Differentiability of the Infinite-Order Scaled Kernel

The infinite-order scaled kernel with finite range $\epsilon$ has an interesting property of its own, namely that there is a certain similarity between the kernel and its derivatives. Every finite-order derivative of the infinite-order scaled kernel function is made out of a certain number of rescaled copies of the kernel itself, concatenated together. This is a consequence of the fact that there is a certain relation between the first derivative of the order-$N$ scaled kernel and the order-$(N-1)$ scaled kernel. After this relation is established it can be iterated, resulting in similar relations for the higher-order derivatives. This property allows one to establish the existence of the $N\to \infty$ limits of all the finite-order derivatives, and thus to prove that the infinite-order scaled kernel is differentiable to all orders.

One can derive the relation between the first derivative of the order-$N$ scaled kernel and the order-$(N-1)$ scaled kernel as follows. If we start with the Fourier expansion of the order-$N$ scaled kernel, written in the form

$$\bar{K}_{\epsilon_{N}}^{(N)}(\theta) = \frac{1}{2\pi} + ... ...ght)} {\left(k\epsilon/2^{n}\right)} \right] \cos(k\theta),$$

we may differentiate once term-by-term and thus obtain

$$\begin{eqnarray*} \frac{d}{d\theta} \bar{K}_{\epsilon_{N}}^{(N)}(\theta) & = ... ...ft[ -2 \sin\!\left(k\epsilon/2\right) \sin(k\theta) \right], \end{eqnarray*}$$

noting that for sufficiently large $N$ all the series involved are absolutely and uniformly convergent. By means of simple trigonometric identities the product of two sines within brackets can now be written as

$$-2 \sin\!\left(k\epsilon/2\right) \sin(k\theta) = \cos\... ...right] - \cos\!\left[k\left(\theta-\epsilon/2\right)\right],$$

so that we have for the derivative of the kernel

$$\begin{eqnarray*} \frac{d}{d\theta} \bar{K}_{\epsilon_{N}}^{(N)}(\theta) & = ... ...)} \right] \cos\!\left[k\left(\theta-\epsilon/2\right)\right]. \end{eqnarray*}$$

Figure 15: The infinite-order scaled kernel (solid line) and its first derivative (dashed line), obtained via the use of their Fourier series, for $\epsilon =0.5$, for a large value of $N$ ($100$), plotted as functions of $\theta$ over the support interval $[-\epsilon ,\epsilon ]$. The derivative was rescaled down to have the same amplitude as the kernel. The dotted lines mark the points where the first derivative is zero.

If we now define $\epsilon'=\epsilon/2$, we may write

$$\begin{eqnarray*} \frac{d}{d\theta} \bar{K}_{\epsilon_{N}}^{(N)}(\theta) & = ... ...t)} \right] \cos\!\left[k\left(\theta-\epsilon'\right)\right], \end{eqnarray*}$$

where we made $n'=n-1$, which implies $n=n'+1$. We see therefore that in this way we recover in the right-hand side the expression of the scaled kernel of order $N-1$ with range $\epsilon/2$, so that we have, writing $\epsilon'$ back in terms of $\epsilon$,

$$\frac{d}{d\theta} \bar{K}_{\epsilon_{N}}^{(N)}(\theta) = ... ...n_{(N-1)}/2}^{(N-1)}\!\left(\theta-\epsilon/2\right) \right].$$

Figure 16: The first (solid line) and second (dashed line) derivatives of the infinite-order scaled kernel, obtained via the use of their Fourier series, for $\epsilon =0.5$, for a large value of $N$ ($100$), plotted as functions of $\theta$ over the support interval $[-\epsilon ,\epsilon ]$. Both derivatives were rescaled down to have the same amplitude as the kernel. The dotted lines mark the points where the second derivative is zero.

Since we already know that the $N\to \infty$ limit of the right-hand side exists, this establishes that the $N\to \infty$ limit of the left-hand exists as well. Taking the $N\to \infty$ limit we end up with the infinite-order scaled kernel on both sides, so that we have the relation

$$\frac{d}{d\theta} \bar{K}_{\epsilon}^{\infty}(\theta) = ... ...\epsilon/2}^{\infty}\!\left(\theta-\epsilon/2\right) \right].$$

Figure 17: The second (solid line) and third (dashed line) derivatives of the infinite-order scaled kernel, obtained via the use of their Fourier series, for $\epsilon =0.5$, for a large value of $N$ ($100$), plotted as functions of $\theta$ over the support interval $[-\epsilon ,\epsilon ]$. Both derivatives were rescaled down to have the same amplitude as the kernel. The dotted lines mark the points where the third derivative is zero.

It follows therefore, as expected, that the infinite-order scaled kernel function is differentiable. Note that, since the kernels on the right-hand side have range $\epsilon/2$, and their points of application are distant from each other by exactly $\epsilon$, each one is just outside the support of the other. Therefore, the derivative is given by the concatenation of two graphs just like the kernel itself, but with the support scaled down from $\epsilon$ to $\epsilon/2$, with the amplitude scaled up by the factor $2/\epsilon$, and with the sign of one of them inverted. This is shown in Figure 15, containing a superposition of the kernel and its rescaled first derivative.

Figure 18: The third (solid line) and fourth (dashed line) derivatives of the infinite-order scaled kernel, obtained via the use of their Fourier series, for $\epsilon =0.5$, for a large value of $N$ ($100$), plotted as functions of $\theta$ over the support interval $[-\epsilon ,\epsilon ]$. Both derivatives were rescaled down to have the same amplitude as the kernel. The dotted lines mark the points where the fourth derivative is zero.

One may now take one more derivative of the expression for the derivative of the order-$N$ scaled kernel, thus obtaining an expression for the corresponding second derivative. After that one may use again the relation for the first derivative, thus iterating that relation, in order to obtain

$$\begin{eqnarray*} \frac{d^{2}}{d\theta^{2}} \bar{K}_{\epsilon_{N}}^{(N)}(\thet... ...on_{(N-2)}/4}^{(N-2)}\!\left(\theta-3\epsilon/4\right) \right]. \end{eqnarray*}$$

Once more the known existence of the $N\to \infty$ limit of the right-hand side establishes the existence of the $N\to \infty$ limit of the left-hand side, and therefore that the infinite-order scaled kernel is twice differentiable. Taking the limit on both sides we get

$$\begin{eqnarray*} \frac{d^{2}}{d\theta^{2}} \bar{K}_{\epsilon}^{\infty}(\theta... ... + \bar{K}_{\epsilon/4}^{\infty}(\theta-3\epsilon/4) \right]. \end{eqnarray*}$$

We now have four copies of the graph of the kernel, with range scaled down to $\epsilon/4$ and amplitude scaled up by $2/\epsilon^{2}$, each one outside the supports of the others, distributed in a regular way within the interval of length $2\epsilon$ around $\theta$. This is shown in Figure 16, containing a superposition of the rescaled first and second derivatives of the kernel. As one can see in the subsequent Figures 17 and 18, the same type of relationship is also true for all the higher-order derivatives. This is so because we can iterate this relation indefinitely, so that any finite-order derivative of $\bar{K}_{\epsilon}^{\infty}(\theta)$ can be written as a finite linear combination of $\bar{K}_{\epsilon}^{\infty}(\theta)$ itself, with a rescaled $\epsilon$ and a rescaled amplitude. Observe that this constitutes independent proof that the infinite-order scaled kernel is a $C^{\infty}$ function.