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.� 67 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 11. Thursday, 12 Mar Quite a lot of new material, but all of it in the various notes and books. So, I will keep it brief. (1) Convex sets and length minimizer The following result does not need the hypothesis of separability of the Hilbert space and allows us to prove the subsequent results – especially Riesz’ theorem – in full generality. Proposition 17. If C ⊂ H is a subset of a Hilbert space which is (a) Non-empty (b) Closed (c) Convex, in the sense that v1, v1 ∈ C implies 1 (v1 + v2) ∈ C2 then there exists a unique element v ∈ C closest to the origin, i.e. such that (11.1) �v�H = inf u∈C �u�H . Proof. By definition of inf there must exist a sequence {vn} in C such that �vn� → d = infu∈C �u�H . We show that vn converges and that the limit is the point we want. The parallelogram law can be written (11.2) �vn − vm�2 = 2�vn�2 + 2�vm�2 − 4�(vn + vm)/2�2 . Since �vn� → d, given � > 0 if N is large enough then n > N implies 2�vn�2 < 2d2 +�2/2. By convexity, (vn +vm)/2 ∈ C so �(vn +vm)/2�2 ≥ d2 . Combining these estimates gives (11.3) n, m > N = ⇒u∈C �u�H ≤ 4d2 + �2 − 4d2 so {vn} is Cauchy. Since H is complete, vn v ∈ C since C is closed. →Moreover, the distance is continuous so �v�H = limn→∞ �vn� = d. Thus v exists and uniqueness follows again from the parallelogram law. If v and v� are two points in C with �v� = �v�� = d then (v + v�)/2 ∈ C so (11.4) �v − v��2 = 2�v�2 + 2�v��2 − 4�(v + v�)/2�2 ≤ 0 =⇒ v = v�. (2) Orthocomplements Proposition 18. If W ⊂ H is a linear subspace of a Hilbert space the (11.5) W ⊥ = {u ∈ H; (u, w) = 0 ∀ w ∈ W } is also a linear subspace and W ∩ W ⊥ = {0}. If W is also closed then (11.6) H = W ⊕ W ⊥ meaning that any u ∈ H has a unique decomposition u = w + w⊥ where w ∈ W and w⊥ ∈ W ⊥. Proof. That W ⊥ defined by (11.5) is a linear subspace follows from the linearity of the condition defining it. If u ∈ W ⊥ and u ∈ W u ⊥ u by the definition so (u, u) = �u�2 = 0 and u = 0. Now, suppose W is closed. If W = H then W ⊥ = {0} and there is nothing to show. So consider u ∈ H, u /∈ W. Consider (11.7) C = u + W = {u� ∈ H; u� = u + w, w ∈ W }.68 LECTURE NOTES FOR 18.102, SPRING 2009 Then C is closed, since a sequence in it is of the form u�n = u+wn where wn is a sequence in W and u�n converges if and only if wn converges. Now, C is non-empty, since u ∈ C and it is convex since u� = u + w� and u�� = u + w�� in C implies (u� + u��)/2 = u + (w� + w��)/2 ∈ C. Thus the length minimization result above applies and there exists a unique v ∈ C such that �v� = infu�∈C �u��. The claim is that this v is perpendicular to W – draw a picture in two real dimensions! To see this consider an aritrary point w ∈ W and λ ∈ C then v + λw ∈ C and (11.8) �v + λw�2 = �v�2 + 2 Re(λ(v, w)) + |λ|2�w�2 . Choose λ = teiθ where the phase is chosen so that eiθ(v, w) = |(v, w)| ≥ 0. Then the fact that �v� is minimal means that (11.9) t(2|(v, w)| + t�w�2) ≥ 0 ∀ t ∈ R = ⇒ |(v, w)| = 0 which is what we wanted to show. Thus indeed, give u ∈/ W we have constructed v ∈ W ⊥ such that u = v + w, w ∈ W. This is (11.6) with the uniqueness of the decomposition already shown since it reduces to 0 having only the decomposition 0 + 0 and this in turn is W ∩ W ⊥ = {0}. � (3) Riesz’ theorem The most important application of these results is to prove Riesz’ rep-resentation theorem (for Hilbert space, there is another one to do with measures). Theorem 7. If H is a Hilbert space then any continuous linear functional T : H −→ C there exists a unique element φ ∈ H such that (11.10) T (u) = (u, φ) ∀ u ∈ H. Proof. (a) Here is the proof I gave quickly in Lecture 10, not using the preceeding Lemma. If T is the zero functional then w = 0 satisfies (11.10). Otherwise there exists some u� ∈ H such that T (u�) = 0 and �then u ∈ H, namely u = u�/F (u�), such that F (u) = 1. Thus (11.11) C = {u ∈ H; T (u) = 1} = T −1({1}) is non-empty. The continuity of T and the second form shows that C is closed, as the inverse image of a closed set under a continuous map. Moreover C is convex since (11.12) T ((u + u�)/2) = (T (u) + T (u�))/2. Thus, there exists an element v ∈ C of minimal length. As in the proof above, it follows that �v + λw�2 ≥ �v�2 for all w ∈ W and λ ∈ C this implies that v ∈ W ⊥. Now continue as in the proof below. (b) Here is the proof I gave in Lecture 11 using the orthocomplement above. Since T is continuous the null space (11.13) W = T −1({0}) = {u ∈ H; T (u) = 0} is a closed linear subspace. Thus (11.14) H = W ⊕ W ⊥� 69 LECTURE NOTES FOR 18.102, SPRING 2009 by Proposition 18 above. Now, if T = 0 is the zero functional then W = H and W ⊥ = {0} and w = 0 works in (11.10). Othewise, W ⊥ � v�, which is not in W, i.e. has T (v�) =� 0 and hence v ∈ W ⊥ with T (v) = 1. Then for any u ∈ H, (11.15) u−T (u)v satisfies T (u−T (u)v) = T (u)−T (u)T (v) = 0 −→ u = w+T (u)v, w ∈ W. Then, (u, v) = T (u)�v�2 since (w, v) = 0. Thus if φ = v/�v�2 then (11.16) u = w + (u, φ)v = T (u) = (u, φ)T (v) = (u, φ).⇒ (4) Adjoints of bounded operators. As an application of Riesz’ theorem I showed that any bounded linear operator on a Hilbert space (11.17) A : H −→ H, �Hu�H ≤ …