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0, x where aK "'" {ak IkE K}.
E
aK}
Define the functional p on X by
p(x),= inf A (x) We shall call p the M inlcowBlci functional of K. 1(, A (x) r,I t/l and 0 'S p(:r:) < 00
'd
Since 0 is an interior point
INTRODUCTION TO NORMED LINEAR SPACES
1.5.12
Lemma.
23
Let K and p be as in the above definition.
Then for
x and y in X,
i. ii. iii.
p(ax) = ap(x) , a ~ 0 p(x + y) ~ p(x) + p(y) p(z) ~ 1 for all Z f K
Proof of (i). It is clear that p(O) a E A(ax), x is in a-1aK and therefore ~
p(x)
inf a- 1a
=
= O. Suppose
a
>
O.
Given
a- 1p(ax)
a.A(a,,)
Thus ap(x)
~
p(ax), a
~
O.
This result implies a>O
or ap(x)
~
p(ax).
Hence p(ax) = ap(x), a
~
O.
Proof of (ii). We first observe that if a and b are nonnegative real numbers, then (a + b)K = aK + bK. Indeed, if x and yare in K, then the convexity of K implies a b a+bx+a+b YEK
Thus ax bK C (a
+ by
E
+ b)K.
(a
a+b~O
+ b)K. Since x and yare arbitrary in K, aK + Obviously, (a + b)K C aK + bK. Hence (a
+ b)K =
aK
+ bK
To conclude the proof of (ii) , we suppose a E A (x) and bE A(y). Then x + y E aK + bK = (a + b)K. Hence p(x + y) ~ a + b. Since a and b are arbitrary in A (x) and A(y), respectively, it follows that p(x + y) ~ p(x) + p(y). Proof of (iii). Suppose Z f K but p(z) < 1. Then there exists an r, such that Z E rK. Since 0 E K and K is convex, it follows that rk = rk + (1 - r)O E K, k E K. Thus Z E rK C K, which is a contradiction.
o ~ r < 1,
Proof of theorem 1.5.10. .Since x , K and K is closed, there exist!;! an r-ball 8, r> 0, such that x + 8 and K are disjoint. Thus k o E K implies x - k o ' -ko K - 8 = K 1• Furthermore, 8 = -8 C K 1, showing that 0 is an interior point of K 1. Since K and 8 are convex, it follows thl1t [(1 is also convex. Hence tj)e Minkowski functional p of K 1 if! do{inn!l. W (I first provo tho thcore~ for tho cl1se when X is 11 real
+
UNBOUNDED LINEAR OPERATORS
24
normed linear space. On the one-dimensional space sp {xo I, where Xo = x - k o, define linear functional f by
Now, f(axo) ::; p(axo); for if a ~ 0, then f(axo) = p(axo) by Lemma 1.5.12. If a < 0, then f(axo) = ap(xo) ::; 0 ::; p(axo). Hence, by Remark 1.5.3 and Lemma 1.5.12, there exists a linear functional F which is an extension of f to all of X such that F(y) ::; p(y), Y E X. Moreover, F is bounded; for if Ilyll = 1, then ±ry ESC K 1• Hence ±F(y) = F(±y) ::; p(±y) ::; r
1
Since Xo = x - k o ¢ K 1 and -k o + k E -k o + K - S = K 1 for all k E K, it follows from (iii) of Lemma 1.5.12 and the definition of p that for each kEK, F(x) - F(k o) = f(xo) = p(xo)
~
1 ;:::: p( -k o
+ k)
;:::: -F(k o)
+ F(k)
Thus F;eO
and
F(x) ;:::: F(k),
kEK
proving the theorem for real normed linear spaces. If X is complex, then as in the proof of the Hahn-Banach extension theorem, let X T be X considered as a real normed linear space. Then by what has just been shown, there exists an fT ;e 0 in such that fT(X) ;:::: fT(k), k E K. Define f on X by
X;
f(x) = fr(x) - ifT(ix) It is easy to verify that f satisfies the demands of the theorem.
For a treatment of separation theorems for convex sets the reader is referred to Dunford and Schwartz [1], pages 409-418.
1.6
CONJUGATE SPACES
1.6.1 Definition. Normed linear spaces X and Yare called equivalent if there exists a linear isometry from X onto Y.
A particular caee of the next theorem ie eSl!lential for the etudy of o} = n~1 En
is countable, since each En is finite. The class of those M for which there exists a solution to Problem 1.7.8 has now been enlarged to those M = sp {u,,}, where {u a } is an arbitrary orthonormal set. Thus, to solve the problem for general closed subspaces M, it remains to show that M is the closure of the span of some orthonormal set. 1.7.14 Lemma. Given any closed supspace M ~ 0 of Hilbert space X, there exists an orthonormal set {u a } such that M = sp {u,,}, Proof. The proof utilizes Zorn's lemma in order to prove the existence of a maximal orthonorm~l set 0 contained in M, maximal in the sense that 0 is not a proper subset of any orthonormal set contained in M. Let (p be the class of all orthonormal subsets of M. (p is not empty since it contains the I-point set {m}, where m eM and Ilmil = 1. Partially order (p by set inclusion. Let 3 be a totally ordered subclass of (p. Since V S is an upper bound in (p for 3, Zorn's lemma asserts that
The requirement that the domain of the operator be dense in X is not as restrictive as might appear at first glance. Suppose ~(T) is not dense in X. Then one can define Xl as the closure of the domain of T and obtain the conjugate T' as a mapping from a subspace of Y' into X~. By inspecting the theorems which require that ~(T) = X, one can usually remove this restriction by considering Xl in place of X. This is done, for example, in the proofs of Theorems V.l.6 and V.l.8. The following simple examples are presented in detail in order to give somo "fr.clilll/;" for I,ho dofinition of It eOlljul/;ItLc opomj,or. Tho conjugl\toF! of (~ort.ltill diITorollt.iul opornl,orfl Itrll ddlll'lllillllll ill Chup. VI.
LINEAR OPERATORS AND THEIR CONJUGATES
Il.2.3
Example. U1
Let X = Y =
= (1,0,0, . . .),
be the unit vectors in
[p.
51
1
[p,
~ p
LINEAR OPERATORS AND THEIR CONJUGATES
59
The theorem shows that state III is impossible for T'. we write T' ¢ lIt.
For brevity,
II.3.2 Definition. If C is a subset of X', then the orthogonal complement of C in X is the set .LC = {x I x E X, x'x = 0 for all x' E C}. II.3.3 Remarks. K.L and .LC are closed subspaces of X' and X, respectively. AlsoK.L = K.L and.LC = .LC.
The proofs of Theorems 11.3.4 to 11.3.8 are left to the reader. II.3.4
Theorem.
If M is a subspace of X, then .L(M.L) =
M.
II.3.5 Theorem. If N is a subspace of X', then (.LN).L :J N. is reflexive, then (.LN).L = N. II.3.6 Remarks. Dieudonne [1] has shown that subspace of X', then (.LN).L = N. Since we later (.LN).L = N, for N finite-dimensional, we shall prove Dieudonne's theorem. Since N C (.LN).t, it follows from Theorem 1.6.4 (1)
dim
.L~
=
dim
(.L~)' =
If X
if N is a reflexive use the fact that this special case of that
dim (.LN).L ;::: dim N
If x~ ,x~, ., x: is a basis for N, then the map A from Xj.LN into Un defined by A[x] =- (x;x, x~x, . . . ,x:x) is 1-1 and linear. Thus (2)
dim
.L~ ~
n = dim N
It follows from (1) and (2) that N = (.LN).L.
