real analysis - Continuity of a general implicit function

I have an implicitly defined function $F(g_1(x_1),\dots,g_n(x_n))=0$, on the $n$-th dimension euclidean space, bounded and continuous in its $n$ arguments. $g_i$ are also real-valued bounded and continuous functions. What I'm trying to figure out is if I additionally need any conditions to guarantee that the implicit relation $x_i=f_i(x_1,\dots,x_{i-1},x_{i+1},\dots, x_n)$ is continuous in its arguments. I know that if $F$ was differentiable in all of its arguments, by the Implicit Function Theorem, the cross partial derivatives would exists and thus I would have continuity. However, in my case it is not necessarily true that $F$ is differentiable in its arguments. My intuition says that as everything is compositions of continuous functions, the implicit relation among variables should also be, but I want to make sure.

Many thanks!

Answer

$\newcommand{\R}{\mathbb{R}}$The standard version of the inverse function theorem (and the implicit function theorem), require that $F$ is continuously differentiable and that its Jacobian at a point of interest is invertible.

There exist, however, generalizations that impose weaker assumptions on $F$.

For example, Theorem 1A.4 in [1] states that for a function $\Phi:\R^m\times \R^n \to \R^n$ such that there is an $L_p \geq 0$ and $L_x \in [0,1)$ such that

$$
\|\Phi(p, x') - \Phi(p, x)\| \leq L_x \|x'- x\|\tag{1}
$$

and

$$
\|\Phi(p', x) - \Phi(p, x)\| \leq L_p \|p'- p\|\tag{2}
$$

Then, the mapping $p \mapsto \{x \in \R^n {}:{} x=\Phi(p,x)\}$ is single-valed for all $p$ and Lipschitz continuous.

You mentioned that since everything is a composition of continuous functions, the result will also be continuous. The issue, and what makes inverse problems very challenging, is that if $f$ is only known to be continuous, $f^{-1}$ is multi-valued. The study of the properties $f^{-1}$ is one of the topics of set-valued analysis.

You can apply this theorem to $\Phi(p,x) = x-F(p,x)$. Assumption (1) might be a little restrictive because $L_x$ must be in $[0,1)$.

See also Theorem 1E.13 in [1]. Lastly, often a function is not differentiable, but it can be shown to be semidifferentiable, so you have a lot of results that allow you to derive Lipschitz continuous inverse functions.

[1] A.L. Dontchev, R.T. Rockafellar, Implicit Functions and Solution Mappings, A View From Variational Analysis, Springer pdf.