Existence of a positively homogeneous function of degree $k$ Question:

For every $k \in \mathbb R$ show that there exists a function $f: \mathbb R^m - \{0\} \to \mathbb R$ of class $C^{\infty}$, positively homogeneous of degree $k$, that is, $f(tx) = t^k f(x)$ , for every $x \in \mathbb R^m - \{0\}$, with $t > 0$, such that $f(x) > 0$ for all $x$ and $f$ is not a polynomial. 

Attempt: I have tried to work with $$f(x) = \frac{1}{k\displaystyle\sum_{j=1}^{m}\frac{1}{x_j^k}}$$   
Then $f(tx) = t^k f(x)$. But I have failed to show that $f(x) > 0$. From $$f(tx) - f(x) = f'(x) \cdot (t-1)x + \rho(x) (t-1) |x|\tag{*}$$
and $$f'(x) \cdot x = \frac{f(x)}{k} $$
I took $t \to 1^+$ and got $k^2 = 1$. 
Any thoughts are welcome. 
 A: If $\lVert\,\cdot\,\rVert$ is any norm on $\mathbb{R}^m$, and $f\colon \mathbb{R}^m\setminus \{0\} \to \mathbb{R}$ a function that is positively homogeneous of degree $k$, then for all $x\in \mathbb{R}^m\setminus \{0\}$ we have
$$f(x) = f\biggl(\lVert x\rVert\cdot \frac{x}{\lVert x\rVert}\biggr) = \lVert x\rVert^k \cdot f\biggl(\frac{x}{\lVert x\rVert}\biggr),$$
so $f$ is determined by its values on the unit sphere $S = \{ x \in \mathbb{R}^m : \lVert x\rVert = 1\}$. And any function $g \colon S \to \mathbb{R}$ uniquely determines a positively homogeneous function $f$ of degree $k$ by setting
$$f(x) = \lVert x\rVert^k\cdot g\biggl(\frac{x}{\lVert x\rVert}\biggr).\tag{$\ast$}$$
$f$ will be strictly positive if and only if $g$ is strictly positive. And $f$ will be continuous if and only if $g$ is continuous.
We want a $C^{\infty}$ function, and to get that conveniently, we should focus on a smooth norm, because then the unit sphere of that norm is a smooth submanifold of $\mathbb{R}^m\setminus \{0\}$. Usually, unless specified otherwise, one endows $\mathbb{R}^m$ with the Euclidean norm, which is smooth. But one could also choose many other smooth norms, a rather prominent family is the family of $\ell^p$-norms $\lVert\,\cdot\,\rVert_p$ where $1 < p < \infty$ (the Euclidean norm is the case $p = 2$; the norms $\lVert\,\cdot\,\rVert_1$ and $\lVert\,\cdot\,\rVert_{\infty}$ are not smooth for $m > 1$).
Anyway, if we use a smooth norm, then the positively homogeneous $f$ induced by $g$ will be smooth if and only if $g$ is smooth.
So, take the Euclidean norm, take any smooth function $g \colon S \to (0,\infty)$, and $f$ defined by $(\ast)$ will be a smooth function, positively homogeneous of degree $k$, and strictly positive. If $k \notin \mathbb{N}$, then $f$ cannot be a polynomial, since it has the wrong growth behaviour. For even $k \in \mathbb{N}$, some choices of $g$ will yield polynomials, but not all. If you choose $g$ in such a way that there is a point $p \in S$ with $g(-p) \neq g(p)$, then $f$ will not be a polynomial.
