 0
is positive definite.
if
Further-
we see that
u'
(Y.x) =
llx-2 and so
(x,x) = lj(x + x)/2112,
is symmetric.
-
2 112
112112
The proof of the linearity of
requires somewhat more work.
Consider any gram Identity in
x,y,z E V. V
we have
Then by the validity of the Parallelo-
13. Hilbert Spaces
378
211x+YII2+211z+YII2=llx+z+2YII2+IIx-z112 211x-Y112+211z-YII2=IIx+z-2y112+IIx-2112, whence, on subtracting the lower expression from the upper, we obtain
2(11x+YII2-IIx-YII2)+2(IIz+Y112-iiz-YII2)
IIx+z+2YII2-IIx+z-2y(l2. Recalling the definition of
(
(x,Y) +
In particular, if
,
14
z = 0,
),
we deduce at once that
x + z,2y)
(x,y,z E V).
then
(x,Y)
(x,y E V).
4x,2Y)
Hence (u + v,y) = Z u + v,2y) _ (u,Y) + (v,Y) and so
is additive in its first argument.
(y,x), x,y E V,
it follows that
(u,v,y E V),
Since
(x,y) _
is additive in both argu-
ments.
Finally, we must show that a E R.
x,y E V,
Since
(ax,y) = a(x,y)
for
is additive, it is obvious that
for any integer
n,
and hence, if
n(n,y) = (nX,Y) = (x,y)
x,y,E V
and
(nx,y) = n(x,y),
n } 0,
(x,Y E V),
that is,
(n,Y) = n x,Y)
(x,y E V).
From these facts we deduce at once that for any rational number
r
379
13.2. Parallelogram and Polarization Identities
(x,y E V).
(rx,y) = r(x,y)
a EIR and let
Now suppose
be a sequence of rationals is continuous, being
Then, since
limkrk = a.
such that
(rk)
defined in terms of the norm on
we see that
V,
(ax,y) = lim(rkx,y) k
lim rk(x,y) k
(x,y E V)
= a(x,y)
because
again follows, as
a EIR,
and
The identity
limkllrkx - axll = 0.
bilinear form on
Thus
(x,y) = (y,x).
is a
V.
is a symmetric positive definite bilinear
Consequently
form, that is, an inner product, on
when
(x,ay) = a<x,y), x,y E V
V,
and the theorem is proved
I = R. Now suppose
I = C.
In this case we define
as
(x,Y) = (x,Y)R + i(x,iy)R
(x,y E V),
where 11x + y112
(x,Y)R =
Ilx
2
12
(x,y E V).
is so defined because
As is easily seen,
(x,Y) =
- ll-2-YII2
-
lJx Z2 + illx2 H 2
We must still, however, show that
- illx--
12
IR,
,
is an inner product.
In the following arguments it is well to keep in mind that is an inner product over
(x,y E V
(
)R
as established in the first part of the
13. Hilbert Spaces
380
proof.
Now some elementary computations reveal that (x,ix)R = 0
(x E V)
and (x,Y E V).
(ix,iy) R = (x,Y)R
The first of these two identities shows at once that (x,x) = (x,x)R + i(x,ix)R = (x,x)R
=
and-that
(x,x) = 0
(x E V)
11x112
if and only if
x = 0.
is positive
Thus
definite.
Next we claim that
is symmetric.
Indeed, utilizing the
second identity, we obtain (Y,x) = (Y,x)R + i(Y,ix)R = (Y,x)R + i(iy,-x)R = (x,Y)R - i(x,iy)R
xZ
The fact that
(x,y E V).
,y)
is additive in both its arguments follows at
once from the additivity of
Hence to show that
bilinear we need only prove, in view of the symmetry of (ax,y) = a(x,y), x,y E V
and
a E C.
First we note that
(ix,y) = (ix,y)R + i(ix,iy) R
= (ix,Y)R + i(x,Y)R = i((x,Y)R - i(ix,Y)R)
is (
that
381
13.2. Parallelogram and Polarization Identities
= i[(x.Y)R,- i(-x,iy) R)
i[(x,Y)R + i(x,iy)R) (x,y E V).
= i(x,y)
Consequently, if
a = c + id E C,
we have
(ax,y) = ((c + id)x,y) = (cx + idx,y)
_ (cx,y) + (idx,y) (cx,y) + i(dx,y)
= c(x,y) + id(x,y) (x,y E V),
= a(x,y)
is a real
where the penultimate equality is valid because bilinear form.
is an inner product on
Thus we have shown that which
(x,x) = 11x112, x E V,
V
for
thereby completing the proof of the
theorem.
Theorem 13.2.3 thus asserts that a normed linear space jis an
inner product space precisely when the Parallelogram Identity is valid for all elements of the space.
This fact also provides us at
once with the next corollary. Corollary 13.2.1. i.
Let
(V,11.11)
be a normed linear space over
If every two-dimensional linear subspace of
product space over Proof.
1,
then
V
V
is an inner
is.an inner product space over
f.
From the hypothesis of the corollary and Theorem 13.2.1
we see that for each
x,y E V
IIx + yll2 + Ilx - y112 = 2IIxiI2 : 2iIY112.
0
13. Hilbert Spaces
382
Some Other General Properties of Hilbert Spaces.
13.3.
In this
section we wish to note some properties of Hilbert spaces that follow immediately from the general theory we have developed in the preceding chapters.
A preliminary and necessary step, however, is to
prove that inner product spaces are uniformly convex (Definition 8.2.1). Theorem 13.3.1.
If
iJxj+
_
Let
be an inner product space over
(V,(.,.))
(x,x)1/2, x E V,
is a uniformly
then
convex normed linear space. Suppose
Proof. 1
6 =
(1
-
0 < e < 1
e2/4)l/2.
and let
.6 > 0
From the Parallelogram Identity (Theorem 13.2.1)
we see that, if x,y E V, (ixil < 1, uyjj < 1,
2 x
I2 =
211212
1
(1 (V, 1I
11)
and -
jIx
jIx
- yll > e, then
yi2
--2--11
s2
1
+ 2
Thus
be such that
4
2
is uniformly convex.
Mil'man's Theorem (Theorem 8.2.1) now provides us with the following corollary: Corollary 13.3.1. Then
(V,IHI)
Let
(V,(.,.))
be a Hilbert space over
is a reflexive Banach space where
§. (x,x)112,
[lxit =
x E V. We now can list a sampling of results about Hilbert spaces that are immediate consequences of the preceding theorem and corollary, as well as the material of the preceding chapters. Let
be a Hilbert space over
t.
Then the following
are valid: (1)
If
K E V
is a nonempty closed convex set, then there
.
383
13.3. Other General Properties of Hilbert Spaces
such that
x0 E K
exists a unique
inf iixli
iixoli =
xEK If
y
x0 E K
4 K, then there exists a unique
such that
0
liY 0
- x o it
=
inf ifY
xEK
- xii'. o
(Theorem 8.2.2 and Corollary 8.2.1).
If
(2)
x* E V*
(V.Tw)
(4)
E C V
closed set in BI = {x
i
x* j 0, then there exists a unique
such that x* (xo) = (ix*ii (Corollary 8.2.2) .
x0 E V, iixoii = 1, (3)
and
is sequentially complete (Theorem 9.3.1(iv)). is compact in
(V,Tw)
if and only if
(V,TW)
In-particular,
and a bounded set in
x E V, iixu < 1)
is a
F.
is weakly compact and weakly sequentially
compact (Corollary 9.4.2 and Theorem 10.3.1). (5)
is separable, then the weak topology
If
restricted to B1
is metrizable (Corollary 9.4.6).
is separ-
is separable if and only if
(6)
Tw
able (Theorem 4.5.1). (7)
If
and only if
K C V KClb61.
K
is closed in
(V,Tw)
for each
is convex, then is closed in
(V,Tw)
if
b > 0 (Corol-
lary 10.2.1). (8)
If
(V,TW)", then
K C V
is a nonempty convex set that is compact in
ext(K) T 4
and
K = co[ext(K)],
can be taken in either the weak topology logy.
In particular,
ext(B1)
and
Tw
where the closure or in the norm topo-
81 = co[ext(BI)) (Theorems
11.2.1 and 9.2.2).
In the next section we shall discuss some results that are more peculiar to Hilbert spaces.
13. Hilbert Spaces
384
The Orthogonal Decomposition Theorem and the Riesz
13.4.
Representation Theorem.
We have two aims in this section: to show
that every closed linear subspace of a Hilbert space
V
is an ortho-
gonal direct summand and to use this result to prove that be identified not only with its second dual space with
V**
can
V
but also
These results are generalizations to Hilbert spaces of
V*.
well-known theorems of linear algebra. We begin with some definitions. Definition 13.4.1. suppose
and
W1
W2
Let
that
For each
be a linear space over
are linear subspaces of
to be the direct sum of (i)
V
and
W1
x E V
W2
V.
Then
I
and
V
is said
if
y E W1
there exist
and
z E W2
such
x = y + z.
(ii)
W1 fl W2 - (03.
In this case we write
V = W1 cD W2.
It follows easily from the definition that, if then the decomposition
x = y + z
is unique.
V = W1 (D W2,
One could also use
this observation as the definition of the direct sum. Definition 13.4.2. over x 1 y,
I.
if
Then
x,y E V
(x,y) - 0.
orthogonal, denoted by and
be an inner product space
Let
are said to be orthogonal, denoted by
Two subsets Eli E2,
if
E1,E2 CV are said to be (x,y) = 0
whenever
x E E1
y E E2.
It is evident that, if
E1
and
E2
are orthogonal, then
E1llE2 c (0):
A criterion for orthogonality is given in the following proposition.
385
13.4. Orthogonal Decomposition Theorem
Proposition 13.4.1. over
I
and let
(i)
If
be an inner product space
4et
x,y E V.
§ = IR,
then
xiy
if and only if
llx+Yll2= 1142+Oil2.
(ii) If
6 = C;,
then x. y
if and only if
llx +.Yll2 = !!x112 + llyl12 and
lix + iyll2 : 1!x!12 + ljYll2. Proof.
The necessity portion of each of the two equivalences
is easily established and is left to the reader.
4 = IR and
Conversely, if
llx + yll2 = 1142 + llyll2, then routine computation shows
that
(y,x) + (x,Y) = 2(x,y) = 0,
lix
whereas if
= U.,
llxll2 + llyll2,
then similar calculations reveal that
llx + Y112 = 11x!12
llYll2,
+
(Y,x) + (x,y) =
and
+
Y112
x-,YT + (x,y)
= 2ae(x,Y) = 0
and
i(Y,x) - i(x,Y)
-i[(x,Y) -
x,Y
_ -i(2ilm(x,Y)] 2Im(x,Y)
= 0. Thus
Re(x,y) = Im(x,y) - 0,
and so in either case
x . y.
=
13. Hilbert Spaces
386
Definition 13.4.3.
over
If
0.
E C V,
be an inner product space
Let
then the orthogonal complement El
of
is
E
defined by
E1 = (x
I
x E V, x1 E).
The correspbndence of the notation for orthogonal complements and for annihilators of subsets of topological linear spaces (Definition 4.6.1) is not accidental.
