Riemann–Lebesgue Lemma

Fourier coefficients die out — ∫f(x)sin⁡(nx) dx→0\int f(x)\sin(nx)\,dx \to 0. Rapid oscillation averages away every bump of ff.

an=1π∫−ππf(x)cos⁡(nx) dx→0a_n = \frac1\pi\int_{-\pi}^{\pi} f(x)\cos(nx)\,dx \to 0
bn=1π∫−ππf(x)sin⁡(nx) dx→0b_n = \frac1\pi\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx \to 0

Examples & Exercises

Compute directly the Fourier coefficients bn=1π∫02πf(x)sin⁡(nx) dxb_n = \frac{1}{\pi}\int_0^{2\pi} f(x)\sin(nx)\,dx of f=1[0,π]f = 1_{[0, \pi]} (on [0,2π][0, 2\pi]), find ana_n as well, and prove completely that an,bn→0a_n, b_n \to 0.

bn=1−(−1)nnπ  →n→∞  0,an=0b_n = \frac{1 - (-1)^n}{n\pi} \;\xrightarrow[n \to \infty]{}\; 0, \qquad a_n = 0
Step mode, n=12n = 12, ∣b∣|b| stems: odd terms sink as ∣bn∣=2nπ|b_n| = \frac{2}{n\pi} and even terms are exactly 0 — ∣b11∣=211π≈0.0579|b_{11}| = \frac{2}{11\pi} \approx 0.0579.
Show proof
  1. Since ff is 1 on [0,π][0, \pi] and 0 on (π,2π](\pi, 2\pi], bn=1π∫0πsin⁡(nx) dxb_n = \frac{1}{\pi}\int_0^{\pi} \sin(nx)\,dx. With ∫sin⁡(nx) dx=−cos⁡(nx)/n\int \sin(nx)\,dx = -\cos(nx)/n, we get bn=1π[−cos⁡(nx)n]0π=1−cos⁡(nπ)nπb_n = \frac{1}{\pi}\left[-\frac{\cos(nx)}{n}\right]_0^{\pi} = \frac{1 - \cos(n\pi)}{n\pi}, and cos⁡(nπ)=(−1)n\cos(n\pi) = (-1)^n gives bn=1−(−1)nnπb_n = \frac{1 - (-1)^n}{n\pi}.
  2. Since ∣1−(−1)n∣≤2|1 - (-1)^n| \le 2, ∣bn∣≤2nπ→0|b_n| \le \frac{2}{n\pi} \to 0 as n→∞n \to \infty. Meanwhile an=1π∫0πcos⁡(nx) dx=sin⁡(nπ)nπ=0a_n = \frac{1}{\pi}\int_0^{\pi} \cos(nx)\,dx = \frac{\sin(n\pi)}{n\pi} = 0 — exactly 0 for every n≥1n \ge 1. Hence an→0a_n \to 0 and bn→0b_n \to 0 are both proved.
  3. The scene's step mode reproduces this formula by numerical quadrature: the odd stems ride the 2/(nπ)2/(n\pi) curve downward while the even stems sit at 0 up to quadrature error. A discontinuous function's coefficients go at rate 1/n1/n — slower than for smooth functions (compare the tent mode).

Prove the Riemann–Lebesgue lemma for a step function s=∑k=1mck 1[xk−1,xk)s = \sum_{k=1}^{m} c_k\,1_{[x_{k-1}, x_k)} (on a partition of [0,2π][0, 2\pi]): an(s),bn(s)→0a_n(s), b_n(s) \to 0.

