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While studying some old exams, I came across this problem, which I can't seem to figure out:

Let $B(r,0) \subset \mathbb{R}^n$ be a ball of radius $r$ centered at $0$. Let $f:B(r,0) \rightarrow \mathbb{R}$ and suppose there exists $\alpha > 1$ such that $|f(x)| \leq ||x||^\alpha$ for all $x \in B(r,0)$. Prove that $f$ is differentiable at $0$. What happens to this result when $\alpha = 1$?

In response to the second part of the question, the function $f(x) = ||x||$ is not differentiable.

Does anyone have any hints for the first part? Thanks.

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2 Answers 2

up vote 3 down vote accepted

Hint: Note that $f(0) = 0$ and $\frac{|f(x)|}{\|x\|} \leq \|x\|^{\alpha - 1}$. Recall the definition of the derivative.

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Of course! Thank you. –  Dane Jul 15 '11 at 21:19

Can you solve a more specific simplier version, such as $f:(-r,r) \rightarrow \mathbb{R}$ having the property that $|f(x)| \leq |x|^2$?

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