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k 0 , p(xn.,x0) < ef2. On the other hand, thesequence {xn} is a Cauchy sequence. Therefore, there exists a number N such that, for n, m > N, PROPOSITION
p(xn, Xm)
e
< 2·
Putting m = nk, k
> k 0 , we obtain
p(xn, x 0) ~p(xn,Xn.)+p(xn., X0)
e
e
< 2+ T =e.
The arbitrarines of e implies the proposition. PROPOSITION
1.4.9. Let (X,
II II) be an
D
F*-space. Let {ei} be an arbitrary 00
2
fixed sequence ofpositive numbers such that the series
ei is convergent.
i=l
If for each sequence {Xi} of elements of X such that
llxill < C:i the series
00
2
Xi is convergent, then the space (X,
II II) is complete.
i=l
Proof. Let {Yn} be an arbitrary Cauchy sequence of elements of the space X. We can choose a subsequence {Ynk} such that llxkll < ek, where Xk 00
=
Ynk+l-Ynk· The assumption implies that the series
2
Xk is convergent
k=l
to an element xeX. We shall show that {Yn} tends to x+y 111 • In fact Sk k
=
2 X;= Ynk-Yn
1
tends to x. Thus Yn. tends to x+yn 1 • By Proposi-
i=l
tion 1.4.8 {Yn} tends to x+yn1 • D A complete F*-space is called an F-space. A complete pre-Hilbert space (Example 1.3.10) is called a Hilbert space. 1.4.10. Let (X, II II) be an F-space. Let Y be a subspace of X. Then the quotient space X/Y (see Section 1.1) is an F-space (i.e. it is complete). THEOREM
Basic Facts on Metric Linear Spaces
23
Proof Let {Zn} be an arbitrary sequence of elements of X/Y such that IIZ11II < 1/211• By the definition ofthe norm in the quotient space there are X11 e Z11 such that llx11ll < 1/211 •The space X is complete, hence the series 00
L X11 is convergent to an element x eX. Let Z denote the coset containing n=l
x. Then, by the definition ofF-norm in the quotient space k
k
112z,--zll~li2x,-xll < i=m i=m
22-1·
00
Therefore, the series
2Z
11
is convergent to Z, and by Proposition 1.4.9
n=l
the space X/ Y is complete.
D
1.4.11. The product (X, II II) of n F-spaces is an F-space. Proof Let xm = (xi) be a Cauchy sequence. Then the sequences {xi}, i = 1, ... , n also is a Cauchy sequence. Since (Xt, II liD are complete, there are Xi eXt such that !!xi-xt 11-+0, i = 1, ... , n. Let x = (xt). Then PROPOSITION
n
l!xm-x!!
=
};!!xi-xt lli-+0.
D
i=l
1.5. COMPLETE METRIC LINEAR SPACES. EXAMPLES 1.5.1. The spaces N(L(Q, L:, p,)) are complete. Proof Let {xn} be a Cauchy sequence in N(L(Q, L:, p)). It is easy to verify that the sequence { 11 } is a Cauchy sequence with respect to the measure (this means that, for each a > 0, PROPOSITION
x
lim p({t: !xn(t)-xm(t)! >a})= 0). n,m~co
Therefore, by the Riesz theorem, the sequence {xn} contains a subsequence {Xnk} convergent almost everywhere to a measurable function x(t). Let e be an arbitrary positive number. Since the sequence {x11 } is a Cauchy sequence, there is a positive integer N such that for n, m > N PN(Xn-Xm)
=
JN(!xn(t)-xm(t)!)dp ~B.
a
24
Chapter I
Put m
=
nk and
let k-+oo. By the Fatou lemma, we obtain
PN(Xn-X)
~e.
This implies that Xn-X E N(L(Q, r, p). Since N(L(Q, r, /1)) is linear, X E N(L(Q, r, /1)). The arbitrariness of e implies that Xn-+X. 0 1.5.2. The space M(Q, £, f.l) is complete. Proof Let {Xn} be a Cauchy sequence in M(Q, £, p). Then the sequence {Xn(t)} is convergent for almost all t. Let x(t) denote the limit of the sequence {xn(t)}.lt is easy to verify that x(t) E M(Q, £, p). Let e be an arbitrary positive number. Since the sequence {xn} is a Cauchy sequence, there is a positive integer N such that for n,m > N PROPOSITION
llxn-Xmll
= esssup lxn(t)-xm(t)l <e. tED
Hence, when m tends to infinity, we obtain llxn-xll = esssup lxn(t)-x(t)l
~e.
tED
The arbitrariness of e implies that Xn-+X.
0
1.5.3. The space C(Q) is complete. Proof Let {xn} be a Cauchy sequence in C(Q). This implies that at each t the sequence of scalars {Xn(t)} is also a Cauchy sequence. Its limit x(t) is continuous as a limit of a uniformly convergent sequence of continuous functions. Let e be an arbitrary positive number. Since the sequence {xn} is a Cauchy sequence, there is a positive integer N such that for n, m > N PROPOSITION
llxn-Xmll = sup ixn(t)-xm(t)i <e. tED
Hence, when m tends to infinity, we obtain llxn-xll = sup lxn(t)-x(t)i
~e.
tED
The arbitrariness of e implies the proposition. 1.5.4. The space C(QjQ0) is complete. Proof C(QjQ0) is a closed subspace ofthe space C(Q).
0
PROPOSITION
0
Basic ! N(l)b for cx:j=fl. Since the set {x"} is non-countable, by Corollary 1.6.2 the space N(L(Q, E, p)) is not separable. 0 1.6.5. A space M(Q, E, p) is separable if and only if the measure J1 is concentrated on a finite number of atoms PI> ... , PTe· Proof. Sufficiency. Suppose that the measure J1 is concentrated on a finite number of atoms PI, .•. , PTe· Then the space M(Q, E, p) is finite dimensional, and thus separable. Necessity. If the measure J1 is not concentrated of a finite number of atoms, then there is a countable family of disjoint sets {An} (n = I, 2, ... ) of positive measure. Let ex= {ni> n2 , • •• } be a subset of the set of positive integers. Let A = Ani u An 2 u . . . The family {A"} is non-countable and for ex =I= {3, p(A"""Ap)+p(Ap""A")> 0. Let x.,. = XAa· Then llxa-xpll = I for ex =F fl. Therefore by Corollary 1.6.2 the space M(Q, E, p) is not separable. 0 PROPOSITION
1.6.6. A space C(Q) is separable if and only if the topology in the compact set Q is metrizable (i.e., it can be determined by a metric d(t, t')). PROPOSITION
Basic Facts on Metric Linear Spaces
29
Proof. Let (Q,d) be a metric compact space. Then for each n = 1, 2, ... , there is a finite system of sets {An,k}, k = 1, ... , Kn, such that An,k (') An,k'
=
fork ::j::. k',
121
Kn
UAn,k= Q, n=l
sup{d(t, t'): t, t' E An,k}
... , un). (2.4.2.ii) This follows immediately from {2.4.2.i); if ua 2 Ua = x, a .s;;; c, then
aEA
E
C): (E[a])
and
L llualla =}; lluallc;? 11J; Ua lie= llxllc· aEA aEA
aeA
(2.4.2.iii). If llxll = 0, then for each positive integer n there exist u~ E r:f(E[a]), a E A = supp x such that ~ n LJ Ua = X
aEA and
2..: ll~lla < ! ·
aEA
Then, for each fixed a, lim lim
u~(a)
llu~lla
= 0. Since Sl(E[a]) is finite-dimensional,
= 0. Since the support A = suppx is finite, x(a) = 0 for all
a E A. Thus x = 0.
An F-norm II II on r:f(E) is called monotone if section A C E.
D
IIPA(x)ll ~ llxll for any
Linear Operators
49
2.4.3 (Kalton, 1977). Suppose that E is tree-/ike and that {II lie. c E E} is a consistent family ofF-norms such that IIPE(aJ (x)llc ~ llxllc for a~ c. Then the limit F-norm II II is monotone. Proof Suppose that A is a section of E. Let x E :!(£). For any B > 0 we can find Ut,Ci, i = I, ... , n such that Ui E :l(E[ct]), x = u1 + ... +u11 and llutllcl+ ·· · +llunllc11 ~ llxll+e. Since A is a section, A n E[ci] = E[at] for some at ~ Ct. Hence PROPOSITION
n
11
IIPA(x)ll
~}.; IIPA(Ut)llc1 ~}.; llutllc1 ~ i=l
llxll+e.
i=l
The arbitrariness of s completes the proof.
0
Let N denote as before the set of all positive integers. By :l(N) we de· note the set of all finitely supported sequences. For each n eN we define en={O,O, ... ,O,I,O, ... }. n-th place
The space :l(E[n]) will be denoted more traditionally by R 11 • Of course R 11 =lin{ e~> ... , e11 }. Let 4J be the family of all F-norms defined on :l(N) and let 4Jn be the family of all F-norms defined on R 11 • Now we shall determine the distance between two F-norms defined on an F-space X as follows. Letp(x), q(x) be two F-norms on X. Let
d*(Ji, q) =sup arctan llog p((x))l· XEX q X x,. 0 and sequences of scalars {rn} and {s11 } such that llrnXnll > b, llsnYnll > (), n = I, 2, ... (2.5.7) Now we define by induction a permutation of positive integers k (n) in the following way: k(n) is the smallest positive integer different than k(l), ... ... , k(n-I) such that I llrnXn-Sk(n)Yk(n) II< 2n · (2.5.8) Formulae (2.5.7) and (2.5.8) imply that the spaces induced by the sequences {Xn} and {Yn} are isomorphic.
2.6.
BASES IN F-SPACES
Let (X, II ID be an F-space. A sequence {en} of elements of the space X is called a Schauder basis (Schauder, I927) (or simply a basis) of the space X if every element x E X can be uniquely represented as the sum of a series (2.6.1)
c
A sequence {g11 }
X is called a basic sequence if it is a basis in the
space Y generated by itself, i.e. it is a basis in Y = lin {gn}. Evidently, if an F-space has a basis, then it is separable. For any x of the form (2.6.1), let n
Pn(x) =
.2; ftet. i=l
2.6.1. The operators Pn are equicontinuous. Proof Let X1 be the space of all scalar sequences y = {ni} such that the
THEOREM co
series }; 'l']t et is convergent. The arithmetical rules of the limit trivially i=l
imply that Let
xl is a linear space. n
IIYII* =sup II n
X'l']tetll·
i=l
Chapter 2
68
The space X1 with the norm II II is an F*-space. We shall show that it is complete, i.e. that it is an F-space. Suppose that a sequence {yk} E X 1 is a Cauchy sequence. Let yk = {nn. Since {yk} is a Cauchy sequence, for an arbitrary e > 0 there is a positive integer m0 such that, for m,k > m0 , n
llym-ykll* = sup n
/I}; (n'[' -nf)et
/J
<e.
(2.6.2)
i=l
Hence n-I
ll(n~-n!)enll ~ 1/l\n'l'-n~)ei 1/ i=l
n
+ Jj}; Cn'l'-nf) ei [/ < 2e i=l
Consequently lim (n';'-n~) m,k-+oo
=
o,
n =I, 2, ...
Thus the sequences of scalars {'11~} are convergent for all n. Write 'l']n
= lim
'11~·
~00
Passing with k to infinity in inequality (2.6.2), we obtain that for m>m0 n
sup\\}; n
(n'!'-nt) et II ~ 2e.
(26.3)
i=l
Let and Taking into account inequality (2.6.3) we obtain
llsr-snll ~ lls';'-s~ll+2e for all m > m0 and all n and r. Fix m1 > m0 • The sequence {s;:''} is a Cauchy sequence. Therefore there is a number n0 such that, for n,r > n0 ,
lls::''-s"/''11 <e.
69
Linear Operators
Hence, for n, r > n0 ,
00
Thus the series 2 rJt et is convergent andy = {nt} e X1 • i-1
From inequality (2.6.3) follows n
sup n
II? (rJi -nt)e,IJ ~ 2e \=1
for m > m0 • Hence the space X1 is complete. Let A be an operator mapping X1 into X defined as follows A(y)
=}; rJtet. i=l
By the definition of the space ~1 the operator A is well-defined on the whole space X 1 • The arithmetical rules of the limit imply that the operator A is linear. Since {en} is a basis, the operator A is one-to-one and maps X1 onto X. The operator A is continuous, because · n
oo
IIA(y)ll = [[}; rJtet II~ sup i=l
n
II}; niet I = IIYII* · i=1
By the Banach theorem (Theorem 2.3.2) the inverse operator A-1 is also continuous. Hence n
IIPn(x)ll
=If LrJtetl[ ~ IIYII* =IIA- {y)ll 1
i=1
and the operators Pn are equicontinuous.
D
Let x eX be reperesented in the form (2.6.1). Let fn(x)
= tn.
It is easy to see thatfn are linear functionals. They are called basis junetionals.
Observe that fn(x)en =· Pn(x)-Pn'- 1(x):
Thus from Theorem 2.6.1 immediately follows
Chapter 2
70
2.6.2. The basisfunctionals are continuous. Suppose we are given two F-spaces X and Y. Let {en} be a basis in X and let {fn} be a basis in Y. We say that the bases {en} and {in} are equivalent COROLLARY
00
00
i=I
i=l
if the series 2 ttet is convergent if and 'only if the series 2 tdi is convergent. Two basic sequences are called equivalent if they are equivalent as bases in the spaces generated by themselves. 2.6.3. If the bases e{n} and {fn} are equivalent, then the spaces X and Yare isomorphic. Proof Let Tn: X-+Ybe defined as follows
THEOREM
oo
Tn(x)
n
= r(~ ttet) = ~ ttfi. i=l
i=l
By Corollary 2.6.2 the operators Tn are linear and continuous. The limit T(x) = lim Tn(x) exists for all x. Thus, by Theorem 2.2.3, T(x) is continuous. Since the bases are equivalent, the operator, Tis one-to-one and maps X onto Y. Thus, by the Banach Theorem (Theorem 2.3.2), the 0 inverse operator r-1 is continuous. The following theorem is, in a certain sense, converse to Theorem 2.6.1. THEOREM 2.6.4. Let (X, II /D be an F-space. Let {en} be a sequence of linearly independent elements in X. Let XI be the set of all elements of X which can be represented in the form
Let n
Pn(x) =
.J; ttet. i=l
be a sequence of linear operator defined on X 1 • If the operators Pn are equicontinuous, then the space XI is complete and the sequence {en} constitutes a basis in this space.
Linear Operators
71
Proof. Since the operators P, are equicontinuous, each element x of X 1
can be expanded in a unique manner in the series
Let X2 be the space of all sequences {ti} such that the series
is convergent. Let n
ll{tt}ll*
=sup n
Jj}; t,e,Jj. i=l
In the same way as in the proof of Theorem 2.5.1, we can prove that II II* is an F-norm and that (X2 , II II*) is an F-space. Observe that
llxll ~ ll{tt}ll*.
(2.6.4)
On the other hand, the equicontinuity of the operators P, implies that if X--+0 then II{ ti}ll*-+0. Hence the space x2 is isomorphic to the space X]. Therefore X 1 is. an F-space. By (2.6.4) the sequence {e,} constitutes a basis in X1 • D 2.6.5. Let X be an F-space. A sequence of linearly independent elements {e,} is a basis in X if and only if
CoROLLARY
(1) linear combination of elements {e,} are dense in X, (2) the operators n
Pn(x)
= }; ttet i=l
are equicontinuous on the set lin{ e,} of all linear combinations of the set
{e,}.
Chapter 2
72
2.6.6. Let X be an F-space with a basis {en}. Let t 1 , t 2 , ••• be an arbitrary sequence of scalars. Let p 1 ,p2, ••• be an arbitrary increasing sequence of positive integers. Let
COROLLARY
n = 1, 2, Let Pn+t
e~=}; t,e,. i=pn+l
Let X1 = {line~} be the space spanned by the elements {e~}. Then the space X 1 is complete and the sequence {e~} constitues a basis in X1 • n
oo
Proof Let x EX, x =
2
ttet. Let Pn(x) =
i=l
2
ttet.
i-1
For ye Xb let n
P~(y)
=}; a,e·,. i=l
Then P~ (y) = PP
n+l
(y) for ally E X1• Therefore, by Theorem 2.6.1, the
operators P~ are equicontinuous and, by Theorem 2.6.4, X1 is complete and {e~} is a basis in X1 • D A basis {e~} of the type described above is called a block basis with respect to the basis {en}· PROPOSITION
2.6.7. Let (X,
II
jl) be an F-spacewith a basis {en}. Let {xn} be
a sequence of elements of X of the form 00
Xk = }; fk,tet
where lim tk,t = 0, i = 1, 2, ...
i=l
k--->ro
If {en} is an arbitrary sequence of positive numbers, then there exist an increasing sequence of indices {Pn} and a subsequence {xkn} of the sequence { Xk} such that Pn+t
llxkn- :L>kn,tetjj <en. i=p.. +l
Linear Operators
Proof (by induction). Let p 1 = 0,
Xk1
=
73
x1. We denote by p 2 an index
satisfying the inequality
Suppose that the element Xkn-l and the index Pn are already chosen. The hypothesis lim fk, t = 0 implies the existence of an element Xkn such k->-oo
that
Let Pn+l be an index satisfying the inequality Pn+t
llxkn-}; tkn,tetll 0} and E~; = {t: u(t)+~v(t) =/= 0} differ by a set of measu~e 0. Proof The sets E n (!J"'-.,E~;) are disjoint for differenU. Since Jl is a-finite, only a countable number of these sets has positive measure. D LEMMA
0 (.Q,
Proof of Proposition 2.7.3. Let qJ e L 0 (.Q, E, p) be a positive function such that qJdJl = 1. The existence of such functions follows from the fact that u Jl is a-finite. Write p(E) = qJdqJ. Let
J
J
E
a= sup{,U({t: u(t) =/= 0}), u e A}.
