Completeness of the Set of Eigenfunctions

Let us show that, up to the degeneracy between the pairs of elements of the basis with the same $k$, the elements of the Fourier basis are the only eigenfunctions of the filter operator, when it is defined within the periodic interval. In order to do this, let us first point out that two eigenvalues, for two different values of $k$, are never equal. We can see this assuming that there are positive values $k$ and $k'$ such that

$$\frac{\sin(k\varepsilon)}{(k\varepsilon)} = \frac{\sin\!\left(k'\varepsilon\right)}{\left(k'\varepsilon\right)}.$$

Since this must stay valid for small changes of $\varepsilon$, we may differentiate with respect to $\varepsilon$ and thus obtain

$$\cos(k\varepsilon) = \cos\!\left(k'\varepsilon\right).$$

This now implies that

$$\sin(k\varepsilon) = \pm\sin\!\left(k'\varepsilon\right).$$

Since $\varepsilon>0$ and both $k$ and $k'$ are positive, we must have

$$\sin(k\varepsilon) = \sin\!\left(k'\varepsilon\right).$$

Since both the cosines and the sines of the two arguments are thus seen to be equal, it follows that the two arguments must be equal, and hence that we must have $k=k'$. Therefore, the eigenvalues for two different values of $k$ are never equal. Let us consider now an arbitrary function $f(x)$ and its expression in the Fourier basis,

$$f(x) = \frac{1}{2}\, \alpha_{0} + \sum_{k=1}^{\infty} \left[ \alpha_{k}\cos(kx) + \beta_{k}\sin(kx) \right],$$

Let us assume that this function is not identically zero, and that it is an eigenfunction of the first-order filter operator, that is

$$\frac{1}{2\varepsilon} \int_{x-\varepsilon}^{x+\varepsilon}dx'\, f\!\left(x'\right) = \lambda f(x),$$

for some real number $\lambda$. We are therefore assuming that it is a normalizable function that is an eigenfunction of the filter operator. Using the expression of the function on the Fourier basis, which is complete to represent almost everywhere any integrable real function on the periodic interval, we get

$$\begin{eqnarray*} \lefteqn { \frac{1}{2\varepsilon} \int_{x-\varepsilon}^{x+... ...{k}\cos(kx) + \sum_{k=1}^{\infty} \lambda \beta_{k}\sin(kx). \end{eqnarray*}$$

Passing all terms to the same side we may write this as the expansion of a certain function in the Fourier basis,

$$\frac{1}{2}\, (1-\lambda) \alpha_{0} + \sum_{k=1}^{\inf... ...(k\varepsilon)} - \lambda \right] \beta_{k}\sin(kx) = 0.$$

This is the expansion of the null function in the Fourier basis, which is unique and therefore implies that all coefficients must be zero. We have therefore

$$\begin{eqnarray*} (1-\lambda) \alpha_{0} & = & 0, \\ \left[ \frac{\sin(k... ...lon)}{(k\varepsilon)} - \lambda \right] \beta_{k} & = & 0, \end{eqnarray*}$$

the last two for all $k>0$. Taking first the case $k=0$, if $\lambda=1$ we may have $\alpha_{0}\neq 0$, but since the other eigenvalues are never equal to $1$, we must have then $\alpha_{k}=0$ and $\beta_{k}=0$ for all $k>0$. On the other hand, if $\lambda$ is equal to one of the eigenvalues with $k>0$, then it is different from all the other eigenvalues, since the eigenvalues for two values of $k$ are never equal. In this case we may have $\alpha_{k}\neq 0$ and $\beta_{k}\neq 0$ for one value of $k$, but all the other coefficients must be zero. Therefore the function $f(x)$ must be either a constant function or a linear combination of $\cos(kx)$ and $\sin(kx)$ for a single value of $k$.