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Use the central second difference ${\Delta }_t^{2}f(x)=f(x+t)-2f(x)+f(x-t)$ and define the second modulus of smoothness byWe first need its scaling inequality. If $q$ is a positive integer, translation by $h$ gives the telescoping identityThe weights sum to $q^2$. For $t>0$, choose $q=\lceil t/\delta \rceil$ and $h=t/q\le \delta$; thenThe 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 asUse the scaling bound and $(1+s{)}^2\le 2(1+s^2)$ to obtainFor $|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$. Bounding the numerator sine by one givesThe bound at $t=0$ holds by continuity. Choosing $\delta =n^{-1/2}$ proves the Fejér second-modulus approximation bound with an explicit universal constant:Finally, for twice continuously differentiable periodic $f$, two applications of the fundamental theorem of calculus give the second-difference integral formulaThe triangular weight has integral $h^2$, so ${\omega }_2(f,\delta )\le {\delta }^2\Vert f^{\prime \prime }{\Vert }_{\infty }$. Therefore

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 $\Vert {\sigma }_{n}f-f{\Vert }_{\infty }\le (1+{\pi }^2)\Vert f^{\prime \prime }{\Vert }_{\infty }/n=O(n^{-1})$.