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, I’ll state it in that form. Theorem 1. Let f be Riemann integrable on [a, b]. Then lim λ→±∞ b a f (t) cos(λt)dt =0 (1) lim λ→±∞ b a f (t) sin(λt)dt =0 (2) lim λ→±∞ b a f (t)e iλt dt =0 (3) Proof. I will prove only the first statement. Since f is integrable, given > 0, there is a partition {a = x 0 ,x 1 ,...,x n = b}, so that /2 > b a f - n 1 m i Δx i 0, where m i is the minimum of f on [x i-1 ,x i ]. But the sum can be written as n 1 m i Δx i = b a g, where g = m i χ [x i-1 ,x i ] , and the inequality takes the form /2 > b a (f - g) 0. Now we use the fact that f - g 0 to get b a f (t) cos(λt)dt b a (f (t) - g(t)) cos(λt)dt + b a g(t) cos(λt)dt (4) b a (f - g)+ (1) m i (sin(λx i ) - sin(λx i-1 )) . (5) The function g has been fixed. Take λ large enough that (1) m i (sin(λx i ) - sin(λx i-1 )) < /2, and we are done. This proof works nearly verbatim for Lebesgue integration and non-compact intervals. 1

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Riemann Lebesgue Proof

Transcript of Riemann Lebesgue Lemma

  • 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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