# Let $f:R\to R$ be a $C^1$

If $f(0)=f'(0)=0$ and $|f(x)|\leq |x|^{1+a}$ then $|f'(x)|\leq c|x|^a$.

I tried to use the mean value theorem and I am pretty sure it is a good way not to prove that.

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who asked you to prove this? It is false. You know too little about the behavior of $f$ to conclude anything about its derivative for nonzero $x$. –  Stefan Smith Apr 24 '13 at 1:10

## 1 Answer

Counterexample: $f(x)=x^{1+a}\sin(x^{-a})$ for $x\neq0$ and $f(x)=0$ for $x\le 0$ and $a>0$.

Differentiability can be shown at $x=0$ using the difference quotient. However, the derivative does not tend to zero as $x\to0$.

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Are you sure this is a proper counterexample? Could you provide a proof that it is continuously differentiable for all $a \ge 0$? –  Cameron Williams Apr 23 '13 at 17:47
It is only continuous differentiable for a>0 –  Moritz Reinhard Apr 23 '13 at 17:49
however, you are right my counterexample is not completely correct. I will edit my answer. –  Moritz Reinhard Apr 23 '13 at 17:52
The other counterexample was right. –  checkmath Apr 23 '13 at 18:04