# If $N$ is a subset of $M$, show that $Lip_0(N)$ is linearly isomorphic to $Lip_0(\bar{N})$

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.

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.