II.3.7 t.
ii.
Theorem (R(T).L = (R(T).L = ;neT') (R(T) = .L;n(T')
In particular, T has a dense range if and only if T' is 1-1.
By interchanging the roles of T and T' in the above theorem, we do not quite obtain the dual type theorems. 11.3.8
. i. ii.
£?:i.
Theorem J.(R(T') ~ ;neT) If ~(T') is total, then ;neT) = .L(R(T') (', :nCT) ~ C ;n(T).L
In pn.rI1'I'lIlnr,
1/ 0 be given.
By hypothesis, there exists an integer
(1)
Since K N is precompact, there exist elements Xl, X2, .'1:; for which
S of X such that given X ~ S, t.hcrn is nn
(2)
., Xm
in the I-ball
78
UNBOUNDED LINEAR OPERATORS
Hence, from: (1) and (2), IIKx - KXil1 ::; IIKx - KNxl1
+ IIKNx -
KNxi/1
+ IIKNXi
-
sse
KXil1 < "3 + "3 + "3
which implies that K is precompact. As an application of the lemma, the following example of a compact operator is given. 1I1.1.6 Example. Let I = [0, 1] and let 1 < p, q < 00, with p' and q' conjugate to p and q, respectively. Suppose k(s, t) is in £r(l X I), where r is the larger of p' and q. Then the linear operator K defined by (Kx)(t) =
10
1
k(s, t)x(s) ds
is compact as a map from £,,(1) into £q(1). Proof. To show, first of all, that K maps £,,(1) into £q(I), suppose x is in £,,(I). Since r' ::; p, we have, hy Holder's inequality, that x is in £r,(I), Ilxll r ' ::; /Ixll", and (1)
IIKxll qq = fa1 dt I fa1 k(s, t)x(s) dslq ::; fa1 dt [( fa1 Ik(s, t)lr ds )q1rllxllr,q] ::; /Ixll"q fa1dt (fa1 I(k, t)/r ds)q1r
Since 0
0 a fJ = fJ(c) > 0 such that for all SE
I,
For such t 1 and t 2 ,
Since {x n } is bounded in "cp(I) , (4) shows that {Yn} is equicontinuous on I. Hence, by the Ascoli-Arzela theorem, {Yn} contains a subsequence converging in C([O, 1]) which in turn implies convergence in"cp([O, 1]). Thus K is compact. Removing the restriction that k be continuous, we nevertheless know by Theorem 0.9 that there exists a sequence {k n } of functions continuous on [0, 1] X [O,l]whichconvergesin"c,(I X l)tok. LetKnd"cp(I),,,cq(I)] be defined by
Then by what was just proved, K n is compact. k in (3) gives, for each x E "cp(I) ,
II(Kn - K)xllq ::; I[xll p (
10[l 10(l [kn(s, t)
Thus K n ~ K in ["cp(I) , "cq(I)]. III. 1.5.
Substituting k n
k(s, t)I'dsdt
-
)1
1,
~0
-
k for
as n -
eX)
Consequently, K is compact by Lemma
In general, if Y is not complete, the limit in [X, Y] of compact operators need not be compact. This may be seen in the following example. IIl.I.7 Example. Let Co be the subspace of l", consisting of the sequences which converge to O. Let To be the operator mapping Co into l2 defined by
T o( (
ak})
=
{~k}
1 ::; k
Define Tto be the operator To considered as amapfromcoonto Y = n,converges in [co, Y] to T. Moreover, (R ( Tn) is finite-dimensional and therefore Tn is compact. Corollary III.1.l0 shows how compact operators may arise from closed operators which do not have a closed range. The following theorem, excluding the assertion that the operator is precompact, is due to Kato [1]. III.I.B Definition. A subspace M of a vector space V is said to have finite deficiency in V if the dimension of V / M is finite. This is written . V d1m M
0, there exists an infinite-dimensional subspace M(e) contained in 'J)(T) such that T restricted to M(e) is precompact and has norm not exceeding e. Proof. The hypothesis implies the existence of an Xl E X such that 1 and IITxll1 < 3- le. There is an x~ E X' such that Ilx~11 = 1 and X~Xl = Ilxlll = 1. Since ;n(x~) has deficiency 1 in X, there exists an X2 E ;n(x~) such that IIx211 = 1 and IITx211 < 3- 2 e. There exists an x; E X' such that Ilx;11 = 1 and X;X2 = IIx211 = 1. Since ;n(x~) (\ ;n(x;) has finite deficiency in X, there exists an Xa E ;n(x~) (\ ;n(x;) such that Ilxall = 1 and IITxal1 < 3- ae. Inductively, sequences {xd and {x~} are constructed in X and X', respectively, with the following properties.
IIxIII =
(1) k-l
(2)
Xk E ( \
;n(x~)
or, equivalently, X~Xk = 0
1
~ i
0 and given M an '/,'nfinite-dimensional subspace of X, there exists an infinite-dimensional subspace N C M such that B restricted to N has norm not exceeding e.
Proof. (i) implies (ii). Suppose B is strictly singular and M is an infinite-dimensional subspace of X. Then B M , the restriction of B to M, is strictly singular. Hence the existence of an N, with the properties asserted in (ii), is a consequence of Theorem III.1.9 applied to B M • (ii) implies (iii). If M is an infinite-dimensional subspace of X, then B does not have a bounded inverse on any subspace having finite deficiency in M; otherwise, B would be precompact and have a bounded inverse on an infinite-dimensional subspace. But this contradicts Theorem III.1.3. Hence (iii) follows by applying Theorem III.1.9 to B M . (i) follows easily from (iii).
The next theorem, due to Lacey [1], implies that if one adds to the above theorem that N has finite deficiency in M, then a characterization of the precompact operators is obtained. lII.2.2 Lemma. If S is a bounded set in X and x~, x~, , x: are in X', then for any e > 0, there exist Xl, X2, . . . ,Xm in S such that given XES, one can find an Xk such that
Proof.
The map A from X to unitnry n-Apo.co definod by A:I' -
(.r;;r, :r~,r, ' , , ,.r;,:r)
85
STRICTLY SINGULAR OPERATORS
is bounded and linear. Hence A is compact. Thus there exist , Xm in S such that given XES, there is an Xk such that
Xl,
X2,
lII.2.3 Theorem. Suppose B E [X, Y]. Then B is precompac~ if and only if for every e > 0 there exists a subspace N having finite deficiency in X such that B restricted to N has norm not exceeding e. Proof. Let B be precompact and let e > 0 be given. There exist X2, . . . ,Xn in the i-ball S of X such that given XES, there is an Xk such that Xl,
(1)
Let y~,
, y: be in the i-sphere of Y' such that y:BXi = flBxill.
Now
n
N = (\ ;n(y:B) has finite deficiency in X.
Suppose X E N (\ Sand Xk
;=1
Since X is in ;n(y~B),
is such that (1) holds.
Thus, from (1) and (2), (3)
Since X was arbitrary in N (\ S, (3) implies that B restricted to N has norm not exceeding e. Conversely, let e > 0 be given. By hypothesis, there exists an N of finite deficiency in X such that B N , the restriction of B to N, has norm not exceeding e. Since IIBNII = IIBNII, we may assume that N is closed. Hence, there exist VI, V2, . • • ,Vn in X such that
X
= N
E9 sp
lVI, . . .
