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I have the following condition of the function $f : \mathbb R \to \mathbb R$: $$\exists\, h:\mathbb R_+ \to \mathbb R_+ \text{ bijective s.t., } \forall s>0, \frac{\partial^2}{\partial^2 s}\left(\left.\frac{\partial^k}{\partial^kt} f\left(h(s)t\right)\right\rvert_{t = -1}\right) >0,\; \forall k \in \{0,\dotsc,n\}.$$

What does it mean for the function $f$?

First developments show that this is equivalent to: $$\exists\, h:\mathbb R_+ \to \mathbb R_+ \text{ bijective s.t., } \forall s>0, \frac{\partial^2}{\partial^2 s}h(s)^kf^{(k)}(-h(s)) >0,\; \forall k \in \{0,\dotsc,n\}.$$

Then, I want to let $g(s) = \sqrt{h(s)}$ to obtain:

$$\exists\, g:\mathbb R_+ \to \mathbb R_+ \text{ bijective s.t., } \forall s>0, \frac{\partial^2}{\partial^2 s}g(s)^{2k}f^{(k)}(-g(s)^2) >0,\; \forall k \in \{0,\dotsc,n\}.$$

And then I'm stuck.

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    $\begingroup$ What is $n$? Is $n = 0$ allowed? $\endgroup$
    – Dieter Kadelka
    Commented Dec 9, 2021 at 21:02
  • $\begingroup$ Well no, if you want to fix $n$, i'll prefer to understand the case $n\to\infty$ than $n=0$. $\endgroup$
    – lrnv
    Commented Dec 10, 2021 at 8:46

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