62.12. THE SPACE M pT (E) 2111
and the expression on the right is measurable because D is countable.Next it is necessary to show this is a norm. It is clear that
||M||M pT (E) ≥ 0
and equals 0 only if
0 = E
(sup
t∈[0,T ]||M (t)||p
)which requires M (t) = 0 for all t for ω off a set of measure zero so that M = 0. It is alsoclear that
||αM||M pT (E) = |α| ||M||M p
T (E) .
It remains to check the triangle inequality. Let M,N ∈M pT (E) .
||M+N||M pT (E) ≡ E
(sup
t∈[0,T ]||M (t)+N (t)||p
)1/p
≤ E
(sup
t∈[0,T ](||M (t)||+ ||N (t)||)p
)1/p
≤ E
((sup
t∈[0,T ]||M (t)||+ sup
t∈[0,T ]||N (t)||
)p)1/p
≡
(∫Ω
(sup
t∈[0,T ]||M (t)||+ sup
t∈[0,T ]||N (t)||
)p
dP
)1/p
≤
(∫Ω
(sup
t∈[0,T ]||M (t)||
)p
dP
)1/p
+
(∫Ω
(sup
t∈[0,T ]||N (t)||
)p
dP
)1/p
≡ ||M||M pT (E)+ ||N||M p
T (E)
Next consider the claim that M pT (E) is a Banach space. Let {Mn} be a Cauchy se-
quence. Then
E
(sup
t∈[0,T ]||Mn (t)−Mm (t)||p
)→ 0 (62.12.49)
as m,n→ ∞. From continuity,
supt∈[0,T ]
||Mn (t)−Mm (t)||= supt∈(0,T )
||Mn (t)−Mm (t)||
Then from theorem 62.5.3 or 62.9.4,
P
(sup
t∈[0,T ]||Mn (t)−Mm (t)||> λ
)≤ 1
λp E (||Mn (T )−Mm (T )||p)