Under good conditions on an even function $f(x)$ we have the Poisson Summation formula ($x>0$):

$$f(0) + 2 \sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \left( \hat{f}(0) + 2 \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) \right)$$

Where $\hat{f}$ is the Fourier transform of $f(x)$.

My question is on the way the partial sum $f(0) + 2 \sum\limits_{n =1}^{P} f(nx)$ converges, uniformly on $\mathbb{R}^+$ or not, for $P \to \infty$ ?

Example:

Lets take a concrete example with $\mu$ a positive real and the function: $f(x)= x^2 e^{-\pi^2 x^2} - \mu (x^2 \mu^2)e^{-\pi^2 \mu^2 x^2} $ With this definition we have $f(0)=0$ and $\hat{f}(0)= \int\limits_{-\infty}^{\infty} f(x) dx =0$ So Poisson formula in this case becomes simply:

$$ S(x)=\sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) = \frac{K}{x} \sum\limits_{n =1}^{\infty} (\frac{2n^2}{x^2}-1) e^{-(\frac{n}{x})^2} -(\frac{2n^2}{\mu^2 x^2}-1) e^{-(\frac{n}{\mu x})^2} $$

As we have $\hat{f}(x)= K (2x^2-1) e^{-x^2}$ (Where $K$ is a constant)

On this example we see clearly that $S(x) \to 0$ for $x \to 0$ and that for all $P$ the partial sum:

$S_P(x)= \sum\limits_{n =1}^{P} f(nx)$, is also such that $S_P(x) \to 0$ for $x \to 0$, with of course simle convergence for all $x$: $$\lim_{P \to \infty} S_P(x)=S(x)$$

But do we have uniform convergence of $S_P(x)$ on $\mathbb{R}^+$ for $P \to \infty$ ?

If we take $\mu=2$ then $S(x)$ can be simplified to: $S(x)=\sum\limits_{n =1}^{\infty} (-1)^n (nx)^2 e^{-\pi^2 (nx)^2}$, and then I can use Abel summation to prove that there is uniform convergence (for $\mu$ an integer it is the same) but in general if $f(x)$ is not composed as above to allow such simplifications ?

This question is linked in a way to another question of uniform convergence I asked here:

Infinite sums with Mobius Inversion : can we have uniform convergence of inversion formula?

Under good conditions on an even function $f(x)$ we have the Poisson Summation formula ($x>0$):

$$f(0) + 2 \sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \left( \hat{f}(0) + 2 \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) \right)$$

Where $\hat{f}$ is the Fourier transform of $f(x)$.

My question is on the way the partial sum $f(0) + 2 \sum\limits_{n =1}^{P} f(nx)$ converges, uniformly on $\mathbb{R}^+$ or not, for $P \to \infty$ ?

Example:

Lets take a concrete example with $\mu$ a positive real and the function: $f(x)= x^2 e^{-\pi^2 x^2} - \mu (x^2 \mu^2)e^{-\pi^2 \mu^2 x^2} $ With this definition we have $f(0)=0$ and $\hat{f}(0)= \int\limits_{-\infty}^{\infty} f(x) dx =0$ So Poisson formula in this case becomes simply:

$$ S(x)=\sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) = \frac{K}{x} \sum\limits_{n =1}^{\infty} (\frac{2n^2}{x^2}-1) e^{-(\frac{n}{x})^2} -(\frac{2n^2}{\mu^2 x^2}-1) e^{-(\frac{n}{\mu x})^2} $$

As we have $\hat{f}(x)= K (2x^2-1) e^{-x^2}$ (Where $K$ is a constant)

On this example we see clearly that $S(x) \to 0$ for $x \to 0$ and that for all $P$ the partial sum:

$S_P(x)= \sum\limits_{n =1}^{P} f(nx)$, is also such that $S_P(x) \to 0$ for $x \to 0$, with of course simle convergence for all $x$: $$\lim_{P \to \infty} S_P(x)=S(x)$$

But do we have uniform convergence of $S_P(x)$ on $\mathbb{R}^+$ for $P \to \infty$ ?

If we take $\mu=2$ then $S(x)$ can be simplified to: $S(x)=\sum\limits_{n =1}^{\infty} (-1)^n (nx)^2 e^{-\pi^2 (nx)^2}$, and then I can use Abel summation to prove that there is uniform convergence (for $\mu$ an integer it is the same) but in general if $f(x)$ is not composed as above to allow such simplifications ?

This question is linked in a way to another question of uniform convergence I asked here:

Infinite sums with Mobius Inversion : can we have uniform convergence of inversion formula?

