702 CHAPTER 21. THE BOCHNER INTEGRAL

and so

R≥∫ b

a||un (t)||pE dt =

Tk

k

∑i=1||un

i ||pE

Therefore, the {uni } are all bounded. It follows that after taking subsequences k times there

exists a subsequence{

unk

}such that unk is a Cauchy sequence in Lp (a,b;W ) . You simply

get a subsequence such that unki is a Cauchy sequence in W for each i. Then denoting this

subsequence by n,

||un−um||Lp(a,b;W ) ≤ ||un−un||Lp(a,b;W )

+ ||un−um||Lp(a,b;W )+ ||um−um||Lp(a,b;W )

≤ η

4+ ||un−um||Lp(a,b;W )+

η

4< η

provided m,n are large enough, contradicting 21.10.49.You can give a different version of the above to include the case where there is, instead

of a Holder condition, a bound on u′ for u∈ S. It is stated next. We are assuming a situationin which ∫ b

au′ (t)dt = u(b)−u(a)

This happens, for example, if u′ is the weak derivative. See [117].

Corollary 21.10.9 Let E ⊆W ⊆ X where the injection map is continuous from W to X andcompact from E to W. Let p≥ 1, let q > 1, and define

S≡ {u ∈ Lp ([a,b] ;E) : for some C, ∥u(t)−u(s)∥X ≤C |t− s|1/q

and ||u||Lp([a,b];E) ≤ R}.

Thus S is bounded in Lp ([a,b] ;E) and Holder continuous into X. Then S is precompact inLp ([a,b] ;W ). This means that if {un}∞

n=1 ⊆ S, it has a subsequence{

unk

}which converges

in Lp ([a,b] ;W ) . The same conclusion can be drawn if it is known instead of the Holdercondition that ∥u′∥L1([a,b];X) is bounded.

Proof: The first part was done earlier. Therefore, we just prove the new stuff whichinvolves a bound on the L1 norm of the derivative. It suffices to show S has an η net inLp ([a,b] ;W ) for each η > 0.

If not, there exists η > 0 and a sequence {un} ⊆ S, such that

||un−um|| ≥ η (21.10.51)

for all n ̸= m and the norm refers to Lp ([a,b] ;W ). Let

a = t0 < t1 < · · ·< tk = b, ti− ti−1 = (b−a)/k.

Now define

un (t)≡k

∑i=1

uniX[ti−1,ti) (t) , uni ≡1

ti− ti−1

∫ ti

ti−1

un (s)ds.