Riemann Lebesgue Lemma
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Transcript of Riemann Lebesgue Lemma
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Riemann-Lebesgue Lemma
December 20, 2006
The Riemann-Lebesgue lemma is quite general, but since we only know Riemann integration, Ill stateit in that form.
Theorem 1. Let f be Riemann integrable on [a, b]. Then
lim
baf(t) cos(t)dt = 0 (1)
lim
baf(t) sin(t)dt = 0 (2)
lim
baf(t)eitdt = 0 (3)
Proof. I will prove only the first statement. Since f is integrable, given > 0, there is a partition
{a = x0, x1, . . . , xn = b}, so that /2 > baf
n1
mixi 0, where mi is the minimum of f on [xi1, xi].
But the sum can be written asn
1 mixi = ba g, where g =
mi[xi1,xi], and the inequality takes the
form
/2 > ba(f g) 0.
Now we use the fact that f g 0 to get baf(t) cos(t)dt
ba(f(t) g(t)) cos(t)dt
+ bag(t) cos(t)dt
(4) ba(f g) +
(1/)mi (sin(xi) sin(xi1)) . (5)The function g has been fixed. Take large enough that(1/)mi (sin(xi) sin(xi1)) < /2,and we are done.
This proof works nearly verbatim for Lebesgue integration and non-compact intervals.
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