When is the periodisation of a function continuous?

Question

Consider a function $f\in\mathcal{C}_0(\mathbb{R})$, where $\mathcal{C}_0(\mathbb{R})$ denotes the space of bounded continuous functions vanishing at infinity. I am interested in the $T$-periodisation of such a function, defined as: $$f_{T}(t)=\sum_{n\in\mathbb{Z}} f(t-nT),\quad \forall t\in \mathbb{R}.$$ As explained in Fischer - On the duality of discrete and periodic functions, $f_{T}$ is a $T$-periodic tempered distribution if $f$ is a rapidly decaying function —i.e. vanishing at infinity faster than any polynomial.

My question concerns the regularity of $f_T$:

For which functions $f\in\mathcal{C}_0(\mathbb{R})$ is the periodised generalised function $f_{T}$ defined above an ordinary, continuous function?

In other words, what should be the assumptions on $f$ so that its periodisation is continuous?

Any lead would be greatly appreciated. Thank you very much in advance!

Answer 1

You just need that $f$ decreases fast enough to make the series uniformly convergent on compact sets. E.g., it would be enough that $|x|^p |f(x)|$ is bounded for some $p>1$. Then you can estimate the terms of the series uniformly on a compact interval $[-a,a]$ for $nT>2a$ by $cn^{-p}$ with a constant $c$.

Answer 2

Short answer: e.g. for Schwartz functions.

Long answer: The Fourier transform of "periodic" is "discrete" and the Fourier transform of "discrete" is "periodic". This is a one-to-one mapping. It is explained in this Fischer - On the duality of discrete and periodic functions.

Analogously, the Fourier transform of "regular" is "local" and the Fourier transform of "local" is "regular". It is another one-to-one mapping. It is explained in Fischer - On the duality of regular and local functions.

The term "regular" refers to ordinary, infinitely differentiable functions which do not grow faster than polynomials. These (regular) functions are so-called multiplication operators for tempered distributions. Their multiplication product with any tempered distribution is again a tempered distribution.

The term "local" refers to tempered distributions which are "local", i.e., they rapidly decay to zero (faster than polynomials). These (generalized) functions are so-called convolution operators for tempered distributions. Their convolution product with any tempered distribution is again a tempered distribution.

The properties of "regular" and "local" fulfill a convolution theorem on tempered distributions.

Now, properties of "periodic", "discrete", "regular" and "local" can be combined. For example, "local+regular" are Schwartz functions and the Fourier transform of Schwartz functions are, again, Schwartz functions ("local+regular"). Moreover, the Fourier transform of "discrete periodic" is again "discrete periodic". It yields the Discrete Fourier Transform (DFT).

Now, the precondition for generalized functions which can be periodized is that they are "local" and the precondition for generalized functions which can be discretized is that they are "regular".

So, back to the original question, in order to periodize an (ordinary or generalized) function, it must be "local" and in order to allow it to be an ordinary function it must be "regular". In other words, Schwartz functions fulfill these two requirements, they are "regular+local".

This property of Schwartz functions of being "regular" and "local" simultaneously, explains their special role as test functions in distribution theory and in quantum physics.

However, there is a difference of "being smooth" in the ordinary and in the generalized functions sense. One may recall, every generalized function is smooth (infinitely differentiable) and, hence, "continuous". To answer this question in the ordinary functions sense, embedded in generalized functions theory, there are more functions beside Schwartz functions. The rectangular function, for example, is smooth in the generalized functions sense but not smooth in the ordinary functions sense. Its periodization, however, yields the function that is constantly 1 for suitable T which is a smooth, ordinary function (in particular continuous). So, obviously, functions which are continuous on an interval [-T/2,+T/2] and such that f(-T/2)=f(+T/2) do also fulfill the requirement.