Fejer means converge uniformly for continuous periodic functions

Let $\epsilon >0$. Because $f$ is continuous on the compact interval $[0,1]$ and one-periodic, it is uniformly continuous modulo $1$. Choose $\delta \in (0,1\mathrm{/}2]$ such that $\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ whenever $x\in \mathbb{R}$ and $t\in [0,\delta ]\cup [1-\delta ,1]$.

For every $x\in \mathbb{R}$, subtract $f(x)$ inside the integral from [L1]: $\mathrm{\mid }{\sigma }_{N}f(x)-f(x)\mathrm{\mid }\le {\int }_0^1\mathrm{\mid }f(x-t)-f(x)\mathrm{\mid }{F}_N(t)dt\mathrm{.}$ Split the integral into the near set $[0,\delta ]\cup [1-\delta ,1]$ and the far set $[\delta ,1-\delta ]$. By step 1.1 and the positivity from [L2], the near part is at most $\epsilon \mathrm{/}2$. The far part is at most $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\delta }^{1-\delta }{F}_N(t)dt\mathrm{.}$

By [L2], choose $N_0$ so large that $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\delta }^{1-\delta }{F}_N(t)dt<\epsilon \mathrm{/}2$ for all $N\ge N_0$. Then step 2.1 gives ${\sup }_{x\in \mathbb{R}}\mathrm{\mid }{\sigma }_{N}f(x)-f(x)\mathrm{\mid }<\epsilon (N\ge N_0)\mathrm{.}$ Since $\epsilon$ was arbitrary, the convergence is uniform.