Lecture 20: Compact Operators and the Spectrum of a Bounded Linear Operator on a Hilbert Space

May 4, 2021 Last lecture, we introduced the concept of a compact operator: an operator A ∈ B(H) (recall that H always denotes a Hilbert space) is compact if K({||u|| ≤ 1}), the closure of the image of the closed unit ball, is compact in H. These operators came up in our discussion of limits of finite rank operators, and we’ll show today that the set of compact operators is indeed the correct closure. Example 208 Some illustrative examples of compact operators include K : `2 → `2 sending a = (a1 , a2 , a3 , · · · ) to ( a11 , a22 , a33 , · · · ), R1 as well as T : L2 → L2 sending f (x) to 0 K(x, y )f (y )dy for some continuous function K : [0, 1] × [0, 1] → R. The latter is particularly important because it comes up in solutions to differential equations: if we take  (x − 1)y 0 ≤ y ≤ x ≤ 1, K(x, y ) = x(y − 1) 0 ≤ x ≤ y ≤ 1, then we can check that u(x) = R1 0 K(x, y )f (y )dy satisfies the differential equation u 00 = f , u(0) = u(1) = 0. Example 209 In contrast, even a simple-looking operator like I on `2 is not compact, because (as we’ve already previously demonstrated, looking at the standard basis vectors) the closed unit ball is not compact. And this argument works to show that the identity is never compact for an infinite-dimensional Hilbert space. Theorem 210 Let H be a separable Hilbert space. Then a bounded linear operator T ∈ B(H) is a compact operator if and only if there exist a sequence {Tn }n of finite rank operators such that ||T − Tn || → 0. (In other words, the set of compact operators is the closure of the set of finite rank operators R(H).) Proof. First, suppose T is compact. Since H is separable, it has an orthonormal basis, and by compactness, {T u : ||u|| ≤ 1} is compact, meaning that it is closed, bounded, and has equi-small tails. In particular, for every ε > 0, there exists some N ∈ N such that X |hT u, ek i|2 < ε2 k>N for all u satisfying ||u|| ≤ 1. We can thus define the partial sums n X Tn = hT u, ek iek : k=1 this is a bounded linear operator because ||Tn u||2 ≤ ||u||2 by Bessel’s inequality, and the range of Tn is contained within the span of {e1 , · · · , en }, so Tn is a finite rank operator for each n. It suffices to show that this choice of Tn does converge to T as n → ∞: indeed, for any ε > 0, we can let N be as above in the equi-small tails condition. Then we have, for any ||u|| = 1, that n ∞ X X ||Tn u − T u|| = hT u, ek iek − hTu , ek iek 2 k=1 k=1 2 , and combining terms and using orthonormality yields 2 = X hTu , ek iek k>n = X k>n |hT u, ek i|2 ≤ X |hT u, ek i|2 < ε2 . k>N Taking the supremum over all u with ||u|| = 1 and then taking a square root yields ||Tn − T || ≤ ε, as desired. For the opposite direction, we will use our second characterization of compact sets (approximating using finitedimensional subspaces). Suppose we know that ||Tn − T || → 0, where each Tn is a finite rank operator. Then {T u : ||u|| ≤ 1} is closed, and because it is contained in the set {v : ||v || ≤ ||T ||}, it is bounded. It suffices to show that for all ε > 0, there exists a finite-dimensional subspace W such that for all u with ||u|| ≤ 1, inf w ∈W ||T u − w || ≤ ε. The idea is to approximate T with Tn : there is some N such that ||TN − T || < ε, and thus we can let W be the range of TN (which is finite-dimensional). We then have, for any ||u|| ≤ 1, ||T u − TN u|| ≤ ||T − TN || · ||u|| ≤ ||T − TN || < ε, and thus inf w ∈W ||T u − w || < ε because TN u is an element of W . This means T is compact. We can also go a bit into the algebraic structure for compact operators, much like we did for finite rank operators: Theorem 211 Let H be a separable Hilbert space, and let K(H) be the set of compact operators on H. Then we have the following: 1. K(H) is a closed subspace of B(H). 2. For any T ∈ K(H), we also have T ∗ ∈ K(H). 