By definition a logarithmically convex function is a positive real-valued function $f(x)$ defined on a convex set such that $\log f(x)$ is convex i.e. $$\forall\alpha\in[0,1]\hspace{0.5cm}\log f(\alpha x+(1-\alpha)y)\leq\alpha\log f(x)+(1-\alpha)\log f(y)$$ I have not been fully successful in proving the following statement about logarithmically convex functions. Hence I would appreciate any suggestions or proposed solutions.

$\textbf{The statement:}$

A real valued function $S(x)$ on the interval $[0,\infty)$ such that $S(0)=0$ is an increasing logarithmic convex function if and only if there is an increasing function $s(t)$ on the interval $[0,\infty)$ such that $$S(x)=\int^{x}_{0}\frac{s(t)}{t}\,dt$$

$\textbf{My attempt:}$

"if direction ($\Rightarrow$) under the assumption that $S(x)$ is twice differentiable"

Let $S(x)$ be an increasing logarithmic convex function. First $S(x)$ is convex since $$S(x)=\exp(\log S(x))$$ and $\exp(\star)$ is an increasing convex function and $\log S(x)$ is convex by assumption. Second given that $\log S(x)$ is increasing then $$\log S(x)\geq \log S(y)$$ whenever $x\geq y$.

Also because $S(x)$ is convex then it is continuous on the specified interval $[0,\infty)$. Under the additional assumption that $S(x)$ is differentiable we could represent $S(x)$ as $$S(x)-S(0)=\int^{x}_{0}\tilde s(t)\,dt$$ for any $x\geq0$. Differentiating both sides of the equality above yields $$S'(x)=\tilde s(x)$$ Using the fact that $S(x)$ is convex then $S''(x)\geq0$ implying $\tilde s'(x)\geq0$. Now set $$\tilde s(x)=\frac{s(x)}{x}$$ where $s(x)$ is a sufficiently increasing function (by sufficiently I mean it dominates the decreasing effect of $\frac{1}{x}$).


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