For continuous $f,g: [0,1] \to [0,1]$ with $f \circ g = g \circ f$ , there exists $x$ such that $f(x)=g(x)$

Question

I have been stuck on this for hours now:

Let $f,g: [0,1] \to [0,1]$ be continuous such that $f \circ = g \circ f$. Show that there exists $x \in [0,1]$ such that $f(x)=g(x)$

My attempt: It is easy to proof that $f,g$ have both a fix point in the interval $[0,1]$. That means there exist $a,b \in [0,1]$ such that $f(a)=a$ and $g(b)=b$. Now we also know that $$f(g(x))=g(f(x)), \text{ for all } x $$ So I can make use of that and say for example that: $$f(g(b))=g(f(b))=f(b) $$ Which shows that $f(b)$ is yet another fix point of $g$. Similarly, by the same argument I'd get: $$f(g(a))=g(f(a))=g(a) $$ and therefore $g(a)$ is yet another fix point of $f$. While this seems great and all but I am very unsure if my next steps are correct:

Define $h: [0,1] \to [0,1]$ such that $h(x)=g(x)-f(x)$. Of course $h$ is continuous, because $f,g$ are. Then I'd obtain: $$h(a)=g(a)-f(a)=g(a)-a \geq 0 \\h(b)=g(b)-f(b)=b-g(f(b))... $$ Where I am not quite sure if those two inequalities are right at all or just misleading me.

Answer 1

Your idea is fine. A priori, $h(x)=g(x)-f(x)$ can take on any value $\in[-1,1]$. But if we assume the claim of the problem statement is wrong, it never is $0$, hence by the IVT is either alsways $>0$ or always $<0$. Wlog. $h(x)>0$ for all $x$. Especially, whenever $a$ is a fixed point of $f$, we have $g(a)>f(a)=a$ as you showed. As you also showed, $g$ maps fixed point sof $f$ to fixed points of $f$. That is $a_0=a$, $a_{n+1}=g(a_n)$ givs us a strictly increasing sequence of fixed points of $f$. As the sequence is bounded, it must convereg to some $\tilde a\in[0,1]$. Then by continuity, $g(\tilde a)=\tilde a$ and $f(\tilde a)=\tilde a$, hence $h(\tilde a)=0$, contradiction.