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Every $f : [a, b] → [a, b]$ has a fixed point and $f$ is continuous (on $[a,b]$). Deduce the intermediate value theorem.

I managed to show the other way, now I'm here.

I know that $f(c)=c$ for some $c\in [a,b]$, and I need to show that for all $x\in [f(a),f(b)]$ there is $f(y)=x$.

I know that $a<f(a),f(b)<b$ and $f(c)=c$.

Can somebody give me a hint?

GRS
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  • Are you saying "$f$ is continuous" or "$f$ has a fixed point at which it is continuous"? – Ian May 08 '16 at 15:59
  • $f$ is continuous on $[a,b]$ – GRS May 08 '16 at 16:00
  • Is this the original wording of the question? – alphacapture May 08 '16 at 16:32
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    We can assume $f$ to be continuous with $f(a) f(b) < 0$ and we need to find $c$ with $f(c) = 0$. This appears tricky as we need to find a function $g$ such that $g(c) = c$ implies $f(c) = 0$. This is possible if $g(x) = f(x) + x$. The challenge is to find an interval $I \subseteq [a, b]$ such that $g(I) \subseteq I$. – Paramanand Singh May 08 '16 at 16:32
  • @alphacapture Exercise. Assume that every continuous $f : [a, b] → [a, b] $ has a fixed point. Deduce the intermediate value theorem. – GRS May 08 '16 at 16:34
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    The word "every" is very important here; also, I believe the problem is not to prove the intermediate value theorem for only functions $f:[a, b] \rightarrow [a, b]$. I think it is to prove the intermediate value theorem in general – alphacapture May 08 '16 at 16:39
  • @alphacapture I see, it's because I did the other implication that confused me, I'm still getting nowhere with this one :( – GRS May 08 '16 at 16:44
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    Notice also that the $[f(a), f(b)] \subset f([a,b])$,and the latter is what you want for the intermediate value theorem – Ant May 08 '16 at 16:48
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    Do you know that every continuous function is bounded? – alphacapture May 08 '16 at 16:59
  • @GRS: Can you use the fact that a continuous function on $[a, b]$ is bounded? – Andrew D. Hwang May 08 '16 at 17:01
  • I can see that every continuous function will be bounded by $[a,b]$ in my case – GRS May 08 '16 at 17:08
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    @GRS: The hypothesis "$f:[a, b] \to [a, b]$" is not enough for the purposes I have (and probably alphacapture has) in mind. As Paramanand Singh notes, we need to "tame" continuous functions that are not known to map some interval to itself. :) – Andrew D. Hwang May 08 '16 at 17:15
  • @AndrewD.Hwang I know that a fixed point exists, then I can say that there exists a contraction since $[a,b]$ is complete. Hence all these functions will map into a domain smaller than original – GRS May 08 '16 at 17:23
  • Existence of a fixed point certainly does not guarantee $f$ is a contraction...! – Andrew D. Hwang May 08 '16 at 17:24
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    Be that as it may, here's a big hint: Show that if $f$ is an arbitrary continuous, real-valued function on $[a, b]$, and if $c$ is a real number between $f(a)$ and $f(b)$, then there exists a non-zero real number $\eta$ such that the function$$g(x) = x + \eta\bigl(f(x) - c\bigr),\quad x \in [a, b],$$maps $[a, b]$ into $[a, b]$. Now apply the fixed-point property to $g$. I've written up the details to my satisfaction (i.e., I'm sure this approach works), but hate to deprive you the pleasure of working things out yourself (which requires some creativity). :) – Andrew D. Hwang May 08 '16 at 17:28
  • @AndrewD.Hwang Thanks a lot :) – GRS May 08 '16 at 17:35
  • thanks to @AndrewD.Hwang. I was trying to find the form of $g(x)$ for sometime, but gave up. – Paramanand Singh May 09 '16 at 10:22

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