Boundness type property of homogenous functions in $\mathbb{R}^n$

Question

Let $f\in C^1(\mathbb{R}^n\setminus \{0\})$ a positive function and $a\in \mathbb{R}$ such that $f(tx)=t^af(x)$ for all $x$ and $t>0$. Prove that there exist $A,~B>0$ such that for all $x\neq 0$:

$$A\|x\|^a\leq f(x) \leq B\|x\|^a.$$

Αttempt. I wonder if it is enough to prove that $$\lim_{|x|\rightarrow +\infty}\frac{f(x)}{\|x\|^a}=0$$

using something like $$f\Big(\frac{x}{\|x\|}\Big)=\frac{f(x)}{\|x\|^a}$$

for $x\neq 0.$

Thank you in advance for the help.

Answer 1

Consider $\varphi:{\bf{S}}^{n-1}\rightarrow[0,\infty)$, $\varphi(x)=f(x)$, then $\varphi$ is continuous and hence $m:=\min f(x)$ and $M:=\max f$ exist, and hence for all $x\ne 0$, \begin{align*} f(x/\|x\|)=\varphi(x/\|x\|)\leq M, \end{align*} so $\|x\|^{-a}f(x)\leq M$, hence $f(x)\leq M\|x\|^{a}$.