,vnl
From the construction of the projection in Theorem II.1.16, it follows that there exist x~, x~, ... ,x: in X' such that every x E X has a unique representation of the form n
(4)
:r
= 'II -I-
_
L X;(X)li; ,-I
86
Since
UNBOUNDED LINEAR OPERATORS
IIBNII
~
€, we have IIBxl1
(5)
~
€llu\l +
~
Ilxll +
n
L Ix:(x)IIIBvi\l
;=1
By (4),
\lull
(6)
n
L Ix:(x) I Ilv,ll
;=1
Combining (5) and (6) and choosing K = max it follows that
IIBxl1
(7)
~
€llxll + €K
n
{IIBv,ll, \lv,ll; 1
~ i ~
n},
n
L Ix:(x)! + K .=1 L Ix~(x)1 .=1
XeX
Given 17 > 0, there exist, by Lemma III.2.2, elements Xl, S such that given v e S, there is an Xk such that
,xm in
X2,
n
L Ix~v -
x:xkl < 17
.=1
Thus, substituting obtain
Since
€
>
0 and 17
v-
>
Xk
for
X
in (7) and noting that
IIv -
Xkl! ~
2, we
0 are arbitrary, it follows that BS is totally bounded.
III.2.4 Theorem. The set SIX, Y] of strictly singular operators and the set of precompaet operators (PX[X,. Y] in [X, Y] are closed subspaces. The set of compact operators xIX, Y] is a subspace (not necessarily closed). Proof. It is obvious that the sets are closed under scalar multi~ plication. Suppose K and L are in xIX, V]. Let {x n } be a bounded sequence in X. Since K is compact, there exists a subsequence {Yn} of {x n } such that {KYn} converges. Since L is compact, there exists a subsequence {Zn} of {Yn} such that {Lz n } converges. Hence {(K + L)z,,} converges, which shows that K + L is compact. Suppose K and L are in (pxlX, Y]. Then K and L are in xIX, f], where f is the completion of Y. Hence, by the above result, K + L is in xlX, f], or, equivalently, K + L is in (pxIX, Y]. Rupposc K find L are in SIX, Y]. Let. M ho ILIl illfinito-climollHionnl !:luhelllWe of X. Them, hy Th(1orom III.2.1, t,horo OXiHI,H nil infiniLo-dilll(\ll~
STRlcnv SINGULAR OPERATORS
87
sional subspace N 1 C M such that K is precompact on N 1 and there exists an infinite-dimensional subspace N C N 1 such that L is precompact on N. Since K is also precompact on N, K + L is precompact on N by what we have just shown. Hence K + L is strictly singular by Theorem 111.2.1. (PX[X, Y) was shown to be closed in Lemma 111.1.5. Let K n ~ K in [X, Y), where each K n is strictly singular. Suppose K has a bounded inverse on subspace M C X. Then there exists a c > 0 such that for all meM,
IIKml1 :2 cllmll
(1)
Choose p so that (2)
Then from (1) and (2),
Thus K p has a bounded inverse on M. Since K p is strictly singular, M is finite-dimensional, whence K is strictly singular. If Y is complete, then the space of precompact operators coincides with the space of compact operators and therefore X[X, Y) is closed. Example 111.1.7 showed that if Y is not required to be complete, then X[X, Y) need not be closed. lII.2.5 Theorem. Let Z be a normed linear space. If K is in X[X, YJ, (PX[X, Y), or S[X, Y), then for A e [Z, X), KA is in X[Z, Y), (PX[Z, Yj, or S[Z, Y), respectively. Similarly, if B is in [V, Z), then BK is in X[X, Z], (PX[X, Z), or S[X, Z), respectively. Pro~f. We shall only prove the theorem for K e S[X, Y). The proofs for K in X[X, Y) or (PX[X, Y) are also easy. Suppose KA has a bounded inverse on a subspace M of Z. Then there exists a c > 0 such that for all x e M
IIKAxl1
:2 clJxll :2 II~ II IIAxll
Thus K has 11 bounded inverso on AM,jl.nd therefore AM is finite-dimenBut A iH 1-1 liIirwo [(A is 1-( Honce ll! is finite-dimensional. COlll:loqllontly, K A is strictly sinll:ullLl'.
MioJla!.
UNBOUNDED LINEAR OPERATORS
88
Suppose BK has a bounded inverse on a subspace N of X. there exists a c > 0 such that for all x E N, IIBllllKxll ~
IIBKxl1
~
Then
cllxll
Thus K has a bounded inverse on N, whence N is finite-dimensional. Therefore BK is strictly singular. III.3
EXAMPLES OF NONCOMPACT, STRICTLY SINGULAR OPERATORS
For the purpose of exhibiting some strictly singular operators, we prove Theorem IlL3.4 and Corollary IlL3,6, which are due to Whitley [1]. Ill.3.1 Definition. A linear operator which takes bounded sequences onto sequences which have a weakly convergent subsequence is called weakly compact.
III.3.2 Remark. Every weakly compact operator is bounded. For suppose B is a weakly compact operator which is unbounded. Then there exists a sequence Ix n } such that I Bx n } converges weakly and II BXn II ---t 00. But this contradicts Theorem Il.1.12. III. 3.3 Theorem. If at least X or Y is reflexive, then every operator in [X, Y] is weakly compact. Proof. The proof follows easily from Theorem 1.6.15 and the continuity of the operator. III.3.4 Theorem. A weakly compact operator on a Banach space which maps weakly convergent sequences onto norm convergent sequencies is strictly singular. Proof. Let BE [X, Y] satisfy the hypotheses of the theorem. Suppose B has a bounded inverse on a closed subspace M eX. Let lmd be a sequence in the I-ball of M. Then there exists a subsequence IVk} of lmd such that lBvd converges weakly to some y E Y. Since BM is closed, y is in BM; otherwise there exists a y' E Y' such that y'y = 1 and y'(BM) = 0, implying that lBvk} does not converge weakly to y. From the assumption that B has a bounded inverse on M, it follows that IVk} converges weakly to BM-ly. From the hypothesis, lBvd converges in norm to y and therefore lvd converges in norm to BM-ly. Hence the I-ball of M is compact and M is finite-dimensional. Therefore B is strictly singular.
The proof of the next theorem requiroR quite fl, bit of intep;mtion theory and will 1I0t ho ilwludod 111'1'0, Thl' rlmdor iH l'efer'l'OlI to J)ulll'Ol'd
STRICTLY SINGULAR OPERATORS
89
and Schwartz [1], Theorem VI.8.12, the remark preceding the theorem on page 508, the remark following Theorem V1.8.14 on pages 510-511, and the representation of the conjugate spaces in tables on pages 374-379. We cite only a few spaces X for which the following theorem holds. 111.3.5 Theorem. Every weakly compact map from X or X' into a Banach space takes weakly convergent sequences onto norm convergent sequences whenever X is anyone of the following Banach spaces. t.
ii. tn.
£1(8,~, fJ.)
and £00(8, ~, fJ.), where (8, ~, fJ.) is a positive measure space. C(S), where 8 is a compact Hausdorff space. B(S), where 8 is an arbitrary set.
Thus every weakly compact operator from X or X' into a Banach space is strictly singular.
