SPECTRAL THEORY for COMPACT SELF-ADJOINT OPERATORS Throughout, H Is a Hilbert Space, and 〈X, Y〉 the Inner Product on H, Whic

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 ∈ B(H) is self-adjoint if hT x, yi = hx, T yi. The ﬁrst three of the following facts follow for self-adjoint T ∈ 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 {xn} 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 {xnj } of {xn} 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 {λj} ⊂ 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 < ∞ . The orthonormal 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 = ∞, 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 |H2 k. Repeating the argument decomposes

H = Nλ1 ⊕ N−λ1 ⊕ Nλ2 ⊕ N−λ2 ⊕ H3 , where again we ignore N±λ2 if it equals {0}, and λ3 = k T |H3 k < λ2 .

Inductively deﬁne {λj} and N±λj . If at some point k T |Hn+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 ∈ N0. To show this, we note that for each n, x is orthogonal to all N±λj with j < n, so x ∈ Hn. This implies that kT xk ≤ λnkxk. This holds for all n, so kT xk = 0.