Lecture 18: The Adjoint of a Bounded Linear Operator on a Hilbert Space

April 27, 2021 We discussed the Riesz representation theorem last time, which states that for a Hilbert space H, we can identify each f ∈ H 0 = B(H, C) with a unique element v ∈ H such that f (u) = hu, v i for all u ∈ H. (In other words, every continuous linear functional on H can be realized as an inner product with a fixed vector.) We can use this to expand on a concept we’ve touched on previously in an assignment: Theorem 185 Let H be a Hilbert space, and let A : H → H be a bounded linear operator. Then there exists a unique bounded linear operator A∗ : H → H, known as the adjoint of A, satisfying hAu, v i = hu, A∗ v i for all u, v ∈ H. In addition, we have that ||A∗ || = ||A||. Proof. We can show uniqueness similarly to how we showed it in the Riesz representation theorem: if hu, A∗1 v i = hu, A∗2 v i for all u, v for two potential candidates A1 , A2 , then hu, (A∗1 v − A∗2 v )i = 0 for all u, v , and we can always set u = (A∗1 v − A∗2 v ) to show that we must have A∗1 v = A∗2 v for all v , meaning that A∗1 and A∗2 were the same operator to begin with. To show that such an operator does exist, first fix v ∈ H, and define a map fv : H → C by fv (u) = hAu, v i. This is a linear map (in the argument u) because for any u1 , u2 ∈ H and λ1 , λ2 ∈ C, we have fv (λ1 u1 + λ2 u2 ) = hA(λ1 u1 + λ2 u2 ), v i = hλ1 Au1 + λ2 Au2 , v i by linearity of A, and then this simplifies to = λ1 hAu1 , v i + λ2 hAu2 , v i = λ1 fv (u1 ) + λ2 fv (u2 ) by linearity in the first argument of the inner product. We claim this is also a continuous linear operator (so that it is actually an element of the dual). Indeed, we can check that if ||u|| = 1, |fv (u)| = |hAu, v i| ≤ ||Au|| · ||v || by the Cauchy-Schwarz inequality, and this is bounded by ||A||·||v ||. Therefore, ||fv || ≤ ||A||·||v || (which is a constant), and thus fv ∈ H 0 . By the Riesz representation theorem, we can therefore find a (unique) element, which we denote A∗ v , of H satisfying hAu, v i = fv (u) = hu, A∗ v i. We now need to show that A∗ is a bounded linear operator. For linearity, let v1 , v2 ∈ H and let λ1 , λ2 ∈ C. We know that for all u ∈ H, hu, A∗ (λ1 v1 + λ2 v2 )i = hAu, λ1 v1 + λ2 v2 i, and now by conjugate linearity in the second variable, this simplifies to = λ1 hAu, v1 i + λ2 hAu, v2 i = λ1 hu, A∗ v1 i + λ2 hu, A∗ v2 i. Pulling the complex numbers back in shows that this is = hu, λ1 A∗ v1 + λ2 A∗ v2 i . The only way for these two boxed expressions to be equal for all u is if the two operators are equal: A∗ (λ1 v1 + λ2 v2 ) = λ1 A∗ (v1 ) + λ2 A∗ (v2 ), which is the desired linearity result for A∗ . We now show that A∗ is bounded with ||A∗ || = ||A||. Take a unit-norm vector ||v || = 1: if A∗ v = 0, then clearly ||A∗ v || ≤ ||A||. Otherwise, we still want to show that same inequality. Suppose A∗ v 6= 0. Then ||A∗ v ||2 = hA∗ v , A∗ v i = hAA∗ v , v i by definition of the adjoint, and now by Cauchy-Schwarz this is bounded by ≤ ||AA∗ v || · ||v || = ||AA∗ v || ≤ ||A|| · ||A∗ v || . Dividing by the nonzero constant ||A∗ v || yields ||A∗ v || ≤ ||A||, as desired, and now taking the sup over all v with ||v || = 1 yields ||A∗ || ≤ ||A||. To finish, we need to show equality. For all u, v ∈ H, we have hA∗ u, v i = hv , A∗ ui = hAv , ui = hu, Av i, so the adjoint of the adjoint of A is A itself (since hu, Av i = hA∗ u, v i = hu, (A∗ )∗ v i). Therefore, we can flip the roles of A∗ and A in this argument to find that ||(A∗ )∗ || ≤ ||A∗ || =⇒ ||A|| ≤ ||A∗ ||, and putting the inequalities together yields ||A|| = ||A∗ || as desired. Let’s see a concrete example of what these adjoint operators look like: Example 186 If our Hilbert space is H = Cn , so that u is an n-dimensional vector, then we know that (Au)i = n X Aij uj j=1 for some fixed complex numbers Aij , and we