A radial function that does not depend on a variable must be null Let $\Omega \subset \mathbb{R}^n$ an open set. A function $u :\Omega \rightarrow \mathbb{R}$ is radial if $u(x) = u(y),$ when |x|=|y|. Suppose that, for exemple, $u$ doesn't depend on the first variable, that is,
$$
u(r, x_2, ..., x_n) = u(s, x_2, ..., x_n), 
$$
for all real numbers $r,s$ such that $(r, x_2, ..., x_n), (s, x_2, ..., x_n) \in \Omega.$
I would like to know if this implies that $u$ is null.
Any help to disapprove this, or with conditions under which this would work is very welcome.
Edit: If we suppose that $\Omega$ is a ball and $u$ is a $C^1$ function in $\overline{\Omega}$ with the conditions above, but zero in the bondary of $\Omega$, then $u$ would be null?
 A: It does not have to be null.
For example, $u(\vec{x}) = 1$ is consistent with the rules you articulated.
First let's consider the case where $\Omega = \mathbb{R}^n$. In this case $\Omega$ must be constant.
the set of values of the form $(\mathbb{R}_{\ge 0}, 0, 0, c\dots)$ hits every possible modulus in $\mathbb{R}_{\ge 0}$ and, additionally, they all must have the same value.
However, $u$ does not have to be constant in general.
Consider $\Omega = (-1, 1)^n \cup (10+n, 11+n)^n$. (Note that one of the components of $\Omega$ is a box and therefore the maximum modulus in the box is the $\sqrt{n}$ by the definition of distance. I will add $n$ to each of the dimensions of the second component to place it far enough away that its set of moduli is distinct.)
Let $u(\vec{x})$ take the value $1$ when $\vec{x}$ is in $(-1, 1)^n$. Let $u(\vec{x})$ take the value $2$ when $\vec{x}$ is in $(10+n, 11+n)^n$. $u$ is still radial, since no modulus is shared between the two components of $\Omega$.
I don't know whether we can force $u$ to be constant by insisting on a natural condition for $\Omega$, like convexity or path-connectedness.
