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Merry Christmas everybody.

Let $a,b\in\mathbb{R}$ and $a<b$. Prove that: If $f: [a,b] \rightarrow [a,b]$ is continuous, then there is a fixed-point in $f$.

So basically, if f is continous I should find a $c \in[a,b]$ so that $f(c) = c$. - Isn't this equivalent to $f(a) = a$ or $f(b) = b$ since a and b $\in[a,b]$?

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    $\begingroup$ To the last question: No. For example, if $a=0, b=1$ and $f(x)=1-x$ then $f(c)=c$ only when $c=\frac{1}2$. In particular $f(0)\neq 0$ and $f(1)\neq 1$. $\endgroup$ Dec 25, 2013 at 2:52
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    $\begingroup$ In fact, without information on $f$, you cannot in general deduce where the fixed point is: Consider $a=0$ and $b=1$, and let $f(x)=c$ where $c$ is your favorite constant between $0$ and $1$. The only fixed point of $f$ is $x=c$. $\endgroup$ Dec 25, 2013 at 2:58
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    $\begingroup$ Possible duplicate of Show that a continuous function has a fixed point $\endgroup$
    – Jigao
    Dec 16, 2018 at 13:13

3 Answers 3

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Let $g(x)=f(x)-x$.

Then $g(a)=f(a)-a\geq0$, since $a$ is a lower bound of $[a,b]$ and $f(a)$ is in $[a,b]$. Likewise $g(b)=f(b)-b\leq 0$.

If equality holds in either case, then we're done. If not, apply the intermediate value theorem! That is, $0$ is between $g(b)$ and $g(a)$, and $g$ is continuous, so $g(c)=0$ for some $c\in(a,b)$.

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  • $\begingroup$ If I may ask, why do you introduce $g(x) = f(x) - x$ for this proof? $\endgroup$
    – Nhat
    Dec 25, 2013 at 3:04
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    $\begingroup$ The cheeky answer is that I knew to do it because I've seen proofs like this before. The real reason is that if you can show $f(c)-c=0$ for some $c$, then add $c$ to both sides and you get $f(c)=c$, i.e a fixed point. $\endgroup$
    – Nick D.
    Dec 25, 2013 at 3:07
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This question is directly followed by Brouwer's fixed point theorem, which states that any continuous function mapping a compact convex set into itself has fixed point.

To show an elementary method, assume there's no fixed point, then $f(x)>x$ or $f(x)<x$ for $x\in[a,b]$ since $f$ is continuous. It follows that $(f(a)-a)(f(b)-b)>0$. However, this is contradiction because $a,b$ is the boundary of the range and thus $f(a)\geq a$ and $f(b)\leq b$.

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This answer is a slight modification of the answer posted by Nick D. On contrary, suppose that there is no fixed point of $f$. Now put $g(x)=\frac{f(x)-x}{|f(x)-x|}$. Then $g$ is a continuous function on a connected set having disconnected image, which is a contradiction.

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