Fejér's Theorem

Start from the Dirichlet kernel $D_k(t) = \sum_{j=-k}^{k} e^{ijt}$.

The partial sum is a convolution: $S_k(f)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-t)\,D_k(t)\,dt$ (orthogonality of the exponentials).

Averaging gives $\sigma_N(f) = \frac{1}{N}\sum_{k=0}^{N-1} S_k(f) = \frac{1}{2\pi}(f * K_N)$ with $K_N = \frac{1}{N}\sum_{k=0}^{N-1} D_k$.

Interchanging the sums, $\sum_{k=0}^{N-1} D_k(t) = \sum_{k=0}^{N-1}\sum_{j=-k}^{k} e^{ijt}$. The frequency $j$ ($|j| \le N-1$) appears in every $k \ge |j|$, i.e. exactly $N - |j|$ times: $\sum_{k=0}^{N-1} D_k(t) = \sum_{|j| \le N-1} (N-|j|)\,e^{ijt}$.

Meanwhile $\left|\sum_{j=0}^{N-1} e^{ijt}\right|^2 = \sum_{j,l} e^{i(j-l)t}$ expands so that the difference $m = j-l$ takes each value with $|m| \le N-1$ exactly $N - |m|$ times — the same sum.

Hence $\sum_{k=0}^{N-1} D_k(t) = \left|\sum_{j=0}^{N-1} e^{ijt}\right|^2 \ge 0$.

The geometric sum gives $\sum_{j=0}^{N-1} e^{ijt} = e^{i(N-1)t/2}\,\frac{\sin(Nt/2)}{\sin(t/2)}$, so $K_N(t) = \frac{1}{N}\left(\frac{\sin(Nt/2)}{\sin(t/2)}\right)^2 \ge 0$ (with the limit value $N$ at $t = 0$).

The integral is easy: $\int_{-\pi}^{\pi} e^{ijt}\,dt$ is $2\pi$ for $j = 0$ and $0$ otherwise, so $\int_{-\pi}^{\pi} K_N = \frac{1}{N}\sum_{k=0}^{N-1}\int_{-\pi}^{\pi} D_k = \frac{1}{N}\cdot N \cdot 2\pi = 2\pi$. Thus $\frac{1}{2\pi}\int K_N = 1$ — the Fejér kernel is a nonnegative approximate identity of mass 1.

Extend $f$ $2\pi$-periodically (still continuous, hence uniformly continuous on all of $\mathbb{R}$: periodicity plus uniform continuity on $[-\pi, 3\pi]$). From the previous problem, $\sigma_N(f)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-t)\,K_N(t)\,dt$ and $\frac{1}{2\pi}\int K_N = 1$, so subtracting $f(x) = \frac{1}{2\pi}\int K_N(t)\,f(x)\,dt$ leaves an error of at most $\frac{1}{2\pi}\int K_N(t)\,|f(x-t) - f(x)|\,dt$ (using $K_N \ge 0$).

Given $\varepsilon > 0$, uniform continuity gives $\delta \in (0, \pi)$ with $|f(s) - f(t)| < \varepsilon/2$ whenever $|s - t| < \delta$.

Split the integral into $|t| < \delta$ and $\delta \le |t| \le \pi$.

The near part is at most $(\varepsilon/2)\cdot\frac{1}{2\pi}\int_{|t|<\delta} K_N \le \varepsilon/2$ (the mass is at most 1).

The tail: on $\delta \le |t| \le \pi$ we have $|\sin(t/2)| \ge \sin(\delta/2) > 0$ (sine increases on $[0, \pi/2]$), so $K_N(t) = \frac{1}{N}\left(\frac{\sin(Nt/2)}{\sin(t/2)}\right)^2 \le \frac{1}{N\sin^2(\delta/2)}$. With $M = \|f\|_\infty$, $|f(x-t) - f(x)| \le 2M$, so the tail is at most $\frac{1}{2\pi}\cdot 2M\cdot\frac{2(\pi-\delta)}{N\sin^2(\delta/2)} \le \frac{2M}{N\sin^2(\delta/2)}$.

Hence for $N > \frac{4M}{\varepsilon\sin^2(\delta/2)}$ we get $|\sigma_N(x) - f(x)| < \varepsilon/2 + \varepsilon/2 = \varepsilon$ for every $x$.

The choice of $N$ is independent of $x$, 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 $S_N$ themselves.

Each $S_N$ is a sum of trigonometric functions, hence continuous. A uniform limit of continuous functions must be continuous — but the square wave $f$ is discontinuous at its jumps.

So $S_N \to f$ cannot converge uniformly on any interval containing a jump.

As the Fourier scene shows numerically, the overshoot $\approx 0.1789$ beside each jump never dies out, no matter how large $N$ gets (the Gibbs phenomenon).

Now fix a compact $K$ at distance $d > 0$ from the jump set $J$.

Reuse the previous proof with $\delta = \min(d/2, \delta_{\text{unif}})$, where $\delta_{\text{unif}}$ is the radius of $\varepsilon/2$-uniform continuity of $f$ on the compact set at distance $\ge d/2$ from $J$ (where $f$ is continuous, hence uniformly continuous).

For $x \in K$ and $|t| < \delta$, the point $x - t$ stays at distance $\ge d/2$ from $J$, so uniform continuity applies; the tail bound is unchanged.

Hence $\sup_{x \in K} |\sigma_N(x) - f(x)| \to 0$ — the Cesàro means recover uniform convergence away from the jumps. This is where Fejér's theorem goes beyond the failure of $S_N$: averaging irons the ringing flat.

Idea level: at a jump itself, the symmetry of $K_N$ makes $\sigma_N$ take half its mass from each side, so for the square wave $\sigma_N(0) = 0$ (the midpoint). In general it converges to the average of the one-sided limits $(f(x^+) + f(x^-))/2$ — proof omitted.