840 CHAPTER 25. NONLINEAR OPERATORS

On the left side of ∗, when you multiply it out, you get

|x|p−|x|p−2 (x,y)−|y|p−2 (x,y)+ |y|p

which is exactly the same thing. Therefore,(|x|p−2 x−|y|p−2 y,x−y

)=

12

(|x|p−2 + |y|p−2

|x−y|p−2

)|x−y|p (**)

+

≥0︷ ︸︸ ︷12

(|x|p−2−|y|p−2

)(|x|2−|y|2

)Suppose first that p≥ 3. Now p≥ 3 and so |x|p−2 is convex. Hence∣∣∣∣x+(−y)

2

∣∣∣∣p−2

≤ 12

(|x|p−2 + |−y|p−2

)and so (

|x|p−2 x−|y|p−2 y,x−y)≥∣∣∣∣x−y

2

∣∣∣∣p−2 1

|x−y|p−2 |x−y|p = 12p−2 |x−y|p

Next suppose p > 2. There is nothing to show if p = 2. Then for a positive integer m, youcan get m(p−2)> 1. Then(

|x|p−2 + |y|p−2)m≥ |x|m(p−2)+ |y|m(p−2) ≥ 21−m(p−2) |x−y|m(p−2)

Thus we can raise both sides of the above to 1/m and conclude

|x|p−2 + |y|p−2 ≥ 21/m−(p−2) |x−y|p−2

Then we use this in ∗∗ to obtain(|x|p−2 x−|y|p−2 y,x−y

)≥ 1

2

(|x|p−2 + |y|p−2

|x−y|p−2

)|x−y|p

≥ 12

12(p−2)−(1/m)

|x−y|p ≥ 12p−1 |x−y|p

Thus, if you have the duality map F for p ≥ 2 for real valued Lp (Ω) to Lp′ (Ω) , it isclear that F f = | f |p−2 f and so

⟨F f −Fg, f −g⟩ =∫

Ω

(| f |p−2 f −|g|p−2 g

)( f −g)dµ ≥ 1

2p−1

∫Ω

| f −g|p dµ

⟨F f −Fg, f −g⟩ ≥ 12p−1 ∥ f −g∥p

Lp(Ω)

A similar result would hold for the duality map from (Lp (Ω))n to(

Lp′ (Ω))n

.