Suppose that $f$ is a function defined in $[a;b]$ to $[a;b]$ and continuous on $[a;b]$. The problem is I haven't the definition of the function, this is more abstract, but even if how can I prove that $f$ would have a fixed point?
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1$\begingroup$ Do you know the intermediate value theorem? $\endgroup$– Kevin ArlinOct 11, 2012 at 19:20
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1$\begingroup$ Hint: define $g(x) = f(x) - x$. $\endgroup$– GMBOct 11, 2012 at 19:25
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$\begingroup$ yes but how can I use here this is very abstract. $\endgroup$– pourjourOct 11, 2012 at 19:26
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$\begingroup$ I need a general proof not a special situation. $\endgroup$– pourjourOct 11, 2012 at 19:38
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$\begingroup$ The hint is not to prove that $f(x)=x$, that is not possible. A fixed point is a root of $f(x)-x$, can you prove that this difference has a root? $\endgroup$– N. S.Oct 11, 2012 at 21:09
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3 Answers
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Note that a fixed point is when $f(x) = x$ or $f(x) - x = 0$. Consider the function $f(x) -x$. Can it be everywhere positive? Can it be everywhere negative? As it is continuous, it thus has to have a zero.
answered Oct 11, 2012 at 19:26
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$\begingroup$ It doesn't really need to be f(x)=x it can have another definition and we can found f(a) = a; even if f(x)!=x; if you understand what I mean. $\endgroup$– pourjourOct 11, 2012 at 19:31
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$\begingroup$ for example f(x)=x^2-3x+4 had a fixed point in 2; f(2)=2 $\endgroup$– pourjourOct 11, 2012 at 19:34
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$\begingroup$ pourjour, you are saying it doesn't have to be $f(x)=x$, and then you are finding that $f(2)=2$. In other words, you are finding that $x=2$ is a solution of the equation $f(x)=x$, which is exactly what Karolis and LevDub and N. S. are telling you to do. $\endgroup$ Oct 12, 2012 at 5:18
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I think you can prove it using the properties of the real line. If $f(a)=a$ or $f(b)=b$ then the result follows. If not then $f(a)\gt a$ and $f(b)\lt b$ try using the continuity of $f$ to show that $\sup\{x\in [a,b]\space|\space f(x)\gt x\}$ is a fixed point for $f$.
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A related problem. Here is what you are looking for Brouwer fixed-point theorem.
In the plane: Every continuous function $f$ from a closed disk to itself has at least one fixed point.
This can be generalized to an arbitrary finite dimension:
In Euclidean space: Every continuous function from a closed ball of a Euclidean space to itself has a fixed point.
A slightly more general version is as follows:
Convex compact set: Every continuous function f from a convex compact subset K of a Euclidean space to K itself has a fixed point.
An even more general form is better known under a different name:
Schauder fixed point theorem: Every continuous function from a convex compact subset K of a Banach space to K itself has a fixed point.
answered Oct 11, 2012 at 21:18