Let $a=f(\frac{1}{3}), \quad b=f(\frac{1}{3})$ So $\frac{1}{2} \leq \frac{1-b}{b-a} \leq2$ (*) and $\frac{1}{2} \leq \frac{b-a}{a} \leq2$ (**) From here we see that $a, b-a$ and $1-b$ has the same sign. Suppose $a > b$. Then $b-a < 0$, $1-b <0$ and $a<0$ so $1 < b < a < 0$ contradiction So $a < b$ and from (**) we get $3a \geq b \geq \frac{3}{2}a$ Using (*) we have $2 \geq 3b-a$ and $1 \leq 3b-2a$ So $2 \geq 3b-a \geq 3 \cdot \frac{3}{2}a -a = \frac{7a}{2} \Rightarrow a \leq \frac{4}{7}$ and $1 \leq 3b- 2a \leq 3 \cdot 3a - 2a = 7a \Rightarrow a \geq \frac{1}{7}$ Comment: I don't know if this bounds can actually be reached by some function $f$; I don't know any nontrivial function that verify hypothesis, except $f(x) = x$; In general, we can prove that $f(r) \in [0,1] \quad \forall r \in \Bbb{Q}$ and $f(r) < f(r') \quad \forall r < r' \in \Bbb{Q}$ For rationals $0 \leq x < z \leq 1$ the hypothesis condition rewrite as $\frac{2f(x) + f(z)}{3} \leq f(\frac{x+z}{2}) \leq \frac{f(x) + 2f(z)}{3} $ Are there nice solutions for a functional equation as $f(\frac{x+z}{2}) = \frac{f(x) + 2f(z)}{3} $?