Let $f:[0,1] \to [0,1]$ such that $f$ has lateral limits at any point, is continuous at $0$ and $1$ and $$\lim_{t\nearrow x}f(t) \leq f(x) \leq \lim_{t \searrow x}f(t), \forall x \in (0,1)$$
Prove that $f$ has at least one fixed point.
I built $A=\{x \in [0,1] \mid f(x) \geq x \}$ and $B=\{x\in [0,1] \mid f(x) \leq x \}$ and tried to prove that $A \cap B \neq \emptyset$. Because $0 \in A$ and $1 \in B$, the sets are not empty and I considered $\alpha = \sup A$ and $\beta = \inf B$. Then I supposed that $\alpha \in A$ or $\beta \in B$ (they can be treated the same).
Suppose $\alpha \in A$. If $\alpha=1$, the problem is solved. Otherwise, because $\alpha=\sup A$, we also know that $\forall 0<\epsilon<1-\alpha$, we have $ \alpha+\epsilon \in B$. So $\alpha+\epsilon \geq f(\alpha +\epsilon)$. Now I tried to use the inequality in the problem's statement to get $f(\alpha+\epsilon) \geq f(\alpha) \geq \alpha$ and then make $\epsilon \to 0$ so that $f(\alpha)=\alpha$, but I didn't manage to do these.
If $\alpha \notin A$ and $\beta \notin A$, I don't even know what to do. I'm also confused because of the condition that $f$ is continuous at $0$ and $1$. I don't see how this helps at all.
Best Answer
You already have $f(\alpha+\epsilon)\le \alpha+\epsilon$ for all sufficiently small positive $\epsilon$, so $$f(\alpha)\le \lim_{\epsilon\to 0^+}f(\alpha+\epsilon)\le \lim_{\epsilon\to 0^+}(\alpha+\epsilon)=\alpha $$ and so $\alpha\in B$.
The final case, $\alpha\notin A$, $\beta\notin B$, cannot occur. In fact, we can show $\alpha\in A$ (and likewise $\beta\in B$): If $\alpha\notin A$, then there exists a sequence $x_n\to \alpha^-$ with $x_n\in A$, so $$f(\alpha)\ge\lim_{x\to\alpha^-}f(x)\ge \lim_{n\to\infty}f(x_n)\ge \lim_{n\to\infty}x_n=\alpha.$$
So if you clean up the above arguments, you do not even really need $B$ because we apparently can show $\sup A\ge f(\sup A)\ge\sup A$ without it.