The real function $f$ such that $\log \cdots \log (f)$ is strictly convex on its domain for any number of $\log$'s

Question

Does there exist a function $f: (a,b) \to \mathbb R$ ($a,b$ are allowed to be infinity) such that $\log \cdots \log (f)$ is strictly convex on its whole domain of definition for an arbitrary number (though finitely many) of $\log$'s?

If such function exists, it must be very very convex ("more convex" than any $\exp \dots \exp (x) $, which will become concave after a finitely many $\log$'s applied on)

The form of $f$ doesn't have to be concrete. It can be infinite series or one can even show its existence or nonexistence.

Answer 1

After realizing the flaw in my previous, affirmative, answer: No such function exists.

If $f$ has a nonempty domain at every point, then it grows arbitrarily large, but since it is continuous, after sufficiently many iterations of log its range will always be $(-\infty,\infty)$. WLOG, take $f$ to be such a function (if any valid $f$ exists, all of its logarithms are valid too). It is strictly monotonically increasing (as it is convex and not bounded below), so we let $c_0<c_1<c_2<\ldots$ be the unique points at which $f$ attains the values $0,1,e,e^e,e^{e^e},\ldots$.

Now consider $g=\log(f)$. We have $g(c_1)=0, g(c_2)=1$, and that $g$ is convex and monotonically increasing. We also have $\lim_{x\to c_0}g(x)=-\infty$. But then consider the line from $(x,g(x))$ to $(c_2,1)$; as $x$ approaches $c_0$ from the positive direction, this line will drop below the point $(c_1,0)$, violating convexity.

It is possible for $f$ and all logs applied to it to be convex on all positive values, as outlined in my other answer, but not over its whole domain.

Answer 2

Define the function $g_n(x) = \text{exp}^{\circ n}(x - n)$ where by $\text{exp}^{\circ n}$ I mean the $n$-fold composition of the function. So, $g_1(x) = \text{exp}(x - 1)$, $g_2(x) = \text{exp}(\text{exp}(x - 2))$ etc.

Then I think $f(x) = g_{[x]}(x)$ satisfies the desired property, where $[x]$ is the ceiling of $x$: suppose that $x$ is such that $[x] = n$, then $\text{log}^{\circ n} f(x) = x - n$ is convex. Applying the $\text{log}$ again is not defined, so such an $x$ is only in the domain of $\text{log}^{\circ n}$ but not $\text{log}^{\circ (n + 1)}$.

EDIT: As mentioned in the comment, this has discontinuities at each integer. A paper that develops the idea of $\text{exp}^{\circ x}$ can be found here.

Answer 3

Edit: This function is only convex on its positive values, as pointed out in the comments. No such function exists - see my other answer.

A solution obtained via some modifications to RJTK's answer:

We set up our function so that it passes through $(0,0), (1/2,1), (3/4,e), (7/8,e^e),(15/16,e^{e^e}),$ and so on. Thus, after $n$ applications of log it will have a domain of $(1-2^{-n},1)$.

To interpolate between these points, we let $a_n=1-2^{-n}$ and define the function from $a_n$ to $a_{n+1}$ to be $\exp(\exp(\ldots\exp(g(\frac{x-a_n}{a_{n+1}-a_n})))$, where there are $n$ applications of the exponential and $g$ is a strictly convex increasing function on $[0,1]$ passing through $(0,0)$ and $(1,1)$.

Since $\log(f(x)) = f(2x-1)$ for all $x$ such that this is defined, we need only check convexity of $f$ and the convexity of each application of log will follow.

Clearly, $f$ is continuous and strictly convex on $(a_n,a_{n+1})$. To see that it is convex at each $a_n$, we want to show that the slope at the start of the interval $(a_n,a_{n+1})$ exceeds that at the end of the interval $(a_{n-1},a_n)$. But $f$ has the property that $f'(x)=2f(x)\cdot f'(2x-1)$. So, by induction and the fact that $f$ is monotonically increasing, it suffices to check convexity at $x=\frac12$.

Here is where $g$ must be chosen; we need that $g'(1) \le 2g'(0)$, by the previous identity. Taking $g(x)=\frac{3x+x^2}{4}$ suffices for this.

(Originally, I tried defining $f(x)$ piecewise between integers, but this forced $g'(1) \le g'(0)$, i.e., that $g$ not be strictly convex. Allowing for the slope to double at each successive piece gets around this, albeit in a somewhat hacky way.)

Answer 4

An interesting example is one defined by Hellmuth Kneser, as an analytic solution $E$ to Abel's equation which I see has been mentionned several times already:

$\forall x>c,E(x+1)=\exp(E(x))$,

where $c$ is a real constant. The function is also strictly increasing and all of its derivatives are positive on (distinct) intervals $(b,+\infty)$, $b\geq c$. This is also the case for $\log \circ \cdot \cdot \cdot \log \circ E$ because of the functional equation above.