24.11. SOME EMBEDDING THEOREMS 693

∥∥∥uXGε∗ψn (x+h)−uXGε

∗ψn (x)∥∥∥

U

=

∥∥∥∥∫Rm

(ũ(x+h−y)XGε

(x+h−y)− ũ(x−y)XGε(x−y)

)ψn (y)dy

∥∥∥∥U

≤∫Rm

∥∥∥(ũ(x+h−y)XGε(x+h−y)− ũ(x−y)XGε

(x−y))∥∥∥

Uψn (y)dy

Changing the variables,

≤∫Rm

∥∥∥∥∥ (ũ(z+h)− ũ(z))XGε(z+h)

+ũ(z)(XGε

(z+h)−XGε(z)) ∥∥∥∥∥

U

ψn (x−z)dz

≤∫Rm

∥∥∥(ũ(z+h)− ũ(z))XGε(z+h)

∥∥∥U

ψn (x−z)dz

+∫Rm∥ũ(z)∥U

∣∣∣XGε(z+h)−XGε

(z)∣∣∣ψn (x−z)dz (24.46)

The first integral

≤(∫

Rm∥ũ(z+h)− ũ(z)∥p

U

)1/p(∫Rm

ψp′n (x−z)dz

)1/p′

You make the obvious change here in case p = 1. Instead of the above, you would have

≤∫Rm∥ũ(z+h)− ũ(z)∥U dz2∥ψn∥∞

Since Lebesgue measure is translation independent, there is a constant Cn such that theabove is ≤ Cn

(∫Rm ∥ũ(z+h)− ũ(z)∥p

U

)1/p< η/2 and this holds for all u ∈ A . As for

the second integral in 24.46, from 24.45, it follows that this term is no larger than

≤(∫

Rm∥ũ(z)∥p

U dz)1/p(∫

Rm

(∣∣∣XGε(z+h)−XGε

(z)∣∣∣ψn (x−z)

)p′

dz)1/p′

and by 24.45, < M η

2M = η

2 . Thus, if δ < dist(Gε ,Ω

C)

and 24.45 holds, then for all u ∈A ,when |h|< δ , ∥∥∥uXGε

∗ψn (x+h)−uXGε∗ψn (x)

∥∥∥U< η

and so the desired equicontinuity condition holds for Aεn. Note that δ does depend on nbut for each n, things work out well.

I also need to verify that the functions in Aεn are uniformly bounded. For x ∈ Gε andu ∈A , ∥∥∥uXGε

∗ψn (x)∥∥∥

V≤∫

Gε

∥u(z)∥ψn (x−z)dz

≤(∫

Ω

∥u(z)∥p dz)1/p(∫

Ω

ψn (x−z)p′)1/p′

≤MCn ■