Relating with fix point theorem and continuity

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Homework Statement


Assume the function f : [0,1] x [0,1] -> [0,1] is continuous and apply the IVT to prove that there is a number c E [0,1] such that f(c,y0) = c for some y0 E [0,1]


The Attempt at a Solution


I tried to break the cube up with the ranging being y0 but I don't know how it maps y0 to [0,1] to be continuous, if I can prove this then I can use the IVT.

thankyou
 
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Let y0 be fixed. Consider the function g(x)=f(x,y0)-x. This function is sometimes [tex]\leq 0[/tex] and sometimes [tex]\geq 0[/tex]. So you can apply the IVT on g.
 
After you apply the IVT on g, you will get that the number between is zero?
 
Yes. You will get that there exists a c such that g(c)=0. This will be the c you're looking for.
 
What do you mean when you mean y0 is fixed?
 
So y0 in the element of [0,1] right? and one side I got g(x) >= 0 >= g(x)-1, Is this right?