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Is there a non negative, convex, and decreasing function $g$ on $[0,\infty)$, with $g(0)=1$, such that $g(s+t)< g(t)g(s)$ for $s,t \in (0,\infty)$?

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  • $\begingroup$ I guess you mean "for all $s,t>0$", since otherwise taking $s=t=0$ would yield $1<1$. Then $g(t)=1+t$ would work. $\endgroup$ Apr 18, 2022 at 13:48
  • $\begingroup$ yes, you are right. $\endgroup$ Apr 18, 2022 at 14:14
  • $\begingroup$ I added the word decreasing in the question. Is there such a function? $\endgroup$ Apr 18, 2022 at 14:28
  • $\begingroup$ Maybe $g(t) =2^{1-t}-1$ $\endgroup$
    – Dattier
    Apr 18, 2022 at 14:47
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    $\begingroup$ $e^{-t^2-2t}$ should work, unless you keep adding more conditions. Btw I think a better place to ask this question would be mathstackexchange $\endgroup$
    – Saúl RM
    Apr 18, 2022 at 16:03

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