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.� 100 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 17. Thursday April 9 was the second test (1) Problem 1 Let H be a separable (partly because that is mostly what I have been talking about) Hilbert space with inner product ( ) and norm � · �.·, ·Say that a sequence un in H converges weakly if (un, v) is Cauchy in C for each v ∈ H. (a) Explain why the sequence �un�H is bounded. Solution: Each un defines a continuous linear functional on H by (17.1) Tn(v) = (v, un), �Tn� = �un�, Tn : H −→ C. For fixed v the sequence Tn(v) is Cauchy, and hence bounded, in C so by the ‘Uniform Boundedness Principle’ the �Tn� are bounded, hence �un� is bounded in R. (b) Show that there exists an element u ∈ H such that (un, v) → (u, v) for each v ∈ H. Solution: Since (v, un) is Cauchy in C for each fixed v ∈ H it is convergent. Set (17.2) T v = lim (v, un) in C. n→∞ This is a linear map, since (17.3) T (c1v1 + c2v2) = lim c1(v1, un) + c2(v2, u) = c1T v1 + c2T v2 n→∞ and is bounded since |T v| ≤ C�v�, C = supn �un�. Thus, by Riesz’ theorem there exists u ∈ H such that T v = (v, u). Then, by definition of T, (17.4) (un, v) → (u, v) ∀ v ∈ H. (c) If ei, i ∈ N, is an orthonormal sequence, give, with justification, an example of a sequence un which is not weakly convergent in H but is such that (un, ej ) converges for each j. Solution: One such example is un = nen. Certainly (un, ei) = 0 for all i > n, so converges to 0. However, �un� is not bounded, so the sequence cannot be weakly convergent by the first part above. (d) Show that if the ei form an orthonormal basis, �un� is bounded and (un, ej ) converges for each j then un converges weakly. Solution: By the assumption that (un, ej ) converges for all j it fol-lows that (un, v) converges as n → ∞ for all v which is a finite lin-ear combination of the ei. For general v ∈ H the convergence of the Fourier-Bessell series for v with respect to the orthonormal basis ej (17.5) v = (v, ek)ek k shows that there is a sequence vk v where each vk is in the finite →span of the ej . Now, by Cauchy’s inequality (17.6) |(un, v) − (um, v)| ≤ |(unvk) − (um, vk)| + |(un, v − vk)| + |(um, v − vk)|. Given � > 0 the boundedness of �un� means that the last two terms can be arranged to be each less than �/4 by choosing k sufficiently large. Having chosen k the first term is less than �/4 if n, m > N by� � � � � � � 101 LECTURE NOTES FOR 18.102, SPRING 2009 the fact that (un, vk) converges as n → ∞. Thus the sequence (un, v) is Cauchy in C and hence convergent. (2) Problem 2 Suppose that f ∈ L1(0, 2π) is such that the constants ck = f(x)e−ikx , k ∈ Z, (0,2π) satisfy 2|ck| < ∞. k∈Z Show that f ∈ L2(0, 2π). Solution. So, this was a good bit harder than I meant it to be – but still in principle solvable (even though no one quite got to the end). First, (for half marks in fact!) we know that the ck exists, since f ∈ L1(0, 2π) and 2 e−ikx is continuous so fe−ikx ∈ L1(0, 2π) and then the con-dition |ck| < ∞ implies that the Fourier series does converge in L2(0, 2π) k so there is a function 1 � ikx (17.7) g = cke .2π k∈C Now, what we want to show is that f = g a.e. since then f ∈ L2(0, 2π). Set h = f − g ∈ L1(0, 2π) since L2(0, 2π) ⊂ L1(0, 2π). It follows from (17.7) that f and g have the same Fourier coefficients, and hence that (17.8) h(x)e ikx = 0 ∀ k ∈ Z. (0,2π) So, we need to show that this implies that h = 0 a.e. Now, we can recall from class that we showed (in the proof of the completeness of the Fourier basis of L2) that these exponentials are dense, in the supremum norm, in continuous functions which vanish near the ends of the interval. Thus, by continuity of the integral we know that (17.9) hg = 0 (0,2π) for all such continuous functions g. We also showed at some point that we can find such a sequence of continuous functions gn to approximate the characteristic function of any interval χI . It is not true that gn χI 1 →uniformly, but for any integrable function h, hgn → hχI in L . So, the upshot of this is that we know a bit more than (17.9), namely we know that (17.10) hg = 0 ∀ step functions g. (0,2π) So, now the trick is to show that (17.10) implies that h = 0 almost everywhere. Well, this would follow if we know that (0,2π) |h| = 0, so let’s aim for that. Here is the trick. Since g ∈ L1 we know that there is a sequence (the partial sums of an absolutely convergent series) of step� � � � � � � 102 LECTURE NOTES FOR 18.102, SPRING 2009 functions hn such that hn g both in L1(0, 2π) and almost everywhere →and also |hn| → |h| in both these senses. Now, consider the functions 0 if hn(x) = 0 (17.11) sn(x) = hn(x) otherwise. |hn(x)| Clearly sn is a sequence of step functions, bounded (in absolute value by 1 in fact) and such that snhn = |hn|. Now, write out the wonderful identity (17.12) |h(x)| = |h(x)| − |hn(x)| + sn(x)(hn(x) − h(x)) + sn(x)h(x). Integrate this identity and then apply the triangle inequality to conclude that |h| = (|h(x)| −|hn(x)| + sn(x)(hn − h) (17.13) (0�,2π) (0,2π) � (0,2π) ≤ (0,2π) (||h(x)| −|hn(x)|| + (0,2π) |hn − h| → 0 as n → ∞. Here on the first line we have used (17.10) to see that the third term on the right in (17.12) integrates to zero. Then the fact that |sn| ≤ 1 and the convergence properties. Thus in fact h = 0 a.e. so indeed f = g and f ∈ L2(0, 2π). Piece of cake, right! Mia culpa. (3) Problem 3 Consider the two spaces of sequences ∞h±2 = {c : N �−→ C; j±4|cj |2 < ∞}. j=1 Show that both h±2 are Hilbert spaces and that any linear functional sat-isfying T : h2 −→ C, |T c| ≤ C�c�h2 for some constant C is of the form ∞T c = cidi j=1 where d : N −→ C is an element of h−2. Solution: Many of you hammered …