COMPACT OPERATORS. 1. Definitions

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1 COMPACT OPERATORS. Definitions S:defi An operator M : X Y, X, Y Banach, is compact if M(B X (0, )) is relatively compact, i.e. it has compact closure. We denote { E:kk (.) K(X, Y ) = M L(X, Y ), M compact P:closu the set of compact operators from X into Y Banach spaces. Proposition.. The set K(X, Y ) is a closed subspace of L(x, y). Proof. Clearly K(X, Y ) is a linear subspace of L(X, Y ). Let M n M in the operator norm, where M n is compact. Fixed ɛ > 0, let n such that M M n L(X,Y ) ɛ 2. Since M n (B(0, )) is relatively compact, then it can be covered by a finite number of balls B Y (y i, ɛ/2) of radius ɛ/2. Then M(B X (0, )) is covered by B Y (y i, ɛ). i As for degenerate maps, M L is compact if one is compact and the other continuous: thus K(X) = K(X, X) is an ideal w.r.t. map composition. We recall that a linear operator M is degenerate if it has finite rank: E:finite (.2) dim(r M ) <. L:hilbecomp Clearly such an operator is continuous if X is Banach, and thus it is compact. We thus have that if M is the limit of a sequence of finite rank operators M n, then it is compact. In Hilbert spaces the converse is true: Lemma.2. If Y is a Hilbert space, then every compact operator is the limit of a sequence of finite rank operators. Proof. Consider a converging of M(B(0, )) with balls of radius ɛ > 0, K = i B(y i, ɛ). Let S = span{y i i, and consider the projector P S. This projection exists because Y is Hilbert. Define the finite rank operator M ɛ = P S M. By construction, if x B X (0, ), then there is y i such that Mx y i < ɛ, so that, since the operator norm of a projection in Hilbert spaces is and P S y i = y i, we have (P S M)x y i < ɛ, It follows that Mx M ɛ x = Mx (P S M)x < 2ɛ,

2 S:adjoint 2 COMPACT OPERATORS 2. Transpose of a linear operator Let X, Y be Banach spaces, with duals X, Y, respectively. Let M : X Y be a bounded linear map. Define the transpose M : Y X by E:tranp (2.) (M ξ)x = ξ(mx). P:adjprop Because of the estimate ξ(mx) ξ Y M L(X,Y ) x, the right hand side is a linear functional over X, which we denote by M ξ. Thus M : Y X is well defined. It is clearly linear and by the above estimate Proposition 2.. If M L(X, Y ), then M L(Y,X ) M L(X,Y ). E:norm3 (2.2) M L(Y,X ) = M L(X,Y ). Moreover, () N M = R M ; (2) N M = R M ; (3) (M + N) = M + N. Proof. The equality (2.2) is an application of Hahn Banach theorem in the space Y. The other relations follow easily from (2.). We now prove that if M is compact, then also its transpose is compact. T:comptrap Theorem 2.2 (Schauder). The operator M K(X, Y ) if and only if M K(Y, X ). Proof. Let ξ n be a sequence in B Y (0, ), and K = M(B X (0, )). Consider the functions φ n (y) = ξ n y C(K), ξ n B Y (0, ). Clearly these functions are equicontinuous (they are Lipschitz continuous with modulus ) and K is compact, so that there is a converging subsequence, which we denote again by φ n. Since φ n is Cauchy, we have ξ n (Mu) ξ m (Mu) = (M ξ n )u (M ξ m )u < ɛ, u B X (0, ), n, m. Hence M ξ n is a Cauchy sequence in M (B Y (0, )). Conversely, if M is compact, then M is compact because of the first part of the proof. It is easy to see that if J X : X X, J Y : Y Y are the canonical immersions, then M (J X x) = J Y (Mx). Since J X (B X (0, )) B X (0, ), then M (J X (B X (0, ))) = J Y (M(B X (0, ))) is relatively compact in Y. Since the canonical immersion J is an isometry, then M(B X (0, )) is relatively compact. S:fredh 3. Fredholm s alternative This section is devoted to the proof of Fredholm s alternative: If M : X X, X Banach, is compact, then either the equation u Mu = v has a unique solution, or u Mu = 0 has n linearly independent solutions, and u Mu = v has a solution if and only if v satisfies the linear conditions { (3.) v (RM) = lv = 0, v RM. From M compact it follows that R M is finite dimensional. T:fredh We prove in fact the following theorem: Theorem 3.. If M : X X, X Banach, is compact, then () N I M has finite dimension; (2) R I M is closed and R I M = N I M ;

