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# The Spectral analysis of compact operators

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

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## The Spectral analysis of compact operators

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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
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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
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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 � > 0 is given then, since an eigenvalue of K, | is ﬁnite, we have that < � 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 �K exists or the equation �K = 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
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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
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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
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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