Fejér's Theorem

The Cesàro means σN\sigma_N of Fourier partial sums converge uniformly for continuous functions — because the Fejér kernel is an approximate identity. Overlay the ringing SNS_N with the smooth σN\sigma_N.

σN=S0+⋯+SN−1N→∥⋅∥∞f\sigma_N = \frac{S_0+\cdots+S_{N-1}}{N} \xrightarrow{\|\cdot\|_\infty} f
KN(t)=1N(sin⁡(Nt/2)sin⁡(t/2))2≥0K_N(t) = \frac1N\left(\frac{\sin(Nt/2)}{\sin(t/2)}\right)^2 \ge 0

Examples & Exercises

For the Fejér kernel KN(t)=1N(sin⁡(Nt/2)sin⁡(t/2))2K_N(t) = \frac{1}{N}\left(\frac{\sin(Nt/2)}{\sin(t/2)}\right)^2 (t≢0(mod2π)t \not\equiv 0 \pmod{2\pi}, KN(0)=NK_N(0) = N), prove KN≥0K_N \ge 0 and 12π∫−ππKN=1\frac{1}{2\pi}\int_{-\pi}^{\pi} K_N = 1. Also show σN(f)=12π(f∗KN)\sigma_N(f) = \frac{1}{2\pi}(f * K_N).

KN(t)=1N∑k=0N−1Dk(t)=1N(sin⁡(Nt/2)sin⁡(t/2))2≥0K_N(t) = \frac{1}{N}\sum_{k=0}^{N-1} D_k(t) = \frac{1}{N}\left(\frac{\sin(Nt/2)}{\sin(t/2)}\right)^2 \ge 0
12π∫−ππKN=1,σN(f)=12π(f∗KN)\frac{1}{2\pi}\int_{-\pi}^{\pi} K_N = 1, \qquad \sigma_N(f) = \frac{1}{2\pi}(f * K_N)
N=8N = 8: the kernel is a spike with K8(0)=8K_8(0) = 8, and σ8\sigma_8 is already smooth — the proof below is why this spike has mass 1.
Show proof
  1. Start from the Dirichlet kernel Dk(t)=∑j=−kkeijtD_k(t) = \sum_{j=-k}^{k} e^{ijt}. The partial sum is a convolution: Sk(f)(x)=12π∫−ππf(x−t) Dk(t) dtS_k(f)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-t)\,D_k(t)\,dt (orthogonality of the exponentials). Averaging gives σN(f)=1N∑k=0N−1Sk(f)=12π(f∗KN)\sigma_N(f) = \frac{1}{N}\sum_{k=0}^{N-1} S_k(f) = \frac{1}{2\pi}(f * K_N) with KN=1N∑k=0N−1DkK_N = \frac{1}{N}\sum_{k=0}^{N-1} D_k.
  2. Interchanging the sums, ∑k=0N−1Dk(t)=∑k=0N−1∑j=−kkeijt\sum_{k=0}^{N-1} D_k(t) = \sum_{k=0}^{N-1}\sum_{j=-k}^{k} e^{ijt}. The frequency jj (∣j∣≤N−1|j| \le N-1) appears in every k≥∣j∣k \ge |j|, i.e. exactly N−∣j∣N - |j| times: ∑k=0N−1Dk(t)=∑∣j∣≤N−1(N−∣j∣) eijt\sum_{k=0}^{N-1} D_k(t) = \sum_{|j| \le N-1} (N-|j|)\,e^{ijt}.
  3. Meanwhile ∣∑j=0N−1eijt∣2=∑j,lei(j−l)t\left|\sum_{j=0}^{N-1} e^{ijt}\right|^2 = \sum_{j,l} e^{i(j-l)t} expands so that the difference m=j−lm = j-l takes each value with ∣m∣≤N−1|m| \le N-1 exactly N−∣m∣N - |m| times — the same sum. Hence ∑k=0N−1Dk(t)=∣∑j=0N−1eijt∣2≥0\sum_{k=0}^{N-1} D_k(t) = \left|\sum_{j=0}^{N-1} e^{ijt}\right|^2 \ge 0. The geometric sum gives ∑j=0N−1eijt=ei(N−1)t/2 sin⁡(Nt/2)sin⁡(t/2)\sum_{j=0}^{N-1} e^{ijt} = e^{i(N-1)t/2}\,\frac{\sin(Nt/2)}{\sin(t/2)}, so KN(t)=1N(sin⁡(Nt/2)sin⁡(t/2))2≥0K_N(t) = \frac{1}{N}\left(\frac{\sin(Nt/2)}{\sin(t/2)}\right)^2 \ge 0 (with the limit value NN at t=0t = 0).
  4. The integral is easy: ∫−ππeijt dt\int_{-\pi}^{\pi} e^{ijt}\,dt is 2π2\pi for j=0j = 0 and 00 otherwise, so ∫−ππKN=1N∑k=0N−1∫−ππDk=1N⋅N⋅2π=2π\int_{-\pi}^{\pi} K_N = \frac{1}{N}\sum_{k=0}^{N-1}\int_{-\pi}^{\pi} D_k = \frac{1}{N}\cdot N \cdot 2\pi = 2\pi. Thus 12π∫KN=1\frac{1}{2\pi}\int K_N = 1 — the Fejér kernel is a nonnegative approximate identity of mass 1.

