Let U and V be metric spaces. Let X be a dense subset of U. Suppose that there exists a function, h: X $\rightarrow$ V, such that h is Lipschitz with constant K. Show that there exists a function f: U $\rightarrow$ V, such that $\left.f\right|_X$ = h, with the same Lipschitz constant. Moreover, show that f is unique.
Attempt at solution
Uniqueness: Suppose that f and g are two functions satisfying the result, i.e., $\left.f\right|_X$ = h and $\left.g\right|_X$ = h. Let u $\in$ U and let {x$_{n}$} $\in$ X be a sequence converging to u (by density of X). Then, since f,g are continuous, it follows that:
f(u) = $\lim\limits_{n \rightarrow \infty}$f(x$_{n}$) = $\lim\limits_{n \rightarrow \infty}$$\left.f\right|_X$(x$_{n}$) = $\lim\limits_{n \rightarrow \infty}$h(x$_{n}$) = $\lim\limits_{n \rightarrow \infty}$$\left.g\right|_X$(x$_{n}$) = $\lim\limits_{n \rightarrow \infty}$g(x$_{n}$) = g(u)
Am I on the right track here? Could I get a hint for the existence part? Any help would be greatly appreciated.
$\underline{\textbf{ If } V \textbf{ is complete }}$:
Let $u\in U$, and let $\{x_n\}\in X$ be a sequence converging to $u$, in particular $\{x_n\}$ is a Cauchy sequence, since $h$ is Lipschitz then the sequence $\{h(x_n)\}$ is a Cauchy sequence, hence converge in $V$ (which is complete). Now we want to show that this limit (the limit of $f(x_n)$ is independent of the choice of the sequence $x_n$ ), let $\{x_n'\}\subseteq X$ another s.t $x_n'\to u$, and showing that $\lim h(x_n)=\lim h(x_n')$.
Since $|h(x_n)-h(x'_n)|\leq K|x_n-x'_n|$, then $\lim(h(x_n)-h(x'_n))=0$, hence $\lim h(x_n)=\lim h(x'_n)$.Now we denote $f(u)=\lim h(x_n)$.
Let $u_,u'\in U$, and $\{x_n\}\subseteq X$ , $\{x_n'\}\subseteq X$ s.t $x_n\to u$ and $x_n'\to u'$, for all $n\in \Bbb N$ we have $|h(x_n)-h(x_n')|\leq K|x_n-x_n'|$ by passing to limit we get $|f(u)-f(u')|\leq |u-u'|$, hence $f$ is Lipschitz and it is continuous.