is $f$ constant or dilation $f:\mathbb{R}^3\to\mathbb{R}, f(rx)=r^{\alpha}f(x)$ for some $\alpha>0$, $\forall x\in\mathbb{R}$, for any $r\in\mathbb{R}$, could any one tell me which of the following is true?
$1.$ If $f(x)=f(y)$ whenever $||x||=||y||=\beta>0$, then $f(x)=\beta ||x||^{\alpha}$
$2.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)= ||x||^{\alpha}$
$3.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)=c||x||^{\alpha}$ for some constant $c$
$4.$ If $f(x)=f(y)$ whenever $||x||=||y||$, then $f(x)$ must be a constant function.
Thank you.
 A: Given you use $\lVert \cdot \rVert$, I assume that $f : \mathbb{R}^n \to \mathbb{R}$.
Note that we can always rescale $f$ by a constant and it will still obey the $f(rx) = r^\alpha f(x)$ and $f(x) = f(y)$ conditions, so there's no way 1) or 2) can hold.
Also, 4) cannot hold because 3) gives an example of a function which is not constant.
So why is 3) true? We have $f(\hat x) = f(\hat y) = c$, say, for normalized vectors $\hat x,\hat y$. But now given any $x \equiv \lVert x \rVert \hat x$ we find
$$f(x) = f(\lVert x \rVert \hat x) = \lVert x \rVert^\alpha f(\hat x) = c\lVert x \rVert^\alpha$$
A: Let $~f(x)=3~\lVert x\lVert^\alpha~.$
For option $\bf {(1)}$, let $~\beta =2~.$
Clearly, $~f(x)=3\cdot2^\alpha=f(y)~,$ whenever $~\lVert x\lVert=\lVert y\lVert=\beta=2\gt0~.$
But $~f(x)=3~\lVert x\lVert^\alpha\ne \beta~\lVert x\lVert^\alpha~$
$\therefore$ option $(1)$ is not true.
For option $\bf {(2)}$, from the above example, whenever $~\lVert x\lVert=\lVert y\lVert=1~,$ $~f(x)=3=f(y)~,$
But $~f(x)=3~\lVert x\lVert^\alpha\ne \lVert x\lVert^\alpha~$
$\therefore$ option $(2)$ is not true.
For option $\bf {(4)}$, from the above example, $~f(x)=f(y)~,~~~~\forall x,y~~$  for which  $~\lVert x\lVert=\lVert y\lVert~.$
But here $~f~$ is not constant.
$\therefore$ option $(4)$ is not true.
As all other options are incorrect, therefore only option $(3)$ is true.