Prove σN(f)→f\sigma_N(f) \to f uniformly for every continuous 2π2\pi-periodic ff. Use the approximate-identity argument: split the integral at δ\delta — uniform continuity on ∣t∣<δ|t| < \delta, and the kernel tail bound KN(t)≤1Nsin⁡2(δ/2)K_N(t) \le \frac{1}{N\sin^2(\delta/2)} on δ≤∣t∣≤π\delta \le |t| \le \pi.

∣σN(x)−f(x)∣≤12π∫−ππKN(t) ∣f(x−t)−f(x)∣ dt\lvert\sigma_N(x) - f(x)\rvert \le \frac{1}{2\pi}\int_{-\pi}^{\pi} K_N(t)\,\lvert f(x-t) - f(x)\rvert\,dt
KN(t)≤1Nsin⁡2(δ/2)(δ≤∣t∣≤π)K_N(t) \le \frac{1}{N\sin^2(\delta/2)} \quad (\delta \le \lvert t\rvert \le \pi)
N=8N = 8: ∥σ8−f∥∞≈0.1101\lVert \sigma_8 - f \rVert_\infty \approx 0.1101, shrinking to ≈0.0639\approx 0.0639 at N=16N = 16 and ≈0.0363\approx 0.0363 at N=32N = 32 — the proof below is the machine behind this uniform convergence.
Show proof
  1. Extend ff 2π2\pi-periodically (still continuous, hence uniformly continuous on all of R\mathbb{R}: periodicity plus uniform continuity on [−π,3π][-\pi, 3\pi]). From the previous problem, σN(f)(x)=12π∫−ππf(x−t) KN(t) dt\sigma_N(f)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-t)\,K_N(t)\,dt and 12π∫KN=1\frac{1}{2\pi}\int K_N = 1, so subtracting f(x)=12π∫KN(t) f(x) dtf(x) = \frac{1}{2\pi}\int K_N(t)\,f(x)\,dt leaves an error of at most 12π∫KN(t) ∣f(x−t)−f(x)∣ dt\frac{1}{2\pi}\int K_N(t)\,|f(x-t) - f(x)|\,dt (using KN≥0K_N \ge 0).
  2. Given ε>0\varepsilon > 0, uniform continuity gives δ∈(0,π)\delta \in (0, \pi) with ∣f(s)−f(t)∣<ε/2|f(s) - f(t)| < \varepsilon/2 whenever ∣s−t∣<δ|s - t| < \delta. Split the integral into ∣t∣<δ|t| < \delta and δ≤∣t∣≤π\delta \le |t| \le \pi. The near part is at most (ε/2)⋅12π∫∣t∣<δKN≤ε/2(\varepsilon/2)\cdot\frac{1}{2\pi}\int_{|t|<\delta} K_N \le \varepsilon/2 (the mass is at most 1).
  3. The tail: on δ≤∣t∣≤π\delta \le |t| \le \pi we have ∣sin⁡(t/2)∣≥sin⁡(δ/2)>0|\sin(t/2)| \ge \sin(\delta/2) > 0 (sine increases on [0,π/2][0, \pi/2]), so KN(t)=1N(sin⁡(Nt/2)sin⁡(t/2))2≤1Nsin⁡2(δ/2)K_N(t) = \frac{1}{N}\left(\frac{\sin(Nt/2)}{\sin(t/2)}\right)^2 \le \frac{1}{N\sin^2(\delta/2)}. With M=∥f∥∞M = \|f\|_\infty, ∣f(x−t)−f(x)∣≤2M|f(x-t) - f(x)| \le 2M, so the tail is at most 12π⋅2M⋅2(π−δ)Nsin⁡2(δ/2)≤2MNsin⁡2(δ/2)\frac{1}{2\pi}\cdot 2M\cdot\frac{2(\pi-\delta)}{N\sin^2(\delta/2)} \le \frac{2M}{N\sin^2(\delta/2)}.
  4. Hence for N>4Mεsin⁡2(δ/2)N > \frac{4M}{\varepsilon\sin^2(\delta/2)} we get ∣σN(x)−f(x)∣<ε/2+ε/2=ε|\sigma_N(x) - f(x)| < \varepsilon/2 + \varepsilon/2 = \varepsilon for every xx. The choice of NN is independent of xx, so the convergence is uniform. This is Fejér's theorem — and the punchline is that it assumes nothing about the behavior of the raw partial sums SNS_N themselves.

