Functional Analysis HW #5 Sangchul Lee October 29, 2015 Contents 1 Solutions........................................ 1 1 Solutions Exercise 3.4. Show that C([0, 1]) is not a Hilbert space, that is, there is no inner product on C([0, 1]) for which the associated norm is the supremum norm. Proof. In view of [Dou98, Proposition 3.11], it suffices to check that the supremum norm invalidates the parallelogram law. To this end, let us choose a very specific function f (x) = cos 2 (2πx) and g(x) = sin 2 (2πx). Then we have f (x) + g(x) = 1 and f (x) g(x) = cos(4πx), hence f + g = f g = 1. On the other hand, we have f = g = 1. Consequently f + g + f g = 2 4 = 2 f + 2 g. Therefore the supremum cannot be induced by an inner product. Exercise 3.7. Give an example of a finite dimensional Banach space containing a closed convex set which contains more than one point of smallest norm.* Proof. For n 2, consider X = C n equipped with the supremum norm (x 1,, x n ) = max{ x 1,, x n }. 1
Now let K be the subset of X defined by K = {(x 1,, x n ) : x 1 = 1}. It is obvious that K is a non-empty, closed and convex subset of X. Moreover, For any x K we have x x 1 = 1, For any x K with max{ x 2,, x n } 1 we have x = 1. Therefore K contains infinitely many norm-minimizers. Exercise 3.9. Let ϕ be a bounded linear functional on the subspace M of the Hilbert space H. Show that there exists a unique extension of ϕ to H having the same norm. Proof. (Existence) Using the uniform continuity of ϕ, we can uniquely extend ϕ to all of clos(m). Let us denote this extension by ϕ. Then use [Dou98, Theorem 3.21] to write H = clos(m) M, and define Φ : H C by Φ(g + h) = ϕ(g) g clos(m) and h M. Then obviously Φ is a well-defined linear functional that extends ϕ and satisfies Φ( f ) Φ = sup f 0 f Φ(g + h) = sup g clos(m) g + h h M = sup g clos(m) h M ϕ(g) g 2 + h 2 = sup ϕ(g) g clos(m) g = ϕ. (Uniqueness) Let Φ be any bounded linear functional on H that extends ϕ and Φ = ϕ. By continuity again, we find that Φ(g) = ϕ(g) for all g clos(m). In order to show the uniqueness, we claim that Φ(h) = 0 for any h M. Indeed, assume otherwise and pick h M such that Φ(h) 0. By multiplying some units if necessary, we may assume that Φ(h) > 0. Now choose {g n } n=1 M such that g n = 1 and ϕ(g n ) ϕ. As before, we may assume that ϕ(g n ) 0. Then for any λ > 0, ϕ Φ(g n + λh) g n + λh = ϕ(g n) + λφ(h) 1 + λ 2 h 2. Taking n, we have ϕ = ϕ + λφ(h) 1 + λ 2 h 2 = ϕ + λφ(h) + O(λ2 ). Thus by choosing λ sufficiently small we can make the right-hand side bigger than ϕ, a contradiction! Therefore Φ is uniquely determined by the construction as above. 2
Exercise 3.10. Let H and K be Hilbert spaces and H K denote the algebraic direct sum. Show that ( h 1, k 1, h 2, k 2 ) = (h 1, h 2 ) + (k 1, k 2 ) defines an inner product on H K, that H K is complete with respect to this inner product, and that the subspaces H {0} and {0} K are closed and orthogonal in H K. Proof. The proof is quite straightforward. (1) To show that H K is an inner product space, we check [Dou98, Definition 3.2] as follows. For α 1, α 2 C and h 1, k 1, h 2, k 2, f, g H K, (α 1 h 1, k 1 + α 2 h 2, k 2, f, g ) = ( α 1 h 1 + α 2 h 2, α 1 k 1 + α 2 k 2, f, g ) For h 1, k 1, h 2, k 2 H K, = (α 1 h 1 + α 2 h 2, f ) + (α 1 k 1 + α 2 k 2, g) = α 1 (h 1, f ) + α 2 (h 2, f ) + α 1 (k 1, g) + α 2 (k 2, g) = α 1 ( h 1, k 1, f, g ) + α 2 ( h 2, k 2, f, g ). ( h 1, k 1, h 2, k 2 ) = (h 1, h 2 ) + (k 1, k 2 ) = (h 2, h 1 ) + (k 2, k 1 ) = ( h 2, k 2, h 1, k 1 ). Complex conjugate linearity with respect to the second argument follows from previous two observations. For h, k H K, ( h, k, h, k )(h, h) + (k, k) 0 with the equality if and only if h = 0 and k = 0, which is equivalent to h, k = 0. (2) Next we show that H K is complete. If ( h n, k n ) is Cauchy, then h n h m 2 + k n k m 2 = (h n h m, h n h m ) + (k n k m, k n k m ) = h n, k n h m, k m 2 shows that (h n ) and (k n ) are also Cauchy. Thus they are convergent with limits h H and k K, respectively. Then h n, k n h, k 2 = h n h 2 2 as n + k n k 0 shows that ( h n, k n ) is convergent. Therefore H K is complete. (3) Finally we show that H {0} and {0} K are closed and orthogonal. To see this, recall the previous calculation in (2). If k n = 0 for all n, then k n 0 and hence ( h n, 0 ) (h, 0). Similar argument shows that ( 0, k n ) (0, k). This shows that both H {0} and {0} K are closed. 3
