SPECTRAL THEORY FOR COMPACT SELF-ADJOINT OPERATORS Throughout, H is a Hilbert space, and hx; yi the inner product on H, which is linear in x and conjugate linear in y. Recall that T 2 B(H) is self-adjoint if hT x; yi = hx; T yi. The first three of the following facts follow for self-adjoint T 2 B(H) by the same proofs as for the finite dimensional case. • The eigenvalues of T are real numbers. • Let λ be an eigenvalue, and Nλ = ker(λI − T ) the corresponding 0 eigenspace. Then Nλ is a closed subspace of H, and Nλ ? Nλ0 if λ 6= λ . ? ? • T : Nλ ! Nλ . ? • If T is compact, then the restriction of T to Nλ is compact. ? The last follows since the image of the unit ball of Nλ is the intersection of ? Nλ with the image of the unit ball of H. Lemma 1. Suppose that T is a self-adjoint operator on H. Then kT k = sup hT x; xi kxk=1 where kT k is the operator norm of T . Proof. Let M = supkxk=1 hT x; xi , so M ≤ kT k : To prove the other direc- tion, note that kT k = supkxk=kyk=1 RehT x; yi, and write 1 Re T x; y = T (x + y); x + y − T (x − y); x − y 4 1 ≤ M kx + yk2 + kx − yk2 4 1 = M kxk2 + kyk2 = M: 2 It follows that either kT k = sup T x; x or kT k = − inf T x; x : kxk=1 kxk=1 Lemma 2. If T is a compact, self-adjoint operator on H, then either ±kT k is an eigenvalue of T . Indeed, if T xn; xn ! ±kT k where kxnk = 1, then a subsequence of fxng converges to an eigenvector for T of eigenvalue ±kT k. 1 2 526/556 LECTURE NOTES Proof. Suppose T xn; xn ! M where M = ±kT k and kxnk = 1. Then 2 2 2 2 0 ≤ T xn − Mxnk = kT xnk + M kxnk − 2M T xn; xn 2 2 ≤ M + M − 2M T xn; xn ! 0 : By compactness of T there is a subsequence fxnj g of fxng so that T xnj ! x. Since T xnj − Mxnj ! 0, then Mxnj ! x 6= 0, and T x = Mx. Theorem 3. Suppose T is a compact, self-adjoint operator on H. Let fλjg ⊂ R be the collection of eigenvalues of T , possibly including 0, with corresponding eigenspaces Nλj . Then the span of the union of the Nλj is dense in H. In particular, there is an orthonormal basis for H consisting of eigenvectors for T . Remark. If 0 is an eigenvalue, then N0 is the same as the kernel of T . This may be infinite dimensional, but for λj 6= 0 then dim Nλj < 1 : The or- thonormal basis for H is obtained by taking orthonormal bases for each Nλj and for N0. By the density statement, the span of the resulting collection of orthonormal vectors is dense, hence the collection is a basis. If 0 is not an eigenvalue, i.e. T is 1-1, then 0 will be in the spectrum of T if and only if dim H = 1, in which case there is a sequence of eigenvalues converging to 0. In this case 0 is in the continuous spectrum of T . Proof. Let λ1 = kT k. By Lemma 2, either ±λ1 is an eigenvalue of T . We decompose H = Nλ1 ⊕ N−λ1 ⊕ H2 ; where N±λ1 are the eigenspaces associated to the eigenvalues ±λ1. (One of the eigenspaces may be 0, in which case we ignore it.) Here, H2 is the orthogonal complement of Nλ1 ⊕ N−λ1 . The norm of T restricted to H2 must be strictly less than λ1 = kT k; otherwise applying Lemma 2 to T on H2 would yield a nonzero eigenvector in H2 with eigenvalue ±λ1. Let λ2 = k T jH2 k. Repeating the argument decomposes H = Nλ1 ⊕ N−λ1 ⊕ Nλ2 ⊕ N−λ2 ⊕ H3 ; where again we ignore N±λ2 if it equals f0g, and λ3 = k T jH3 k < λ2 : Inductively define fλjg and N±λj . If at some point k T jHn+1 k = 0, then Hn+1 = ker T , and H = Nλ1 ⊕ N−λ1 ⊕ · · · ⊕ Nλn ⊕ N−λn ⊕ N0 : Otherwise there is a countable set of nonzero eigenvalues, of the form ±λj, where λj ! 0. To see that the span of the N±λj together with N0 is dense in H, we show if x is orthogonal to all N±λj , then x 2 N0. To show this, we note that for each n, x is orthogonal to all N±λj with j < n, so x 2 Hn. This implies that kT xk ≤ λnkxk. This holds for all n, so kT xk = 0. .