Let {Un} be a sequence of elements of A such that an= ,U({t: un(t) =/= 0})-+a.
By Lemma 2.7.4 there are reals ~2 ,
... ,
~n
such that
En= {t: lu1(t)l+ ... +lun(t)l > 0} = {t: Vn(t)=/=0},
Chapter 2
80
where Vn(t)
=
u1(t)+;2u2(t)+ ... +;nun(t).
(2.7.4)
Observe that p(En) increases to a. 00
Let EA =
D"'-.U
En. The setEA is the required unfriendly set. Indeed,
n~l
suppose that EA is not an unfriendly set. Then there is a function u(t)E A such that 7-t({t E EA: u(t) #- 0});;:::, b
> 0.
By the definition of {un} we can find an Un such that a-an< b.
Using Lemma 2.7.4, we can construct a function of the form u+;un such that
f,t({t: u(t)+;un(t) =F 0}) = b+an >a, which contradicts to the definition of a. Now we shall show that EA is a maximal unfriendly set up to the sets of measure 0. Indeed, P, (EA) = 1-a. Hence E A is unique up to the sets of p-measure 0. Thus it is unique up to the sets of .a-measure 0. D PROPOSITION 2.7.5 (Aronszajn and Szeptycki, 1966). Let a measure space (Q, .E, .u) be a-finite. Let A C L 0(Q, .E, .u) be a solid F-space. Let EA be a maximal unfriendly set for A. Then there is a v e A such that EA = {t: v(t) = 0}.
Proof Let {un} be chosen as in the proof of Proposition 2.7.3. Let positive numbers Mn and An be chosen so that
,U({t: lun(t)l > Mn} < 2-n,
(2.7.5)
where p(E) is defined as in the proof of Proposition 2. 7.3, and
ll;unll < for;,
1~1
2-n
< An.
Having { Mn} and {An} we shall choose by induction two sequences {;n} and {1Jn} in the following way. We define~~ = 1, 0 < 171 ~ 1. Sup-
Linear Operators
81
pose that ~to ••• , ~nand 'f/~o ... , 'f}n are chosen. Using Lemma 2.7.4, we can choose c;n+1 such that 0 < ~n+l < min(A.n+l• (2Mn+I)-~n) (2.7.6) and
{t: Vn(t)+~n+IUn+I(t) ::1= 0}
= {t:
lvn(t)l+lun+l(t)l
> 0},
where Vn(t) is given by formula (2.7.4). We take
0
0, a sequence En ~o, and a sequence Un E C such that
llunXE.. II >
(2.7.10) Since Un E C C Xa, we can construct by induction subsequences { Un.i}, { EnJ} such that s.
8
llun,XE,.1 II < 2
fori <j.
(2.7.11)
By (2.7.10) we get fori <j.
Hence the set C is not compact. Sufficiency. Let {vk} be any sequence of elements of C. By the compactness of C in L 0(Q, 1:, Jl) and the Riesz theorem we can find a subsequence {un} = {Vkn} of C tending to u E C almost everywhere. Now we shall show that {un} tends to uin X. Without loss of generality we may assume that X does not have an unfriedly set. Then by Proposition 2.7.5 there is a rp EX positive almost everywhere. Let En,k
=
{r
E
[J:
suplum(t)-u(t)l;:;:::, m~n
+rp(t)}.
Linear Operators
83
Observe that, En, k ~0 for each fixed k andfor any e > 0 we can find by (2.7.9.ii) an index such that llvxn,kll
e
nk, v e C,
(2.7.12)
where, for brevity, we write Xn,k = XEn,.· Then llun-ull ~ llxn,k(un-u)ll+ll(l-xn,k)(un-u)ll ~ llxn,kUnll+llxn,kull(1-xn,k)(un-u)ll
~ ; + ; + I ~ ~II <e. provided that k is chosen so that
I ~ ~11 < ; .
0
2.7.7 (monotone convergence theorem, Luxemburg and Zaanen, 1963). Let XC £ 0(!2, 1:, p) be a solid F*-space. Then u e Xa if and only iffor each sequence {vn} such that PROPOSITION
(2.7.13) tending to zero almost everywhere, llvnll-+0. Proof Let u eX, En ~0, Vn = UXEn· By our hypothesis llvnll-+0. Thus ueXa. Conversely, let u e X a and let {Vn} be a sequence satisfying (2. 7.13) and tending to 0 almost everywhere. Let Em,n
=
{t:
Vn(t)
~ ~
u(t) }·
Let Xm,n denote the characteristic function of Em,n· For a fixed m, !2"'-._Em,n ~0 and II(I-xm,n)ull-+0 since u e Xa. We have llvnll
~II~ Xm,nUII +11(1-xm,n)ull ~II;
I
+11(1-xm,n)ull.
(2.7.14)
Chapter 2
84
For each s
> 0 there is an m such that
II; II El> J1.1) such that {un} tends to u in the space L 0(Q1 , E 1 , Jl.J. By (2.7.17), {uk} contains a subsequence {u;} = {uk,} su«h that 00
}; llu~+l-u~llm,n
jt1) simultaneously to u0 and to u1 • Thus u0 = u1• Hence l/un-u0 1/ 1 = l/un-u0 1/+l/un-U0 1/vx-+0.
Therefore (X, 1/ 1/J is complete, and by Corollary 2.3.3 the two norms are 0 equivalent. This implies the continuity of K. THEOREM 2.7.12 (Aronszajn and Szeptycki, 1966). Let (X, 1/ 1/x) CcL0(f:J 1 ,l:t>Jit) and (Y,/1 I/Y)CcL0(Q,l:,Ji) be given F-spaces. If XC Dx and KX C Y, then the operator K: X-+Yis continuous. Proof. We introduce a new norm on X, llxll
= llxl/x+IIK(x)l/y,
and, in the same way as in the proof of Theorem 2.7.11, we show that 1/ //)is complete. We take a fundamental sequence {un} in (X, II //).It is alsofundament~lin(X, II llx). Thus there is a ue X such that llu11 -ul/x-+0. By Theorem 2.7.11 llun-ullvx-+0 and by Theorem 2.7.10 Kun-+Ku in L 0 (Q, 1:, Jl). Suppose that 1/Kun-VIIy-+0. Since Y C c L 0 (Q, 1:, Jl), Ku11 -+v in L 0 (Q, 1:, Jl). This implies that Ku = v. Hence (X, II //)is complete. Thus, by Corollary 2.3.3, the norms llxll and llxl/x are equivalent. 0 This implies the continuity of Ku. (X,
PROPOSITION 2.7.13. (Aronszajn and Szeptycki, 1966). Let K be an integral transformation. Then (Dx)a = Dx. Proof. Letfe Dx. Let Et~0. Thus Jl(Qn,t n E,)--+0,
n = 1,2, ...
Therefore
n = 1,2, ...
88
Chapter 2
By Proposition 2.7.7, IKIIIXE,I tends to 0 almost everywhere and, by the definition of the F-pseudonorm llxllm,n,
II/XEI!Jm,n ~ 0 •
D
Example 2.1.14 Let D = [0,27t] with the Lebesgue measure. Let D 1 = Z be the set of all integers with the discrete measure
Pl{y}) Let k(t, s)
=
1,
y
=
= e'"'. Then Dx =
Example 2. 7.15 Let D = Z and let D 1
0, +I, ±2, ... L(Z)
= fl.
= [0,27t]. Let k(t,s) =
eil8 •
Then Dx
= £1[0,27t].
Example 2.1.16 Let D = D 1 = R with the standard Lebesgue measure. Let k(t,s) = eilB. Then Dx = £1( -oo, +oo).
Chapter 3
Locally Pseudoconvex and Locally Bounded Spaces
3.1.
LOCALLY PSEUDOCONVEX SPACES
Let X be a metric linear space. A set A C X is said to be a starlike (starshaped) set if tA C A for all t, 0 < t ~ I. The modulus of concavity (Rolewicz, I957) of a starlike set A is defined by c(A)
= inf {s > 0:
A+A C sA},
with the convention that the infimum of empty set is equal to +oo. A starlike set A with a finite modulus of concavity, c(A) I such that rx ¢ A+A. Then A+A ¢ c(A) A r '
because x ¢ c(A)A. Since r >I, we obtain a contradiction of the defiD nition of the modulus of concavity. Observe that we always have c(A) ;::;::: 2. A set A is said to be convex if x, yEA, a, b;::;::: 0, a+b = I imply ax+bye A. Of course, for each convex set A the modulus of concavity c(A) of the set A is equal to 2. The condition c(A) = 2 need not imply convexity, but the following proposition holds : 89
90
Chapter 3
3.1.2. Let A be an open starlike set. If c(A) = 2, then the set A is convex. Proof Let x,y EA. Since A is an open starlike set, there is a t > 1 such that tx,ty EA. Since c(A) = 2, A+A E 2tA. Thus tx+ty E 2tA. PROPOSITION
This implies that (x-;-y) EA. Therefore, for every dyadic number r,
rx+(1-r)y EA.
(3.1.2)
The set A is open. Then the intersection of A with the line
L = {tx+(1-t)y: t real} in open in L. Therefore there is a positive numbers such that
x 0 = t0 x+(1-t0 )y E A
for
ltol < E
and Applying formula (3.1.2) for x = x 1 andy= x 0 , we find that, for every a such that Ia - rl < s, ax+(l-a)y EA. (3.1.3) Since r could be an arbitrary dyadic number, (3.1.3) holds for an arbitrary real a, 0 ~ a ~ I. Thus the set A is convex. D PROPOSITION 3.1.3. Let A be a starlike closed set. If c(A) = 2, then the set A is convex. Proof Since the set A is closed, 2A = sA. Thus c(A) = 2 implies
n
•>2
x+y that A+A C 2A. Therefore, if x,y E A, then - 2- EA. This implies (3.1.2) for every dyadic number r. Since A is closed, (3.1.3) holds. D A metric linear space X is called locally pseudoconvex if there is a basis of neighbourhoods of zero { Un} which are pseudoconvex. If moreover c(Un) ~ 2 11P, we say that the space X is locally p-convex (see Turpin, 1966; Simmons, 1964; Zelazko, 1965). THEOREM 3.1.4. Let X be a locally pseudoconvex space. Then there is a sequence of Pn-homogeneous F-pseudonorms {II lin}, i.e. such that
lltxlln = ltiPnllxlln,
(3.1.4)
Locally Pseudoconvex and Locally Bounded Spaces
91
determining a topology equivalent to the original one. If the space X is locally p-convex, we can assume Pn = p (n = 1,2, ... ). Proof Let {Un}be a basis ofpseudoconvex neighbourhoods ofO. Without loss of generality we may assume that the sets Un are balanced (cf. Section 1.1). From the definition of pseudoconvexity it follows that there are positive numbers Sn such that Un+Un C SnUn.
Let (q = 0, ±1, ±2, ... ). t
For every dyadic number r > 0,-, = }; at 2i, where at is equal either to i=B
0 or to 1, we put Un(r) = asUn(28)+ ... +atUn(21). In the same way as in the proof of Theorem 1.1.1, we show that Un(r 1 +r2)
Un(r 1 )+ Un(r 2)
)
and U11 (r) are balanced. Moreover, the special form of U11 (r) implies U11 (2qr) = s~ U11 (r). Let //xi/~= inf {r > 0: x E Un(r) }. The properties of the sets U11 (r) imply the following properties of 1/xl/~: . (1) 1/x+y//~ ~ 1/x/I~+IIYII~ (the triangle inequality), (2) 1/ax/1~ = 1/x/1~ for all a, /a/ = 1, (3) 1/s~x/1~ = 2ql/xl/~.
Let log 2 Pn =log sn" Let 1/xl/n = sup 1>0
1/tx/1~ . fPn
By (3) 1/xl/n is well determined and finite since 1/xl/n = sup 1>0
1/tx/1~ fPn
=
sup 1,;;;1,;;;sn
1/tx/1~ • fPn
92
Chapter 3
We shall show that llxlln are F-pseudonorms. Indeed, if lal = 1, then
llaxlln =sup lltaxll~ =
sup
t>O
t>O
fP•
lltxll~
= llxlln
fP•
and (n2) holds. Let x,y e X. Then II x + y II 11 = sup llt(x+y)ll~ P• = sup t
t>O
lltxll~
=S;; sup
tPn
I.,;;t.,;;sn
llt(x+y)lln
I.,;;t.,;;s,.
+ sup I.,;;;t.,;;sn
lltyll~ tP•
t
P•
=S;; llxlln+IIYIIn
and (n3) holds. Observe that lltxlln = sup llrtxll~ = sup llrltlxll~ r>O
jtjP•
r>O
lf/P• sup llsxll~
=
sP•
•>O
(r It i)P•
=
lt/P•
/t/P•
llxlln.
Hence llxll is Pn-homogeneous. This implies (n4)--(n6). Now we shall prove that the system of pseudonorms {llxlln} yields a topology equivalent to the original one. Indeed, from the definition of llxlln, llx[[~ =S;; llxlln, in other words, {x: llxlln
O
tP•
I.,;;t.,;;sn
tP•
in other words Un(r)C {x: llxlln
< 2r}.
(3.1.6)
Formulae (3.1.5) and (3.1.6) imply the first part of the theorem. If the space X is locally p-convex, then by the definition there is a basis of neighbourhoods of 0 {Un} such that c(Un) =S;; 2 11P. Since Un are open it implies that Un+UnC2 1 iPUn.
Locally Pseudoconvex and Locally Bounded Spaces
93
Putting s11 = 21 /P and repeating the construction above, we obtain a system of p-homogeneous pseudonorms determining a topology equivalent to the original one. D 3.1.5. Let X be a locally pseudoconvex space with a topology given by a sequence of Pn·homogeneous F-pseudonorms {II 11 11 }. A set A C X is bounded if and only if PROPOSITION
On=
sup {JJxJJn:
XE
A}
0
< 1}.
Thus I tn
On~-.
Conversely, let U be an arbitrary neighbourhood of 0. Then there are e > 0 and n such that {x: JJxJJn
< e}
C U.
By (3.1.7) I
e-4c u. On
0
A topological linear space X is called locally convex if there is a basis of convex neighbourhoods of 0. The construction given in Theorem 3.1.4 'leads to the fact that if X is a locally convex metric linear space, then the topology could be determined by a sequence of homogeneous (i.e. !-homogeneous) F-pseudonorms. Homogeneous F-pseudonorms will be called briefly pseudonorms whenever no confusion results. Locally convex metric linear spaces are called B~-spaces. As we shall see later, for locally convex spaces there is also another much simpler construction of a sequence of homogeneous pseudonorms determining the topology. A complete B~-space is called a B0 -space. Let X be an F*-space. If the topology in X can be determined by a sequence of Pn·homogeneous pseudonorms, then the space X is locally
Chapter 3
94
pseudoconvex. In the particular case where the pseudonorms are homo-
geneous, the space X is locally convex. Indeed, the sets - 1- Kn, where m Kn
= {x: //x//n < 1},
constitute a basis of neighbourhoods of 0. The modulus of concavity ofthe set Kn is not greater than 2Pn. Indeed, if x,y E Kn, then llx+y//n ~ 2. Hence
i.e. x+ y E 2P• Kn. Thus Kn +Kn C 2P• Kn. In the particular case where the pseudonorms are homogeneous the sets Kn are convex. It is easy to verify that the following spaces are B0-spaces: LP(Q,E,J1) for 1 ~p ~+oo, C(Q), C(QjQ0 ), the Hilbert space, 0 (D), C 00 (Q), c:S(En). We say that a set A in a metric linear space X is absolutely p-convex, 0
2r.
+ ...
PN(n(tn) (xn, 1 +xn,k))> r and n(tn (xn, 1 + ... +xn,k) ¢A.
Since n(t?i)-+0, we do not have A+A+ ... +ACKA
for any K> 0.
This implies c(A) = +oo. Now we shall consider the case where R = lim N(t) < +=. t-+co
Let !21 be a subset of Q on which the measure ll is non-atomic. Let A be an arbitrary open set in N(L(Q, E, It)) such that r = suppN(x) ze.A
0 such that (3.4.1) holds for u >a. We say that M(u), N(u) are equivalent at 0 if there are b,A,B > 0 such that (3.4.1) holds for 0 < u 0, then the spaces M(L(Q,E,f.-l)) and N(L(Q,E,f.-l)) are identical as sets of functions and, moreover, PM(xn)--*0 if and only if PN(Xn)--*0, i.e. the topologies in this spaces are equivalent. Proposition 3.4.2 and Theorem 3.4.1 imply THEOREM 3.4.3 (Rolewicz, 1959). Let M(u)
=
0
M 0(uP),
00
Un
(3.4.3) there is a sequence {un}--*oo such that N(un)IJp --*OO. Let kn be the greatest integer such that kn ~ N(un), kn = [N(un)]. Since the measure f.-l is not purely atomic for sufficiently small s > 0 there are disjoint s sets An,i, i = 1, 2, ... , kn, such that J.l(An,t) = 1 +kn ·
Chapter 3
114
Let Xn,t =
UnXAn,t•
Then
0, such that for each u, 0 < u < b, there is an infinite family of disjoin sets {An,u} such that 8
rN(u)
0.
