Does local gradient collinearity imply factorization

Question

Assume that we have two analytic functions $f(x,y)$ and $g(x,y)$ defined near $(0,0)$ on the plane. Let us also assume that $\nabla f =k \nabla g$, where $k=k(x,y)$ is also analytic and $\nabla f(0,0)\neq 0$, $\nabla g(0,0)\neq 0$.

Does it imply that there exists a locally defined analytic function h=h(u) of one variable, such that $g(x,y)=h(f(x,y))$ locally?

In the reverse direction the conclusion is easily checked using the chain rule.

Answer 1

By continuity we may assume near $(0,0)$ neither $\nabla f$ and $\nabla g$ vanish. Let $u = \nabla f(0,0)$, extended as a constant vector field on $\mathbb{R}^2$. Let $v$ be a non-vanishing vector field orthogonal to $\nabla f$ (and hence also to $\nabla g$). Note that in this case $v$ is well-defined up to the choice of direction and length.

Observations:

1. $f$ and $g$ are both constant along the integral curves of $v$.
2. $\nabla f$ is locally transversal to the integral curves of $v$, hence the integral curves can be locally parametrized by the value of $f$. In particular, there is a function $h$ such that $g = h\circ f$ (since $g$ is constant on the integral curves).
3. Constant the one dimensional functions $\hat{f}(s) = f(s u)$ and $\hat{g}(s) = g(s u)$. Both are by assumption real analytic. As $\hat{f}'(s) \neq 0$, by the real analytic inverse function theorem $\hat{f}$ has a real analytic inverse $p$. One can check then that $h$ (from previous step) is equal to $\hat{g}\circ p$. Hence $h$ is real analytic.