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Given a curve $\gamma$ in a Banach space $X$ and a function f defined along the curve s.t. $$\big\Vert f(\gamma(t))-f(\gamma(s))\big\Vert\\leq L\big\Vert\gamma(t)-\gamma(s)\big\Vert$$ is it possible to extend the Lipschitz functions to the whole of $X$?

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That depends more on the set of values $f$ can take than on the domain metric space, so you failed to provide the most relevant information here: what is the range space of $f$? – fedja Sep 5 '12 at 2:43
up vote 4 down vote accepted

It is not always possible to extend when $X$ is a Banach space. Take a Banach space $Y_n$ which contains an $n$ dimensional subspace $E_n$ such that every projection from $Y_n$ onto $E_n$ has norm at least $C_n$ with $C_n\to \infty$. ($Y_n$ can e.g. be $L_1$ and $E_n$ the span of $n$ IID gaussian random variables; then $C_n$ is of order $n^{1/2}$.) Let $X_n = Y_n \oplus_2 E_n$. For the curve in $Y_n$ take any curve in the unit sphere of $E_n \oplus \{0\}$ that contains an $\epsilon_n$ net $A_n$ of the unit sphere of $E_n \oplus \{0\}$. For $f_n$ take the natural isometry from $E_n \oplus \{0\}$ onto $ \{0\} \oplus E_n $ restricted to the curve. Let $F_n$ be an extension of $f_n$ to a Lipschitz mapping on $X_n$; WLOG $F_n$ maps into $ \{0\} \oplus E_n $ since this is a norm one complemented subspace of $X_n$. Let $G_n$ be the positively homogeneous extension of the restriction of $F_n$ to the unit sphere of $X_n$. Then the Lipschitz constant of $G_n$ is at most three times the Lipschitz constant of $F_n$. Compose $G_n$ with the obvious isometry from $ \{0\} \oplus E_n $ onto $E_n \oplus \{0\}$. The restriction of this map to $Y_n$ gives a positively homogenous mapping from $Y_n$ into $E_n$ that is the identity on $A_N$. By the arguments in $$ $$ Johnson, William B.(1-OHSN); Lindenstrauss, Joram(IL-HEBR) Extensions of Lipschitz mappings into a Hilbert space. Conference in modern analysis and probability (New Haven, Conn., 1982), 189–206, Contemp. Math., 26, Amer. Math. Soc., Providence, RI, 1984 $$ $$

we conclude that if $\epsilon_n$ is sufficiently small, there is a projection from $Y_n$ onto $E_n$ whose norm is no worse than something like ten times the Lipschitz constant of $G_n$.

All of this shows that you cannot get Lipschitz extensions with controlled norms. Take an infinite direct sum to get an example where you cannot get any Lipschitz extension.

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If you mean a real-valued function $f$, yes, and keeping the same constant $L$, by a simple construction. Check the last mentioned property listed here.

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The basic extension result for Lipschitz functions is the theorem of Kirszbaum. This works for functions with values in $\mathbb{R}^n$ and is expounded in Federer's book on Geometric Measure Theory. I think that it even works for functions with values in Hilbert space but can't trace a reference.

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Kirszbaum's theorem is for mappings from a subset of a Hilbert space into a Hilbert space. – Bill Johnson Sep 5 '12 at 11:48
And his last name is Kirszbraun. – Mateusz Wasilewski Sep 5 '12 at 15:23
@Bill Johnson. Sorry, you are right, of course. The theorem I should have quoted was the MacShane-Whitney result that you can extend any Lischitz function from a subset of a metric space retaining the Lipschitz constant, but for the real-valued case which I assume, by default, is what the questioner intended. Sorry about misspelling the name. Not sure about the etiquette in this forum. Will this mea culpa suffice or should (can) I edit my response? hanks again. – jbc Sep 5 '12 at 16:14
You can edit your answer and this is one of the arch-typical reasons for being able to do so. – BSteinhurst Sep 5 '12 at 21:18

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