A twist in your normal functional equations

Nice problem Anqi! :D

Nice problem ! First note that for all the functions $f \in \mathcal{B}$ , we have the following series of inequalities valid for all $x \in \mathbb{R}_+$ :

$f(3x) \geq f(f(2x)) + x \geq \tau f(2x) +x \geq 2\tau^2 x + x= (1+2\tau^2)x$ ,

i.e., $f(x) \geq \frac{(1+2\tau^2)}{3}x, \forall x \in \mathbb{R}_+$

Hence by definition of $\tau$ , we have $\tau \geq \frac{(1+2\tau^2)}{3}$ , i.e. $\frac{1}{2}\leq \tau \leq 1$ . Now to show that $\tau=\frac{1}{2}$ , we simply note that the function $g(x)=\frac{1}{2}x, \forall x\in \mathbb{R}_+$ belongs to the set $\mathcal{B}$ .