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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 deveoppements 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.

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.

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

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

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

And then I'm stuck.

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 deveoppements 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.

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 $f$?

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

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

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

And then I'm stuck.