Pelczynski [2] proved that the strictly singular operators and the weakly compact operators are the same in the following spaces: t.
ii.
[C(8), Y], Y an arbitrary Banach space and 8 a compact Hausdorff space. [£1(8,~, fJ.1), £1(8, ~, fJ.2)], 8 a topological space, ~ the field of all Borel subsets of 8, and fJ.1 and fJ.2 nontrivial measures (for details of these spaces, see Dunford and Schwartz [1], Chap. IV).
111.3.6 Corollary. Let X be anyone of the spaces in the above theorem. Then every bounded linear map from X or x' into a reflexive space is strictly singular. Proof.
Theorems III.3.3 and III.3.5.
The rest of this section gives examples of strictly singular operators which are not compact. 111.3.7 Example. Let X = l1 and let Y be any infinite-dimensional separable reflexive Banach space. By Corollary 11.4.5, there exists a bounded linear operator B which maps X onto Y. Hence B is strictly singular but not compact by Corollaries II1.3.6 and III.1.12. Without referring to Corollary III.3.6, B can be shown to be strictly singular based on the fact that weak convergence is the same as norm convergence in l1, as remarked in 1.6.14. B' is not strictly singular, since it has a bounded inverse by Theorem IIAA. Now B" maps X" into reflexive space Y", and X" may be identified with the conjugate of loo. Thus, by Corollary III.3.fi, B" is strictly singular. This example shows, in particular, that t,he eonjup;nte of 11 strictly singular operator need not be strictly singular lLll0
and CR(T) is closed, then T is closed.
Proof. LetT be the 1-1 operator induced by T. Then ('1')-1 is continuous with domain a closed subspace of Y. Thus ('1')-1 is closed, and therefore '1' is also closed. Hence T is closed by Lemma 11.4.7. IV.l.8 Theorem. Let 'JL(T) be closed and let 'J)(T) be dense in X. "(T) > 0, then "(T) = "(T') and T' has a closed range.
If
Proof. Let Y 1 = CR(T) and let T 1 be the operator T considered as a map onto Y 1• The 1-1 operator '1'1 is surjective and has a bounded inverse, since "(T 1) = "(T) > O. Hence, by the State diagram 11.3.14, ('1'1)' is also in 11' Before evaluating "(T'), we need the following observations.
i.
By Theorems 11.3.7 and 1.6.4, Y' /'JL(T') = Y' /CR(T)J. and Y' /ffi(T)J. is equivalent 1.0 Y: under the map [V'] - y~, whore 11;/ ifil thC1 TI1Al.ridion of 11'10 VI. III plI.l'lil'1I111r, 11111'111 ... 111I~IIl.
OPERATORS WITH CLOSED RANGE
ii. iii.
99
It is clear that II(T)'y'll = II(T1)'y~ll. Thus, for y' E D(T'), 11
°
and T' is closed.
The above theorem does not hold if "{(T) is not required to be positive. Indeed, Diagram 11.3.14 shows that state (1 2 , 111 1 ) can exist. However, if T is assumed to be closed, then the following corollary due to Kato [1] is obtained. IV.l.9
Corollary.
If X and Yare complete and T is closed, then
"{(T) = "{(T').
Proof. By Theorems IV.1.6 and IV.1.2, the statements "{(T) = 0, CR(T) is not closed, CR(T') is not closed, and "{(T') = are all equivalent. If "{(T) > 0, then by the theorem just proved, "{(T) = "{(T').
°
The remaining portion of the section gives some conditions under which a closed operator has a closed range. IV.l.10 Theorem. Let X and Y be complete and let T be closed. A sufficient condition that T have a closed range is that T maps bounded closed sets onto closed sets. If ;n(T) is finite-dimensional, then the condition is also necessary. Proof. Suppose T maps bounded closed sets onto closed sets. If CR(T) were not closed, then "{(T) = 0, which implies the existence of a sequence {[xnll in XJ(T) j;n(T) such that (1)
II[xn]11
=
1
and
We show that the existence of such a sequence leads to a contradiction. Consequently, CR(T) has to be closed. Let {znl be a sequence such that Zn E [x n] and Ilznll ~ 2. Then TZ n = TX n -... O. If IZnl has no convergent subsequence, it is a closed bounded set. Thus, by hypothesis, I TZ n I is a closed set, whence TZ N = 0 for !:lome N. nut tlwn III;I'NIII = II[ZN]/I--= 0, which contradicts (1). If .:,,' -. t for !'IOm(1 HII1IHII(\I\(1I1I'1' Iz", I, 1.111'11 'I'z ~ 0, HiIH'1' 'I'z", ) 0 al1d '/' iii
100
UNBOUNDED LINEAR OPERATORS
closed.
Hence
which again contradicts (1). Suppose ;)"[(1') is finite-dimensional and (1') n 8 and y = Tx = T(x - z) E 1'8. Thus TS is closed. IV.l.ll Lemma. Let Y be complete. Suppose that M and N are linearly independent closed subspaces of Y such that Y = M EEl N (cf. Definition 1.7.16). Then Y' = M1. EEl N1.. Proof. Clearly, M1. n N1. = (0). By Theorem I1.1.14, there exists a projection P from Y onto M with ;)"[(P) = N. Giveny' E Y', let y~ = y'P and let y; = y'(I - P). Then y' = y~ + y;, y~ E N1. and y~ E M 1.. IV.l.12 Theorem. Let X and Y be complete and let l' be closed. Suppose N is a closed subspace of Y such that (1'0) = 1>(1') X N C X X y
To(x, n)
=
Tx
+n
The linear operator To is closed. Suppose (Xlr, nk) ~ (x, y) E X X Yand To(xk, nk) ----> z. Then Xk ~ x, nk ~ y and TXk + nk ~ z. Hence yEN and TXk ----> Z - y. Since Tis closed, x is in 1>(1') and Tx = Z - y. Thus (x, y) E 1>(1') X N = 1>(1'0) and To(x, y) = Tx + y = z. By hypothesis, dim N Then there exists an m ;;e 0 in M such that
Lemma.
(thus dim N
O.
Let B be bounded with ;nCB)
~
IIBII < 'Y(T),
aCT + B) :::; aCT) dim Y j Lemma V.1.1, aCT + B) :::; aCT).
II[x]l\ = d(x,
meT)), and therefore, by
Proof oj (ii). Let XI = ~!)('1') Itncl let B I bo B reHtl'ictou to ~D(T). ConRidclI'ili1/: '/' Illid HI /1.101 OpPI'/I(,OI'H with dOT1millfol c\PIHI(1 in XI. t,hn 0011-
PERTURBATION THEORY
111
jugates T' and B~ exist with domains in Y' and ranges in X~. Theorem IV.1.8. -yeT')
=
-yeT)
> IIBII 2': IIBIII =
By
IIB~II
Hence, it follows from (i) applied to T' and B~ and from Theorem IV.2.3, that dim ----:::=TiiFy=;=~ = dim 0 such that for
IAI < p, (2)
K(T
+ AB)
= K(T)
anrt
a(T I
+ AB
I)
... K(T I ) "" a('l'I)
116
UNBOUNDED LINEAR OPERATORS
Now, (1) implies that ;neT particular,
+ XB)
= ;n(Tl
+ XB 1)
for X ~ O.