can represent A as a finite-dimensional matrix. To determine the adjoint of A, we need to figure out the operator B that satisfies hAu, v i = hu, Bv i. Towards that, notice that hAu, v i = n X X (Au)i v i = Aij uj v i i=1 i,j and switching the order of summation yields = n n n X X X uj Aij vi = uj (A∗ v )j , j=1 i=1 j=1 where the adjoint of A acts on v as (A∗ v )i = n X Aji vj . j=1 So for matrices, the adjoint is also representable by a martix, and it is the conjugate transpose of A. Example 187 Now consider the space `2 , in which an operator is described with a double sequence {Aij }∞ in C so that X N X N X |Aij |2 = lim N→∞ i,j |Aij |2 < ∞. i=1 j=1 Specifically, we define A : `2 → `2 via (Aa)i = ∞ X Aij aj . j=1 We can check by the Cauchy-Schwarz inequality that this is a bounded linear operator as long as P i,j |Aij |2 is satisfied (the order of summation does not matter because all terms in the double sum are nonnegative). So A ∈ B(`2 , `2 ), and for all a, b ∈ `2 , we have ! hAa, bi`2 = XX i Aij aj bi = j X aj X Aij bi = ha, A∗ bi, i j where we define the adjoint similarly to in the finite-dimensional case: (A∗ b)i = ∞ X Aji bj . j=1 Finally, we can try doing an integral instead of an infinite sum: Example 188 Let K ∈ C([0, 1] × [0, 1]), and define the map A : L2 ([0, 1]) → L2 ([0, 1]) via 1 Z Af (x) = K(x, y )f (y )dy . 0 We can then check that the adjoint A∗ is defined as Z 1 ∗ A g(x) = K(y , x)g(y )dy , 0 so we’re again flipping the indices and taking a complex conjugate. Theorem 189 Let H be a Hilbert space, and let A : H → H be a bounded linear operator. Then (Ran(A))⊥ = Null(A∗ ), where Ran(A) is the range of A (the set of all vectors of the form Au), and Null(A∗ ) is the nullspace of A∗ (the set of all vectors for which A∗ u = 0). In particular, if we know that the range of A is a closed subspace, then always being able to solve Au = v (surjectivity) is equivalent to knowing that that the adjoint is one-to-one (injectivity), sine the range of A is then the orthogonal complement of the zero vector, which is the whole space. Proof. Note that v ∈ Null(A∗ ) if and only if hu, A∗ v i = 0 for all u ∈ H, which is equivalent to hAu, v i = 0 for all u ∈ H. So v is orthogonal to all elements in Ran(A), and that’s equivalent to saying that v ∈ Ran(A)⊥ . (All steps here go in both directions, so this shows the equivalence of the two sets.) This is essentially an infinite-dimensional version of rank-nullity, and we want to see if we can say similar things about the solutions to linear equations that we could in the finite-dimensional case (our input needs to satisfy certain linear relations, and then our final solution is unique up to a linear subspace). But before we get to that, these operators that we’ll solve solvability for have particular important properties on bounded sequences. We take for granted that a bounded linear operator takes bounded sets to bounded sets in finite-dimensional spaces, and so we can find a convergent subsequence using Heine-Borel. So the point is that there is some compactness hidden in here in Rn and Cn , so we need to study some facts about how compactness and Hilbert spaces before we can talk about solvability of equations. Definition 190 Let X be a metric space. A subset K ⊂ X is compact if every sequence of elements in K has a subsequence converging to an element of K. Example 191 By the Pigeonhole Principle, all finite subsets are compact. As just described, we also have the following result from real analysis: Theorem 192 (Heine-Borel) A subset K ⊂ R (also Rn and Cn ) is compact if and only if K is closed and bounded. Examples on the real line include closed intervals and also the set {0} ∪ { n1 : n ∈ N}. We know this doesn’t hold for arbitrary metric spaces or even Banach spaces, and in fact it’s still not true for Hilbert spaces: Example 