3 COMPACT OPERATORS 3 (3) N I M = 0 is equivalent to R I M = X; (4) the dimension of N I M is equal to the dimension of N I M. In particular, the index of the operator I M, M compact, is 0: Proof. Point () follows from the observation that ind(i M) = dim N I M dim R I M = 0. N I M = M(N I M ), and since M is compact, the space N I M is locally compact, hence finite dimensional. Let u n Mu n = y n y. Since N I M has finite dimension, there is v n N I M which minimize u n v n = inf u n v. v N I M and (u n v n ) M(u n v n ) = y n. By dividing the above equation by u n v n, one sees that if u n v n, then the sequence w n = (u n v n )/ u n v n satisfies w n + Mw n 0, w n = Since M is compact, then we can extract a subsequence Mw n w, so that w + Mw = 0, but w =, This is a contradiction, because w / N I M. It follows that u n v n remains bounded. Thus up to subsequences we have that u n v n converges. This prove that R I M is closed. Since for a closed subspace Y, the Hahn Banach theorem implies (Y ) = Y, then (2) follows. To prove (3), assume that N I M = {0, and R I M = X X, then for v R I M Mv = M ( I M ) x = ( I M ) Mx X, X closed. The operator M K(X ), so that we can consider again X 2 = (I M)(X ) = (I M) 2 (X) X, because I M is injective and X = (I M)(X). Proceding in this way we find a sequence of subspaces X n = (I M) n (X), and thus we can find points x n X n such that We have for n m x n y 2, y X n. M(x n x m ) = u n u m + (I M)u m (I M)u n = u n y nm, y nm X n. Hence M(x n x m ) /2, but this contradicts the assumption M compact. Conversely, if R I M = X, then we have N I M = {0, and thus using the first part R I M = X. Using again Proposition 2., we conclude N I M = {0. This concludes (3). Since M K(X), then M K(X ), so that both kernels have finite dimension. Assume that d = dim N I M < d = dim N I M. Since N I M is finite dimensional, then there is a continuous projector from X into N I M. Using the fact that R I M has finite codimension, there is a continuous projector on a linear complement E of R I M. By assumption, there is Λ : N I M E which is injective but not surjective. Define S = M + Λ P NI M. Then S K(X), because Λ has finite rank. Moreover N I S = {0. From point (3) it follows that R I S = X, but this contradict the fact that S is not surjective. We have thus proved that d d. Using the above result, it follows that for M dim N I M dim N I M dim N I M. Since N I M J(N I M ), we have proved (4).

