Identification du dual topologique de C[a, b] Louis ... - CQFD - EPFL
Identification du dual topologique de C[a, b] Louis ... - CQFD - EPFL
Identification du dual topologique de C[a, b] Louis ... - CQFD - EPFL
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4. IDENTIFICATION DE C[a, b] ′ 21<br />
et tel que le pas <strong>de</strong> la subdivision {y o , y 1 , . . . , y n−1 , y n } soit < δ, où on a<br />
posé y 0 := a et y n := b<br />
Ainsi,<br />
∫ b<br />
∫ b<br />
∣ f(x)dp(x) − f(x)dp ∗ (x)<br />
∣ =<br />
a<br />
a<br />
∫ b<br />
n−1<br />
∑<br />
n−1<br />
∑<br />
∫ b<br />
f(x)dp(x) + f(ξ<br />
∣<br />
k )(p(y k+1 ) − p(y k )) − f(ξ k )(p(y k+1 ) − p(y k )) − f(x)dp ∗ (x)<br />
a<br />
k=0<br />
k=0<br />
a<br />
∣ ≤<br />
∫ ∣<br />
b<br />
n−1<br />
∑<br />
∣∣∣∣ ∫ b<br />
n−1<br />
f(x)dp(x) − f(ξ<br />
∣<br />
k )(p(y k+1 ) − p(y k ))<br />
∣ + ∑<br />
f(x)dp ∗ (x) − f(ξ k )(p(y k+1 ) − p(y k ))<br />
∣<br />
a<br />
k=0<br />
Comme le pas <strong>de</strong> {y 0 , . . . , y n } est < δ, on sait que<br />
∫ b<br />
n−1<br />
∑<br />
f(x)dp(x) − f(ξ<br />
∣<br />
k ) ( p(y k+1 ) − p(y k ) )∣ ∣ ∣∣∣<br />
< ɛ 3<br />
De plus,<br />
Or<br />
a<br />
∣ ∣∣∣∣ ∫ b<br />
a<br />
k=0<br />
n−1<br />
∑<br />
f(x)dp ∗ (x) − f(ξ k ) ( p(y k+1 ) − p(y k ) )∣ ∣ ∣∣∣<br />
≤<br />
k=0<br />
n−1<br />
a<br />
k=0<br />
∫ b<br />
∑<br />
f(x)dp ∗ (x) − f(ξ<br />
∣<br />
k ) ( p ∗ (x k+1 ) − p ∗ (x k ) )∣ ∣ ∣∣∣<br />
+<br />
a<br />
k=0<br />
n−1<br />
∑<br />
f(ξ<br />
∣ k ) ( p ∗ (x k+1 ) − p ∗ (x k ) ) n−1<br />
∑<br />
− f(ξ k ) ( p(y k+1 ) − p(y k ) )∣ ∣ ∣∣∣<br />
≤<br />
k=0<br />
∣ ∣∣∣∣ ∫ b<br />
a<br />
k=0<br />
n−1<br />
∑<br />
f(x)dp ∗ (x) − f(ξ k ) ( p ∗ (x k+1 ) − p ∗ (x k ) )∣ ∣ ∣∣∣<br />
< ɛ 3<br />
k=0<br />
Par conséquent,<br />
n−1<br />
∑<br />
f(ξ<br />
∣ k ) ( p ∗ (x k+1 ) − p ∗ (x k ) ) n−1<br />
∑<br />
− f(ξ k ) ( p(y k+1 ) − p(y k ) )∣ ∣ ∣∣∣<br />
=<br />
k=0<br />
k=0<br />
n−1<br />
∑<br />
f(ξ<br />
∣ k ) ( p ∗ (x k+1 ) − p ∗ (x k ) − p(y k+1 ) + p(y k ) )∣ ∣ ∣∣∣<br />
=<br />
k=0<br />
n−1<br />
∑<br />
f(ξ<br />
∣ k ) ( (p ∗ (x k+1 ) − p(y k+1 )) − (p(y k ) − p ∗ (x k )) )∣ ∣ ∣∣∣<br />
≤<br />
k=0<br />
n−1<br />
∑<br />
|f(ξ k )| { |p ∗ (x k+1 ) − p(y k+1 )| + |p(y k ) − p ∗ (x k )| } ≤<br />
k=0<br />
n−1<br />
∑ {<br />
ɛ<br />
|f(ξ k )|<br />
6n · (1 + max{f(ξ k ) : 1 ≤ k ≤ n}) +<br />
ɛ<br />
}<br />
≤ ɛ 6n · (1 + max{f(ξ k ) : 1 ≤ k ≤ n}) 3<br />
k=0<br />
Par conséquent<br />
∫ b<br />
∣ f(x) dp(x) −<br />
a<br />
∫ b<br />
a<br />
f(x) dp ∗ (x)<br />
∣ < ɛ 3 + ɛ 3 + ɛ 3 = ɛ