75.2. DUALITY MAPS 2535
Hence
⟨Fu−Fv,u− v⟩ = ∥u∥2 +∥v∥2−⟨Fu,v⟩−⟨Fv,u⟩≥ ∥u∥2 +∥v∥2−2∥u∥∥v∥ ≥ 0
Now suppose ∥x∥= ∥y∥= 1 but x ̸= y. Then⟨Fx,
x+ y2
⟩≤∥∥∥∥x+ y
2
∥∥∥∥< ∥x∥+∥y∥2= 1
It follows that12⟨Fx,x⟩+ 1
2⟨Fx,y⟩= 1
2+
12⟨Fx,y⟩< 1
and so⟨Fx,y⟩< 1
For arbitrary x,y, x/∥x∥ ̸= y/∥y∥
⟨Fx,y⟩= ∥x∥∥y∥⟨
F(
x∥x∥
),
(y∥y∥
)⟩It is easy to check that F (αx) = αF (x) . Therefore,
|⟨Fx,y⟩|= ∥x∥∥y∥⟨
F(
x∥x∥
),
(y∥y∥
)⟩< ∥x∥∥y∥
Now say that x ̸= y and consider
⟨Fx−Fy,x− y⟩
First suppose x = αy. Then the above is
⟨F (αy)−Fy,(α−1)y⟩ = (α−1)(⟨F (αy) ,y⟩−∥y∥2
)= (α−1)
(⟨αF (y) ,y⟩−∥y∥2
)= (α−1)2 ∥y∥2 > 0
The other case is that x/∥x∥ ̸= y/∥y∥ and in this case,
⟨Fx−Fy,x− y⟩= ∥x∥2 +∥y∥2−⟨Fx,y⟩−⟨Fy,x⟩
> ∥x∥2 +∥y∥2−2∥x∥∥y∥ ≥ 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 ′