If the mean of the commutator is zero, then what we can say about the commutator itself?

Question

Suppose we have

\begin{equation} \langle[H,N]\rangle=0 \tag{1} \end{equation} where both $H$ and $N$ are hermitian. Under which assumption I can claim that then $$[H,N]=0~?\tag{2}$$

Answer 1

Let us see what this means: assuming that the commutator is properly defined, $\langle \psi,[H,N]\psi\rangle =0$ simply states that the image (range/codomain) of the commutator is orthogonal to its domain. If the domain of the commutator is dense everywhere in a complex separable Hilbert space, then the only vector orthogonal to every vector in the domain is the 0 vector. The only operator which sends an arbitrary vector into the 0 vector is the 0 operator. Thus a necessary condition to have [H,N] =0 is that the domain of the commutator be dense everywhere. In particular, the two operators could be bounded.

EDIT:

Statement: "If the domain of the commutator is dense everywhere in a complex separable Hilbert space, then its orthogonal complement is the 0 vector".

PROOF: The statement reads. "If $\bar{D_A} = \mathcal{H}$, then $ D_A ^{\perp} = \{0\}$". This follows trivially form the weak continuity of the scalar product. We prove that $D_A^\perp \perp \bar{D_A}$. Let $(s_n)\in D_A$ a sequence of vectors in $D_A$ which converges to $s\in\bar{D_A}$. Then $\forall x\in D_A^\perp, \langle x,s\rangle = \langle x,\lim_n s_n\rangle = \lim_n \langle x,s_n\rangle =0$. Therefore we have shown that $D_A^\perp \perp \mathcal{H}\Rightarrow D_A^\perp =\{0\} $.