Nice one, but a bit easy for TST (unless my proof is wrong). Putting $x=0$ in the $1$st relation, we conclude that $f(0)^2\le 0\implies f(0)=0$, so $f(x)\le x^2$ trivially holds for $x=0$. Now suppose there exist some $x>0$ such that $f(x)>x^2$. We seek to obtain a contradiction from that. Define the sequence $\{u_n\}_{n=0}^{\infty}$ as $u_n=2^n-2n-1$.Then it is very easy to verify that $u_{n+1}=2u_n+2n-1$ holds for all nonnegative integers $n$. We claim that $f\left(\frac{x}{2^n}\right)\ge x^2\times 2^{u_n}$ holds for all $n\in\mathbb{N}_0$. We shall establish our claim by induction. The base case $n=0$ is just our assumption. Now suppose our claim holds for some $n$. Then putting $\frac{x}{2^n}$ for $x,y$ in the first relation, we get $$f\left(\frac{x}{2^n}\right) ^2\le 2\cdot\left(\frac{x}{2^n}\right) ^2 f\left(\frac{x}{2^{n+1}}\right)\implies f\left(\frac{x}{2^{n+1}}\right)\ge \frac{f\left(\frac{x}{2^n}\right) ^2}{\frac{x^2}{2^{2n-1}}}.$$By the inductive assumption, $f\left(\frac{x}{2^n}\right)\ge x^2\times 2^{u_n}$, so the above inequality gives $$f\left(\frac{x}{2^{n+1}}\right)\ge \frac{\left(x^2\times 2^{u_n}\right)^2}{\frac{x^2}{2^{2n-1}}}=x^2\times 2^{2u_n+2n-1}=x^2\times 2^{u_{n+1}}.$$This completes our induction step. Now clearly for large enough $n$, $\frac{x}{2^n}$ becomes smaller than $1$, and $x^2\times 2^{u_n}=x^2\times 2^{ 2^n-2n-1}$ becomes larger than $2016$. This yields a contradiction to the $2$nd condition, as desired. $\blacksquare$

Ankoganit wrote: Nice one, but a bit easy for TST (unless my proof is wrong). $\blacksquare$ It's just a practice quiz, that's why it's not too difficult.

By the way, can anyone give an example of such a function?

Ankoganit wrote: By the way, can anyone give an example of such a function? $f \equiv 0$ I think $f(x) = 0.001x^2$ also.

v_Enhance wrote: I think $f(x) = 0.001x^2$ also. Thank you. Following your example, I find that all functions of the form $f(x)=cx^2$ work, where $c\in \left[ 0,\frac 12\right]$. In fact we can prove a stronger result than the OP: Suppose that the function $f:[0,\infty)\to[0,\infty)$ satisfies: (1)$\forall x,y \geq 0,$ we have $f(x)f(y)\leq y^2f\left(\frac{x}{2}\right)+x^2f\left(\frac{y}{2}\right)$; (2)$f(x)$ is bounded from above on $[0,1]$. Prove that $f(x)\leq cx^2$ for all $x\geq 0$, and $c\in \left(\frac 12,\infty\right)$.

Sketch of proof