2648 CHAPTER 77. STOCHASTIC INCLUSIONS

Thus0≥

∣∣∥∥uµ

∥∥−∥u∥∣∣p−∥u∥p− pµ |G(0)|∥∥uµ

∥∥This requires that there is some constant C such that

∥∥uµ

∥∥≤C∥u∥+C. The details follow.Let a =

∥∥uµ

∥∥ ,b = ∥u∥ . Then they are both positive and

0≥ |a−b|p−bp−αa

where α = pµ |G(0)|. Want to say a ≤Cb+C for some C. This is the conclusion of thefollowing lemma.

Lemma 77.8.7 Suppose 0 ≥ |a−b|p− bp−αa for a,b ≥ 0 and α > 0. Then there existsa constant C such that

a≤Cb+C

Proof: If b ≥ a, then there is nothing to show. Therefore, it suffices to show that thedesired inequality holds for a > b. Thus from now on, a > b.

0≥ (a−b)p−bp−αa

Suppose a > nb+n. Let x = b/a. Then for x ∈ [0,1] ,

0 ≥ (1− x)p− xp−α1

ap−1

≥ (1− x)p− xp−α1

(nb+n)p−1

≥ (1− x)p− xp−α1

(n)p−1

Now for all n large enough, the right side is a decreasing function of x which is positive atx = 0 and negative at x = 1. Thus x corresponds to the place where this function is negative.Taking a limit as n→ ∞, it follows that we must have

x≥ δ , δ ∈ (0,1)

It is where (1− x)p− xp = 0. Thus x = ba ≥ δ . Then, since a > nb+n,

1δ

b≥ a > nb+n

Now this is a contradiction when n is taken increasingly large. Hence, for large enoughn,a≤ nb+n.

It follows that∥∥uµ

∥∥≤C∥u∥+C for some C. Hence,

∥∥Gµ u∥∥ ≤ 1

µ p−1

∥∥uµ −u∥∥p−1 ≤ 1

µ p−1

(∥∥uµ

∥∥+∥u∥)p−1

≤ 2p−2

µ p−1

(∥∥uµ

∥∥p−1+∥u∥p−1

)