let $U$ be a convex open set, a function $f \; : \; U \to \mathbb{R}^{+}$ is said to be $\log$-convex when $$f\left(\frac{x}{p}+\frac{y}{q}\right) \leq f(x)^{1/p}f(y)^{1/q}$$ for any $x,y \in U$ and $p,q \in [1,\infty]$ such that $1/p + 1/q = 1$ (equivalently $\log(f)$ is a convex function).
A class of $\log$-convex functions are the functions of the type $$f(x) := \int_{X}{g(t)^{x}d\mu}$$ where $(X,\mu)$ is a measurable space, if for example one takes $g(t) = t$ and $d\mu = \frac{e^{-t}}{t}dt$ one recovers the Gamma function.
of course the $\log$-convexity of $f$ is a trivial application of Holder-Inequality.
I was wondering if, at least locally, any $\log$-convex function can be represented in this form, so I was wondering if the following conjecture is true
Conjecture : let $f \; : \; U \to \mathbb{R}^+$ be a $\log$-convex function with (WLOG) $0 \in U$, ,then there exists a $\delta > 0$, a measure space $(X,\mu)$ and a measurable function $g \; : \; X \to \mathbb{R}^+$ such that $B(0,\delta) \subset U$ and for any $x \in B(0,\delta)$ $f(x) = \int_{X}{g(t)^x d\mu}$
Is this true? Has this been observed before?
You probably need more conditions.
Consider $x_1$, $\ldots$, $x_n$ points in the domain $U$ of $f$. The matrix
$$\left(f(\frac{x_i+x_j}{2})\right)_{1\le i,j\le n}$$
is positive semidefinite ( check that). For $n=2$ this means logarithmically convex ( the midpoint), but then $n$ can be anything.
Now we need such an $f$ that does not satisfy the condition. A possible approach. Start with an $f_1$ that has a representation. Now it may happen that $f\colon = f_1^{\alpha}$ will not have a representation ( a determinant of order $3$ perhaps not positive) for some $0 < \alpha < 1$. Note that $f$ is still log convex.
$\bf{Added:}$
If $g$ is a positive valued function on some space $X$ with a measure $\mu$ then we have the change of variable formula
$$\int_X g(t)^x d\mu = \int_{(0, \infty)} t^x d \mu_1$$
where $\mu_1$ is the push forward of $\mu$ by $g$. We can also write the second integral as
$$\int_{\mathbb{R}} e^{x t} d \eta(t)$$
where $\eta$ is the push of $\mu_1$ by $\log$. Therefore $f$ is either a (generalized) moment function or a (generalized) Laplace transform ( with the Laplace transform: care that $\eta$ has apriori support in $\mathbb{R}$, not only in $[0, \infty)$. So now we can bring all that we know about moment functions or Laplace transforms. Note that $f$ will be a standard Laplace transform ( of a positive measure) if an only if $0< g \le 1$. In any case the condition for the matrix above to be positive semidefinite is a classical one ( see the Hamburger moment problem or the Stieltjes moment problem)
A concrete example of a log convex function that is not a moment function. Start with $f_1 (x) = (a^x + b^x)$, $a>b>0$. Take any $\alpha\in (0, 1)$. Then the function $f(x) \colon = f_1(x)^{\alpha}$ is not a moment function. Indeed, consider a $3\times 3$ matrix $ \left ( f(\frac{x_i+x_j}{2}) \right )_{1\le i,j \le 3}$ for distinct $x_1$, $x_2$, $x_3$. Its determinant is negative ( see example )
$\bf{Added:}$
There are also differential inequalities that are satisfied by function $$f(x) = \int_I e^{x t} d\eta(t)$$
where $I$ is a subset of $\mathbb{R}$, and $\eta$ a positive measure on $I$.
Assume that $P(t)$ is a polynomial of $1$ variable that is $\ge 0$ on $I$. We have
$$P(D) f(x) = \int_I e^{x t} P(t) d \eta(t) \ge 0$$
($D = \frac{d}{d x}$ is the derivative operator).
In particular, since every square of a square of a real polynomial is $\ge 0$ everywhere, we conclude
$$\sum f^{(i+j)}(x) a_i a_j \ge 0$$
for every $a_0$, $\ldots$, $a_n$ real numbers. We conclude that for every $x\in D$ and $n\ge 0$, the matrix
$$(f^{(i+j)}(x))_{0\le i,j\le n}$$ is positive semidefinite. In particular all derivatives of even order are $\ge 0$.
These differential inequalities are similar to the inequalities in Bernstein's theorem.