1--+<Xl
Choose{) > 0 and T > 0 such that N(t)/t ;;?.: {) for t > T and Tb Then for cot > T, 0 < co < 1, N(t) coN(t). = cot1 -;;?.: Tb
>
>
eq0 •
eq 0 •
Since cot ;;?; Timplies t;;?; T, N(cot) ::'( CcoN(t)forcot Let (Ncot) sup-for t > T, O<wo;;;I CO P(t) = wt>T
>
T, 0 "(co::'( 1.
1
for 0 ::'( t ::'( T.
N(t)
Let 0
< a ::'( 2, at > P(at) =
T. Then
sup
O<wo;;;l
N(coat) N(coat) =a sup co O<wo;;;l coa
wat~T
wat~T
N(st) ::'(a sup - 0T
s
= a P(t).
(3.4.6)
Chapter 3
118
Formula (3.4.6) implies that the function P(t)ft is non-decreasing for t > T. Indeed, let T < t' < t"
t' ) t' ( p -" t" I I PUU) t ::::::::: _t_-:--_ t' """ t'
P(t') t'
Let
l
P(t)
Q(t)
=
for t
>
for 0
< t~T
;(T)
tT2
P(t")
=--ru-··
T,
and u
M(u)
=
JQ(t)dt. 0
Since Q(t) is a non-decreasing function, the function M(u) is convex. We shall show that the functions M(u) and N(u) are equivalent at infinity. To begin with, we shall prove that P(t) and N(t) are equivalent at infinity. Indeed, P(t);;?: N(t) and for sufficiently large t, N(wt)/w ~ CN(t), hence P(t) ~ CN(t). Since P(t) is equivalent to N(t) at infinity and N(t) satisfies condition (A 2), P(t) also satisfies that condition, i.e. there is a positive constant K such that for sufficiently large t, P(2t) ~ KP(t). The function Q(t) is non-decreasing, whence~ for sufficiently large t, M(t)
< tQ(t) =
P(t)
and M(t);;?o
J'Q(s)ds;;?o ~ Q(~)=P(~). t/2
Hence, for sufficiently large t, P(t)
~ KP ( ~) ~ KM(t)
and the
functionsM(t) andP(t) are equivalent at infiriity. Therefore the functions M(u) and N(u) are equivalent at infinity. 0 THEOREM 3.4.10 (Mazur and Orlicz, 1958). Suppose that there is a constant K > 0 and an infinite family of disjoint sets {Et} such that 1 ~JJ.(E,)
~
K.
Locally Pseudoconvex and Locally Bounded Spaces
119'
If the space N(L(Q,E,p.)) is locally convex, then the function N(u) is equivalent to a convex function at 0. Proof. Since the space N(L(Q,E,p.)) is locally convex, there is a positive e such that PN(Xk) < e, k = 1,2, ... , n, implies
PN ( x 1 + ·~· +xn)
~
1.
Let p be an arbitrary positive integer and let
{~
xk(s) =
We get PN(Xk) and
~
for sEEk+tn (i=0,1, ... ,p-1), otherwise. (k=1,2, ... ,n)
KpN(t)
__.. pn N ( - t ) ~PN n
(x + ...n +Xn) · 1
Hence N(t)~ efKp implies N(t)/n ~ 1/pn. Let 0 < N(t) < efK, and let p be chosen so that efK(p+1) ~ N(t) < efKp. Then N(tfn) ~ 1/pn ~ 2Kfe N(t)fn. Given p > 0. Choose a q ~ 1 such that N(tfq) < e for it I < p. Condition (LlJ implies that there is a constant D~ such that N(qt) ~DqN(t)for Jti 0. fl
a
Since, if a space X is not locally convex, there are continuous functions which are not integrable, it is reasonable to ask what classes of functions are integrabl• in what classes of spaces.
Chapter 3
124
An important class of functions is exemplified by the class of analytic functions. We say that a function x(t) with values in an F-space X is analytic if, for any t 0 , there is a neighbourhood U of the point t0 such that, forte U, the function x(t) can be represented by a power series 00
x(t)
=;}; (t-t )nxn, 0
(3.5.3)
ii,xn EX.
·n=O
If the space X is locally pseudoconvex, then the topology can be determined by a sequence of Pn-homogeneous pseudonorms II lin· This implies that if the series (3.5.3) is convergent at a point t 1 , then it is convergent for all t such that !t-t0 ! bJ)· N,ow. we ~hall consider the Rie-
j=l
mann integral of the func:ion x(t) on the interval [a1,bj]. Let e be an
Locally Pseudoconvex and Locally Bounded Spaces
arbitrary positive number. Let n0 be such an index that Mk/2no Let x(t) = Yn 0 (t)+zn 0 (t), ai ~ t ~ b,, where
...
Yno (t)
125
< e/2.
00
=}; (t-Sj)n Xn,
Zn0 (t)= }; (t-SJ)nxn. n=n0 +1
n=O
The function Yno(t) is integrable, because it has values in the finite-dimensional space spanned by elements x 0 , ••• , Xno· For any subdivision of the interval [aJ>bJ] a1
=
< '11 < ...
bi] with respect to the pseudonorm II Ilk· Since there is only a finite number of the sets V8~, those sums converge on the whole interval [a,b]. This holds foreverypseudonorm II Ilk, whence the completeness of the space X implies that the function x(t) is integrable. 0 Let C be a complex plane. Let D be an open set in C. Let x(z) be a function defined on D with values in a locally pseudoconvex F-space X. In the same way as for the function of the real argument, we say that x(z) is analytic in D, if for each z 0 ED, there is a neighbourhood Vz0 of the point z 0 such that the function x(z) can be represented in Vz0 as a sum of power series 00
x(z)
=}; (z-z )nxn, 0
n=O
XnEX.
Chapter 3
126
As in the real-argument case, we can define the Riemann integral of a function x(t) on a curve contained in D. In the same way as in the proof of Theorem 3.5.2 we can show that, if the curve Tis smooth and the function x(z) is analytic, then the integral x(z)dz exists. r Using the same arguments as in the classical theory of analytic functions, we can obtain
J
THEOREM 3.5.3. Let Then
r
be a smooth closed curve. Let x(z) be analytic.
Jx(z)dz = 0, _
(3.5.4)
J
r
x(z) - dZ
1
( 0) - XZ 21t
r
z-z 0
Or each Zo inside the domain SUrrounded by
x Where Zo
(n) (
) _
Zo -
1
21t
J
x(z)
(z-zo)n+l
d
r,
z,
(3.5.5) (3.5.6)
r and Fare as in (3.5.5).
As a consequence of (3.5.6) we obtain the Liouville theorem: THEOREM 3.5.4. Let x(z) be an analytic function defined on the whole complex plane C, with values in a locally pseudoconvex space X. If x(z) is bounded on the whole plane, then it is constant. Turpin and Waelbreock (1968, 1968b) generalized Theorem 3.5.2 to m dimensions and to a more general class of functions. We shall present their results here without proofs. Let U be an open domain in an m-dimensional Euclidean space Rm. Let Nm denote the set of all vectors k = (k1 , ••• , km), where kt are non-negative integers, (i = 1 ,2, ... , m). We adopt the folowing notation
Jxl = Jkl =
sup
l.,;t.;;m m
!xtl,
.L lktl' i=l
Locally Pseudoconvex and Locally Bounded Spaces
n
127
m
xk =
x~·.
i=l m
k! =
flkd, i=l
where x = (x1, .•. , Xm) E Rm, k = (kl> ... , km) E Nm. We say that a functionf(x) mapping U into an F-space X is of class Cr(U,X), where r is a positive number, if, for any k E Nm such that lkl ~ r, there are continuous mappings Dk f of U into X and Pr,k f of Ux U into X such that D0 f=J, Pr,kf(x,x) = 0 and
Dd(y)=
~
.L.J
Dk+hf(x)•
lhl~r-lkl
(y-x)h 'h! +iy-xir-lklpr,kf(x,y) •
(see Turpin and Waelbroeck, 1968). THEOREM 3.5.5 (Turpin and Waelbroeck, 1968b). Let X be a locally p-convex space. Let U be an open set in Rm. If f(x) E C,(U,X) for r > m(l-p)fp is a boundedfunction, then it is Riemann integrable. We say that a functionf(x) is of class C00 (U, R) if it is of class Cr(U, X) for all positive r .
. THEOREM 3.5.6 (Turpin and Waelbroeck, 1968b). If the space X is locally pseudoconvex, then each bounded function f(x) E Coo (U,X) is Riemann integrable. Indeed, Turpin and Waelbroeck obtained an even stronger result, namely that the integration is a continuous operator from the space C,( U, X) (Coo ( U, X)) with topology of uniform convergence on compact sets all Dkf, lkl ~r, andpr,kf(all Dkfand allpr,kf) into the space X.
3.6. VECTOR VALUED MEASURES Let (X, II II) be an F-space. Let Q be a set, and let .E be a a-algebra of subsets of Q. We consider a function M(E) defined for E E .E with
128
Chapter 3
values in X such that, for any sequence of disjoint sets E1o ... , En, ... belonging to E M(E1 u
... uEnu ... ) = M(EJ+ ... +M(En)+ ...
The function M (E) is called a vector valued measure (briefly a measure). By the variation of the measure Mona set A C Q we shall mean the number M(A) =sup {liM(H)Ii: He A, HeE}.
THEOREM 3.6.1 (Drewnowski, 1972). The variation M of a vector valued measure has the following properties : M(e) = 0,
(3.6.1.i)
if A C B, then M(A) M(Q) 1+IIM(Hl)ll+ ··· + IIM(Hn-t)li.
The sets Hn = Hn"'-(H1 u
... UHn-t)
A
are disjoint and IIM(Hn)li
ro
> 1. Thus the series };
n=l
M(Hn) is not con-
vergent, and this leads to a contradiction. ro
(3.6.1.iv). Let A be an arbitrary set contained in
U Hn.
n=l
Let An
Locally Pseudoconvex and Locally Bounded Spaces
129 00
= (A" (H1 u
... u Hn-J) () Hn The sets
An are disjoint and A =
U An. n-1
Then 00
jjM(A)IJ
00
00
II}; M(An) II~}; IJM(A,.)IJ ~}; M(Hn).
=
n=l
n=l
n=l
00
The set A is an arbitrary subset of the set
U Hn.
n=l
Thus 00
M{H1 UH2u ... ) = sup{IJM(A)IJ: AC
U Hn} n=l
00
~}; M(Hn). n=l
(3.6.1.v). Let Hn E l: be an increasing sequence of sets. Thus Hn C H and by (3.6.l.ii) M(Hn) ~ M(H), n = 1,2, ... This implies that lim M(Hn) ~ M(H). Let e be an arbitrary positive number. Let A C H, A e 1: be such that M(H)-e
0}.
Observe that the n-th coordinate of M(En) = 1/TC. This implies that the sequence M(En) does not contain any convergent subsequence. Thus the measure M(E) is not compact. Let f(t) be a scalar valued function defined on Q. We say that the· functionf(t) is measurable if, for any open set U contained in the field of scalars, the setf-1(U) belongs to .E. A sequence of measurable functions {fn(t)} is said to tend to a measurable function f(t) almost everywhere if lim fn(t) = f(t) for all t except a set E such that M(E) = 0. n->oo
PROPOSITION 3.6.7 (Drewnowski, 1972).
If fn
tends almost everywhere to
J, then for each e > 0 00
lim k->oo
M(U
En (e)) = 0,
n=k
where En(e) = {x: //fn(x)-f(x)// ~ e}. Proof The sequence
is a decreasing sequence of measurable sets. Then by (3.6.1.v) lim M(An) =
M (lim
An) = M (lim supEn(e)). n->oo
Observe that if x
E
limsupEn(e), then the sequence fn(x) is not convern
gent to f(x). Thus lim supEn(e) C A= {x: fn(x) is not convergent tof(x)}. By our hypothesis M(A) = 0. Thus, by (3.6.l.ii),
M(lim sup En(e))~ M(A) =
0.
D
11->CO
As a consequence of Proposition 3.6.7, we obtain extentions of the classical theorems of Lebesgue and Egorov
Chapter 3
134
3.6.8 (Lebesgue). If {fn(x)} tends to f(x) almost everywhere, then, for each e > 0,
THEOREM
lim M({x: 1/fn(x)-f(x)// ~ e})
=
0
n---+ro
3.6.9 (Egorov). If {/n(x)} tends to f(x) almost everywhere, then for each e > 0 there is an FE .E such that {fn(x)} tends to f(x) uniformly on F. Proof By Proposition 3.6.7, for each k there is an index Nk such that THEOREM
(3.6.2)
(3.6.3)
M(D"'F) =
M((J
0 En(-k ))~t M( 0 En(-k 1
k-l n-N•
k=l
n-N•
1 ))
OC>
~2
;k =e.
k=l
We shall show that {fn(x)} tends uniformly to f on th set F.· Indeed, let 'YJ be an arbitrary positive number, let k be such that 1/k < 'YJ· If x E F, then
and //fn(x)- f(x)//
or n
~
1
< -k
0 there is a set A such that M(D"'A) < 'YJ and {gn} converges uniformly on A. Since g is a simple function, there is an 'YJ > 0 such that for each setH such that M(H) < 'YJ we have I\
JsgdMII
lsg(t)l for tEA. The arbitrariness of e and simply (3.6.8.iv). (3.6.8.v). To begin with, we shall assume that ft and / 2 are simple. Let g(t) be an arbitrary simple measurable function such that lg(t)l ~ l.h(t)+};(t)l. Let gt(t)
=
J
l
g,(t) lft(t)j
~ft(t)l+l/2(t)l
if lft(t)l+lf2(t)l =fo 0, i = 1,2. if lfl(t)l+lf2(t)l = 0.
The functions g 1(t) and g 2(t) are simple and g(t) = gl(t)+g2(t).
Hence II
JgdMII~II nJgldMII+II nJg2dMII~M·(fl)+M·(.t;).
n
The arbitrariness of g implies
Suppose now that.ft and,/; are not simple. Then there are two sequences of simple mesurable functions {.h,n} and {h,n} tending almost everywhere to / 1 and}; and such that lfi.nl ~I AI, i = 1,2. Therefore, by (3.6.8.iv),
Let B(f) = (
~
limsup M· (ft.n+An)
~
supM ·(ft,n)+supM ·(h,n)
~
M ·(h)+M·(JJ.
n
n
JgdM: g simple measurable functions, n
0 igl
~Ill}
Locally Pseudoconvex and Locally Bounded Spaces PROPOSITION
145
3.6.18. The set B(f) is bounded if and only if
1imM ·(if) = 0.
(3.6.9)
~0
Proof If the set B(f) is bounded, then limM· (if)= lim (sup{llxll: x t--+0
E
B(tf)})
t-->-0
= lim(sup{llxll: x
E
tB(f)}) = 0.
(3.6.10)
t-->-0
Conversely, if (3.6.9) holds, then by (3.6.10) the set B(f) is bounded. 0 Let X denote the set of those /for which B(f) is bounded. By (3.6.8.v) and (3.6.9), M· (f) is an F-pseudonorm on X. We shall now use the standard procedure. We take the quotient space X/{f: M·(/)=0}. In this quotient space M ·(f) induces an F-norm. We shall take completion of the set induced by simple functions. The space obtained in this way will be denoted by D(D,E, M). Since, for each simple function g,
II
JgdM[J ~ M·(g).
{l
We can extend the integration of simple functions to a linear continuous operator mapping D(D,E,M) into X.
3.7.
INTEGRATION WITH RESPECT TO AN INDEPENDENT RANDOM MEASURE
Let (D0,E0,P) be a probability space. Let Q be another set and let E be a a-algebra of subsets of D. We shall consider a vector valued measure M(A), A E E, whose values are real random variables, i.e. belong to X= L 0(D0 1:0 P). We say that M(A) is an independent random measure if, for any disjoint system of sets {AI> ... , An}, the random variables M(At) EX, i = 1,2, ... , n are independent. We recall that a vector measure M is called non-atomic if, for each A E D such that M(A) =1=- 0, there is a subset A0 C A such that M(A 0) =1=- M(A).
Chapter 3
146
Let X(w) be a random variable, i.e. X(w) we denote Fx(t) = P({w: X(w) ~ t}). The function Fx(t) is non-decreasing, lim
E
L 0(£2 0 ,.E0 ,P). By Fx(t)
Fx(t)
= 0, limFx(t) = I.
t-+- ro
I->-+ ro
It is called the distribution of the random variable X(w).
Let Q = [0, 1], and let .E be the algebra of Borel sets. We say that a random measure M is homogeneous if for any congruent sets A 1 A 2 C C [0, 1] (i.e. such that there is an a such that a+A 1 = A2 (mod 1)) the random variables M(A 1) and M(A 2) have identical distribution. It is not difficult to show that, if M is a homogeneous independent random measure, then for each n we can represent M(A) as a sum M(A) = M(A 1)+ ... +M(An), where the random variables M(Ai), i = 1,2, ... , n, are independent and have the same distribution (Prekopa, 1956). Let X(w) be a random variable. The function +ro
fx(t) =
J
eits dF(s).
