In this section will always denote a compact operator Theorem 61 If 0 then either is an eigenvalue of or Proof Suppose that 0 is not an eigenvalue of We show that The proof of this is in several stages a For some c we have that that 955I ID: 22997 Download Pdf

In this section will always denote a compact operator Theorem 61 If 0 then either is an eigenvalue of or Proof Suppose that 0 is not an eigenvalue of We show that The proof of this is in several stages a For some c we have that that 955I

Download Pdf

Download Pdf - The PPT/PDF document "The Spectral analysis of compact operato..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.

Page 1

6 The Spectral analysis of compact operators. In this section will always denote a compact operator. Theorem 6.1 If = 0 then either is an eigenvalue of or Proof. Suppose that = 0 is not an eigenvalue of . We show that ). The proof of this is in several stages. (a) For some c > , we have that that λI k for all ∈ H Suppose this is false. Then the inequality fails for for = 1 ··· . Therefore there is a sequence of unit vectors such that λI k that is, (( λI 0. Applying the condition that is compact, there is a subsequence ( ) such that ( Kx ) is convergent.

Call its limit . Then (( λI Kx and so ( . Since ( ) is a sequence of unit vectors, = 0. But then, λI = lim λI = 0 This contradicts the fact that is not an eigenvalue, so (a) is established. (b) ran( λI ) = Let = ran( λI and write . It follows from (a) using Lemma 5.1 that ) is a sequence of closed subspaces. Also λI +1 ⊇ ··· Note that, if then Ky = (( λI λy so that We now use the compactness of to show that the inclusion +1 is not always proper. Suppose, on the contrary that ⊃ ··· Using Lemma 1.7 , for each we can ﬁnd a unit vector

such that and +1 . We show that ( Kx ) cannot have a Cauchy subsequence. Indeed, if m > n Kx Kx = ( λI λx Kx λx + [( λI Kx λx 31

Page 2

where +1 Kx +1 follows from m > n ]. Thus Kx Kx ≥ | and so ( Kx ) has no convergent Cauchy subsequence. Therefore, the inclusion is not always proper. Let be the smallest integer such that +1 . If = 0 then choose . Then ( λI +1 and so, for some λI = ( λI +1 = ( λI where = ( λI . Now 6 so = 0 and λI )( ) = 0 contradicting the fact that is not an eigenvalue. Therefore = 0, that is ran(

λI ) = (c) Completing the proof. This is done exactly as in Theorem 5.5 (i). For any ∈ H , there is a unique ∈ H such that = ( λI . Deﬁne ( λI Then k so λI k showing that ( λI ∈ B ) (i.e. it is continuous). Thus 6 ). Lemma 6.2 If are eigenvectors of corresponding to diﬀerent eigenvalues , then is a linearly independent set. Proof. This is exactly as in an elementary linear algebra course. Suppose the state- ment is false and is the ﬁrst integer such that ,x ··· ,x is linearly dependent. Then =1 = 0 and = 0. Also, by hypothesis Kx

with the ’s all diﬀerent. Now =1 (where / ) and so 0 = ( =1 showing that ,x ··· ,x is linearly dependent, contradicting the deﬁnition of Theorem 6.3 \{ consists of eigenvalues with ﬁnite-dimensional eigenspaces. The only possible point of accumulation of is Proof. Let be any non-zero eigenvalue and let Kx λx be the eigenspace of . If is not ﬁnite-dimensional, we can ﬁnd an orthonormal sequence ) of elements of [apply the Gram-Schmidt process (Theorem 3.4 ) to any linearly independent sequence]. Then Kx Kx λx λx = 2 which is impossible, since is

compact. 32

Page 3

To show that ) has no points of accumulation other than (possibly) 0, we show that > } ) is ﬁnite for any δ > 0. Suppose this is false and there is a sequence of distinct eigenvalues ( ) with > for all . Then we have vectors with Kx Let = span ,x ··· ,x . Then, since is a linearly independent set, we have the proper inclusions ⊂ ··· It is easy to see that and ( . Choose, as in The- orem 1.1 a sequence of unit vectors ( ) with and . Then, for n > m Ky Ky [( Ky Since ( and Ky , the vector in square brackets is in . Therefore, since Ky Ky > showing

that ( Ky ) has no convergent subsequence. Corollary 6.4 The eigenvalues of are countable and whenever they are put into a sequence we have that lim = 0 Proof. [The set of all eigenvalues is (possibly) 0 together with the countable union of the ﬁnite sets of eigenvalues = 1 ··· ). If ² > 0 is given then, since an eigenvalue of K, | is ﬁnite, we have that < ² for all but a ﬁnite number of values of . Hence ( 0. ] Corollary 6.5 If is a compact selfadjoint operator then equals its eigenvalue of largest modulus. Proof. This is immediate from Theorem 5.5 (ii). The Fredholm

alternative. For any scalar , either µK exists or the equation µK = 0 has a ﬁnite number of linearly independent solutions. (Fredholm formulated this result for the speciﬁc operator ( Kf )( ) = x,t dt In fact, he said : EITHER the integral equation x,t dt 33

