25.2. DUALITY MAPS 839

if y = αx for α ̸= 1.Thus the desired result holds in the case that one vector is a multiple of the other. The

other case is that neither vector is a multiple of the other. Thus, in particular, x/∥x∥ ̸=y/∥y∥ , and in this case, it follows from 25.2.10

⟨Fx−Fy,x− y⟩= ∥x∥p +∥y∥p−⟨Fx,y⟩−⟨Fy,x⟩

> ∥x∥p +∥y∥p−∥x∥p−1 ∥y∥−∥y∥p−1 ∥x∥

≥ ∥x∥p +∥y∥p−(∥x∥p

p′+∥y∥p

p

)−(∥y∥p

p′+∥x∥p

p

)= 0

Thus F is strictly monotone as claimed.Another useful observation about duality maps for p = 2 is that

∥∥F−1y∗∥∥

V = ∥y∗∥V ′ .This is because

∥y∗∥V ′ =∥∥FF−1y∗

∥∥V ′ =

∥∥F−1y∗∥∥

V

also from similar reasoning,⟨y∗,F−1y∗

⟩=⟨FF−1y∗,F−1y∗

⟩=∥∥F−1y∗

∥∥2V = ∥y∗∥2

V ′

You can give specific inequalities in certain cases. Here is a nice little inequality whichwill allow this.

Theorem 25.2.8 Let p≥ 2 then for x,y ∈ Rn,(|x|p−2 x−|y|p−2 y,x−y

)≥ 1

2p−1 |x−y|p (*)

Proof: We have (x,y) = 12

(|x|2 + |y|2−|x−y|2

). Consider the following.

12

(|x|p−2 + |y|p−2

|x−y|p−2

)|x−y|p + 1

2

(|x|p−2−|y|p−2

)(|x|2−|y|2

)multiplying this out gives

12

(|x|p−2 + |y|p−2

)(|x|2 + |y|2−2(x,y)

)+

12

(|x|p−|x|2 |y|p−2 + |y|p−|x|p−2 |y|2

)thus this yields

12

[|x|p + |y|p−2 |x|2 + |x|p−2 |y|2 + |y|p−

(2(x,y) |x|p−2 +2(x,y) |y|p−2

)]+

12

(|x|p + |y|p−

(|x|2 |y|p−2 + |x|p−2 |y|2

))It simplifies to

|x|p + |y|p−2(x,y)(|x|p−2 + |y|p−2

)