I was trying to solve this problem:
For a continuous function on $f$ on $\mathbf{S}^1$, consider $$ (S_n(f))(\theta) = \frac{1}{2\pi}\int_0^{2\pi} f(\theta + x) D_n(x) \, dx, $$ where $D_n$ is the Dirchelet kernel. Let $\phi_n: C(\mathbf{S}^1) \to \mathbf{R}$ be defined by $\phi_n(f) = S_n(f)(0)$. Prove that $\|\phi_n\|_{C(\mathbf{S}^1)^\star} = \frac{1}{2\pi} \|D_n\|_{L^1(\mathbf{S}^1)}$, for every $n \in \mathbf{N}$.
I know it suffices to establish the following equality: $$ \sup_{\|f\|_{\infty} = 1} \left| \int_0^{2\pi} f(x) D_n(x)\, dx \right| = \int_{0}^{2\pi} |D_n(x)| \, dx . $$ I can prove $\leq$ easily, but I am struggling to show $\geq$. I know that the idea is to restrict to a sequence which achieves the integral on the RHS but I'm really struggling to make the details precise. Any hints/suggestions would be helpful!
Let $x_0=0 < x_1 < x_2 < \cdots < x_k < 2\pi=x_{k+1}$ be the interior zeros of $D_N(\theta)$, and let $\epsilon > 0$ be small. Define a piecewise linear and continuous function $f_{\epsilon}$ on $[0,2\pi]$ that is $1$ on $[x_{k-1}+\epsilon,x_{k}-\epsilon]$ if $D_N > 0$ on that interval, and that is $-1$ on $[x_{k-1}+\epsilon,x_{k}-\epsilon]$ if $D_N < 0$ on the interval, and that is linear and continuous on the remaining parts of the interval. Then $|f_{\epsilon}| \le 1$ on $[0,2\pi]$, and $$ \left|\int_{0}^{2\pi}f_{\epsilon}(\theta)D_N(\theta)d\theta\right| \le \int_{0}^{2\pi}|D_N(\theta)|d\theta \\ \left|\int_{0}^{2\pi}f_{\epsilon}(\theta)D_N(\theta)d\theta-\int_{0}^{2\pi}|D_N(\theta)|d\theta\right| \rightarrow 0 \mbox{ as } \epsilon\rightarrow 0. $$ The desired result then follows from combining the above: $$ \int_{0}^{2\pi}|D_N(\theta)|d\theta= \sup_{\|f\|_{\infty}=1}\left|\int_{0}^{2\pi}f(\theta)D_N(\theta)d\theta\right| \le \int_{0}^{2\pi}|D_N(\theta)|d\theta. $$