Indija TS 2012

Put $x=y=t$ ($t>0$). We get $$4f(t)\le f(2t),f(2t)\ge 4f(t),$$ 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 $$\mathrm{tg}(t)+2\mathrm{tg}(2t)\le (3t/2)g(3t).$$ 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 $f(x)=a{x}^2$, where $a\ge 0$. It is easy to verify that this gives indeed a solution to our equations.