Let $f(x,y)$ be a $C^1$ function where $f(0,0)=0$. What conditions on f guarantee $f(f(x,y),y))=0$ can be solved for $y$ as a $C^1$ function of $x$ near $(0,0)$
Hello, I've been stuck on this problem for a bit and was wondering if anybody could help me out. I tried using the chain rule on the composed function to try and see when the implicit function theorem conditions will hold, but got stuck. Any help/hints are appreciated!
Define $g: \mathbb{R}^2 \to \mathbb{R}^2$ by $g(x, y) = (f(x, y), y)$ and define $F: \mathbb{R}^2 \to \mathbb{R}$ by $F = f \circ g$ i.e. $F(x, y) = f(g(x,y)) = f(f(x,y), y)$. Our goal is to use the implicit function theorem on $F$ to show that in some neighborhood $U$ of $0$, there exists a function $\tilde{y}: U \to \mathbb{R}$ such that $\tilde{y}(0) = 0$ and $F(x, \tilde{y}(x)) = 0$ for all $x$ in $U$. To apply the implicit function theorem in this manner, $F$ must satisfy $\det \frac{\partial F}{\partial y}(0,0) \neq 0$. In this case, since $\frac{\partial F}{\partial y}$ is a scalar, this reduces to $\frac{\partial F}{\partial y}(0,0) \neq 0$.
Let us compute $DF$ to see what conditions on $f$ guarantee this to be the case. By the chain rule: \begin{align} DF(x,y) &= Df(g(x,y)) Dg(x,y)\\ &= \begin{pmatrix} \frac{\partial f}{\partial x}(g(x, y)) & \frac{\partial f}{\partial y}(g(x, y)) \end{pmatrix} \begin{pmatrix} \frac{\partial f}{\partial x}(x, y) & \frac{\partial f}{\partial y}(x, y) \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} \frac{\partial f}{\partial x}(g(x, y)) \frac{\partial f}{\partial x}(x, y) & \frac{\partial f}{\partial x}(g(x, y)) \frac{\partial f}{\partial y}(x, y) +\frac{\partial f}{\partial y}(g(x, y)) \end{pmatrix} \end{align} Thus, $\frac{\partial F}{\partial y}(x,y) = \frac{\partial f}{\partial x}(g(x, y)) \frac{\partial f}{\partial y}(x, y) +\frac{\partial f}{\partial y}(g(x, y))$. Evaluating at $(0, 0)$, we find: \begin{align*} \frac{\partial F}{\partial y}(0,0) &= \frac{\partial f}{\partial x}(0,0) \frac{\partial f}{\partial y}(0, 0) +\frac{\partial f}{\partial y}(0,0) \\ &=\left(\frac{\partial f}{\partial x}(0,0) + 1\right) \frac{\partial f}{\partial y}(0,0) \end{align*}
For $\frac{\partial F}{\partial y}(0,0) \neq 0$, we must have $\frac{\partial f}{\partial x}(0,0) + 1 \neq 0$ and $\frac{\partial f}{\partial y}(0,0) \neq 0$.