Page 4

has a unique solution, OR the associated homogeneous equation x,t dt = 0 has a ﬁnite number of linearly independent solutions.) We now turn to compact selfadjoint operators. For the rest of this section will denote a compact selfadjoint operator. Note that every eigenvalue of of is real. This is

immediate from Theorem 5.5 , but can be proved much more simply since if Ax λx , where is a unit eigenvector, Ax, x x,Ax x,x Ax,x λ. Lemma 6.6 Distinct eigenspaces of are mutually orthogonal. Proof. Let and be eigenvectors corresponding to distinct eigenvalues and Then, x,y Ax, y x,A x,Ay = x,y x,y (since is real) and so x,y = 0. Theorem 6.7 If is a compact selfadjoint operator on a Hilbert space then has an orthonormal basis consisting of eigenvectors of Proof. Let ( =1 ··· be the sequence of all the non-zero eigenvalues of and let be the eigenspace of . Take an orthonormal basis of

each and an orthonormal basis of = ker . Let ( ) be the union of all these, put into a sequence. It follows from Lemma 1.6 that this sequence is orthonormal. Let for all . Then, if we have that ,Ay Ax ,y , y = 0 and so . Therefore with its domain restricted to is a compact selfadjoint operator on the Hilbert space . Clearly this operator is selfadjoint [ Ax,y x,Ay for all x,y ∈ H so certainly for all x,y ]. Also it cannot be have a no-zero eigenvector [for then = (0) for some k > 0]. Therefore, by Corollary 1.5 , it is zero. But then . But also and so = (0). Therefore ( ) is a basis.

Corollary 6.8 Then there is an orthonormal basis of such that, for all Ah =1 h,x Proof. Let ( ) be the basis found in the Theorem and let Ax ,x (this is merely re-labeling the eigenvalues. The from Theorem 3.3 (iii), for any ∈ H =1 h,x 34

Page 5

Acting on this by , since is continuous and Ax we have that Ah =1 h,x Theorem 6.9 If is a compact selfadjoint operator on a Hilbert space then there is an orthonormal basis of such that =1 where the series is convergent in norm. Proof. Let be the basis found as above so that Ax and Ah =1 h,x Note that ( 0. Let =1 We need to show that k

0 as Now +1 h,x and, using Theorem 3.3 (v) +1 h,x 〉| sup +1 +1 | h,x 〉| sup +1 =1 | h,x 〉| = sup +1 Thus k sup +1 , and so since ( 0, we have that k as Alternatively, Theorem 4.4 may be used to prove the above result. Let and and be as above and let =1 Then, since is a basis, Theorem 3.3 (iii) shows that ( ) converges pointwise to the identity operator . Since AP , Theorem 4.4 shows that ( ) converges to in norm. 35

Page 6

Exercises 6 1. Let be a compact operator on a Hilbert space and let = 0 be an eigenvalue of . Show that λI has closed range. [Hint : let

= ker( λI ) and let . If ran( λI ), show that = lim λI with . Now imitate the proof for the case when is not an eigenvalue.] 2. Find the norm of the compact operator deﬁned on [0 1] by V f )( ) = dt Hints: Use Corollary 5.4 and the fact that the norm of the compact selfadjoint opera- tor is given by its largest eigenvalue. Now use the result of Exercises 2 Question 6 to show that if satisﬁes V f λf then it satisﬁes λf 00 = 0 (1) = 0 , f (0) = 0 [You may assume that any vector in the range of (being in the range of two integrations) is twice

diﬀerentiable (almost everywhere).] Note that a direct approach to evaluating seems to be very diﬃcult (try it !). 3. Let be an orthonormal basis of and suppose that ∈ B ) is such that the series =1 Tx converges. Prove that (i) is compact, (ii) =1 Ty converges for every orthonormal basis of and for the sum is the same for every orthonormal basis. Note : an operator satisfying the above is called a Hilbert-Schmidt operator. Hints: (i) write ∈ H as a Fourier series, =1 where h,x . Deﬁne =1 Tx and show that +1 Tx ≤ k +1 Tx (ii) Take an orthonormal basis of

consisting of eigenvectors of the compact operator . Observe that if T then T , T 0. Now use the spectral theorem for to prove that if for any orthonormal basis =1 Tx converges then =1 Tx =1 Tx ,x =1 Note that for a double inﬁnite series with all terms positive, the order of summation may be interchanged. 36

Â© 2020 docslides.com Inc.

All rights reserved.