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Let $f : [a , b] \rightarrow [a ,b]$ , and $f$ is continuous over $[a,b].$ Show that there is a $k \in [a , b] $ such that $f(k) = k.$ Is this it unique ?

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Consider the function $g(x):= f(x)-x$. Note that $g(a)=f(a)-a\geq 0$ and $g(b)=f(b)-b\leq 0$. Can you conclude from here?

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  • $\begingroup$ It sounds like Intermediate value theoreme :-) . thanks . of course the uniqueness of $~k~~$ depends on wether $~f~$ is monotonous or not . $\endgroup$ – Hilbert Dec 26 '16 at 17:30
  • $\begingroup$ You're welcome. Also you have the right idea about uniqueness, but be careful: there are non-monotone $f$ for which $k$ is unique and vice-versa. The big point here is that the IVT does not give any idea of the number of roots so we cannot deduce uniqueness. You can write down counterexamples quite quickly. $\endgroup$ – Matt Dec 26 '16 at 19:11

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