1. ## Geometric proof for theorem about continuous functions

Hello

I'm trying to prove that if f : [a,b] ---> [a,b] is continuos, there exists an x in [a,b] such that f(x) = x

I've formed a sort of geometric proof - if you take the square [a,b] and draw the diagonal down the middle representing the identity function, then any function defined on the interval [a,b] has to intersect the line at least once.

I'm having trouble formalizing it though. Help would be greatly appreciated...

2. Try experimenting with the new function g(x) = f(x) - x, and think about the Intermediate Value Theorem.

3. Okay, I see that if g intersects with the x axis, then at that point the value of f is x. But I don't understand why the point of intersection has to be x...and why can't the values of f all be greater than x, so that there is no intersection?

4. Originally Posted by moses
I'm trying to prove that if f : [a,b] ---> [a,b] is continuos, there exists an x in [a,b] such that f(x) = x
Notice that $\displaystyle f(a)\ge a~\&~f(b)\le b.$
So if $\displaystyle g(x)=f(x)-x$ then $\displaystyle g(b)\le 0\le g(a)$.

5. shouldn't it be g(a) <= 0 <= g(b) ?

Anyway I see now why g gets the value zero, from the Intermediate Value Theorem, but why does this have to happen at x?

6. Originally Posted by moses
shouldn't it be g(a) <= 0 <= g(b) ?
No indeed. $\displaystyle f(a)\ge a$ implies $\displaystyle f(a)-a\ge 0$
And $\displaystyle f(b)\le b$ implies $\displaystyle f(b)-b\le 0$.

7. Got it!!! I think...if g(b) <= 0 <= g(a) then from the i.v.t there exists c in [a,b] such that g(c) = 0, and then f(c) - c = 0 and then f(c) = c

Whoo thanks