[proofplan] We sum the geometric series $\sum_{n=-N}^N e^{inx}$ in closed form by factoring out $e^{-iNx}$, applying the finite geometric series formula, and simplifying via the identity $e^{i\theta} - e^{-i\theta} = 2i\sin\theta$. The convolution representation follows by exchanging the finite sum and integral in the definition of $S_N f$. [/proofplan] [step:Sum the geometric series to obtain the closed form of $D_N$] For $x \notin 2\pi\mathbb{Z}$, $e^{ix} \neq 1$. Factor out $e^{-iNx}$ and reindex: \begin{align*} D_N(x) = \sum_{n=-N}^N e^{inx} = e^{-iNx}\sum_{m=0}^{2N} e^{imx} = e^{-iNx} \cdot \frac{e^{i(2N+1)x} - 1}{e^{ix} - 1}. \end{align*} To simplify, write numerator and denominator in terms of sines. Multiply top and bottom by $e^{-ix/2}$: \begin{align*} \frac{e^{i(2N+1)x} - 1}{e^{ix} - 1} = \frac{e^{i(2N+1)x/2}}{e^{ix/2}} \cdot \frac{e^{i(2N+1)x/2} - e^{-i(2N+1)x/2}}{e^{ix/2} - e^{-ix/2}} = e^{iNx} \cdot \frac{2i\sin((N+\tfrac{1}{2})x)}{2i\sin(x/2)}. \end{align*} Multiplying by the prefactor $e^{-iNx}$: \begin{align*} D_N(x) = \frac{\sin((N + \tfrac{1}{2})x)}{\sin(x/2)}. \end{align*} At $x = 2\pi k$, each term $e^{inx} = 1$, so $D_N(2\pi k) = 2N + 1$. [/step] [step:Derive the convolution representation $S_N f = \frac{1}{2\pi}\int f(x-t)\,D_N(t) \, d\mathcal{L}^1(t)$] For $f \in L^1(\mathbb{T})$ with Fourier coefficients $\hat{f}(n) = \frac{1}{2\pi}\int_{-\pi}^\pi f(t)\,e^{-int} \, d\mathcal{L}^1(t)$: \begin{align*} S_N f(x) &= \sum_{|n| \leq N} \hat{f}(n)\,e^{inx} = \sum_{|n| \leq N} \frac{1}{2\pi}\int_{-\pi}^\pi f(t)\,e^{-int} \, d\mathcal{L}^1(t) \cdot e^{inx} \\ &= \frac{1}{2\pi}\int_{-\pi}^\pi f(t) \sum_{|n| \leq N} e^{in(x-t)} \, d\mathcal{L}^1(t), \end{align*} where the exchange of sum and integral is justified because the sum is finite. The inner sum is $D_N(x - t)$. Substituting $s = x - t$ and using the $2\pi$-periodicity of the integrand gives $S_N f(x) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x - t)\,D_N(t) \, d\mathcal{L}^1(t)$. [/step]