$f$ is twice differentiable and bijective. It is given $f(a)=f(f(b))=1/3, f(1/2)=1/2$, $f''(x)>0$ for $x\in[0,0.5)$ and $f''(x)<0$ for $x\in(0.5,1]$.
Prove that $b>a$ implies $f$ is increasing.
From the given information I figured out that since $f$ is bijective, $a=f(b)$ Also, $f'(x)$ is increasing in $[0,0.5)$ and decreasing in $(0.5,1]$ but I can't figure out how to conclude whether $f$ is increasing or decreasing.
I suppose that the domain of $f$ is $I=[0,1]$. As $f$ is continuous and bijective, $f$ is strictly monotonic. If we suppose that $f$ is strictly decreasing, as $f(1/2)=1/2$, one has $f([0,1/2])=[1/2,1]$ and $f([1/2,1])=[0,1/2]$. As $f(a)=1/3$, we have $a\in [1/2,1]$. As $f(b)=a$, we have $b\in [0,1/2]$, hence $b\leq a$, a contradiction. Problem: I have not used all the hypothesis. Is there any problem in this proof ? or is the domain not correct ?