Prove a pointwise version of Fejer's theorem: If $f\in \mathscr{R}$ and $f(x+),f(x-)$ exist for some $x$, then $$\lim \limits_{N\to \infty}\sigma_N(f;x)=\frac{f(x+)+f(x-)}{2},$$ where $\sigma_N(f;x)=\frac{s_0+s_1+\cdots+s_N}{N+1}$.
Proof: Let $f(x-)=l_1$ and $f(x+)=l_2$. Let $\varepsilon>0$ be given then: $$\left|\sigma_N(f;x)-\frac{l_1+l_2}{2}\right|=\left|\frac{1}{2\pi}\int \limits_{-\pi}^{\pi}\left[f(x-t)-\frac{l_1+l_2}{2}\right]K_N(t)dt\right|$$ since $\frac{1}{2\pi}\int_{-\pi}^{\pi}K_N(t)dt=1$ and $\sigma_N(f;x)$ is convolution of $f$ and $K_N(t)$ - Fejer's kernel. We can wrote $\int_{-\pi}^{\pi}=\int_{-\pi}^{0}+\int_{0}^{\pi}$ and making transform $t\mapsto -t$ in first integral we get: $$\frac{1}{2\pi}\left| \int \limits_{0}^{\pi}K_N(t)\left[f(x+t)+f(x-t)-l_1-l_2\right]dt\right|\leq$$$$\leq\frac{1}{2\pi} \int \limits_{0}^{\pi}K_N(t)\left|f(x+t)+f(x-t)-l_1-l_2\right|dt.$$ Since $f(x-)=l_1$ and $f(x+)=l_2$ then $\exists \delta>0$ such that for any $t\in (0,\delta)$ such that $|f(x-t)-l_1|<\varepsilon/2$ and $|f(x+t)-l_2|<\varepsilon/2$. Hence by triangle inequality $|f(x+t)+f(x-t)-l_1-l_2|<\epsilon$ for $t\in (0,\delta)$.
Also on $[\delta,\pi]$ where $\delta>0$ we have $K_N(t)\leqslant \frac{2}{N+1}\cdot \frac{1}{1-\cos \delta}$. Hence $$\int\limits_{0}^{\delta}K_N(t)\left|f(x+t)+f(x-t)-l_1-l_2\right|dt<\epsilon \int\limits_{0}^{\delta}K_N(t)dt\leq \frac{2\epsilon \pi}{2}$$ since $\frac{1}{2\pi}\int_{-\pi}^{\pi}K_N(t)dt=1$ and $$\int\limits_{\delta}^{\pi}K_N(t)\left|f(x+t)+f(x-t)-l_1-l_2\right|dt\le (2M+|l_1-l_2|)\dfrac{2(\pi-\delta)}{(N+1)(1-\cos \delta)}$$ where $M=\sup |f|$. Then $$\left|\sigma_N(f;x)-\frac{l_1+l_2}{2}\right|<\frac{\varepsilon}{2}+\frac{K}{(N+1)(1-cos \delta)}$$ and the second term would be $\leq \epsilon/{2}$ for large $N$.
Is my proof correct? Can anyone check my proof please.
I would be thankful for help!