62.2. KOLMOGOROV ČENTSOV CONTINUITY THEOREM 2057

≤ maxk≤2n+1

(12

∥∥X(tn+12k−1

)−X

(tn+12k

)∥∥+ 12

∥∥X(tn+12k−1

)−X

(tn+12k−2

)∥∥)≤Mn+1

Denote by ∥·∥∞

the usual norm in C ([0,T ] ,E) ,

maxt∈[0,T ]

∥Z (t)∥ ≡ ∥Z∥∞,

Then from what was just established,

E(∥Xn+1−Xn∥α

∞

)=∫

Ω

∥Xn+1−Xn∥α

∞dP≤ E

(Mα

n+1)=C2−nβ

which shows that

∥Xn+1−Xn∥Lα (Ω;C([0,T ],E)) =

(∫Ω

∥Xn+1−Xn∥α

∞dP)1/α

≤C(

2(β/α))−n

Also, for m > n, it follows from the assumption that α ≥ 1,

∥Xm−Xn∥Lα (Ω;C([0,T ],E)) ≤

∞

∑k=n

C(

2(β/α))−k≤C

(2(β/α)

)−n

1−2(−β/α)=C

(2(β/α)

)−n(62.2.12)

Thus {Xn} is a Cauchy sequence in Lα (Ω;C ([0,T ] ,E)) and so it converges to some Y inthis space, a subsequence converging pointwise. Then from Fatou’s lemma,

∥Y −Xn∥Lα (Ω;C([0,T ],E)) ≤C(

2(β/α))−n

. (62.2.13)

Also, for a.e. ω, t→ Y (t) is in C ([0,T ] ,E) . It remains to verify that Y (t) = X (t) a.e.From the construction, it follows that for any n and m≥ n

Y (tnk ) = Xm (tn

k ) = X (tnk )

Thus

∥Y (t)−X (t)∥ ≤ ∥Y (t)−Y (tnk )∥+∥Y (tn

k )−X (t)∥= ∥Y (t)−Y (tn

k )∥+∥X (tnk )−X (t)∥

Now from the hypotheses of the theorem,

P(∥X (tn

k )−X (t)∥α > ε)≤ 1

εE(∥X (tn

k )−X (t)∥α)≤ C

ε|tn

k − t|1+β

Thus, there exists a sequence of mesh points {sn} converging to t such that

P(∥X (sn)−X (t)∥α > 2−n)≤ 2−n