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For $f:[a,b]\rightarrow [a,b]$ is a continuous . Prove that $f$ has a fixed point . Is that true if we change $[a,b]$ by $[a,b)$ or $(a,b)$.

Martin Sleziak
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Hung Nguyen
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3 Answers3

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Another idea without using Intermediate Value Theorem:

Proof: Firstly, let us pick any point from $[a,b]$, say $x_0$. Assume we have defined $x_n$, let $x_{n+1}=f(x_n)$. Thus we abtain a sequence $\{x_n: n\in \Bbb N^+\}$ from $[a,b]$. Since $[a,b]$ is compact, then for any infinite subset of $[a,b]$ there exists an limit point in $[a,b]$, say $x$.

Then we can claim that $f(x)=x$. Since $f$ is continuous, $f(x)=f(\lim x_n)=\lim f(x_n)=\lim x_{n+1}=x$. This complete the proof.

The interval $[a,b)$ or $(a,b)$ is not enough, Since $x$ maybe fell on the point $b$, which is no longer in the origin interval.

Added: The answer seems right. However it is not right, just as Landscape points; he gives us a counterexample in the comments. Many times if the answer is wrong, I will delete it immediately. However this is a special case. From it, I learn some lessons. Something cannot be of course. So I hope to keep it as long as I can.

