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Suppose that $f:[a,b] \to [a,b]$ is continuous. Prove that there is at least one fixed point in $[a,b]$ - that is, $x$ such that $f(x) = x$.

I haven't a clue where to even start on this one.

Nick
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    http://math.stackexchange.com/questions/416126/prove-that-f-has-a-fixed-point, http://math.stackexchange.com/questions/13582/show-that-a-continuous-function-has-a-fixed-point, http://math.stackexchange.com/questions/425033/show-that-any-continuous-f0-1-rightarrow-0-1-has-a-fixed-point-zeta, http://math.stackexchange.com/questions/255403/proof-of-that-a-continuous-function-has-a-fixed-point – Jonas Meyer Sep 24 '13 at 04:13

2 Answers2

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Consider the function $g(x) = f(x) - x$. Then $$g(a) = f(a) - a \ge a - a = 0$$ while $$g(b) = f(b) - b \le b - b = 0$$ So by the intermediate value theorem, ...?

  • I don't really understand how you can just assert those two inequalities. Why does that have to be true? – Nick Sep 24 '13 at 04:39
  • I followed a proof in my book almost exactly, tell me if this sounds reasonable, starting from where you left off- Consequently, by the modified Bolzano’s theorem, there is a c within (a,b) such that g(c) =0. Therefore 0 = g(c) = f(x)-x, and f(x)=x. – Nick Sep 24 '13 at 04:43
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    Since $f:[a,b]\to[a,b]$ then for all $x\in[a,b]$ we have $a\le f(x)\le b$. In particular $a\le f(a)\le b$ and $a\le f(b)\le b$. So $f(a)-a\ge0$ and $f(b)-b\le0$. – user71352 Sep 24 '13 at 04:44
  • @user71352, that makes sense. What do you think about my ending of the proof? noted above. – Nick Sep 24 '13 at 05:01
  • Looks okay to me. – user71352 Sep 24 '13 at 05:10
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If $f(a)=a$ or $f(b)=b$ then we are done. Otherwise assume that $f(a)>a$ and $f(b)<b$. Consider $g(x)=f(x)-x$. $g(a)>0$ but $g(b)<0$ hence there exists $c\in(a,b)$ such that $g(c)=0$. Hence $f(c)=c$.

user71352
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