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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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  • $\begingroup$ 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. $\endgroup$ Dec 8, 2010 at 23:28
  • $\begingroup$ Here is an interesting answer: quora.com/… $\endgroup$
    – PinkyWay
    Jan 15, 2020 at 15:16

8 Answers 8

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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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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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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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Consider $x-f(x)$ and use Intermediate Value Theorem.

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If $f(0) = 0$ we are done. If the former is not true then, since $f: [0,1] \to [0,1]$, 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 $g: [0,1] \to [0,1]$ such that $g(x) = f(x) - x$. If $0$ belongs to $\text{Image($g$)} $ then we are done. If it does not, then $\text{Image($g$)} $ is contained in the union the following disjoint sets: $(-\infty,0)$ and $(0,+\infty)$.

These two open sets disconect $\text{Image($g$)} $, a contradiction since $g$ is continuous and $[0,1]$ is connected.

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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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You could also nuke the mosquito: in particular, this is a special case of Kakutani's fixed point theorem.

This works because:

  • By the closed graph theorem, the compactness of the codomain of $f : [a,b] \rightarrow [a,b]$ is sufficient to deduce the closedness of its graph.
  • Given $x \in [a,b]$, the element $f(x)$ can be identified with the corresponding singleton set $\{f(x)\}$, and every singleton set is non-empty and convex.
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One way is to use the property of continuous function on compact set. Take g(x) =|x - f(x)| on [a,b]. Clearly g is continuous so image of g must be compact, hence it will achieve it's minimum value. Can you show now that minimum will be zero ?

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  • $\begingroup$ How to show the minimum will be zero? $\endgroup$
    – Hamilton
    Dec 9, 2021 at 0:35

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