Let $z=F(x,y)$ be a function from $\mathbb R^d\times \mathbb R$ to $\mathbb R$ and $z=F(x,y)$ is Lipschitz continuous. Assume that for any $x\in\mathbb R^d$, there is a unique $y$ such that $F(x,y)=0$. Let $y=y(x)$ be a function that is produced by the implicit function $F(x,y)=0$. Can we prove that $y=y(x)$ is also Lipschitz continuous? Thank you!


Motivation of this question:
Most of analysis of implicit functions uses strong assumptions to prove the uniqueness of $y$ given $x$. These assumptions are often violated in applications. If we already know the existence of the implicit function, can we directly show that the implicit function is Lipschitz continuous?



Lipschitz continuity of an implicit function
The answer is **no**.



> 
> **Theorem.** Let $f:\mathbb{R}^d\to\mathbb{R}$ be any continuous function. Then there is a Lipschitz function $F:\mathbb{R}^d\times\mathbb{R}\to\mathbb{R}$ such that for every $x\in\mathbb{R}^d$, $F(x,y)=0$ if and only if $y=f(x)$.
> 
> 
> 


*Proof.* Let $G\subset\mathbb{R}^{d}\times\mathbb{R}$ be the graph of $f$. Then the set $G$ is closed, and the function $F:\mathbb{R}^d\times\mathbb{R}\to\mathbb{R}$ defined by
$$
F(x,y)=\operatorname{dist}((x,y),G) 
$$
is $1$-Lipschitz continuous. Since $F$ vanishes exactly on $G$,we have that for every $x\in\mathbb{R}^d$, $F(x,y)=0$ if and only if $(x,y)\in G$, i.e. $y=f(x)$. $\Box$


There are examples of Lipschitz functions $F(x,y)$ as in your problem, with the zero set $y=f(x)$ being given by a discontinuous function $f$.



> 
> **Example.** Let $f:\mathbb{R}\to\mathbb{R}$, $f(x)=1/x$ if $x\neq 0$, and $f(0)=0$. Note that the graph $G$ of $f$ is a closed subset of $\mathbb{R}^2$. Therefore, the above construction gives $F(x,y):\mathbb{R}^2\to\mathbb{R}$ with the zero zet being the graph of $y=f(x)$ which is a discontinuous function. Note however, that the function $y=f(x)$ is unbounded.
> $\Box$
> 
> 
> 



> 
> **Theorem.** Suppose that $F:\mathbb{R}^n\times\mathbb{R}^m\to\mathbb{R}$ is continuous and $M>0$ is a constant. If for every $x\in\mathbb{R}^n$ there is a unique $y=y(x)\in\mathbb{R}^m$ such that
> $$
> |y(x)|\leq M
> \quad
> \text{and}
> \quad
> F(x,y(x))=0,
> $$
> then the function $\mathbb{R}^n\ni x\mapsto y(x)\in\mathbb{R}^m$ is continuous.
> 
> 
> 


*Proof.* Suppose that $x\_i\to x$. We need to prove that $y(x\_i)\to y(x)$. Suppose to the contrary that this is not the case so there is $\varepsilon$ and a subsequence such that
$$
(\*)
\quad \quad \quad
|y(x\_{i\_j})-y(x)|\geq\varepsilon.
$$
Since $|y(x\_{i\_j})|\leq M$ is bounded, we can select a convergent subsequence $y(x\_{i\_{j\_k}})\to y$, $|y|\leq M$. Continuity of $F$ yields
$$
0=F(x\_{i\_{j\_k}},y(x\_{i\_{j\_k}}))\to F(x,y).
$$
That is $|y|\leq M$ and $F(x,y)=0$, which by the assumptions in the theorem mean that $y=y(x)$, so $y(x\_{i\_{j\_k}})\to y(x)$ and we arrive at a contradiction with (\*). $\Box$


