Infinitely Soft Singularities

Consider the following power series, which has coefficients that converge monotonically to zero from positive values,

$$S_{z} = \sum_{k=1}^{\infty} \frac{1}{\,{\rm e}^{\sqrt{k}}}\, z^{k}.$$

If we apply the ratio test to it, we get

$$\begin{eqnarray*} R & = & \frac{\,{\rm e}^{\sqrt{k}}\vert z\vert^{k+1}}{\,{\r... ...ghtarrow \\ \ln(R) & = & \ln(\rho) + \sqrt{k}-\sqrt{k+1}, \end{eqnarray*}$$

where $\vert z\vert=\rho$. In the large-$k$ limit we have

$$\begin{eqnarray*} \lim_{k\to\infty} \ln(R) & = & \ln(\rho) + \lim_{k\to\in... ... \lim_{k\to\infty} \frac{1}{2\sqrt{k}} \\ & = & \ln(\rho). \end{eqnarray*}$$

It follows that in the limit we have $R=\rho$ and therefore the conditions of the test are satisfied if and only if $\rho<1$. This establishes the open unit disk as the maximum disk of convergence of the series. Within this disk the series converges to an inner analytic function $w(z)$, and we may write

$$w(z) = \sum_{k=1}^{\infty} \frac{1}{\,{\rm e}^{\sqrt{k}}}\, z^{k}.$$

Since the maximum disk of convergence of the series is the open unit disk, this function must have at least one singularity on the unit circle. Note that since the series is monotonic with step $1$, we already know that $w(z)$ has a single dominant singularity on that circle, located at $z=1$. Consider now the logarithmic derivatives of this function. Using the series we have within the open unit disk, for the $n^{\rm th}$ logarithmic derivative of $w(z)$,

$$w^{n\!\mbox{\Large$\cdot$}\!}(z) = \sum_{k=1}^{\infty} \frac{k^{n}}{\,{\rm e}^{\sqrt{k}}}\, z^{k}.$$

This notation includes the original series as the case $n=0$. All these series converge on the open unit disk, of course. Let us now consider the corresponding series of absolute values, for $\rho=1$,

$$\overline{S}^{\,n\!\mbox{\Large$\cdot$}\!}_{v} = \sum_{k=1}^{\infty} \frac{k^{n}}{\,{\rm e}^{\sqrt{k}}}.$$

The terms of this sum can be bounded, for $k>k_{m}$ and some minimum value $k_{m}$ of $k$, by the function $1/k^{2}$, since we have that it is always possible to find a value $\xi_{m}$ of a variable $\xi$ such that for $\xi>\xi_{m}$ we have

$$\,{\rm e}^{-\xi} < \frac{1}{\xi^{2n+4}},$$

since the exponential goes to zero faster than any inverse power. Making $\xi^{2}=k$ and therefore $\xi=\sqrt{k}$ we have

$$\begin{eqnarray*} \,{\rm e}^{-\sqrt{k}} & < & \frac{1}{k^{n+2}} \;\;\;\Right... ...\\ \frac{k^{n}}{\,{\rm e}^{\sqrt{k}}} & < & \frac{1}{k^{2}}, \end{eqnarray*}$$

thus proving the assertion. This implies that all these series are absolutely and uniformly convergent over the whole unit circle, to continuous functions. Therefore, all these series must have a soft singularity on the unit circle, at $z=1$. This is one example in which we may differentiate as many times as we will, without the singularity ever becoming hard. Therefore, that singularity is necessarily an infinitely soft one.

Note that in this case the DP Fourier series on the unit circle converge to $C^{\infty}$ functions, although there are singularities on that circle. Although these real functions are $C^{\infty}$, in the real sense of this concept, the complex function $w(z)$ cannot be $C^{\infty}$ on the unit circle, in the complex sense of the concept. One may ask how can a restriction of the complex function $w(z)$, which is $C^{\infty}$ in the whole interior of the unit disk, be $C^{\infty}$ at the boundary of the disk while $w(z)$ itself is not.

The answer is that the $C^{\infty}$ condition in the real sense is a weaker condition than the $C^{\infty}$ condition in the complex sense. While in the case of the real functions on the unit circle only the derivatives with respect to $\theta$ must exist, in the case of the complex function the derivatives in the perpendicular direction, that is those with respect to $\rho$, must also exist, and in fact must give the same values as the derivatives in the direction of $\theta$.

Unlike real functions over one-dimensional domains, which can be folded around at will, complex analytic functions over two-dimensional domains are rigid objects. If one restricts such a function to a one-dimensional domain and then folds that domain around, the resulting real function over it may no longer be the restriction of a complex function to the new one-dimensional domain resulting from the folding process.