Let $X$ be an $n$dim (compact, if needed) Alexandrov space with curvature $\geq k$, with $k\geq0$, and let $Y$ be an Alexandrov space with curvature $\leq0$ globally. Given any bounded nonempty $E\subset X$ and $f:E\mapsto Y$ Lipschitzian. Can we extend $f$ to the whole $X$, with $lip(f\big_X)\leq C(n,k,Y)\cdot lip(f\big_E)$, for some constant $C(n,k,Y)$? Here $lip(f\big_E):=\sup_{x,y\in E}\frac{d_Y(f(x),f(y))}{d_X(x,y)}$. Thank you !
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1$\begingroup$ Surely there are topological obstructions. Start with an isometric map from a loop in $X$ to a loop in $Y$. If the loop bounds a disc in $X$ but not in $Y$, then there will be no continuous extension, let alone Lipschitz extension. $\endgroup$ – Nik Weaver Dec 13 '15 at 15:20

2$\begingroup$ Are you familiar with the results of Lang, Pavlovic and Schroeder on extensions of Lipschitz maps to Hadamard spaces? link.springer.com/article/10.1007%2FPL00001660 and math.psu.edu/petrunin/papers/alexandrov/… $\endgroup$ – Misha Dec 14 '15 at 1:02

1$\begingroup$ @NikWeaver Just to clarify, the more widely used description of the space $Y$ would be that it is CAT(0), or that it is a Hadamard space. In particular, it is contractible, so there are no topological obstructions. $\endgroup$ – John Harvey Dec 14 '15 at 9:08

$\begingroup$ Oh, I see. I don't know, then. $\endgroup$ – Nik Weaver Dec 14 '15 at 12:54