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Riemann–Lebesgue lemma for continuous functions on a compact interval

Statement

Let a<b and let f:[a,b]R be continuous. For positive integer n, put

In:=abf(t)sin(nt)dt,Jn:=abf(t)cos(nt)dt.

Then

limnIn=0,limnJn=0.

In particular, for every continuous g:[0,π]R, limn0πg(x)sin(nx)dx=0.

Facts & Assumptions

Given: Reals a<b, a continuous f:[a,b]R, and a real ε>0.

[L1]

For ab, every continuous real function on [a,b] is a uniform limit of polynomials (Polynomials are uniformly dense in C([a,b],R) for every closed interval).

[L2]

If integrable functions u,v satisfy u(x)v(x)η between endpoints, then uvη times the endpoint distance (Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

[L3]

If u,v are differentiable on [a,b] with integrable derivatives, then abuv=u(b)v(b)u(a)v(a)abuv (If u,v are differentiable on [a,b] with u,v integrable, then abuv=u(b)v(b)u(a)v(a)abuv).

[L4]

The derivative formulas for sine and cosine, together with the chain rule, give (sinnt)=ncosnt and (cosnt)=nsinnt for positive integers n (The derivatives of sine and cosine are cosine and minus sine, The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then fg is differentiable at c with (fg)(c)=f(g(c))g(c)).

[L8]

For every real η>0 there is a positive integer N with 1/N<η (For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε).

[L9]

For every real x, sinx1 and cosx1 (Parity and the Pythagorean identity for sine and cosine).

[L10]

The number π=2γ is positive because the smallest positive zero of cosine satisfies γ(0,2) (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two).

[L12]

A real sequence converges to zero when, for every positive rational ε, its terms are eventually smaller than ε in absolute value (Limits and Cauchy sequences of reals).

Proof

technique · direct
1.1

By [L1], choose a polynomial p with f(t)p(t)<ε/(4(ba)) for every t[a,b].

givenL1choose
1.2

Put Cp:=p(a)+p(b)+abp(t)dt. Using v(t)=cos(nt)/n in [L3], and then [L6], [L7], [L9], and [L11], gives abp(t)sin(nt)dtCp/n. Using v(t)=sin(nt)/n gives the identical bound for the cosine integral.

L3L4L5L6L7L9L11algebra
1.3

Apply [L8] to ε/(2(Cp+1)) and choose a positive integer N with (Cp+1)/N<ε/2. Then Cp/n<ε/2 whenever nN.

L8choosealgebra
2.1

The functions f(t)sin(nt) and p(t)sin(nt) are integrable, and [L2] with [L6] and [L9] gives ab(f(t)p(t))sin(nt)dtε/4<ε/2 for every positive integer n; the same estimate holds with cosine.

step 1.1L2L5L6L9algebra
3.1

For every nN, linearity [L7] splits each integral into its (fp) part and its p part. Steps 2.1, 1.2, and 1.3 make the absolute value of each integral less than ε, for sine and for cosine.

step 2.1step 1.2step 1.3L7algebra
4.1

Since ε>0 was arbitrary, step 3.1 is exactly convergence of both sequences of integrals to zero by [L12]; [L10] permits the substitution a=0, b=π, and g for f in the stated special case.

step 3.1L10L12

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