4 4 COMPACT OPERATORS 4. Spectral analysis S:spect If M L(X), then the resolvent set of M is { E:resolv (4.) ρ(m) = λ C : (λi M) L(X). The spectrum of M is λi M E:spectr (4.2) σ(m) = C \ ρ(m) = λ C : λi M λi M E:solve not injective injective but not surjective injective, surjective but with not continuous inverse For bounded operators the last case cannot occur, because of the open mapping theorem. The values λ such that the first case holds are the eigenvalues of M. The space N λi M {0 is the eigenspace associated to λ, and its elements are the eigenvectors of M. Proposition 4.. If M L(X), then E:contr (4.3) σ(m) { λ M L(X). S:sefladj Proof. The proof follows from the fact that the series + n=0 converges strongly and it is the inverse of λi M. Mn λn+ For compact operators the spectrum has a precise form. Theorem 4.2. Let M K(X), with X infinite dimensional Banach. Then 0 σ(m); λ σ(m) \ {0 is an eigenvalue; σ(m) \ {0 is either empty, or finite, or it is a sequence of eigenvalues converging to 0. Proof. The first point follows because M cannot exists, otherwise M M(X) = X is compact. To prove point (2), we just use the (3) implication of Theorem 3., which gives a contradiction if N λi M = {0. To prove the last point, we consider a sequence λ n σ(m) \ {0. converging to some λ. For all eigenvalues λ n, let e n N λn I M with norm. It is easy to verify that N λn I M N λm I M = {0 if n m, so that all e n are different. Define E n = span {e, e 2,..., e n, and consider u n E n such that u n =, u n y /2 for y E n. We have for m < n Mu n Mu m λ n λ m = u n u m + (λu n I M)u n (λ m I M)u n λ n λ m 2, since (λu n I M)u n E n. λ n Since Mu n has a converging subsequence, then λ n 0. This shows that the set σ(m) { λ /n has at most a finite number of eigenvalues. 5. Spectral decomposition of compact self adjoint operators We say that M L(H), H Hilbert space is self adjoint if E:selfadj (5.) (Mx, y) = (x, My), x, y H. Proposition 5.. Let M L(H) be self adjoint, and define E:minmax (5.2) m = inf (Mu, u), M u = Then σ(m) [m, M], m, M σ(m). = sup (Mu, u). u =

5 COMPACT OPERATORS 5 Proof. If λ C \ R or λ > M, then ( λi M)u, u ) = λ u 2 (Mu, u) 0, since (Mu, u) R. Moreover, (λi M)(H) is a subspace of H, which is closed because from the above relation ( λ M + Iλ ) u (λi M)u. T:deco S:exerc The same argument implies that (λi M)(H) = H. This proves that σ(m) (, M). For λ = M, then we have as in the proof of Schwartz inequality that (Mu Mu, v) (Mu Mu, u) /2 (Mv Mv, v) /2 If u n, u n =, is a maximizing sequence (Mu, u) M, it follows that (MI M)u n converges to 0. If now M ρ(m), then u n = (MI M) (MI M)u n 0, which contradicts u n =. Replacing M with M, we obtain the other part of (). In particular, if σ(m) = {0, then (Mu, u) = 0, and Mu = 0. Theorem 5.2. If M K(H), H Hilbert, is self adjoint, then there exists an Hilbert base generated by eigenvector of M. Proof. The result follows if we can prove that H = N M λ n 0 N λn I M. In fact, the orthonormal base is just the union of the orthonormal bases of each eigenspace. Moreover, as in the finite dimensional space, one sees that the eigenvalue of M are real, and the spaces N λni M are orthogonals each other. To prove that the vector space Y generated by N M and {N λn I M n is dense in H, we first observe that Y is invariant for M, so that M(Y ) Y, because M is self adjoint. The operator M Y is self adjoint and compact, and by construction σ(m Y ) = {0. It follows M Y = 0 and Y N M. 6. Exercises () Let M : X Y, X Banach, Y reflexive. Show that if x n x, then Mx n Mx. (2) Define the adjoint of M : H H, H Hilbert space, by (x, M y) = (Mx, y). Prove that Proposition 2. holds for the adjoint operator. (3) Prove that if Y X, X Banach, is a subspace, then Ȳ = (Y ). (4) On l, consider the linear operator Su(n) = u(n + ). Compute the spectrum of S (consider the functions λ n ). (5) Consider the Hilbert space l 2 and a sequence of real numbers x n 0. Define Mu(n) = x n u(n). Show that T is compact and find its spectrum. (6) Find the eigenvalues and eigenvectors of the orthogonal projection P M on M subspace of H Hilbert. Is P M compact? (7) Fixed g(t, s) C ([0, ] 2, C), consider the linear operator M : C([0, ], C) C([0, ], C), Mu(t) = 0 g(t, s)u(s)ds. Discuss its spectrum. (8) Let H be a separable Hilbert space, K C a compact set in C, {λ n n a countable dense sequence in K.

6 6 COMPACT OPERATORS Show that there is a unique bounded linear operator M L(H) such that Me n = λ n e n. Show that σ(m = K, but the eigenvalues of M are {λ n n. Prove that for λ K \ {λ n n, then R λi M is dense in H.