Lipschitz property and Lipschitz extension Is there a Lipschitz function $f$ from a subset of a metric space $U$ to a complete  metric space $V$ that has no Lipschitz extension to the whole space $U$? 
 A: The answer is “yes”. There exist $A\subset U$, $V$ and $f:A\to V$ where $U$ is a metric space, $A$ is a subset of $U$, $V$ is a complete metric space, and $f$ is a Lipschitz map such that $f$ cannot be extended to a Lipschitz map from $U$ to $V$. 
Here is the reason why this is the case. Let $U = L_1[0,1]$ and $V$ be a subspace of $U$ isometrically isomorphic to $\ell_2$. Note that space $V$ is non-complemented in $U$. Let $A= V$ and $f$ be the identity map on $V$. Suppose that $\hat f:U \to V$ is a Lipschitz extension of $f$. Note that $\hat f$ is linear on $U$. It turns out that then we can assume that $\hat f$ is also a linear map on  whole $U$ (i.e. if a Lipschitz extension $\hat f$ exists then there exists another Lipschitz extension $\tilde f$, which is linear; I don't prove that). That is, $\hat f$ is a continuous linear projection of $U$ on $V$. This is impossible since $V$ is not complemented in $U$.
There are several important cases when every Lipschitz map from a subset of $U$ to $V$ can be extended to a Lipschitz map from $U$ to $V$.


*

*If $U\subset \ell_2$ and $V=\ell_2$ (Kirszbraun's theorem),

*If $V = \mathbb{R}$ or $V = \ell_{\infty}$, $U$ is an arbitrary metric space (McShane's theorem),

*If $V$ is a finite dimensional normed space, $U$ is an arbitrary metric space (follows from McShane's theorem),

*If $U \subset L_p$ and $V = L_q$ with $1 < q < 2 < p <\infty$ (theorem of Naor, Peres, Schramm, and Sheffield).


In cases (1) and (2), there exists an extension $\hat f$ with $\|\hat f\|_{Lip} \leq \|f\|_{Lip}$; in cases (3) and (4), there might be no extension $\hat f$ with $\|\hat f\|_{Lip} \leq \|f\|_{Lip}$ (i.e. we might need to increase the Lipschitz constant when we extend $f$).
A: I add two finite-dimensional examples. 


*

*(With a topological obstruction.) Let $U=[0,1]$ (interval), $V=\{0,1\}$ (two-point set). The identity map $\{0,1\}\to \{0,1\}$ does not extend to a continuous map $U\to V$ since any continuous map $U\to V$ must be constant. 

*(Without a topological obstruction.) Replace $V$ in the first example by a snowflake that contains the points $0$ and $1$. Now the identity map $\{0,1\}\to \{0,1\}$ does extend to a continuous map $U\to V$. However, there is no Lipschitz extension since any Lipschitz map $U\to V$ must be constant. (One way to prove the latter is to observe that any nontrivial subarc of the snowflake has infinite length, while the length of a Lipschitz image of $U$ must be finite.)
