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


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

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.