Under good conditions on an even function $f(x)$ we have the Poisson Summation formula ($x>0$):

$$f(0) + 2 \sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \left( \hat{f}(0) + 2 \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) \right)$$

Where $\hat{f}$ is the Fourier transform of $f(x)$.

My question is on the way the partial sum $f(0) + 2 \sum\limits_{n =1}^{N} f(nx)$ converges, uniformly or not?

Example:

Lets take a concrete example with $\mu$ a positive real and the function: $f(x)= x^2 e^{-\pi^2 x^2} - \mu (x^2 \mu^2)e^{-\pi^2 \mu^2 x^2} $ With this definition we have $f(0)=0$ and $\hat{f}(0)= \int\limits_{-\infty}^{\infty} f(x) dx =0$ So Poisson formula in this case becomes simply:

$$ S(x)=\sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) = \frac{K}{x} \sum\limits_{n =1}^{\infty} (\frac{2n^2}{x^2}-1) e^{-(\frac{n}{x})^2} -(\frac{2n^2}{\mu^2 x^2}-1) e^{-(\frac{n}{\mu x})^2} $$

As we have $\hat{f}(x)= K (2x^2-1) e^{-x^2}$ (Where $K$ is a constant)

On this example we see clearly that $S(x) \to 0$ for $x \to 0$ and that for all $P$ the partial sum:

$S_P(x)= \sum\limits_{n =1}^{P} f(nx)$, is also such that $S_P(x) \to 0$ for $x \to 0$, with of course simle convergence for all $x$: $$\lim_{P \to \infty} S_P(x)=S(x)$$

But do we have uniform convergence ?

If we take $\mu=2$ then $S(x)$ can be simplified to: $S(x)=\sum\limits_{n =1}^{\infty} (-1)^n (nx)^2 e^{-\pi^2 (nx)^2}$, and then I can use Abel summation to prove that there is uniform convergence (for $\mu$ an integer it is the same) but in general if $f(x)$ is not composed as above to allow such simplifications ?

This question is linked in a way to another question of uniform convergence I asked here:

Infinite sums with Mobius Inversion : can we have uniform convergence of inversion formula?

Under good conditions on an even function $f(x)$ we have the Poisson Summation formula ($x>0$):

$$f(0) + 2 \sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \left( \hat{f}(0) + 2 \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) \right)$$

Where $\hat{f}$ is the Fourier transform of $f(x)$.

My question is on the way the partial sum $f(0) + 2 \sum\limits_{n =1}^{P} f(nx)$ converges, uniformly on $\mathbb{R}^+$ or not, for $P \to \infty$ ?

Example:

Lets take a concrete example with $\mu$ a positive real and the function: $f(x)= x^2 e^{-\pi^2 x^2} - \mu (x^2 \mu^2)e^{-\pi^2 \mu^2 x^2} $ With this definition we have $f(0)=0$ and $\hat{f}(0)= \int\limits_{-\infty}^{\infty} f(x) dx =0$ So Poisson formula in this case becomes simply:

$$ S(x)=\sum\limits_{n =1}^{\infty} f(nx)= \frac{1}{x} \sum\limits_{n =1}^{\infty} \hat{f}(\frac{n}{x}) = \frac{K}{x} \sum\limits_{n =1}^{\infty} (\frac{2n^2}{x^2}-1) e^{-(\frac{n}{x})^2} -(\frac{2n^2}{\mu^2 x^2}-1) e^{-(\frac{n}{\mu x})^2} $$

As we have $\hat{f}(x)= K (2x^2-1) e^{-x^2}$ (Where $K$ is a constant)

On this example we see clearly that $S(x) \to 0$ for $x \to 0$ and that for all $P$ the partial sum:

$S_P(x)= \sum\limits_{n =1}^{P} f(nx)$, is also such that $S_P(x) \to 0$ for $x \to 0$, with of course simle convergence for all $x$: $$\lim_{P \to \infty} S_P(x)=S(x)$$

But do we have uniform convergence of $S_P(x)$ on $\mathbb{R}^+$ for $P \to \infty$ ?

If we take $\mu=2$ then $S(x)$ can be simplified to: $S(x)=\sum\limits_{n =1}^{\infty} (-1)^n (nx)^2 e^{-\pi^2 (nx)^2}$, and then I can use Abel summation to prove that there is uniform convergence (for $\mu$ an integer it is the same) but in general if $f(x)$ is not composed as above to allow such simplifications ?

This question is linked in a way to another question of uniform convergence I asked here:

Infinite sums with Mobius Inversion : can we have uniform convergence of inversion formula?