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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

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

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If smooth means $C^\infty$ then $x \mapsto \sqrt{|x|}$ is obviously not smooth – Mercy King Jun 25 '12 at 19:10
ah, ok yes my fault, it means all derivatives must be continuous. ok, can you please give me a hint how to prove that the fraction is bounded for smooth $\phi$? – Stefan Jun 25 '12 at 19:12

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 $0$.

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(Or use the Mean Value Theorem.) – Micah Jun 25 '12 at 19:27

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 $[-1,1]$).

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