Symmetry in partial derivatives. I was wondering how the relationship $$x_j \partial_i f(x) = x_i \partial_j f(x)$$ means, that a function has rotational symmetry? I mean with rotational symmetric, that the value of $f$ at a point $x$ depends only on $\|x\|_2$.
Or more precisely: Is there any $C_1(\mathbb{R}^n \backslash \{0\}; \mathbb{R})$ function, that has no rotational symmetry, but fulfills this equation for all $x \in \mathbb{R}^n\backslash\{0\}$?
 A: What you are asking is: Does the relation imply $\nabla f(x) \cdot y = 0$ whenever $y$ is orthogonal to $x$? 
Assume without loss of generality that $x = (1, 0, \ldots, 0)$. Take $j \neq 1$ and $i = 1$. Then the relation says $0 = \delta_{1 i} \partial_j f(x) = \partial_j f(x)$. Hence $\partial_j f(x) = 0$ for all $j \neq 1$. Or: $\nabla f(x) \cdot y = 0$ for all $y$ orthogonal to $x$.
Now take a path $x(t)$ lying in any sphere (of positive radius) centered at the origin. Then $(f(x(t)))' = \nabla f(x(t)) \cdot \dot{x}(t) = 0$, because $x(t) \cdot \dot{x}(t) = 0$. Why? Differentiate $\|x(t)\|_2^2 = x(t)^T x(t)$.
Hence the answer is yes, the relation does imply rotational symmetry.
On second thinking, it is not clear that no generality is lost by assuming $x = (1, 0, \ldots, 0)$. Take any rotation matrix $U$, and $g(x) := f(U x)$. Then $\nabla g(x) \cdot y  = \nabla f(U x) \cdot U y$. Or, reversing the substitution, $\nabla g(U^T x) \cdot U^T y  = \nabla f(x) \cdot y$. Hence yes: rotating $x$ and the deviation $y$ by the same matrix does not change the value of the quantity that we are after.
