totally monotonic

[bo198214]

As Jay serveral times mentioned there is a condition that all odd derivatives are positive. This condition occured also in Szekeres paper [1] however in a slightly different context:
Definition. We call \( f(x) \) totally monotonic at \( x_0 \) if it has derivatives of any order and \( (-1)^{k+1}f^{(k)}(x_0)>0 \) for every \( k>0 \).

Then he shows that if the inverse of a function \( f \) (\( f \) real analytic for \( x\ge 0 \), \( f(x)>x \), \( f'(x)>0 \) for \( x>0 \) and \( f(x)=x+ax^2+\dots \), \( a>0 \)) is totally monotonic then the regular Abel function is also totally monotonic and is uniquely determined by this property.

\( x\mapsto e^x-1 \) meets the criteria and its inverse is \( x\mapsto \log(x+1) \) and is totally monotonic. Hence the regular Abel function is also totally monotonic.

In our case however the situation is a bit different. The function \( e^x \) has no fixed point. \( \text{slog}_e \) is an Abel function for it but is not totally monotonic, but the inverse of slog is (/seems to be) totally monotonic.

I would bet there is no proof for the uniqueness claim by total monotonicity, though it sound quite plausible.

[1] G. Szekeres, Fractional iteration of exponentially growing functions, 1961.

[Catullus]

In our case however the situation is a bit different. The function \( e^x \) has no fixed point. \( \text{slog}_e \) is an Abel function for it but is not totally monotonic, but the inverse of slog is (/seems to be) totally monotonic.

\(e\uparrow x\) has fixed points at \(-\text{W}_n(-1)\forall n\in\Bbb Z\).

[bo198214]

In the context of monotony you typically consider real valued functions, because there is no suitable "less-than"/"grater-than" comparison on the complex numbers.
So what help is the statement that there are a lot of non-real fixed points???

Well, I am glad that MphLee gave you already some good advice in the "new super moderators thread" about checking the context before you make a correcting comment.

[JmsNxn]

This may be slightly irrelevent, but there was a large thread on here, I want to say 7 years ago about totally monotonic solutions, and whether that was a uniqueness criterion. It evolved into a thread on MO, in which a couple of things were revealed.



One thing of interest that always sticks in my mind is the following.



If \(0 < \lambda < 1\) and \(f(x)\) is a smooth function (could be analytic). Such that:



\[

\begin{align}

f(0) &= 1\\

f(x)\,\,&\text{is totally monotonic}\\

f(x+1) &= \lambda f(x)\\

\end{align}

\]



Then necessarily \(f(x) = \lambda^x\).



The second fact is a bit more involved. But if \(g(z) = b^z\) where \(g\) has an attracting fixed point at \(x_0\) and \(1 < b < \eta\); while additionally \(g'(x_0) = \lambda\)--then the inverse Schroder function \(\Psi\) has the form:



\[

\Psi(x) = x_0 - \sum_{j=1}^\infty a_j x^j\\

\]



Where \(a_j > 0\). Then, we can introduce:



\[

\text{tet}_b(x+c) = \Psi(\lambda^x) = x_0 - \sum_{j=1}^\infty a_j \lambda^{jx}\\

\]



We managed to identify that totally monotonic could be a uniqueness condition, provided you could show a boundedness/derivative condition on the above. And the proof followed very similar to the proof that \(f(x)\) is unique.



This is to mean, the conjecture would stand. The function \(F\) such that:



\[

\begin{align}

F(0) &= 1\\

F(x)&\,\,\text{is totally monotonic}\\

F(x+1) &= b^{F(x)}\\

\end{align}

\]



Is unique. All that was needed was something small to adapt the lemma for the case \(\lambda^x\) to an infinite series. I forget because it was so long ago, but I believe Sheldon and I and the original poster agreed on the answer. I cannot remember his name for the life of me, he used to be on here occasionally. It was a long russian name, I want to say Reshetnikov, or something close to that. Damn my memory. I'll try to find the answer. But I believe if you're willing to put the work in this answer was affirmed in the positive.



The result involved Borel measures, I remember that, as it was required to prove the above uniqueness on \(f\).

---

Edited for clarity.

The proof went about as follows. Since \(f\) is totally monotonic, there exists a borel measure \(\mu\) such that:

\[
f(x) = \int_0^\infty e^{-xt}\,d\mu(t)\\
\]

Since:

\[
\lambda f(x) = \int_0^\infty e^{-(x+1)t}\,d\mu(t)\\
\]

We must have that \(\mu(t)\) is a mole at \(t_0\) such that \(e^{-t_0} = \lambda\). This is covered by the uniqueness of Borel measures under the laplace transform.

A similar argument can be made for \(F(x)\), as there exists a Borel measure \(\sigma\) such that:

\[
F(x) = \int_0^\infty e^{-xt}\, d \sigma\\
\]

From here you subbed in the Taylor series, and used this Laplace representation to deduce that for the function to be totally monotonic:

\[
\sigma(t) = \delta(t)x_0 - \sum_{j=1}^\infty a_j\delta(t - j\log(\lambda))\\
\]

For \(\delta\) the dirac function. Then, this is where I believe there was a missing key in the proof somewhere to hammer the nail home.

[Gottfried]

This may be slightly irrelevent, but there was a large thread on here, I want to say 7 years ago about totally monotonic solutions, and whether that was a uniqueness criterion. It evolved into a thread on MO, in which a couple of things were revealed.
(...)

I forget because it was so long ago, but I believe Sheldon and I and the original poster agreed on the answer. I cannot remember his name for the life of me, he used to be on here occasionally. It was a long russian name, I want to say Reshetnikov, or something close to that. Damn my memory. I'll try to find the answer. But I believe if you're willing to put the work in this answer was affirmed in the positive.

(...)

I think it has been this thread (Jan. 2017) and his author's name is Vladimir Reshetnikov

The basic idea was around to take "q-binomials" as basis for the representation of the tetration as series. Some of the involved members of this forum have deleted their accounts, which measure unfortunately anonymized their contributions; even more V.Reshetnikov reduced his presence in MO shortly after that with no more answers and only few questions, and so this thread had no "children".

Funny thing...

Gottfried

p.s. perhaps it's worth to transfer the contents of the thread to the forum here in some format? I think it's well written material...