(3.7.3)
-ro
is called the characteristic function of the random variable X(w ). If the random variables X1 , ••• , Xn are independent, then fxi+ ... +Xn(t)
= fxit)· ··· -fxn(t).
(3.7.4)
Directly from the definition of the characteristic function !ax(t) =fx(at).
(3.7.5)
We say that a random variable X(w) is infinitely divisible if for each positive integer n there is a random variable xn such that X= Xn+ ... +Xn,
(3.7.6)
in other words, by (3.7.4). fx = (/xn)n.
(3.7.7)
If X is an infirute-divisible random variable, then its characteristic function/x can be represented in the form
fx(t) =
exp(iyt+
]"'(eitx_1- 1 ~xx2 ) 1 ~2x2 dG(x)),
-00
(3.7.8)
Locally Pseudoconvex and Locally Bounded Spaces
147
where y is a real constant, G(x) is a non-decreasing bounded function and we assume that at x = 0 the function under integration is equal to - t 2/2. This is called the Levy-Kchintchin formula (see for example Petrov (1975)). If X is a symmetric random variable (which means that X and -X have this same distribution), by (3.7.8) we trivially obtain co
fx(t)
= exp
J
l+u2 (costu-1)----zt2 dG(u).
(3.7.9)
0
Of course, without loss of generality we may assume that G(O) = 0. Suppose that M(A) is a non-atomic independent random measure with symmetric values. Then by (3.7.9) co
fM(t)
= exp
J
1+u2 (cos tu-1)----uz-dGA(u).
(3.7.10)
0
For an arbitrary real number a we obtain by (3.7.5) and (3.7.10) co
/aM(A)
=
exp
J
1+u2 (cos tu-1) ----zt2 dGA(u).
(3.7.11)
0
Let A 1 ,A 2 E I be two disjoint sets. By (3.7.4) (3.7.12) Formulae (3.7.11) and (3.7.12) imply that for a simple real-valued function h(s) fjh(s)dM(t) u
= exp(-
JTM(th(s))ds),
(3.7.13)
0
where (3.7.14) In the sequel we assume Q = [0, 1]. Now se shall prove some technical lemmas.
Chapter 3
148
Let co
UM(x) =
J
min ( xll,
-~ll) {1 +ull)dGM(u)
(3.7.15)
0
and for x
> 0,
(3.7.16)
for x =0. It is easy to see that the two functions Uy(x) and 'Py (x) are equal
to 0 at 0, continuous and increasing. LEMMA 3.7.1. For all x;;;: 0, a;;;: 0, there are positive numbers c1(a) and ell such that max TM(v)::::;;;c 1(a)UM(x) (3.7.17) O~v~ax
and 1
f TM(xt)dt ;;;: c UM(x).
(3.7.18)
2
0
Proof. Observe that
1-cosuv = 2sinll
z;;:::::;;;-} u v
2 2
and 1-cosuv:::::;;; 2. Hence, for c1{a)
=
max(2,-} all), we have
max (1-cosuv):::::;;; c1 (a)min(1,x2u2). O~v~ax
Therefore ( ) . ( ll 1 ) max 1-cosuv :::::;;; c1 a mm x , - 2 • 2
o,;;;v,;;;ax
U
(3.7.19)
U
Integrating (3.7.19) with respect to 1+ull dGy we obtain {3.7.17). ull Observe that sinz . (z, 2 1) 1---;;;: cllmm ,
z
Locally Pseudoconvex and Locally Bounded Spaces
149
where · ( mm . ( 1·- sinz) . z- 2( 1- sinz)) c2 =mm - , mm - . Z
Z>1
0(x)
=
(3.7.22)
G;;) du.
1/X
Then the function ([>(yx) is concave. Proof We apply a change of variable v = l/u 2 in formula (3.7.22), and obtain ([> (yx) =
f G~~) jG( ylv) du =
1/y;
dv.
0
Thus the derivate of ([>(yx) is equal to G(I/yx). It is non-increasing, since G is non-decreasing. Therefore the function ([>(yx) is concave. D 3.7.3 (Urbanik and Woyczyiiski, 1967). Let {In} be a sequence of simple measurable functions belonging to L 0 [0, 1]. The sequence
LEMMA
1
Jfn(s)dM(s) tends to 0 in L [0, 1] 0
0
if and only if 1
J'l'M(/fn(s)l)ds = 0.
lim
n-+oo 0
Proof We shall use the classical statement that a sequence {hn} of functions belonigng to L 0 [0, I] tends to zero in L 0[0, I] if and only if the logarithms of its characteristic functions tend uniformly to 0 on 1
each compact interval. Thus by (3.7.13) {
Jfn(s)dM(s)} tends to zero in 0
1
L 0 [0, 1] if and only if { jTM(fn(s))ds} tends to Ouniformly on each como
pact interval [0, T]. 1
If~
JTM(tfn(s))ds} tends uniformly on each compact interval, then 0
1
lim
1
J JTM(ifn(s))dtds =
n-oo 0
0
1
lim ~oo
1
J JTM(ifn(s))dsdt = 0
0
0.
Locally Pseudoconvex and Locally Bounded Spaces
151
Hence by (3.7.18) 1
lim
JUM(Ifn(s)l)ds = 0.
(3.7.23)
n-+oo 0
Since UM(x) and IJfM(x) are equivalent at infinity, 1
lim
J'PM(Ifn(s)l)ds = 0.
(3.7.24)
x--+co 0
Conversely, if (3.7.24) holds, then (3.7.23) holds. By (3.7.17) we find 1
that { JrM(tfn(s))ds} tends to Ouniformly on each compact interval. 0 0
THEOREM 3.7.4 (Urbanik and Woyczyiiski, 1967). Le M be a symmetric homogeneous independent random measure with values in the space L 0[0, 1]. Then there exists a continuous function N(x) defined on [O,+oo) such that N(O) = 0, N(J./i) is concave and such that the set of all real 1
functions for which the integral J fdM exists is the space N(L). 0
Conversely, if N(yx) is a concave continuous function such that N(O) = 0, then there is a symmetric independent homogeneous random measure M with values in the space L 0 [0, 1], such that the set of real val1
ued functions f for which the integral J fdM exists is precisely the space 0
N(L). Proof The first part of the theorem follows immediately from Lemmas 3.7.2 and 3.7.3. Conversely, if N(yx) is concave, then it can be represented in the integral form 0. Let M be a symmetric independent homogeneous random measure such that
152
Chapter 3
(3.7.10) holds. To complete the proof it is enough to show that the functions N(x) and :~:,
00
NG(x)
J G~~)
=
du
=
1/Z
J
min (I ,q(u))du
0
are equivalent at infinity. Since q(u) is non-increasing, q(u) ~ q(1) for u > 1. Thus z
1
0 there is an index N such that for each sequence either 0 or 1 00
l
2;
BnXn
n=.N+l
I
2
Xn
is unconditionally convergent. Then
n=l
0 there is an index N such that
k+m
IIJ:
enXn
.n=k
fork> N, m
>
I
K and s ;;::;: 0 r+s
II~ Xp(n) I! = n=r
>
K, p(n)
>
N. Then for
q
114 I < 8 tXt
(3.8.2)
8•
•=p
where p = inf {p(n): r ~ n ~ r+s}, q =sup {p(n): r ~ n ~ r+s}, t = {1 if i = p(n) (r ~ n ~ r+s), 8 0
otherwise.
The arbitrariness of 8 implies that the series
2"'
Xp(n)
is convergent.
n=l
Suppose now that the series 2"'
Xn
is not unconditionally convergent.
n=l
This means that there are a positive number tJ and a sequence {8n}, taking the value either 1 or 0 and a sequence of indices {rk} such that r•+l
il ~
8 nXn n=r•+l
8
I > J.
Now we shall define a permutation p(n). Let m be the number of those 8n, n = rk+ 1 , ... , rk+I• which are equal to 1. Let p(rk+v) = n(v), where n(v) is such an index that 8n(v) is a v-th 8i equal to 1, rk < i ~ rk+I• Q < v ~ m. The remaining indices rk < n ~ rk+1 we order arbitrarily. Then r.+m
r•+l
n=r•
n=r.+l
II~ Xp(n)ll =II~
This implies that the series
8
nXnll >
J.
2"'
is not convergent.
xp(n)
D
n=l
A measure M induced by a series "'
2
Xn
is L"' -bounded if and only if
n=l
for each bounded sequence of scalars {an} the series 2"'
anXn
is con-
n=l
-vergent. The series with this property will be called bounded multiplier convergent.
Locally Pseudoconvex and Locally Bounded Spaces
155
THEOREM 3.8.3 (Rolewicz and Ryll-Nardzewski, 1967). There exist an 00
F-space (X,
II II) and an uncoditionally convergent series}; Xn
of elements
n=1
of X which is not bounded multiplier convergent.
The proof is based on the following lemmas. LEMMA 3.8.4. Let X beak-dimensional real space. There exists an open symmetric starlike set A in X which contains all points PI> ... , p 3k of the type (e~> ... , ek), where ei equals 1 or 0 or -1, such that the set Ak-1 =A+ ... +A (k-1)-fold
does not contain the unit cube C=
{(a1,
... ,
ak):
lail
... , Ps•· Obviously the set A~- 1 is (k-1)-dimensional. Therefore there is a positive number e such that the set (A 0 +A.)k-\ where A. denotes the ball of radius e (in the Euclidean sense), has a volume less than 1. Thus the set A= A 0 +A. has the required property. D
LEMMA 3.8.5. There is a k-dimensional F-space (X, II II) such that IIPill ~I, i = 1,2, ... , 3k and there is a point p of the cube C such that IIPII;;:, k-1. Proof We construct a norm II II in X in the way described in the proof of Theorem 1.1.1, putting U(l) =A. Since PtE A = U(I), IIPill ~ 1. Furthermore, since Ak- 1 = U(k-I) does not contain the cube C, there is a point p E C such that p E U(k-1). This implies that II PII ;;:, k-1. D Proof of Theorem 3.8.3. We denote by (Xk, space constructed in Lemma 3.8.4. Let 1
II
II~) the 2k dimensional
'
llxllk = 2k llxllk· Let X be the space of all sequences a
= {an} such that
00
lllalll
=}; ll(a2•-•+t• ... , a2~)/lk-1+la1l 6/2, the sequence {tn} tends to 0. On the other hand, (zp} is a C-sequence. Therefore, by (3.10.3), {z~} is also a C-sequence. co
This means that if tn-+0, then the series}; tpz~ is convergent. Therefore, p=l
the basis z~ is equivalent to the standard basis of c0 • Thus, by Theorem 2.6.3, the space X 0 is isomorphic to the space c0 • D PROPOSITION 3.10.4. Let (X, II II) be a locally bounded space. with a basis {en}. Let Y be a subspace of the space X. Suppose Y is not a C-space. Then the space Y contains a subspace Y0 isomorphic to the space c0 • Proof Without loss of generality we may assume that I II is p-homogeneous. Using the same construction as in the proof of Theorem 3.10.2, we find elements {zp} and {z~}. By formula (3.10.3) and Theorem 3.2.16 the sequence {zp} is a basic sequence. The rest of the proof is the same as in Theorem 3.10.2. D Let X be a non-separable F-space. If Xis not a C-space, then it contains an infinite dimensional separable subspace X which is not a c~space. Indeed, if X is not a C-space, then there is a C-sequence {xn} such that 00
the series };
Xn
is not convergent. Let
X= lin{xn}. The space X has the
n=l
required properties. By the classical theorem of Banach and Mazur (1933), space C[O, I] is universal for all separable Banach spaces. The space C[O, 1] has a basis (Example 2.6.10). Thus we have PROPOSITION 3.10.5 (Bessaga and Pelczynski, 1958). If X is a Banach .space which is not a C-space, then X contains a subspace isomorphic to c0 • Now we shall prove
Locally Pseudoconvex and Locally Bounded Spaces
THEOREM 3.10.6 (Schwartz, 1969). The spaces LP(D,E,p), 0 ~ p are C-spaces.
169
< +oo,
The proof is based on the following propositions. PRoPOSITION 3.10. 7 (Kolmogorov-Kchintchin inequality ; see also Orlicz, 1933b, 1951, 1955). Suppose that in the space L 0(D,E,p,) a C-sequence {x,.(t)} is given. Then on each set D 0 of finite measure the series 00
}; lx,.(t)l 2 convergent almost everywhere. n=l
Proof(Kwapien, 1968). Let {r,.(s)} = {sign(sin2n211s)} be a Rademacher system on the interval [0, 1]. Since under our hypothesis {x11 (t)} is a C-sequence, the set of elements
"
{ Yn,s(t)
=}; rt(s) Xt(t):
n = 1, 2, ... , 0
~ s ~ 1}
i=l
is, by Proposition 3.10.1, bounded in the space L 0 (D,E,p). This implies that for any positive B there is a constant C such that, for all n = 1,2, ... and all s, 0 ~ s ~ 1, ~t({te
D 0 : IYn,s(t)i ~ C}) ~ ~t(D0)-B.
(3.10.4)
Let n be fixed al!d let
At= {s: IYn,s(t)i ~ C}.
(3.10.5)
By the Fubini theorem and (3.10.4) we find that the set E of those t for which the Lebesgue measure IAel of the set At is greater than l-ye has a measure greater than ~t(D0)-ye, i.e. if E
= {t e D0 : jAel
~
2-y'e},
then ~t(E) ~ ~t(D 0)-ye.
Lett e E. Formula (3.10.5) implies n
JI}; rt(s) Xt(t)r ds ~ C
Ao
i=l
2•
Chapter 3
170
Thus n
J _2/xt(t)/ 2 -2 J ( _2 A, i=l
A,
Re(xt(t)x,(t))rt(s) r;(s)) ds
~ C2 •
(3.10.6)
l~i<j~n
Applying the Schwarz inequality to (3.10.6), we obtain n
C ;;?: (1-y'B)}; /xt(t)/ 2 2
i=l
-2 (
_2
/xt(t)x1(t)J2) 112 (
l~i<j~n
_2
( Jrt(s)r1(s)dsrf' 2.(3.10.7)
l~i<j~n
A,
Since rt is a Rademacher system, J rt(s) r;(s)ds = A•
(3.10.8)
J ri(s) r,(s)ds, 1'\..A•
where I denotes the interval [0, 1]. The system {ri(s)rJ(s)}, 1 ~ i <j, is an orthonormal system in the space £2[0, 1]. The Fourier coefficients of the characteristic function X['\..A• of the set /~At are equal to fri(s)r,(s)ds, thus equalto- Jri(s)rJ(s) 1'\..A•
A,
ds, by (3.10.8). Therefore, by the Parseval inequality. _2
(
l~i<j~n
f rt(s) r,(s)dsr ~/!"'At/ nr-1 such that
Jlxnlt)l dp ~ 2~,
Br
and
J lxnr(t)IP dp >b.
(3.10.13)
t:J"-Br
Formula (3.10.13) implies that there is an index m, and a set A, C Qmr disjoint with the sets A 0 , ••• , Ar-1 such that
Jlxnr(t)IPdp >b.
(3.10.14)
Ar
In this way we have constructed by induction the required sequence. Let {en} be a sequence of positive numbers tending to 0 such that co
,2;c: = +oo.
(3.10.15)
n=1
Then r
r
r
f\2: CkXn.(t)\P dp ~ 2 J12: CkXn.(t)\p dp (J
k=O
j=O A1
r
k=O
r
J(cJixnlt)IP- L cflxn.(t)IP)dp
~};
j=OA1
k=O k#j
r
~b
2
j=O
r
cJ- C
2 2;~k
j,k=O
r
~ b}; cJ-4C,
(3.10.16)
;~o
where C =sup c!. n co
Then, by (3.10.15), the series}; Ck xn. is not convergent in LP(Q,E,p). k=1
Therefore, {xn} is not a C-sequence and we obtain a contradiction.
0
LEMMA 3.10.10 (Schwartz, 1969). If {xn} is a C-sequence in a space LP(Q,E,p), 1 ~p ..• , Zn) be an analytic function ofn variables defined on an open set U containing the set a(x)
=
{(x1(t), ... ,Xn(t)): 0
~
t ~ 2n}
a(x) = {x1(z), ... , Xn(z)): lzl ~ 1}).
Then the function <J>(xl> ... , Xn) belongs to NF (or respectively, to NH).
We shall not give here an exact proof. The idea is the following. Replacing the Cauchy integral formula by the Weyl integral formula, we can define analytic functions of many variables on complete locally bounded algebras. ·
3.12.
LAW OF LARGE NUMBERS IN LOCALLY BOUNDED SPACES
Let (Q, .E, P) be a probability space. Let (X, [[ I[) be a locally bounded space. Let the norm II II be p-homogeneous. As in the scalar case, a measurable function X(t) with values in X will be called a random variable. We say· that two random variables X(t), Y(t) are identically distributed if, for any open set A C X P({t: X(t)
E
A})= P({t: Y(t)
E
A}).
