Short note on compact operators - Monday 24 th March, 2014 Sylvester Eriksson-Bique 1 Introduction In this note I will give a short outline about the structure theory of compact operators. I restrict attention to K(X) = {K : X X compact}. The case of K(X Y ) involves other technicalities, since we can not do spectral theory. First I will give a few examples. Example 1: A : l 2 l 2 given by A( x i e i ) = i=0 a i x i e i is compact iff lim i a i = 0. This means that compact operators correspond, roughly speaking, with c 0. Example 2:Take the unit circle T. Consider the map from L 2 (T) L 2 (T) given by f F N f (the Fejer kernel). This is bounded and compact. In fact it is finite rank. Note that for any even nominally smooth function T C α, we have that the integral operator T f = T (x y)f(y)dy, T will be comapt. Thus mollification is compact. Note that the convolution is given by a multiplication in the frequency space, i.e. the Fourier transform diagonalized a convolution operator. Thus one can think of compact operators as mollifying infinities. Side note: This is actually a remarkably interesting class of examples, because it is exactly the method used to find linear representations of compact topological groups (Peter-Weil-Neumann theory). The idea is to look at compact operators arising as mollifiers, and their eigenspaces. Other tools are Haar-measures, and approximations of unities. With another significant i=0 1
step this leads to a solution to Hilbert s 5th problem and a large number of powerful corollaries in geometry follow. Example 3: I lied a little bit, more is true. If K(x, y) L 2 (T T), then the integral operator T f = K(x, y)f(y)dy, T will be compact. This is a good exercise to think about. 1 The idea here is that integral operators, however wild, usually make functions a bit nicer. Averaging helps a lot, and if there is good control on the kernel then the operator will be very nice (here even compact). One reason for this is cancelling out big frequencies. Example 4: In partial differential equations certain operators of the above form arise naturally. Consider the function 1 (n 2)ω n 1 n > 2 x y n 2 N(x) =. 1 log x y n=2 2π This is the Newton kernel, and ω n is the surface area of the n-sphere. The Newton kernel is the fundamental solution to the Laplacian, and corresponds to the potential field generated by a point mass/charge. Take a domain with sufficiently smooth boundary D R n 2. And consider the operators mapping C( D) C(D) given by T 1 f(x) = N(x y)f(y)dy, D and T 2 f(x) = ( ν N)(x y)f(y)dy. D These correspond to the single and double layer potentials. The names derive from physics; T 1 computes the potential for a charge distribution given by f on the boundary, whereas T 2 computed the potential distribution given 1 Hint: Finite rank condition (easier way), weak convergence defintion (slightly harder). 2 Sufficiently nice=c 2 to make your life easy, C 1 if you are brave, and Lipschitz if you have had a few beers. The Lipschitz case is kind of hard, but I managed to convince myself that the errors can be bounded... 2
by infinitesimal dipoles placed normally to the boundary with charge distribution f (or small bar magnets, if you think in terms of magnetic fields). T 1 corresponds to soliving the Neumann problem and T 2 for the Dirichlet problem. Take the boundary values of ν T 1 and T 2, which gives S 1, S 2 : C( D) L ( D) (It is important to take non-tangential limits). These can not be taken naively, because there is a so called jump condition. We get in fact S 1 = 1/2I T 2, S 2 = 1/2I + T 2 The wonderful fact is that T 2 is a copact operator mapping L 2 L 2 (see the third example). It is called a weakely singular operator. Now since T 1 f, T 2 f will be clearly harmonic in the interior, we are left to study solvability of the equation 1/2I ± T 2 f = φ. Here the Fredholm alternative and some basic maximum principle arguments will take you home. Thus you get a proof of existence of solutions for the Dirichlet/Neumann problems for sufficiently nice boundaries. Example 5: 3 I am compelled to present another type of beautiful example, while I won t give details. Consider a domain of the euclidean space D, with nice boundary D. Further consider the non-homogenous wave equation ( t 2 1 )u = c 2 c u = 0. u t=0 = u 0 C 0 (D), t u t=0 = 0 A problem in inverse-tomography, or more precisely photo-aucustic tomography involves reconstructing the signal u 0, or it s singularities, from measurements along the boundary D (or part of it, but I will assume we have complete data). Singularities are important, because they represent boundaries of tissue. There is a beautiful theory about how the singularities of the wave equation propagate. In fact, they propagate along geodesics for the metric 1/c 2 dx 2, with unit speed! The main concept here is the so called Wave-front set, and 3 Recently published in more general form by Gunther Uhlmann, Plamen Stefanov. See Thermoacoustic tomography with variable sound speed. 3
