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Suppose that $a < b$ and $f: [a,b] \to\mathbb{R}$ is a continuous function such that the range of $f$ contains $[a,b]$. Prove that $f$ has a fixed point.

I did a similar proof where f contains the range of $f$ instead of the other way around. How to go about doing this one?

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Let $f:[a,b]\rightarrow \mathbb R$ be a continuous function so that $f\{[a,b]\}\supset [a,b]$

define $g:[a,b]\rightarrow \mathbb R$ be such that $g(x)=f(x)-x$, we want to show that $g$ is zero in at least one point

Then there is $a'$ so that $f(a')=a$ and there is $b'$ so that $f(b')=b$

Suppose that $a'<b'$. Then consider the interval $[a',b']$. Notice $g(a')=a'-a\geq 0$ and $g(b')=b'-b\leq0$. SO by the intermediate value theorem there is an $x$ in $[a',b']$ so that $g(x)=0$ as desired.

if $a'>b'$ do the same thing with the interval $[b,a]$

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  • $\begingroup$ for the one where the range is contained in [a,b] I did the following: g(x)=f(x)-x x=a : g(a)=f(a)-a>=0 x=b : g(b)=f(b)-b<=0 then by IVM, there exist some c such that g(c)=0 hence f(c)=c $\endgroup$ – user296855 Dec 11 '15 at 3:49
  • $\begingroup$ Perfect, the same argument works on the second case. $\endgroup$ – Jorge Fernández Hidalgo Dec 11 '15 at 14:09
  • $\begingroup$ The argument works the same regardless of which contains which? $\endgroup$ – user296855 Dec 11 '15 at 18:49
  • $\begingroup$ Oh my, I made a mistake, Let me fix it. $\endgroup$ – Jorge Fernández Hidalgo Dec 11 '15 at 22:24
  • $\begingroup$ Fixed, I'm so sorry, I had misread your question at first. $\endgroup$ – Jorge Fernández Hidalgo Dec 11 '15 at 22:30
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HINT: Think about the function $f(x)-x$. Can you find some value where this is guaranteed to be non-positive? Non-negative?

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