Strong Divergence

In order to discuss this case we must go all the way back to the case of plain DP trigonometric series, without any consideration of whether or not they are Fourier series. Hence, let us suppose that we have an arbitrary pair of FC trigonometric series,

$$\begin{eqnarray*} S_{\rm c} & = & \sum_{k=1}^{\infty} a_{k}\cos(k\theta), \\ S_{\rm s} & = & \sum_{k=1}^{\infty} a_{k}\sin(k\theta). \end{eqnarray*}$$

Without any further assumptions, we may then construct the complex series $S_{v}$ and $S_{z}$,

$$\begin{eqnarray*} S_{v} & = & \sum_{k=1}^{\infty} a_{k}\,{\rm e}^{\mbox{\bol... ...$}k\theta}, \\ S_{z} & = & \sum_{k=1}^{\infty} a_{k}z^{k}. \end{eqnarray*}$$

Let us suppose that $S_{z}$ fails to converge at a single point $z_{1}$ located strictly inside the open unit disk. Then it follows from the basic convergence theorem that it cannot converge at any point outside the circle centered at zero with its boundary going through the point $z_{1}$. This is so because, if the series did converge at some point outside this disk, them by that theorem it would converge everywhere in a larger disk, which contains the point of divergence $z_{1}$. This is absurd, and therefore we conclude that the series must be divergent at all points strictly outside the closed disk defined by the point of divergence $z_{1}$.

It follows therefore that in this case the $S_{z}$ series is divergent everywhere over the whole unit circle. The same is then true for $S_{v}$, of course, and similar conclusions may be drawn for the associated DP trigonometric series. Note that while the convergence of $S_{v}$ at a given point implies the convergence of both $S_{\rm c}$ and $S_{\rm s}$ at that point, the lack of convergence of $S_{v}$ implies only that at least one of the two DP trigonometric series diverges. The other one might still converge. We may immediately conclude that at least one of the two series in the FC pair is divergent. But in fact, one can see that both must be divergent almost everywhere, by the following argument.

If we consider the absolute value of the terms of $S_{z}$ at $z_{1}$, we have $\left\vert a_{k}z_{1}^{k}\right\vert=\vert a_{k}\vert\rho_{1}^{k}$, where $\vert z_{1}\vert=\rho_{1}$ and $\rho_{1}<1$. Since $S_{z}$ is divergent at $z_{1}$, $\vert a_{k}\vert\rho _{1}^{k}$ cannot go to zero faster than $1/k$, because a limit to zero as $1/k^{1+\varepsilon}$ with any strictly positive $\varepsilon$ would be sufficient for the series of the absolute values of the terms of $S_{z}$ to converge, since in this case it can be bounded from above by a convergent asymptotic integral, as is shown in Section B.1 of Appendix B. This would imply that $S_{z}$ is absolutely convergent and hence convergent, which is absurd. This means that we must have

$$\vert a_{k}\vert > \frac{1}{k\rho_{1}^{k}},$$

for $k>k_{m}$ and some minimum value $k_{m}$ of $k$. In other words, $\vert a_{k}\vert$ must in fact diverge to infinity exponentially with $k$, in fact just about as fast as $1/\rho_{1}^{k}$, which goes to infinity as $k\to\infty$ since $\rho_{1}<1$. Since these coefficients are those of both $S_{\rm c}$ and $S_{\rm s}$, and a limit of the coefficients to zero as $k\to\infty$ is a necessary condition for convergence, both these series must diverge almost everywhere, that is, everywhere except possibly for a few special points, such as the sine series at $\theta=0$.

Therefore, in this case there are no trigonometric series at all that converge almost everywhere on the unit circle. We say that such series are divergent almost everywhere. Besides, since any Fourier series is a particular case of trigonometric series, in this case there are no almost-everywhere convergent Fourier series as well. This is true whenever the maximum disk of convergence of $S_{z}$ around $z=0$ is smaller than the open unit disk, which also means that the function $w(z)$ to which $S_{z}$ converges must have at least one singularity strictly within that disk.

In conclusion, given any DP trigonometric series, be it a Fourier series or not, if the power series built from its coefficients is divergent anywhere strictly within the unit disk, then the trigonometric series is divergent almost everywhere in the periodic interval. The same follows if it can be verified that the function $w(z)$ has a singularity strictly within the open unit disk. We refer to this situation as one of strong divergence.

Note that in this case we cannot assert whether or not the trigonometric series is a Fourier series. In this case it is not possible to take any limit from the interior of the maximum disk of convergence to the unit circle, which becomes disconnected from the analytic region of the $S_{z}$ series. In fact, it is an interesting question whether or not, in this situation, any real functions can exist that give the coefficients of the trigonometric series by means of the usual integrals. While a negative answer to this question seems to be compelling, for now we must leave this here as a simple conjecture.