the amazing Lax-parametrix. It is astonishing that we can give such a precise description of the singularities, while at the same time being so intuitive. Now assume that the velocity is a smooth function and D is geodesically convex without trapped geodesics. That is, sound propagates out of the domain and does not return (for example, for simplicity just take c=1). consider a very large time T. Since the singularities of u 0 will propagate along shortest paths (when c=1, along straight lines), they must leave the domain D by the time T, if it is chosen large enough. In fact the map u 0 u(t, ) D will be regularizing, sending the Sobolev space H 2 1 to smooth functions C (D). Being regularizing and using some functional analysis arguments, it must be compact! This is important in proving that a certain iterative scheme for the solution converges. For example, from compactness, the norm of the operator must actually be strictly less than 1, because some energy must flow out. Otherwise, the norm being 1, would mean that there was a u that flows first out of D and then back into D (this is Tataru s unique continuation result). Lemma 1. D is a smooth bounded domain. Let T : L 2 (D) L 2 (D) be a bounded operator and T f H1 2 for all f L 2, then in fact the induced operator S : L 2 H1 2 is bounded, and moreover T is compact. Proof: The latter follows from the first, because by T = i S, where i : H1 2 L 2 is the inclusion and by Rellich is compact. For the first claim, apply closed graph. The graph of S is {(x, Sx) = (x, T x) x L 2, Sx H 2 1}. We prove that this is closed in L 2 H 2 1. Assume x i x and Sx i = T x i y, where the convergence is in H 1 2 norm. But we know that T x i T x in L 2, so y = T x by L 2 continuity, which completes the proof! Example 6: Consider the operator T : i=0 x ie i i=0 a ix i e i+1, where lim i a i = 0 is compact, but has no eigenfunctions. 2 Spectrum of compact operator Recall λ σ(t ) iff λi T is not invertible. I.e. one of the following occurs: 4
λi T does not have dense image, and is injective, i.e. λ is in the residual spectrum σ r (T ). λi T is injective but not bounded from below, λ is in the approximate point spectrum σ ap (T ). λi T is surjective, but not injective, in which case λ is an eigenvalue/is in the point spectrum σ p (T ). λi T is surjective and injective by open mapping theorem it is invertible and λ σ(t ). It turns out that for compact operators there is (essentially) no residual spectrum. Further approximate eigenvalues, are real eigenvalues (modulo the degeneracy at 0, where anything can happen). Lemma 2 (No approximate spectrum). For compact operators 0 λ σ ap (K) iff λ σ p (K). Proof: In Hilbert spaces, and reflexive and separable cases this is easier. Assume by scaling that λ = 1. Bounded below means that (I K)x c x, so since we don t have this bound we can take x i = 1 such that x i Kx i 1/i. Take a weakely convergent subsequence x i y. Then K(x i ) K(y) strongly, and since Kx i 1, we have y = 1. Further take any element in the dual λ X, and consider: 0 = lim < λ, x i K i >= lim < λ, y Ky >. Thus y = Ky, because λ was arbitrary. The problem here was that I used relfexivity, and separability to extract converging sequences. So another argument is needed. Take the x i as above, and assume that there is no subsequence that converges. This is only possible, if there exists a subsequence x nk and an ɛ > 0 such that x ns x nt > ɛ. But now also Kx ns Kx nt > ɛ/2 for all s, t big enough. But this contradicts compactness, since the image sequence does not have a convergent subsequence. 5