3. For any T ∈ K(H) and A, B ∈ B(H), we have AT B ∈ K(H). In other words, the set of compact operators is also a star-closed, two-sided ideal in the algebra of bounded linear operators. Proof. Point (1) is clear because K(H) is the closure of R(H) (from above). For (2), notice that if T ∈ K(H), then there exists a sequence of finite-rank operators with ||Tn − T || → 0, meaning that ||Tn∗ − T ∗ || → 0 (since the operator norm of the adjoint and the original operator are the same). Since Tn∗ is finite rank for each n, this means T ∗ is indeed a compact operator. Finally, for (3), we also assume we have a sequence Tn → T of finite rank operators. Since we’ve already shown that condition (3) is satisfied by finite rank operators, we have ATn B a finite rank operator for each n, and thus ||ATn B − AT B|| = ||A(Tn − T )B|| ≤ ||A|| · ||Tn − T || · ||B| → 0 (because the operator norms of A and B are finite). Thus ATn B is a sequence of finite rank operators converging to AT B, and thus AT B is compact. We’ll now turn to studying particular properties of our operators: some of the most important numbers we associate with matrices are the eigenvalues. In physics, the eigenvalues of the Hamiltonian operator (which may not be finite rank) give us the energy levels of the system, and we’ll explain formally how we make that definition now, making a generalization of what we encounter in linear algebra. Proposition 212 Let T ∈ B(H) be a bounded linear operator. If ||T || < 1, then I − T is invertible, and we can compute its inverse to be the absolutely summable series (I − T ) −1 = ∞ X T n. n=0 We did this proof ourselves, and we can also use it to prove this next result: Proposition 213 The space of invertible linear operators GL(H) = {T ∈ B(H) : T invertible} is an open subset of B(H). Proof. Let T0 ∈ GL(H). Then we claim that any operator T satisfying ||T − T0 || < ||T0−1 ||−1 is invertible. Indeed, ||T0−1 (T − T0 )|| ≤ ||T0−1 || · ||T − T0 || < 1, so I − T0−1 (T − T0 ) is invertible by Proposition 212, meaning that I − T0−1 T + I = T0−1 T is invertible, meaning that T is invertible as well. Thus T0 has an open neighborhood completely contained in GL(H), meaning that GL(H) is open. The reason we’re talking about invertible linear operators here is that symmetric, real-valued matrices can be diagonalized, and we find those diagonal entries (eigenvalues) by trying to study the nullspace of A−λI. So eigenvalues are basically impediments to the invertibility of A − λI, and that’s how we’ll define our spectrum here: Definition 214 Let A ∈ B(H) be a bounded linear operator. The resolvent set of A, denoted Res(A), is the set {λ ∈ C : A − λI ∈ GL(H)}, and the spectrum of A, denoted Spec(A) is the complement C \ Res(A). Notice that if A − λI ∈ GL(H), then we can always uniquely solve the equation (A − λI)u = v for any v ∈ H. We will often write A − λI as A − λ for convenience. Example 215 " 2 2 Let A : C → C be the linear operator given in matrix form as A = λ1 0 # " λ1 − λ 0 # . Then A − λ = 0 λ2 0 λ2 − λ is not invertible exactly when λ = λ1 , λ2 , so the spectrum of A is {λ1 , λ2 } (and Res(A) = C \ {λ1 , λ2 }). In other words, the spectrum behaves as we expect it to for finite-dimensional operators, but there is an extra wrinkle for infinite-dimensional operators which we’ll see soon. Definition 216 If A ∈ B(H) and A − λ is not injective, then there exists some u ∈ H \ {0} with Au = λu, and we call λ an eigenvalue of A and u the associated eigenvector. Example 217 If we return to our compact operator T : `2 → `2 sending a 7→ ( a11 , a22 , a33 , · · · ), then the nth basis vector en is an eigenvector of T with eigenvalue n1 , so the spectrum contains at least the set { n1 : n ∈ N}. But there’s also an additional eigenvalue for T that we missed in this argument: it turns out that 0 is also in the spectrum, despite there being no nonzero vectors satisfying T v = 0. This is because while the operator T − 0 = T is indeed injective, it is not surjective and thus not invertible – in particular, the inverse of T would need to map a 7→ (a1 , 2a2 , 3a3 , · · · ), and this is not a bounded linear operator. So 0 is not in the resolvent, and thus it is in the spectrum, and this is an additional complication because of the infinite-dimensional structure. (The