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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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    Well, since $f(1)-f(0) = 1-0$, the Lipschitz constant must be at exactly one. Using contraction seems like overkill.2012-11-09

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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 $f(x)=x$.

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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 $1$.