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Consider a regular $n$-simplex, and a map from the vertices to $L^1$.

How can we find the minimum Lipschitz constant of an extension of this map to the entire simplex?

Is there any literature or insight on this problem? Already something on the case of finite dimensional $L^1$ would do if that's any easier.

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    $\begingroup$ IIRC if you remove the word "regular" then we don't know if the constant is bounded when $n$ goes to infinity. (Keith Balls extension problem.) See section 1.6 here web.archive.org/web/20140826001040id_/http://… $\endgroup$
    – Vladimir Zolotov
    Commented Aug 19, 2022 at 8:14
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    $\begingroup$ In the regular case you can get some small constant like $6$. And the strategy is: Let WLOG say that the map of vertices is non expanding. Say that the image points are $x_1,\dots, x_n$. There is a point $W$ such that the distance from any of $x_i$ is less or equal to $1$. (For example you can take $W = x_1$). We take the intersection of our simplex with $1/3$-ball around it's first vertex and map this thing into $[x_1, W]$. Similar with other vertices. And shove the remaining part into $W$. $\endgroup$
    – Vladimir Zolotov
    Commented Aug 19, 2022 at 9:01
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    $\begingroup$ Actually I think if you take $1/2$-balls instead than constant will be $2$. $\endgroup$
    – Vladimir Zolotov
    Commented Aug 19, 2022 at 9:10
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    $\begingroup$ I think that by mapping vertexes into Rademacher basis(?) (for $n=3$ this is $x_1 = (1,1,1,1,0,0,0,0), x_2 = (1,1,0,0,1,1,0,0), x_3 = (1,0,1,0,1,0,1,0)$) one gets to the situation that $\min_W max_i \frac{||W - x_i||}{||x_1 - x_2||} \rightarrow 1$, when $n \rightarrow \infty$. Then one can get an asymptotic lower bound of $\sqrt{2}$ by extending the map to the baricenter of vertices only. $\endgroup$
    – Vladimir Zolotov
    Commented Aug 19, 2022 at 10:18

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