differentiable function in higher dimensions

Question

$f:\mathbb R^{n}\to\mathbb R^{k}$ is a function with $\|f(x)\| \leq \|x\|^{2}$ , $x$ is an element of $\mathbb R^{n}$. Show that $f$ is differentiable at $0$. I already found that $f(0)=0$. I thought that $f(x)= o(|x|)$ (little o-notation). And f is differentiable in 0 if and only if there is a linear map A such that $f(x)= f(0) + A(x) + o(|x|)$. So we take for A the linear map that projects everything on zero. Is this correct or totally wrong? Thanks

Answer 1

Yes. You have $$ \frac{\|f(x)-f(0)-0\|}{\|x\|}=\frac{\|f(x)\|}{\|x\|}\leq\frac{\|x\|^2\|}{\|x\|}=\|x\|. $$ This implies that $$ \lim_{x\to0}\frac{\|f(x)-f(0)-0\|}{\|x\|}=0, $$ which says that $f$ is differentiable at $x=0$.