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Let $f_0>0$, $\alpha>0$, $p>1$, $T>0$ and $\left[\frac{t_0}2,t_0\right]\subseteq[0,T]$ for some $t_0>0$. Suppose $f:[0,T]\to\mathbb R$ satisfies $f\ge\frac{f_0}2$ and $f'>0$ on $\left[\frac{t_0}2,t_0\right]$. Moreover, assume $f''\ge\alpha|f^p|$ on $[0,T]$.

How can we conclude that $f'(x)>\sqrt{\frac\alpha{p+1}}\sqrt{f^{p+1}(x)-f^{p+1}\left(\frac{t_0}2\right)}$ for all $x\in\left(\frac{t_0}2,t_0\right)$?

I've played around with this and I guess it is really easy, but I'm not able to find the correct approach to show this. The closest I was able to obtain is considering \begin{equation}\begin{split}\frac{\rm d}{{\rm d}x}f^{p+1}(x)&=\frac{\rm d}{{\rm d}x}f^p(x)f(x)\\&=\left(\frac{\rm d}{{\rm d}x}f^p(x)\right)f(x)+f^p(x)f'(x)\\&=(p+1)f^p(x)f'(x)\\&\le\frac{p+1}\alpha f''(x)f'(x).\end{split}\end{equation}

Remark: Somehow this reminds me on what's written about "first integrals" here. Or it might be an application of the mean value inequality ...

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Yes, this is a first integral problem and you are most of the way there.

Note that $f\geq\frac{f_0}{2}\geq0$, so we can the drop absolute values in $$f''\geq\alpha|f|^p$$

You are now missing an integrating factor: multiplying by $2f'$, we have $$2f'f''\geq2\alpha f^pf'$$ The left-hand side is the derivative of $f'^2$; the right-hand side the derivative of $\frac{2\alpha}{p+1}f^{p+1}$. Now integrate and solve.

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  • $\begingroup$ Thank you for your answer. I don't know why I didn't saw that in the first place! Now I'm still curious how I can obtain the inequalities I've assumed here in the first place: math.stackexchange.com/q/4149374/47771. Maybe you can answer that question as well. Would be great if you could take a look! $\endgroup$
    – 0xbadf00d
    May 26, 2021 at 6:21
  • $\begingroup$ Oh, aren't you missing a factor of $2$ on the right-hand side of $2f'f''\geq \alpha f^pf'$? $\endgroup$
    – 0xbadf00d
    May 26, 2021 at 6:41
  • $\begingroup$ @0xbadf00d: I was indeed. $\endgroup$ May 28, 2021 at 3:41

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