Indeed, an immediate conse-
quence of the Riesz Representation Theorem (Theorem 13.4.2) is that the two notions are identical in a Hilbert space.
In view of this
observation the next proposition should come as no surprise.
The
proof is left to the reader. Proposition 13.4.2. over 4 (i)
and let E1
E c V.
Let
be an inner product-space
Then
is a closed linear subspace of
V.
(E1)1
(ii)
E11
=
(iii)
If
E
-D E.
is a linear subspace, then
E n E1 = (0).
We turn next to the Orthogonal Decomposition Theorem. Theorem 13.4.1 (Orthogonal Decomposition Theorem). be a Hilbert space over linear subspace, then Proof.
x0 E V z0 E W1.
If
w c v
is a closed
V = W cD W.L.
Since WfW1 = (o],
can be expressed as But, given
I.
Let
it suffices to show that each
xo = yo + z0,
x0 E V,
where
ya E W
and
it is evident that
x0 - W
is a
nonempty closed convex subset of
V,
and hence the results of the
preceding section (or Theorem 8.2.2) reveal the existence of unique
z0 E x0 - W
such that
IIz0II =
inf 11x1j. xEx0-W
a
387
13.4. Orthogonal Decomposition Theorem
yo = X0 - zo, we see at once that
Setting
yo E W
and
xo
.yo + zo'
z0 E W1.
It only remains to show that
However, recalling the proof of the Cauchy-Schwarz Inequality (Theorem 13.1.1), we know that for each
there exists some
x E W
such that
a E t
I (x,
zo)11
Ilxllzllzoll2 - ll zo - axll2lix{l2
=
zo - ax = x0 - yo - ax E xo - W
Moreover,
by-the choice of
and so
llzo
- axll > llzoll
Thus we deduce that
z0.
1 (x, zo) 12
K
Ek =
such that
is a linear subspace of
Moreover,
W
((akn))) c W
Suppose
Then for any
(ak) E V
W
is not a Hilbert space.
V
V
n
such that
is closed, as we see from the is a sequence that converges to
be a positive integer for which and any
Iak/k = 0.
ak = O,k > K.
we have
M
I E ak/kI
ak/ki
k-1
k=1 I E
k=1
a(n))/k1 + I E a(n)/kI
(a
k
k,
k=1
k
N rkE 11/k211/ZrkE flak 11
`
0
-
akn)IZ 11/2 M
1
IT )11211(ak)
- (akn)}112
is given, let
N1
+
+
J
I E a, (n)/kI
k=1
I E akn)/k k=1
be a positive integer such that,
n > Nil
WII(ak) - (akn))112 < z(62 and then choose such an
N2
that, if
M > N2, then
M I E a(N1)/kI
max(N2;K) M I
E ak/kI = I E
k=1
k=1
ak/ki
389
13.4. Orthogonal Decomposition Theorem
g
i
2 + 2
= C. Since
is arbitrary, it follows that
c
We claim, however, that
Wl = (0).
then, given a positive integer ck = 0,k # n
defined by
Clearly
a,,
is a closed linear subspace of
W
Hence
[ak) E W.
J"k
and
let
n,
V.
Indeed, suppose [ck)
(bk) E W.
denote the sequence
k # n + l,cn = n,cn
-(n + 1). + 1
(ck) E W, and
E ckk
0 = ([ck),(bk)) =
k=1
= n E - (n + 1)bn + I . Thus we see that bI # 0,
t;ien
fact that
b
n+
1 0 [n/(n + 1)]bn,n = 1,2,3,,..,
bn 4 O,n = 1,2,3,...
and so, if
This, however, contradicts the
.
(bk) E W1 C V.
Consequently
Wj = (0),
and
W
W # V.
W1
It is readily apparent from the Orthogonal Decomposition Theorem that, given a closed linear subspace
W
of a Hilbert space
there exists precisely one linear subspace V
W ® X
and W. X,
namely,
X = Wl,
is itself closed (Proposition 13.4.2).
X
of V
V,
such that
and that this subspace
X
If, however, we drop the
orthogonality requirement, then there may exist many linear subspaces
X
W = ((x,0)
for which ,
x E]R),
V ='W ® X.
For instance, if V = IR2
then any straight line
X
and
that passes through
the origin and is distinct from the real axis is a linear subspace
of
set
gig
such that I = W E) X.
( (0,y)
In this case
W1
is, of course, the
y E IR).
Let us now see how the Orthogonal Decomposition Theorem provides
us with a description of the continuous linear functionals,on a
13. Hilbert Spaces
390
is an inner product
First, suppose that
Hilbert space.
x E V.
space and let
Then define
x*
by
V
on
(y E V).
x*(Y) = (Y,x)
Since the inner product is linear in its first argument, we see at once that
x* E V',
and the Cauchy-Schwarz Inequality (Theorem 13.1.1)
asserts that J x* (Y) J = I (y, x> I
x* E V*
Hence
and V
into
V*.
lix*ij
=
since
jixli
In this way we obtain an isometric
x*(x) = (x,x) = llxll2, x E V.
mapping from
Actually
(fix*jj < Ijxlj.
(y E V) .
< liyliiixh
The mapping, however, is not linear
since
ax*(Y) = a(y,x) = (v,ax)
(y E V; a E 0).
= (ax)*(y)
The mapping is easily seen to be antilinear.
The preceding development shows that the correspondence is an antilinear isometry from product space. V
if
V
V*
into
when
is a Hilbert space.
be a Hilbert space over (i)
(ii)
to
Then the following are equivalent:
t.
There exists a unique x
x*(y) = (y,x), y E V, V*.
Let
x* E V.
Moreover, the mapping tion
is an inner
The important point is that the mapping is surjective
Theorem 13.4.2 (Riesz Representation Theorem).
V
V
x - x*
x*
x E V from
such that
V to
V*,
x*(y) = (y,x), y E V.
defined by the equa-
is a surjective antilinear isometry from
13.4. Orthogonal Decomposition Theorem
Proof.
391
Evidently the theorem will be proved if we can show
that part(i) implies part(ii), as the remaining assertions are apparSo let
ent from the previous discussion. x* = 0
If
x*(y) _ (y,x), y E V.
then
x = 0,
is the only such element of
x = 0
more,
and
x* E V*.
V
Further-
0 = x*(x) _ (x,x)
since
llxl(2 Suppose then that x*.
subspace of
and let
V = W d) W.
xo # 0.
Clearly
y E V
W
Wl
Since
x*(xo) 4 0,
T
as
(0},
there exists some
x0 t N(x*) = W,
we can form the element
Evidently
the kernel of
W = N(x*),
is a proper closed linear
and so, by the Orthogonal Decomposition Theorem,
V,
we have
each
x* T 0
From Theorem 3.3.2 we see that
x0 E W,
and thus for
[x*(y)/x*(x0)]x0 E V
y -
y - [x*(y)/x*(x0)Jx0 E W,
and hence
(Y - [X* x )]x0,xo) = 0, 0
that is,
(Y,x0) =
](xo,x0
Consequently we see that
x*(x x* ( Y) = (Y, I 111
and we set
) 11x Oil 2 0
(y E V) ,
IX0
x = [x* (x0 /IIx0II2Ix o .
The uniqueness of
x
follows at once from the isometric nature
of the mapping, and thus the first hypothesis of the theorem implies the second one.
0
Thus, for example, suppose lary 8.3.2 we know that for each corresponds a unique,
V = L2([-n,n],dt/2n).
From Corol-
x* E L2([-n,TT],dt/2n)*, there
g E L2([-n,n],dt/2n)
such that
13. Hilbert Spaces
392
(f E L2([-n,n],dt/2n)).
x*(f) = Zn fnnf(t)g(t) dt
is the one that appears in the Riesz Representa-
g - x*
The mapping tion Theorem.
Combining Theorems 13.4.2 and 4.6.1, we easily obtain the next The details are left to the reader.
result.
be a Hilbert space over W11 = W. is a closed linear subspace, then
Corollary 13.4.1. If
W
V
Let
f.
(V,(.,..))
This corollary can, of course, also be proved by direct arguments without using Theorems 13.4.2 and 4.6.1. Orthogonal Projections.
13.5.
In this section we wish to show
that there exists a one-to-one correspondence between the closed linear subspaces of a Hilbert space and certain continuous linear transformations on the Hilbert space called orthogonal projections. To begin we consider a Hilbert space linear subspace
W
of
From the Orthogonal Decomposition Theorem
V.
(Theorem 13.4.1) we know that
there exist a unique
y E W
We define
by
P
:
V - V
Thus for each
V = W d3 WI.
and a unique
P(x) = y.
linear transformation from denoted as usual by
and a closed
V
z E W1
x E V
such that
It is evident that
P
x = y + z. is a
to itself and that the range of
R(P), is precisely
we see from Proposition 13.4.1 that
W.
11x112
Moreover, since
= Ily +
z112 = IIY112 +
P, y 1 z,
IIzII2,
and so
llp(x)112
whence
P E L(V)
and
1(P11 < 1. 11P11 >_
= IIY1I2
< I1XII2,
If W # (0), then sup Upwil
xEW IIXII=1
IIPII = 1
since
393
13.5. Orthogonal Projections
sup IIxII xEW IIxII =1 = 1.
The fact that
P(x) = x, x E W,
utilized in the preceding inequa2
lities, also shows at once that
P
P
= P; that is,
2
(x) = P(x),
x,y E V.
(P(x),y) = (x,P(y)), x,y E V, arguing
We claim furthermore that
y = z + w,
where
and
x
as follows: Suppose u,z E W
and
v,w E W1.
Then, since
u i w
and
we see that, on the one hand,
z. v,
(P(x),Y) _ (P(u + v),z + w) = (u,z) + (u,w) = (u,z),
and, on the other hand, (x,P(Y)) = (u + v,P(z + w)) = (u,z) + (v,z) = (u,z).
(P(x),y) = (x,P(y)), x,y E V.
Hence
Finally, we claim that the transformation the only element of
L(V)
such that
(P(x),y) _ (x,P(y)), x,y E V. these properties.
(x
I
P
defined here is
P2 = P, R(P) = W, and
Indeed, suppose
Q E L(V)
also has
Then we note first that
x E V, P(x) = x) = R(P) = W = R(Q) _ {x
I
x E V, Q(x) = x).
This follows easily on noting,'for example, that, if
x = P(y) E R(P),
13. Hilbert Spaces
394
(x
In a similar vein we claim that
P(x) = P2(y) = P(y) = x.
then
x E V, P(x) = 0) = R(P)1 = W1 = R(Q)1 = (x
I
The argument is as follows: If, for example, and hence
(P(y),x) = (y,P(x)) = 0, y E V, (x
I
x E V, P(x) = 0) C W1.
Conversely, if
W1 = (x
Thus
But then for any
P(x) = 0,
then
x 1 R(P); that is,
x E W1 = R(P)1,
then
(P(x),P(x))
x E V, P(x) = 0).
I
x E V,
x E V, Q(x) = 0).
and so, in particular
(y,P(x)) _ (P(y),x) = 0, y E V, IIP(x)II2 = 0.
I
if
x = y + z, y E W,
P(x) = P(y + z) = y = Q(y + z) = Q(x),
that is,
z E W1,
we have
P = Q.
