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 $X$? 



Extension of lipschitz functions along a curve
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
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$$
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
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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.


