positively homogeneous function that differentiable at $0^k$

Question

Let $f:\Bbb R^k \to \Bbb R$ be continious function and positively homogeneous function with degree $\alpha$ in $\Bbb R^k \setminus {0^k} $

(the definition of positively homogeneous function in our course is $f(\lambda\cdot x)=\lambda^\alpha \cdot f(x)$ for $\lambda > 0$ and x $\neq 0^k$)

if $\alpha > 1$ and $f(0^k)=0$ prove that f is differentiable at $0^k$

note: the definition of differentiability at $0^k$ in our course is $lim_{h\to 0^k}$ $f(0^k+h)-f(0^k)-\nabla f(0^k)\cdot h \over |h|$ = $0$

I succeeded in proving that $∇f(0^k) = 0^k$, which means that with the data $f(0^k) = 0$ What is left to prove is $lim_{h\to 0^k}$ $f(0^k+h) \over |h|$ = $0$

The question is taken from an old exam in Calculus 3

Answer 1

Are you sure it is not $f(\lambda x) = \lambda^\alpha f(x)$ and $\alpha >1$ ? Without my first suggestion, the definition is weird. Furthermore, with the first suggestion but not the second, the function $|x|^\alpha$ should satisfy the hypotheses, but it is not differentiable at $0$.

By making those two changes, you can say that if $x$ is not equal to $0$, then $|f(x)| =||x||^\alpha |f(x / ||x||)|$. Furthermore, $f$ is continuous, and is hence bounded on the unit ball since it is a compact since we work in finite dimension. Hence, $|f(x)| \leq M ||x||^\alpha$. Thus, $|f(x) - f(0)| = |f(x)| = o(||x||)$, which is the definition of being differentiable at $0$ with null derivative.

Answer 2

Since $f$ is continuous and homogeneous of order $\alpha>1$, $f(\mathbf{0})=0$.

$$f(\mathbf{0}+\mathbf{h})=f(\mathbf{h})=|\mathbf{h}|^1|\mathbf{h}|^{\alpha-1}f\big(\frac{1}{|\mathbf{h}|}\mathbf{h}\big) $$

This can be rewittten as $$f(\mathbf{0}+\mathbf{h})=f(\mathbf{0})+\mathbf{0}\cdot\mathbf{h}+r(\mathbf{h})$$ with $\lim_{\mathbf{h}\rightarrow\mathbf{0}}\frac{r(\mathbf{h})}{|\mathbf{h}|}=0$. FRom the definition of derivative, we find that $f$ is differentiable at $\mathbf{0}$ and that $f'(\mathbf{0})=\mathbf{0}$.