We can summarize this discussion by saying that a closed linear subspace
W
of a Hilberrt space P2 = P, R(P) = W, and
such that.
determines a unique
V
P E L(V)
(P(x),y) = (x,P(y)), x,y E V.
The
converse of this observation is also valid, as will be demonstrated by Theorem 13.5.1.
First, however, we introduce a name for such
transformations. Definition 13.5.1. over
§.
(i)
(ii)
then
P
If
P E L(V)
Let
be an inner product space
is such that
PZ = P
(P(x),y) = (x,P(y))
(x,y E V),
is said to be an orthogonal projection.
The source of the name will become apparent from Proposition 13.5.1.
Theorem 13.5.1. and let (i)
(ii)
that
W C V. W
Let
be a Hilbert space over
is a closed linear subspace of
V.
There exists a unique orthogonal projection
R(P) = W.
f
Then the following are equivaleft:
P E L(V)
such
13.5. Orthogonal Projections
395
The implication from part (i) to part (ii) is contained
Proof.
in the foregoing discussion.
Suppose part (ii) holds.
We only need to show that
is then a linear subspace.
Clearly W
W
is closed,
and to this end, in view of Proposition 13.4.2(i), it suffices to prove that
W = X1
mapping on
V,
I
that is,
If
denotes the identity
I
then it is apparent that
I(x) = x, x E V,
We set. X = R(I - P).
- P E L(V). If
X CV.
for some set
x E W = R(P),
then there exists some
P(u) = x,
and if
for which
(I - P)(v) = y.
y E X = R(I - P),
u E V
such that
then there exists some
v E V
Hence
(x,Y) = (P(u),(I - P)(v)) = (u,P(I - P)(v)) _ (u,(P - P2)(v)) = 0,
as
P2 = P.
Thus
W C X1.
Convert ely, suppose
x E X.
Then for any
y E V
we have
(I - P) (y) iE X, and so.-
0 = (x, (I - P)(y))
= (x,Y - P(y)) (x,Y) - (x,P(Y)) (x,Y) - (P(x),Y)
=((I-P)(x),Y). In particular, if
y = (I - P)(x),
then
0 =
I1(I -
P)(x)112; that is,
x = P(x) E R(P) = W.
Therefore
W = X1,
and
W
is closed.
0
13. Hilbert Spaces
396
It is easily verified that, if P E L(V), I
- P
then
V
is a Hilbert space and
is an orthogonal projection if and only if
P
is an orthogonal projection, and that if
projection, then
R(P) = R(I - P)1
and
R(I
-
is an orthogonal
P P)
= R(P)
The
These observations, combined with
details are left to the reader.
the Orthogonal Decomposition Theorem (Theorem 13.4.1), immediately yield the next proposition. Proposition 13.5.1. If
and
be a Hilbert space over
Let
is an orthogonal projection, then
P E L(V)
R(P) i R(I
§.
V = R(P) d? R(I - P)
- P).
This proposition should shed some light on the reason for the name "orthogonal projection".
There are several equivalent formulations of the notion of orthogonal projection.
We collect some of these in the next pro-
position, whose proof is left to the reader. Proposition 13.5.2.
be a Hilbert space over .
Let
Then the following are equivalent: (i)
P E L(V)
is an orthogonal projection.
(ii)
P E L(V)
is such that
R(P)1 = {x (iii)
13.6.
P E L(V)
P2 = P
I
is such that
and
x E V, P(x) = 01. P2 = P
Complete Orthonormal Sets.
and
R(P).LR(I - P).
The Riesz Representation
Theorem proved in Section 13.4 is obviously a generalization to Hilbert spaces of a standard result of linear algebra.
In this
section we wish to develop another generalization to Hilbert spaces of a fundamental result for finite-dimensional linear spaces: the notion of an orthonormal basis. To be more concrete, suppose
V = M P.
Then it is well-known
397
13.6. Complete Orthonormal Sets
that there exists a basis for
consisting of vectors
V
e1,e2,...,en
such that
(ek,e.) = 1
fpr
k = j,
(ek,e.) = 0
for
k
j.
Our goal is to obtain a counterpart in an arbitrary Hilbert space
for such a basis inch.
The notion of basis that is most appropri-
ate for our concerns is not the usual algebraic ,ie, but rather a concept utilizing the fact that a Hilbert space comes equipped with a natural topology (the norm topology), which thus allows us to consider infinite linear combinations of elements of the space as well as finite linear combinations.
The precise definition of basis we
require will appear only after some preliminary development.
We
begin with a definition and a fundamental theorem. Definition 13.6.1. over
be an inner product space
Let
A set of nonzero vectors
t.
Ix
a E A) C V
I
is said to be
of
an orthogonal set. if -(xa,x0) = 0, a,S E A, a # P; it is said to be
an orthonormal set if it is an orthogonal set such that a E A.
An orthonormal set of vectors in
in no other orthonormal set in
V
V
(x
x o) = 1,
that is properly contained
is said to be a complete ortho-
normal set.
Theorem 13.6.1. and let Proof. in
V.
Let
Clearly
0
be an inner product space over
Let
V # (0).
Then
V
contains a complete orthonormal set.
denote the collection of all orthonormal sets
0 j 0
because, if
We define a partial ordering in EI,E2 E 0.
If
U = (E(Y
I
0
a E A)
then it is easily verified that Consequently we may apply
11xjj
by setting
= 1,
then
EI > E2
if
(x) E 0. E1 -D E2,
is a linearly ordered subset of
0,
is an upper bound for
U.
E0 = UCY E0
Zorn's Lemma [DS1, p.61 to deduce the
existence of a maximal element ordering.
x E V,
E
in
0
with respect to the partial
13. Hilbert Spaces
398
From the preceding definition we see at once that
is a
E
complete orthonormal set.
Thus every inner product space contains complete orthonormal Moreover, the proof of Theorem 13.6.1 can be modified in an
sets.
obvious way to obtain the following more general result.
The details
are left to the reader. .Corollary 13.6.1. over
is an orthonormal set, then there exists a
F C V
If
E.
be an inner product space
Let
complete orthonormal set
E C V
such that
E D F.
An alternative description of complete orthonormal sets is provided by the next theorem. Theorem 13.6.2. i
be an inner product space over
Let
E CV is an orthonormal set.
and suppose
Then the following are
equivalent: (i)
is a complete orthonormal set.
E
x E V
If
(ii)
is such that
Suppose
Proof.
is such that
xiE
x = 0.
is a complete orthonormal set.
E
x # 0,
and
then evidently
orthonormal set that properly contains the fact that
then
x t E,
E,
E U (x/IjxII)
x E V
-is an
thereby contradicting
is a complete orthonormal set.
E
If
Thus
x.LE
implies
x = 0. Conversely, suppose
x i E
implies
x = 0.
If
E
were not a
complete orthonormal set, then there would exist some orthonormal set
F C V
such that
then (lxii = 1 implies
and
F
x i E,
properly contained
E.
Hence, if
x E F - E,
contrary to the assumption that x i E
x = 0.
Therefore is finished.
E
is a complete orthonormal set, and the proof
0
399
13.6. Complete Orthonormal Sets
We shall ultimately see that a complete orthonormal set in a Before coming
Hilbext space is an orthonormal basis for the space.
to grips with this, though, we need to review some facts concerning the notion of noncountable sums in a normed linear space. is a normed linear space over
Suppose E A) c V.
[X
We denote by
and
I
the collection of all finite
R
of
subsets
n
nI > n2
if
R
in
of the index set
A,
nI D n2, ni,n2 E R.
is a directed system, and
and partially order
by setting
ti
Clearly, with this partial ordering
is a net of elerents
(E E nxa)n E R a
With these notational conventions agreed on, we make the
V.
following definition:
Definition 13.6.2. t
and suppose
to
(xa I a E A) e V.
if the net
x E V
[E of
given
then
e > 0,
be a normed linear space over
Let
E nxa)n E R
there exists some
EAxa
Then
is said to converge
converges to
x; that is,
such that, if
no E R
n > nO
l x- Ea E nx l< e. Standard arguiberts reveal that, if
EAxa = x
and
E
y
A a = y,
then (i)
(ii)
EA(xa + ya) = x + Y.
EA(axa) = ax
(a E I).
Although our main concern will be with sums, there is one general fact about nets in Banach spaces that we shall have need of: the sequential completeness of Banach spaces implies net completeness. More precisely, we have the following proposition: Proposition 13.6.1. If
that
(x(Y)a E A
lim x
as
Proof. if
a,0 > n,
Let
be a Banach space over
is a Cauchy net, then there exists some
x E V
4.
such
= x.
For each positive integer then
l'xa - x01 < 1/n.
n
let n E A
be such that,
This is possible because
13. Hilbert Spaces
400
Then set
is a Cauchy net.
(xa}a E A
n
n
1
a > a , a > a
such that
now show that
such that
x E V
n-1 , an > n, n = a
of
E A
2,3,4,... V,
.
and so
Routine arguments
limnxoP = x.
0
= x.
x
lira
and choose
al = al
is a Cauchy sequence in
(xOn }
It is then apparent that there exists some
n
2
of
An important fact about the sum
is the following pro-
EAxa
position:
Proposition 13.6.2. over
§
and suppose
such that of
Let
(x
I
xa = 0
then
EAxa = x,
(V,ll'II) a E A} C V.
be a normed linear space If there exists some
x E V
for all but a countable number
a.
Proof.
n
For each positive integer
'if n > nn, then
a countable subset of
6 E A
If
A.
and
whence we see that for each
n = 1,2,3,...,
11x
nn E U
Clearly
- EaEnxall < 1/2n.
IIx
let
S f n
nm
m
,
then
be such that,
Un
= Inn
is
6 $ nn'
n
11
11
II
a E nn U (O}xa
aE
E aEnUWxa n
x11
I
1
2n
2n
- IIx -
E x0jI
or Enn
n,
as
n U (o) > n Therefore
n
and
x0 = 0
n
n
> n
if
n
.
0 f n.-
0
If the normed linear space in question is just the scalar field t,
then we have the following corollary.
The first portion of the
result is proved by noting that a series of numbers converges absolutely if and only if every rearrangement of the series converges.
401
13.6. Complete Orthonormal Sets
Corollary 13.6.2.
Ej+ak
absolutely to a if and only if Q+ _ (1,2,3,...).
a E A,
(ak),
then
a E A) C §, a
17k
= laak
where
a E
and
> 0,
a -
converges absolutely to a.
we shall often write
V = §
In the case that
a
converges a,
and if this countable number of indices is enumerated as
the sequence
(a
ak
=
for all but a countable number
as = 0
Then
2
converges to
and suppose there exists some
I.& E A) C i
a
EAaa = a.
such that of
[a
Let
(ii)
Then
be a sequence.
(ak) C §
Let
(i)
E a
A a
EAaa < co
converges to some
when
a E C.
We can now make a meaningful definition of orthonormal basis. Definition 13.6.3. over
§.
(i)
,(ii)
that
(ea
A set of vectors
orthonormal basis for (e
a
I a E A)
V
f
or E A) C V
is said td be an
if
is an orthornormal set.
there exist some
x E V
For each
be an inner product space
Let
aa(x) E t, a E A, such
x = EAaa(x)ea.
