Proving a Continuous Function has a Fixed Point

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Discussion Overview

The discussion revolves around proving that a continuous function has a fixed point, specifically through the examination of the function g(x) = f(x) - x. Participants explore the conditions under which g(x) crosses the X-axis and the implications of its behavior.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the squeeze theorem to approach the problem but expresses difficulty in applying it.
  • Another participant introduces the function g(x) = f(x) - x and questions whether it crosses the X-axis, noting that g can be both positive and negative.
  • A subsequent reply emphasizes that g is not necessarily negative, providing examples where g takes the value 0, but questions how to show that g has a root for all functions f.
  • Another participant agrees that g must take the value 0 somewhere, suggesting this is sufficient for the proof, while also clarifying the definitions of positive and negative in this context.
  • One participant apologizes for any confusion regarding the definition of positivity in the context of the function.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which g(x) has a root, and there is no consensus on how to prove that g has a root for all functions f. The discussion remains unresolved regarding the general case.

Contextual Notes

Participants have not fully resolved the assumptions regarding the behavior of g(x) across all continuous functions f, nor have they clarified the necessary conditions for g to have a root.

JellyFox
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This is a question from the exam for the calculus class I took last semester:

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It looks like it might be able to be done with squeeze theorem, but I can't work it out. Please help me with this, before I descend into madness.
 
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Consider the function g(x)=f(x)-x. Does this ever cross the X-axis? (HINT: the function g is sometimes positive and sometimes negative)
 
micromass said:
Consider the function g(x)=f(x)-x. Does this ever cross the X-axis? (HINT: the function g is sometimes positive and sometimes negative)

But it isn't necessarily negative. If f(x) = 1, then the minimum value of g is is 0, or if f(x) = 1/3 + 2x/3 then the minimum value of g is 0 again. And they show that there's a fixed point, but I don't know how to show that g has a root for all functions f. Thanks.
 
JellyFox said:
But it isn't necessarily negative. If f(x) = 1, then the minimum value of g is is 0, or if f(x) = 1/3 + 2x/3 then the minimum value of g is 0 again.
Correct, but g necessarily takes the value 0 somewhere, right? That's all you need.

(Or we could say that micromass's hint is correct, provided that 'positive' resp. 'negative' is defined as [itex]\geq 0[/itex] resp. [itex]\leq 0[/itex], instead of [itex]> 0[/itex] resp. [itex]<0[/itex] :) )
 
Yes, I'm sorry, a function is positive for me if it is [tex]\geq 0[/tex]. Sorry for the confusion...
 

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