5 Compact linear operators

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1 5 Compact linear operators One of the most important results of Linear Algebra is that for every selfadjoint linear map A on a finite-dimensional space, there exists a basis consisting of eigenvectors. In particular, with respect to this basis the operator A can be represented by a diagonal matrix. The situation turns out to be much more complicated for operators on infinite dimensional spaces (in fact, self-adjoint operators may have no eigenvectors at all). Nonetheless, a satisfactory theory can be developed for so-called compact operators which we now introduce. Definition 5.1 (Compact linear operator). A linear operator T : X Y between normed spaces X and Y is called a compact linear operator if for every bounded sequence (x n ) n 1 in X, the sequence (T x n ) n 1 has a convergent subsequence. We note that every compact operator T is bounded. Indeed, if T =, then there exists a sequence (x n ) n 1 such that x n 1 and T x n. Then (T x n ) n 1 cannot have a convergent subsequence. Hence, T <. Example. 1. Consider the linear operator T N : l 2 l 2 defined by T N x = (x 1, x 2,..., x N, 0, 0,... ). We claim that T N is compact. Given any bounded sequence (x (n) ) n 1 in l 2, the sequence (T N x (n) ) n 1 is bounded in F N l 2. Since every bounded sequence in R N or C N has a convergent subsequence, it follows that T N is compact. 2. The identity operator I : l 2 l 2 is not compact. To prove this, consider the bounded sequence (e n ) n 1 where e n = (0,..., 0, 1, 0,...). Then for any n m, Ie n Ie m 2 = e n e m 2 = 2. This implies that any subsequence of (Ie n ) n 1 cannot be Cauchy and, hence, cannot converge. So that I is not compact. 3. (integral operators) Let K C([0, 1] 2 ) and T : L 2 ([0, 1]) L 2 ([0, 1]) is defined by T f(x) = 1 0 K(x, y)f(y)dy. Then the operator T is compact. We will not prove this in class, and it could be a possible topic for a level-m presentation. We show that limits of compact operators is also compact. Theorem 5.2. Let (T n ) n 1 be a sequence of compact linear operators from a normed space X into a Banach spaces Y. If T n T (that is, T n T 0), then the limit operator T is compact. Proof. Since T 1 is a compact operator, we know that the sequence (T 1 (x n )) has a convergent (hence Cauchy) subsequence (T 1 (x 1,m )), where (x 1,m ) is a subsequence of the original sequence (x n ). The subsequence (x 1,m ) is bounded, so we can repeat the argument with T 2 to produce a subsequence 1

2 (x 2,m ) of (x 1,m ) with the property that (T 2 (x 2,m )) converges. We continue in the same way, and then define a sequence (y m ) = (x m,m ). Notice that (y m ) is a subsequence of (x n ), so it is bounded, say by y n c, and it has the property that for every fixed n, the sequence (T n (y m )) is convergent, and hence Cauchy. We claim that (T (y m )) is a Cauchy sequence in Y. Let ɛ > 0. Since T n T 0, there is some p N such that T p T < ɛ 3c. Also, since (T p (y m )) is Cauchy, there is some N > 0 such that T p (y j ) T p (y k ) < ɛ 3 whenever j, k > N. Therefore, for j, k > N, we have T (y j ) T (y k ) T (y j ) T p (y j ) + T p (y j ) T p (y k ) + T p (y k ) T (y k ) T (y j ) T p (y j ) + T p (y j ) T p (y k ) + T p (y k ) T (y k ) < T T p y j + ɛ 3 + T T p y k < ɛ 3c c + ɛ 3 + ɛ 3c c = ɛ, which proves that (T (y m )) is Cauchy. Since Y is a Banach space, it is by definition complete, so (T (y m )) converges. We have thus produced, for an arbitrary bounded sequence (x n ) X, a convergent subsequence of its image under T. Therefore, T is compact. Example. Let T : l 2 l 2 be an operator defined by T x = (λ n x n ) n 1 for a sequence λ n 0. We claim that T is compact. To show this, we approximate T by compact operators T N such that T N x = (λ 1 x 1,..., λ N x N, 0,...). As in the previous example we observe that T N is a compact operator. For x l 2, ( ) 1/2 T N x T x 2 = λ n x n 2 ( ) sup λ n x 2. ( ) ( ) 1/2 sup λ n x n 2 This shows that T N T sup λ n. Since λ n 0, it follows that T N T 0. Hence, T is compact by the previous theorem. We note that for the operator T in the previous example there exists a basis of eigenvectors e n with eigenvalues λ n. Remarkably, every self-adjoint compact operator on a Hilbert space is of this form: it can be diagonalised with respect to suitable orthonormal set. The following is the main result of this lecture: Theorem 5.3 (spectral theorem). Let H be a Hilbert space and T : H H a compact self-adjoint operator. Then there exists an orthonormal set 2