We need make no explicit mention of the linear independence in the definition, since an orthornormal set of vectors is always linearIndeed, every orthogonal set of vectors is linearly
ly independent.
independent, as shown by the following:
Suppose
(x
an orthogonal set and, for example, that Ek= lakx
a
a E A)
= 0,
is
where
(Yk
ak E §.
Then we see that n
0 = (kZ lakxak,xaj) = aj(xaj,xaj) whence
a. = 0,
j
= 1,2,...,n.
Thus
(xa I
(j = 1,2,...,n),
a E A)
is linearly
independent.
Furthermore, if
a ( A) is an orthonormal basis for V, (ea then it is easy to compute the coefficients (aa(x)) for each x E V: if
x = EAaa(x)ea,
I
then it follows at once from the orthonormality
13. Hilbert Spaces
402
of
(e
and the continuity of the inner product that
a E A]
(E aa(x)ea,e A = E aa(x)(ea,ea) A
(0 E A) .
= 8A (x) Hence, if
(e
space
then
V,
is an orthonormal basis for an inner product
a E A)
(
x=E(x,e)e A a
(xEV).
Cr
The next three theorems are concerned with the fundamental properties of orthonormal sets, and the last of the three contains detailed descriptions of orthonormal bases for Hilbert spaces. Theorem 13.6.3. and let (i)
(e
be an inner product space over
Let
be an orthonormal set.
( a E A)
Given
x E V,
if
Then
n E II, then
inf 1kx -
aaEf
E asea+j
aEn
sEn occurs when
aa = (x E
E V,
then
(x,e a) f 0
for at most a coVntable
nwnber of a E A.
(iv)
E
Proof.
(x,y E V).
< jIx1j((y(j Let
Then for any
x E V.
E 4
a
we have
of
11x-
E aea12=(x- E ae,x - E ae)
aEn a
aEn a a
0 En
naa(eax) -
= (x,x) -
aE
E
9En
(x,ee) +
E aaa0(ea,e
a,OEn
403
13.6. Complete Orthonormal Sets
II42
'
-
IIxIl2 +
E a x,e a aEn
-
E a (x,e ) + E la
E Iaa - (x,ed,2 -
aEn
aEn a
a
aETi a
Of
F.
2 1
I(x,edl2.
aETT
It is now apparent that
Ea aEn a
inf llx -
aaEf
aEn occurs when
as = (x,ea),
thereby proving part (i),
E
I(x,ea)I2
< 11x112.
aEn
Since this last estimate remains valid for each EAI(x,ea)I2 < m
argument shows that
and that
n E R,
an easy
and
E I (x,ea) 12 < jjxjl2. Thus part (ii) holds, and part (iii) follows immediately from Proposition 13.6.2 or Corollary 13.6.2(ii). x,y E V,
Furthermore, if and
(y,e
)
a
the sequence
= 0
then, by part (iii),
(x,e ) = 0
for all but a countable number of a E A,
a
say for
Then from Corollary 13.6.2(ii), the Cauchy-
(ak).
Schwarz Inequality for series and the second part of the present theorem we see that
E I(x,e ) y,e a
I
=
E I(x,e
k=1
)
E I(x,e
k=1 _
yy,
I
cyk
°7c
)I21 /21
E I(y.e )121/2 °'k
k=1
)121/2 [E I(x e `121/2(E llA I(y,e a A J
13. Hilbert Spaces
404
no
e > 0
n' > no,
and
x E V then
such that
as = (x,e(Y ), a E A.
EA`aaI2
< m. Then, since only countably ((Proposition
are onzero
verified that given if
then the following are equivalent:
or E A) C t,
n A n' = (n - n') U (n' - n)
13.6.2), it is easily there exists some
then E E
n A
n'Eaa1
Consequently, if
n0 E R 2
< c,
n > Ti
0
such that,
where and
we have
11
E a e -. E a a aEn'
aEn o' °
II
E a aEnAn' a
2
n' > n ,
0
405
13.6. Complete Orthonormal Sets'
2
E
aEnAT'Iaal < c, since
is an orthonormal set.
j a ( A)
(e of
is a Cauchy net in there exists some
Thus (a E
naaea)n E R
and so from Proposition 13.6.1 we see that
V,
x E V
such that
Hence part (i)
x = EAaaea.
implies part (ii).
Conversely, suppose
x = EAaaea.
Then the continuity of the
inner product (Corollary 13.1.2) and Bessel's Inequality (Theorem 13.6.3(ii)) combine to show that
XAIa01 I2
< w.
Therefore part (ii) implies part (i), and the proof is complete.
Note that Theorem 13.6.4 does not assert that every can be expressed as (ea `
x = EA(x,ea)ea
a E A); that is, it does not assert that
orthonormal basis for given
x E V,
V.
x E V
for a given orthonormal set [e
a
a E A)
I
is an
All we can deduce at this point is that,
there exists some
y E V
for which
y = EA(x,ea)ea
From the continuity of the inner product we know, however, that (x - y,ea) = 0, a E A,
and so
x = y
provided
(e
(
a E A)
is a
of
complete orthonormal set (Theorem 13.6.2).
This observation proves
the first implication of the next theoreih.
Theorem 13.6.5.
and suppose
(ea
I
be a Hilbert space over
Let
a E A)
is an orthonormal set.
§
Then the follow-
ing are equivalent: (i)
[e
a E A)
is a complete orthonormal set.
or E A)
is an orthonormal basis for
01
(ii)
(iii)
(ea
The linear subspace of
norm dense it
.V
spanned by
(e
V.
a
V.
(iv)
If
x E V,
(v)
If
x,y E V,
then then
EAl (x,ea)l2 = jIxl`2
EA(x,ea) yy,e ) = (x,y).
a E A)
is
13. Hilbert Spaces
406
Proof.
The implication from part (i) to part (ii) was estab-
lished in the paragraph preceding the theorem, and that f;.om part (ii) to part (iii) is obvious. (e
a k A)
j
Suppose that part (iii) holds and that
is not a complete orthonormal set.
a exist, by Theorem 13.6.2, some' x
E V,
lix
0
0
Ij
Ea
which j(xo -
IIX0
such that and
i ea,
x
0
it E 11
for
But Theorem 13.6.3(i) then shows that
1/2.
En
= 1,
as E §
However, by part (iii) there exist
E A.
Then there would
-
-
E aaea aEn E (x aEn
e )e all o
aa
= 1,
which is clearly absurd.
Hence part (iii) implies part (i), and
parts (i) through (iii) are equivalent. It is evident that the implication from part (iii) to part (i) can also be established by an appeal to the Riesz Representation Theorem (Theorem 13.4.2) and Corollary 4.2.8 to the Hahn-Banach Theorem.
a E A) is an orthonormal basis for a Then we know from the comments following Definition 13.6.3 that
Now suppose that
x = EA(x,edea
[e
for each
I
x E V.
Appealing to the continuity of the
inner product (Corollary 13.1.2), we conclude that IIx1I2
= (x,x) _ (n (x,ea)ea,E (x,e0)e1)
= E (x,ea) x,e A = E I(x,ea)IA and
o part (ii) implies part (iv
V.
(ea,e
407
13.6. Complete Orthonormal Sets
A similar argument shows that part (ii) implies part (v), x = y
obviously part (v) implies part (iv) on setting
xo E V
Finally, suppose part (iv) holds and let
xo 1 ea, a E A.
x
0
be such that
Then llxoll2
and so
and
in (v).
= 0.
= 0,
= E I(xo,ea)1 A
Thus part (iv) implies part (i).
Therefore all five parts of the theorem are equivalent.
In particular, we see from the preceding theorem that every
Hilbert space has an orthonormal basis and that any complete orthonormal set can serve as such a basis. Parts (iv) and (v) of the theorem are generally referred to as Parseval's Identity.
Let us now look more closely at the remark preceding the Suppose that
kiesz-Fischer Theorem (Theorem 13.6.4). is a Hilbert space over
§
complete orthonormal set. over
where
t,
sure on
E
E E S, E
then
A
a E A) C V
µ(E)
a E A
that is, is
is plus infinity cp
:
y (x)
(x,ea).
V
L2(A,S,p)
is that element That
y (x) E L2(A,S,µ),
is an immediate consequence of Bessel's Inequality.
Clearly p
is a linear mapping, and the two forms of Parseval's
Identity show that
p
products; that is,
(x,y) = (p(x),cp(y)), x,y E V.
Theorem shows that theorem:
L2(A,S,µ)
is equal to the number
Then define the mapping
whose value at
is a
is counting mea-
y
and
µ(E)
is finite, and
y (x) = ((x,ea)), x E V,
L2(A,S,p)
x E V,
when
is infinite.
by setting of
E
tea I
Consider the Hilbert space
is all subsets of
A; that is, if
of points in when
S
and that
cp
is an isometry and that
is surjective.
cp
preserves inner The Riesz-Fischer
Hence we have the following
408
13. Hilbert Spaces
Theorem 13.6.6.
be a Hilbert space over
Let
(e a I a E A) e V
and suppose
A
and
p
Then
is a complete orthonormal set.
is isometrically isomorphic to all subsets of
$
L2(A,S,p),
is counting measure on
where
S
is
A.
Thus we see that Hilbert spaces are essentially just
L2-spaces.
A number of other rather easily derived results about Hilbert spaces are obtainable from the foregoing development. however, merely to mention one of these.
We wish,
The details are left to
the reader.
First, in analogy with the case of finite-dimensional spaces, we make the following definition: Definition 13.6.4. If
(e.g r
sion of of
a E A) c :V V,
Let
(V,(.,.))
be a Hilbert space over
t.
is a complete orthonormal set, then the dimen-
denoted by
dim(V),
is defined to be the cardinality
A.
On the surface this definition would seem to depend on the choice of the complete orthonormal set.
because we can show that, if
However, this is not the case,
(e
I
a E A)
and
(f,
0 E A')
are
of
complete orthonormal sets in A'
are equal.
V,
then the cardinalities of
A
and
Thus the notion of dimension in a Hilbert space is
well-defined.
One useful result involving this concept is the following theorem; Theorem 13.6.7.
Let
be a Hilbert space over
'.
Then the following are equivalent: (i)
(ii)
dim(V) = o. (V,1I.11)
is separable.
Consequently a Hilbert space is. separable if and only if it has
a countable orthonormal basis.
Fourier Analysis in
13.7.
409
L2([-n;n],dt/2n)
13.7. Fourier Analysis in
We now wish to see
L2([-n,n],dt/2n).
what the development of the preceding section can tell us about the Hilbert space
Recall that the inner product in
L2([-n,rr],dt/2n).
this space is defined by (f,g) = Tn fn_, f(t)g(t) dt
(f,g E L2([-TT,n],dt/2n)
The first thing we do is to examine a particular orthonormal basis
for L2 ([-n,rr],dt/2n) . Theorem 13.7.1.
contained in Proof.
The family of functions
(elk,
I
k E Z]
is a complete orthonormal set.
L2([-n,n],dt/2n)
The fact that the complex exponentials form an ortho-
normal set in
L2([-n,nj,dt/2n)
is a routine exercise in calculus.
