Fixed Point for Function f

Mark just gave us this problem, I figured we could discuss it a bit.

Suppose $f:[a,b] \rightarrow [a,b]$ continuously. Show f has a fixed point.

Thoughts so far?

Either $a \leq 0$ and $b \geq 0$ (So $0 \in \lbrack a, b \rbrack$) or $\exists$ $g: \lbrack a,b \rbrack + c \rightarrow \lbrack a,b \rbrack + c$ so $g: \lbrack a + c, b + c \rbrack \rightarrow \lbrack a + c, b + c \rbrack \ni a + c \leq 0, b + c \geq 0$. Then $f(0) = 0$ or $g(0) = 0$. Since $g(x) = f(x) + c$ By the Intermediate Value Theorem, we can WLOG suppose that there is a function, either $f$ itself or a related function, $g$ with the fixed point $0$. Or something.

Mark gave this to us early as the Brouwer FPT, which we could research on the google, but also we can think of what it means to to be a fixed point graphically which would be when our given function intersects with the line $y=x$,

We could then do some iterations by starting with an arbitrary $x_0$, then finding $f(x_0)$ and then setting $x_1=f(x_0)$ and we can continue this until we find our fixed point. Doing this could give an image like,

Sorry this isn't proof-y but it can help to give an understanding to the problem

@kmaclean Unfortunately, this is just a special case - the full Brower Fixed Point theorem is about a billion times harder to prove.

@jmacdona Hmm... Try $g(x)=f(x)-x$ and forget the cases.

@mark so something like let $f: A \rightarrow \mathbb{R}$, and suppose $f$ is continuous on $A$. Let $g(x) = f(x) - x$. Since $f(x)$ is given to be continuous and $x$ is uniformly continuous on $\mathbb{R}$ it follows that $g(x)$ is also continuous on $A$. Then, by the I.V.T $\exists$ $u \in \mathbb{R}$ satisfying $f(a) < u < f(b)$ and $c \in A$ s.t. $f(c) = u$. It also follows that $c - g(a) > x > c - g(b)$ so $c - f(a) + x > x > c - f(b) + x$ Hence $c - f(a) > 0 > c - f(b)$. Not sure where to go from here.

@jmacdona, Since for all $x$ in $[a,b]$ we have $a < f(x) < b $ and we let $g(x) = f(x) - x$, more specifically we can say:
$$g(a) = f(a) -a \geq a-a = 0$$ and
$$g(b) = f(b) - b \leq b-b = 0$$
So by the IVT, there exists some $c \in [a,b]$ such that $g(c) = 0$.
Since $g(c) = f(c) - c = 0$, then $f(c) = c$.
Is this sound?

@cseagrav I really like what you did but technically you're supposed to have $a < f(x) < b $ for the IVT. You can just note if $a=b$ then clearly $f : [a,a] \rightarrow [a,a]$ has a fixed point, i.e. $f(a)=a$. Then we have strict inequalities with $g(a)>0$ and $g(b)<0$ which implies $\exists c\in (a,b)$ s.t. $g(c)=0$. I know its subtle but the IVT is usually worded so the $c$ exists on an open interval. You can see people fight over this on StackExchange here.

Thanks @jmincey, no arguments here. I edited my post.....