## Pointwise gradient estimate of function

$\mathbf{Problem:}$ Suppose $f\in C^\infty_c(\mathbb{R})$, and $f\geq 0$, then

$\displaystyle |f'(x)|^2\leq 2f(x)\max|f''|$, $\forall\, x\in\mathbb{R}$

$\mathbf{Proof:}$ Let $f_\epsilon=f+\epsilon>0$, then $f'_\epsilon=f'$ and $f''_\epsilon=f''$.

Consider $\displaystyle \frac{|f'_\epsilon|^2}{f_\epsilon}\in C^\infty_c(\mathbb{R}^n)$, it must has maximum at some point $x_0$,

$\displaystyle \max\frac{|f'_\epsilon|^2}{f_\epsilon}=\frac{|f'_\epsilon|^2}{f_\epsilon}(x_0)$.

Then at $x=x_0$, $\displaystyle \left(\frac{|f'_\epsilon|^2}{f_\epsilon}\right)'(x_0)=0$. Since we have

$\displaystyle \left(\frac{|f'_\epsilon|^2}{f_\epsilon}\right)'(x_0)=\frac{f'_\epsilon(2f_\epsilon f_\epsilon''-f_\epsilon'^2)}{f_\epsilon^2}(x_0)=0$

If $\displaystyle f'(x_0)=0$, then obviously $\displaystyle |f_\epsilon'(x)|^2\leq 2f_\epsilon(x)\max|f_\epsilon''|$.

Otherwise we have $2f_\epsilon(x_0) f_\epsilon''(x_0)-f_\epsilon'^2(x_0)=0$, that is

$\displaystyle \max\frac{|f'_\epsilon|^2}{f_\epsilon}=\frac{|f'_\epsilon|^2}{f_\epsilon}(x_0)=f''_\epsilon(x_0)$

This also means $\displaystyle |f_\epsilon'(x)|^2\leq 2f_\epsilon(x)\max|f_\epsilon''|$.

Letting $\epsilon\to 0$, we have $\displaystyle |f'(x)|^2\leq 2f(x)\max|f''|$, $\forall\, x\in\mathbb{R}$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ This proof is due to Ming Xiao.