3
$\begingroup$

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

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $t\to\infty$.

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.