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, $$ 0<c<|\mu_n(F)|(\ln n)^n<C<\infty,\quad\forall n\in\mathbb{N}. $$ 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.

  • 3
    $\begingroup$ 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 $\rho = \delta_0)$. $\endgroup$ Apr 26, 2019 at 4:02

1 Answer 1


$\newcommand{\R}{\mathbb{R}} \newcommand{\si}{\sigma} \newcommand{\supp}{\operatorname{\mathrm supp}} \newcommand{\cch}{\operatorname{\mathrm cch}} $ If $F\in L^2$, then the condition \begin{equation*} |\mu_n(F)|(\ln n)^n<C<\infty\quad\forall n\in\mathbb{N} \tag{1} \end{equation*} implies that $F=0$ almost everywhere (a.e.) on $[0,1]$.

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<C<\infty\quad\forall n\in\mathbb{N}_0:=\{0,1,\dots\}, \tag{1a} \end{equation*} 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 \begin{equation} 0<c<|\mu_n(F)|(\ln n)^n<C<\infty \quad\forall n\in\mathbb{N} \tag{3} \end{equation} can actually never hold as stated, because the inequalities $0<c<|\mu_n(F)|(\ln n)^n$ 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 \begin{equation} 0<c<|\mu_n(\si)|(\ln n)^n \quad\forall n\in\{2,3,\dots\}. \tag{4} \end{equation}

  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.

  • $\begingroup$ 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. $\endgroup$ Apr 29, 2019 at 14:53
  • $\begingroup$ 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. $\endgroup$
    – Bedovlat
    Apr 30, 2019 at 4:53
  • $\begingroup$ @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 $f(-iz)$.) $\endgroup$ Apr 30, 2019 at 14:49
  • $\begingroup$ 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. $\endgroup$
    – Bedovlat
    Apr 30, 2019 at 16:42
  • $\begingroup$ @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. $\endgroup$ Apr 30, 2019 at 19:27

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