= Solution
Use the central second difference $\Delta_t^2f(x)=f(x+t)-2f(x)+f(x-t)$ and define the <second modulus of smoothness> by
$$
\omega_2(f,\delta)=\sup_{|h|\le\delta}\|\Delta_h^2f\|_\infty.
$$
We first need its scaling inequality. If $q$ is a positive <integer>, translation by $h$ gives the telescoping identity
$$
\Delta_{qh}^2f(x)=\sum_{j=-q+1}^{q-1}(q-|j|)\Delta_h^2f(x+jh).
$$
The weights sum to $q^2$. For $t>0$, choose $q=\lceil t/\delta\rceil$ and $h=t/q\le\delta$; then
$$
\omega_2(f,t)\le\left(1+\frac t\delta\right)^2\omega_2(f,\delta).
$$
The same bound covers $t=0$ trivially.
In the paper's normalization, the <Fejér kernel> is even and nonnegative, with $\pi^{-1}\int_{-\pi}^{\pi}F_n=1$. Pairing the positive and negative translations expresses the <Fejér sum> error as
$$
\sigma_nf(x)-f(x)=\frac1{2\pi}\int_{-\pi}^{\pi}F_n(t)\Delta_t^2f(x)\,dt.
$$
Use the scaling bound and $(1+s)^2\le2(1+s^2)$ to obtain
$$
\|\sigma_nf-f\|_\infty
\le\omega_2(f,\delta)\left[1+\frac1{\pi\delta^2}
\int_{-\pi}^{\pi}t^2F_n(t)\,dt\right].
$$
For $|t|\le\pi$, $\sin(|t|/2)\ge|t|/\pi$. Bounding the numerator sine by one gives
$$
t^2F_n(t)\le\frac{\pi^2}{2n},\qquad
\frac1\pi\int_{-\pi}^{\pi}t^2F_n(t)\,dt\le\frac{\pi^2}{n}.
$$
The bound at $t=0$ holds by continuity. Choosing $\delta=n^{-1/2}$ proves the <Fejér second-modulus approximation bound> with an explicit universal constant:
$$
\boxed{\|\sigma_nf-f\|_\infty\le(1+\pi^2)\omega_2(f,n^{-1/2}).}
$$
Finally, for twice continuously differentiable periodic $f$, two applications of the <fundamental theorem of calculus> give the <second-difference integral formula>
$$
\Delta_h^2f(x)=\int_{-|h|}^{|h|}(|h|-|v|)f''(x+v)\,dv.
$$
The triangular weight has integral $h^2$, so $\omega_2(f,\delta)\le\delta^2\|f''\|_\infty$. Therefore
$$
\boxed{\|\sigma_nf-f\|_\infty\le\frac{(1+\pi^2)\|f''\|_\infty}{n}
=O(n^{-1}).}
$$
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