# Homogeneous function must be polynomial

Let $u:\mathbb{R}^n\setminus \{0\} \rightarrow \mathbb{R}$ be a $C^1$-function which is positive homogeneous of degree $k \neq 0$, then by Euler's theorem $x \cdot \nabla u = k u$. Homogeneous polynomials of degree $k$ satisfy this equation, can I evoke a uniqueness theorem to conclude that the these homogeneous functions are necessarily sum of Homogeneous polynomials of degree $k$? In case this is true, is there any other simple proof of this fact?

Uniqueness theorems always come with some requirements, with their exact nature depending on the nature of the PDE. In the case of a first-order equation like $x \cdot \nabla u =k u$, this is provided by the method of characteristics: solutions to the PDE are uniquely determined by their values on a surface that intersects each characteristic exactly once. Since the characteristics here are just the rays from the origin, the most natural surface to use is the unit sphere $$\mathbb S^{n-1} = \{ x \in \mathbb R^n : |x| = 1\}.$$
Indeed, any function $u$ defined on the sphere can be extended to $\mathbb R^n \setminus \{0\}$ by $$u(x) = |x|^k u\left(\frac{x}{|x|}\right),$$ which is homogeneous of degree $k$; so there are as many $k$-homogeneous functions as there are functions on the sphere. In particular, there is an infinite-dimensional vector space of such functions, while the homogeneous polynomials of degree $k$ form a finite-dimensional space.
The function $f: \mathbb{R}^n \backslash \{0\} \to \mathbb{R}$ given by $f(x) = |x|^k$ is homogeneous of degree $k$ but not a polynomial.