If $\| f(x) \| \leq \| x \|^{\alpha}$, for $\alpha > 1$ then f is differentiable in $p=0$

Let $f: \mathbb{R}^k \rightarrow \mathbb{R}^m$, show that if $\| f(x) \| \leq \| x \|^{\alpha}$, for $\alpha > 1, \forall x \in \mathbb{R}^k$, then f is differentiable in $p=0$ with differential $df_0 = 0$.

I guess I'd simply have to show that:

$$\lim_{h \rightarrow 0} \frac{\| f(x+h)-f(x) \|}{\|h\|} =\lim_{h \rightarrow 0} \frac{\| f(h)-f(0) \|}{\|h\|} = 0$$,

but I'm not sure how to do this and I don't know how the inequality is supposed to be of any help. Any hints?

• How about the triangle inequality? – 5xum Jun 15 '15 at 10:20
• what is $p$???? – tattwamasi amrutam Jun 15 '15 at 10:20
• @5xum I would get a greater than inequality by using the triangle inequality, wouldn't I? $\|f(h)-f(0)\| \geq \| \|f(h)\| - \|f(0)\| \|$, I don't see how this would be of any help. – eager2learn Jun 15 '15 at 10:27
• @tattwamasiamrutam $p=0$, a point in $\mathbb{R}^k$, as it was stated in the question. – eager2learn Jun 15 '15 at 10:27
• @eager2learn Yes, but you also get $||f(h) - f(0)|| \leq ||f(h))|| + ||f(0)||$. And you can use the inequality to calculate what $f(0)$ is equal to. – 5xum Jun 15 '15 at 10:29

Yes, that is pretty much it. The inequality gives $f(0) = 0$. So: $$0 \leq \frac{\|f(h)\|}{\|h\|} \leq \|h\|^{\alpha - 1} \implies 0 \leq \lim_{h \to 0}\frac{\|f(h)\|}{\|h\|} \leq 0 \implies \lim_{h \to 0}\frac{\|f(h)\|}{\|h\|} = 0,$$ because $\alpha - 1 > 0$.