real analysis - Proving that the Cesaro means of Fourier sums for continuous $f$ converge uniformly to $f$

Let $f$ be a $1$-periodic and continuous function on $\mathbb{R}$. I would like someone to verify that the following proof is correct. I prove that $\|\sigma_n(f)-f\|_\infty\to0$, where $\sigma_n(f)$ denotes the $n$-th Cesaro mean of the partial sums of the Fourier series of $f$. That being said, $\sigma_n(f)=f*K_n$, where $K_n$ is Fejer's kernel: $K_n(x)=\displaystyle\frac{1}{n+1}\biggl(\frac{\sin((n+1)\pi x)}{\sin(\pi x)}\biggr)^2$ (for $x\not\in\mathbb{Z}$; for $x\in\mathbb{Z}$ it is $K_n(x)=n+1$).

Now a popular and sharp estimate of Fejer's kernel is the following: for all $x\in[-1/2, 1/2]$ it is $K_n(x)\leq\Phi_{\frac{1}{n+1}}(x)$, where $\Phi_{\frac{1}{n+1}}(x)$ is a piecewise function: For $|x|<\frac{1}{2(n+1)}$ it is $\Phi_{\frac{1}{n+1}}(x)=n+1$ and for $\frac{1}{2(n+1)}\leq|x|<\frac{1}{2}$ it is $\Phi_\frac{1}{n+1}(x)=\frac{1}{4(n+1)x^2}$.

I am not going to prove the estimate; as I said, it's popular.

Now for our goal: It is $\|\sigma_n(f)-f\|_\infty\leq\displaystyle\sup_{x\in[-\frac{1}{2},\frac{1}{2})}|\sigma_n(f)(x)-f(x)|$. As proved here, since $f$ is $1$-periodic and continuous it is uniformly continuous; Let $\varepsilon>0$ and $\delta>0$ be the corresponding quantity. Let $x\in[-\frac{1}{2},\frac{1}{2})$. We have $|\sigma_n(f)(x)-f(x)|=|(f*K_n)(x)-f(x)|=|\int_{(-1/2,1/2)}f(x-y)K_n(y)dy-\int_{(-1/2,1/2)}f(x)K_n(y)dy|\leq\displaystyle{\int_{(-1/2,1/2)}|f(x-y)-f(x)|K_n(y)dy}\;\;(1)$

Now we can break the integral on RHS in two parts: The first, $I_1$, being the integral over $(-\delta,\delta)$ and the second, $I_2$, the integral over the rest of $(-1/2,1/2)$.

But $I_1=\displaystyle{\int_{(-\delta,\delta)}|f(x-y)-f(x)|K_n(y)dy\leq\varepsilon\int_{(-\delta,\delta)}K_n(y)dy\leq\varepsilon}$ (we used the fact that $\int_0^1K_n(y)dy=1$)

As for $I_2$, for $n$'s large enough so that $\frac{1}{2(n+1)}<\delta$, we have that $I_2=2\displaystyle{\int_{(\delta,\frac{1}{2})}|f(x-y)-f(x)|K_n(y)dy\leq4M\int_{(\delta,\frac{1}{2})}K_n(y)dy\leq 4M\int_\delta^\frac{1}{2}\frac{1}{4(n+1)y^2}dy=\frac{C_\varepsilon}{(n+1)}},$ a quantity that also gets arbitarily small as we increase $n$.

Everything is independent of $x$; therefore the $\infty$-norm gets small as $n$ gets large.

Is there something wrong with my proof?