By (1), and take $x = y = 0$, we have $f(0) = 0$.

Now, suppose there exists $x_0 > 0$ such that $f(x_0) > x_0^2$. By using (1), we have
$$f\left(\frac{x_0}{2}\right) > \frac{1}{2}x_0^2.$$
In fact, by mathematical induction, and use (1) repeatedly, it is easy to show that
$$f\left(\frac{x_0}{2^k}\right) > 2^{2^k - 2^{k-1}} x_0^2$$
for all positive integers $k$. However, since $x_0$ is a constant, we can choose $k$ sufficiently large such that $x_0/2^k \in [0, 1]$ and $f(x_0/2^k) > 2016$, which contradicts (2).