A random variable X(t) is called symmetric if X(t) and -X(t) are identically distributed. Random variables X 1(t ), ... , Xn(t) are called independent if, for arbitrary open sets AI> ... , An
n n
P({t: Xt(t) EAt, i
=
1, ... , n})
=
P({t: Xt(t) EAt}).
i=l
A sequence of random variables {Xn(t)} is called a sequence of independent random variables if, for each system of indices n1o ... , nk, the random variables Xn 1 (t), ... , Xn.(t) are independent.
Chapter 3
184
3.12.1 (Sundaresan and Woyczynski, 1980). Let X be a locally bounded space. Let II II be a p-homogeneous norm determining the topology in X. Let {Xn(t)} be a sequence of independent, symmetric, identically distributed random variables. Then
THEOREM
E(IIX1ID
=
JIIXl(t)lldP < +oo
(3.12.1)
a
if and only if (3.12.2)
almost everywhere. Proof. Necessity. To begin with we shall show it under an additional hypothesis that X1 takes only a countable number of values x1,x2 , ... Since Xn are identically distributed, all Xn admit values x 1, ... For each positive integer m we shall define new random variables Xk'(t)
= { ~k(t)
if Xk(t) = x 1 , elsewhere.
... , Xm,
By Rk'(t) we shall denote Xk(t)-xr(t). For each fixed m, {IIXrll} constitutes a sequence of independent identically distributed symmetric random variables taking real values. Moreover, (3.12.3)
The random variables {xr} takes values in a finite-dimensional space. Thus we can use the strong law of large numbers for the one-dimensional case (see for example Petrov, 1975, Theorem IX.3.17). Let an = n11P. Then CX)
.}; ak-2 n=k
CX)
= .}; n-2/p n=k = O(n- 2/P+l) =
O(na;: 2 ).
(3.12.4)
Having (3.12.3) and (3.12.4), we can use the strong law of large numbers by coordinates (here we use the fact that xr takes values in a finite dimensional space).
Locally Pseudoconvex and Locally Bounded Spaces
185
Thus (3.12.5) almost everywhere. At the same time, IIR::'II is a sequence of independent identically distributed real random variables with finite expectation, so that, by the classical strong law of large numbers,
IIR~II+ ... +IIR::'II -+E(IIR~II) n
(3.12.6)
almost everywhere. Since IIR~II tends pointwise to 0 as m tends to infinity and IIR~II ~ !lXIII, by the Lebesgue dominated convergence theorem E(IIR~II) tends to 0. The set D 0 of those t for which (3.12.5) and (3.12.6) converge at t is of full measure, i.e. P(D0 ) = P(D) = 1 Let 8 > 0. Choose an m > 0 such that m
8
E(IIR1 ID < 4· For any t E D 0, we can find anN= N(8,t,m) such that
lln-liP(X~(t)+ ... X~(t))ll < ; for n > N. Thus, for t E D 0 and n
(3.12.6)
> N,
This completes the proof under the condition that Xn are countable valued. To complete the proof of necessity we shall use the standard approximation procedure. For each 8 > 0, there is a symmetric Borel function T, taking values in a countable set in X such that
IITs(x)-xll
0:
llxllu
~ E u}.
Evidently, (4.1.1)
llxllu:?::-0 and U
= {x: llxllu < 1}.
Moreover lltxllu
= tllxllu for t > 0
(positive homogeneity)
(4.1.2)
and llx+yllu ~ llxllu+IIYIIu.
(4.1.3)
Formula (4.1.2) is trivial. We shall prove formula (4.1.3). Let e be an arbitrary positive number. The definition of llxllu implies and The set U is convex; therefore llxllu x IIYIIu Y (1-e) llxllu+II.YII~ llxllu- +(1-e) llxllu+II.Ylk7" IIYIIu
x+y
= (1-e) llxllu+IIYIIu
E
U.
Since e is an arbitrary positive number, we obtain llx+yllu =:;::: 1 llxllu+IIYIIu ""' ' and this implies (4.1.3). A functional satisfying conditions (4.1.1)-(4.1.3) is called a Minkowski functional. If we replace (4.1.2) by llxllu
= ltlllxllu
for all scalars t,
(4.1.2')
then a Minkowski functional becomes a homogeneous pseudonorm (see Section 3.1). Let us remark that llxllu is a homogeneous pseudonorm if and only if the set U is balanced.
Existence of Continuous Linear Functionals and Operators
189
Let X be an F*-space. Let f(x) be a linear functional defined on X. If there is an open convex set U containing 0 such that Jf(x)J
< llxllu,
then the functionalf(x) is continuous. On the other hand, if f(x) is a continuous linear functional and
{x: Jf(x)J < 1 }.
U=
then lf(x)J
< llxllu.
THEOREM 4.1.1 (Hahn 1927, Banach 1929). Let X be a rea/linear space. Let p(x) be a real-valued functional (generally non-linear) such that: (1) p(x+y) 0 (positive homogeneity). Let X 0 be a subspace of the space X. Let fo(x) be a linear functional defined on X 0 such that fo(x)
< p(x).
Then there is a linear functionalf(x) defined on the whole space X such •that f(x) =fo(x)
for x
f(x) 0.
t Then in the space N(L(Q,E,p.)) there are non-trivial continuous linear functionals. Proof Corollaries 4.1.2 and 4.1.3 imply that it is enough to show that. in the space N(L(D,E,p.)) there is an open convex set U different from the whole space. Let E E E, where p.(E) = a, 0 < a < +oo. Since 1->-00
lim inf N(t) 1->-00
> 0, there are a positive constant a and a posii:ive number
t
T such that, for t > T, N(t) >at. Let U be a convex hull of the set {x: PN(x) < 1}. The set U is an open convex set. We shall show that it is different from the whole space N(L(Q,E,p.)). Let x1 , ••• , Xn be arbitrary elements such thatpN(xt) ~ 1, i = 1,2, ... , n Let Bt = {s: lxi(s)l > T}. Let Xt(s) xi(s) = { 0
for SE Bt, elsewhere,
and
Let
,
xi+ ...
+x~
,
xi'+ ...
+x~'
Xo = --=----and xo = --=-------'-'n n Since lx;'(s)l~ T(i= l,2, ... ,n),lx"(s)I~T. It implies that (lx~'(s)ldp. il ~
Ta. Moreover a
Jlxi(s)ldp. ~JE N(lxi(s)ldJi.~PN(Xt) < 1
E
(i= 1,2, ... ,n).
Chapter 4
194
Hence
J/x~(s) /dp < 1/a. Therefore, the function z(s) = (T+ 1/a)xE does E
D
not belong to the set U.
If the measure f-l has an atom E 0 of finite measure, then there is a non-trivial continuous linear functiomil in the space N(L(Q,E,p). Indeed, by the definition of measurable function, x(t) is constant on E0 p-almost everywhere, x(t) =c. Let us put f(x) = c. It is obvious that f(x) is a continuous linear functional. If the measure Jl is atomless, the following theorem, converse to Theorem 4.2.1, holds :
THEOREM 4.2.2 (Rolewicz, I959). Let Jl be an atomless measure. N(t) 0 . . fI1m1n -= ,
(4.2.1)
t
t-->-00
If
then there are no non-trivial continuous linear functionals in the space N(L(Q,E,p)). Proof Lets be an arbitrary positive number. Suppose that (4.2.1) holds.
Then there is a sequence {tm} tending to infinity such that N~:m) -J>-0. Let km be the smallest integer greater that N(tm)/s. Let E be an arbitrary set belonging to E of the finite measure. Since the measure Jl is atomless, there are measurable disjoint sets E 1 , ••. , Ek., such that km
E=
U Et i=l
(i
and
= I, 2, ... , km).
Let xi,.(s)
for s E Et (i = I, 2, ... , km), elsewhere.
tm
= {0
Obviously, for sufficiently large m, PN(x:J:::::::;; s. On the other hand, for seE Ym (s)
I
km
\1
t
tm
= k...::.... Xm(s) = k' m
i=l
m
Existence of Continuous Linear Functionals and Operators
Since
~= -+oo, every function
195
of the type axE belongs to the convex hull
U of the set {x: PN(x) <e}. The setEis an arbitrary setoffinitemeasure. Hence all simple functions with suports of finite measures belong to U. The set U is open and convex. Therefore U = N(L(Q,E,J1)).
4.2.S (Day, 1940). In the spaces LP(Q,E,p), 0 nk,
(4.4.4)
(4.4.6)
Since the functionals fn are linearly independent and only a finite number of them are continuous with respect to II Ilk such a construction is possible. · 00
Condition (4.4.6) implies that the series}; Xn is convergent. Let us ~
n=l
denote its sum by x 0 • Conditions (4.4.4) and (4.4.5) imply that x 0 has the required property. D COROLLARY 4.4.8. Let {a 11 } be an arbitrary sequence of numbers and let {tn} be an arbitrary sequence of points of the interval (0, 1). Then there is an infinitely differentiable function x(t) defined on the interval [0, 1], vanishing together with all its derivatives at point 0 and 1 and such that x(tn) =an. Proof. The functionals fn(x) = x(tn) defined on the space C000 ([0, 1]), wchih is a su bsqace of the space C 00 ([0, 1]) of functions vanishing together
206
Chapter 4
with all their derivatives at 0 and I, satisfy the hypotheses of Theorem 4.4.7. [] 4.4.9. For any B0 -space X which is not a Banach space there is a continuous linear operator T mapping X onto (s).
COROLLARY
4.5.
NON-EXISTENCE OF NON-TRIVIAL COMPACT OPERATORS
Let X, Y be two F-spaces. An operator T mapping X into Y is called compact if there is an open set U such that the closure of its image,
T( U), is compact. An operator T mapping X into Y is called finite-dimensional if dim T(X) -f-oo. Of course, each finite-dimensional operator is compact. If there is a non-trivial linear continuous functional f defined on X, then there are finite-dimensional operators mapping X into Y, namely operators of the form T(x) = f(x)y, y E Y, y :;i= 0. Kalton and Shapiro (1975) showed that there is an F-space X with a trivial dual, such that there is a compact operator K :;i= 0 mapping X into itself. Pallaschke (1973) proved that in certain spaces N(L(Q, l:,Jl) there is no compact operator different from 0 acting in those spaces.
SUPPN(XXA)
0, and a sequence {Xn} C K and a sequence of sets {An} such that p(An)--+0 and P.v(XnXA.) :;;:: s0 • Now we shall choose by induction subsequences {xn.} and {An) in the following way. As Xn 1 we shall take an arbitrary element of the sequence. Suppose that the elements {xn 1 , ••• , xn.} are chosen. Of course, there is a number IJ > 0 such that for each set A, p(A) < IJ, we have
i = l, 2, ... , k.
(4.5.1)
Now take as An.+> a set from the sequence {An} such that p(An.+) < IJ and as xn.+, the corresponding element. By the property of the sequence {xn} and (4.5.1) for i-::j::j. Formula (4.5.2) implies that the set K is not compact.
(4.5.2)
0
Chapter 4
208
THEOREM 4.5.3 (Pallaschke, 1973). If a measure J1 is non-atomic and . . f 1 Itmtnn-+co
n
N(N-l(n)) =a> n
0
,
(4.5.3)
then each compact operator T mapping N(L(Q,E,p)) into itself is equal to 0. Proof Let T =F 0 be a compact operator mapping N(L(Q,E,p)) into itself. Let r > 0 be such a number that T(B(O, r)), where B(O, r) = {x: PN(x) < r}, is compact. Since simple functions are dense in N(L(Q,E,p)), there exists a set A, 0 0 be chosen so that the set B = {t E Q: lz(t)l > c} has a positive measure, p(B) > 0.
Let . {sgnz(t) J(t) = c- 1 sgnz(t)
for for
tE
Q""-. B,
tE B,
where sgnz = 0 for z = 0 and sgnz =
z JZT for z =F 0 and a denotes the
number conjugate to a. Suppose that T1(x) = j(t) T(x)t, i.e. that T1 is a composition of the compact operator T and the operator of multiplication by the function j(t). Of course, T1 also maps B(O, r) into a compact set K1 • Observe that (4.5.4) Let oc = p(A). We take a partition of the set A, An, 1 , ••• , An,n, such that oc p(An, i) =
n-.
Let
Then N- 1(n)
1
---XA = -(Yn,l+ ··· -l-yn,n). n n
(4.5.5)
Existence of Continuous Linear Functionals and Operators
Let
In,k
{ te Q; JT1(yn,k)Jtl
=
N- 1
209
(n)} ·
> -n-
Then by (4.5.4) and (4.5.5) N- 1(n)
( N- 1(n)
)I
-n-Xn(t)~ T 1 -n-XA t =
~ Tl(yn,t)lt n1 -61_
and (4.5.6)
Let
M =sup
J N(JxJ)dJl.
xEK 1 0
Since the set K 1 is compact, M is finite. By the choice of the set A, PNCYn,i) < r, i = 1,2, ... , n. Hence
N-l(n)) N ( - n - Jl(Jn,k) ~ M and
M
p(ln,>),;;
N( N-~(n)) .
(4.5.7)
Let e < {- a(l(B), where a is defined by (4.5.3). The set K 1 is compact, hence by Lemma 4.5.2 there is a (J > 0 such that
PN(ITI(Yn,k)IXA)
<e.
provided Jl(A) < b. By (4.5.7) and (4.5.3) we can choose an m0 such that for m Jl(lm,k) < b, k = 1,2, ... , m and _I N(N-l(m))
m Thus by (4.5.6)
m
>}:_a. 3
> m0
Chapter 4
210
Hence
2 1 (N-m(m)) ~
3a~mN
1
s p(B)
a
0, such that Ref(x) ~ c-e for x E K and Ref(p) ~ c. But the hypothesis implies that the functional f(x) is continuous in COROLLARY
the topology -r2 • Hence p could not belong to the closure K 2 of the set K in the topology -r2 • This implies that K 2 = K.
226
Chapter 5
If K is a convex set closed in the topology -r2 , then using the same 0 arguments we infer that it is closed in the topology -r1 •
5.2.
WEAK TOPOLOGIES. BASIC PROPER TIES
Let X be a linear space over real or complex numbers. Let X' denote the set of all linear functionals defined on the space X. A subset of the set X' is called total if f(x) = 0 for all f E F implies that X = 0 (compare Section 4.2). Let F be a total linear set of functionals. By the F-topology of the space X we shall mean a topology determined by the neighbourhoods of the type
r
N(p,fi, ... ,Jk; a 1 , ... , ak) = {x: Jfi(x-p)J < ai (i = 1, 2, ... , k)}, where at> 0 andfi E F. Obviously, the space X with a F-topology is a locally convex space. We say that a set A C X is F-closed (F-compact) if it is closed (compact) in the F-topology. The closure of a set A in the F-topology will be called the F-closure. A functional f(x) is said to be F-continuous if it is continuous in the r-topology. Let X be a locally convex topological space. Let X* be the set of all continuous linear functionals defined on X (conjugate space). The X*-topology is called the weak topology. Obviously, the weak topology is not stronger than the original one. Let X be the space conjugate to a locally convex space X-. Let us recall that in the conjugate space we have the topology of bounded convergence. Each element X-EX- induces a continuous linear functional F on the space X by the formula F(x) = x(x_).
(5.2.1)
We shall indentify the set of functionals defined by formula (5.2.1) with the space X-. The X--topology in X is called the weak topology of functionals or the weak-*-topology. Since we always have X*:::> X-, the weak topology offunctionals is not stronger than the weak topology.
Weak Topologies in Banach Spaces
227
5.2.1. Let X be a linear space. Let r be a total linear set of functionals. A linear functional f(x) is continuous in the F-topology if and only iffe PROPOSITION
r.
The proof is based on the following lemma : LEMMA 5.2.2. Let X be a linear space. Let g, ];_, ... .fn be linear functionals defined on X. If
ji(x) = 0,
i = 1, 2, ... , n,
implies g(x)
=
0,
then g(x) is a linear combination of the functionals ft, ... .fn. Proof Without loss of generality we can assume that the functionals ft, ... .fn are linearly independent. Let X0
=
{xe X: Ji(x)
=
0, i
=
1, 2, ... , n}.
(5.2.2)
Let X be the quotient space XJX0 • The functionals ];_, ... ,fn induce linearly independent functionals ];_, ... ,fn on X. The assumption about g(x) implies that the functional g (x) also induces a linear functional g(x) defined on X. Then space X is n-dimensiona1, therefore, g (x) is a linear combination of];_, ... ,f,,
-
-
i=
-
-
ad1 + ... +an/n.
It is easy to verify that g
=
ad1 + ... +anfn·
D
Proof of Proposition 5.2.1. Sufficiency. From the definition of neighbourhoods in the r-topology it trivially follows that each functional fer is r-continuous (i.e. continuous in the r-topology). Necessity. Let g(x) #- 0 be a functional continuous in the T-topology. Then there is a neighbourhood of zero U in that topology such that sup jg(x)l < 1. But the neighbourhood U is of the type XEU
U= {xeX: i.fi(x)i
i· Then
max ly*{yt)l
max ly*(y,)-yt(Yt)+y7{yt)l
=
l~i~m
r=:;;;i~m
for ally* E F of norm one.
