Suppose we have a not necessarily linear operator $A:X\to X'$, where X is a real Hilbert space that can be written as the inner direct sum $X=V \oplus V^{\perp}$. Consider $u\in X$, so $u$ can be written as $u=x+y$, where $x\in V$ and $y\in V^{\perp}$. Furthermore, assume that $A$ is monotone ($\langle Au-Av, u-v\rangle\geq 0$ for all $u,v\in X$), hemicontinuous (for all $u,v\in X$ the mapping $t\mapsto \langle A(u+tv),v\rangle$ is continuous) and bounded by $$\|Av\|_{X'} \leq K(\|v\|_X + 1),$$ where $K>0$.
Given the hypotheses above, is there any way to guarantee that $Au = Ax + Ay$? I don't know if it's possible to establish that, so if anyone can help me with this, I would really appreciate it.
No. Fix a nonzero element $x'\in X'$ and set $Ax=x'$ for all $x$. Then this satisfies all of the conditions, but $A(x+y) \neq Ax+Ay$ for any $x, y$.