# Find a function $F$ on $[0,1]$ with moments decaying as $(\ln n)^{-n}$

Let $$F:[0,1]\to\mathbb{R}$$ be a measurable function such that $$\mu_n(F)=\int_0^1F(t)t^ndt\sim\frac1{(\ln n)^n}\quad\mbox{as}\quad n\to\infty.$$ More precisely, $$0 Note that in this case the series $$\sum_{n=0}^\infty\mu_n(F)z^n$$ represents an entire function. The generating function $$\sum_{n=0}^\infty\frac{\mu_n(F)z^n}{n!}$$ is an entire function of minimal exponential type. I am not aware of any explicit description of such a function, so this is already a sign.

Question: Can we find an explicitly given such function $$F$$? Here by "explicitly given" I mean in a "closed form", e.g., no series.

I think that by convexity arguments it can be shown that $$F$$ cannot be non-negative (non-positive). In fact, I do not need $$F$$ be a function; $$F(x)dx$$ may be a signed measure on $$[0,1]$$, just given explicitly.

Thank you.

• Indeed it is clear that these can't be the moments of a positive measure since then $\int t^n \, d\rho(t) \gtrsim c^n$ for some $c>0$ (unless $\rho = \delta_0)$. Apr 26, 2019 at 4:02


Indeed, let $$\begin{equation*} f(z):=\sum_{n=0}^\infty\frac{\mu_n(F)(iz)^n}{n!} =\int_0^1 F(t)\,dt \sum_{n=0}^\infty\frac{(itz)^n}{n!} =\int_0^1 F(t)e^{itz}\,dt \end{equation*}$$ for all complex $$z$$. By (1), for any natural $$k\ge2$$ $$\begin{equation*} |f(z)|\le O(1+|z|^{k-1})+C\sum_{n=k}^\infty\frac{(|z|/\ln n)^n}{n!} \le O(1+|z|^{k-1})+C\sum_{n=0}^\infty\frac{(|z|/\ln k)^n}{n!} \le c_k e^{|z|/\ln k} \end{equation*}$$ for some real $$c_k>0$$ and all complex $$z$$. So, $$f(z)$$ is an entire function of exponential type $$a$$ for any real $$a>0$$.

Hence, by a Paley--Wiener theorem (more specifically, see e.g. Theorem 19.3 on page 375), for each real $$a>0$$ there is an $$L^2$$ function $$F_a$$ such that for all complex $$z$$ $$\begin{equation*} f(z)=\int_{-a}^a F_a(t)e^{itz}\,dt \end{equation*}$$ Taking the inverse Fourier transform, we see that for each $$a\in(0,1)$$ and all complex $$z$$ $$\begin{equation*} f(z)=\int_0^a F(t)e^{itz}\,dt. \end{equation*}$$ Thus, for all complex $$z$$ $$\begin{equation*} 0=f(z)=\int_0^1 F(t)e^{itz}\,dt. \end{equation*}$$ So, indeed $$F=0$$ a.e. on $$[0,1]$$.

Consider now the more general setting when $$F(t)\,dt$$ is replaced by $$\rho(dt)$$, where $$\rho$$ is a signed measure over $$\R$$ with support $$\supp\rho\subseteq[0,1]$$. Then condition (1) is replaced by $$\begin{equation*} |\mu_n(\si)|(\ln(n+2))^n where $$\begin{equation*} \mu_n(\si):=\int_0^1 t^n\si(dt)=\int_0^1 t^{n+2}\rho(dt)\quad\text{and}\quad\si(dt):=t^2\rho(dt). \end{equation*}$$ So, $$\supp\si\subseteq[0,1]$$. Also, (1a) implies $$\mu_0(\si)<\infty$$, that is, the signed measure $$\si$$ is finite. So, by reasoning quite similar to that in the above $$L^2$$ case, we see that under condition (1a) $$\begin{equation*} f(z):=\int_\R e^{itz}\,\si(dt) \end{equation*}$$ is an entire function of exponential type $$a$$ for any real $$a>0$$.

For real $$b>0$$ and $$t\in\R$$, let $$\begin{equation*} G_b(t):=(g_b*d\si)(t):=\int_\R g_b(t-s)\,\si(ds) =\frac{\si([t-b,t+b])}{2b}, \end{equation*}$$ where $$g_b:=\frac1{2b}\,1_{[-b,b]}$$. Then $$G_b\in L^2(\R)$$, since the signed measure $$\si$$ is finite with $$\supp\si\subseteq[0,1]$$. Also, $$\begin{equation*} \hat g_b(z):=\int_\R e^{itz}g_b(t)\,dt=\frac{\sin bz}{bz} \end{equation*}$$ for $$z\ne0$$, and hence $$\begin{equation*} f_b(z):=\int_\R e^{itz}\,G_b(t)\,dt =\int_\R e^{itz}\,(g_b*d\si)(t)\,dt=\hat g_b(z)f(z) \end{equation*}$$ is of exponential type $$b+a$$ for all real $$a>0$$. Thus, by the cited Paley--Wiener theorem, $$\supp(g_b*d\si)\subseteq[-b-a,b+a]$$ for all real $$a>0$$ and hence $$\begin{equation*} \supp(g_b*d\si)\subseteq[-b,b]. \end{equation*}$$

