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Another Analysis question: continuity and compactness 
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#1
Oct2708, 12:35 AM

P: 16

Let I = [0,1] be the closed unit interval. Suppose f is a continuous mapping from I to I. Prove that for one x an element of I, f(x) = x.
Proof: Since [0,1] is compact and f is continuous, f is uniformly continuous. This is where I'm stuck. I'm wondering if I can use the fact that since max {xy = 1} if xy = 1 then f(x)  f(y) < max {episolin}. This of course only occurs when WLOG x=1 y=0. Stuck as far as the rest of it goes. 


#2
Oct2708, 03:14 AM

P: 367

Consider the function [itex]g : I \to \mathbb{R}[/itex] given by [itex]g \left( x \right) = f \left( x \right)  x[/itex]. What happens if [itex]g \left( x \right) = 0[/itex] for some [itex]x[/itex]? How can you show that this has to happen in [itex]I[/itex]? I'm sure you must know a nice theorem that you can use here.



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