Invariance of Averages Over the Period

Let us determine the effect of the filter on the Fourier coefficient $\alpha_{0}$. We start with the coefficient of $f(x)$, which is given by

$$\alpha_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi}dx\, f(x),$$

where we arbitrarily chose $[-\pi ,\pi ]$ as the periodic interval. The Fourier coefficient $\alpha_{\varepsilon,0}$ of $f_{\varepsilon}(x)$ is similarly given by

$$\begin{eqnarray*} \alpha_{\varepsilon,0} & = & \frac{1}{\pi} \int_{-\pi}^{\p... ... \int_{x-\varepsilon}^{x+\varepsilon}dx'\, f\!\left(x'\right), \end{eqnarray*}$$

where we used the definition of $f_{\varepsilon}(x)$ in terms of $f(x)$. We now make the change of variables $x'=x''+x$ on the inner integral, implying $dx'=dx''$ and $x''=x'-x$, leading to

$$\begin{eqnarray*} \alpha_{\varepsilon,0} & = & \frac{1}{2\varepsilon\pi} \in... ...repsilon}dx''\, \frac{1}{\pi} \int_{-\pi}^{\pi}dx\, f(x''+x). \end{eqnarray*}$$

Since the integral on $x$ is over the whole period, we may now change variables on it without changing the integration limits, using $x=x'-x''$, with $x'=x''+x$ and $dx=dx'$, and thus obtaining

$$\begin{eqnarray*} \alpha_{\varepsilon,0} & = & \frac{1}{2\varepsilon} \int_{... ...psilon}^{\varepsilon}dx''\, \alpha_{0} \\ & = & \alpha_{0}, \end{eqnarray*}$$

were we recognized the form of $\alpha_{0}$. In this way we show that $\alpha_{\varepsilon,0}=\alpha_{0}$, that is, the filter does not change $\alpha_{0}$ at all. Another way to state this is to say that the filter does not change the average value of $f(x)$ over the periodic interval.