Periodicity of Filtered Functions

Let us show that if $f(x)$ is a periodic function, with a period that we choose arbitrarily to be $2\pi$, then $f_{\varepsilon}(x)$ is also periodic, with the same period. It suffices to simply calculate

$$\begin{eqnarray*} f_{\varepsilon}(x+2\pi) & = & \frac{1}{2\varepsilon} \int_... ...{x-\varepsilon}^{x+\varepsilon}dx''\, f\!\left(x''+2\pi\right), \end{eqnarray*}$$

where we changed variables to $x''=x'-2\pi$, so that $x'=x''+2\pi$. Since $f(x)$ is periodic with period $2\pi$, we now have

$$\begin{eqnarray*} f_{\varepsilon}(x+2\pi) & = & \frac{1}{2\varepsilon} \int_... ...lon}dx''\, f\!\left(x''\right) \\ & = & f_{\varepsilon}(x), \end{eqnarray*}$$

so that we may conclude that $f_{\varepsilon}(x)$ is periodic with period $2\pi$.