, 2012

Put $x = y = t$ ($t > 0$). We get
$$4f(t) \le f(2t), \quad f(2t) \ge 4f(t),$$
for all $t > 0$. Hence $f(2t) = 4f(t)$, for all $t > 0$. By induction
$$f(2^m t) = 2^{2m} f(t), \text{ for all } t > 0.$$
Let $g(x) = f(x)/x$, $x > 0$, $g(0) = 0$. We show that $g(nt) = ng(t)$ for all $n \in \mathbb{N}$ and $t \ge 0$. We have proved this for $n = 2^m$. We observe that $g(x+y) \le g(x) + g(y)$ for all $x \ge 0$, $y \ge 0$. Hence by induction $g(nt) \le ng(t)$ for all $n \in \mathbb{N}$ and $t \ge 0$. Choose $m$ such that $2^{m-1} \le n < 2^m$. Then
$$2^m g(t) = g(2^m t) \le g(nt) + g((2^m - n)t) \le ng(t) + (2^m - n)g(t) = 2^m g(t).$$
Hence equality holds and $g(nt) = ng(t)$.

Next we show that $g$ is a non-increasing function. We have
$$f(t) + f(2t) \le \frac{f(t+2t)}{2} = \frac{f(3t)}{2}.$$
Hence
$$\operatorname{tg}(t) + 2\operatorname{tg}(2t) \le (3t/2)\operatorname{g}(3t).$$
This gives
$$g(t) + 4g(t) \le (9/2)g(t).$$
Hence $g(t) \le 0$, for all $t \ge 0$. Thus for any $0 < x \le y$, we have
$$g(x) \ge g(y) + g(x - y) \ge g(y).$$
Hence $g$ is a non-increasing function.

Let $g(1) = a \le 0$. We show that $g(t) = at$ for all $t > 0$. Suppose $g(t) < at$ for some $t > 0$. Choose positive rational $p/q$ such that $(p/q) > \text{and } g(t) < a(p/q)$. But $g(nt) = ng(t)$ for all $n \in \mathbb{N}$. Hence
$$g\left(\frac{p}{q}\right) = \frac{p}{q}g(1) = \frac{pa}{q}.$$
Hence
$$g(t) \ge g\left(\frac{p}{q}\right) = \frac{pa}{q},$$
a contradiction to $g(t) < a(p/q)$. Similarly we can show that $g(t) > at$ is also not possible. We conclude that $g(t) = at$ for all $t > 0$. This gives $f(x) = ax^2$, where $a \ge 0$. It is easy to verify that this gives indeed a solution to our equations.