'F' for Functions!

Assuming $\alpha$ is a well-defined extended real number (i.e. real or $\pm \infty$ )

Since $f(x) = \frac{1}{2} x$ satisfies $f(3x) = f(f(2x)) + x$ , we see that $\alpha \leq \frac{1}{2}$ . Also, we note that $f(x) \geq 0 = 0\cdot x$ for all $f\in F$ , so $\alpha \geq 0.$

Let $\beta > \alpha.$ Then there is some $f \in F$ and some $y \in \mathbb{R}^{+}$ such that $f(y) < \beta \cdot y.$ If we set $y = 3x,$ then $\begin{aligned} 3\beta \cdot x &= \beta \cdot (3x) \\ &> f(3x) \\ &\geq f(f(2x)) + x \\ &\geq \alpha\cdot f(2x) + x \\ &\geq \alpha^2 \cdot (2x) + x \\ &= (2\alpha^2 + 1)\cdot x. \end{aligned}$ By dividing by $x > 0$ , $3\beta > 2\alpha^2 + 1,$ and then letting $\beta \to \alpha^{+}$ , $3\alpha \geq 2\alpha^2 + 1 \implies \frac{1}{2} \leq \alpha \leq 1$

To finish, we note that the only solution to $\alpha \leq \frac{1}{2} \leq \alpha \leq 1$ is $\alpha = \boxed{\frac{1}{2}}$

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Remark : As we noted above, the proof follows from the assumption $\alpha$ is well-defined, but we remove this assumption by instead considering $\alpha^{*} = \sup \{a \geq 0 : f(x) \geq a\cdot x, \forall f\in F\}$ Then the proof I gave above implies $\alpha^{*} = \frac{1}{2}.$ Therefore the only thing we'd need to show to imply $\alpha$ is well-defined is that $f(x) \geq \alpha^{*}\cdot x, \forall f \in F,$ as then $\alpha$ is well-defined to be $\alpha^{*}$ . To show this, suppose otherwise. Then there is some $f_0 \in F, x_0\in \mathbb{R}^+$ such that $f_0(x_0) < \alpha^{*} \cdot x_0$ However, by the definition of $\alpha^{*}$ as the supremum, there must be some $a \in \left(\frac{f_0(x_0)}{x_0}, \alpha^{*}\right]$ such that $f_0(x_0) \geq a \cdot x_0,$ and by putting this together, we have $\frac{f_0(x_0)}{x_0} < a \leq \frac{f_0(x_0)}{x_0}$ which is a contradiction. It then follows that $\alpha = \alpha^{*}$ is well-defined.