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Given that $f:[a,b]\to[a,b]$ is a real continuous one-to-one function, and that neither $a$ or $b$ are fixed points of $f$, show that there exists a fixed point in $(a,b)$.

Proof:

Since $f(a)\ne a$ and $f(b)\ne b$, and $f$ is one-to-one on $[a,b]$, by IVT, $\exists c_1, c_2 \in (a,b)$ such that $f(c_1)=a$ and $f(c_2)=b$. Since $a$ and $b$ are boundary points, $\exists d_1, d_2\in (a,b)$, with $d_1\ne d_2$ (WLOG, let $d_1 < d_2$), such that $f'(d_1)=f'(d_2)=0$. Because of this and since $f$ is continuous, $\exists k\in (d_1,d_2)$ such that $\left| f'(k)\right|=1$. This implies that $f(x)$ coincides with $g(x)=x$ at least once, which implies that a fixed point exists in $(a,b)$.

I think that rigour suffers someplace in this proof. Would appreciate some feedback and suggestions.

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  • $\begingroup$ 1-1does not mean "there is $x$ such that $f(x)=a$" $\endgroup$ – Nick Nov 10 '16 at 1:28
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    $\begingroup$ This proof seems to assume that $f$ is also onto, but we are not told that it is. The range of $f$ can be a proper subset of $[a, b]$, and then there isn't a $c_{1}$ with $f(c_{1}) = a$. $\endgroup$ – avs Nov 10 '16 at 1:29
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    $\begingroup$ Draw a picture: Since $f(a) \neq a$ and $f(b) \neq b$, it follows that $f(a) > a$ and $f(b) < b$. Therefore, the points $(a, f(a))$ and $(b, f(b))$ lie on different sides (half-planes) of the diagonal $y = x$. The graph of the function is a continuous curve connecting these two points... $\endgroup$ – avs Nov 10 '16 at 1:33
  • $\begingroup$ @avs, drawing a picture will help, but how do you prove that $(a, f(a))$ and $(b, f(b))$ are in different half-planes? $\endgroup$ – sequence Nov 10 '16 at 2:46
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    $\begingroup$ The continuous function $g(x) = f(x) - x$ is negative at $x=a$ and positive at $x=b$. By the IVT, $g(x)$ has a zero somewhere in $(a, b)$. A zero of $g$ is a fixed point of $f$. $\endgroup$ – avs Nov 10 '16 at 4:23
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I don't really understand your point about $d_1, d_2$, but I'd say $f(x) = a+b-x$ is a counterexample. Plus, you don't even know if $f$ is differentiable.

Consider now $g(x) =f(x) - x$. $f$ is a continuous bijection, so it's monotone, and you can then easily check that $f(a) =b$ and consequently $f(b) =a$, so $g(a) > 0, g(b) < 0$ and you just need to apply Bolzano.

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  • $\begingroup$ But this is just a special case. We need to prove this for the general case. $\endgroup$ – sequence Nov 10 '16 at 2:45
  • $\begingroup$ What is the "general" case? And why are you downvoting our answers? $\endgroup$ – Momo Nov 10 '16 at 3:39
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You don't even need $f$ to 1:1 in order to prove the fixed point Continuity is enough. Just apply the intermediate value theorem for $g(x)=f(x)-x$

If you know that your function is increasing, you don't even need continuity. In this case your fixed point is $c=\sup\{x:f(x)\geq x\}$ This last result is an application of the Knaster-Tarski theorem.

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  • $\begingroup$ Why my answer is being downvoted? $\endgroup$ – Momo Nov 10 '16 at 1:58

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