A kernel 'more analytic' than $\exp(-x^2)$

Question

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.

Answer 1

$\newcommand{\eps}{\varepsilon}$

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.

Answer 2

There is no such function. As already suggested by Aleksei in a comment, this follows by putting together some facts about entire functions. I'm almost certainly using machinery that is too heavy here.

First of all, the condition on the Taylor coefficients implies that $F$ is entire of order $\le 1/2$: if, say, $\sum |c_n| (2n)! < \infty$, then we can estimate $|F(z)|\le \sum |c_n| |z|^n$ by cutting off the sum at $N\simeq |z|^{1/2}$.

In particular, $\log_+ |F(x)| \lesssim 1+|x|^{1/2+\epsilon}$ for real $x$, so $\int_{-\infty}^{\infty} \frac{\log_+ |F(x)|}{1+x^2}\, dx<\infty$, and thus the Riesz-Fejer theorem shows that $F=GG^{\#}$, for some entire $G$ of exponential type (and $G^{\#}(z)=\overline{G(\overline{z})}$). [Note that I'm ignoring most of the statement of the theorem, I just use it as a crutch to take a square root of $F$ of sorts.]

Since also $G\in L^2(\mathbb R)$, the Paley-Wiener theorem shows that the Fourier transform $\widehat{G}\in L^2(-R,R)$ is compactly supported, and $G$ has non-zero type ($=R$, if $R$ was minimal). Then the same is true of $F$ since its Fourier transform is also compactly supported, so $F$ has order $1$, but we already saw above that our basic assumption gives $F$ order $\le 1/2$.

Finally, and again as already pointed out by Aleksei, literally the same argument shows that $\sum |c_n|\epsilon^n (An)!<\infty$ is not possible for any $A>1$ (and here we can interpret $(An)!$ as a quantity with the asymptotics suggested by the Stirling formula).

Answer 3

If "more analytic" means "faster decrease of Tylor coefficients", you can have a kernel "more analytic then normal". Your condition implies that $F$ is entire, of order $1/2$. Such functions cannot be integrable on the real line.

On the other hand, $\exp(-x^2)$ has order $2$, for it $|c_n|=1/(n/2)!$. The smallest order of entire function which allows bounded (or integrable) functions on the real line is $1$. So you can have a kernel with coefficients like $1/n!$, for example, $$\frac{\sin^2(x/2)}{x^2}.$$ The formula for the order of an entire function $$\rho=\limsup\frac{n\log n}{\log(1/|c_n|)}$$ can be found, for example in Levin's book Distribution of zeros of entire functions, Chap. I, sect. 2, Theorem 2.