Suppose $M$ is a pointed Banach space. Denote $$Lip_0(M)= \{ f:M \rightarrow \mathbb{R}: f(0)=0, f - Lipschitz \}$$

If $N$ is a subset of $M$, show that $Lip_0(N)$ is linearly isomorphic to $Lip_0(\bar{N})$, where $\bar{N}$ denotes the closure of $N$.

I know that if $f:X \rightarrow \mathbb{R}$ is a Lipschitz function, then $g:\bar{X} \rightarrow \mathbb{R}$ defined by $$g(y) = \inf_{x \in X}(f(x) - \| f \|_{Lip}d(x,y))$$ is an extension of $f$ with the same Lipschitz constant as $f$.

I am wondering is this useful to show the linear isomorphism.


1 Answer 1


You don't need to resort to the extension map that you mention; extention by continuity does the job here because Lipschitz maps are uniformly continuous and thus extensible to the topological completion of their domain of definition.

Let us use the fundamental isomorphism theorem to solve your problem. Consider the restriction map $r : Lip_0 (\bar N) \to Lip_0 (N), r(g) = g \big| _N$. Note that surjectivity has been explained in the above paragraph. It remains to find $\ker r$: if $g \big| _N = 0$ then its (unique) extension by continuity to $\bar N$ is the zero map, so the kernel is trivial. The fundamental isomorphism theorem now ends the proof.


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