Consider the square wave ff (jump discontinuities). Prove the raw partial sums SNS_N do NOT converge uniformly (Gibbs ringing — shown numerically in the Fourier scene). In contrast, prove σN→f\sigma_N \to f uniformly on every compact set bounded away from the jumps, by localizing the previous argument. State the behavior at the jumps themselves at idea level only.

Gibbs: sup⁡x∣SN−f∣≳0.17 for all N\text{Gibbs: }\sup_x \lvert S_N - f\rvert \gtrsim 0.17 \text{ for all } N
⟹ no uniform convergence\Longrightarrow\ \text{no uniform convergence}
σN→f uniformly on K⋐[−π,π]∖{jumps}\sigma_N \to f \text{ uniformly on } K \Subset [-\pi,\pi] \setminus \{\text{jumps}\}
For the triangle wave (continuous), the sup error of S8S_8 (≈0.0252\approx 0.0252) is smaller than that of σ8\sigma_8 (≈0.1101\approx 0.1101) — Cesàro's real victory shows up only where there are jumps (see the square-wave scene).
Show proof
  1. Each SNS_N is a sum of trigonometric functions, hence continuous. A uniform limit of continuous functions must be continuous — but the square wave ff is discontinuous at its jumps. So SN→fS_N \to f cannot converge uniformly on any interval containing a jump. As the Fourier scene shows numerically, the overshoot ≈0.1789\approx 0.1789 beside each jump never dies out, no matter how large NN gets (the Gibbs phenomenon).
  2. Now fix a compact KK at distance d>0d > 0 from the jump set JJ. Reuse the previous proof with δ=min⁡(d/2,δunif)\delta = \min(d/2, \delta_{\text{unif}}), where δunif\delta_{\text{unif}} is the radius of ε/2\varepsilon/2-uniform continuity of ff on the compact set at distance ≥d/2\ge d/2 from JJ (where ff is continuous, hence uniformly continuous). For x∈Kx \in K and ∣t∣<δ|t| < \delta, the point x−tx - t stays at distance ≥d/2\ge d/2 from JJ, so uniform continuity applies; the tail bound is unchanged.
  3. Hence sup⁡x∈K∣σN(x)−f(x)∣→0\sup_{x \in K} |\sigma_N(x) - f(x)| \to 0 — the Cesàro means recover uniform convergence away from the jumps. This is where Fejér's theorem goes beyond the failure of SNS_N: averaging irons the ringing flat.
  4. Idea level: at a jump itself, the symmetry of KNK_N makes σN\sigma_N take half its mass from each side, so for the square wave σN(0)=0\sigma_N(0) = 0 (the midpoint). In general it converges to the average of the one-sided limits (f(x+)+f(x−))/2(f(x^+) + f(x^-))/2 — proof omitted.

8. Series of Functions & Fourier