 Solution

Use the centered second modulus of smoothnessThe relevant modulus properties are ${\omega }_2(f,t)\le (1+t/\delta {)}^2{\omega }_2(f,\delta )$ for $t\ge 0$, $\delta >0$, and ${\omega }_2(f,\delta )\le {\delta }^2\Vert f^{\prime \prime }{\Vert }_{\infty }$ when $f^{\prime \prime }$ is continuous. The latter follows from the second-difference integral formulaThe Fejér kernel is even, nonnegative and has integral one. Averaging the convolution at $t$ and $-t$ therefore givesApply the second modulus of smoothness bound and $(1+s{)}^2\le 2(1+s^2)$:For $|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$, while ${\sin }^2(nt/2)\le 1$. ThusThe removable value at $t=0$ causes no problem. With $\delta =n^{-1/2}$, the Fejér second-modulus approximation bound isIf $f^{\prime \prime }$ is continuous, this also yields $\Vert {\sigma }_{n}f-f{\Vert }_{\infty }\le (1+{\pi }^2)\Vert f^{\prime \prime }{\Vert }_{\infty }/n=O(n^{-1})$.