Cesàro mean diverges Suppose $f$ is an integrable $2\pi$-periodic function.
Show that if $\lim\limits_{h\to 0} f(x_0+h)+f(x_0-h)=\infty$ then $\sigma_n f(x_0)=\infty$, where $\sigma_n f$ represents the Cesàro mean, the average of the first $n+1$ Fourier series
My thoughts: if $\lim\limits_{h\to 0} f(x_0+h)+f(x_0-h)=\infty$ then does it imply $f(x_0)=\infty$?
Any tips on this question?
 A: Recall that $\sigma_n (f) (x_0)=\int_{-\pi}^{\pi} f(x_0-y)K_n(y) dy$ where $$K_n (s) = \begin{cases} \frac{1}{n+1} \left(\frac{\sin \left(\frac{n+1}{2} s\right) }{\sin \left(\frac{s}{2}\right)}\right)^2 & \text{ if } s\neq 0 \\ n+1 & \text{ if } s=0.\end{cases}$$
$K_n$ has several nice properties: 1) $K_n(s)\geq 0$ for $s\in [-\pi,\pi]$, 2) For any $\delta>0$, $K_n\to 0$ uniformly on $[-\pi,\pi]-[-\delta,\delta]$, 3) $\frac{1}{2\pi} \int_{-\pi}^{\pi} K_n(s) \, ds =1$, and 4) $K_n(-s)=K_n(s)$. 
With these four properties and splitting the formula for $\sigma_n (f)(x_0)$ into an integral over $[-\pi,\pi] -[-h,h]$ and an integral over $[-h,h]$, playing with the integral over $[-h,h]$ should lead to your desired result.
**all of these results except property 4 were lifted directly from Koerner's Fourier Analysis.
ADDED: So using property 4) and my suggestion, we have that
$$\begin{align*}
\sigma_n(f)(x_0) = \left [ \int_{-\pi}^{-h} f(x_0-y) K_n(y) dy + \int_{h}^{\pi} f(x_0-y) K_n(y) dy \right ] + \\ \int_{0}^{h} f(x_0-y) K_n(y) dy+\int_{-h}^{0} f(x_0-y)K_n(y) dy 
\end{align*}$$
Call the stuff in '[...]' $\epsilon_{h,n}$. Changing the variable z=-y in the last integral and invoking property 4) allows us to see that
$$
\sigma_n(f)(x_0) = \epsilon_{h,n} +\int_{0}^{h} \left(f(x_0-y)+f(x_0+y)\right) K_n(y) dy.
$$
Now we have two choices to make: $h$ sufficiently small that $f(x_0-y)+f(x_0+y)$ is big on $[0,h]$ and $n$ sufficiently large such that $K_n(y)$ is sufficiently concentrated on $[-h,h]$. The properties of $K_n$ and integrability of $f$ allow for our chooses to preform double duty to control $\epsilon_{h,h}$, in fact $|\epsilon_{h,n}| \leq \|K_n\|_{L^{\infty}([\pi-h,\pi])} \|f\|_{L^1([-\pi,\pi])}  $
