Proving a Continuous Function has a Fixed Point

[JellyFox]

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

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.

 

[micromass]

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)

 

[JellyFox]

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.

 

[Landau]

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 \geq 0 resp. \leq 0, instead of > 0 resp. <0 :) )

 

[micromass]

Yes, I'm sorry, a function is positive for me if it is \geq 0. Sorry for the confusion...