In
(4)
It is clear from (2), (3) and (4) that OI(T + XB) and (j(T + XB) are constant in the annulus 0 < 171.1 < p under the assumpt'on OI(T) < 00. If OI(T) = 00, then (j(T) < 00 by hypothesis. As in the proof of Theorem V.1.5, we apply the result just proved to the appropriate conjugate operators in order to prove the corollary when OI(T) = 00. V.I.B Theorem. Let X and Y be complete and let B be continuous with S:>(T) C S:>(B). Define U to be the set of X for which T + XB is normally solvable and has an index. Then i. ii.
U is an open set. If C is a component of U, that is, a largest connected subset of U, then on C, with the possible exception of isolated points, OI(T + XB) and (j(T + XB) have constant values nl and n2, respectively. At the isolated points,
and Proof of (i). of T.
For X E U, apply Theorem V.1.5 to T
+ XB
in place
Proof of (ii). The component C is open, since any component of an open set in the space of scalars is open. Let 01(71. 0) = nl be the smallest integer which is attained by 01(71.) = OI(T + XB) on C. Suppose 01(71.') ~ nl. Owing to the connectivity of C, there exists an arc r lying in C with endpoints 71. 0 and X'. It follows from Corollary V.1.7 and the fact that C is open, that about each IL E r there exists an open ball S(IL) contained in C such that 01(71.) is constant on the set S(IL) with the point IL deleted. Since r is compact and connected; three exist points
on r such that
, S(X n ) cover rand SeX,)
n
S(X;+I) ~ cJ> O:::;i:::;n-1
We assert that 01(71.) = 01(71. 0) on all of S(X o). Indeed, it follows from Theorem V.1.5 that 01(71.) :::; 01(71. 0) for Xsufficiently close to 71.0. Therefore, since 01(71. 0) is the minimum of 01(71.) on C, 01(71.) = 01(71. 0) for X sufficiently close to 71. 0• Since 01(71.) is constant for all X ~ 71.0 in 8(71.0), thiA COIIHtl1nt must be OI(XO). Now OI(X) is I:OIIH(,ILII t on tho Aot 8 (X,) wi (,h tho poi 11 I, X,
117
PERTURBATION THEORY
deleted, 1 ~ i ~ n. Hence, it follows from (1) and the observation a(X) = a(X o) for all X E S(X o), that a(X) = a(X o) for all X ~ X' in S(X') and a(X') > nl. The result just obtained can be applied to T' + XB~, as in Theorem V.1.6, in order to prove the analogous results for (j(T
+ XB) = a(T' + XB~)
V.l.9 Remarks. The following examples show that Theorem V.1.6 does not hold in general even though IIBII = 'Y(T). If we take T to be the identity operator mapping an infinite-dimensional Banach space onto itself, then 'Y(T)
=
1
=
and
IITII
a(T) = (j(T) = 0
However, T perturbed by - T is the zero operator which has 00 as its kernel index and deficiency index. Less trivially, in Example IV.1.5, choose Xl, X2, • • • to be a strictly decreasing sequence of numbers converging to a positive number r. Then a(T) = (j(T) = 0, 'Y(T) = r, but 'Y(T - rI) = O. Hence, T - rI does not have a closed range, and therefore (j(T - rI) = 00 by Corollary IV. 1.6. V.2
PERTURBATION BY STRICTLY SINGULAR OPERATORS
If B is strictly singular with no restriction on its norm, we obtain the following important stability theorem due to Kato. V.2.1 Theorem. Let X and Y be complete and let T be normally solvable with a(T) < 00. If B is strictly singular and 5)(T) C 5)(B), then i. ii. m.
T + B is normally solvable. K(T B) = K(T) a(T XB) and (j(T XB) have constant values nl and n2, respectively, except perhaps for isolated points. At the isolated points,
+ +
00
>
a(T
+
+ XB) >
nl
and
(3(T
+ XB) > n2
Proof of (i). Since a(T) < 00, there exists a closed subspace M of X such that X = M EEl 'J'L(T). Let T M be the operator T restricted to M n 5)(T). Then T M is closed with m(T M ) = m(T). Suppose that T + B does not have a closed range. Then it follows from Lemma V.1.5 that the e1mlP.d oprmtor T M dool(not have a closed l'fLnge. Hence thC1TO l"XiRtR, hy Corollary TTT.l.lO, n. dORnel infinit.n-dimoTlRinnfLI BuhApnr.O
+n
118
UNBOUNDED LINEAR OPERATORS
M 0 contained in :n(TM )
= :n(TM
+ B) such that
II(T M + B)xll < 'Y(;M)
IIxll
xfiM o
Thus, since T M is 1-1, it follows that for all x in M o,
which shows that B has a bounded inverse on the infinite-dimensional space Mo. This, however, contradicts the hypothesis that B is strictly singular. We now prove a(T + B) < 00. There exists a closed subspace N I such that
;n(T
(1)
+ B)
= ;n(T
+ B) (\ ;neT)
E9 N I
Let T I be the operator Trestricted toNI. By Lemma IV.2.9, CR(T I ) = TN! is closed. Moreover, T I is 1-1. Hence its inverse is bounded. Since B = - T I on N I and B is strictly singular, N I must be finite-dimensional. Therefore (1) implies that ;n(T + B) is finite-dimensional.
Proof of (i,t). We have shown above that for all X, T + XB is normally solvable and a(T + XB) < 00. Thus, applying Theorem V.1.6 to T + XB in place of T, it follows that cp(X) = K(T + XB) is continuous from [0,1) into Z, where [0,1) has the usual topology and Z is the topological space consisting of the integers and - 00 with the discrete topology. Hence cp is a constant function. In particular, K(T) = cp(O) = cp(l) = K(T
+ B)
Proof of (iii). Since T + XB is normally solvable for all X, (iii) is an immediate consequence of Theore~ V.1.8. V.2.2 Corollary. Let X and Y be complete with :neT) dense in X. Let T be normally solvable and have an index. Suppose :neT) C :nCB) and both Band B' are strictly singular; for example, B is compact. Then ~.
n. m.
T + B is normally solvable. K(T + B) = K(T) a(T + XB) and (J(T + XB) have constant values nl and n2, respectively, except perhaps for isolated points. At the isolated points, and
Proof.
Apply Theorolll V.2.1 {,o 7"
(J(T R,lld
+ XB) > n2
11'
Whllll
(J(T)
O.
Ilxlll "'" /1,II.r.11 + bllTxl1
PERTURBATION THEORY
123
The idea of the proof is to apply Theorem V.1.6 to T 1 and B I in order to obtain the assertions about T + B. We therefore want to know how 'Y(T 1) compares with 'Y(T). This can easily be obtained in the following manner for a > 0 and b > O. Given x in D I , d(x, ;n(TI)) where
=
inf
nE~(T)
Ilx - nlll = a
[xl E <J)(T)/;n(T).
inf
nE~(T)
Thus = inf XED.
Ilx - nil + bllTxll = all[xlll + bllTxl1
IITxl1 all[xlll + bllTxl1
IITxll/ll[xlll
= inf XED.
a
+ bIITxll/ll[xlll
'Y(T)
a
+ b'Y(T) >
1
Since IIBIII ~ 1 < 'Y(T I) and T I is bounded on the Banach space D I, it follows that the conclusions of Theorem V.1.6 hold for T I and B I in place of T and B, respectively. However, m(T I + B I) = m(T + B), and ;n(T I + B I) = ;neT + B). Thus Theorem V.1.6 holds for T and B, where a> O. If a = 0, choose E > 0 so that E + b"{(T) < 'Y(T). Then IIBx!1 ~ Ellxll + bllTxll, and the theorem is proved by what was just shown. V.3.7
Theorem.