193 Let H be an infinite-dimensional Hilbert space. Then the closed ball F = {u ∈ H : ||u|| ≤ 1} is a closed and bounded set, but it is not compact. This is because we can let {en }∞ n=1 be a countably infinite orthonormal subset of H (it doesn’t need to be a basis), which we find by Gram-Schmidt, so that all elements en are in F , but ||en − ek ||2 = ||en ||2 + ||ek ||2 + 2Rehen , ek i = 2. So the distance between any two elements of the sequence is 2, so there is no convergent subsequence (since it cannot be Cauchy). Motivated by this, we know that all compact sets are closed and bounded, and thus we want to figure out an additional condition guarantees compactness for a Hilbert space (so that we can verify compactness without using the subsequence definition). And this is in fact related to something that we can discuss in 18.100B in a different context when thinking about the space of continuous functions. Definition 194 Let H be a Hilbert space. A subset K ⊂ H has equi-small tails with respect to a countable orthonormal subset {en } if for all ε > 0, there is some n ≥ N so that for all v ∈ K, we have X |hv , ek i|2 < ε2 . k>N We know that the sequence for any given v converges by Bessel’s inequality, so that the inequality above will eventually hold for some N for each v . But this equi-small tails requirement is a more “uniform” condition on the rate of convergence – we need to be able to pick an N that works for all v ∈ K at the same time. Example 195 Any finite set K has equi-small tails with respect to any countable orthonormal subset (we can take the maximum of finitely many Ns). The motivation for this definition is that, as mentioned above, finite sets are always compact, so we should hope that this additional uniformity gives us compactness. We won’t get to that result today, but here’s some more motivation for why this is the correct condition to add, building on the {0} ∪ { n1 : n ∈ N} example from above: Theorem 196 Let H be a Hilbert space, and let {vn }n be a convergent sequence with vn → v . If {ek } is a countable orthonormal subset, then K = {vn : n ∈ N} ∪ {v } is compact, and K has equi-small tails with respect to {ek }. Proof. Compactness will be left as an exercise for us. For equi-small tails, the idea is that for sufficiently large n, vn will be close to v , so we can use v to take care of all but finitely many of the points in our sequence. Let ε > 0: since vn → v , there is some M ∈ N so that for all n ≥ M, we have ||vn − v || < 2ε . We choose N large enough so that for this fixed v , X |hv , ek i|2 + k>N X max 1≤n≤M−1 |hvn , ek i|2 < k>N ε2 . 4 (There are only finitely many terms here, and we can choose our N large enough so that it makes the n = 1 term smaller than ε2 8, the n = 2 term smaller than ε2 16 , and so on.) We claim that this N uniformly bounds our tails: indeed, X |hv , ek i|2 < ε2 < ε2 , 4 |hvn , ek i|2 < ε2 < ε2 . 4 k>N and for all 1 ≤ n ≤ M − 1 we also have X k>N So we just need to check the condition for n ≥ M: Bessel’s inequality tells us that !1/2 X 2 |hvn , ek i| !1/2 X = k>N 2 |hvn − v , ek i + hv , ek i| , k>N and this is the `2 norm of the sum of two sequences indexed by k, so by the triangle inequality this is boudned by !1/2 ≤ X k>N 2 |hvn − v , ek i| !1/2 + X k>N 2 |hv , ek i| . The second term is at most 2ε , and then the first term is bounded by Bessel’s inequality by ||vn − v ||. Since we chose N large enough so that that norm is less than 2ε , we indeed have that this is bounded by < ε ε + = ε, 2 2 as desired. Next time, we’ll prove that if we have a subset of a separable Hilbert space which is closed, bounded, and has equi-small tails with respect to an orthonormal basis (which we know exists), then we have compactness, and then we’ll rephrase that fact in a way that doesn’t involve Hilbert spaces and go from there.