[proofplan] We bound $\|D_N\|_{L^1(\mathbb{T})}$ from below and above by replacing $\sin(x/2)$ with its linear bounds ($x/\pi \leq \sin(x/2) \leq x/2$ for $x \in [0, \pi]$). The substitution $u = (N+\tfrac{1}{2})x$ transforms the integral into $\int_0^{(N+1/2)\pi} |\sin u|/u \, d\mathcal{L}^1(u)$, which decomposes over half-periods as a sum $\sim \frac{2}{\pi}\sum_{k=1}^N 1/k = \frac{2}{\pi}H_N \sim \frac{2}{\pi}\log N$. [/proofplan] [step:Reduce to a one-sided integral using evenness of $D_N$] Since $D_N(x) = \sin((N+\tfrac{1}{2})x)/\sin(x/2)$ is even: \begin{align*} \|D_N\|_{L^1(\mathbb{T})} = \frac{1}{2\pi}\int_{-\pi}^\pi |D_N(x)| \, d\mathcal{L}^1(x) = \frac{1}{\pi}\int_0^\pi \frac{|\sin((N+\tfrac{1}{2})x)|}{\sin(x/2)} \, d\mathcal{L}^1(x). \end{align*} [/step] [step:Establish a logarithmic lower bound] Since $\sin(x/2) \leq x/2$ for $x \geq 0$: \begin{align*} \|D_N\|_{L^1(\mathbb{T})} \geq \frac{2}{\pi}\int_0^\pi \frac{|\sin((N+\tfrac{1}{2})x)|}{x} \, d\mathcal{L}^1(x). \end{align*} Substitute $u = (N+\tfrac{1}{2})x$: \begin{align*} \frac{2}{\pi}\int_0^\pi \frac{|\sin((N+\tfrac{1}{2})x)|}{x} \, d\mathcal{L}^1(x) = \frac{2}{\pi}\int_0^{(N+1/2)\pi} \frac{|\sin u|}{u} \, d\mathcal{L}^1(u). \end{align*} On each half-period $[k\pi, (k+1)\pi]$: $\int_{k\pi}^{(k+1)\pi} |\sin u| \, d\mathcal{L}^1(u) = 2$ and $u \leq (k+1)\pi$. Therefore: \begin{align*} \int_{k\pi}^{(k+1)\pi} \frac{|\sin u|}{u} \, d\mathcal{L}^1(u) \geq \frac{2}{(k+1)\pi}. \end{align*} Summing from $k = 0$ to $N - 1$: \begin{align*} \int_0^{N\pi} \frac{|\sin u|}{u} \, d\mathcal{L}^1(u) \geq \frac{2}{\pi}\sum_{k=1}^N \frac{1}{k} = \frac{2}{\pi}H_N, \end{align*} where $H_N \sim \log N$. Hence $\|D_N\|_{L^1} \geq \frac{4}{\pi^2}H_N \geq c\log N$. [/step] [step:Establish a logarithmic upper bound] Using $\sin(x/2) \geq x/\pi$ for $x \in [0, \pi]$ (from the concavity bound $\sin\theta \geq 2\theta/\pi$ for $\theta \in [0, \pi/2]$): \begin{align*} \|D_N\|_{L^1(\mathbb{T})} \leq \int_0^\pi \frac{|\sin((N+\tfrac{1}{2})x)|}{x} \, d\mathcal{L}^1(x). \end{align*} The same substitution $u = (N+\tfrac{1}{2})x$ and half-period decomposition (now bounding $u \geq k\pi$ for $k \geq 1$) gives: \begin{align*} \int_{k\pi}^{(k+1)\pi} \frac{|\sin u|}{u} \, d\mathcal{L}^1(u) \leq \frac{2}{k\pi} \quad \text{for } k \geq 1. \end{align*} The $k = 0$ term contributes at most $\pi$ (since $|\sin u|/u \leq 1$). Summing: \begin{align*} \int_0^{(N+1/2)\pi} \frac{|\sin u|}{u} \, d\mathcal{L}^1(u) \leq \pi + \frac{2}{\pi}\sum_{k=1}^N \frac{1}{k} \leq \pi + \frac{2}{\pi}(\log N + 1). \end{align*} Hence $\|D_N\|_{L^1} \leq C\log N$. [/step] [step:State the asymptotic result] Combining the lower and upper bounds: $\|D_N\|_{L^1(\mathbb{T})} = \frac{4}{\pi^2}\log N + O(1)$ as $N \to \infty$. In particular, $\|D_N\|_{L^1} \to \infty$, so the operator $f \mapsto S_N f$ is not uniformly bounded on $C(\mathbb{T})$ or on $L^1(\mathbb{T})$. [/step]