I am struggling with a very concrete situation: I am given a quadrilateral, i.e. a connected compact set $K\subset \mathbb{R}^2$ whose boundary consists of four straight line segments. Further, let $g:\partial K\rightarrow \mathbb{R}^2$ be continuous.
The question I am interested in is whether there exists a Lipschitz continuous function $f:K\rightarrow \mathbb{R}^2$ with $f(P)=g(P)$ for all boundary points $P\in\partial K$.
In my concrete case, I proved the existence of such a Lipschitz function by explicitly constructing it. But I must admit that it is not a very elegant solution. So I am wondering if there is an abstract (or elegant) way of proving the existence of such a Lipschitz function $f$? Is there a more general argument or theory which can be used here?
I would appreciate your help very much!
Best wishes
Best Answer
First consider the case of $g:\mathbb{R}^2\to \mathbb{R}$.
Such an extension exists if and only if $g$ is a Lipschitz function on $\partial K$, and in fact you can take $f:\mathbb{R}^2\to \mathbb{R}$, as $$f(x)=\inf\{ f(y)+M|x-y|: y\in \partial K\},$$ where $M=\sup_{z,w\in \partial K} \frac{|g(z)-g(w)|}{|z-w|}.$ In this case we also have $$ \sup_{x,y\in \mathbb{R}^2}\dfrac{|f(x)-f(y)|}{|x-y|}=M. $$
Now for $g:\mathbb{R}^2\to \mathbb{R}^2$, you can extend each coordinate to get an extension that unfortunately doesn't satisfy that it has the same Lipschitz constant as $g$. This can be achieved, but is a little more involved (see Kirszbraun's Theorem)
As a final remark, the construction (for scalar functions) works for extending any Lipschitz function on any set, as long as they satisfy a Lipschitz condition.