I am looking for an analytic function $F: \mathbb{R} \rightarrow (0,\infty)$ with $\int\_{\mathbb{R}} F(x) \, dx = 1$ and the property, that $\sum\limits\_{k=0}^{\infty} |c\_k| \varepsilon^k (2k)! < \infty$ for some $\varepsilon > 0$, if $\sum\limits\_{k=0}^{\infty} c\_k x^k$ is the series representation of $F(x)$ around zero.


$F$ being symmetric around zero would be a nice bonus.


Any help is much appreciated!




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Second Edit: As Iosif Pinelis correctly pointed out, the below arguments only work for positive definite kernels $F$ so there may still be hope.




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Edit: Sorry, this might be impossible. By Bochner's Theorem the first two properties $F: \mathbb{R} \rightarrow (0,\infty)$ and $\int\_{\mathbb{R}} F(x) \, dx = 1$ hold if and only if the (generalized) Fourier transform $\hat{\mu}$ of $F$ is a probability distribution on $\mathbb{R}$. We can then write
\begin{align\*}
& F(x) = \int\_{\mathbb{R}} e^{i2\pi \xi x} \, d\hat{\mu}(\xi) = \sum\limits\_{k=0}^\infty x^k \underbrace{\frac{(i2\pi)^k}{k!} \overbrace{\int\_{\mathbb{R}} \xi^k \, d\hat{\mu}(\xi)}^{\mathbb{E}[X^k]}}\_{c\_k} \ .
\end{align\*}
Let $X$ be a random variable with distribution $\hat{\mu}$, then the last property is
\begin{align\*}
\infty & \overset{!}{>} \sum\limits\_{k=0}^\infty \frac{|2\pi|^k}{k!} \big|\mathbb{E}[X^k] \big| \varepsilon^k (2k)!\\
& \geq \sum\limits\_{m=0}^\infty \frac{|2\pi|^{2m}}{(2m)!} \mathbb{E}[X^{2m}] \varepsilon^{2m} (4m)!
\end{align\*}
Hoever Jensens inequality gives $\mathbb{E}[X^{2m}] \geq \mathbb{E}[X^2]^{m}$. Since $\mathbb{E}[X^2]>0$, the above sum always diverges.



A kernel 'more analytic' than $\exp(-x^2)$
$\newcommand{\eps}{\varepsilon}$


I will show that if $\sum |c\_n| C^n n! < \infty$ for all $C > 0$ then there's no such function $F$ (if we only know it for some fixed $C$ then there are such functions). Note that this condition is equivalent to having a minimal exponential type. The proof is based on the following version of the Phragmén–Lindelöf principle:


$\textbf{Lemma}$ Assume that the function $F$ is analytic in the closed upper half-plane $\{z: Im(z) \ge 0\}$, is in $L^1(\mathbb{R})$ and is of minimal exponential type, that is for all $\eps> 0$ we have $|F(z)| \le C\_\eps e^{\eps |z|}$. Then $F$ is bounded on $\{z: Im(z) \ge 1\}$.


Classical Phragmén–Lindelöf principle assumes that $F\in L^\infty(\mathbb{R})$ and gives us that $F$ is bounded in $\{z: Im(z) \ge 0\}$.


To get from this lemma and the classical principle our result is easy: by the lemma $F$ is bounded on $\{ z: Im(z) = 1\}$, applying the classical Phragmén–Lindelöf principle to the half-planes $\{z: Im(z) \ge 1\}$ and $\{z: Im(z) \le 1\}$ we get that $F$ is bounded on $\mathbb{C}$, but then by the Lioville's theorem it is constant, but the constants are not in $L^1(\mathbb{R})$.


So, it remains to prove the lemma, whose proof will mimic that of the Phragmén–Lindelöf principle. We will actually just prove that $F$ is bounded on $\{z: Im(z) = 1\}$ and then invoke the classical Phragmén–Lindelöf principle to finish it off. Consider $G(z) = F(z)e^{iz}$. It is bounded (and in $L^1$) on $\{iy: y \ge 0\}$ and is in $L^1(\mathbb{R})$. We will deal with $\{z: Re(z) < 0\}$ and $\{z: Re(z) > 0\}$ separately. I will only cover the first case, the other is entirely similar. Note also that the extra factor $e^{iz}$ does not affect boundedness on $\{z: Im(z) = 1\}$.


So, we have a function $G$ which is defined in the sector of angle $\frac{\pi}{2}$ and of minimal exponential type ($|e^{iz}|\le 1$ in all of our domain) and which is $L^1$ on its boundary and we want to get an upper bound for the values at least $1$ away from said boundary.


By rotation we assume that the angle is $|\arg(z)| \le \frac{\pi}{4}$ and consider $G\_\eps(z) = G(z)e^{-\eps z}$. This function tends uniformly to $0$ when $|z|\to \infty$ since $G$ is of minimal type (note that $|e^{-\eps z}| \le e^{-\eps |z|/2}$). We fix $z\_0$ which is at least $1$ away from the rays $\arg(z) = \pm \frac{\pi}{4}$. Consider the domain $\Omega$ bounded by these rays and the arc of a circle of radius $R$.


We want to use subharmonicity of $\log |G\_\eps(z)|$. Here, unfortunately, we would need one little extra knowledge beyond the usual proof of the Phragmén–Lindelöf principle -- what is the harmonic measure. We want to bound the value of $\log |G\_\eps(z\_0)|$ via the values on the boundary of our domain, which is achieved by integrating these values at the boundary against the harmonic of $z\_0$ with respect to $\Omega$, so we have to somehow bound it.


Harmonic measure on the arc doesn't matter for us since $G\_\eps$ on it is at most $1$ anyway if $R$ is big enough in terms of $\eps$. As for the rays, here is a nice cheat: if we enlarge the domain then harmonic measure can only increase. So, for a ray $\arg z = \frac{\pi}{4}$, we can consider the whole half-plane $-\frac{3\pi}{4} \le \arg z \le \frac{\pi}{4}$, on which the harmonic measure is just a Poisson measure, which is uniformly bounded by the Lebesgue measure for all $z\_0$ which are at least $1$ away from said ray. And the same for the other ray.


It remains to note that $||\log\_{+} |G\_\eps| ||\_{L^1} \le ||\log\_{+}|G|||\_{L^1} \le ||G||\_{L^1} < \infty$, so we get a uniform upper bound for $G\_\eps(z\_0)$ by sending $R$ to infinity and this bound does not depend on $\eps$. Now, sending $\eps\to 0$ we get a uniform upper bound on $G(z\_0)$ which is exactly what we wanted.


If for some reason you checked the above argument very closely, we actually didn't cover the segment $[-1 + i, 1 + i]$, but it is a compact segment and $F$ is obviously bounded on it.


It turned out to be a bit more verbose than I expected, but what we really did is we just repeated a textbook proof of the Phragmén–Lindelöf principle with a little change using the harmonic measure to cover $L^1$ instead of the $L^\infty$ assumption.


