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The problem that I posted was a much generalized form of what I had in my mind. All I want to know the literature of Hahn-Banach type extension of Lipschitz map. I know only about the result by Kirszbrzun and very few of its development. The main problem is:

Let $X$ be a real Banach space, $Y$ a subspace of it and $f$ a $1$-Lipschitz map from $Y$ to $\mathbb{R}$. Is it possible to get an extension of $f$ from $X$ to $\mathbb{R}$ with Lipschitz constant $1$?

The case when $X$ is a metrizable tvs and $Y$ a finite dimensional subspace follows from MR0737400(86a:46018). But can we say anything about a infinite dimensional $Y$? Let us assume $X=C(K)$, $K$ is compact, $T_2$ (assume metrizable too if you want) and $Y=\{f\in X:f|_D=0\}$ (i.e. an M-ideal) of $X$. Now can a real valued $1$-Lipschitz map from $Y$ necessarily have a similar extension? Will it be unique ?

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    $\begingroup$ The question doesn't make sense --- there is no notion of distance in an LCTVS. $\endgroup$
    – Nik Weaver
    Commented Jul 25, 2015 at 16:28
  • $\begingroup$ Okay, I mean a metrizable lctvs but not necessarily normed linear space. $\endgroup$
    – Tanmoy Paul
    Commented Jul 25, 2015 at 18:05
  • $\begingroup$ I knew only about the result by Kirszbrzun but not any further development.Here is what I would like to ask.Let $X$ be a real Banach sp,$Y$ a subspace of it and $f$ a 1-Lip map from $X$ to $\mathbb{R}$.Is it possible to get an extn of $f$ from $X$ to $\mathbb{R}$ with Lip constant 1? The case when $X$ is a metric sp and $Y$ a finite dim. subsp, the result follows from MR0737400(86a:46018).But let us assume $X=C(K)$,$K$ is cpt $T_2$(can assume metrizable also)and $Y=\{f:f|_D=0\}$ (ie an M-ideal) of $X$.Now can a real valued 1-Lip map from $Y$ necessarily has a similar extn ? Can it be unique ? $\endgroup$
    – Tanmoy Paul
    Commented Jul 26, 2015 at 6:48
  • $\begingroup$ I suggest that you "unaccept" the answer below, otherwise people will think that your question has been fully answered and move on without looking $\endgroup$
    – Yemon Choi
    Commented Jul 26, 2015 at 15:55
  • $\begingroup$ Lipschitz maps defined on subspaces of Hilbert spaces and taking values in an arbitrary Banach space extend to Lipschitz maps (with the same Lipschitz constant) on the whole space. However, I shall deeply appreciate to know if the result extends to type 2 Banach space in place of Hilbert spaces, this time with the mappings also taking values in a Hilbert space. The linear analogue of this statement, with linear maps being used in place of Lipschitz maps is a well known theorem of B. Maurey. $\endgroup$
    – M A Sofi
    Commented Sep 18, 2017 at 10:24

1 Answer 1

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If $X$ is any metric space, $Y$ any subset of $X$, and $f: Y \to \mathbb{R}$ a Lipschitz function, then there is an extension $\tilde{f}: X \to \mathbb{R}$ of $f$ with the same Lipschitz constant. This easy result (just extend the function one point at a time) has been rediscovered many times. See Theorem 1.5.6 of my book Lipschitz Algebras.

The result does not follow from the paper of Johnson and Lindenstrauss that you mention. The case when $X$ is "metrizable" is meaningless as you need an actual metric to have a notion of Lipschitz map. The answer by MKO is basically unrelated to the question as far as I can tell.

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