Consider the operator $S_n: \mathcal{L}^1(\mathbb{T}) \to \mathcal{L}^1(\mathbb{T})$ defined by $S_nf(x)=\sum_{|k|\le n} e^{i k x}\hat f_k = f * D_n(x),$ where $\hat{f}_k$ are the Fourier coefficients and $D_n(x)= \frac{\sin((n+\frac12)x)}{2\pi\cdot \sin(\frac12x)}$ is the Dirichlet kernel. Find the operator norm $\|S_n\|$.
The first bound that came to my mind is $\|f*D_n\|_{\mathcal{L}^1}\leq \|f\|_{\mathcal{L}^1}\|D_n\|_{\mathcal{L}^1},$ because of the Young's inequality, that immediately implies $\|S_n\|\le \|D_n\|_{\mathcal{L}^1}.$
Willing to show that the norm is actually $\|D_n\|_1,$ I took $f_k=k\cdot\mathbb{I}_{\left[-\frac k2,\frac k2\right]}$ so that $f_k*D_n\to_k D_n$ pointwisely and then conclude that $\|f_k*D_n\|_1 \to \|D_n\|_1$ by dominated convergence.
Is it correct? Is there a more smart and direct way to conclude that? Is it true in general that the norm of the convolution operator with an $\mathcal{L}^1$ function $g$ is the $\mathcal{L}^1$ norm of $g?$