GeorgeRP wrote: The function $f: A \rightarrow A$ is such that $f(x) \leq x^2 \mbox{ and } f(x+y) \leq f(x) + f(y) + 2xy$ for any $x, y \in A$. a) If $A = \mathbb{R}$, find all functions satisfying the conditions. b) If $A = \mathbb{R}^{-}$, prove that there are infinitely many functions satisfying the conditions. (With $\mathbb{R}^{-}$ we denote the set of negative real numbers.) a) Let $P(x,y)$ be the assertion $f(x+y) \leq f(x) + f(y) + 2xy$. $P(0,0)$ implies that $f(0)\leq 2f(0)$, then $f(0)\geq 0$. Combined with $f(0)\leq 0^2=0$, we have $f(0)=0$. Hence from $P(x,-x)$ we get $0=f(x-x)\leq f(x)+f(-x)-2x^2\leq x^2+(-x)^2-2x^2=0$. The equality must hold, which means $\boxed{f(x)\equiv 0,x\in\mathbb{R}}$. b) In fact, let $f(x)=kx, x<0$, where $k$ is a positive constant. This satisfies the condition: Firstly, $f(x)=kx<0<x^2$. Besides, $f(x+y)=k(x+y)<kx+ky+2xy=f(x)+f(y)+2xy$ stands true because $x,y<0, 2xy>0$. The problem is proved. $\square$