Let $F$ be the set of all convex functions $f\colon[0,\infty)\to[0,\infty)$ with $f(0)=0=f'_+(0)$ and $f_+(\infty-)=\infty$, where $f'_+$ is the right derivative of $f$. For any function $f\in F$, its Legendre–Fenchel transform $g_f\colon[0,\infty)\to[0,\infty)$ (also known as the convex conjugate of $f$) is defined by the formula $$g_f(y):=\sup_{x\ge0}(xy-f(x))$$ for real $y\ge0$.

E.g., if $f(x)=x^p/p$ for a real $p>1$ and all real $x\ge0$, then $f\in F$ and $g_f(y)=y^q/q$ for $q:=1/(1-1/p)$ and all real $y\ge0$.

A couple of other pairs $(f,g_f)$ of "explicit" functions with $f\in F$ can be obtained from this table, including the one with $f(x)=e^x-1-x$ for all real $x\ge0$ and $g_f(y)=(1+y)\ln(1+y)-y$ for all real $y\ge0$.

Are there any pairs $(f,g_f)$ of "explicit" (say elementary, in some sense) functions with $f\in F$ such that $f$ increases faster than any exponential function: $f(x)/e^{cx}\underset{x\to\infty}\longrightarrow\infty$ for any real $c$?

The question can restated in the following, essentially equivalent, and possibly more transparent, form: Does there exist a continuous strictly increasing function $a\colon[0,\infty)\to[0,\infty)$ such that $a(0)=0$, $a(u)/e^{cu}\underset{u\to\infty}\longrightarrow\infty$ for each real $c$, and the functions $f\colon[0,\infty)\to[0,\infty)$ and $g\colon[0,\infty)\to[0,\infty)$ given by $$f(x):=\int_0^x a(u)\,du,$$ \begin{aligned} g(y)&:=\int_0^y a^{-1}(v)\,dv \\ &=\int_0^{a^{-1}(y)}u a'(u)\,du \\ &=a^{-1}(y)y-\int_0^{a^{-1}(y)}a(u)\,du \\ &=a^{-1}(y)y-f(a^{-1}(y)) \\ &=(f')^{-1}(y)y-f((f')^{-1}(y)) \end{aligned} for all real $x,y\ge0$ are elementary, in some sense? Here $a^{-1}$ is the function inverse to $a$.

So, it is enough to find an elementary convex function $f\colon[0,\infty)\to[0,\infty)$ such that $f(0)=f'(0+)=0$, $f(x)/e^{cx}\underset{x\to\infty}\longrightarrow\infty$ for each real $c$, and $(f')^{-1}$ is elementary function.