D
Proof of Theorem 5.3.1. A_,.B. Let {an} be a sequence of elements of A. Let X 0 denote the space spanned by {an}· The set X0 is weakly
closed as a linear subspace. Therefore, the intersection of the weak closure w(A) of the set A with the space X0 is a weakly compact set in X. By Lemma 5.3.2 the weak topology on w(A) n X 0 is metrizable, because the space X 0 is separable. Therefore, there is a subsequence {an.} weakly convergent to an element a E X0 , i.e. lim f(an.)
=
f(a)
(5.3.1)
k-Ht:J
xr
for all f E Hence this is also true for f E X*. B_,.C. This is clear. C_,.A. Suppose that a set A satisfies condition C. Then, for each continuous linear functional f(x), the set of scalars f(A) = {f(x): x E A} also satisfies this condition. Therefore it is bounded. Hence the BanachSteinhaus theorem implies that the set A is also bounded. Let.n denote the natural embedding X into X**. Let w*(n(A)) be the closure of the .set n(A) in the X*-topology. Now we shall show (5.3.2)
w*(n(A)) C n(X).
Let X** be an arbitrary element of the set w*(n(A)). Let xt be an .arbitrary element of the space X* of norm one. Since x** belongs to the set w*(n(A)), there is a point a1 E A such that l(x**-n(ai))(xi)l
< 1.
(5.3.3)
The space spanned by x** and x**-n(a1) is two-dimensional. Hence by Lemma 5.3.3 there are points x:, ... , x: each of norm one, xt EX*,
Weak Topologies in Banach Spaces
i
233
= 2, 3, ... , n (2), such that max jy**(x*)l
>
!..lly**ll
(5.3.4)
2
2 {-lly**ll
n(2)<m..;n(3)
for ally** belonging to the space spanned by the elements x**, n(a1 ), n(a2). Once more using the fact that x** is in w*(n(A)), choose a 2 E A so that max I(x**-n(a3))(x!)l < 1/3 and continue the construction. I.,;m..;n(3)
Then we obtain : (1) an increasing sequence of positive integers {n(k)}, (2) a sequence of elements X, {xi> x 2, ••• } , (3) a sequence of elements of A, {a1oa 2, •• •} such that max
jy**(x;)j
>
(5.3.5)
{-lly**ll
I.,;m.,;n(k)
for ally** belonging to the space xk spanned by the elements x, n(al), ... . . . , n(ak_ 1). And max
jx**-n(ak))(xm)l
1
< -k.
(5.3.6)
I.,;m..;n(k)
Let X 0 be the space spanned by the elements x, n(a1), ... By hypothesis there is a point x 0 EX which is a cluster of the sequence {an} in the weak topology of X. Since X 0 is weakly closed as a closed subspace of X, x 0 EX. Formula (5.3.5) implies that max jy**(xm)l ~ flly**ll for y E X 0 • m
In particular, for x**-n(x0) we have max I(x**-n(x0)) (x!)j
>
m
fllx**-n(x 0)11.
On the other hand, formula (5.3.6) implies that for m j(x**-n(ak))(x!)i
x2 , ... }, y = {Y~>Y2• ... }.From the definition of the norm llx+yll it follows that for any positive 8 there is an increasing sequence of indices PI> ... , p 2n+l such that n
Jlx+yJJ ~ ( .2; (xp,i-1+YP•i-1-Xp,;-yp,i) 2 +(xp,n+>+ i=l +YP•nH)2]1/2+8 n
~ [}; (xp.t-1-xp.;)2+(xp•n+>)2Y' 2 i=l n
+(
L (ypai-1 -yp,;)2+(YP•n+>)2r' +8 ~ llxll + JJyJJ+8 · 2
i=l
Thus the arbitrariness of 8 implies that
llx+yli ~ llxii+IIYII. Let zn
= {0, ... , 0, 1, 0, ... }. ~
Chapter 5
236
It is easy to verify that the linear combinations of the elements zn are dense in the whole space B, because lim Xn = 0. Moreover, for all posn-->-oo
itive integers n and p,
Then Theorem 3.2.15 implies that {zn} is a basis in B. We shall show now that the space B is not reflexive. For this purpose we shall prove that the closed unit ball in B is not weakly compact. Let Yn = z1 + ... +zn. Of course, IIYnll = 1. If the closed unit ball is weakly compact, then {Yn} converges to a y E B. Since {zn} is a basis, y ought to be of the form (1, 1, ... ). This is impossible, because, for all x €: B, lim Xn = 0. n--->-00
Now we shall describe all functionals f belonging to the second conjugate space B**. Let {gn} be the basis functionals with respect to the basis {en}. According to Lemma 5.4.1, {gn} constitute a basis in the conjugate space B*. Let F be a functional from B**. Then the functional F is of the following form: there is a sequence of real numbers 00
00
{Fi} such that F(f) = 2 Fi}i for any /E B, f = 2 figi. Let us calcui=I
i=l
late the norm of the functional F. We have n
n
I}; Fi/il
=
n
If(}; Fizi)l ~ 11/11[12 Fizill
i=l
i=l
i=l
n'
- 11/11 sup -
[
"\' L..J (Fp,i-1- Fp,;) 2 +(Fq,n'+>)2]1/2 ,
i=l
where the supremum is taken over all positive integers n' and increasing sequences of indices PI> ... , p 2n'+l with Fpk replaced by 0 if Pk > n. The arbitrariness of n implies n
IIF II~ sup [}; (Fp,i-1-Fp.t)2 +(Fp,n'+t)2 i=l
r. (5.4.2)
Weak Topologies in Banach Spaces
237
n
Let us now fix nand let un =
1: F;zi. Let us define a linear functional i=l
f on the space Yn spanned by the elements un: zn+l, zn+2, ... in such a way that and
f(zi) = 0
fori= n+l, n+2, ...
Then 00
L
jJ(aun+
00
atzi)j=JiaunJI~jjaun+};
i=n+l ~
Thus
11/11 =
atzijj.
i=n+l CD
f = }; figi be an extension of the functional f to
1. Let
i=l
whole space B of norm one. Thenfi = 0 fori> nand oo
F(f) =
n
j}; Ftfi\j = j}; Ftfij = /f(un)/ = i=l
//unJI
~ //F//.
i=l
Hence, calculating the norm of un, we obtain (5.4.3) for all positive integers n and all finite increasing sequences of integers .. • ,p 2n+l· C~mbining (5.4.2) and (5.4.3), we obtain
p1,
(5.4.4) where the supremum is taken over all positive integers n and finite increasing sequences p 1 , ••• , p 2n+l· The norm IIF II is finite if and only if there is a limit lim Fn. Since the space B is not reflexive, B** contains n=co
an element which does not belong to n(B). Then the only possibility is B** that n(B) is a subspace of codimension 1, i.e. that dim n(B) = 1. There-
(BXB)** X** fore, dim n(B X B) = 2, and since dim fl(X) is an invariant of an isomorphism, the space B is not isomorphic to its Cartesian square (see Bessaga and Pelczynski, 1960b).
Chapter 5
238
Pelczyiiski and Semadeni (1960) showed another example of a space which is not isomorphic to its square. Their example is of the type C(.Q). An example of a reflexive Banach space non-isomorphic to its square was given by Figiel (1972). Problem 5.4.3. Does there exist a Banach space non-isomorphic to its Cartesian product by the real line?
The answer is positive for locally convex spaces, as will be shown in Corollary 6.6.12. Rolewicz (1971) gave an example of normed (non-complete) space X non-isomorphic to its product by the real line. Dubinsky (1971) proved that each B 0-space contains a linear subset X which is not isomorphic to its product by the real line. Bessaga (1981) gave an example of a normed space which is not Lipschitz homeomorphic to its product by the real line.
5.5.
EXTREME POINTS
Let X be a linear space over the real or the complex numbers. Let K be an arbitrary subset of X. We say that a point k e K is an extreme point of the set Kif there are no two points k1 , k 2 e K and no real number a, 0 < a < 1 such that k = ak1 +(1-a)k2 •
(5.5.1)
The set of all extreme points belonging to K is denoted by E(K). A subset A of the set K is called an extreme subset if, for each k e A the existence of k 1 ,k2 , 0 A. Since the set K is compact, the intersection of a decreasing family of closed sets is a closed non-void set, and obviously it is also an extreme set, provided the members of the family belong to m.
Weak Topologies in Banach Spaces
239
Then, by the Kuratowski-Zorn lemma, there is a minimal element A 0 of the family m. We shall show that the set A 0 contains only one point. Indeed, let us suppose that there are two different points p, q e A 0 • Then there is a functional x* e X* such that Rex*(p) o:F Rex*(q). (5.5.2) Let (5.5.3) A1 = {xe A0 : Rex*(x) = inf Rex*(y)}. yEA,
Since the set A 0 is compact, the set A1 is not empty, Moreover, formula (5.5.2) implies that the set A1 is a proper part of the set A 0 • Let k1ok 2 be points of K such that there is an a, 0 < a < 1, such that ak1 +(1-a)k2 eA 1 •
(5.5.4)
Since A 0 is an extreme subset, k1 and k 2 belong to A 0 • Since (5.5.4) aRex*(k1)+(1-a)Rex*(k2) = inf Rex*(y). yE.Ao
This is possible if and only if Rex*(k1) = Rex*(k 2) = inf Rex*(y). yE.A,
This implies that k1 ,k2 e A1 • Hence A1 is an extreme set. Thus we obtain a contradiction, because A 0 is a minimal extreme subset. Therefore, A 0 is a one-point set, A 0 = { x 0 } and, from the definition, x 0 is an extreme point. D THEOREM 5.5.2 (Krein and Milman, 1940). Let X be a locally convex topological space. Let K be a compact set in X. Then conv E(K) ) K.
(5.5.5)
Proof Suppose that (5.5.5) does not hold. This means that there is an element k e K such that k ¢ conv E(K). Then there are a continuous linear functional x* and a constant c and positive e such that
Rex*(k) ~ c
(5.5.6)
and Rex*(x);;, c+e
for x e conv E(K.).
(5.5.7)
Let K 1 = {xe K: Rex*(x)
= infRex*(y)}. yEK
(5.5.8)
Chapter 5
240
Since the set K is compact, the set K1 is not empty. By a similar argument to that used in the proof of Proposition 5.5.1, we can show that K1 is an extreme set. By formula (5.5.7) the set K1 is disjoint with the set E(K). This leads to a contradiction, because, by Proposition 5.5.1, K1 contains an extremal point. 0 5.5.3.
COROLLARY
If a set K
is compact, then
conv K = conv E(K), 5.5.4. For every compact convex set K,
COROLLARY
K
= conv E(K).
5.5.5. Let X be a locally convex topological space. Let Q be a compact set in X such that the set convQ is also compact. Then the extreme points of the set conv Q belong to Q. Proof Letp be an extreme point of the set convQ. Suppose thatp does not belong to the set Q. The set Q is closed. Therefore, there is a neighbourhood of zero U such that the sets p+ U and Q are disjoint. Let V be a convex neighbourhood of zero such that PROPOSITION
v-vc u. Then the sets p+ V and Q+ V are disjoint. This implies that p E Q+ V. The family {q+ V: q E Q} is a cover of the set Q. Since the set Q is compact, there exists a finite system of neighbourhoods of type qi+ V, 11
i = 1,2, ... , n, covering Q, QC
U (qi+V).
i=l
Let Kt
= conv ((qi+ V) n Q).
The sets Ki are compact and convex; therefore conv(K1 u ... u Kn) = conv (K1 u ... u Kn) = convQ. Hence n
p =
.J; atki, i=l
n
at~ 0,
.J; at =
1, kt E K t.
i=l
Since p is an extreme point of conv Q, all at except one are equal to 0.
Weak Topologies in Banach Spaces
241
This means that there is such an index i that pEKt C Q+V, which leads to a contradiction.
0
REMARK 5.5.6. In the previous considerations the assumption that the space X is locally convex can be replaced by the assumption that there is a total family of linear continuous functionals T defined on X. Indeed, the identity mapping of X equipped with the original topology into X equipped with the T-topology is continuous. Thus it maps compact sets onto compact sets. Therefore, considering all the results given before in the space X equipped with the T-topology we obtain the validity of the remark.
5.6. EXISTENCE OF A CONVEX COMPACT SET WITHOUT EXTREME POINTS Roberts (1976, 1977) constructed an F-space (X, II ID and a convex compact set A C X, such that A does not have extreme points. The fundamental role in the construction of the example play a notion of needle points (Roberts, 1976). Let (X, II ID. be an F-space. We say that a point x 0 EX, x 0 ::j::. 0, is a needle point if for each c; > 0, there is a finite set F C X such that x 0 EconvF,
(5.6.1)
sup{llxll: xEF}<s,
(5.6.2)
conv {0, F} c conv {0, x 0 }+B.,
(5.6.3)
where, as usual, we denote by B. the ball of radius c;, B.= {x: llxll < s}. A point x 0 is called an approximative needle point if, for each c; > 0, there is a finite set F such that (5.6.2) and (5.6.3) hold, and moreover x 0 E conv F+B•.
(5.6.4)
Since c; is arbitrary, it is easy to observe that x 0 is a needle point if and only if it is an approximative needle point. Let E denote the set of all needle points. The set Eu {0} is closed.
242
Chapter 5
From the definition of needle points and the properties of continuous linear operators we obtain PROPOSITION 5.6.1. Let X, Y be two F-spaces. Let T be a continuous linear operator mapping X into Y. If x 0 EX is a needle point and T(x0) #- 0, then T(x0 ) is a needle point. We say that an F-space (X, II II) is a needle point space if each x 0 EX, x 0 #- 0 is a needle point. The construction of the example is carried out in two steps. In the first step we shall show that in each needle point space there is a convex compact set without extreme points, in the second step we shall show that a large class of spaces (in particular, spaces LP, 0 < p < 1) are needle point spaces. THEOREM 5.6.2 (Roberts, 1976). Let (X, II II) be a needle point F-space. Then there is a convex compact set E C X without extreme points. Proof Without loss of generality we may assume that the norm II II is non-decreasing, i.e. that lltxll is non-decreasing for t > 0 and all x EX. Let {sn} be sequence of positive numbers such that co
,2;
Sn
< +oo.
(5.6.5)
n~o
Let x 0 #- 0 be an arbitrary point of the space X. We write £ 0 = conv({O,x0}). Since X is a needle point space, there is a finite set F = E 1 = {x~, ... , x~} such that (5.6.1)-(5.6.3) holds for s = s0 • For each xt, i = 1, ... , n~> we can find a finite set Ft such that x}Econv({O} u Fi), sup{llxll: XE Fl}
(5.6.6)1
< ~,
(5.6.7)1
n1
conv({O} u Ft) C conv{O, xl}+B~.
(5.6.8)1
n,
Observe that (5.6.8)1 implies conv({O} u E 2) C conv({O} u E1 )+Be,,
(5.6.9)1
Weak Topologies in Banach Spaces
243
where (5.6.10\ The set E 2 is finite, and thus we can repeat our construction. Finally, we obtain a family of finite sets En such that for each x E En we have XE
conv({O} u En+ 1),
(5.6.6)n
sup{JJxJJ: XEEn} <en, conv({O}
U
(5.6.7)n
En+ 1 ) C conv({O} U En)+Ben·
(5.6.9)n
Let <X>
K 0 = conv(
U En U n=O
{0}).
The set K 0 is compact, since it is closed and, for each e a finite e-net in K0 • Indeed, take n0 such that
>
0, there is
00
.2; en< e. n=n0 no
By (5.6.9)n the set
U En constitute an e-net in the set K
0•
n=O
Observe that no x =I= 0 can be an extremal point of K0 , since 0 is the 00
unique point o£ accumulation of the set
U En,
and, by construction,
n=O
no x E En is an extremal point of K 0 • Thus the set K 0 - K 0 does not have extremal points. D Now we shall construct a needle point space. Let N(u) be a positive, concave, increasing function defined on the interval [O,+oo) such that N(O) = 0 and lim N(u) = 0. u
11-+00
(in particular, N(u) could be uP, 0 < p < 1). Let Q = [0, 1]"' be a countable product of the interval [0, 1] with the measure p, as the product Lebesgue measure. Let E be a u-algebra induced
Chapter 5
244
by the Lebesgue mesurable sets in the interval by the process of taking product. 1
Take now any function f(t)
E
L co[O, I] such that
J f(t)dt = I. We 0
shall associate with the function fa function St(j) defined on· [0, I]w by the formula St(f)lt = f(tt),
where t = {tn}. Observe that the norm of Si(/) in the space N(L(Q,E,Jl.)) is equal to 1
IISt(f)JI =
f N(f(t))dt,
i~
(5.6.11)
I, 2, ...
0
Of course Si(f) can be treated as an independent random variable. Thus, using the classical formula n
£2
n
2; (Xt- E(Xt)) = 2; E (Xt- E(Xt)) 2
i=l
i=l
n
we find that, for at ;:?o 0 such that }; ai = I, i=l
n
n
f [2; at(St(f)-I)r dJi. = 2; af a~(St(f)-1)2dJ1
a
i=l
i=I n
:::=;::;a
2; at J (St(/)-1) dJ1. 2
a
i=I 1
=
J(f(t)-1) dt,
a
(5.6.12)
2
0
where a = max {aI>
••• ,
an}.
