1. ## Differentiability Question

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

2. Originally Posted by JohnLeee

I don't understand how to prove this considering f(x) is not specifically given....
At zero, $f(0)\leq ||0||^a\implies f(0)=0$. So $\frac{|f(x)-f(0)|}{||x-0||}\leq||x||^{a-1}$.

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

If $a=1$, all you know is that $|f'(0)|\leq1$.

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

4. A good counterexample is $f:\mathbb{R}\longrightarrow\mathbb{R}$ with $f(x)=x$.

5. Originally Posted by redsoxfan325
A good counterexample is $f:\mathbb{R}\longrightarrow\mathbb{R}$ with $f(x)=x$.
But $f$ is differentiable at $0$, I think we need a function such that $\vert f(x) \vert \leq \Vert x \Vert$ and it's not diff. at $0$ (If such a counter-example exists)

6. Oh, in that case how about $x\sin(1/x)$?