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

Dag_Nilsen
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1 Answers1

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

Kavi Rama Murthy
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  • I see, if you allow me I have one more question. I know that for $X =\mathbb{R}$ $C_c^{\infty }(\mathbb{R}$) is a subspace of $\mathcal{D} (L)$, but $C_c^{\infty }(\mathbb{R}) $ is dense in $L^2(\mathbb{R})$, does this prove that $\mathcal{D} (L)$ is dense? – Dag_Nilsen May 19 '22 at 08:53
  • Yes, it does prove that $\mathcal D(L)$ is dense in $L^{2}(\mathbb R)$. @Dag_Nilsen – Kavi Rama Murthy May 19 '22 at 08:56
  • But how did you prove that $C_c^{\infty} $ is dense in $\mathcal D(L)$? You need some assumption on $m$, not just measurability. – Kavi Rama Murthy May 19 '22 at 09:02
  • You are right, I was thinking of a special case for $m$ where this is true. – Dag_Nilsen May 19 '22 at 09:22