MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.116 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 20. Thursday, April 23: Spectral theorem for compact self-adjoint operators Let A ∈ K(H) be a compact operator on a separable Hilbert space. We know of course, even without assuming that A is compact, that (20.1) Nul(A) ⊂ H is a closed subspace, so Nul(A)⊥ is a Hilbert space – although it could be finite-dimensional (or even 0-dimensional in the uninteresting case that A = 0). Theorem 15. If A ∈ K(H) is a self-adjoint, compact operator on a separable Hilbert space, so A∗ = A, then null(A)⊥ has an orthonormal basis consisting of eigenfunctions of A, uj such that (20.2) Auj = λj uj , λj ∈ R \{0}, arranged so that |λj | is a non-increasing sequence satisfying λj → 0 as j → ∞ (in case Nul(A)⊥ is finite dimensional, this sequence is finite). Before going to the proof, let’s notice some useful conclusions. One is called ‘Fredholm’s alternative’. Corollary 4. If A ∈ K(H) is a compact self-adjoint operator on a separable Hilbert space then the equation (20.3) u − Au = f either has a unique solution for each f ∈ H or else there is a non-trivial finite dimensional space of solutions to (20.4) u − Au = 0 and then (20.3) has a solution if and only if f is orthogonal to all these solutions. Proof. This is just saying that the null space of Id −A is a complement to the range – which is closed. So, either Id −A is invertible or if not then the range is precisely the orthocomplement of Nul(Id −A). You might say there is not much alternative from this point of view, since it just says the range is always the orthocomplement of the null space. � Let me separate off the heart of the argument from the bookkeeping. Lemma 14. If A ∈ K(H) is a self-adjoint compact operator on a separable (possibly finite-dimensional) Hilbert space then (20.5) F (u) = (Au, u), F : {u ∈ H; �u� = 1} −→ R is a continuous function on the unit sphere which attains its supremum and infi-mum. Furthermore, if the maximum or minimum is non-zero it is attained at an eivenvector of A with this as eigenvalue. Proof. So, this is just like function in finite dimensions, except that it is not. First observe that F is real-valued, which follows from the self-adjointness of A since (20.6) (Au, u) = (u, Au) = (A∗u, u) = (Au, u). Moreover, continuity of F follows from continuity of A and of the inner product so (20.7) |F (u) − F (u�)| ≤ |(Au, u) − (Au, u�)| + |(Au, u�) − (Au�, u�)| ≤ 2�A��u − u�� since both u and u� have norm one.117 LECTURE NOTES FOR 18.102, SPRING 2009 If we were in finite dimensions this finishes the proof, since the sphere is then compact and a continuous function on a compact set attains its sup and inf. In the general case we need to use the compactness of A. Certainly F is bounded, (20.8) F (u) sup (Au, u)| | ≤ �u�=1 | | ≤ �A�. F (u−inf F. The weak compactness of the unit sphere means that we can pass nnsup F and another u−to a subsequence in each case, and so assume that u±� u± converges weakly. nThus, there is a sequence u such that F (u ) such that + + →n n) → Then, by the compactness of A, Au±then we can write nAu± converges strongly, i.e. in norm. But → (20.9) |F (u±n) − F (u±)| ≤ (A(u±n|(A(u±nn− u±), u±n) ) ( (± ± ± ±, A− u , u u un)n(Au±, u±− u±))− u±)+| | |2±Au≤ �n) are respectively the sup and inf of F. Thus − Au±�+= | | | | to deduce that F (u±) lim F (u±indeed, as in the finite dimensional case, the sup and inf are attained, as in max and min. So, suppose that Λ+ = sup F > 0. Then for any v ∈ H with v ⊥ u+ the curve n(20.10) Lv : (−π, π) � θ �−→ cos θu+ + sin θv lies in the unit sphere. Computing out (20.11) F (Lv(θ)) = (ALv(θ), Lv(θ)) = cos2 θF (u +) + 2 sin(2θ) Re(Au+ , v) + sin2(θ)F (v) we know that this function must take its maximum at θ = 0. The derivative there (it is certainly continuously differentiable on (−π, π)) is Re(Au+, v) which must therefore vanish. The same is true for iv in place of v so in fact (20.12) (Au+ , v) = 0 ∀ v ⊥ u + , �v� = 1. Taking the span of these v’s it follows that (Au+, v) = 0 for all v ⊥ u+ so A+u must be a multiple of u+ itself. Inserting this into the definition of F it follows that Au+ = Λ+u+ is an eigenvector with eigenvalue Λ+ = sup F. The same argument applies to inf F if it is negative, for instance by replacing A by −A. This completes the proof of the Lemma. � Proof of Theorem 15. First consider the Hilbert space H0 = Nul(A)⊥ ⊂ H. Then A maps H0 into itself, since (20.13) (Au, v) = (u, Av) = 0 ∀ u ∈ H0, v ∈ Nul(A) =⇒ Au ∈ H0. Moreover, A0, which is A restricted to H0, is again a compact self-adjoint operator – where the compactness follows from the fact that A(B(0, 1)) for B(0, 1) ⊂ H0 is smaller than (actually of course equal to) the whole image of the unit ball. Thus we can apply the Lemma above to A0, with quadratic form F0, and find an eigenvector. Let’s agree to take the one associated to sup FA0 unless supA0 < − inf F0 in which case we take one associated to the inf . Now, what can go wrong here? Nothing except if F0 ≡ 0. However, Lemma 15. In general for a self-adjoint operator on a Hilbert space (20.14) F ≡ 0 ⇐⇒ A ≡ 0. =118 LECTURE NOTES FOR 18.102, SPRING 2009 Proof. In principle F is only defined on the unit ball, but of course we can recover (Au, u) for all u ∈ H from it. Namely, if u = 0 it vanishes of course and otherwise (20.15) (Au, u) = �u�2F ( u ). �u� Thenxs we can recover A by ‘polarization’. Since (20.16) 2(Au, v) = (A(u + v), u + v) + i(A(u + iv, u + iv). Thus if F ≡ 0 then A ≡ 0. � So, we know that we can find an eigenvector unless A ≡ 0 which would imply Nul(A) = H. Now we proceed by induction. Suppose we have found N mutually or-thogonal eigenvectors ej for A all with norm 1 and eigenvectors λj – an orthonormal set of eigenvectors and all in H0. Then we consider (20.17) HN = {u ∈ H0 = Nul(A)⊥; (u, ej ) = …