A suprising fact about compact operators is given by the following lemma. Lemma 3. If λi K is injective, then it is surjective for λ > 0. Proof: Again, let λ = 1 and T = I K. Consider V i = Im(T i ). Clearly V i+1 V i and T (V i ) V i+1. Take vectors x i V i such that d(x i, V i+1 ) = inf x Vi+1 x i x = 1/2 and x i = 1. These are called generalized eigenvectors. Now x i is one-separated, and T (x i ) V i+1. However for j > i Kx i Kx j = x i x j T (x i ) + T (x j ) 1/2. This last follows since x j T (x i ) + T (x j ) V i+1, but this contradicst compactness of K. Corollary 4. If λi K is surjective, then it is injective for λ > 0. Proof: The dual of T = I K is T = I K, which is injective if and only if T is surjective. Lemma 5 (Fredholm alternative). For compact operators K, σ r (K) {0}. More is true, Im(λI K) is closed for all λ 0 and codim(x) = dim(x/im(λi K)) = dim(ker(λi K)) Proof: The lemma states that λi K is Fredholm with index 0. First of all define T = λi K, S = ker(t ). First note that S is finite dimensional. This is because K S = λi and λi is a compact operator only on finite subspaces. Since S is finite dimensional, it is possible to find a closed subspace V such that X = S V, possibly with slightly distorted norm (Hint: consider first dim(s) = 1). Denote ImV = V. The operator T V : V V is compact. Also it satisfies for some C > 0: 1/C x T x C x. 6
But then if T x i is a Cauchy sequence, then also x i must be a Cauchy sequence, so x i y and T x i T y, hence the image is closed. Recall that the adjoint of the compact operator K is compact and hence T = λi K is also Fredholm and we can repeat the previous argument to get that kert = S is finite dimensional. Thus we have two decomopositions of X = S V = S V. Further we note that T is invertible if and only if T is invertible. We prove that dim(s) = dim(s ). Assume dim(s) dim(s ). Take your favourite map S S which is an injection, call it F. Now K + F is compact. Also T = λi (K + F ) is injective, and thus by the previous corollary it is surjective. But this means that dim(s ) dim(s). By contradiction we have affirmed the equality. On the other hand, assume dim(s) dim(s ). Then take an F operator mapping S surjectively onto S and consider again T = λi (K + F ). This is surjective and by previous corollary injective, so again the result follows. Theorem 6. The spectrum of a compact operator is countable and can have only 0 as an accumulation point. Proof: Every non-zero element in the spectrum must be in the point spectrum by the above. Thus we have λ i λ, and Kx i = λ i x i, x i > 1. Take V i = span(x 1,... x i ), and as above choose y i = a i x i + v i 1, v i 1 V i 1, such that y i = 1, and d(y i, V i 1 ) > 1/2. Note that y i y j > 1/2, and again for j > i Ky i Ky j = λ i y i λ j y j λ j λ j λ i. Thus this doesn t have any convergent subsequences. Theorem 7. For compact self-adjoint operators on a separable Hilbert space all the above holds, but also the space splits into orthogonal eigenspaces. 7
Proof: As we discussed. One uses a variational principle to prove that the eigenvalues can be obtained by a maximizing procedure, and iteratively extracting eigenspaces. By separability one will be done after countable operations (use a Zorns-lemma argument by taking the union of all orthonormal sets that diagonalize the operator). 3 Final remarks In the non-selfadjoint case the main problems arise because we may not have reflexivity (can t use weak convergence), and because we cant induct with orthogonality. In the self-adjoint case we can simply use a variational principle to extract eigenvalues and use orthogonality to get smaller and smaller subspaces. For general compact operators it is harder, since there may be no eigenvalues, i.e. the case when the spectral radius satisfies lim n ( Kn ) 1/n = 0. Thus we don t have many invariant subspaces. The proof still uses the same idea of invariant subspaces, but needs to deal with approximation and errors. One has to deal with the fact that T may also contain shifts like in example 6. Also the spectral character of 0 could be very wild. Thus a general form of the spectral theorem is technical without self-adjointness. If one tries to work without compactness all hell breaks loose. Self-adjointness salvages most of the theorems. The tools will change to use operator algebras, and functional calculus. This time one doesn t have eigenvalues, but approximate eigenvalues. Instead of splitting the space into eigenspaces one can split the space into orthogonal parts where the operator looks like multiplication by λ. This is a continuous spectral resolution. Further one can divide the spectrum depending on it s measure theoretic properties into singular, point and absolutely continuous spectrum. For unbounded operators the theorem is the same, but with more complications. 8