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][1] $g_f\colon[0,\infty)\to[0,\infty)$ (also known as the [convex conjugate ][2]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][3], 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 $c$? 

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If it helps, the question can restated in the following, essentially equivalent 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 
$$f(x):=\int_0^x a(u)\,du,\quad g(y):=\int_0^y a^{-1}(v)\,dv$$
for all real $x,y\ge0$ are elementary, in some sense? Here $a^{-1}$ is the function inverse to $a$.  


  [1]: https://en.wikipedia.org/wiki/Legendre_transformation
  [2]: https://en.wikipedia.org/wiki/Convex_conjugate
  [3]: https://archive.org/details/convexanalysisno00borw_812/page/n61/mode/2up