∣1π∫xk−1xksin⁡(nx) dx∣≤2nπ  →n→∞  0\left|\frac{1}{\pi}\int_{x_{k-1}}^{x_k} \sin(nx)\,dx\right| \le \frac{2}{n\pi} \;\xrightarrow[n \to \infty]{}\; 0
Step mode, n=24n = 24: each of the finitely many intervals contributes at most 2/(nπ)2/(n\pi), so the total ∣bn∣|b_n| goes to 0 — the numerical version of the general step-function proof.
Show proof
  1. By linearity of the integral, bn(s)=∑k=1mck bn(1[xk−1,xk))b_n(s) = \sum_{k=1}^{m} c_k\,b_n(1_{[x_{k-1}, x_k)}). Each interval indicator contributes bn(1[xk−1,xk))=1π∫xk−1xksin⁡(nx) dx=cos⁡(nxk−1)−cos⁡(nxk)nπb_n(1_{[x_{k-1}, x_k)}) = \frac{1}{\pi}\int_{x_{k-1}}^{x_k} \sin(nx)\,dx = \frac{\cos(nx_{k-1}) - \cos(nx_k)}{n\pi}.
  2. Since ∣cos⁡(nxk−1)−cos⁡(nxk)∣≤2|\cos(nx_{k-1}) - \cos(nx_k)| \le 2, ∣bn(1[xk−1,xk))∣≤2nπ\left|b_n(1_{[x_{k-1}, x_k)})\right| \le \frac{2}{n\pi} — a uniform bound independent of the interval's position. Hence ∣bn(s)∣≤∑k=1m∣ck∣ 2nπ=Cn|b_n(s)| \le \sum_{k=1}^{m} |c_k|\,\frac{2}{n\pi} = \frac{C}{n} (with C=2π∑∣ck∣C = \frac{2}{\pi}\sum |c_k| independent of nn), which goes to 0 as n→∞n \to \infty. The same computation with cos⁡\cos in place of sin⁡\sin sends an(s)→0a_n(s) \to 0.
  3. The point is finiteness: with only finitely many intervals, adding up the O(1/n)O(1/n) contributions still gives O(1/n)O(1/n). This is the platform the density argument to continuous functions (problem 3 below) stands on.

First prove that a continuous ff on [0,2π][0, 2\pi] can be uniformly approximated by step functions (via uniform continuity), then use it to prove bn(f)→0b_n(f) \to 0 (and an(f)→0a_n(f) \to 0) completely by an ε/3\varepsilon/3 argument.

∣bn(f)∣≤∣bn(f−s)∣+∣bn(s)∣≤2ε+∣bn(s)∣  →n→∞  2ε|b_n(f)| \le |b_n(f - s)| + |b_n(s)| \le 2\varepsilon + |b_n(s)| \;\xrightarrow[n \to \infty]{}\; 2\varepsilon
Tent mode, n=24n = 24: continuous functions are uniformly approximable by step functions, so ∣an∣→0|a_n| \to 0 too — here sinking even faster as ∣an∣=4/(π2n2)|a_n| = 4/(\pi^2 n^2) (odd nn).
Show proof
  1. Lemma (uniform approximation by step functions): if ff is continuous on [0,2π][0, 2\pi] it is uniformly continuous, so for every ε>0\varepsilon > 0 there is δ>0\delta > 0 with ∣x−y∣<δ⇒∣f(x)−f(y)∣<ε|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon. Split [0,2π][0, 2\pi] into intervals I1,…,ImI_1, \dots, I_m of width <δ{} < \delta and define the step function ss taking the constant value f(xk−1)f(x_{k-1}) (left endpoint) on each IkI_k. For x∈Ikx \in I_k, ∣x−xk−1∣<δ|x - x_{k-1}| < \delta, so ∣f(x)−s(x)∣<ε|f(x) - s(x)| < \varepsilon — i.e. ∥f−s∥∞<ε\|f - s\|_\infty < \varepsilon. Lemma proved.
  2. Fix ε>0\varepsilon > 0 and pick a step function ss with ∥f−s∥∞<ε\|f - s\|_\infty < \varepsilon by the lemma. By the triangle inequality ∣bn(f)∣≤∣bn(f−s)∣+∣bn(s)∣|b_n(f)| \le |b_n(f - s)| + |b_n(s)|. The first term satisfies ∣bn(f−s)∣≤1π∫02π∣f−s∣ ∣sin⁡(nx)∣ dx≤1π ε⋅2π=2ε|b_n(f - s)| \le \frac{1}{\pi}\int_0^{2\pi} |f - s|\,|\sin(nx)|\,dx \le \frac{1}{\pi}\,\varepsilon \cdot 2\pi = 2\varepsilon — independent of nn.
  3. The second term ∣bn(s)∣→0|b_n(s)| \to 0 as n→∞n \to \infty by problem 2. So choose NN large enough that ∣bn(s)∣<ε|b_n(s)| < \varepsilon for n≥Nn \ge N; for those nn, ∣bn(f)∣<2ε+ε=3ε|b_n(f)| < 2\varepsilon + \varepsilon = 3\varepsilon. Since ε\varepsilon is arbitrary, bn(f)→0b_n(f) \to 0. The same argument with ∣cos⁡(nx)∣≤1|\cos(nx)| \le 1 gives an(f)→0a_n(f) \to 0. Every step is complete — nothing is left at idea level.

8. Series of Functions & Fourier