1424 CHAPTER 43. INTERPOLATION IN BANACH SPACE

The above is dominated by∫R

2p−1 (||Y (s)||p + ||Y (s)||p)X[−2T,2T ] (s)ds

=∫ 2T

−2T2p−1 (||Y (s)||p + ||Y (s)||p)ds < ∞

Therefore,

limn→∞

∫ 2T

−2T

(∫R||Y (γn (t)+ s)−Y (t + s)||pE ds

)dt = 0

by the dominated convergence theorem. Now Fubini. The above equals∫R

∫ 2T

−2T||Y (γn (t)+ s)−Y (t + s)||pE dtds

Change the variables on the inside.∫R

∫ 2T+s

−2T+s||Y (γn (t− s)+ s)−Y (t)||pE dtds

Since γn (t− s)+ s is within 2−n of t and Y (t) vanishes if t /∈ [0,T ] , this reduces to∫R

∫ 2T

−2T||Y (γn (t− s)+ s)−Y (t)||pE dtds.

This converges to 0 as n→ ∞ as was shown above. Therefore,∫ T

0

∫ T

0||Y (γn (t− s)+ s)−Y (t)||pE dtds

also converges to 0 as n→ ∞. The only problem is that γn (t− s) + s ≥ t − 2−n and soγn (t− s)+ s could be less than 0 for t ∈ [0,2−n]. Since this is an interval whose measureconverges to 0 it follows∫ T

0

∫ T

0

∣∣∣∣Y ((γn (t− s)+ s)+)−Y (t)

∣∣∣∣pE dtds

converges to 0 as n→ ∞. Let

mn (s) =∫ T

0

∣∣∣∣Y ((γn (t− s)+ s)+)−Y (t)

∣∣∣∣pE dt

Then

P([mn (s)> λ ])≤ 1λ

∫ T

0mn (s)ds.

It follows there exists a subsequence nk such that

P([

mnk (s)>1k

])< 2−k