Action on the Fourier Coefficients with $k>0$

Let us determine the effect of the filter on the Fourier coefficients for $k>0$. We start with the Fourier coefficients of $f(x)$, which are given by

$$\begin{eqnarray*} \alpha_{k} & = & \frac{1}{\pi} \int_{-\pi}^{\pi}dx\, f(x)... ...k} & = & \frac{1}{\pi} \int_{-\pi}^{\pi}dx\, f(x) \sin(kx), \end{eqnarray*}$$

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

$$\begin{eqnarray*} \alpha_{\varepsilon,k} & = & \frac{1}{\pi} \int_{-\pi}^{\p... ...ac{1}{\pi} \int_{-\pi}^{\pi}dx\, f_{\varepsilon}(x) \sin(kx). \end{eqnarray*}$$

Let us work out only the first case, since the work for the second one in essentially identical. Using the definition of $f_{\varepsilon}(x)$ in terms of $f(x)$ we have

$$\begin{eqnarray*} \alpha_{\varepsilon,k} & = & \frac{1}{\pi} \int_{-\pi}^{\p... ...psilon\pi k} \int_{-\pi}^{\pi}dx\, \sin(kx) f(x-\varepsilon), \end{eqnarray*}$$

where we integrated by parts and where there is no integrated term due to the periodicity of the integrand on the domain. We now change variables in each integral, using $x'=x\pm\varepsilon$, in order to obtain

$$\begin{eqnarray*} \alpha_{\varepsilon,k} & = & -\, \frac{1}{2\varepsilon\pi ... ...ft[ \sin(kx'+k\varepsilon) - \sin(kx'-k\varepsilon) \right], \end{eqnarray*}$$

where the integration limits did not change in the transformations of variables due to the periodicity of the integrand on the domain. We are left with

$$\begin{eqnarray*} \alpha_{\varepsilon,k} & = & \frac{1}{2\varepsilon\pi k} \... ...t_{-\pi}^{\pi}dx'\, f\!\left(x'\right) \cos\!\left(kx'\right). \end{eqnarray*}$$

Since we recover in this way the expression of the Fourier coefficients $\alpha_{k}$ of $f(x)$, we get

$$\alpha_{\varepsilon,k} = \left[ \frac{\sin(k\varepsilon)}{(k\varepsilon)} \right] \alpha_{k},$$

and repeating the calculation for the other coefficients one gets

$$\beta_{\varepsilon,k} = \left[ \frac{\sin(k\varepsilon)}{(k\varepsilon)} \right] \beta_{k}.$$

Once again we see the sinc function of the variable $(k\varepsilon)$ appearing here. Since $\sin(k\varepsilon)$ is a limited function, in terms of the asymptotic behavior of the coefficients, for large values of $k$, the net effect of the filter is to add a factor of $k$ to the denominator.