By the Schwartz inequality we have n
n
( j 12; a,S(fi)-IJdJl.r::::;::; f 2; at(St(f)-1) dJ1. 2
a
i=1
a
i=1
(5.6.13)
Weak Topologies in Banach Spaces
245
The function N(u) is concave, hence the following inequality results directly from the definition (compare the Jensen inequality for convex functions) n
n
(5.6.14)
N(); atut} ) ' ) ; atN(ut). i=1
i=1
As an intermediate consequence of formula (5.6.14), we infer that for each g E N(L(Q,E,p)) n L(Q,E,p), we have
JlgJJ ~ N(
j
(5.6.15)
JgJ dp).
!1
By (5.6.12), (5.6.13) and (5.6.15) we obtain n
1
J[); atSt(f)-1[[ ~N( a( J(f(t)-1) dtr' 2
i=1
2
(5.6.16)
).
0
Now we shall introduce the notion of IJ-divergent zone. Letfe L 00 [0, 1] 1
be such that
I
f(t)dt = 1 and let IJ
> 0. An interval [a,b], 0 ... , an such that a 1 + ... +an ~ 1 and at ~ b, i = 1,2, ... , n. Thus n ~ m and, by (5.6.11) and the triangle inequality, we obtain m
1
II}; atSt(/)11 ~m j N(if(t)i)dt ... , xn such that n
A C
U (xt+U). i=l
K
conv({x1, •.. , xn}).
{6.1.3)
Let =
The set K is bounded and finite-dimensional, therefore, there is a finite system of points y 1 , ... , Ym such that m
K C
U (Yi+U). i=l
Thus, by (6.1.3), conv A c conv(K+ U) Therefore m
convA C
=
(6.1.4)
K+ U. m
U (yd-U)+U = i=l U (Yt+2U). i=l
The arbitrariness of U and Proposition 6.1.1 implies the proposi~.
D
Since Proposition 6.1.1 holds for complete topological spaces, Proposition 6.1.3 holds for complete locally convex spaces. Therefore, in the case of complete spaces, we can omit in Corollary 5.5.5 the assumption that the set conv Q is compact.
6.2. MONTEL SPACES In the preceding section we proved that each locally compact space is finite-dimensional. In finite dimensional spaces each bounded closed set is compact. There are also infinite-dimensional spaces with this property. Examples of such spaces will be given further on. F-spaces in which each closed bounded set is compact are called Monte/ spaces.
Chapter 6
252
PROPOSITION 6.2.1. Locally bounded Monte/ spaces are finite-dimensional. Proof Let X be a locally bounded Montel space. Since X is locally bounded, there is a bounded neighbourhood of zero U. The space X is a Montel space, hence the closure U of the set U is compact. Therefore, the space X is locally compact and, by Proposition 6.1.2, finite-dimensional. D PROPOSITION 6.2.2 (Dieudonne, 1949; Bessaga and Rolewicz, 1962). Every Monte/ space is separable.
The proof of the theorem is based on the notion of quasinorm (Hyers, 1939; Bourgin, 1943), similar to the notion of pseudonorm. Let X be an F-space. By ~ we denote the class of all open balanced set. Let A E ~. The number [x]A=inf{t>O:
~ EA}
is called the quasinorm of an element x with respect to the set A. Quasinorms have the following obvious properties: (a) [tx]A = It I[x]A,, (b) if A) B, then [x]A :::( [xB], (c) the quasinorm [x]A is a homogeneous pseudonorm (i.e., satisfies the triangle inequality) if and only if the set A is convex. PROPOSITION 6.2.3. [x+y]A+B :::( max((x]A, [y]B). Proof Let us write r = max([x]A, (y]h). Let e be an arbitrary positive number. By the definition of the quasinorm,
XE(1+e)rA
and
yE(1+e)rB.
Hence
x+yE(1+e)r(A+B). Therefore [x+y]A+B :::( (1 +e)r. The arbitrariness of e implies the proposition. D PROPOSITION 6.2.4. Let a sequence {An} C ~ contitute a basis of neighbourhoods of zero. Then a sequence {xm} tends to 0 if and only if lim [xm]An = 0 ~0
(n = 1, 2, ... ).
(6.2.1)
Monte! and Schwartz Spaces
253
Proof. Necessity. Suppose that Xm-+0; then, for arbitrary s > 0, and n, there is an m 0 dependent on n and s such that for m > m 0 , Xm E sAn, whence [xm]A,. :::;;;; s.
Sufficiency. Let us suppose that (6.2.1) holds; then for every n there is an m 0 dependent on n such that, for m > m 0 , [xm]A,. :::;;;; 1. This means that Xm E An. Since {An} constitutes a basis of neighbourhoods of zero,
D
~~
PROPOSITION 6.2.5. Let {An} C m: constitute a basis of neighbourhoods of zero. A set K is bounded if and only if there is a sequence of numbers {Nm} such that sup[x]A .. :::;;;; N,. XEK
Proof. Sufficiency. Suppose that a sequence {Nn} with the property
described above exists. Let {xm} be an arbitrary sequence of elements of K and let {tm} be an arbitrary sequence of scalars tending to 0. Then [tmXm]A,.:::;;;; ltmiNn-+ 0.
Hence, by Proposition 6.2.4, the sequence {tmxm} tends to zero. Therefore, the set K is bounded. Necessity. Suppose that there is an index n such that sup[x]A. = +=. xEK
Then there is a sequence {xm} C K such that 0 < [xm]A .. -+oo. Let tm = 1/[xm]A ..· The sequence {tm} tends to 0. On the other hand [tmxm]A .. = 1, l).ence, by Proposition 6.2.4, the sequence {tmxm} does not tend to 0. This implies that the set K is not bounded. D Proof of Proposition 6.2.2. Let X be a non-separable F-space with the F-norm llxll· Since the space X is non-separable, there are a constant ~ > 0 and an uncountable set Z such that liz -z'll > ~ for z, z' E Z, z -i= z'. Let Kn = {x: llxll < 1/n}, and le~ us write briefly [x]n = [x]x..·
Since the set Z is uncountable, there is a constant M 1 such that the set Z 1 = Z n {x EX: [x] 1 :::;;;; M 1 } is also uncountable. Then there is a constant M 2 such that the set Z 2 = Z 1 n {x EX: [x] 2 :::;;;; M 2 } is uncountable. Repeating this argumentation, we can find by induction a sequence of uncountable sets {Zn} and a sequence of positive numbers {Mn} such that the set Zn is a subset of the set Zn-1 and sup[x]t:::;;;; Mn for XEZn
i= 1,2, ... ,n.
Chapter 6
254
Let us choose a sequence {zn} such that Zn E Zn and Zi =I= Zk fori =I= k. Then sup [zn]k ,;;; max (Mk, [z1]k, ... , [zk-t]k) n
t5 for i =F k. This implies that the D set {.Zn} is not compact. Therefore, X is not a Montel space. The following question has arised: is it sufficient for separability if we assume that each bounded set is separable? The answer is negative. Basing on the continuum hypothesis Dieudonne (1955) gave an example of a non-separable -space in which each bounded set is separable. For F-spaces such an example was .given by Bessaga and Rolewicz (1962). For D0 -spaces it was given by Ryll-Nardzewski.
D:
PROPOSITION 6.2.6 (Ryll-Nardzewski, 1962). There is a non-separable D0 -space in which all bounded sets are separable. Proof Let S denote the class of all sequences of positive numbers. We introduce in S the following relation of order ~. We write that {In} ~ {gn} ifln < gn for sufficiently large n. A subclass S1 of class Sis called limited if there is a sequence {hn} E S such that {fn} ~ {hn} for all sequences {In} E S1 • Let us order class Sin a transfinite sequence {I:} of type w 1 (here we make use of the continuum hypothesis). Now we define another sequence of type w 1 as follows: {g~} is the first sequence (in the previous order) which is greater in the sense of relation ~ than all {I~} for oc < {3. It is easy to see that no non-countable subclass of this sequence is limited. Let us consider the space X of all transfinite sequences {x"} (oc < w1) of real numbers such that x" vanishes except for a countable number of indices and ,, p
ll{xp}lln = L.J gnjxpj < +oo,
n
=
1, 2, ...
P<w,
Let us introduce a topology in the space X by the sequence of pseudonorms ll{xp}lln· The space X with this topology is a Dri -space. We shall
Monte! and Schwartz Spaces
255
show that the space X is complete. Let {xn} be a fundamental sequence of elements of the space X. Let A= {a: x::;t=o for certain n}.
The definition of the space X implies that the set A is countable. Therefore, the subspace X 0 spanned by the elements {x!}, {x!}, ... is of the type V(am,n). Thus it is complete. Therefore, the sequence {x:} has the limit {xa} E X 0 C X. To complete the proof it is enough to show that each bounded set Z C X is separable. If the set Z is bounded, then there is a sequence of positive numbers {Mn} such that sup ll{xa}/ln ~ Mn,
n = 1, 2, ...
{x..)EZ
Let k be a positive integer. By h we denote the set of all such indices that there is an x ={xu} E Z and such anindex p such that [xp] > 1/k. Then we 1
have kg~< ll{xa}lln ~ Mn. Hence, for
P E h, we have {g!} -3
{Mn} and 00
by the property of the class {g~} the set h is countable. Let I= ,._..
U Ik. k=l
Then the set I is, of course, also countable. Let y be the smallest ordinal greater than all the terms of the set I. Then from the definition of the set I it follows that if {xa} E Z, then X 0 = 0 forb > y. Thus the set Z is separable. 0
6.3.
SCHWARTZ SPACES
Let X be an F*-space. We say that a set K is totally bounded with respect to a neighbourhood of zero U, if, for any positive c:, there is a finite system of points xi> ... , Xn. such that ~ C
00
U (xi+c:U). A set which is totally i=l
bounded with respect to all neighbourhoods of 0 is called totally bounded or precompact. Proposition 6.1.1 implies that if a set K is closed and totally bounded with respect to all neighbourhoods of zero, then it is compact.
Chapter 6
256
An F-space X is called a Schwartz space if, for any neighbourhood of zero U, there is a neighbourhood of zero V totally bounded with respect to U. 6.3.1. Let X be a Schwartz space. Then its completion also a Schwartz space. PROPOSITION
X
Proof Let U0 be a neighbourhood of 0 in X. Let U = U0 n X. Since X is a Schwartz space, there is a neighbourhood of zero V C X such that for any s > 0 there is a system of points x1 , ••• , Xn such that n
V C
U (xi+sU). i=l •
Thus n
V C
n
U (xi+sU) c U (xi+2sU). i=l
D
i~l
6.3.2. Every Schwartz F-space is a Monte/ space. Proof Let K be a closed bounded set. Let U be an arbitrary neighbourhood of zero. Since we consider a Schwartz space, there is a neighbourhood of zero V totally bounded with respect to U. The set K is bounded. Then there is a positive a such that K C a V. The neighbourhood Vis totally bounded with respect to U, and so there is a finite PROPOSITION
system of points Then Kc aV c
Y~> ... , Yn
such that V C
0 1
(Yi+
~). Let
Xi
= ayi.
co
U (xi+U). i=l
Since the set K is closed and U is arbitrary, the set K is compact (see Proposition 6.1.1). D There are also Mantel spaces which are not Schwartz spaces. An example will be based on PROPOSITION 6.3.3. Let am, n ~ am+I,n. The space M(am, n) (or LP(am, n)) (see Example 1.3.9) is a Schwartz space if and only. if, for any m, there
Monte! and Schwartz Spaces
257
is an index m' such that
. am.n O 1Im --= '
(6.3.1)
am•,n
lnl~oo
where lnl = lnii+In2l+ ... +lnkl· Proof. Sufficiency. Let U be an arbitrary neighbourhood of zero. Let U0 be such a neighbourhood of zero that U0 + U0 C U. Let m be such
an index that the set
~
{x: llxllm
{x: l!xm•ll < b}. Let A be the set of such indices n that amn bm' ' - >2- . am•,n Since (6.3.3) holds, the set A is infinite. Let yn =
{y~},
where
fork= n, fork =I= n. It is obvious that yn E V (n = 1,2, ... ). On the other hand, if n,n' E A, n-=/= n', then llyn-yn'll > Mm.f2 (resp. Pm(yn-yn') > [Mm,f2]P). Since the set A is finite, this implies that Vis not totally bounded with respect to U. The arbitrariness of V implies the proposition. D Example 6.3.4 (Slowikowski, 1957) Example of a Monte! space which is not a Schwartz space. Let k, m, n~> n2 be positive integers. Let
ak,m,n,,n, = n~max(l,n~-n·).
Let X denote the space of double sequences x = {xn1,n 2} such that l!xllk.m = sup ak,m,n,,n, Jxn,.n,l < nhna
+ 00
with the topology determined by the pseudonorms l!xl!k,m· X is a B0 -space of the type M(am, n). The space X is not a Schwartz space. Indeed, let us take two arbitrary pseudonorms l!xl!k,m and l!xllk',m'· Let n~ > m,m'. Then lim n 2~co
ak,m,n~.n. = (n?)k-k'
Therefore lim sup ak,m,n,,n, lnl-+o:>
> 0.
ak',m',nY,na
ak','11'tnl,nl
> 0,
and from Proposition 6.3.2 it follows that the space X is not a Schwartz space.
Monte) and Schwartz Spaces
259
Now we shall show that the space X is a Montel spac~. Let A be a bounded set in X. Since X is a space of the type M(am, n), it is enough to show that lim ak,m,n.,n,sup lxn,,n,l = 0.
lnl--+oo
(6.3.4)
XEA
Let us take any sequence {(n~,nD}, such that lim ln~l+ln~l
= +oo.
We have two possibilities : (1) n~-+oo,
(2) nr is bounded. Let us consider the first case. Let x = {Xn 1, n 2} E A and let k' > k, m' > m. Since the set A is bounded, there is a constant Mk',m' such that
Then for sufficiently large
n~
(6.3.5) Let us consider the second case. Let m' Then
> m and m' >
n~,
k'
>k
Therefore, by (6.. 3.5) and (6.3.6) formula (6.3.4) holds. This implies that X is a Monte! space. D
6.4.
CHARACTERIZATION OF SCHWARTZ SPACES BY A PROPERTY OFF-NORMS
In the previous section we introduced the notion of Schwartz spaces. Now we shall give a characterization of those spaces by a property of F-norms. Let Y be an arbitrary F*-space with the F-norm llxll and let 8 be an arbitrary positive number. We write c(Y, 8, t)
= inf {lltxll : x E Y, llxll = 8}
260
Chapter 6
if there is such an element x c(Y, e, t)
if sup
llxll
k
nt
U U {xj} constitutes i=l j=l
an r~c-net in K 0 , because Am C A~c for m > k. Since the set K0 is closed, the arbitrariness of 8 implies that the set K is compact. Let K 1 = {[x]: xe K 0}. Since the set K0 is compact, the set K1 is also compact. Moreover, [~] oo
= Z! and the set
n•
U U {Zf} is dense inK. Hence K = k=l i=l
'
K1•
D
Monte) and Schwartz Spaces
267
Proposition 6.5.4 implies the following fact. Let The a continuous linear operator mapping an F-space Y onto an F-space X. Then for every compact set Kin X there is a compact set K 0 in Y such that T(K0) = K. Indeed, let Z = {x: Tx = 0}. Then the space X is isomorphic to the quotient Y/Z and the operator T induces the operator T' mapping y E Y into the coset [y] E Y/Z. 6.5.5. Let X and Y be two F-spaces. linear operator T mapping Y onto X, then
PROPOSITION
If there
is a continuous
M(X) C M(Y). Proof To begin with, les us remark that if Tis a linear operator, then M(A, B,
e)~
M(T(A), T(B), e).
Hence M(A, B) C M(T(A), T(B)).
Since the inverse image of an open set is always open and in our case, by the Banach theorem (Theorem 2.3.1 ), the image of an open set is open,
n M(A, B) c n M(T(A), B).
BE(j
BEQ
BcX
BeY
On the other hand, for any compact set K c X there is a compact set K 0 C Y such that T(K0) = K. Therefore M (X)= COROLLARY
nn M(A, B) c nn
AE':J BEO
AE':f BEO
AcXBcX
AcYBcY
6.5.6. If
codim1 X~ codim1 Y (codiml X= codim1 Y), then, M(X) ::> M(Y)
(resp. M(X) = M(Y)).
By a simple calculation we obtain PROPOSITION
6.5. 7. If X is ann-dimensional space, then
M(X) = {tp(e): lim entp(e) = n~oo
+=}.
_
M(A, B)- M(Y).
268
Chapter 6
PROPOSITION 6.5.8. Let X be an F*-space. If X is not a Schwartz space, then the set M(X) is empty. Proof If X is not a Schwartz space, then there is a neighbourhood of zero U such that, for any neighbourhood of zero V, there is a positive number ev such that
M(V, U, ev) = +oo.