3 (e n ) n 1 consisting of eigenvectors of T with eigenvalues λ n such that T x = λ n x, e n e n, x H. n=1 The proof of this theorem require several auxilary results. Theorem 5.. Let T be a self-adjoint operator on a Hilbert space H. Then T = sup{ T x, x : x = 1}. Proof. Let m = sup{ T x, x : x = 1}. For x H with x = 1, This shows that m T. For any x, y H, T x, x T x x T x 2 = T. Re T x, y = 1 ( T (x + y), x + y T (x y), x y ). This formula can be checked by expanding the right hand side. We use that T u, u m u 2 for all u H. Then from the above formula, we obtain Re T x, y 1 ( T (x + y, x + y + T (x y), x y ) 1 (m x + y 2 + m x y 2 ) = m 2 ( x 2 + y 2 ), where we used the parallelogram identity. Replacing x by λx where λ = 1 and λ is chosen so that λ T x, y = T x, y is real and non-negative, we obtain T x, y m 2 ( x 2 + y 2 ). Suppose that T x = 0 and take y = x T x T x. This gives x T x m x 2. Hence, T x m x. This inequality also obviously holds when T x = 0. We conclude that T m. Theorem 5.5. Let T be a compact self-adjoint operator on a Hilbert space H. Then either T or T is an eigenvalue. Proof. We assume that that T 0. By the previous theorem there exists x n H with x n = 1 such that T x n, x n T. Since T = T, T x n, x n = x n, T x n = T x n, x n, so that T x n, x n is real. Passing to subsequence we may assume that T x n, x n λ where λ = T or λ = T. Then T x n λx n 2 = T x n 2 2λ T x n, x n +λ 2 x n 2 2λ 2 2λ T x n, x n 0. 3

4 This shows that T x n λx n 0. Since T is compact, passing to a subsequence we may assume that T x n y for some y H. Tnen it follows that λx n y. Since T is continuous, λt x n T y, but λt x n λy. Hence, we conclude that T y = λy. We claim that y 0. Indeed, by the triangle inequality, T x n λx n T x n λx n = λ T x n λx n λ = T > 0. Hence, y = lim T x n > 0. We have proved that λ is an eigenvalue of T. Theorem 5.6. Let T be a compact self-adjoint operator on a Hilbert space H and λ n s are eigenvalues of T with linearly independent eigenvectors x n. Then λ n s are real, and for every c > 0 there are only finitely many n s such that λ n c. Proof. We have λ n x n 2 = T x n, x n = x n, T x n = λ n x n 2. Since x n 0, it follows that λ n = λ n and λ n is real. Suppose that λ n λ m. Then λ n x n, x m = T x n, x m = x n, T x m = λ m x n, x m. This implies that x n, x m = 0, that is x n and x m are orthogonal. If λ n s are repeating, then consider N λ = {n : λ n = λ}. We note that linear combinations of x n s with n N λ are also eigenvectors with eigenvalue λ. Using the Gram-Schmidt orthogonalisation algorithm, we can construct orthonormal eigenvectors for λ n with n N λ. Now we can assume that x n s are orthonormal. Suppose that for infinitely many n s λ n c. Then for such indices n m, T x n T x m 2 = λ n x n λ m x m 2 = λ n 2 + λ m 2 2c 2. This gives a sequence of the form (T x n ) which does not contain any Cauchy subsequence. This contradicts compactness of the operator T. We note that Theorem 5.6 implies that if we order the eigenvalues of T as λ 1 λ 2, then λ n 0. Theorem 5.7. Let T be a self-adjoint operator on a Hilbert space H, and U is a closed subspace of H such that T (U) U. Then T (U ) U. Proof. Let x U. Then for any y U, T y U, and T x, y = x, T y = 0. Hence, T x U. Now we are ready for the proof of the main Theorem 5.3. The idea is to apply Theorem 5.5 inductively.

5 Proof of Theorem 5.3. By Theorem 5.5, there exists an eigenvector e 1 with eigenvalue λ 1 = ± T. Note that if T x = λx with x 0, then T x = λ x T x, so that λ T. Hence, λ 1 is maximal among eigenvalues. After rescaling we can assume that e 1 = 1. The subspace U 1 = span(e 1 ) is closed and T -invariant. By Theorem 5.7, U1 is also T -invariant. Now we can apply the same construction to the operator T 2 : U1 U 1, which is the restriction of the operator T to U 1, to construct a unit eigenvector e 2 U1 with eigenvalue λ 2 such that λ 2 = T 2. Since T 2 T, we have λ 2 λ 1. Next, we consider T 3 : U2 U 2 where U 2 = e 1, e 2, and so on... After n steps we produce an orthonormal set {e 1,... e n } consisting of eigenvectors with eigenvalues λ 1,..., λ n such that λ 1 λ n. Setting U n = e 1,... e n, we have the orthogonal decomposition H = U n Un. For a vector x H, we set y n = x n i=1 x, e i e i. Since e i s are orthonormal, it is easy to check that y n Un. Hence, we have the orthogonal sum x = n i=1 x, e i e i + y n, and by the Pythagoras theorem, x 2 = n i=1 x, e i 2 + y n 2. In particular, y n x. We observe that ( ) n T x n λ i x, e i e i T x x, e i e i i=1 i=1 = T n+1 y n T n+1 y n λ n+1 x. Since by Theorem 5.6, λ n+1 0, this proves that the operator T can be represented as T x = i=1 λ i x, e i e i, x H. 5