I am for example(s) of an invertible Convex or concave function $\phi: [0,\infty)\to [0, \infty)$ such that $\phi(0)=0$ and there exists $\theta>0$ and for all $s\leq t$ we have

\begin{align}\label{EqI}\tag{I} \phi\big(\theta \frac{s}{t}\big) \leq \frac{\phi(s)}{\phi(t)} \qquad\text{or equaly} \qquad \theta \leq \phi^{-1}\big(\frac{s}{t}\big)\frac{\phi^{-1}(t)}{\phi^{-1}(s)} \end{align}

The most simple class consists of polynomial functions of the form $\phi(t)= ct^p$ with $c>0$ and $p>0$.

Question: Are there other possible non-polynomial examples satisfying $\eqref{EqI}$?

As an attempt with $\phi(t)= e^{t^\alpha}-1$, I wonder if there is a constant $c>0$ such that

$$ \ln(t+1)\ln\left(\frac{s}{t}+1\right)\geq c\ln(s+1),\qquad \text{for all $0\leq s\leq t$}.$$


1 Answer 1


With $f:=\phi$, $u:=s/t$, and $c:=\theta$, the desired inequality can be rewritten as $$f(cu)f(t)\le f(tu) \tag{1}$$ for $u\in[0,1]$ and real $t\ge0$.

Let us show that (1) holds with $c=1$ if $f(x)\equiv\ln(1+x)$. That is, we have to show that $$g(t):=\ln(1+u)\ln(1+t)-\ln(1+tu)\le0$$ for $u\in[0,1]$ and real $t\ge0$. We have $g(0)=0$ and $$g'(t)=\frac{\ln(1+u)}{1+t}-\frac u{1+tu} \le \frac u{1+t}-\frac u{1+tu}\le0$$ for $u\in[0,1]$ and real $t\ge0$. So, the desired result follows.

More generally, any concave function $f$ such that $f(0)=0$ and $0<f'\le1/c$ satisfies (1); here $f'$ denotes the right derivative of $f$. Indeed, then $0\le f(cu)\le u$ for all real $u\ge0$ and hence for $$h(t):=f(cu)f(t)-f(tu)$$ we have $h(0)=0$ and $$h'(t)=f(cu)f'(t)-uf'(tu)\le uf'(t)-uf'(tu)\le0$$ for $u\in[0,1]$ and real $t\ge0$.

So, taking any positive decreasing function $f_1$ on $[0,\infty)$ and then letting $c:=1/f_1(0)$ and $$f(x):=\int_0^x f_1(y)\,dy,$$ for real $x\ge0$, one has (1) satisfied.

The inequality $$ \ln(t+1)\ln\left(\frac{s}{t}+1\right)\ge c\ln(s+1)\qquad \text{for all $0\le s\le t$} \tag{2}$$ in your post does not hold for any real $c>0$ if $s$ is small and $t$ is large, because for such $s$ and $t$ we have $\ln\left(\frac{s}{t}+1\right)\sim\frac{s}{t}$ and $\ln(s+1)\sim s$, whereas $\ln(t+1)/t\to0$ as $t\to\infty$.

  • $\begingroup$ Thank you for this nice answer. can we have an example with a convex function instead of a concave? In fact, I am working with Young functions "which are convex"... I just forgot to mention it in my question... $\endgroup$
    – Guy Fsone
    Commented Apr 20, 2021 at 19:22
  • $\begingroup$ @GuyFsone : I think such convex functions exist. However, that would be a different question -- please ask it in a separate post. $\endgroup$ Commented Apr 20, 2021 at 20:55
  • $\begingroup$ I will ask not in this room (that won't be serious) but the in stack-exchange: please see here math.stackexchange.com/q/4109486 $\endgroup$
    – Guy Fsone
    Commented Apr 20, 2021 at 21:24
  • $\begingroup$ Or I can just edit the current version and add convex and concave so that you edit yours? See the Edit. please edit your answer. This room is very strict regarding posts.. $\endgroup$
    – Guy Fsone
    Commented Apr 20, 2021 at 21:30
  • $\begingroup$ @GuyFsone : I don't understand why a post with an additional condition would be less serious than the original one, without the additional condition. Also, I don't think editing one's question to invalidate a valid answer is encouraged on MO. $\endgroup$ Commented Apr 20, 2021 at 21:43

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