Lemma 6

Let $g_1$ and $g_2$ be $\mu$-measurable functions on X.

a) The partial operator $m_{g_1}+m_{g_2}$ is closable and its closure is the multiplication operator $m_{g_1+g_2}$;

b) One has $m_{g_1}\circ m_{g_2}\subset m_{g_1g_2}$;

c) Suppose that $g_1$ is bounded. Then one has $m_{g_2}\circ m_{g_1}=m_{g_1g_2}$. Moreover, the domain of $m_{g_2}$ is contained in the domain of $m_{g_1g_2}$, and $m_{g_1}\circ m_{g_2}$ is the restriction of $m_{g_1g_2}$ to dom($m_{g_2}$).

It is elementary that $m_{g_1+g_2}$ is an extension of $m_{g_1}+m_{g_2}$; the latter operator is therefore closable, and $\overline{m_{g_1} + m_{g_2}}\subset m_{g_1+g_2}$.

Let $f\in \mathscr{L}^2(X, \mu)$ be such that the function $h= (g_1+g_2)f$ belongs to $\mathscr{L}^2(X, \mu)$. For every integer $n\geqslant 1$, let $X_n$ denote the set of $x\in X$ such that $|g_1(x)|+|g_2(x)|\leqslant n$, and let $\varphi_n$ be the characteristic function of $X_n$. We have $\varphi_nf\in$ dom($m_{g_1}+m_{g_2}$). Since $(\varphi_nf)(x)$ tends to $f(x)$ for every $x\in X$, and since for every $n\in \mathbf{N}$ one has $|\varphi_nf|\leqslant |f|$ and $|(g_1+g_2)\varphi_nf|\leqslant$ $|(g_1+g_2)f|=|h|$, with $h\in \mathscr{L}^2(X, \mu)$ by hypothesis, Lebesgue's theorem (INT, IV, p. 137, § 3, No.$^o7$, Theorem 6) implies that the sequence of pairs of classes in $L^2(X, \mu)\times L^2(X, \mu)$ of $(\varphi_nf,(g_1+g_2)\varphi_nf)$, which belong to the graph of $m_{g_1}+m_{g_2}$, tends to the pair of classes of $(f, h)$ in $L^2(X, \mu)$. The closure of $m_{g_1}+m_{g_2}$ is therefore indeed equal to $m_{g_1+g_2}$.

It is elementary that $m_{g_1}\circ m_{g_2}\subset m_{g_1g_2}$. Suppose that $g_1$ is bounded, so that $m_{g_1}$ is a continuous linear mapping on $L^2(X, \mu)$. Then the domain of $m_{g_2}\circ m_{g_1}$ is the set of classes of functions $f\in \mathscr{L}^2(X, \mu)$ such that $g_2(g_1f)\in \mathscr{L}^2(X, \mu)$, that is to say the domain of $m_{g_1g_2}$. Hence $m_{g_2}\circ m_{g_1}=m_{g_1g_2}$.

It is likewise elementary that dom($m_{g_1}\circ m_{g_2}$) $=$ dom($m_{g_2}$) is contained in dom($m_{g_1g_2}$), and that the restriction of $m_{g_1g_2}$ to this space is equal to $m_{g_1}\circ m_{g_2}$.

One must beware of believing that $m_{g_1}\circ m_{g_2}=m_{g_1g_2}$ in general (exercise 10 of IV, p. 347). Nevertheless, one has the following partial result: