why is $\frac{\phi(x)-\phi(-x)}{x}$ for smooth $\phi$ bounded at $x=0$

Question

Why is $\frac{\phi(x)-\phi(-x)}{x}$ for smooth $\phi$ bounded at $x=0$? If i set $\phi(x) = \sqrt{|x|}$, it definitely not bounded. I saw this on page 293 of

http://www.math.ucdavis.edu/~hunter/book/pdfbook.html

where in example it 11.7 it is claimed the integrand (which is an expression like mine) is bounded?

Answer 1

Using fundamental theorem of calculus: $$\phi(x)-\phi(-x)=\int_{-x}^x\phi'(t)dt=x\phi'(x)+x\phi'(-x)-\int_{-x}^xt\phi''t)dt,$$ hence for $0<|x|\leq 1$, we have $$\left|\frac{\phi(x)-\phi(-x)}x\right|\leq 2\sup_{|s|\leq 1}|\phi'(s)|+2\sup_{|s|\leq 1}|\phi''(s)|,$$ which shows boundedness, since both supremum are finite ($\phi'$ and $\phi''$ are continuous on $[-1,1]$).

Answer 2

If $\phi$ is smooth (or even $C^1$), then we can write:

$$2\phi'(0)=\lim_{x \to 0} \frac{\phi(x)-\phi(0)}{x} + \lim_{x \to 0} \frac{\phi(-x)-\phi(0)}{-x} = \lim_{x \to 0} \frac{\phi(x)-\phi(-x)}{x}\, .$$

So your quotient can be extended to a function which is continuous at $0$, which means it must be bounded near $0$.