Norm of Fourier Sum Operator at 0 with continuous functions as domain

Given the operator $$F_n : C(\mathbb{T}) \rightarrow \mathbb{C}, f \mapsto S_nf (0)$$ where $$C(\mathbb{T})$$ are continuous functions on $$\mathbb{T} = (-\pi, \pi)$$ and $$S_n$$ is the Fourier Sum operator $$S_n f(x) := \frac{1}{\sqrt{2\pi}} \sum_{k=-n}^n \hat{f}(k) e^{i k x}$$, I want to calculate the operator norm $$\|F_n\|$$.

I already got that $$|F_n f| = |S_n f(0)| = \frac{1}{2\pi} | \int_{-\pi}^\pi D_n(x) f(x) dx| \leq \frac{1}{2\pi} \|D_n\|_1 \|f\|_\infty$$ with the Dirichlet kernel $$D_n(x) := \sum_{k=-n}^n e^{ikx}$$.

So, $$\|F_n\| \leq \frac{1}{2\pi} \|D_n\|_1$$.

How do I now get equality? I tried letting $$f(x):=\frac{\overline{D_n(x)}}{|D_n(x)|}$$ which would work apart from the fact that this f is not continuous.

I would appreciate any help and ideas, thanks.

Fix $$\def\e{\varepsilon}\e>0$$. Since $$f(x)=\frac{\overline{D_n(x)}}{|D_n(x)|}$$ is measurable, by Lusin's Theorem there exists $$g\in C(\def\TT{\mathbb T}\TT)$$ and $$E\subset\TT$$ with $$\|g\|_\infty=1$$, $$g=f$$ on $$E$$ and $$m(E^c)<\e$$. Then$$\def\abajo{\\[0.3cm]}$$ \begin{align} 2\pi|F_ng|&≥\int_E|D_n|-\int_{E^c}D_n g≥\int_E|D_n|-\e\|D_n\|_1\abajo &=\|D_n\|_1-\int_{E^c}|D_n|-\e\|D_n\|_1\abajo &≥\|D_n\|_1-m(E^c)^{1/2}\|D_n\|_2-\e\|D_n\|_1\abajo &≥\|D_n\|_1-\e^{1/2}\|D_n\|_2-\e\|D_n\|_1. \end{align} As $$\e$$ was arbitrary, this shows that $$\|F_n\|≥\frac1{2\pi}\|D_n\|_1.$$