To prove the completeness of this orthonormal set is, however, less trivial.
Recalling Theorem 13.6.2, we see that it suffices to show
that, if
f E L2((-n,n],dt/2n)
f(t) = 0
almost everywhere.
Suppose
(f,elk.) = 0, k E Z,
ftn f(s) ds
2n
L2([-n,n],dt/2n) c L1([-n,n],dt/2n),
(t E
F E L2([-rr,n],dt/2n).
Define
G
of
G E L2([-n,n],dt/2n), G. exists and equals
indefinite integral of
f
f
G
by
(G',etk )
(t E [-n,n]).
is continuous, and the derivative
almost everywhere, because [Ry, pp. 10,5-107].
integration by parts, we see that 0 - (f,eik.)
F
[-n,n](Ry, p. 106]
G(t) = F(t) - ?n Inn F(s) ds Then
-n,n]).
it is evident that,
is a well-defined absolutely continuous function on In particular,
then
is such a function and define
f
F(t) =
Since
and
F
is the
Consequently, utilizing
13. Hilbert Spaces
410
1
=
2n
fnn G(t)e -ikt
ik G(t)e -ikt nn + Zn
dt
cos(kn) [G(n) - G(-n)] + ik(G,elk.)
=
(k E Z).
whence we con-
G(n) - G(-n) = 0,
On setting
k = 0, .we see that
clude that
(G,elk.) = 0, k E Z, k # 0.
k = 0, we
However, when
have (G,1) = ?n f'-;n G(t) dt
= 2n fnn F(t) dt - 2n fnn F(t) dt
= 0.
(G,elk.> = 0, k E Z.
Thus
{eik,
Now, considering functions defined on
I
F = (z
the linear subspace of routinely verified that
C(F)
I
as a family of continuous
k E Zj
we denote by
z E (, IzI - 1), (elk*
spanned by
I
W, is a subalgebra of
C(1)
W
It can be
k E Z).
that s%parates
points, is closed under complex conjugation, and is such that Z(W) = (z
I
Hence by the Stone-Weierstrass
z E T, h(z) = 0, h E W) = Q.
W
Theorem (Theorem 11.4.1) we ccnclude that
is dense in
It is apparent that this is equivalent to asserting that the linear subspace of
C([-n,n])
(elk,
spanned by
the space of continuous functions
I
h
on
is norm dense in
k E Z)
[-n,n]
such that
h(n) _
h(-n).
Suppose that and
G(n) = G(-n),
and let
IIGII2 f 0
c > 0.
Since
the preceding paragraph shows that there exists
some finite linear combination of the elements of tall it
h,
G E C([-n,n])
such that
JIG
-
hII m
< c.
2n jr n G(t h(t) dt =
Evidently
G,h
= 0.
(elk. I
k E Z),
(G,elk.) = 0, k E 71,
as
elk
the
411
L2([-n,n],dt/2n)
13.7. Fourier Analysis in
and
is a finite linear combination of
h
Hence
.
(IIGII2
=
Zn fnn G(t)G(t) dt
2n f'nn G(t)[G(t) - h(t)] dt 2n fnn IG(t IIG(t) - h(t)I dt
dt
< 2n fTn IG(t)I < C, JIG 112'
the last step being valid because of the Cauchy-Schwarz Inequality (Theorem 13.1.1).
Consequently
trary, we must conclude that
Thus
IIGII2 < c,
IICj2 = 0,
and since
is arbi-
a
contrary to assumption.
and so
IIGII2 = 0,
0(t) = F(t) - 2n fnn F(s) ds
(t E [-n,n]),
= 0 G
as
have
is continuous.
But, since
F'(t) - f(t) - 0
is an indefinite integral, we
for almost all t.
(elk*
Therefore
F
I
k E zi)
is a complete orthonormal set.
Theorem 13.7.1 has as a corollary the fact that the Fourier transform on
L1([-n,n],dt/2n)
is injective, a result remarked on
in Section 7.8. w
Corollary 13.7.1.
Let
f
f = 0.
E
Proof.
It is evident from the proof of Theorem 13.7.1 that
G
13. Hilbert Spaces
412
G(t) = F(t) - 2n fnn F(s) ds
defines an element
(t E [-n,n])
G E C([-n,n]) C L2([-n,n],dt/2n),
where, as
before, (t E [-n,n]).
F(t) = ftn f(s) ds G' = F' = f
Since
deduce once again that 13.7.1,
(G,eik-) = 0, k E Z,
is identically zero.
G
f(k) = 0, k E Z,
almost everywhere and
Hence
we
and so, by Theorem
f = 0.
0
The proof of the next theorem is obtained by quoting the approWe establish the notational
priate result of the preceding section. convention that f(k) = (f,eik.)
(k E Z; f E L2([-n,n],dt/2n))
This is consistent with previous uses of Sections 3.1, 6.6, 7.8, and 9.7.
f
in, for instance,
It also provides a rationale for
the Fourier coefficient terminology mentioned in the preceding section.
Theorem 13.7.2. over
Consider the Hilbert space
L2([-n,u],dt/2n)
E.
(i)
If
f E L2([-n,n],dt/2n),
If
f,g E L2([-n,n],dt/2n),
IIf!I2 = [ (ii)
then
f = E$f(k)elk-
f (k) 12J1/2
1
Zn
n
f-n
f(t)g(t) dt =
then
E f(k)g(k) k = -cc
and 1
2n
n
f -n
OD
f(t)g(t) dt =
E
k = -w
A
f(k)g(-k).
and
(ak
If
(iii)
k E F) C C
I
then there exists a unique
f = E akeik
and
is such that
Fmk= -.Jak12 < CO,
f E L2([-n,n],dt/2n)
such that
ak = f(k), k E Z.
The mapping p
(iv)
413
L2([-n,n],dt/2n)
13.7. Fourier Analysis in
f E L2([-n,n],dt/2n)
:
defined for each
L2([-n,n],dt/2n) -. L2(Z)
by p(f)(k) = f(k), k E Z,
is a surjective
isometric isomorphism.
The results of this theorem have valid analogs in the more general situation in which
space
where p
L2(G,µ),
G,
G,
and one considers the Hilbert
is Haar measure on
complete orthonormal set in characters of
is replaced by an arbitrary
[-n,n]
compact Abelian topological group
L2(G,p)
G.
In this. case a
is provided by the continuous
that is, by all continuous homomorphisms of
into the circle group
F = (z
the continuous characters on
(
G
z E C, JzJ = 1].
G
The proof that
form a complete orthonormal set
hinges on the fact that these functions separate the points of
G.
This is a highly nontrivial fact (see, for instance, [HR1, pp. 335-355; R, pp. 23-26]).
Furthermore, the second portion of implication (i),
as well as implications (ii) and (iv) of Theorem 13.7.2 can be generalized to the context of locally compact Abelian topological groups.
Part (iv) of Theorem 13.7.2 is generally referred to as
Plancherel's Theorem.
The interested reader may consult, for instance,
[HR2' pp. 225 and 226; HSt, pp. 411-413; Ka, pp. 139-142; Lo, pp. 141-146; R, pp. 128-130; Rul, pp. 26 and 271. In particular, Theorem 13.7.2(i) says that the Fourier series FZf(k)elk. of
of
f E L2([-n,n],dt/2n)
L2([-n,n],dt/2n).
converges to
f
in the norm
It was conjectured by Lusin [Lu] in 1915 that
the Fourier series of such a function actually converges almost everywhere.
This conjecture was given an affirmative answer by
Carleson'[Ca] in 1966.
Subsequently the result was extended to
Lp([-n,n],dt/2n), 1 < p < w,
by Hunt [Ht].
A discussion of these
results is considerably beyond the scope of this book, and the reader is referred either to the original papers or to (M] for an exposition.
However, we would like to prove a result of Kolmogorov
13. Hilbert Spaces
414
[Ko], which asserts that certain subsequences of the symmetric partial sums of the Fourier series of an L2-function converge almost everywhere.
We need one
act that we do not prove: if
f E L2([-n,n],dt/2n),
then, as indicated in Section 9.1, the nth Cesaro mean of the Fourier series of
f
can be written as n
E
an(f)(t) =
(I
-
n +yI)f(k)eikt
(t E [-n,nl)
k = -n It is the case that the sequence where to
f.
(an(f))
converges almost every-
This follows at once from Carleson's result, as con-
vergence almost everywhere implies convergence of the Cesaro means. However, a simpler and more direct proof is also available [El, p. 96). One definition is necessary before we can state Kolmogorov's theorem.
Definition 13.7.1.
(nk)
A sequence
to be a Hadamard sequence if
infk[nk
+
l/nk] > 1.
Theorem 13.7.3 (Kolmogorov's Theorem). space
L.,([-n,n],dt/2n)
then for each
over
E.
f E L2([-n,n],dt/2n)
verges almost everywhere to
f
(nk)
If
of positive integers is said
Consider the Hilbert is a Hadamard sequence,
the sequence
(snk (f))
con-
where nk
sn (f)(t) = E f()eimt k m = -nk Proof.
(t E [-n,n]).
We first observe the following immediate consequence
of the Monotone Convergence Theorem [Ry, pp. 84 and 227]: Suppose (gk)
is a sequence of nonnegative integrable functions on
such that co
pn
k= 1J -n
gk(t)
dt
0
(x2 E cl(W1)),
E Tk(x2)11 < e k=0
being arbitrary, that
lnm IIn; 1 Therefore
n
n k EOTk(x2)I1 =
([Ek _ 0Tk(x)]/n + 1)
(x E cl(W1))
converges to
P(x), x E V,
and the proof is complete.
It should be noted that the sequence
[(E
.
0Tk)/n + 1)
not, in general, converge in the norm topology to
P.
does
439
13.11. Mean Ergodic Theorem
If we combine the Mean Ergodic Theorem with our discussion of the unitary transformation p
preserving mapping
on
induced by a measure
L2(X,S,µ)
on
T
we easily deduce the validity of the
X,
next theorem. Let
Theorem 13.11.2. let
be a measure preserving mapping on
cp
f E L2(X,S,µ),
then for each
fo E L2(X,S,µ)
such that
(i)
1 lim 11n. I
n
(ii)
If
X.
there exists some
f E L2(X,S,µ)
n
E. Tk(f) - f0II2 = 0. k=0
f0
is the image of
under the orthogonal projection
f
onto W= [g] g E L2(X,S,N), T(g) = g o 9 = g).
L2(X,S,µ)
It is not difficult to extend this result to vided we assume that
If
Let
and let
µ(X) = 1
cp
f0 E L1(X,S,µ)
Jim
(i) (ii)
Let
f E L1(X,S,p)
we see that
is norm dense in
exists some
we have
f E LI(X,S,µ)
there
such that
T(f0) = f0.
L2(X,S,µ)
1Jh11l
then for each
X.
k (f )- f o l 1 l= 0.
µ(X) = 1,
that
be a measure preserving mapping on
n `l n+ 1 k= O T
Proof.
p(X) - 1,
pro-
be a positive measure space such
(X,S,p)
T(f) = f a cp, f e L1(X,S,p),
exists some
LI(X,S,µ)
has finite total mass.
p
Theorem 13.11.3. that
T(f) = f o cp,
T(fo) = fo.