Paul
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  • Your argument does not work, because $x_n$ may not converge, despite that is has convergent subsequence. – 23rd Jun 10 '13 at 07:51
  • @Landscape: Yes. $x_n$ may not converge, however there must exists an accumulation of the sequence. – Paul Jun 10 '13 at 07:54
  • Then how do you define $x$ and how do you conclude $f(x)=x$? – 23rd Jun 10 '13 at 08:04
  • @Landscape: Pick any one accumulation point with corresponding subsequence ${y_n}$ of ${x_n}$. – Paul Jun 10 '13 at 08:09
  • Please consider my comments seriously. If you define $x$ is this way, $f(x)=x$ does not hold in general. – 23rd Jun 10 '13 at 08:18
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    @Landscape: We can replace ${x_n}$ with ${y_n}$. – Paul Jun 10 '13 at 08:25
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    I think there must be a little correction (don't take $,{x_n},$ but a subsequence of it that converges to some $;x\in [a,b];$), but imo Paul's answer's basically correct. and the downvote seems too harsh. – DonAntonio Jun 10 '13 at 09:29
  • @DonAntonio: I cannot agree with you. How could you find $x_0$, such that $x_n$ is convergent? For me, I think the easiest way is to find a fixed point $x_0$, but then Paul's argument will make no sense. Please notice that for an arbitrary $x_0$, the sequence $x_n$ could be very complicated, at least it could be dense in $[a,b]$. – 23rd Jun 10 '13 at 11:12
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    @Landscape, I think it's the other way around: let $,{x_{n_k}:=f\left(x_{n_k-1}\right)},$ be a subsequence of $;{x_0, x_1:=f(x_0),x_2:=f(x_1),\ldots};$ converging to some $;x\in [a,b],$ (exists such subseq. by compactness) , and now use continuity of $,f,$ to show $;x;$ is a fixed point. We do not begin with "a fixed point" which existence we still have to prove! – DonAntonio Jun 10 '13 at 11:26
  • @DonAntonio: Once you have defined $x_n=f(x_{n-1})$ for every $n\ge 1$, you cannot redefine $x_{n_k}:=f(x_{n_{k-1}})$. – 23rd Jun 10 '13 at 11:30
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    I agree with you, @Landscape, as the indexes could be wild... though they must be an ascending sequence on the naturals, but still you have the subsequence $,{x_{n_1},x_{n_2},...} ={f(x_{n_1-1}),,,f(x_{n_2-1}),...};$ I'm not saying these indexes are consecutive, and not even the nonsense I wrote in my past comment: they're just indexes...For example, it could be $${x_2=f(x_1), x_{11}=f(x_{10}), x_{1334}=f(x_{1333}),...}$$ – DonAntonio Jun 10 '13 at 11:38
  • @DonAntonio: However, if you only know that $x_{n_k}$ is convergent, you cannot say anything about whether its limit is a fixed point or not. Here is a simple example to show the absurdity of Paul's argument: $f(x)=1-x$ on $[0,1]$ and $x_0=0$. I didn't want to show such examples because it gives one some illusion that the situation is not too bad. Actually, as I mentioned before, there are many examples of $f$ which are so called topologically transitive, for those maps the orbits of many initial points are dense in $[a,b]$. – 23rd Jun 10 '13 at 11:47
  • I still don't get your point, @Landscape, and even less I understand your example, which has $,x=0.5,$ as a fixed point. What I say is $$x_{n_k}\xrightarrow [k\to\infty]{}x\implies \lim_{k\to\infty}f(x_{n_k})=f(\lim_{k\to\infty}x_{n_k})=f(x);\ldots\ldots$$ I can't see a flaw here... – DonAntonio Jun 10 '13 at 11:52
  • Oh, I think I get it now...! From $,f(x_{n_k})=x_{n_k-1};$ I can not deduce that also the subsequence $,{x_{n_k-1}},$ converges to the same $,x,$ as the subseq. $,x_{n_k},$...nor even that it converges at all! Great (though I still don't get your example above), and thanks a lot,@Landscape – DonAntonio Jun 10 '13 at 11:54
  • Many times I do delete the nonsenses I write in comments but this time I will leave all this (1) for me and others to learn a very nice thing, and (2) to teach me (oh, yeah: once again!) to be more careful next time. – DonAntonio Jun 10 '13 at 11:56
  • @DonAntonio: There is nothing wrong in the evaluation of your limit, but you cannot conclude $f(x)=x$. In my example, $x_{2n}=0$ and $x_{2n+1}=1$. Then all the limit points of $x_n$ are $0$ and $1$; you cannot get $0.5$. – 23rd Jun 10 '13 at 11:56
  • @Paul , I'm afraid I've been convinced duly that your argument is wrong. – DonAntonio Jun 10 '13 at 11:57
  • Yes, of course @Landscape...in fact, that line with limits in my comment is perfectly correct...but I can't tell anything about the fixed point. – DonAntonio Jun 10 '13 at 11:58
  • @DonAntonio: Here is a further comment. Unfortunately, the topological transitivity of a given map is usually not so easy to show, so I didn't mention that before. If you are interested, you may look at $f(x)=4x(1-x)$ on $[0,1]$ for example. If you allow change the domain of $f$ to a circle, say $S^1:=\mathbb R/\mathbb Z$, then there are simpler examples: it is easy to see that for any irrational number $\alpha$, the map $f:S^1\to S^1$, $f(x)=x+\alpha$ is topologically transitive. – 23rd Jun 10 '13 at 12:00
  • @DonAntonio: Thanks for your perfect comments. I've learnt a lot. I will take some time to consider the answer. – Paul Jun 10 '13 at 12:02
  • @DonAntonio: Are you Ok if I delete the answer? – Paul Jun 10 '13 at 12:03
  • @Landscape: Are you OK if I delete the answer, then your comments will also be deleted? – Paul Jun 10 '13 at 12:05
  • I'll be fine if @DonAntonio also agrees. By the way, it seems that users of 10k+ reputations could view deleted answers. – 23rd Jun 10 '13 at 12:05
  • @Landscape: if downvotes are not too many to the answer, maybe I will keep as long as I can. – Paul Jun 10 '13 at 12:08
  • @Paul, it's up to you. As you say, people with enough points can see all deleted questions so it doesn't really matter, yet I think this could really teach some people several nice things. I know I learned something. – DonAntonio Jun 10 '13 at 12:11
  • @Paul: If you want to delete, please feel free to do that; if not, you may either edit your answer to indicate that your argument is flawed or replace your argument with a correct one if possible. Thank you for your consideration to ask my opinion. :) – 23rd Jun 10 '13 at 12:15
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Hint: Apply Intermediate Value Theorem on the function $g(x) = f(x) - x$.

The closed interval is needed, because of the conditions of IVT. For example, the function $f:(0,\infty) \rightarrow ( 0, \infty)$ given by $f(x) = x + 1$ has no fixed point. You can modify this slightly if you want an interval $[a,b)$ or $(a,b)$.

Calvin Lin
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I have an ideal .

1) We have $f:[0,1]\rightarrow [0,1]$ ($f$ is continuous mapping ) has a fixed point . ( from $x-f(x)=g(x)$ with $g(z)=0\Rightarrow z=f(z)$.

2) We need prove that : $X,Y$ be homeomorphic topological spaces. then each continous mapping $h:X\rightarrow X$ has a fixed point if and only if $k:Y\rightarrow Y$ has a fix point .

$Proff.$ We have $h:X\rightarrow X$ , $g:Y\rightarrow X$,$f:X\rightarrow Y$ , $k:Y\rightarrow Y$.Suppose $h$ has a fixed point , then $h=g.k.f$ is continuous $z\in X : h(z)=z$. Let $w=f(z)$. We have $k(w)=k(f(z))=fg(k(f(z)))=f(h(z))=f(z)=w.$. Thus $w$ is fixed point of $k$.

Hung Nguyen
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