[proofplan] We derive the closed form of the Fej\'er kernel $F_N$ by summing the Dirichlet kernels using a telescoping product-to-sum identity, revealing that $F_N$ is a non-negative squared sine ratio. We verify the three properties of an approximate identity (positivity, unit mass, concentration at the origin), then use them to bound $|\sigma_N f(x) - f(x)|$ uniformly by splitting the convolution integral into a near-origin piece (controlled by uniform continuity) and a far piece (controlled by concentration). [/proofplan] [step:Derive the closed form $F_N(x) = \frac{1}{N+1}\cdot\frac{\sin^2((N+1)x/2)}{\sin^2(x/2)}$] By the [Dirichlet Kernel Formula](/theorems/581), $D_n(x) = \sin((n+\tfrac{1}{2})x)/\sin(x/2)$. Apply the product-to-sum identity $2\sin A\sin B = \cos(A-B) - \cos(A+B)$ with $A = (n+\tfrac{1}{2})x$ and $B = x/2$: \begin{align*} 2\sin(x/2)\,D_n(x) = 2\sin(x/2)\sin((n+\tfrac{1}{2})x) = \cos(nx) - \cos((n+1)x). \end{align*} Summing from $n = 0$ to $N$, the right side telescopes: \begin{align*} 2\sin(x/2)\sum_{n=0}^N D_n(x) = 1 - \cos((N+1)x). \end{align*} Using $1 - \cos\theta = 2\sin^2(\theta/2)$ with $\theta = (N+1)x$: \begin{align*} \sum_{n=0}^N D_n(x) = \frac{\sin^2((N+1)x/2)}{\sin^2(x/2)}. \end{align*} Dividing by $N+1$ gives $F_N(x) = \frac{1}{N+1}\cdot\frac{\sin^2((N+1)x/2)}{\sin^2(x/2)}$. [/step] [step:Verify the three approximate identity properties of $F_N$] **(i) Positivity.** From the closed form, $F_N(x) \geq 0$ since it is a ratio of squares. **(ii) Unit mass.** By definition $F_N = \frac{1}{N+1}\sum_{n=0}^N D_n$, and $\frac{1}{2\pi}\int_{-\pi}^\pi D_n \, d\mathcal{L}^1 = 1$ for each $n$ (only the $k = 0$ Fourier mode contributes). Averaging: $\frac{1}{2\pi}\int_{-\pi}^\pi F_N \, d\mathcal{L}^1 = 1$. **(iii) Concentration.** For $|x| \geq \delta > 0$: $\sin^2(x/2) \geq \sin^2(\delta/2)$ and $\sin^2((N+1)x/2) \leq 1$, so: \begin{align*} \sup_{\delta \leq |x| \leq \pi} F_N(x) \leq \frac{1}{(N+1)\sin^2(\delta/2)} \to 0 \quad \text{as } N \to \infty. \end{align*} [/step] [step:Bound $|\sigma_N f(x) - f(x)|$ uniformly by splitting the convolution integral] Using $\frac{1}{2\pi}\int F_N = 1$: \begin{align*} \sigma_N f(x) - f(x) = \frac{1}{2\pi}\int_{-\pi}^\pi [f(x-t) - f(x)]\,F_N(t) \, d\mathcal{L}^1(t). \end{align*} Let $\varepsilon > 0$. Since $f$ is uniformly continuous on $\mathbb{T}$ (continuous on a compact set), choose $\delta > 0$ with $|f(x-t) - f(x)| < \varepsilon$ for $|t| < \delta$ and all $x$. **Near-origin piece ($|t| < \delta$):** Using $F_N \geq 0$ and $\frac{1}{2\pi}\int F_N \leq 1$: \begin{align*} \frac{1}{2\pi}\int_{|t| < \delta} |f(x-t) - f(x)|\,F_N(t) \, d\mathcal{L}^1(t) < \varepsilon. \end{align*} **Far piece ($|t| \geq \delta$):** Using $|f(x-t) - f(x)| \leq 2\|f\|_{L^\infty}$ and the concentration bound: \begin{align*} \frac{1}{2\pi}\int_{|t| \geq \delta} |f(x-t) - f(x)|\,F_N(t) \, d\mathcal{L}^1(t) \leq \frac{2\|f\|_{L^\infty}}{(N+1)\sin^2(\delta/2)}. \end{align*} Choose $N_0$ large enough that this is less than $\varepsilon$. For all $N \geq N_0$ and all $x$: $|\sigma_N f(x) - f(x)| < 2\varepsilon$. Since $\varepsilon$ was arbitrary, $\sigma_N f \to f$ uniformly. Each $\sigma_N f$ is a trigonometric polynomial, so trigonometric polynomials are uniformly dense in $C(\mathbb{T})$. [/step]