Clearly of
be a positive measure space and
(X,S,p)
and
e > 0
be given.
L2(X,S,µ) C LI(X,S,µ),
g E L2(X,S,N)
and moreover
LI(X,S,µ) [Ry, P.244]. such that
Now, since
Thus there
hf - gill < e/3.
Since
we see from the Cauchy-Schwarz Inequality (Theorem 13.1.1) 11h112, h E L2(X,S,o),
and hence for each
n,m = 0,1,2,...
13. Hilbert Spaces
440
m
1 E Tk(f) - m+1
+
II
T (f)II1
k=0
n
k=0
n
n
E Tk(f) -
< IIn + 1 i
E Tk (g) ] II 1
k=0
k=0
n
m
E Tk (g) -
+ IIn + 1
k=0
E Tk(g)
-
k=0 1
n+
1
M+1
E Tk (g) I(1 k=0
E Tk(f)]II1
k=0
E IITk(f - g) II1
k=0
n +
II
+
m+1
n
+
1
m
E Tk(g) - m+ 1 E Tk(g)II2 k=0 k=0 E IITk (g -
k=0
f) II 1
i Tk(g) 3 +IIn+ 1 k=0 T
since
is an isometry on
- m+ 1
by Proposition 13.11.1.
L1(X,S,µ)
From the Mean Ergodic Theorem we see that a Cauchy sequence in ([
=
OTk(f)]/n + 1)
LZ(X,S,µ),
some
f E L1(X,S,µ) 0
([q =
0Tk(g)]/n + 1)
L1(X,S,µ)
L1(X,S,µ)
is a Banach space, there exists
such that
n
1nm IIn +
1 k E 0Tk (f)
foli l
= 0.
Moreover,
IIT(fo)
- folll = limn n 4
k0
IIT[ E Tk(f)] -
= lira n+ 1 IITn+ 1(f) n = 0,
is
from which it follows at once that
is a Cauchy sequence in
Consequently, since
E Tk(g)IIZ, k=0
fill
=
k=0
Tk(f)Q1
13.12. A Theorem About
as
T
441
lit
is an isometry.
Therefore
T(fo) = fo,
Obviously, if
and the proof is complete.
then from Theorem 13.11.3 we can
f E LI(X,S,µ),
deduce the existence of a subsequence of converges almost everywhere to
G
fo.
{[`k= 0Tk(f)]/n + 1}
that
Actually it can be shown that
the sequence itself converges almost everywhere to
f0.
This is
an immediate consequence of the following general theorem due to C. D.Birkhoff:
Theorem 13.11.4 (Individual Ergodic Theorem). a a-finite positive measure space and let mapping on
(ii)
lim n n
I
fo E L1(X,S,µ)
Tk(f)(t) = fo(t) 1
be
(X,S,µ)
be a measure preserving
T(f) = f o p, f E LI(X,S,µ),
there exists some
f E LI(X,S,p) (i)
If
X.
y
Let
theh for each such that
for µ-almost all
t E X.
k=0
T(fo) = f0.
We do not give the proof of this theorem, as it would involve too long a digression from our main interests.
For proofs of the
theorem and other material on ergodic theory we refer the reader to [F, Gr, Hal].
It is also of interest to compare the development of
this section with the.discussion of fixed point theorems in Chapter 12.
13.12.
A Theorem About H2.
In this section we wish to estab-
lish an analog of a classical theorem about analytic functions for the elements of a certain subspace of the Hilbert space
L2([-n,n],dt/2n).
The main Hilbert space tool we shall employ will be the Riesz Representation Theorem (Theorem 13.4.2).
Before stating the theorem we
discuss some preliminary material that we hope will illustrate the connection between a certain subspace `the theory of analytic functions. mentioned in this discussion.
H2
of
L2([-n,n],dt/2n)
and
We do not prove any of the results
13. Hilbert Spaces
442
The Hardy space
Definition 13.12.1.
H2 = (f
I
f E L2([-n,nJ,dt/2n), f(k) _ (f,eik.) = 0, k E Z, k < 0).
It is easily verified that the Hilbert space space.
is defined by
H2
is a closed linear subspace of
H2
and so is itself a Hilbert
L2([-n,n],dt/2n)
There are other descriptions of
tionship between
that exhibit the rela-
H2
on the open
and certain analytic functions
H2
unit disk in the complex plane.
We outline these here in order to
further motivate the theorem to be established later in the section. No proofs are given.
The interested reader is referred instead to
[Dn, pp. 1-38; Ho, pp. 27-39; R, pp. 84-89; Z, pp. 271-285].
Consider the linear space in
H2
of functions
g
that are analytic
(rest It E [-n,n], 0 < r < 1)
the open unit disk
there exists a constant
depending on
Mg,
g(relt)12
Y J_n
I
and such that
for which
g,
dt < Mg
(0 0) .
0
But we also know that {h(t)+2e-ikt
(1h12)^(k) = 2n fn,
tt 1
n
f-n
dt
ekth(t)h(t)
dt
=
h(-k)
(kEZ,k>0)'.
=0 Thus
(1h+
)
13.12.2 that
(k) = 0, k E IL, k # 0,
is a constant.
Ih12
Ihj E L2([-n,nl,dt/2n)
implies
whence we conclude from Lemma The lemma is applicable because
IhJ2 E L1([-n,n],dt/2n)
by the
Cauchy-Schwarz Inequality (Theorem 13.1.1). However,
h E W
implies that
ishes almost everywhere on h
E,
h,
and hence also
Ih12,
a set of positive measure.
must vanish almost everywhere, and so therefore
van-
Hence
g(0) = (g,h) = 0,
g E W. Consequently
13.13.
W = (0),
and the proof is complete.
Some Basic Results in Spectral Theory.
As already indi-
cated (for example, in Section 3.1), the action of a linear transformation
T
from a finite-dimensional linear space V
to itself can
always be realized as matrix multiplication of some matrix and the coefficient vectors of the elements of of course, depends on the choice of a basis for
V. V.
A = (a .k)
The matrix For certain
A,
447
13.13. Some Basic Results in Spectral Theory
types of such transformations one can choose the basis of
V
T(x) =
is either
1FP
A
This means that, for exam-
only has nonzero entries on the diagonal. ple, in the case where
in
is a diagonal matrix -- that is,
A
such a way that the matrix
V
or e,, 'one can write
m E (x,ek)akek'
(x E V),
k=1 are certain linearly independent orthonormal
el,e2,...,em
where
vectors in
A1,A2,...,Am
V;
are the not necessarily distinct-nonzero
diagonal entries of the matrix product;
and
m < n.
denotes the usual inner
A;
This is the case, for instance, whenever
T
is self-adjoint -- that is, whenever the matrix corresponding to
T
is equal to its conjugate transpose.
Our ultimate goal in the remaining sections is to establish a counterpart of this result for certain self-adjoint linear transformations on an arbitrary Hilbert space over
C.
This'result will
Section 13:15 when we prove a spectral decomposition
appear in
theorem for compact self-adjoint linear transformations.
The result
for finite-dimensional linear spaces discussed at the beginning of this section will be a special case of this theorem.
However, before
we can establish the indicated result, we must develop some aspects of the general spectral theory of linear transformations on Hilbert spaces.
For reasons that will subsequently become apparent we re-
strict our attention to Hilbert spaces over Definition 13.13.1. and let
Then
E L(V).
if there exists some (AI
for
- T)(x) = 0; T
and
Thus we see that
be a Hilbert space over C
Let
A E C
x E V, x
is said to be an elgenvalue for 0,
x E V, x
if there exists some A E s
0,
A E C
such that
such that
T(x)
is an eigenvalue for
AI - T
for
then one can always find a vector
T(x) = Ax.
T(x)
is not injective.
Ax,
T
that is,
is said to be an eigenvectoT
only if T,
C.
Evidently, if
Ax.
T E L(V) A
if and
is an eigeivaiu,
x f V, tixI} a I. such that
13. Hilbert Spaces
'448
o(T), the spectrum of
(i)
such.that
A E C
is the set of
T,
is either not injective or not surjective.
Al - T
the point spectrum of
PC(T),
(ii)
the continuous spectrum of
Ca(T),
(iii)
A E C
is the set of
T,
is not injective.
Xi - T
such that
is the set of
T,
is injective, but not surjective, and such
XI - T
such that
A E C
c
Then
T E L(V).
and let
be a Hilbert space over
Let
Definition 13.13.2.
that R(AI - 1)- is dense in. V. the residual spectrum of
RC(T),
(iv)
is injective, but not surjective, and such that
such that
XI - T
R(AI - T)
is not dense in
Jt is evident that
C = o(T) U p(T)
o (T)
L
A E p(T),
If
T.
=
Clearly
pairwi'se disjoint.
values of
V
liflear mapping from
1'
Thus
XI - T
and that
(T) U Ca (T) U RC (T)
PC(T)
then
consists precisely of the eigenAl - T
is a bijective continuous
to itself, and so by Corollary 7.2.2 to the
Open Mapping Theorem we see that L(V).`
(Al - j-1
exists and belongs to
has a bounded inverse defined-on all of
we shall often write
used only when
T.
(Al - T)-I = R(A,T)
Note that thenotation (Al
-
V.
not generally. hold.
T((ak))
such that
When
A E p(T),
and call .R(A,T) for
R(A,T)
the T)-1
(XI -
is
T)-I E L(V). PC(T) = a(T)
It should be pointed out that the equality
defined by
A E C
is precisely the set of points
p(T)
resolvent of
A E C
Moreover, the various sets defined here are
T E L(V).
for-each
is the set of
T,
is bijective.
Al - T
such that
V.
the resolvent set of
p(T),
(v)
A E C
is the set of
T,
For example, the shift transformation (bk)
where
bl = 0
is clearly injective, but not surjective.
and
bk =. ak - 1,
Hence
does
T E Lfe2), k = 2,3,...,
0 E a(T) - PC(T).
Actually it is possible to construct examples of continuous linear
449
13.13. Some Basic Results in Spectral Theory
transformations Ra(T), and
such that each or any pair o` the sets
T
CC(T)
Pa(T),
is empty, with the one exception that not all can
he simultaneously empty -- that is, it cannot be the case that a(T) =
We shall prove this last assertion shortly.
P._
Since ulti-
mately we shall be mainly concerned with compact self-adjoint linear a(T) -
transformations, where
Pa(T)
is either
V
not pursue further an investigation of the sets Our first concern is to examine
p(T)
or
Ca(T)
and
R(X,T)
we do
(0),
and
Ra(T).
a bit more
thoroughly.
Lemma 13.13.1. let
Let
ix - kal < l/!IR(ko,T)ii, (1)
I
-
(V,(.,.))
and suppose
T E L(V),
()L
be a Hilbert space over
ko E p(T).
If
is bijective.
(ii)
[I
-
(ko - )L)R(ko,T)]-1 E L(V).
(iii)
[I
-
()L o
Proof.
is such that
then
o - k)R(ko,T)
-
X E C
t,
k)R(ka,T)]-1
=_ 0[(ko -
k)R(Xo,T)]k.