root of what’s going on is that the range of T can be dense but not closed.) Also in contrast to the finite-dimensional case, it’s also possible for an operator to have no eigenvalues at all: Example 218 Let T : L2 ([0, 1]) → L2 ([0, 1]) be defined via T f (x) = xf (x). Then T has no eigenvalues, but the spectrum is Spec(T ) = [0, 1] (so again we see a discrepancy between eigenvalues and the spectrum). Theorem 219 Let A ∈ B(H). Then Spec(A) is a closed subset of C, and Spec(A) ⊂ {λ ∈ C : |λ| ≤ ||A||}. In particular, this means the spectrum is a compact subset of the complex numbers – this is another way we can understand that if { n1 : n ∈ N} is in our spectrum, the limit point 0 must also be. Proof. It is equivalent to show that the resolvent set Res(A) is open and contains the set {λ ∈ C : |λ| > ||A||}. To show openness, let λ0 ∈ Res(A), meaning that A − λ0 is invertible. Since GL(H) is open, there exists some ε > 0 such that ||T − (A − λ0 )|| < ε =⇒ T ∈ GL(H). Now because |λ − λ0 | < ε =⇒ ||λI − λ0 I|| < ε =⇒ ||(A − λ) − (A − λ0 )|| < ε, this means that for all λ in an ε-neighborhood of λ0 , we have λ ∈ Res(A), and that shows openness. We now want to show that if |λ| > ||A||, then A − λ is invertible. Indeed, for any |λ| > ||A|| (in particular λ is nonzero), we have I − λ1 A invertible by Proposition 212. Thus   1 A − λ = −λ I − A λ is indeed invertible, and thus λ ∈ Res(A) as desired. Remark 220. It makes sense to ask whether the spectrum can be empty, and the answer is no. This requires some complex analysis – if the spectrum were empty, then for all u, v ∈ H, f (λ) = h(A − λ)−1 u, v i is a continuous, complex differentiable function in λ on C. As λ gets large, the operator norm of (A−λ)−1 goes to 0, but now Liouville’s theorem tells us that because f (λ) → 0 as |λ| → ∞, our function must be identically zero, which means that (A − λ)−1 = 0, a contradiction. For our purposes going forward, though, we’ll focus on self-adjoint operators, and it’ll be useful to have a better characterization of them. Theorem 221 If we have a self-adjoint operator A ∈ B(H), meaning that A = A∗ , then hAu, ui is real for all u, and ||A|| = sup||u||=1 |hAu, ui|. Proof. The first fact is easy to show: notice that hAu, ui = hu, Aui = hu, A∗ ui = hAu, ui using the definition of the inner product and the adjoint. For the second fact, let a = sup||u||=1 |hAu, ui|. For all ||u|| = 1, we have (by Cauchy-Schwarz) |hAu, ui| ≤ ||Au|| · ||u|| ≤ ||A|. So taking a supremum over all u, we find that a is a finite number, and a ≤ ||A||. To finish, it suffices to prove the other inequality. For any u ∈ H satisfying ||u|| = 1 such that Au 6= 0 (there is some u for which this is true, otherwise A is the zero operator and the result is clear), we can define the unit-length vector v = ||Au|| = Au ||Au|| , and hAu, Aui = hAu, v i = RehAu, v i, ||Au|| and we can verify ourselves that this can be written as = 1 Re [hA(u + v ), u + v i − hA(u − v ), u − v i + i (hA(u + i v ), u + i v i − hA(u − i v ), u − i v i)] . 4 Now the i (hA(u + iv ), u + iv i − hA(u − iv ), u − i v i) part is purely imaginary, since hA(u ± i v ), u ± i v i are real by the first part of this result, and thus those two terms drop out when we take the real part. We’re left with = 1 (hA(u + v ), u + v i − hA(u − v ), u − v i) , 4 and now using the fact that hAu, ui ≤ a for any unit-length u, meaning that hAu, ui ≤ a||u||2 for all u, we can bound this as 1 (a||u + v ||2 + a||u − v ||2 ), 4 and by the parallelogram law this simplifies to (because ||u|| = ||v || = 1) ≤ = a · 2(||u||2 + ||v ||2 ) = a. 4 Thus ||Au|| ≤ a for all u, meaning that ||A|| ≤ a as desired. Remark 222. In quantum mechanics, observables (like position, momentum, and so on) are modeled by self-adjoint unbounded operators, and the point is that all things measured in nature (the associated eigenvalues) are real. So there are applications of all of our discussions here to physics! We’ll discuss more about the spectrum of self-adjoint operators next time, seeing that it must be contained in R and also within certain bounds involving hAu, ui.