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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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. |
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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. |
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