You are given that which implies .
Use the hint so and .
If either of those is equal to 0 you are done. WHY?
Now finish.
PROBLEM:
Let a and b be real numbers with a<b. Suppose that f:[a,b]-->R is a continuous function that satisfies f([a,b]) is a set of [a,b].
Prove that f has a fixed point.
{The hint is: g=f(x)-x and use IVT}
ATTEMPT AT PROOF:
so using IVT if there is a and b then there must be a c so, a<c<b
which means f(a)<f(c)<f(b) or f(a)>f(c)>f(b). if f([a,b]) is a set of [a,b] then at most they are the same interval. if f is a fixed point then f(x)-x = 0 so, g=0.
OKay here is where i am stuck ..... if g(a)>0 and g(b)<0 then there is a fixed point and if g(a)<0 and g(b)>0 there is a fixed point but is that all they can ever be? .... is there an instant where g(a)<0 and g(b)<0 and still be existent with a fixed point in this interval.
WHY MUST THERE BE A FIXED POINT IN THIS INTERVAL?
Also can someone please give an example of what f:[a,b] --> R actually looks like. Thank you for your time
I think you have forgotten that "f([a,b]) is a set of [a,b]". If f(a)= a, you are done- a is a fixed point. So you can assume that . And since f(a) must be in [a,b], f(a)> a. If f(b)= b you are done so you can assume and that implies that f(b)< b.
Defining g(x)= f(x)- x, then, g(a)= f(a)-a> 0 and g(b)= f(b)- b< 0.
For example, f:[0,1]-> [0,1] defined by f(x)= 1- x is such a function. It's fixed point, of course, is 1/2.WHY MUST THERE BE A FIXED POINT IN THIS INTERVAL?
Also can someone please give an example of what f:[a,b] --> R actually looks like. Thank you for your time
On the other hand f(x)= x+ 1 is a continuous function from [0,1]--> R which does NOT have a fixed point. The theorem does not hold here because this function does not map an interval [a,b] into itself which is a crucial hypothesis.
(Blast! Plato got in just ahead of me again! )