The question is the following:
Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces. A map $f:X \to Y$ is called Lipschitz continuous if there exists $C \geq 0$ such that
$$
d_Y(f(x),f(y)) \leq Cd_X(x,y),\ \forall x,y \in X
$$
The smallest possible constant for the above inequality is called the Lipschitz constant $Lip(f)$ of $f$.Let $\tilde{X}$ be a metric space and $X$ be a dense subset of $\tilde{X}$. Let $Y$ be a complete metric space. Then,
(a) For any Lipschitz map $f:X \to Y$, prove that there exists a unique continuous map $\tilde{f}: \tilde{X} \to Y$ such that $\tilde{f}|_X = f$. Moreover $\tilde{f} $ is Lipschitz and $Lip(\tilde{f}) = Lip(f)$.
(b) Show that one cannot replace "Lipschitz" in (a) by "continuous".
I come up with the following construction (could be wrong):
$$
\begin{cases}
f(x) \quad &\text{if }x \in X \\
\lim_{n \to \infty} f(a_n) \quad &\text{if } x \in \tilde{X} \setminus X,\ a_n \in X,\ a_n \to x
\end{cases}
$$
based on the definidion of $X$. But I am not sure that what I need to prove to make this function well-defined. I may know how to approach the continuity, but I have no idea to prove the uniqueness. Any help is appreciated.
Best Answer
Define directly that $\widetilde{f}(x)=lim f(a_{n})$ for $a_{n}\in X$ that $a_{n}\rightarrow x$.
For $(a_{n}),(b_{n})$ both $a_{n}\rightarrow x$ and $b_{n}\rightarrow x$, consider $c_{2n+1}=a_{n}$ and $c_{2n}=b_{n}$, then $c_{n}\rightarrow x$ and $d(f(c_{n}),f(c_{m}))\leq Cd(c_{n},c_{m})\rightarrow 0$. As $Y$ is complete, $f(c_{n})\rightarrow y$. But $(f(a_{n}))$ and $(f(b_{n}))$ are subsequences of $(f(c_{n}))$, so $y=\lim f(a_{n})=\lim f(b_{n})$ and hence it is well-defined.
Consider the constant sequence $(x,x,...)$, $x\in X$, then $\widetilde{f}(x)=f(x)$.
To prove the continuity, note that $d(\widetilde{f}(x),\widetilde{f}(y))=\lim d(f(a_{n}),f(b_{n}))\leq\lim d(a_{n},b_{n})=d(x,y)$ for $a_{n}\rightarrow x, b_{n}\rightarrow y$.