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Given strictly concave function $f(t)$ that satisfies $f'(t)>0$, $f'(t)=o(1)$ (i.e. $\lim\limits_{t\to\infty}f'(t)=0$) , and $f'(t)=\omega\left(t^{-1}\right)$ (i.e. $\lim\limits_{t\to\infty}tf'(t)=\infty$) , we denote $F=\exp\left(f\left(t\right)\right)$. Let $H(t)$ be a solution of $$ \dot{H}=F $$ Can we approximate H by F?

Specifically, I want to show that $$ \lim_{s\to\infty}\frac{H^{-1}(s)}{F^{-1}(s)}=1\,, \ \ \ \ (a) $$ where $F^{-1}$ and $H^{-1}$ are the inverse functions of $F$ and $H$, respectively.

If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

* In this question, $(a)$ was proved for $f'(t)=\Omega(1)$ (without the assumption that $f(t)$ is strictly concave). I'm interested in the case that $f'(t)=o(1)$.

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  • $\begingroup$ By $\dot{H}$ you mean the derivative of $H$, or something else ? $\endgroup$
    – N. de Rancourt
    Commented Oct 18, 2018 at 19:19
  • $\begingroup$ Yes, the derivative of $H$. $\endgroup$
    – Mor
    Commented Oct 18, 2018 at 19:22

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Not a full answer, but a hint at a negative answer: take $f(t)=\log t$, which doesn't satisfy $\lim_{t\to\infty}tf'(t)=\infty$, critically, but is nearer to the proven case (where $f'(t)=\omega(1)$ ).

In this case you have $F(t)=t$, $H(t)=t^2/2$, $F^{-1}(s)=s$, $H^{-1}(s)=\sqrt{2s}$.

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    $\begingroup$ Thank you for your comment. I agree that eq. (a) doesn't hold for $f(t)=\log(t)$ which doesn't satisfy the condition $f'(t)=\omega(t^{-1})$. However, I think that with this condition the equation should be true. We have that $\dot{H}(t)=F(t)=\exp(f(t))$. If $\exp(f(t))\approx\exp(f(t))f'(t)=\frac{d}{dt}\exp(f(t))$ then I think we can approximate $H\approx \exp(f) = F$. For this to hold we need $\log(f’(t))=o (f (t))$ (i.e. $\lim\limits_{t\to\infty}\frac{\log(f’(t))}{f(t)}=0$). Note that $f(t)=\log(t)$ doesn’t satisfy this requirement, but $f(t)=\log^\epsilon(t)$ for $\epsilon>1$ does. $\endgroup$
    – Mor
    Commented Oct 20, 2018 at 11:46
  • $\begingroup$ Yes, it's more subtle than I thought... $\endgroup$
    – Jean Duchon
    Commented Oct 24, 2018 at 16:16

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