Is there a characterization of all continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying:


* $f(0)=0$
* $f$ is monotonically increasing
* $f$ is concave


My intuition is that $f$ should admit an integral representation:
$$
f(t) = \int\_0^t\, g(s)\,ds
$$
for some positive continuous function to satisfy 1 and 2, but I don't know what would be needed for 3?



Representation of continuous, monotone, concave functions
$\newcommand\R{\mathbb R}$Since $f$ is nondecreasing, concave, and continuous on $[0,\infty)$, it has a nonnegative nonincreasing right derivative $g=f'\_+$ on $[0,\infty)$, which is of course right continuous. So, there is a limit $a\_g:=\lim\_{u\to\infty}g(u)\in[0,b\_g]$, where $b\_g:=g(0)<\infty$.


Let $\mu\_g$ be the finite Lebesgue--Stieltjes measure on $(0,\infty)$ defined by the condition $\mu\_g((u,\infty))=g(u)-a\_g$ for $u\in(0,\infty)$; such a measure exists because $g$ is right continuous. Then for all $x\in[0,\infty)$
$$
\begin{aligned}
f(x)&=\int\_0^x du\,g(u) \\ 
&=\int\_0^x du\,\Big(a\_g+\int\_{(u,\infty)}\mu\_g(dv)\Big) \\ 
&=a\_g x+\int\_{(0,\infty)}\mu\_g(dv)\int\_0^{\min(x,v)} du \\ 
&=a\_g x+\int\_{(0,\infty)}\mu\_g(dv) f\_v(x), 
\end{aligned}
$$
where
$$f\_v(x):=\min(x,v).$$


Vice versa, since for any $v\in(0,\infty)$ the function $f\_v$ on $(0,\infty)$ is nondecreasing, concave, and continuous, and $f\_v(0)=0$, we have the following: if
$$f(x)=a x+\int\_{(0,\infty)}\mu(dv) f\_v(x) \tag{1}\label{1}$$
for some real $a\ge0$, some finite measure $\mu$ on $(0,\infty)$, and all real $x\ge0$, then the function $f$ on $(0,\infty)$ is nondecreasing, concave, and continuous, and $f(0)=0$.


We conclude that the functions $f$ on $[0,\infty)$ that are nondecreasing, concave, and continuous, and with $f(0)=0$ are precisely the functions of the form \eqref{1}.


