Bounds of the Scaled Kernels

Figure 12: The scaled kernel for $\epsilon =0.5$ and $N=1$, plotted as a function of $\theta$ over the support interval $[-\epsilon ,\epsilon ]$. The five invariant points are marked by stars, and the support interval of the next filter in the construction sequence is shown. The dashed lines mark the center and the two ends of the support interval, and hence three of the five invariant points. The dotted lines mark the sectors where the function is defined in a piece-wise fashion.

Let us observe that since the filtered function $f_{\epsilon}(\theta)$ is defined as an average of the function $f(\theta)$, it can never assume values which are larger than the maximum of the function it is applied on, or smaller than its minimum, without regard to the value of the range $\epsilon$. Therefore, since the first kernel we start with in the process of construction of the infinite-order scaled kernel, that is the kernel $\bar{K}_{\epsilon_{1}}^{(1)}(\theta)$, with $N=1$ and range $\epsilon/2$, is bound within the interval $[0,1/\epsilon]$ for all values of $\theta$, so is the next one, the kernel $\bar{K}_{\epsilon_{2}}^{(2)}(\theta)$. We may now apply the same argument to this second kernel, and conclude that the third one in the sequence is also bound in the same way, and so on. It follows that, for all values of $N$, we have

$$0 \leq \bar{K}_{\epsilon_{N}}^{(N)}(\theta) \leq \frac{1}{\epsilon},$$

for all values of $\theta$ within the periodic interval $[-\pi ,\pi ]$, and in particular for all values of $\theta$ within the support interval $[-\epsilon ,\epsilon ]$. It also follows that, if the $N\to \infty$ limit of the sequence of kernel functions exists, then it is also bound in the same way.