Multiplication operator is densely defined.

Question

I saw the question Multiplication Operator on $L^2$ is densely defined

Disintegration by parts answers using the following argument.

If $f \perp \mathcal{D} (L)$ then $\frac{1}{m^2+1} f \in \mathcal{D} (L)$.

I can't understand why this is true, $f \perp \mathcal{D} (L)$ means $\langle f,g \rangle_{L^2} =0$ for all $g \in \mathcal{D}(L)$. Can you explain why does it follow that $\frac{1}{m^2+1} f \in \mathcal{D} (L)$.

Answer 1

$\frac 1{m^{2}+1}f \in \mathcal D(L)$ because $m (\frac 1{m^{2}+1}f) \in L^{2}$: note that $f \in L^{2}$ and $|\frac m{m^{2}+1}| \leq 1/2$. Since $f \perp \mathcal{D} (L)$ it follows that $ \int \overline f \frac 1{m^{2}+1}f =0$ which implies $f=0$ a.e. Hence, $\mathcal D(L)$ is dense.