The proof is a straightforward application of the remarks
following Theorem 6.3.3, combined with the observation that the series
Z;= 0[(ko - k)R(ko,T)jk
converges in
L(V)
since
Iko - kl < 1/IIR(k0 ,T)il.
The details are left to the reader.
O
This lemma allows us to establish the following important result concerning the existence of Theorem 13.13.1. let 1),
Let
T E L(V), and suppose
- X01 < 1/l!R(X0,T)II,
(i) (ii)
R(k,T): (V,(.,.))
X0 E p(T).
be a Hilbert space over If
A E C
is such that
then
k E p (T) R(k,T z R(ko,T)Fk = 0[(ko - A)R(Xo,T)]k.
Proof.
By Lemma 13.13.1 we see that
R(kaT) E [(ko - k)R(ko,T)]k = R(ko,T)(I
k=0
-
(ko - X)R(ko,T)
C,
13. Hilbert Spaces
450
Moreover, T)J-
(Al
- T)R(Ao)T)[I -
(ko - k)R()L
0
= [(A - Xo)I + Xol - TJR(X0,T)(I - (A0 _
[(A - ao)R(AO,T) + I][I -
(Ao
a)R(XOPT)]-I
X)R()L
OPT)]-I
= I.
Similarly, since
R(A0,T)[I - (ko -
X)R(A0,T)J-1
=
_
E ()o -
X)k[R(Ao,T)]k* 1
k=0 [I
(Ao - A)R(XOPT)]-IR(L0T),
-
wt see that
R(ao,T)[I - (ao - A)R(A. ,T)]-1(XI - T) = I. It then follows immediately that 00 1
R(.,T) = (XI - T) and
k=0
[(A0
0
T E L(V),
then
closed subset of
p(T)
ba a Hilbert space over
(V,(.,.))
Let
is an open subset of
Theorem 13.13.2. T E L(V), Proof.
Q:,
and
a(T)
C.
is a
C.
We are now in a position to prove that
If
X)R(X0IT)]I k
A E p(T).
Corollary 13:13.1. If
= R(A0,T) E
then
a(T)
is always nonempty.
be a Hilbert space over
Let
C.
o(T) j o.
We first make some preliminary observations based on
Theorem 13.13.1.
For each
x E V
complex-valued function fx,x*
on
and p(T)
x* E V* by
we define the
451
13.13. Some Basic Results in Spectral. Theory
fx,x(k) = x*(R(k,T)(x)) is analytic on
fx,xk
We claim that each such function
(k E P(T)).
Since
p(T).
is open, to prove this assertion it suffices to show that
p(T)
has a power series expansion about each point in X0 E.p(T),
p(T).
fx,x*
But, if
then,from Theorem 13.13.1 we see that
k) k fx,x*(k) = E x*((R(ko,T)) k+I (x))(ko -
k=0
for all
k E C
such that
is analytic on
IX -
0
k E C
Furthermore, we note that, if
fx,x*
is such that.
Ikl > IITII,
k E p(T), as, in this case, we see at once that the series
then
Ek
Thus each
1 < 1/IIR(ko,T)II.
p(T).
0T k/kk + l
converges in L(V),
=
(Al
- T) E
as
IITII/.I kI < 1.
lim()L I
- T) E k=0 k
n
k
k=0 k
n
k
n
1 im [ E- E k=0k
Inrrt rrI
k
-Tc3
n
n
Hence 1
kI
k=0 A
Tn + 11 - kn + 1
= I,
since
limn(IITII/IxI)n+ 1 = 0.
Similarly Tk
k=0 kk+1 and so
R(k,T) - z;= 0 Tk/kk + l . IIR(k,T)I1
from which it follows easily that x E V
and each
x* E V*.
Consequently IITII IxIk+1
k-0 =
T) = I,
(Al
k (1
-I IT[IT"
limk
mlfx,x*(k)I = 0
for each
13. Hilber Spaces
452
the function
x" E V*,
f
is analytic on
x,x*
ceding paragraph shows that each
fx
is bounded.
x*
Hence by
R(A,T) = 0, A E C,
Therefore
fx'x*
is
Thus by a consequence.of the Hahn-Banach
C.
Theorem (Corollary 4.2.6) we have and hence
that is,
p(T) = C,
Liouville's Theorem [Ru2, p. 213] we conclude that each identically zero on
and
Moreover, the conclusion of the pre-
is an entire function.
fx,x*
x E V
We see that for each
o(T) = 0.
Now suppose that
R(A,T)(x) = 0, A E C
x E V,
and
which is clearly absurd.
a(T) j t. Cl
We have restricted our attention in the present discussion to in order to ensure the
Hilbert spaces over the complex numbers C
If one considers Hilbert spaces
validity of the preceding result. over
then the theorem may very well be false -= provided, of
IR,
course, one recognizes that in any case a(T)
must be, a subset of
the scalar field of the Hilbert space in question. could not meaningfully discuss the transformation
Otherwise, one XI - T.
for instance, if one considers the linear transformation
Thus, on
T
IR2
determined by the matrix A
((0 -1 -l1
then it is readily verified that
0
Al - T
only if the determinant of the matrix is,
A
is a root of
Hilbert space over
13.14.
mations.
JR.
A2 + 1 = 0).
is not injective if and
Al - A
Since
IR2
it is apparent that
vanishes at
k
(thaw
is a finite-dimensional a(T)
Some Spectral Theory Results for Self-adjoint Transfor-
In this section we wish to discuss some special results in
the spectral theory of self-adjoint continuous linear transformations. The utility of these results will become apparent in the next section. We begin by defining the notion of the numerical range of an element of
1(V).
453
13.14. Spectral Theory of Self-adjoint Transformations
Definition 13.14.1. T E L(V).
and let
be a Hilbert space over
Let
w(T)
Then the numerical range
of
T
1
is defined
by
w(T) = ((T(x),x)
I
x E V, llxil = 1).
From Proposition 13.10.1 we see that, if 'T E L(V) since
w(T) C IR,
adjoint, then
is self-
An immediate
(T(x),x) E II;, x E V.
consequence of this observation and the next theorem is that whenever
T
is self-adjoint.
is self-adjoint, then
T E L(V) Proof.
Clearly
be a Hilbert space over
Let
13.14.1.
Theorem If
o(T) C IR
Suppose
d > 0.
cl[w(T)]
A
Moreover, if
x E V
o(T)
C.
cl[w(T)].
and let
d = info
and
= 1,
IIxii
E
cl[w(T)]IA - I
then by the
Cauchy-Schwarz Inequality (Theorem 13.1.1) and the definition of
w(T)
we see that d < IA - (T(x),x)I I(()LI - T)(x),x)I
= II (AI - T) (x) II IIxII
II(AI - T)(x)ll. Thus it follows easily that for any
x E V
dllxll < II (AI - T) (x)I! . Consequently by Proposition 3.2.3 we have
(Al - T)-I E L(R(AI - T),V).
An elementary argument, whose details we leave to the reader, shows that
R(AI - T)
R(AI - T) = V
is a closed linear subspace of by the following argument: If
V.
We claim that
(XI - T) # V,
then
by a consequence of the Hahn-Banach Theorem (Theorem 4.2.3) and the Riesz Representation Theorem (Theorem 13.4.2) there exists some y E V,
Ilyll = 1,
such that
((AI - T)(x),y) = 0, x E V.
Then, in
13. Hilbert Spaces
454
((XI - T)(y),y) = 0,
particular, we would have
k = (T(y),y) E w(T),
at once that k $ cl[w(T)].
Hence
(XI
X E p(T),
Therefore
Corollary 13.14.1. If
T E L(V)
from which we deduce
contrary to the assumption that
- T) = V,
whence
(XI - T)-1 = R(X,T) E L(V).
and
o(T) C cl[w(T)].
0
be a Hilbert space over
Let
is.self -adjoint, then
C.
c(T) C III.
An easy consequence of Corollary 13.14.1 is that the eigenvectors corresponding to distinct eigenvalues of a self-adjoint continuous linear transformation are orthogonal. Corollary 13.14.2. and let that
T E L(V)
k1 # k2
(V,(.,.))
be self-adjoint.
and if
T(xk) = kkxk, k = 1,2, Proof.
Let
x1,x2 E V then
If
be a Hilbert space over C k,1,k2 E Pa(T)
are such
are nonzero vectors such that
xl1 x2.
Clearly we have k1(xl,x2) _ (k1xl,x2) (T(xI),x2) (x1,T(x2)) (xl,k2x2)
= k2(xl,x2), o(T) c IR
since
by Corollary 13.14.1.
Thus
(k1 - 42)(xl,x2) = 0,
and so x1 1x2. If
T E L(V),
then it is apparent that for each
x E V, ixII
we have (T(x),x)l 0
is infinite.
In this
If this were not true, then there
such that
n,m = 1,2,3,..., n f m,
[k ) k
Jkkl > s, k = 1,2,3,...,
and
we would have
IIT(en) - T(em)112 = Ilanen - amemll2
> 0, as
(ek)
is an orthonormal set.
the compactness of
T,
convergent subsequence.
This, however, clearly contradicts
since the sequence Hence
limkXik = 0,
(T(ek))
could have no
which proves part (iv)
of the theorem. If
x E V,
then by Bessel's.Inequality (Theorem 13.6.3(ii).) and
the Riesz-Fischer Theorem (Theorem 13.6.4) it is apparent that the
466
13. Hilbert Spaces
infinite sum
l
=1("'k
converges to some element of
)ek
we can define
xm = x - E;. (,,,e
sequently for each
x E V
It is obvious that
x..i ek, k = 1,2,3,...,
k = 1,2,3,... to
Vk
T(x.) = Tk(xm),
Thus
.
of T.. But then, since
where
IITkil = IAk +
(Ak),
the construction of the sequence
IIT (xm) II
and so
Con-
V.
k )e k'
x00 E Vk,
is the restriction
Tk
by
11, k - 1,2,3,...,
we see that
= IITk (x.) II IITkIiIIxJI
(k = 1,2,3,...).
_ .IAk+IIIIxjI Hence
as
T(x ) = 0,
Thus we conclude that
link-"k = 0.
(x E V) ,
E (x,ek)Akek
T(x) =
k=1 as
T E L(V).
is not finite-dimensional, we know from
V
Moreover, since
Proposition 13.15.1 that
injective, whence we deduce from
then
T(x,,) = 0, x E V,
that
is a complete orthonormal set in It remains only to show that, if
then
o(T) _ (AkIU(0).
it is apparent that We claim that
X0
V V
when
PC(T).
Ak' k - 1 , 2 , 3 , .
.
. ,
is
Were
0 f PC(T).
(ak)U (0) c a(T).
limkXk = 0
such that
0,
T
is not finite dimensional,
when
d = inf(IA0 - AkI, IAol
X0
x0 E V, xo
Clearly Since
X0 E C - ((Ak).U(0)).
some
If
Hence it follows from Theorem 13.6.5 that
x = 1 = 1(x,ek)ek, x E V. (ek)
0 t PC(T),
0 E e(T).
I
X0 E PC(T), T(xo) = Aoxo.
we know from
(A k)
So suppose is infinite,
k = 1,2,3,...] > 0.
then there would exist However, since
Corollary 13.14.2 that
x0 1 ek,
Consequently
k =
Aoxo = T(xo) =
E (x0 ek)kkek - 0,
k=1 which is a contradiction because is,
A0I - T
is injective.
