# 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 $U$?

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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 $V$. Is this what you are asking about? – Yury Nov 16 '12 at 15:43
Yes, it is what I'm asking. – Alex Nov 16 '12 at 15:45
Do you want to preserve the Lipschitz constant? If yes. Then, the answer is in general negative. See Federer, geometric Measure Theory. 2.10.44. If the range space is R. Then, the answer is positive. This is the content of classical Mcshane lemma. – Saugata Nov 16 '12 at 15:49

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 $V$.

1. If $U\subset \ell_2$ and $V=\ell_2$ (Kirszbraun's theorem),
2. If $V = \mathbb{R}$ or $V = \ell_{\infty}$, $U$ is an arbitrary metric space (McShane's theorem),
3. If $V$ is a finite dimensional normed space, $U$ is an arbitrary metric space (follows from McShane's theorem),
4. 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 $f$).

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Regarding 3) there may not be any extension with same Lipschitz constant if the norm of V is not induced by an inner product. – Saugata Nov 16 '12 at 16:20
Yes, in (3) and (4) the Lipschitz constant of $\hat f$ may be larger than the Lipschitz constant of $f$. – Yury Nov 16 '12 at 16:22

1. (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.
2. (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.)