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For constructing another proof I need two functions explicitly and therefore

I was wondering whether there exists a function that has nowhere a fixed point and a function that (maybe depending on the closed interval $[a,b] \subset \mathbb{R}$ where it is defined) has always somewhere a fixed point for each interval $[a,b]$?

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    $\begingroup$ $y=e^x$ has no fixed point; $y=x$ has some in every nonempty interval. $\endgroup$ – Gerry Myerson Jul 11 '13 at 10:15
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Sure -- $f(x)=x-1$ and $g(x)=x$. I think the latter is the only such function with no restrictions on the interval; for the former, any function which does not intersect $y=x$ works.

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    $\begingroup$ $g(x)=x$ is far from the only example. If $g(x)=x$ for all rational $x$, and is whatever-you-like for all irrational $x$, that works. $\endgroup$ – Gerry Myerson Jul 11 '13 at 12:59
  • $\begingroup$ Hmm, good point. if $a=b$ then either $g(a)=a$ is a fixed point, or $g(a) \notin [a,b]$ and the statement is vacuously true. And if $a < b$ then the interval contains a rational. $\endgroup$ – Sneftel Jul 11 '13 at 13:37

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