Non-Analyticity of the Infinite-Order Scaled Kernel

From the construction described in the previous section for the order-$n$ derivatives of the infinite-order scaled kernel, which can all be written in terms of the kernel itself, and from the fact that the infinite-order scaled kernel is zero at the two ends of its support interval, it follows at once that all the order-$n$ derivatives of the kernel are also zero at these two points. Therefore the kernel and all its order-$n$ derivatives, for all $n\geq 0$, are zero at the two ends $\theta=\pm\epsilon$ of the support interval, as in fact we already knew, since this is also a consequence of the fact that the kernel is a $C^{\infty}$ function over its whole domain.

In a similar way, we may also determine other points within the support interval where almost all the order-$n$ derivatives of the kernel are zero. For example, at the central point, although the kernel itself is not zero, we see that its first derivative is, and in fact iterating the construction one can see that all the higher-order derivatives are zero there. Therefore all the order-$n$ derivatives of the kernel, for all $n\geq 1$, are zero at the central point $\theta=0$ of the support interval. An examination of the situation at the two inflection points $\theta=\pm\epsilon/2$ reveals that at those two points all the order-$n$ derivatives of the kernel, for all $n\geq 2$, are zero.

Table 1: Points where almost all derivatives of the infinite-order scaled kernel are zero. The integer $n$ is the order of the first null derivative, and $m$ gives the orders of all derivatives which are zero at the corresponding set of points.

Order	Number of points	Null Derivatives
0	$2$	$m\geq 0$
1	$3$	$m\geq 1$
2	$5$	$m\geq 2$
3	$9$	$m\geq 3$
4	$17$	$m\geq 4$
n	$2^{n}+1$	$m\geq n$

The iteration of this process of analysis can be continued indefinitely, with the result that there are sets of increasing numbers of points regularly spaced in the support interval where all derivatives above a certain order are zero. This can be systematized as shown in Table 1. We see therefore that there is a set of $2^{n}+1$ points regularly spaced within the support interval where all derivatives with order $n$ or larger are zero. In the $n\to\infty$ limit this set of points tends to be densely distributed within the support interval. Outside the support interval the derivatives of all orders are zero at all points, of course, since the kernel is identically zero there.

In we assemble the Taylor series of the kernel function around one of the points where all the derivatives of order $n$ and larger are zero, we obtain a convergent power series, which is in fact a polynomial of order $n-1$. Since the kernel function is obviously not such a polynomial, it is therefore not represented by its convergent Taylor series around this reference point, at any points other than the reference point itself. Since in order to be analytic the kernel function would have to be so represented within an open set around the reference point, it follows that it is not analytic at any of these points. Since this set of points tends to become densely distributed within the support interval, we may conclude that the kernel function is not analytic at all points of the support interval.

One can try to extend this argument to show in a somewhat heuristic way that the kernel function cannot be represented by a convergent power series around any point of the support interval, whether or not it is in the dense subset. Let us consider a point where the kernel function has non-zero derivatives of arbitrarily high orders, and where the Taylor series built from them converges in an open neighborhood of that reference point. This implies that the point at issue is not in the dense subset. Note that since the kernel function is $C^{\infty}$ we know that all its derivatives at the point exist, whether or not they are zero. Since the subset of points discussed above is dense in the support interval, there is at least one point of the dense subset within this open neighborhood. Therefore there is another Taylor series around this point, which is also convergent.

Since both are Taylor series of the same function and converge in a common domain, we must be able to transform each one into the other by a transformation of coordinates that is a simple shift of the argument of the series. Note that all the derivatives of the kernel function, of all orders, are themselves continuous and differentiable functions. However, no such transformation of variables can transform the second series, which is a polynomial of finite order, into a series such as the first one, with no upper bound to the powers present in it. This seems to produce an absurd situation. Therefore, one is led to think that either the first series cannot be a convergent series, or it must converge to some function other than the kernel function. In any case, it follows that the kernel function is not represented by this Taylor series either, and once again that it cannot be analytic at the point under discussion.

One may wonder about whether the real infinite-order kernel function can be extended analytically to the complex plane. It is clear that this cannot be done in the usual way, with the simple exchange of its argument by a complex variable. In addition to this, we know that it can be obtained as the limit to the unit circle of an inner analytic function, and that the inner analytic function has a densely distributed set of singularities on the support of the kernel. It is therefore reasonable to think that this is not possible, but no complete proof of this is currently available.