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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 an elementary function.

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  • $\begingroup$ For $a(u) = u\exp(u)$ the inverse $a^{-1}$ is the Lambert-W function and $g$ can also be expressed in terms of Lambert-W functions - but this is probably not elementary enough? $\endgroup$ – Dirk Oct 12 at 9:23
  • $\begingroup$ @Dirk : Thank you for your comment. I'd agree that Lambert's function is almost elementary. From the latest edit of the post, it should now be clear that what is mainly needed is that both $f$ and $(f')^{-1}$ be elementary functions. $\endgroup$ – Iosif Pinelis Oct 12 at 15:42

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