= Solution
Use the <modulus of continuity> on the circle, $\omega(f,\delta)=\sup_{|h|\le\delta}\|f(\cdot+h)-f\|_\infty$. Splitting a displacement into steps of length at most $\delta$ and applying the <triangle inequality> gives
$$
\omega(f,t)\le\left(1+\frac t\delta\right)\omega(f,\delta),\qquad t,\delta>0.
$$
The <Fejér kernel> is nonnegative and has <integral> one, so
$$
\|\sigma_nf-f\|_\infty
\le\int_{-\pi}^{\pi}\omega(f,|t|)F_n(t)\,dt
\le\omega(f,\delta)\left(1+\frac1\delta\int_{-\pi}^{\pi}|t|F_n(t)\,dt\right).
$$
We therefore need its first absolute moment. The inequalities $|\sin(nt/2)|\le n|\sin(t/2)|$ and $\sin(|t|/2)\ge|t|/\pi$ on $[-\pi,\pi]$ imply
$$
0\le F_n(t)\le\min\left\{\frac{n}{2\pi},\frac{\pi}{2nt^2}\right\}.
$$
Splitting at $1/n$ gives
$$
\int_{-\pi}^{\pi}|t|F_n(t)\,dt
\le2\int_0^{1/n}t\frac{n}{2\pi}\,dt
+2\int_{1/n}^{\pi}t\frac{\pi}{2nt^2}\,dt
=\frac1{2\pi n}+\frac\pi n\ln(\pi n)
\le C\frac{\ln n}{n}\quad(n\ge2).
$$
Taking $\delta=\delta_n=(\ln n)/n$ in the preceding error bound proves the <Fejér first-moment approximation bound>:
$$
\boxed{\|\sigma_nf-f\|_\infty\le c\,\omega\left(f,\frac{\ln n}{n}\right),\qquad n\ge2.}
$$
The constants do not depend on $f$ or $n$. By the <Heine-Cantor theorem>, a <continuous function> on the compact circle is <uniformly continuous>, hence $\omega(f,\delta_n)\to0$ as $\delta_n\to0$. \b[Thus the Fejér sums converge uniformly to $f$.] The index restriction is necessary: at $n=1$ the printed scale is zero, while $\sigma_1f$ is the mean of $f$ and generally differs from it.
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