Is it possible to define explicitly a Lipschitz function $f:[a,b]\times[c,d]\rightarrow \mathbb{R}$ in term of $f(a,\cdot)$, $f(b,\cdot)$, $f(\cdot,c)$, $f(\cdot,d)$ if I know these functions and they are Lipschitz? Thanks.
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4$\begingroup$ Yes, it is possible. But it looks an awful lot like homework... $\endgroup$– Igor RivinMay 3, 2011 at 14:47
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1$\begingroup$ I am not so sure this is homework. Could the original poster please expand on how they came to consider this question, which partial results they have already found, and so forth? $\endgroup$– Yemon ChoiMay 3, 2011 at 14:58
2 Answers
If $a=c=0$ and $b=d=1$, define $f$ by affine interpolation between $f(x+y,0)$ and $f(0,x+y)$ if $x+y\leq 1$ respectively $f(x+y-1,1)$ and $f(1,x+y-1)$ if $x+y\geq 1$.
The general case can be reduced to the previous one by an affine transformation.
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1$\begingroup$ Nice answer, why somebody downvoted it? $\endgroup$ May 3, 2011 at 17:56
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$\begingroup$ I guess I could have left it as a comment: It was really more of an exercice. $\endgroup$ May 3, 2011 at 18:28
Also note that the data define a Lipschitz function on the boundary of the square, and recall that any real-valued Lipschitz function defined on a subset of a metric space always admits a Lipschitz extension to the whole metric space, with the same Lipschitz constant. There exists the minimal and the maximal such extension (here's the maximal one).