Here is just a little curiosity. Assume the $f : [0,1] \to [0,1]$ is a Lipschitz function that maps 0 to 0 and 1 to 1. If we impose that the Lipschitz constant of $f$ is $\le 1$, can $f$ be anything other than the identity map?
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1Nice question! A first observation is that the constant cannot be strictly less than $1$ because otherwise we would be in contradiction with the contraction lemma. – Giuseppe Negro Nov 09 '12 at 22:21
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Thanks, what do you mean by contraction lemma? – passerby51 Nov 09 '12 at 22:36
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You're welcome. Here's what I mean: http://en.wikipedia.org/wiki/Banach_fixed-point_theorem – Giuseppe Negro Nov 09 '12 at 22:54
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1Well, since $f(1)-f(0) = 1-0$, the Lipschitz constant must be at exactly one. Using contraction seems like overkill. – copper.hat Nov 09 '12 at 22:59
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No. For $x\in[0,1]$ you get $f(x)=\lvert f(x)-f(0)\rvert\le\lvert x-0\rvert=x$ and $1-f(x)=\lvert f(1)-f(x)\rvert\le\lvert 1-x\rvert=1-x$, so $f(x)\ge x$. Thus $f(x)=x$.
Harald Hanche-Olsen
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If $(x,y)\in[0,1]^2$ but is not on the diagonal line, then either the line from that point to $(0,0)$ or the line from that point to $(1,1)$ has slope more than $1$.