How do you show that there is a solution to a function that may not be continuous using a similar concept to IVT?
Let $A = \{x\in [0,1]\mid f(x)\geq x\}$. This set is bounded (by $1$), and non-empty (since $0 \in A$), and thus has a least upper bound, say $a = \sup A$. I now disprove the following two statements by contradiction:
Thus we are forced to conclude that $f(a) = a$.
If $f(0) = 0$ or $f(1) = 1$, we are done, so suppose this is not the case.
Then, $f(0) > 0$ and $f(1) < 1$.
Now, consider the function $g(x) = x – f(x)$. Here, $g(0) < 0$ and $g(1) > 0$.
Then, either $g(y) = 0$ for some value of $y \in (0,1)$ or there exists $y \in (0,1)$ such that
$\forall \epsilon > 0, \exists x$ such that $y-\epsilon < x < y$ and $g(x) < 0$ while $g(y) > 0$.
Let $g(y) = \delta > 0$ and pick $\epsilon < \delta$. Then if we pick $x$ with $y-\epsilon < x < y$ and $g(x) < 0$, we have
$f(x) > x > y-\epsilon > y-\delta = f(y)$ which contradicts the fact that $f$ is increasing
Suppose that it isn’t true, so there’s no $x\in[0,1]$ for which $f(x)=x$. Define $A=\{x\in[0,1]\colon f(x)>x\}$ and $B=\{x\in[0,1]\colon f(x)<x\}$. Both these sets aren’t empty ($0\in A$, $1\in B$). Consider $z=\inf B$.
If $z\in A$ then $f(z)>z$. But since $z=\inf B$ then there exists $x\in B$ such that $z<x<z+(f(z)-z)=f(z)$ for which $f(x)<x<f(z)$ which contradicts the fact that $f(x)$ is increasing.
If $z\in B$ then $z>0$, and $f(z)<z$. Similarly, there exists $x\in A$ such that $f(z)<x<z$, for which $f(x)>x>f(z)$ (again, contradiction).
Let $g(x)=f(x)-x$. Then $g(0)=f(0)>0$ (assuming $f(0)\ne 0$ and $f$ is increasing) and $g(1)=f(1)-1<0$ (assuming $f(1)\ne 1$ and $f$ is bounded above by $1$) which suggests that $g(x)=f(x)-x=0$ has at least one root in the interval $(0,1)$$\implies f(x)=x$ is satisfied at least once in $(0,1)$.