Let $$f:\mathbb{R}^n\rightarrow \mathbb{R}^n$$ be radial. Then can $$f$$ necessarily be represented as $$A(\|x\|)x = f(x),$$ for some function $$z\mapsto A(z)$$ from $$\mathbb{R}$$ into $$SO(n)$$?
Here, by radial I mean that there exists some $$f_0\in C^1(\mathbb{R},\mathbb{R}^{n})$$ such that $$f(x)=f_0(\|x\|)$$. So I'm curious if we can assume that $$f_0$$ is acting by matrix multiplication...
• As stated, no. Let's use $n=2$, and $f=\vec{e}_{1} =f_0$, simply a constant unit vector in 1-direction. Then, for $x=\vec{e}_{1}$, $A$ is the identity map, whereas for $x=-\vec{e}_{1}$, $A$ is minus the identity map. However, $A$ must be the same in these two cases, contradiction. This seems too simple - perhaps there's still an error in the description? – Michael Engelhardt Oct 5 at 17:19
• not clear: in the given definition, $f$ is radial iff its restriction on any sphere centered at the origin is constant, whereas $A(\|x\|)\cdot x$ is not. – Pietro Majer Oct 5 at 19:19
• The given definition of "radial" immediately answers the question in the negative: for constant $\lVert x\rVert$, the action of a constant matrix on $x$ will not give a constant result. And what is meant by $f_0$ "acting"? (On what?) – gmvh Oct 6 at 13:52