Theorem V.2.1 holds if B is T-strictly singular.
Proof. Let D I , T 1, and B I be as in the proof of the preceding theorem with a = b = 1, and apply Theorem V.2.1. To see that T + B is closed, let E be the identity map from <J)(T) onto D I. Then E is closed and T + B = (T 1 + BI)E. Thus T + B is closed by Theorem IV.2.7. V.3.B
Corollary.
Suppose T and B are closed and B is T-compact.
Then t.
Given E > 0, there exists a constant K f E <J)(T) ,
IIBfl1 n. m.
~ Kllfll
= K(s) such that for all
+ Ell Tfll
If Y is complete, then T + B is closed. If X and Yare complete, then T is normally solvable with finite kernel index if and only if T + B is normally solvable with finite kernel index. In this case, K(T) = K(T + B).
Proof of (i). Assume (i) is false. Then for some positive illtol/;cr n, thoro 0xiRtA '!til i . E ~D(T) su(~h thltt ~-
(I)
II/U.. II > /I. II!.. I + ell'I:t'"II
E
>
0 and each
124
UNBOUNDED LINEAR OPERATORS
Let gn = fn/llfnIlT, where
II
liT is the T-norm.
Then from (1),
(2)
Now, IlgnllT = 1 and B is T-compact. Hence there exists a subsequence {v n } of {gn} such that {Bvn} converges in Y. Since Ilvnll < IIBvnll/n and {IIBvnll} is bounded, Vn ~ 0 in X. From the assumption that B is closed it follows that BV n ~ O. Thus TUn ~ 0 by (2). But this is impossible, since 1 = IlvnllT = Ilvnll + IITvnl1 ~ O. Proof of (ii).
+ B is closed by (i) and Lemma V.3.5. Any set which is T + B-bounded is T-bounded, since
T
Proof of (iii). (i) implies the existence of a constant K such that for all
f IITfl1 S II(T
E '.D(T)
=
+ B)fll + IIBfl1 s
'.D(T
II(T
+ B) + B)fll + Kllfll + ill Tfll
whence IlfilT S 2(K
+ 1)llfllr+B
Therefore the T-compactness of B implies the T + B-compactness of B. Thus if T + B is normally solvable with OI(T + B) < 00, then T = (T + B) - B is also normally solvable and K(T + B) = K(T) by Theorem V.3.7. In particular, OI(T) < 00. The converse statement is a particular case of Theorem V.3.7. The next corollary generalizes a result due to Beals [2]. V.3.9 Corollary. Let X and Y be complete and let T be a Fredholm operator with domain dense in X. If B is T-strictly singular and B' is T'-strictly singular, then (T + B)' = T' + B'. Proof. We first note that if iJ and V are linear operators with equal, finite indices and V is an extension of U, then U = V. This follows from the relations OI( U) S OI(V)
{3(U)
~
{3(V)
and
which imply OI(U) = OI(V) and (3(U) = (3(V). Now, (T + B)' is clearly an extension of T' follows from Theorems V.3.7 and IV.2.3 that K«T
Thus ('I'
+ B)')
+ B)'
=
-K(T
= 7"
+ R'.
+ B) =
-K(T)
=
K(T')
K(U) = K(V)
+ B'. =
K(T'
Moreover, it
+ B')
chapter VI
APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS
In this chapter we shall consider properties of certain linear operators which arise from ordinary differential expressions of the form
where D = d/dt and the coefficients ak are complex-valued functions of a real variable, subject to certain conditions on an interval I. A differential expression T may give rise to many operators T which have their domains in ,cp(l) and ranges in ,cq(l), where 1 ~ p, q ~ 00 For example, T may be defined as follows:
n
Tf = Tf =
L akDkf k-O
Thus the operators determined by T are distinguished by means of their dOlHl1ins. The ones which we. present are those which are of recognized importlUwo und to whioh the gCIl{)rn.1 theory developed in the preceding (\hnptors npplioH.
UNBOUNDED LINEAR OPERATORS
126
When there is no possibility of confusion, a function in £1' °(1) and the element in £1'(1) which it determines are used interchangeably. Unles8 mention is made regarding the nature of I, the interval is arbitrary.
The main results, with somewhat different proofs, in Sees. VI.I to VI.5, are due to Rota [1] for 1 ::; p = q ::; 00.
VI.I
CONJUGATES AND PRECONJUGATES OF DIFFERENTIAL OPERATORS
In order to determine some properties of a differential operator l' with domain a subspace of, say, £",(I) and range in £q(1), 1 < q ::; OCJ, we introduce what we shall call the preconjugate of T. Since the domains of the operators l' which we consider are not dense in £",(1), the preconjugate '1' of l' is introduced to compensate for the nonexistence of 1". The idea is to consider l' as a map from a subspace of the conjugate space £",(1) = £~(I) into the conjugate space £q(I) = £;,(1) and to define '1' in such a way that, in certain cases, Tis the conjugate of '1'. VI.I.I Definition. Let X and Y be normed linear spaces and let l' be a linear operator with domain a total subspace of Y' and range in X'. The preconjugate '1' of Tis defined as follows. '£)('T) is the set of those x E X for which there exists ayE Y such that Ty'x = y'y for all y' in '£)(1'). Let 'Tx = y. Since '£)(1') is total, '1' is unambiguously defined. It follows that '1' is a linear map from a subspace of X into Y and Ty'x = y'('Tx)
X E '£)('1'),
y' E '£)(1')
VI.I.2 Lemma. Let l' be a linear operator with domain a total subspace of Y' and range in X'. Then the preconjugate of l' is closed. Proof. Suppose X n -----+ x and ''Fx n -----+ y. Then for each y' E '£)(1'), Ty'x n -----+ Ty'x and Ty'x n = y'('Tx n ) -----+ y'y. Thus Ty'x = y'y, y' E '£)(1'), which means that x E '£)('1') and 'Tx = y. VI.I.3 Lemma. Let l' be a linear operator with domain a total subspace of Y' and range in X'. Suppose '£)('1') is dense in X. Then ('1')' is an extension of 1', and therefore l' is closable. If, in addition, X and Yare reflexive, then '1' = ('1')', where '1' is the minimal closed linear extension of T. Proof. If y' E '£)(1') and x E '£)('1'), then y'('Tx) = Ty'x. Thus y' E '£)«'1')') and ('T)'y' = Ty', y' E '£)(1'). Since the conjugate operator ('1')' is closed, Tis closable. Let X and Y be reflexive. Suppose '1' ¢ ('1')'. Then therp. exists some (v', ('T)'v') j! G(T) = (J('J'). Thus there exists !1 z" E (Y' X X')'
127
APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS
such that z"(v', ('T)'v') r!' 0, and z"(y', Ty') = 0, y' E ~(T). Define y" E Y" and x" E X" by y"y' = z"(y', 0) and x"x' = z"(O, x'). Then (1)
y"y'
+ x"Ty'
(2)
= z"(y', Ty') = 0
y"v'
+ x"('T)'v'
y'
E
~(T)
r!' 0
Thus, since X and Yare reflexive, there exist x E X and y (3)
y'y
(4)
+ Ty'x v'y
= 0
+ ('T)'v'x
y' ¢'
E
E
Y such that
~(T)
0
Now, (3) shows that x is in ~('T) and 'Tx = -v. But then (4) implies that v'y + v'('Tx) = v'y - v'y r!' 0, which is absurd. Hence 'i' = ('T)'. VI.l.4 Lemma. Let A be a closable linear operator with domain dense in X and range in Y. Then A = '(A').
