476 CHAPTER 17. BANACH SPACES

13. ↑ The Hahn Banach theorem can be used to establish separation theorems. Let Mbe an open convex set containing 0. Let x /∈ M. Show there exists x∗ ∈ X ′ suchthat Rex∗(x) ≥ 1 > Rex∗(y) for all y ∈M. Hint: If y ∈M,ρ(y) < 1. Show this. Ifx /∈M, ρ(x)≥ 1. Try f (αx) = αρ(x) for α ∈ R. Then extend f to the whole spaceusing the Hahn Banach theorem and call the result F , show F is continuous, then fixit so F is the real part of x∗ ∈ X ′.

14. A Banach space is said to be strictly convex if whenever ||x||= ||y|| and x ̸= y, then∣∣∣∣∣∣∣∣x+ y2

∣∣∣∣∣∣∣∣< ||x||.F : X → X ′ is said to be a duality map if it satisfies the following: a.) ||F(x)|| =||x||. b.) F(x)(x) = ||x||2. Show that if X ′ is strictly convex, then such a duality mapexists. The duality map is an attempt to duplicate some of the features of the Rieszmap in Hilbert space which is discussed in the chapter on Hilbert space. Hint: Foran arbitrary Banach space, let

F (x)≡{

x∗ : ||x∗|| ≤ ||x|| and x∗ (x) = ||x||2}

Show F (x) ̸= /0 by using the Hahn Banach theorem on f (αx) = α||x||2. Next showF (x) is closed and convex. Finally show that you can replace the inequality in thedefinition of F (x) with an equal sign. Now use strict convexity to show there is onlyone element in F (x).

15. Prove the following theorem which is an improved version of the open mapping the-orem, [42]. Let X and Y be Banach spaces and let A ∈L (X ,Y ). Then the followingare equivalent.

AX = Y,

A is an open map.

There exists a constant M such that for every y ∈ Y , there exists x ∈ X with y = Axand

||x|| ≤M ||y||.

Note this gives the equivalence between A being onto and A being an open map. Theopen mapping theorem says that if A is onto then it is open.

16. Suppose D⊆ X and D is dense in X . Suppose L : D→ Y is linear and ||Lx|| ≤ K||x||for all x ∈ D. Show there is a unique extension of L, L̃, defined on all of X with||L̃x|| ≤ K||x|| and L̃ is linear. You do not get uniqueness when you use the HahnBanach theorem. Therefore, in the situation of this problem, it is better to use thisresult.

17. ↑ A Banach space is uniformly convex if whenever ||xn||, ||yn|| ≤ 1 and ||xn+yn|| →2, it follows that ||xn− yn|| → 0. Show uniform convexity implies strict convexity(See Problem 14). Hint: Suppose it is not strictly convex. Then there exist ||x||