# Does $f'>0$ and $f''\ge\alpha f^p$ on $[t_0/2,t_0]$ imply $f'(x)>\sqrt{\frac\alpha{p+1}}\sqrt{f^{p+1}(x)-f^{p+1}\left(\frac{t_0}2\right)}$?

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 $$$$\begin{split}\frac{\rm d}{{\rm d}x}f^{p+1}(x)&=\frac{\rm d}{{\rm d}x}f^p(x)f(x)\\&=\left(\frac{\rm d}{{\rm d}x}f^p(x)\right)f(x)+f^p(x)f'(x)\\&=(p+1)f^p(x)f'(x)\\&\le\frac{p+1}\alpha f''(x)f'(x).\end{split}$$$$

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

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.
• Oh, aren't you missing a factor of $2$ on the right-hand side of $2f'f''\geq \alpha f^pf'$? May 26, 2021 at 6:41