Proof. Since A is closable, ~(A') is total by Theorem II.2.11. Hence the preconjugate of A' is defined. Now, '(A') is an extension of A; for if x E ~(A), then A'y'x = y'Ax, y' E ~(A'). Thus, by the definition of '(A'), x is in ~('(A')) and '(A')x = Ax. Since '(A') is closed by Lemma VI.1.2, it remains to prove that G('(A')) C G(A) = G(A). By an argument analogous to the one given in Lemma VI.1.3, it follows that G('(A'))
C G(A).
VI.l.S Lemma. T = ('T)'.
If T is the conjugate of a closable operator, then
Proof. Suppose T = A', where A is closable. VI. 1.4, A = '(A') = 'T. Hence
Then, by Lemma
T = A' = (A)' = ('T)' by Theorem II.2.11. VI.l.6
Definition.
Let T be a differential expression of the form
where each ak is a complex-valued function on I. For each positive integer n, define A,.(l) to be the 8et of' complex-valued functions f on I for which ! 0 be given. Since G is uniformly continuous on I X I, there exists a 0 = O(E) > 0 such that for all s E I, (3)
Now
Thus (3) and Holder's inequality intply
IYk(t 2)
-
Yk(t l )!
~
q Mlt l - t 2 I l/ 'llfkllq
+ e II~JL (b
- a)1/Q'llfkllq
Since (11!kIIQ} is bounded, it follows that (yd is equicontinuous. Therefore, by the Ascoli-Arzela theorem, (Yk} has a subsequence converging in C(I) which, in turn, must converge in £p(I). Hence T-l is compact. The following example exhibits very simple maximal operators which do not have closed range.
VI.3.4 Example. Let I = [0, 00) and let n = D - ~e-u, where>. is a SCalftf. We show thnt fOf 7\ = 'r'A,P,f/' (p, q) admiRRihlo, nnd nil >., 'fA
APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS
147
does not have a closed range, or equivalently, by Theorem VI.2.11, T), is not surjective. Givenflocally integrable on I, the general solution in A1(I) for which TAy(t) = y'(t) - Ae-ity(t) = f(t) a.e. is given by y(t) = ee-
dim CB = K(T I) - K(To) = -K(T~)
+ K(T~)
= K(T*) - K(T*o)
If 1 00
>
0, there exists a constant K depending only on s, p, and the length of I (l may be unbounded), such thatforallf E Wn,p(l),
IIJ leI) > e, then I is the union of nonoverlapping intervals II and I 2, where l(11) = ke/2 for some positive integer k and e ~ l(1 2) ~ e/2. Applying (3) to f on the interval II with L = e/2 yields
If
(8)
Similarly, taking L
=
l(12) in (3) gives
(9)
1I!'11"'.1, ~
I; Ilfll"'.1, + ellf"II"'.1.
Thus, from (8) and (9), (10)
For e >
Ilf'II",.1 ~
I; Ilfll"'.1 + ell!"I1",.1
°fixed, define 12 K(l(1»
=
leI) 12
1e
leI) ~ l(1)
e
>e
Then K is a nonincreasing function of the length of I, and for (11)
11!'11",.1
~
00
>
leI),
Kllfll",.1 + ell!"II",.1
by (7) and (10). If leI) = 00, then (11) still holds upon choosing L = e in (3). Hence the lemma is proved for n = 2 and p = 00. By making use of (6), a similar argument proves the lemma for n = 2 and 1 ~ p < 00. Assume the lemma to be true for n = j. Suppose f is in W i +l. p (1), 1 ~ p < 00. Then for any compact subinterval J of I, f(j-ll is in W(2.V) (1). Now, the lemma holds for n = 2. Hence, given 1/ > 0, there exists a K o depending only on 1/, p, and the length of J such that f(jl is in £p(J) and (12)
Given 1/1 > 0, there exists, by the induction hypothesis, a K 1 depending only on 1/1, p and the length of J such that (13) Thus, from (12) and (13),
UNBOUNDED LINEAR OPERATORS
160
Choosing '71 so'that 1 - K 0'71
i, it follows from (14) that
;::::
(15)
Since '7 and '71 are arbitrarily small and K o and K 1 can be chosen to be nonincreasing functions of the length of J, the lemma for 11. = j + 1 follows from (13) and (15). By induction, the lemma is proved. The proof when p = 00 is the same, with I II", replacing II II~. VI.6.2
Theorem.
Let r be the differential expression r =
L" cl.I - TD,p) = 1 and Re"l>. < 0
(2)
Suppose Re "l>. > O. By Theorem V1.2.11, H - TD,p = TX-D,v is surjective if and only if THD,pl is surjective. Given g E £p,(J)
is a solution in AI(J) (\ £p,(J) to the equation "l>.y + y' = g by Theorem 0.11. Thus TX+D,pl is surjective and consequently (3(H - TD,p) = O. Since eXt is a solution to "l>.y - y' = 0 on every bounded subinterval of J and eXt is not in £p(J), it follows that a(H - TD,p) = O. Thus (3)
Re"l>.
>0
Suppose Re"l>. = O. Then"l>. is not in Pe(TD,p) = Fp(TD,p); otherwise it would follow from Theorem V.1.6 that K(p,[ - Tv,p) = K(H - Tv,p) for all p, sufficiently close to"l>.. But this contradicts the fact that both (2) and (3) hold. If p = 00, then by Theorem V1.2.7, (T.(TT,«» = (Te(T,-,l), which was just shown to be the set (p*(z) IRe z = OJ, where p* is the polyn
nomial
1 (-I)kakzk
n
1 (-I)k akDk.
corresponding to T* =
k=O
Since
k=O
P*(z) = P( -z) for all complex numbers z, it follows that U.(TT,«» = (P(z) IRe z = O}
Since (H - TT'P)' = TO,X-T-,P" 1 ~ p < 00, Theorem 11.3.7 implies that CR(H - TT,P)iS dense in £p([O, 00», 1 ~ p < 00. Thus (i) is proved. The rest of the theorem follows from (2), (3), and Theorem 1V.2.14. The classical method of solving the Euler differential equation n
1 bktkDkf = g, where each bk is a constant, is to introduce a change of k=O
variable, thereby transforming the given equation into one of the form vft = gl, where v is a differential expression with constant coefficients. A similar procedure is used in the proof of the next theorem. n
VI.7.3
Theorem. Let T =
1 bktkDk, where each bk is a constant, and k-O
let P be the polynomial
APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS
165
with lip = 0 when p => 00. The maximal operator TT,p corresponding to (T, p, p) and the interval [1, 00) has the following properties. i,
U-.(TT'P) = (P(z) IRe z = O} For A E U.(TT,p), CR(Al - TT'P) is a proper dense subspace of £p([I, 00».
ii. iii.
For A E P.(TT,P)' AI - TT,P is surjective, 1 If 1 ~ p < 00 and A is in P.(TT,P)' then
~
p
~
00.
is the number of zeros of A - P(z), counted according to their multiplicity, which lie in the half plane Re z < O.
