Lipschitz function

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.

-
Yes, it is possible. But it looks an awful lot like homework... – Igor Rivin May 3 '11 at 14:47
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? – Yemon Choi May 3 '11 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.

-
Thank you. I don't know why I didn't get it! – 888 May 3 '11 at 15:17
Nice answer, why somebody downvoted it? – Pietro Majer May 3 '11 at 17:56
I guess I could have left it as a comment: It was really more of an exercice. – Roland Bacher May 3 '11 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).

-