17.7. EXERCISES 477

and ||y|| both equal to 1 and∣∣∣∣ xn+yn

2

∣∣∣∣ = 1 consider xn ≡ x and yn ≡ y, and use theconditions for uniform convexity to get a contradiction. It can be shown that Lp isuniformly convex whenever ∞> p> 1. See Hewitt and Stromberg [64] or Ray [109].

18. Show that a closed subspace of a reflexive Banach space is reflexive. Hint: Theproof of this is an exercise in the use of the Hahn Banach theorem. Let Y be theclosed subspace of the reflexive space X and let y∗∗ ∈ Y ′′. Then i∗∗y∗∗ ∈ X ′′ and soi∗∗y∗∗ = Jx for some x ∈ X because X is reflexive. Now argue that x ∈ Y as follows.If x /∈ Y , then there exists x∗ such that x∗ (Y ) = 0 but x∗ (x) ̸= 0. Thus, i∗x∗ = 0. Usethis to get a contradiction. When you know that x = y∈Y , the Hahn Banach theoremimplies i∗ is onto Y ′ and for all x∗ ∈ X ′,

y∗∗ (i∗x∗) = i∗∗y∗∗ (x∗) = Jx(x∗) = x∗ (x) = x∗ (iy) = i∗x∗ (y).

19. We say that xn converges weakly to x if for every x∗ ∈ X ′, x∗(xn)→ x∗(x). xn ⇀ xdenotes weak convergence. Show that if ||xn− x|| → 0, then xn ⇀ x.

20. ↑ Show that if X is uniformly convex, then if xn ⇀ x and ||xn|| → ||x||, it follows||xn− x|| → 0. Hint: Use Lemma 17.2.9 to obtain f ∈ X ′ with || f ||= 1 and f (x) =||x||. See Problem 17 for the definition of uniform convexity. Now by the weakconvergence, you can argue that if x ̸= 0, f (xn/ ||xn||)→ f (x/ ||x||). You also mighttry to show this in the special case where ||xn||= ||x||= 1.

21. Suppose L ∈L (X ,Y ) and M ∈L (Y,Z). Show ML ∈L (X ,Z) and that (ML)∗ =L∗M∗.

22. Let X and Y be Banach spaces and suppose f ∈ L (X ,Y ) is compact. Recall thismeans that if B is a bounded set in X , then f (B) has compact closure in Y. Show thatf ∗ is also a compact map. Hint: Take a bounded subset of Y ′,S. You need to showf ∗ (S) is totally bounded. You might consider using the Ascoli Arzela theorem onthe functions of S applied to f (B) where B is the closed unit ball in X .