Ao # 0.
Thus
).o
PC(T),
that
467
13.15. A Spectral Decomposition Theorem
some
y E V,
this we shall construct, given (k0 I
is even surje-tive.
X0I - T
We claim, moreover, that
To see
such that
x EiV
We note that, if such an equation were valid,
- T)(x) = y.
then we would have (y,ek) _ ((MOI - T)(x),ek) X0(x,ek) - (T(x).ek) = ao(x,ek) - (x,T(ek)) (k = 1,2,3,...),
= (AO - kk)(x,ek) and so
because
(x,ek) T
k = 1,2,3,...,
Xk),
(y,ek)/()Lo -
X0 # kk, k = 1,2,3,...
is self-adjoint and
m
.
With this
and consider the sum
y E V
observation in mind, Jet
which is valid
(y,ek)
E
2
k=1 ko - kk Since
it follows at once from the estimate
(x0 - kkI > d > 0,
(Y)2
I(Y,ej),2
0, x E V.
if
is posi-
T
Prove each of the
following: (a)
is positive and invertible, that is,
T
if and only if
T-1 E L(V), (c)
If
formation and *64.
T E L(V).
Let
mT > 0,
mT_ 1 =
then (MLI.)
(V,(.,.))
Prove that
-1
where
E w(T)).
mT = inf(§ (b)
mT > 0,
is positive if and only if
T
T
exists and
mT > 0. T-1
and
is a posi`Cive self-adjoint trans-
MIT 1 = (ml)'.
be a'Hilbert space over C -and let
w(T),
the numerical range of
T,
is convex.
This result is called the Toeplitz-Hausdorff Theorem. *65.
T E L(V). 66.
T E L(V)
be a Hilbert space over C
Let
Prove that Let
a(T)
(V,(.,.))
be self-adjoint.
and let
cl(w(T)].
be a Hilbert space over C Prove that
and let
cljw(T)) = co[o(T));
that
is, the closure of the numerical range is the smallest closed convex set that contains the spectrum of.
T.
This result is also true for
normal transformations (see Problem 37).
13. Hilbert Spaces
482
If
is compact and self-adjoint, prove that
T E L(V)
be a Hilbert space over
Let
*68.
T E L(V).
Prove that
verges in
(V,I1.tI)
(Tn) C L(V)
Prove that
there exists some
if
such that
that is, prove that
is the limit in
T
where
T = A + iB,
Corollary 13.15.1 71.
T E L(V).
If
Let
72.
and let that
T
is also compact.
T
and let
s > 0
is finite dimensional
R(Te)
is compact if and only
T
and
T
(Hint: Recall that B
can be applied to
A
and
B.)
be a Hilbert space over C
is compact, prove that
(V,(.,.))
can be written
are compact and self-adjoint.
T*
V.
and let
is also compact.
be a Hilbert space over
be an orthonormal set in
(en)
If
C,
T
let
T E L(V),
is compact, prove
limn(T(en),en) = 0. 73.
T E L(V)
Let
(V,(.,.))
may write
T
T =
as T,
T E L(V)
Let
TS = ST,
By Corollary 13.15.1 we S E L(V),
If
and let
prove that
if and only if
S
S
com-
commutes with
.
(V,(.,.))
be compact.
and only if
- IXkPk.
that is,
Pk, k = 1,2,3.... 74.
be a Hilbert space over C
be compact and self-adjoint.
mutes with
A } 0
A
(V,(.,.))
Let
Tn
If
of a sequence of transformations
with finite-dimensional ranges. as
and suppose
is compact if and only if for every
T
Te E L(V)
- TeII < e;
JIT
C
be a Hilbert space over C
(V,(.,.))
con-
ha IT - Tn11 = 0.
prove that
n = 1,2,3,...,
Let
T E L(V).
[x n)
are such that
T E L(V)
and
and let (T(xn))
(V,TW).
be a Hilbert space over
is compact for 70.
converges in
C.
0 E P0(T).
C
is compact if and only if
T
whenever
Let
69.
and
be a nonseparable Hilbert space over
(V,(.,.))
Let
67.
be a Hilbert space over C
Prove that, if
X E Po(T*).
cannot be dropped.
A # 0,
then
and let A E P0(T)
if
Give an example to show that the assumption
483
13.16. Problems
75.
T E L(V) (x
J
be compact.
If
x E V, T(x) = Xx) 76.
be a Hilbert space over
(V,(.,.))
Let
X E Pa(T), X # 0,
be a Hilbert space over V - V
If
T E L(V)
and
T
77.
Let
V = L2
T
:
and let
is a finite-dimensional linear subspace of
Let
xo E V*.
C
prove that
is defined by
C
and let
T(x) = xo(x)xo,
x0 E V,
prove that
is compact.
and define
T E L(t2)
by
T((ak)) = (ak/k).
Prove each of the following: (a)
T is compact and self-adjoint.
(b)
Pa(T) _ (1/n
I
n = 1,2,3,...)
(Hint: Problem 12.6.1 is useful.)
and
CG(T) _ (0).
V.
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W. Hurewicz, Lectures on Ordinary Differential Equations, M.I.T. Press, Cambridge, Mass., 1958.
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K. A. Ross, Fourier Series and Integrals, Yale University Department of Mathematics, New Haven, Conn., 1965.
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INDEX
A
A(D), 10
approximate identity, 272
arithmetic mean, 175
Abelian semigroup of linear transformations, 90
B
absolutely sumeable series, 30
Bn, 171
absorbing set, 19 closed in rhys, 29 open in rays, 29
BW, 171
adjoint, 96,425
80
8k' 19 19
)L
affine mapping, 354
B(X), 7
annihilator, 99
BV([0,1)), 170
applications of Banach-Alaoglu Theorem, 263-282,303-309,362-365 Banach-Steinhaus Theorem, 154-169 Closed Graph Theorem, 191-196, 241-242 fixed point theorems, 357-365 Hahn-Banach Theorem, 90-116, 209-211,243,277-282,303-309, 322-324,450-452 Krein-Mil'man Theorem, 326-333, 337-345 Krein-$mulian Theorem, 303-309 Open Mapping Theorem, 196-199, 277-282,303-309,362-365 Stone-We}erstrass Theorem 333-337,409-411 Uniform Boundedness Theorem, 154-169,24*047,254-257, 303-309
B(IR), 121
Baire Category Theorem, 147 Baire function, 171
balanced set, 2 Banach-Alaoglu Theorem, 254 applications of, 263-282, 303-309,362-365 Banach ririt, 96,264 Banach space, 4
Banach-Steinhaus Theorem, 150,154 applications of, 154-169 Banach-Stone Theorem, 342 Bessel's Inequality, 404
489
Index
490
bidual, 208 bilinear form, 372 nonnegative, 372 positive definite, 372 symmetric, 372 bornological, 54
co (E) , 301 bounded linear transformation, 61 canonical embedding, 208 bounded set in a seminormed linear space,
category 1, 146
15
in a topological linear space,
category II, 146
41
Cauchy-Schwarz Inequality, 373 bounded variation, 105 Cesaro mean, 175,269 bounded weak* topology, 291 circled set, 2 Brouwer Fixed Point Theorem, 350 closed convex hull, 301 C
Closed Graph Theorem, 188, 189 applications of, 191-196, 241-242 closed mapping, 178,200-201 closed under complex conjugation, 330 closure of a set, 41
codimension one, 68 C'(X), 6
compact linear transformation, 458 CR(X), 7 CR(X), 7 CR (X) , 7
C(X)*, 109 Cn([a,b)), 7
C ([a,b]), 7
complete, 46
completion of a nosed linear space, S
complete orthonormal set, 397 characterization of, 40S conjugate space, 56 continuous linear functional, 56
Ca(T), 448
continuous linear transformation, SS
491
Index
continuous spectrum, 448 contraction mapping, 3S2 contractive mapping, 368 convergence
in F-topology, 236
E EA.
E
,
99,386
(E*), 99
E(C), 10
in a seminormed linear space, Eberlein-mulian Theorem, 303 25 in the strong operator topology, eigenvalue, 447 155
in the weak operator topology, 476 in the weak topology, 239 in the topology, 241 weak*
convex cone, 118 convex hull, 301 closed. 301 convex get, 2
convolution, 102,269,274 counting measure, 9
eigenvector, 447 equicontinuous family in -equivalent norms, 12
Euler-Knopp matrix, 176 extension of a positive linear functional, 118-119 extreme point, 317
in Lp(IR,dt), 317 in
M(X), 338
extremal subset, 320
F
Dn, 165
dimension of a Hilbert space, 408
f, 59
f * h, 102, 269
direct sum, 205,384 f * µ, 274
direct summand, 205
IIfIIn .
7
Dirichlet kernel, 166
Ilfli. p 8 discontinuous linear functional,
250-251 dual space, 56
V*, 314
IIfIIW, 6,8 F-topology, 236
Fejer kernel, 269 fixed point, 350
Index
492
fixed point property, 350
Helly's Theorem, 112
Fourier analysis, 198-199,202,267-274,333-337, 409-417,441-446
Hilbert relation, 480
Fourier coefficients, 404
Hilbert-Schmidt transformation, 460
Frechet space, 46
Hilbert space, 11,375 adjoint, 475
function lower semicontinuous, 147 Rademacher, 417-418 Walsh, 419
set, 334
hyperplane, 127 real, 127
I
G
int(E), 41
G(f), 177 Individual Ergodic Theorem, 441 gauge, 20
graph of a mapping, 177
inner product, 11,372 space, 11,372 interior of a set, 41
H
invariant under H2, 442 H2, 442
T, 472
isometric mapping, 101 isometry, 101
Flaar measure, 365 K
Hadamard sequence, 414 Hahn-Banach Theorem analytic form, 84 applications of, 90-116, 209-211,243,277-282,303-309, 322-324',450-452
kernel Dirichlet, 166 Fejer, 269
of a linear functional, 68 of a linear transformation, 123 Poisson, 1f9-121
equivalence of analytic and geometric forms, 136 geometric form, 134 Kolmogorov's Theorem, 414 half-space, 130
Krein-Mil'man Theorem, 322 applications of, 326-333,337-345
Hardy space, 442 Helly's Selection Theorem, 261
Krtein-gmulian Theorem, 299 applications of, 303-309
493
Index
Hilbert-Schmidt, 460 normal, 477 orthogonal, 430 positive, 478 self-adjoint, 429 unitary, 430
L
p, 9
1,
9
Lp(X,S,µ) - Lp(X,p) = Lp(1,), 8 L.(X, S,1+) a Lm(X,1+) - Lm(1i) ,
linear variety, 127 real, 127
8
Lipschitz condition, 360
.L(V1,V2), 55 locally bounded, 54
1 (Vl,V2) ,
55
locally convex topological linear space, 38 bornological, 54
L1(X,S,µ)*, 228 Lp(X,S,1+)', 1 < p < m, 227
local membership in is weakly sequentialL (X,S,N) lly complete, 247 L (X,S,µ), 1