On the other hand, because $$\supp g_b$$ and $$\supp\si$$ are both compact, by Theorem 4.3.3 on page 117, $$\cch\supp(g_b*d\si)=\cch\supp g_b+\cch\supp\si$$, where $$\cch$$ denotes the closed convex hull. Since $$\cch\supp g_b\ni0$$, we conclude that $$\cch\supp\si\subseteq\cch\supp(g_b*d\si)\subseteq[-b,b]$$, whence $$\supp\rho\subseteq\supp\si\cup\{0\}\subseteq[-b,b]$$, for any real $$b>0$$. Thus, condition (1a) implies $$\begin{equation*} \supp\rho\subseteq\{0\}. \tag{2} \end{equation*}$$

Vice versa, trivially (2) implies (1a). Thus, (1a) holds iff $$\supp\rho\subseteq\{0\}$$.

Response to the second comment by the OP: Let us try to sort all this out.

1. Your condition $$$$0 can actually never hold as stated, because the inequalities $$0 will always be false for $$n=1$$. So, of course, I assumed (3) with $$\mathbb{N}$$ replaced by $$\{2,3,\dots\}$$.

2. In your post here, you did not even mention "analytical distributions" or Ehrenpreis. Instead, you wrote: "Let $$F:[0,1]\to\mathbb{R}$$ be a measurable function" and then you also wrote "In fact, I do not need $$F$$ be a function; $$F(x)dx$$ may be a signed measure on $$[0,1]$$".

3. If $$F(x)dx$$ is replaced by a signed measure $$\rho(dx)$$, then the corrected version of your condition (3) can be rewritten as the conjunction of my condition (1a) and the condition $$$$0

4. It was shown in this answer that, under condition (1a), the support of the signed measure $$\rho$$ must be contained in the set $$\{0\}$$. (Method-wise, this was first done in the case when $$\rho(dx)=F(x)dx$$ with $$F\in L^2$$, and then extended to the general case of any signed measure $$\rho$$ on $$[0,1]$$.)

5. It then immediately follows that there is no signed measure $$\rho$$ on $$[0,1]$$ satisfying (1a) and (4) -- or, equivalently, satisfying the corrected version of your condition (3). In other words, your conditions do result in a contradiction. This completely answers your posted question.

• I have added an extension to the more general setting when $F(t)\,dt$ is replaced by $\rho(dt)$, where $\rho$ is a signed measure. Apr 29, 2019 at 14:53
• Paley-Wiener theorem (e.g., Theorem 19.3 in Ruding you refer to) requires $f(z)$ be $L^2$ on the real line, which is not true, I think (there are reasons why it shouldn't). Same is true for $f_b(z)$ in your second application of Paley-Wiener, I think. But I can see your point. If $f$ is of minimal exponential type and $F$ (or $\rho$) its Fourier transform, then $F$-s support is $\{0\}$. Instead of classical Paley-Wiener you can use other things (f.i., a la Ehrenpreis). Thank you. Apr 30, 2019 at 4:53
• @Bedovlat : Since $F\in L^2(\mathbb R)$, by the Plancherel theorem $f=\hat F$ is also $L^2$ on $\mathbb R$. See e.g. mathworld.wolfram.com/PlancherelsTheorem.html or en.wikipedia.org/wiki/Plancherel_theorem . So, I see no problem here. (You may have confused my $f(z)$ with your (moment) generating function $f(-iz)$.) Apr 30, 2019 at 14:49
• Dear Prof. Pinelis, your $f(-iz)$ is also an entire function of minimal exponential type, and thus cannot be $L^2$ on the real line. No one cancelled Plancherel theorem, of course, but what I am saying is that your assumption $F\in L^2$ is already in contradiction with your (1). In fact, $F$ cannot be even a distribution in the classical sense. It can be viewed as an "analytical distribution" in the sense of Ehrenpreis, as I wrote before. Apr 30, 2019 at 16:42
• @Bedovlat : In my answer, I have added a detailed response to your latter comment. I hope this will help. Here I wanted to add that whatever may be said about $f(-iz)$ is irrelevant to my answer, dealing with $f(z)$. More importantly, we indeed get a contradiction. But this contradiction results from your own assumptions, not mine. In the answer, I did not make any assumptions in addition to yours. Apr 30, 2019 at 19:27