How to Prove a Function Has a Fixed Point on a Closed Interval?

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SUMMARY

This discussion focuses on proving the existence of fixed points for continuous functions on closed intervals, specifically using the Intermediate Value Theorem (IVT). The first problem involves two continuous functions, f and g, defined on the interval [a,b], where f(a) ≤ g(a) and f(b) ≥ g(b), leading to the conclusion that there exists a point c in [a,b] such that f(c) = g(c). The second problem asserts that any continuous function f mapping [0,1] to [0,1] must have a fixed point, f(x) = x, which can be demonstrated by showing that f intersects the line g(x) = x. The hint provided emphasizes the use of the function h(x) = f(x) - g(x) in conjunction with the IVT.

PREREQUISITES
  • Understanding of the Intermediate Value Theorem (IVT)
  • Knowledge of continuous functions and their properties
  • Familiarity with fixed point theorems
  • Basic algebraic manipulation of functions
NEXT STEPS
  • Study the application of the Intermediate Value Theorem in various contexts
  • Explore fixed point theorems, including Brouwer's Fixed Point Theorem
  • Practice proving properties of continuous functions on closed intervals
  • Investigate the implications of continuity in real analysis
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Mathematicians, students studying real analysis, and anyone interested in the properties of continuous functions and fixed point theory.

SomeRandomGuy
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1.) Suppose f:[a,b]->R and g:[a,b]->R are continuous such that f(a)<=g(a) and f(b)=>g(b). Prove that f(c)=g(c) for some c in [a,b].

I started out by using the Intermediate Value Property for some c1 and c2 with f(c1)=L1 and g(c2)=L2. I am trying to conclude that L1=L2. This was one approach. Then, I also tried setting up a ratio where f(a)/g(a) > 1 and f(b)/g(b) < 1 by assumption. These are continuous if g !=0, so there must exist a point where f/g = 1. But, what if g(x)=0. This method seems to fall apart as well, and now I'm stuck.

2.) Let f:[0,1]->[0,1] be a continuous function. Prove that f must have a fixed point; that is, show f(x)=x for some x in [0,1]. I'm trying to show that f must intersect the graph g(x)=x by using the IVP again. Don't quite know how, though. Seems that if I figure out the first problem, this one would be very similar.

Thanks for any help, I appreciate it.
 
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Hint for 1. Consider the function h(x)=f(x)-g(x) and apply the intermediate value theorem.

2 is easy once you know 1.
 
Wow... I tell ya, sometimes I just feel like a moron with how obvious some of these proofs are.

Thanks for your help, both of these problems were cake.
 

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