65.14. A TECHNICAL INTEGRATION BY PARTS RESULT 2269
in the above formula 65.14.25 is dominated by
Cε
E
((∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
)1/2
∧δ
)
+P
([(∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
)1/2
> δ
])
≤ Cδ
ε+P
([(∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
)1/2
> δ
])(65.14.27)
Let η > 0 be given. Then let δ be small enough that the first term is less than η . Fix sucha δ .
Consider the second of the above terms.
P
([(∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
)1/2
> δ
])
≤ 1δ
(E(∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
))1/2
and this converges to 0 because ( I−Pn)X (s) is assumed to be bounded and converges to0. Next consider 65.14.26. By similar reasoning, we end up with having to estimate
1δ
(E(∫ T
0||Z (s)||2
∣∣∣( I−Pn)X lk (s)
∣∣∣2 ds))1/2
.
But this is dominated by
2δ
(E(∫ T
0||Z (s)||2
∣∣∣X lk (s)−X (s)
∣∣∣2 ds))1/2
+2δ
(E(∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
))1/2
The first term is no larger than η provided k is large enough, independent of n thanks to thepointwise convergence and the assumption that X is bounded. Thus, there exists K suchthat if k > K, then the term in 65.14.26 is dominated by
2η +2δ
(E(∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
))1/2
It follows that for k > K, the sum of 65.14.25 and 65.14.26 is dominated by
3η +3δ
(E(∫ T
0||Z (s)||2 |( I−Pn)X (s)|2 ds
))1/2