Function positively homogeneous of degree $1$

Question

Let $f:U \to \mathbb{R}$ be a differentiable, positively homogeneous of degree $1$ in an open $U \subset \mathbb{R}^m$ containing $0$. Show that $f$ is a restriction to $U$ of a linear transformation from $\mathbb{R}^m$ to $\mathbb{R}$. Conclude that the function $f:\mathbb{R}^2 \to \mathbb{R}$ given by$$ f(x,y) = \begin{cases} \dfrac{x^3}{x^2 + y^2}; & (x, y) \neq (0, 0)\\ 0; & (x, y) = (0, 0) \end{cases} $$ is not differentiable in $0$.

I do not know how to start this question. In previous questions, I proved:

• Euler's relations for positively homogeneous functions of degree $k$.

• For all $k \in \mathbb{R}$ there exists a function $f \in C^{\infty}(\mathbb{R}^{m}\setminus \{0\})$ positively homogeneous of degree $k$ such that $f(x)>0$ for all $x \in \mathbb{R}^m$ but is not a polynomial.

But, I do not know if it helps. I would like some hints. Thanks in advance.

Answer 1

$\def\vec{\boldsymbol}$For $\vec{x} = (x_1, \cdots, x_m)^{\mathrm{T}} \in U$, define$$ g(\vec{x}) = f(\vec{x}) - \sum_{k = 1}^m x_k \frac{\partial f}{\partial x_k}(\vec{0}), $$ then $g$ is also positively homogeneous of degree $1$. Note that $f(\vec{0}) = 0$ and $f$ is differentiable at $\vec{x} = \vec{0}$, then$$ \lim_{\vec{y} \to \vec{0}} \frac{g(\vec{y})}{\| \vec{y} \|} = 0. \tag{1} $$ Now suppose $B(\vec{0}, δ) \subseteq U$. For any $\vec{x} \in U\setminus \{0\}$, note that $t \vec{x} \in U$ and $g(\vec{x}) = \dfrac{g(t \vec{x})}{t}$ for any $0 < t < \dfrac{δ}{\| \vec{x} \|}$, then from (1) there is$$ g(\vec{x}) = \lim_{t \to 0^+} \frac{g(t \vec{x})}{t} = \| \vec{x} \| · \lim_{t \to 0^+} \frac{g(t \vec{x})}{\| t \vec{x} \|} = 0. $$ Also, $g(\vec{0}) = f(\vec{0}) = 0$. Therefore $g = 0$, i.e.$$ f(\vec{x}) = \sum_{k = 1}^m x_k \frac{\partial f}{\partial x_k}(\vec{0}). \quad \forall \vec{x} \in U $$

Answer 2

Proof assuming that $Df$ is continuous at the origin: we have $\frac {||f(x+h)-f(x)-Df(x)(h)||} {||h||} \to 0$ as $||h|| \to 0$. Replace $x$ by $tv$ and $h$ by $tu$ where $u$ and $v$ are fixed. Using homogeinity we get $\frac {||f(v+u)-f(v)-Df(tv)(u)||} {||u||} \to 0$ as $t \to 0$. Assuming continuity of $Df$ at $0$ this gives $f(v+u)-f(v)=Df(0) (u)$ for all $u$ and $v$. Put $v=0$ to get $f(u)=f(0)+Df(0)(u)$ for all $u$. Homogeinity implies that $f(0)=0$ so $f$ is the linear map $Df(0)$.