692 CHAPTER 24. THE BOCHNER INTEGRAL

Proof: Let ∞ > M ≥ supu∈Lp(Ω;V ) ∥u∥pLp(Ω;V )

. Let {ψn} be a mollifier with support inB(0,1/n). I need to show that A has an η net in Lp (Ω;W ) for every η > 0. Supposefor some η > 0 it fails to have an η net. Without loss of generality, let η < 1. Then by24.43, it follows that for small enough ε > 0,Aε ≡

{uXGε

: u ∈A}

fails to have an η/2

net. Indeed, pick ε small enough that for all u ∈ A ,∥∥∥uXGε

−u∥∥∥

Lp(Ω;W )< η

5 . Then if{ukXGε

}r

k=1is an η/2 net for Aε , so that ∪r

k=1B(

ukXGε, η

2

)⊇ Aε , then for w ∈ A ,

wXGε∈ B

(ukXGε

, η

2

)for some uk. Hence,

∥w−uk∥Lp(Ω;W ) ≤∥∥∥w−wXGε

∥∥∥Lp(Ω;W )

+∥∥∥wXGε

−ukXGε

∥∥∥Lp(Ω;W )

+∥∥∥ukXGε

−uk

∥∥∥Lp(Ω;W )

≤ η

5+

η

2+

η

5< η

and so {uk}rk=1 would be an η net for A which is assumed to not exist.

Pick this ε in all that follows. By compactness, Lemma 24.11.1, there exists Cη suchthat for all u ∈V,

∥u∥pW ≤

η

50(2p−1)M∥u∥p

V +Cη ∥u∥pU (24.44)

Let Aεn consist of Aεn ≡{

uXGε∗ψn : u ∈A

}. I want to show that Aεn satisfies the

conditions for Theorem 24.11.3.

Lemma 24.11.5 For each n, Aεn satisfies the conditions of Theorem 24.11.3.

Proof: First consider the equicontinuity condition of that theorem. It suffices to showthat if η > 0 then there exists δ > 0 such that if |h|< δ , then for any u ∈A and x ∈ Gε ,∥∥∥uXGε

∗ψn (x+h)−uXGε∗ψn (x)

∥∥∥U< η

Always assume |h|< dist(Gε ,Ω

C), and x∈Gε . Also assume that |h| is small enough that(∫

Rm

∣∣∣(XGε(x−y+h)−XGε

(x−y))

ψn (y)∣∣∣p′ dz

)1/p′

=

(∫Rm

∣∣∣(XGε(z+h)−XGε

(z))

ψn (x−z)∣∣∣p′ dz

)1/p′

2M(24.45)

This can be obtained because by Holder’s inequality,(∫Rm

∣∣∣(XGε(z+h)−XGε

(z))

ψn (x−z)∣∣∣p′ dz

)1/p′

≤(∫

Rm

∣∣∣XGε(z+h)−XGε

(z)∣∣∣2p′

dz) 1

2p′(∫

Rmψn (x−z)

2p′ dz) 1

2p′

which is small independent of x for |h| small enough, thanks to continuity of translation inL2p′ (Rm). Then

692 CHAPTER 24. THE BOCHNER INTEGRALProof: Let o > M > supyerp(a:v) lllcoca.vy: Let {y,,} be a mollifier with support inB(0,1/n). I need to show that has an 7) net in L? (Q;W) for every 1 > 0. Supposefor some 7 > 0 it fails to have an 7 net. Without loss of generality, let 7 < 1. Then by24.43, it follows that for small enough € > 0,.% = {u 2G; Due o@} fails to have an n /2net. Indeed, pick € small enough that for all u € &/, |WG, —u| < q. Then ifLP(Q;W)r{m2}, is an 1/2 net for %, so that Ur_,B (u 2G 1) > @&e, then for w € &,w2e€B (u 2G 4) for some u;. Hence,| —uellaromy S| —w ee yy t Mae me Poe7 1,7<tytitviaw)2 512155"LP L?(Q:W)+ Up 2G — Uxand so {uz };_, would be an 77 net for .</ which is assumed to not exist.Pick this € in all that follows. By compactness, Lemma 24.11.1, there exists Cy suchthat for all u € V,7lelliy SSH copay ag Mall + Cn lel (24.44)Let %, consist of Wp = {u 2G W, ue a. I want to show that .%, satisfies theconditions for Theorem 24.11.3.Lemma 24.11.5 For each n, &en satisfies the conditions of Theorem 24.11.3.Proof: First consider the equicontinuity condition of that theorem. It suffices to showthat if 7 > 0 then there exists 6 > 0 such that if |h| < 6, then for any u € & and x € Gg,wre VY, (w@th)-—u%G*yV, (2)|| <7Always assume |h| < dist (Ge, QC) , and « € Gz. Also assume that |h| is small enough thatp' 1/p’U, tc) =R”Uh. (2c +h) — %eq(2)) Walw—2)This can be obtained because by Holder’s inequality,(2acle@-y +h) - Xe(e—y)) Wn (y)p! 1/p! ndz <= (24.45)2Mpl y"dz< (/.. 2a (z+h) - %el2)|" az) ’ (/,. W, (@— 2)" as) iwhich is small independent of x for || small enough, thanks to continuity of translation inLP (R”). Then(/.. | (2a-( +h)— %@-(2)) y,, (x@—2)