(6.5.1)
Let Vn = {x: llxll < 1/2n}, where llxll denotes, as usual, the F-norm in X. Let us write, for brevity, en = evn· Let rp(e) be an arbitrary positive function. Formula (6.5.1) implies that there is a system of points x~, ... , x~n such that mn > rp (en), x£ E Vn, i = I, mn
oo
2, ... , mn, xj-x~¢enUforj=f= k. LetK =
U U {x£}. The set Khas a n=l i=l
unique cluster point 0. Then the set K is compact. On the other hand, M(K, U, en)> mn > rp(en). Therefore rp(e) does not belong to M(K, U). The arbitrariness of rp(e) implies that the set M(X) is empty. D The Kolmogorov definition of approximative dimension has some disadvantages. Simply there are "too many" open and compact sets. This is the reason why in many cases it is more convenient to consider a definition of approximative dimension introduced by Pelczyri.ski (1957). The Pelczynski approximative dimension of a space X can be defined (after certain modifications) as the set of functions M'(X)=
Do lJOnpOive M(nV,
U).
integers
In other words, M'(X) = {rp(e): for each open set U there is an open . 'P(e) set Vsuch that, for all n, hm M( V U ) = +oo}. £--* 0 n , ,e Let { Ui} be a basis of neighbourhoods of zero. Let M~(X) = {rp(e): for each i, for almost allj, for all n, lim M( e--->0
~(e) U
n i+i•
)
i, e
=
+=}.
6.5.9. For all F*-spaces, M'(X) = M~(X). Proof The definitions of M'(X) and M~(Y) imply that M'(Y) C M~(Y). Suppose that rp(e) ¢ M'(Y). This means that there is a.neighbourhood of
PROPOSITION
Monte! and Schwartz Spaces
269
zero U such that for all neighbourhoods of zero V there is a positive integer n such that lim M( '1'~8)U ) < +=. Let Ut be an arbitrary e~o n ' ,8 neighbourhood contained in U. Putting V = Ut+i(j = 1, 2, ... ),we find that rp(8) does not belong to M 0(X) C M'(X). D In applications the most useful class is the class M~(X), because it is described by a countable family of functions. PROPOSITION
6.5.10. Let X be a Schwartz space. Then the set M'(X) is not
empty. Proof Let us choose a decreasing basis of neighbourhoods of zero { Ut} in such a manner that M(nUt+~o Ui, 8) < for all positive integers n. Such a choice is possible, because X is a Schwartz space.
+=
Let
n n
rp(8)
=
_.!._
M(nUi+I, Ut, 8)
8 i=l
1 1 for--
rp(en).
This implies t4at the function rp(s) does not belong to M(X). Hence M'(X) = Mv (X)) M(X).
0
Since it is not known whether the equality M'(X) = M(X) holds in general, we shall prove for M'(X) propositions and corollaries similar to Propositions and Corollaries 6.5.1-6.5.6. PROPOSITION 6.5.14. Let X andY be two Schwartz spaces. If the spaces X and Yare isomorphic, then
M'(X) = M'(Y).
Proof The above follows immediately from the definition of M'(X) and · the fact that the image (the inverse image) of an open set under an isomorphism is an open set. D
Chapter 6
272
6.5.15. Let X be an F*-space and let Y be a subspace of the space X. Then PROPOSITION
M'(X) C M'(Y). Proof The proof is the same as the proof of Proposition 6.5.2. CoROLLARY
0
6.5.16. If
diml X:;:;;; dimlY
(dimlX =dimlY),
then M'(X)
~
M'(Y)
(resp. M'(X) = M'(Y)).
PROPOSITION 6.5.17. Let X andY be two F-spaces. LetT be a continuous linear operator mapping X onto Y. Then
M'(X) C M'(Y). Proof The Banach theorem (Theorem 2.3.1) implies that the image of an open set is an open set. In general, the inverse image of an open set is an open set. The rest of the proof is the same as the proof of Proposition 6.5.6. 0 COROLLARY
6.5.18. If
codiml X:;:;;; codiml Y
(codimlX = codimlY),
then M'(X)
~
M'(Y)
(resp. M'(X) = M'(Y)).
It is easy to calculate approximative dimensions of spaces M(am,n). For the calculation of M'(X) it is enough to know the functions M(Ui+1• Ui, s) for a basis of neighbourhoods of zero {Un} (see Proposition 6.5.9). PROPOSITION
6.5.19. Let X= M(am,n) be a real Schwartz space. Let
Ut = {x: JJxllt < 1}. Then (6.5.4) where E'(r) is the greatest integer less than r.
Monte! and Schwartz Spaces
273
Proof Letx = {xn} andy= {Yn} be two elements of U;+i· Letx-y E eU;. Then there is an index n such that a;,nlxn-Yni >e. On the other hand, ai+i•nXn and ai+i•nYn are less than 1. Since there may exist no E'(1+l:__at.n )numbersbksuch thata;+f,nlbkl < 1 and e ai+j,n . Jbk-bk,Ja;,n > e fork-=!= k', and systems with this number of elements
more than
exist, the proposition holds.
D
Proposition 6.5.13 and formulas (6.5.4) and (6.3.1) imply that M'(M(am,n))
=
M(M(am,n)).
If X= M(am,n) is a complex space, then tp(e) E M(X) if and only if
y cp(e) belongs to M(Xr), where Xris a real space M(am,n).
Propositions 6.5.19 and Corollaries 6.5.12', 6.5.16, 6.5.18 imply 6.5.20 (Pelczynski, 1957). There is no Schwartz space universal (or co-universal) for all Schwartz spaces. Proof Since for every Schwartz space X the set M'(X) is non-empty, it is sufficient to show that for any functionf(e) there is a Schwartz space X1 such thatf(e) ¢; M'(X,). Let a!= 2/k for nk-l < n ~ nk, where nk = logd(l/k)+l. Let X1 = M(am,n), where am,n = (a!) 11m. By proposition 6.3.3 the PROPOSITION
space X1 is a Scl].wartz space. On the other hand,
M( U;+i, U;, !)= ~
i1 E'(1+k(a~) ~- i:i) [1 E'(l+ka~) ~ > 1(!).
Therefore,J(e) <j M'(XJ).
2n•
D
Let us observe that, for any sequence {Xn} of Schwartz space, there is a Schwartz space X universal for the sequence {Xn}. Indeed, let X be the space of all sequences x = {xn}, Xn E Xn with the F-norm 00
~ 1
llxll = ;21
2n
llxnlln l+JJxnlln'
Chapter 6
274
where JjxJJn denotes the F-norm in Xn. It is easy to verify that X is a Schwartzspace and that it is universal and co-universal for all spaces Xn.
6.6.
DIAMETRAL DIMENSION
In this section we shall consider another definition of approximative dimension, so-called diametral approximative dimension or briefly diametral dimension (see Mityagin; 1960, Tichomirov, 1960; Bessaga, Pelczynski and Rolewicz, 1961, 1963). Let A, B be arbitrary sets in a linear space X. Let B be balanced. Let L be a subspace of X. We write c5(A, B, L)
= inf(e > 0: L+eB >A).
Let us write c5n(A, B)= infc5(A, B, L),
where the infimum is taken over all n-dimensional subspace L. Let c5(A, B) denote the class of all sequences t = {tn} of scalars such that lim
tn c5n(A, B)
n-+co
=
0.
The following properties of the class c5(A, B) are ob':ious: if A' C A and B c5(aA, bB)
~
B', then c5(A', B') C c5(A, B);
= c5(A, B) for all scalars a, b different from 0.
(6.6.1) (6.6.2)
Let X be an F-space. Let 0 denote the class of open sets and c;J the class of compact sets. Let c5(X) =
U U c5(B, U).
UE() BE':f
The class c5(X) is called the diametral approximative dimension (briefly diametral dimension) of the space X. PROPOSITION
6.6.1. Let X andY be two isomorphic F-spaces. Then
c5(X)
~
c5(Y).
Monte! and Schwartz Spaces
275
Proof The proposition immediately follows from the fact that the classes of open sets and compact sets are preserved by an isomorphism. D In many cases diametral dimension is easier to calculate than approximative dimension. Unfortunately we do not know the answer to the following question: do we have b(X) C o(Y) is X is a subspace of an F-space Y? As we shall show later, the answer is affirmative under certain additional assumptions. PROPOSITION 6.6.2 (Mityagin, 1961). Let X andY be two F-spaces. LetT be a continuous linear operator mapping X onto Y. Then
b(X) :::> b(Y). Proof The definition trivially implies that, for arbitrary A, B C X and an arbitrary subspace L, b(A, B, L) ?= o(T(A), T(B), T(L)). Since dim T(L)
~
dimL, this implies
bn(A, B)?= On(T(A), T(B))
and
o(A, B) :::> o(T(A), T(B)).
The inverse image of an open set is an open set. For any compact set K C Y there is a compact set K 0 C X such that T(K0 ) = K (cf. Proposition 6.5.4). Then
o(X) =.
U BEO U o(A, B) :::> U BEO U b(T(A),
AE~
~
:::>
AE~
T(B))
AE~
A eX BeX
AeX Be X
U b(A, B)= o(Y).
BEO
AeY BeY
COROLLARY 6.6.3.
/f
codim1 X~ codim1 Y
(codim 1X= codim1 Y),
then (resp.b(X)
b(X) C o(Y)
=
o(Y)).
PROPOSITION 6.6.4 (Bessaga, Pelczynski and Rolewicz, 1963). Let 3(X) =
U
n
UEO VEO
UeX VeX
b(V, U).
D
Chapter 6
276
Then b(X) = b(X). Proof Let B be an arbitrary compact set and V, U arbitrary neighbourhoods of zero. Since the set Bis compact, there is a positive number a such that BcaV. Then by (6.6.1) and (6.6.2) we obtain r5(B, U)Cb(V, U) for all B and V. Hence
U b(B, U) C
BE~
BcX
n
b(V, U).
VEO VeX
Therefore
u ub(B, u) c u n b(v, u),
UEO BE~
UEO VEO
i.e. b(X) C '"" b(X). Now we shall show the converse inclusion. Let U0 be a neighbourhood of zero and let
{tn}E
n
b(V, Uo).
VEQ
Let { V1c} be a decreasing countable basis of neighbourhoods ofzero. Fort mula (6.6.3) implies that~~ bn(V:, Uo) = 0 (k = 1, 2, ... ).Then we can choose an increasing sequence of indices kb k 2 , tn lim =0 11--+CO bn( vkn, Uo) . For every V1cn there is a finite Zn C V1cn such that
•••
such that
1
bn(Zn, Uo) ~ 2b(VTcn, Uo). co
Let B =
U Zn. Since the set B has a unique cluster point 0, it is compact. n=l
On the other hand, JtnJ ~ JtnJ ~ 2JtnJ bn(B, Uo) ""' bn(Zn, Uo) ""' bn(Vkn, Uo)
This means that {tn}
E
b(B, U0 ). Then {tn}
E
O -+
b(X).
·
D
As an obvious consequence of Proposition 6.6.4 and (6.6.1) we obtain
Montel and Schwartz Spaces
277
PROPOSITION 6.6.5 (Bessaga, Pelczynski and Rolewicz, 1961). Let { Un} be a basis of neighbourhoods ofzero in an F-space X. Then
n b(Um, Un). co
00
b(X) =
U
n~l m~l
CoROLLARY 6.6.6 (Bessaga, Pelczynski and Rolewicz, 1961). If X is a Schwartz space, then there is a sequence {t;} tending to 0 such that
{t:} ¢b(X). Proof If X is a Schwartz space, then there is a basis of neighbourhoods of zero { Un} such that lim bn( Ui+I• UJ) = 0 (j = 1, 2, ... ). Since we have
only a countable number of sequences bn( Ui+I• UJ), it is not difficult to construct a sequence {t;} tending to 0 such that . I1m
X
In
n~oo bn(Ui+l• Uj)
= +oo,
j= 1, 2, ... ,
Proposition 6.6.5 implies that {t;} ¢ b(X).
D
PROPOSITION 6.6.7 (Mityagin, 1961). Let X be a B 0 -space with the topology given by a sequence of Hilbertian pseudonorms, i.e. the pseudonorms [[x[[n = (x,x)n, where (x,x)n are inner products. Let Y be a subspace of the space X. Then b(Y) C b(X). Proof Let Un = {x: [[x[[n 0 there is a finite set H C V n L such that (6.6.3)
Vn L C H+aU.
Then each bounded set B such that lim bn(B, U) = 0 for all open balanced enighbourhoods of zero U is totally bounded. Proof Let W be an arbitrary neighbourhood of zero. Let U be a balanced neighbourhood of zero such that
U+Uc W. By our hypothesis there is a finite-dimensional space L such that B C L+ +U. Hence Bc(B+U)nL+U. Without loss of generality we may assume that U nL is bounded. Hence for sufficiently small a> 0 1
Bc(B U)nL+Uc-Vn L+U. a
(6.6.4)
Thus by (6.6.3) and (6.6.4) there is a finite set H such that 1
1
Be -H+U+UC -H+W. a a
(6.6.5)
Monte! and Schwartz Spaces
279
Therefore, for all a' > 0, there is a finite setH such that B C H+a'W.
(6.6.6)
0
REMARK 6.6.11. In Lemma 6.6.9 the hypothesis that H is finite can be replaced by the hypothesis that it is totally bounded.
Indeed, if His totally bounded, then for each a' > 0 and a neighbourhood of zero W there is a finite set H 0 such that H C H 0 +a'W.
(6.6.7)
Thus by (6.6.6) and (6.6.7) we obtain B C H 0 +a'W+a'W.
6.6.12. Let X be an F*-space without arbitrarily short lines. Then each bounded set B such that lim (Jn(B, U) = Ofor an arbitrary hal-
PROPOSITION
anced neighbourhood of zero U is totally bounded. Proof The proposition follows immediately from Remark 6.6.11 and Lemma 2.4.6. D
As an immediate consequence of Proposition 6.6.12 we obtain 6.6.13. Let X be a locally bounded space. Let B C X be a bounded set such that lim (Jn(B, U) = 0 for each open balanced neigh-
PROPOSITION
bourhood of zero U. Then the set B is totally bounded. PROPOSITION 6.6.14. Let X be an F*-space with a topology given by a sequence of F-pseudonorms {II lin} (see Section 1.3). Suppose that for each n there is an an > 0 such that ,for all x such that' llxlln # 0,
sup lltxlln > an.
(6.6.8)
tER
Let B be a bounded set such that, for each open balanced set V, lim (Jn(B, V) = 0. Then the set B is totally bounded. n-+ro
Proof Let U be an arbitrary neighbourhood of zero. Then there are n and
280
Chapter 6
a number a'
> 0 such that
{x: llxlln O
Thus we have (6.6.12) and since Lis finite-dimensional the hypotheses of Lemma 6.6.10 hold. D THEOREM 6.6.18 (Turpin, 1973). In the space N(L(Q, E, fl.)) each bounded set B such that lim bn(B, U) = 0 for each balanced neighbourhood of zero ,__,.co U is totally bounded. Proof We shall show that the hypothesis of Lemma 6.6.17 holds. Lets be an arbitrary positive number. Let ,h, ... ,fk be measurable functions. Suppose that there are sequences {sn,i}, i = 1, ... , k such that sn,i >0,
Monte! and Schwartz Spaces
283
Sn i-1 sn,i-+oo,-'- -+oo and for all n Sn,i n
k
PN(l' Sn,ifi)
=
i=l
f N(ll' Sn,Jil)d-,u <e.
(6.6.15)
i=l
Q
Let k
A=
U {t: .fi(t)::;i:O}. i=l k
For each tEA,
I};
S11 ,i./i(t)
I tends to infinity.
Thus, by (6.6.15) and
i=l
the Fatou lemma,
J
A
N(u)d',u ~ e,
sup
O-00
The class b'(A, B) has the following properties: if A' C A, B':) B, then b'(A', B') :) b'(A, B), (6.6.1') b' (aA, bB) = b' (A, B) for all scalars a, b different from 0. (6.6.2') Let b'(X) =
n n b'(K, U).
KE':JUeO
By a similar argument to that used for b(X) we find that b'(X) is an invariant of linear co dimension, which means that if codimzX ~ codimzY
(codimzX = codimzY),
then b' (X) :) b' (Y)
(resp. b' (X)= b' (Y)).
Let J'(X) =
U U b'(V, U).
UeO VeO
By a similar argument as to that used in the proof of Proposition 6.6.4 we obtain '
PROPOSITION
-
6.6.22. b (X) = b'(X).
As a consequence of Proposition 6.6.22 we obtain the following proposition, in a certain sense dual to Proposition 6.6.20.
Monte] and Schwartz Spaces PROPOSITION
6.6.23./f a space M(am,n) is regular, then
()' (M(am,n))
=
{
{tn}: for all p there is a q such that tn
CoROLLARY
285
ap,n aq,n
-+
o}.
6.6.24. Let {an}, {bn} be sequences ofreals tending to infinity.
Then (j'(M(b;;I/m)) ()'(M(a~))
=
M(b;;l/m),
= M-(a?;:) = {{tn}: tna;;m -+0 for some m }.
In the same way as in Corollary 6.6.7 we can prove 6.6.25. Let X be a B0 -space with the topology given by a sequence of Hilbertian pseudonorms. Let Y be a subspace of the space X. Then ()'(X):::> ()'(Y).
COROLLARY
The following, natural question arises : when does the equality of diametral approximative dimensions imply isomorphism? We shall show that this holds for an important class of spaces called Kothe power spaces. Let an be a sequence of positive numbers tending to infinity. The space M(a?;:) is called a Kothe power space of the infinite type, the space M (a~Ifm) is called a Kothe power space of the finite type. To begin with, we shall show that two Kothe power spaces of infinite type (of the finite type) induced by sequences {an} and {~n} are equal as the sets if and only if there are two positive constant A, B such that
A< logan