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.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: