Fejér's theorem

Step 1 (Fejér kernel is a positive approximate identity). Averaging $D_0,\ldots,D_N$ and using a telescoping trigonometric identity gives the closed form $F_N(x) = \frac{1}{N+1}\left(\dfrac{\sin\left(\tfrac{N+1}{2}x\right)}{\sin(x/2)}\right)^2 \ge 0$. Since $F_N = \frac{1}{N+1}\sum_{k=0}^N D_k$ and $\frac{1}{2\pi}\int D_k = 1$ for every $k$, also $\frac{1}{2\pi}\int F_N = 1$; being nonnegative, $\frac{1}{2\pi}\int |F_N| = 1$ too, so the $L^1$ norm never blows up.

Step 2 (mass concentrates at $0$). For any fixed $\delta>0$, on $\delta \le |x| \le \pi$ the denominator $\sin(x/2)^2$ is bounded below by a positive constant, so $F_N(x) = O(1/N)$ uniformly there; the mass outside $[-\delta,\delta]$ tends to $0$ as $N\to\infty$.

Step 3 (approximate-identity estimate). Write $\sigma_N f(x) - f(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} F_N(y)\big(f(x-y)-f(x)\big)\,dy$ using $\frac{1}{2\pi}\int F_N = 1$. Split the integral into $|y|<\delta$ and $\delta\le|y|\le\pi$. On the first piece, uniform continuity of $f$ makes $|f(x-y)-f(x)|<\varepsilon$ for $\delta$ small, and $F_N\ge0$ with total mass $1$ bounds that piece by $\varepsilon$. On the second piece, $F_N=O(1/N)$ and $f$ is bounded, so that piece $\to 0$ as $N\to\infty$. Both bounds are uniform in $x$, so $\sigma_N f\to f$ uniformly.