25.2. DUALITY MAPS 833

Duality maps exist. Here is why. Let

F (x)≡{

x∗ : ||x∗|| ≤ ||x||p−1 and x∗ (x) = ||x||p}

Then F (x) is not empty because you can let f (αx) = α ||x||p . Then f is linear and definedon a subspace of X . Also

sup||αx||≤1

| f (αx)|= sup||αx||≤1

|α| ||x||p ≤ ||x||p−1

Also from the definition,f (x) = ||x||p

and so, letting x∗ be a Hahn Banach extension, it follows x∗ ∈ F (x). Also, F (x) is closedand convex. It is clearly closed because if x∗n→ x∗, the condition on the norm clearly holdsand also the other one does too. It is convex because

||x∗λ +(1−λ )y∗|| ≤ λ ||x∗||+(1−λ ) ||y∗|| ≤ λ ||x||p−1 +(1−λ ) ||x||p−1

If the conditions hold for x∗, then we can show that in fact ||x∗|| = ||x||p−1. This isbecause

||x∗|| ≥∣∣∣∣x∗( x

||x||

)∣∣∣∣= 1||x|||x∗ (x)|= ||x||p−1 .

Now how many things are in F (x) assuming the norm on X ′ is strictly convex? Supposex∗1, and x∗2 are two things in F (x) . Then by convexity, so is (x∗1 + x∗2)/2. Hence by strictconvexity, if the two are different, then∣∣∣∣∣∣∣∣x∗1 + x∗2

2

∣∣∣∣∣∣∣∣= ||x||p−1 <12||x∗1||+

12||x∗2||= ||x||

p−1

which is a contradiction. Therefore, F is an actual mapping.What are some of its properties? First is one which is similar to the Cauchy Schwarz

inequality. Since p−1 = p/p′,

sup||y||≤1

|⟨Fx,y⟩|= ||x||p/p′

and so for arbitrary y ̸= 0,

|⟨Fx,y⟩| = ||y||∣∣∣∣⟨Fx,

y||y||

⟩∣∣∣∣≤ ||y|| ||x||p/p′

= |⟨Fy,y⟩|1/p |⟨Fx,x⟩|1/p′

Next we can show that F is monotone.

⟨Fx−Fy,x− y⟩ = ⟨Fx,x⟩−⟨Fx,y⟩−⟨Fy,x⟩+ ⟨Fy,y⟩≥ ||x||p + ||y||p−||y|| ||x||p/p′ −||y||p/p′ ||x||