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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

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?

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    If smooth means $C^\infty$ then $x \mapsto \sqrt{|x|}$ is obviously not smooth2012-06-25
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    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 $\phi$?2012-06-25

2 Answers 2