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Question: Let $a, b \in \mathbb{R}$ with $a < b$ and let $f: [a,b] \rightarrow [a,b]$ continuous. Show: $f$ has a fixed point, that is, there is an $x \in [a,b]$ with $f(x)=x$.

I suppose this has to do with the basic definition of continuity. The definition I am using is that $f$ is continuous at $a$ if $\displaystyle \lim_{x \to a} f(x)$ exists and if $\displaystyle \lim_{x \to a} f(x) = f(a)$. I must not be understanding it, since I am not sure how to begin showing this... Should I be trying to show that $x$ is both greater than or equal to and less than or equal to $\displaystyle \lim_{x \to a} f(x)$ ?

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I don't know what lim_{x \to x} f(x) is supposed to mean. Try drawing some examples of f (say with a = 0, b = 1) to convince yourself that this is plausible, then think about the intermediate value theorem. –  Qiaochu Yuan Dec 8 '10 at 23:28
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6 Answers

up vote 7 down vote accepted

Let $g(x) = f(x) - x$. Then $g$ is continuous and $g(a) \geq 0$ while $g(b) \leq 0$. By the Intermediate Value Theorem, $g$ has at least one zero on $[a, b]$.

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Please do not give complete answers to homework problems. –  Qiaochu Yuan Dec 8 '10 at 23:35
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My answer is correct and on-topic. I disagree that it is "complete" in the sense that any teacher asking this level of question is also likely to require more detail than I've provided. –  Alex Basson Dec 9 '10 at 0:26
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@Qiaochu: I think the ecological effect of scolding each answerer, leading to further metadiscussion in many cases (as we see here) as well as larding the comment threads with "do nots", should be considered. If there were a [task] tag or the like, it could just be silently attached to the question with all the filtering and answering implications that might have, and without homework metadiscussion. –  T.. Dec 9 '10 at 6:14
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Consider $x-f(x)$ and use Intermediate Value Theorem.

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Consider $g(x)=f(x)-x$. $f(a)\ge a$ so $g(a)=f(a)-a\ge 0$. $f(b)\le b$ so $g(b)=f(b)-b\le 0$. By the Intermediate Value Theorem, since $g$ is continuous and $0\in[g(b),g(a)]$ there exists $c\in[a,b]$ such that $g(c)=f(c)-c=0$, so $f(c)=c$ for some $c\in[a,b]$.

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Please do not give complete answers to homework problems. –  Qiaochu Yuan Dec 8 '10 at 23:36
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@Qiaochu Yuan: Where exactly does it say that this is a homework problem? –  Isaac Dec 8 '10 at 23:37
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@Isaac: this is a standard homework problem in real analysis, and it is stated as a command. All of the other questions the OP has posted also look like homework problems. –  Qiaochu Yuan Dec 8 '10 at 23:39
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@Qiaochu Yuan: it's also a common demonstration example for the IVT in calculus. Sure, I'll admit, it might be homework, but it is by no means a certainty. –  Isaac Dec 8 '10 at 23:41
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@Qiaochu: superseding the [homework] tag with an often accurate, but speculative, personal assessment is an innovation that wasn't noted in the meta thread, which is why I asked. I don't have a specific scenario in mind but imagine it would come up indirectly in many ways, such as users of could-be-homework getting close votes on their postings and moderators sometimes deciding the fate of the questions. –  T.. Dec 9 '10 at 8:35
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For a different approach then the ones above, let us take $a = 0$ and $b = 1$. So assume $f:[0,1] \to [0,1]$ has no fixed point. Then $[0,1] = \{x \in [0,1] : f(x) < x\} \cup \{x \in [0,1] : f(x) > x \}$. Now argue that this is not possible.

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Please do not give complete answers to homework problems. This is also not a different approach; it's the formal proof of the intermediate value theorem. –  Qiaochu Yuan Dec 8 '10 at 23:37
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@Qiaochu: I edited it a bit. I was not aware that it is homework. With different approach I mean more "topological". –  Jonas Teuwen Dec 8 '10 at 23:44
    
I disagree that it is not a different approach. There is more than one approach to proving the intermediate value theorem, and one need not use connectedness to prove it. E.g., one could instead first prove than an intersection of nested closed bounded intervals is nonempty and iteratively bisect intervals to prove IVT. Maybe I use "approach" too loosely, but I think that appealing directly to connectedness gives a nice perspective that the OP may not have thought of in light of the answer in the other 3 posts. –  Jonas Meyer Dec 9 '10 at 0:03
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You could argue on the contradiction by assuming your given function has a fixed point. By definition a function has a fixed point iff $f(x) = x$. If you substitute your function into the definition it would be clear you get an impossible mathematical equality, thus you have proved by contradiction that your function does not have a fixed point. Hope this helps. Try with $f(x) = 1+ e^x$, because as already stated it is one of the functions which has no fixed point.

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If f(0) = 0 we are done. If the former is not true then, since f: [0,1] -> [0,1], it must be that f(0) > 0. Applying the same reasoning we conclude that f(1) = 1 (and we are done) or f(1) < 1. Lets asume then that f(0)>0 and f(1)<1 .

Take g: [0,1] -> [0,1] such that g(x) = f(x) - x. If 0 belongs to Image(g) then we are done. If it does not, then Image(g) is contained in the union the following disjoint sets: (-infinity,0) and (0,+infinity).

These two open sets disconect Image(g), but g was continuous and [0,1] was connected. Absurd

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