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This post asked, essentially, for an example of a "non-polynomial" invertible increasing function $f\colon[0,\infty)\to[0,\infty)$ such that $f(0)=0$ and \begin{equation} f(cu)f(t)\le f(tu) \tag{1}\label{1} \end{equation} for some real $c>0$, all $u\in[0,1]$, and all real $t\ge0$; the question in that post is a bit rephrased here. An answer to that question was given.

Then the OP said in a comment that the condition that $f$ be a Young function was missing in that post. A simple example of a Young function $f$ satisfying the above conditions will be given in the answer below.

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  • $\begingroup$ I need More examples (2 or 3) of invertible non-polynomial Young functions 𝑓 satisfying the condition: There exists 𝑐>0 such that 𝑓(𝑐𝑢)𝑓(𝑡)≤𝑓(𝑡𝑢) for all 𝑢∈[0,1] and 𝑡≥1. $\endgroup$
    – Guy Fsone
    Commented Feb 21, 2022 at 16:18
  • $\begingroup$ @GuyFsone : I have added two families of such examples. $\endgroup$ Commented Feb 21, 2022 at 19:47
  • $\begingroup$ @GuyFsone : Do you have a response about the added families of examples? $\endgroup$ Commented Feb 24, 2022 at 4:22
  • $\begingroup$ @losif Pinelis, Sorry I completely forgot the bounty... Thanks all the effort made.. This problem has been puzzling me for long time... In fact wanted an example of such functions with exponential growth... all the example I have so far are only, of polynomial growth $\endgroup$
    – Guy Fsone
    Commented Mar 17, 2022 at 0:02
  • $\begingroup$ @GuyFsone : You did not say "with exponential growth". Instead, you said "More examples (2 or 3) of invertible non-polynomial Young functions", which is what I provided. $\endgroup$ Commented Mar 17, 2022 at 3:40

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Consider the Young function $f$ defined by \begin{equation*} f(x):=\ln(1 + x) - x + x^2/2. \tag{$*$}\label{*} \end{equation*} Take $c=1$ and then rewrite \eqref{1} as \begin{equation*} g(t):=f(u)f(t)-f(tu)\le0. \tag{2}\label{2} \end{equation*} We have $g(0)=0$ and \begin{equation*} g'(t)=f(u)f'(t)-uf'(tu) =\frac{t^2}{2 (1 + t)}\,g_1(u), \end{equation*} where \begin{equation*} g_1(u):=2 \ln(1+u)-\frac{u \left((t+2) u^2+(2 t-1) u+2\right)}{1+t u}. \end{equation*} Next, $g_1(0)=0$ and \begin{equation*} g'_1(u)=-\frac{2 u^2 \left(t^2 u (u+2)+t \left(2 u^2+3 u+3\right)+3 u+2\right)} {(1+u) (1+t u)^2}, \end{equation*} which is manifestly $\le0$ for $u\in[0,1]$ and $t\ge0$. So, $g_1\le0$ and hence $g$ is decreasing. Thus, \eqref{2} follows, and hence \eqref{1} follows as well.


User Guy Fsone has now stated, in a comment to the question above, "I need More examples (2 or 3) of invertible non-polynomial Young functions $f$ satisfying the condition: There exists $c>0$ such that $f(cu)f(t)\le f(tu)$ for all $u\in[0,1]$ and $t\ge1$."

Let us therefore provide an entire series of examples generalizing the example above. To do that, take any natural $m$ (so that $m\ge1$) and let \begin{equation} f(x):=f_m(x):=\ln(1+x)-\sum_{k=1}^{2m}\frac{(-1)^{k-1}}k\,x^k \tag{3}\label{3} =\int_0^x\frac{a^{2m}\,da}{1+a} \end{equation} for real $x\ge0$, so that the function $f$ given by \eqref{*} is $f_1$. Moreover, $f=f_m$ is an invertible non-polynomial Young function.

Take $c=1$ and then rewrite \eqref{1} as \eqref{2}, as was done previously. We have $g(0)=0$ and \begin{equation*} g'(t)=f(u)f'(t)-uf'(tu) =\frac{t^{2m}}{1 + t}\,g_1(u), \end{equation*} where \begin{equation*} g_1(u):=f(u)-\frac{u^{2m+1}}{1 + ut}\,(1+t). \end{equation*} Next, $g_1(0)=0$ and \begin{equation*} g'_1(u)=-\frac p{(1+u) (1+t u)^2}, \end{equation*} where \begin{equation*} p:=u^{2 m} \left(u^2 \left((2 m-1) t^2+2 m t\right)+u \left(2 m t^2+(4 m-1) t+2 m+1\right)+(2 m+1) t+2 m\right), \end{equation*} so that $g'_1(u)$ is manifestly $\le0$ for $u\ge0$ and $t\ge0$ (and $m\ge1$). So, $g_1\le0$ and hence $g$ is decreasing, from $g(0)=0$. Thus, \eqref{2} follows, and hence \eqref{1} follows as well.


Yet another family of such examples is given by the formula \begin{equation*} f(x):=f_b(x):=\frac{x^b}{1+x} \end{equation*} for real $b>2$. Here all the desired conditions can be checked easily.

More generally, for each real $c>0$, consider the following set of functions (indexed by three parameters $b,p,q$): \begin{equation*} F_c:=\big\{f_{c;\,b,p,q}\colon b\in(2,\infty), p\in(0,c^b], q\in(0,c^b]\big\}, \end{equation*} where the function $f_{c;\,b,p,q}$ is defined by the formula \begin{equation*} f_{c;a,p,q}(x):=\frac{x^b}{p+qx} \end{equation*} for real $x\ge0$. Then it is easy to check that every function $f\in F_c$ is an invertible non-polynomial Young function satisfying condition \eqref{1} for all real $u,t\ge0$.

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