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I don't understand how to prove this considering f(x) is not specifically given....(Wondering)

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- Oct 25th 2009, 08:51 PMJohnLeeeDifferentiability Question
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I don't understand how to prove this considering f(x) is not specifically given....(Wondering) - Oct 25th 2009, 09:13 PMredsoxfan325
At zero, $\displaystyle f(0)\leq ||0||^a\implies f(0)=0$. So $\displaystyle \frac{|f(x)-f(0)|}{||x-0||}\leq||x||^{a-1}$.

Take the limit of both sides as $\displaystyle x\to0$ and you get $\displaystyle f'(0)=0$.

If $\displaystyle a=1$, all you know is that $\displaystyle |f'(0)|\leq1$. - Oct 25th 2009, 09:16 PMJose27
Remember that for a function to be diff. at $\displaystyle 0$ there exists a linear transformation $\displaystyle D_f(0)$ such that $\displaystyle \vert \frac{f(h)-f(0)-D_f(0)h}{ \Vert h \Vert } \vert \rightarrow 0$ as $\displaystyle h \rightarrow 0$. Take $\displaystyle D_f(0) \equiv 0$ then $\displaystyle \vert \frac{f(h)-f(0)}{ \Vert h \Vert } \vert \leq \frac{ \Vert h \Vert ^a}{ \Vert h \Vert } = \Vert h \Vert ^{a-1}$ and since $\displaystyle a-1>0$ we have that this goes to $\displaystyle 0$ as $\displaystyle h$ goes to $\displaystyle 0$. This proves that $\displaystyle f$ is diff. at $\displaystyle 0$ and $\displaystyle D_f(0) \equiv 0$. When $\displaystyle a=1$ we can only bound by $\displaystyle 1$, but I can't think of a counter-example at the moment for this case.

- Oct 25th 2009, 09:19 PMredsoxfan325
A good counterexample is $\displaystyle f:\mathbb{R}\longrightarrow\mathbb{R}$ with $\displaystyle f(x)=x$.

- Oct 25th 2009, 09:27 PMJose27
- Oct 25th 2009, 09:31 PMredsoxfan325
Oh, in that case how about $\displaystyle x\sin(1/x)$?