= Solution
{c}
Write the <Fourier coefficients> as $c_k=(2\pi)^{-1}\int_{-\pi}^{\pi}u(t)e^{-ikt}dt$. The partial sum through frequencies $-n,\ldots,n$ is therefore a <convolution> with
$$
D_n(\theta)=\sum_{k=-n}^ne^{ik\theta}
=\frac{\sin((n+1/2)\theta)}{\sin(\theta/2)}.
$$
The last expression follows by summing a finite <geometric series>, with its removable value $2n+1$ at zero. Periodicity and the evenness of $D_n$ give
$$
\boxed{s_n(x)=\frac1{2\pi}\int_{-\pi}^{\pi}
u(x+\theta)D_n(\theta)d\theta.}
$$
This also proves linearity and that the image lies in $F_n$. We use the continuous <periodic function> space, consistent with the premise and the displayed maximum <norm>.
The <convolution> formula gives the upper bound
$$
\|S_nu\|_\infty\leq\frac{\|u\|_\infty}{2\pi}
\int_{-\pi}^{\pi}|D_n(\theta)|d\theta.
$$
To prove sharpness within continuous functions, take
$$
u_\epsilon(\theta)=\frac{D_n(\theta)}
{\sqrt{D_n(\theta)^2+\epsilon^2}}.
$$
It is continuous and periodic, with <norm> at most one. At $x=0$, dominated convergence gives $(S_nu_\epsilon)(0)\to(2\pi)^{-1}\int|D_n|$. Hence that upper bound is the actual <operator norm>. Evenness yields
$$
\boxed{\|S_n\|=\frac1\pi\int_0^\pi
\left|\frac{\sin((n+1/2)\theta)}{\sin(\theta/2)}\right|d\theta.}
$$
The approximating continuous signs avoid assuming that the discontinuous sign of the kernel belongs to the domain space.
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