Moreover, for h, 0 H {0} and 0, k {0} K, we have ( h, 0, 0, k ) (h, 0) + (0, k) = 0. Therefore H {0} and {0} K are orthogonal. Exercise 3.12. Show that the w -closure of the unit sphere in an infinite dimensional Hilbert space is the entire unit ball. Proof. Let H be an infinite-dimensional Hilbert space. We want to show that any f H with f 1 is a w -limit of elements in the unit sphere in H. To this end, notice that the orthogonal complement M := (C f ) is a closed subspace of H which is also infinite-dimensional (since the codimension is finite). With the inner product inherited from H, M is also a Hilbert space and we can choose a countable orthonormal subset {e n } n=1 of M. (For example, choose any orthonormal basis of M and pick a countable subset of it. Or we can utilize the Gram-Schmidt process, which is proved in the next problem.) Clearly {e n } n=1 lies in the unit sphere in H. We claim that {e n } n=1 converges to 0 in w -topology. Indeed, for any g H we have (g, e k ) 2 = 2 (g, e k )e k 2 (g, e k )e k + g 2 (g, e k )e k = g 2 since n (g, e k )e k and g n (g, e k )e k are orthogonal. This shows that (g, e n ) 0 as n and hence e n 0 in w -topology. We summarize what we have shown so far: for any f H with f 1, we have constructed a sequence {e n } n=1 of unit vectors such that f, e 1, e 2, are orthogonal. Then we define a sequence ( f n ) n=1 by f n = f + (1 f 2 ) 1/2 e n. Since f and e n are orthogonal, we have f n = 1. Moreover it is clear that f n converges to f in w -topology. This completes the proof. Exercise 3.15 (Gram-Schmidt). Let { f n } n=1 be a subset of the Hilbert space H whose closed linear span is H. Set e 1 = f 1 / f 1 and assuming {e k } n to have been defined, set ( ) / e n+1 = f n+1 ( f n+1, e k )e k f n+1 ( f n+1, e k )e k, (1.1) 4
where e n+1 is taken to be zero if f n+1 = ( f n+1, e k )e k. (1.2) Show that {e n } n=1 is an orthonormal basis for H. Proof. We inductively prove the following claim. Claim 1.1. Define K n = span{ f 1,, f n } for n 1. Then (a) e 1,, e n are orthogonal, (b) span{e 1,, e n } = K n. The case n = 1 is obvious. Next, given n 1, let us assume that (a) and (b) are true for n. Assume first that (1.2) holds. Then e n+1 = 0 and thus part (a) is trivial. Also, (1.2) implies that f n+1 K n. So we have both span{ f 1,, f n+1 } = span{ f 1,, f n } and span{e 1,, e n+1 } = span{e 1,, e n }, which proves part (b). Now assume that (1.2) does not hold. Let g = n ( f n+1, e k )e k ck n. We claim that f n+1 g is orthogonal to K n. Indeed, for any n α k e k K n we have and hence ( f n+1 g, e k ) = ( f n+1, e k ) ( f n+1 g, = ( f n+1, e k ) ) α k e k = ( f n+1, e j )(e j, e k ) j=1 ( f n+1, e j )δ jk = 0 j=1 ᾱ k ( f n+1 g, e k ) = 0. This shows that f n+1 g is orthogonal to K n and hence part (a) follows. Moreover, for any constants α 1,, α n+1 C, n+1 α k f k = ( ) ) α n+1 f n+1 g e n+1 + (α n+1 g + α k f k span{e 1,, e n+1 } 5
and hence K n+1 span{e 1,, e n+1 }. Since span{e 1,, e n+1 } K n+1 is obvious from the definition (1.1), we have equality and hence part (b) is proved. This completes the proof of the induction step and hence the claim follows. From this claim, we know that {e 1, e 2, } is an orthonormal subset of H and span{ f 1, f 2, } = span{ f 1,, f n } = span{e 1,, e n } = span{e 1, e 2, }. n=1 n=1 Therefore, upon deleting the zero terms from {e 1, e 2, }, it is an orthonormal basis of H. References [Dou98] R.G. Douglas. Banach Algebra Techniques in Operator Theory. Banach Algebra Techniques in Operator Theory Series. Springer New York, 1998. 6