Let us take the exponential function $\lambda^z$ where $0 < \lambda < 1$. There are many great uniqueness conditions this holomorphic function satisfies. For example, it is the only function holomorphic in the right half plane, bounded in the right half plane, that interpolates $\lambda^{z} \Big{|}_{\mathbb{N}}$. It is also the only function to satisfy the differential equation $\frac{d}{dz}f(z) = \log(\lambda)f(z)$ when $f(0) =1$. Also it is the only function such that $f(z)f(w) = f(z+w)$ and $f(1) = \lambda$ and $f:\mathbb{R}^+\to\mathbb{R}$. It is also the only function holomorphic such that $f(z)f(w) = f(z+w)$ and $f(1) = \lambda$ and $f$ has a period of $2 \pi i /\log(\lambda)$.

I'm asking this question because of this similar question on tetration An explicit series representation for the analytic tetration with complex height However when approaching the uniqueness of this tetration function on the real positive line, what is barring our path is a lemma (if you will) about the exponential function. This lemma is rather simple to state, and leads me to believe there must be a solution of this question.

I am not conjecturing this lemma to be true by calling it a lemma (I have a feeling its true); it'd just be nice to see a proof in the positive. If someone has references or suggestions or even a straight proof about why this must be true/not true, that'd be outstanding. The lemma has to do with completely monotonic functions, and is rather tricky. It can be stated plainly as

If a function $F$ is analytic on $\mathbb{R}^+$ and satisfies

$$F(1) =\lambda$$ $$\lambda F(x) = F(x+1)$$ $$(-1)^n\frac{d^n}{dx^n}F(x) >0$$

must $F(x) = \lambda^x$?

This can also be stated, if $F(x) = \lambda^{x}\phi(x)$ for some $1$-periodic function $\phi(x)$, and $(-1)^n\frac{d^n}{dx^n}F(x) > 0$, must $\phi(x)$ be constant?

It can also be stated in the manner I think is most likely to have an answer ready at hand. I also believe this to be the most valuable way of stating the question.

Letting $F$ be analytic on $\mathbb{R}^+$, if $F(1) = e$ and $$\frac{d^n}{dx^n}F(x) > 0$$ $$e\cdot F(x)= F(x+1)$$ must $F(x) = e^{x}$?


The answer is yes. Let me address the last stated form of the question. Functions $F$ with $F^{(n)}\geq 0$ are called totally monotone, and S. Bernstein proved that all such functions are Laplace transforms of positive measures, that is $$F(x)=\int e^{xt}d\mu(t).$$ Your functional equation $F(x+1)=eF(x)$ then implies $$\int e^{xt}(e^td\mu(t))=\int e^{xt}(ed\mu(t)).$$ From this follows $e^td\mu(t)=ed\mu(t)$ by the uniqueness theorem for the Laplace transform. This means that $\mu$ is an atom at the point $1$. This means that $F(x)=ce^x$, and your normalization implies $c=1$.

The references for Laplace transform are Widder, Laplace transform, and Feller, An introduction to probability... vol. II. Widder has Bernstein's theorem and uniqueness theorem.

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  • $\begingroup$ I dont get it :( plz say more. How can 2 different things be equal when we have uniqueness ? $\endgroup$ – mick Feb 23 '17 at 8:49
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    $\begingroup$ @mick: it follows from uniqueness theorem that these two things are actually not different but equal. $\endgroup$ – Alexandre Eremenko Feb 1 '18 at 3:30
  • $\begingroup$ If $e^t u(t) = e u(t) $ does that not mean $u(t) = 0 ?? What does " atom at the point 1 " mean ?? Sorry Im not good at this. $\endgroup$ – mick Feb 4 '18 at 17:01
  • $\begingroup$ @mick: "atom" at $1$ means a measure $\mu$ defined by $\mu(E)=0$ if $E$ does not contain $1$ and $\mu(E)=k$ when $E$ contains $1$. Here $k$ is some positive constant. $\endgroup$ – Alexandre Eremenko Feb 4 '18 at 17:06

Here's a proof that also uses Bernstein's representation theorem for totally monotone functions (I wondered if that result was needed, and Alexander Eremenko's answer now makes me think, that it is).

If $F:\mathbb{R}_+\to\mathbb{R}_+$ satisfies $F(x+1)=eF(x)$, it is of the form $F(x)=\exp(x+h(x))$, with $h$ a $1$-periodic function. But a totally monotonic function is logarithmically convex (see below) and $x+h(x)$ can only be convex if $h$ is a constant (actually zero, if also $F(0)=1$ or $F(1)=e$).


Totally monotonic functions are logarithmically convex:

From $F(x)=\int_{\mathbb{R}_+}e^{\lambda x}d\mu(\lambda)$, differentiating under the integral sign and applying the Cauchy-Schwarz inequality $$F'(x)^2=\bigg(\int_{\mathbb{R}_+}\lambda e^{\lambda x}d\mu(\lambda)\bigg)^2=\bigg(\int_{\mathbb{R}_+}\lambda e^{\lambda x/2}\cdot e^{\lambda x/2}d\mu(\lambda)\bigg)^2\le$$$$ \int_{\mathbb{R}_+}\lambda^2 e^{\lambda x}d\mu(\lambda)\int_{\mathbb{R}_+}e^{\lambda x}d\mu(\lambda)=F''(x)F(x)\ . $$

Therefore $(F'/F)'\ge0,$ that is $\log F(x)$ is convex.

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  • $\begingroup$ Of course one has to use Bernstein's representation. How else can you deal with the conditions like $f^{(n)}\geq 0$ or $(-1)^nf^{(n)}\geq 0$? $\endgroup$ – Alexandre Eremenko Jan 24 '17 at 16:48
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    $\begingroup$ I don't know. If we see the Bernstein's representation as an instance of Choquet theorem, then, for instance, for the implication totally convex $\Rightarrow$ logarithmically convex the Krein-Mil'man thm suffices, and it is a little less than Choquet thm (but always functional analysis, of course) $\endgroup$ – Pietro Majer Jan 24 '17 at 17:52
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    $\begingroup$ perhaps you are right. (For myself, Krein-Milman and Choquet are "higher mathematics" in comparison with Bernstein's thm:-) $\endgroup$ – Alexandre Eremenko Jan 24 '17 at 19:38

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