Proof. Let J 0 = [0, 00) and J 1 = [1, 00). Then £p(J 1) is equivalent to £p(J 0) under the map 7J: £p(J 1) - 7 £p(J 0) defined by O~s
0, there exists a K, depending only on S, the length of I, and p, 1 < P < 00, such that for all b locally in £p(I) and all f in the domain of the maximal operator T corresponding to (D, p, p),
Ilbfll~.r =::; (sllf'II:.r
+ Kllfll:,r) 18,8+1] sup j8+1 Ib (t)I P dt CI 8
K may be chosen to be nonincreasing as a function of the length of I. Proof. Let II and I 2 be nonoverlapping subintervals of I such that I = II U h with II "to the left" of h For 7J > 0 such that t 7J and t - 7J are in I for all t E II and I 2 , respectively, choose cp E C' ([0, 7JJ) so that 0 =::; cp(t) =::; 1 on [0, 7Jl, cp(O) = 1, and cp(7J) = O. For f E 'J:J(T) and t E II,
+
f(t) - -
[(~
d (cp(s)f(t
jll (s l
+ H»
dB ... - jo,r.~ cp(s)f'(t
+ s) ds
- Iu~ cp' (8)f(t + 8) dB
168
UNBOUNDED LINEAR OPERATORS
Letting M = max 1~'(8)1, we obtain from Holder's inequality o:::;.:::;~
(1)
IJ(t)!:::; fo~ If'(t
+ s)1 ds + M fo~ IJ(t + s)1 ds :::; 7/1/P' [( fo~ If'(t + s)!p ds t + M (fo~ IJ(t + s)lp ds t :::; 27/I/p' (fo~ If' (t + s) Ip ds + Mp fo~ If(t + s) Ip ds t P
P ]
P
Taking 7/ sufficiently small, it follows that there exists a K I depending only on e, p, and the length of I such that If(t)lp :::; ~ fo~ !f'(t
(2)
+ s)lp ds + K
e {t+~ = 2 }t 1f'(s)lp ds
+K
I
fo~ IJ(t
(t+~
I
}t
+ s)lp ds
If(s)lp ds
From the conditions put on 7/ we see that as the length of I increases, 7/ can be kept fixed so that (1) still holds. Thus K 1 can be chosen to be nonincreasing as a function of the length of I. Letting a be the left endpoint of I I, Fubini's theorem and (2) imply that for tEl I and 7/ sufficiently small, (3)
h, Ib(t)f(t)!p dt :::; h, f+~ Ib(t)lp G[f'(s)[p + KtIf(s)[p) ds dt :::;
;;~ 1f'(s)lp + Kt\f(s)lpds J~aX(8-~,a) Ib(t)lpdt
: :; (-2e 11f'11~.I + KII\f!I~'I)
sup
[8,8+11CI
(8+1 Ib(t)lp dt
}8
For t E h J(t)
= -
(~
d
}o ds (~(s)J(t - s)) ds
Thus, by the argument used to establish (2),
As in (3), (4)
h. Ib(t)J(t)lp dt :::; £~ IJ'(s)lp + KIIJ(s)lp ds t+~ Ib(t)lp dt : :; (-2e 1\f'II~,I + Kl[lfl[~'I) [•.•sup /."+1 Ib(t)lp dt +IICI
The lemma now follows from (3) nod (4).
APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS
169
Suppose bk is locally in £p(J) and
Proof of the theorem.
O.::s;k.::s;n-l
Let J 1 be any subinterval of J of length greater than 1. Since any nonnegative numbers e and d satisfy (e P + d p )1lp .::s; e + d, it follows from Lemmas VI.8.2 and VI.6.1 and Theorem VI.6.2 that given J.l > 0 and 1/ > 0, there exist constants K o and K 1, depending only on p and J.l and p and 1/, respectively, such that for all f E 'neT),
Using equations (1) and (2), a simple computation shows the existence of a K 2, depending only on p and e, such that for all f E 'neT), (3)
It follows from Theorem VI.6.2 that there exists a constant C, depending only on p, such that for all f E 'neT), (4)
We may derive from (3) and (4) an inequality of the form n-l
(5)
IllIfl!p,J,.::s;
lllbk!(kl l\p.J ~ 1
Kllfllp.J,
+ ellTfllp,J,
f
E
'neT)
k=O
where K depends only on p and e. It is clear from (5) that 'neT) is contained in 'n(B) and that B is T-bounded. To prove TT+. = T + B, it suffices to show that 'n(TT+.) = 'neT). If f E 'neT), Tf and IIf are in £p(J) by (5), which implies f E 'n(TT+.)' On the other hand, if f is in 'n(TT+.), then (T + lI)f is in £p(J). Now, f(j), 0 .::s; j .::s; n - 1, is continuous on each compact subinterval J 1 of J. Hence IIf and Tf = (T + lI)f - IIf are in £p(J 1). By considering only those J 1 of length greater than 1 we have from (5) and the triangular inequality that
whence (0)
O<e.
f
~
p,(T) implies
k
~
n
174
UNBOUNDED LINEAR OPERATORS
Proof. In view of Theorem VI.8.1, it suffices to prove the corollary for l = T + bJ)n. Theorem VI.6.2 implies that :oCT) = :0 (L). For any 13calar X, n-l
X-
(1)
l =
X-
= (1
T
+ b (X - + n
T
an
bn) +a (X
- T)
.
L akDk -
k=O
+ n~ ~
1
k=O
n
bn
ak -
an
X)
Dk - bn -
X
an
Let AA be the maximal operator corresponding to (1 + bn/an)(X - T). It follows from (1) and Theorem VI.8.1 that 2). Each aa is in C~(n); that is, aa is a restriction to Q of some function in C~(e), where (') is an open set in R" which eonittills n.
THE DIRICHLET OPERATOR
187
Vll.2.6 Definition. Let Wo"""(U) be the closure in W"',"(U) of Cooo(U) and let Tp be the linear map from a subspace of £p(U) into £p(U) defined by
T p is called a Dirichlet operator. VIl.2.7
i. ii.
iii.
Remarks m(Tp) is indeed a subspace of £p(U), since aa is continuous on nand Dau is in £p(U), lal ~ 2m, u € W2m,p(u). If u is in C'Xl(En) ',D(Tp), then Df3u = 0 on au, 113/ < m, in the classical sense; that is, (ai-1/api-1)u = 0, j = 1, 2, . . . , m, where a/ap is the normal derivative. For a characterization of ',D(Tp), see Browder [3J, Lemma 9.
n
The next theorem is a special case of a result due to Agmon, Douglis, and Nirenberg [IJ, and Browder [3]. VIl.2.8 Theorem. all u € ',D(Tp),
For n
> 2,
VIl.2.9
For n
>
Corollary.
there exists a constant K p such that for
2, T p is a Fredholm operator.
The proof that (3(T p ) < 00 is quite involved. See, for example, Browder [3], Theorem 6. The idea is to construct a linear operator S which maps £p(U) into Wm,p(u) such that TpS = I + K, where K is compact. Since m(TpS) C m(Tp), one obtains
We shall only prove that T p is normally solvable and has finite kernel index.
n
Proof. Let ',Dp be ',D(Tp) with norm(" 112m,p. Since Wom,p(u) W2m,p(u) is a closed subspace of Banach space W2m,p(u), ',Dp is complete.
Moreover,
and tho idontity map from :Op onto '.D(T 1') with the T p-norm is continuous. H
