Proposition 5

The multiplication operator $\widetilde{m}_g$ from $L^{p_1}_K(X, \mu)$ into $L^{p_2}_K(X, \mu)$ is a closed operator with dense domain.

Let us first prove that the partial operator $\widetilde{m}_g$ is closed. Let $(f_n, h_n)_{n\in\mathbf{N}}$ be a sequence in $\mathscr{L}_K^{p_1}(X, \mu)\times \mathscr{L}_K^{p_2}(X, \mu)$ such that the sequence $(f_{\widetilde{n}},\widetilde{h}_n)$ of the classes of $f_n$ and $h_n$ belongs to the graph of $\widetilde{m}_g$ and converges in $L^{p_1}_K(X, \mu)\times L^{p_2}_K(X, \mu)$ when $n$ tends to infinity. Let $(f, h)$ be a pair in $\mathscr{L}_K^{p_1}(X, \mu)\times \mathscr{L}_K^{p_2}(X, \mu)$ such that the pair $(\widetilde{f} ,\widetilde{h})$ of their classes is the limit of $(\widetilde{f}_n,\widetilde{h}_n)$.

There exists a sequence $(f_{n_k})_{k\in\mathbf{N}}$ extracted from the sequence $(f_n)_n$ such that $f_{n_k}(x)$ converges to $f(x)$ for $\mu$-almost every $x$ (INT, IV, p. 131, § 3, n$^o4$, th. 3). This implies that $h_{n_k}(x) =g(x)f_{n_k}(x)$ converges $\mu$-almost everywhere to $g(x)f(x)$. Moreover the sequence $(h_{n_k})$ converges to $h$ in the space $\mathscr{L}_K^{p_2}(X, \mu)$. The functions $h$ and $gf$ are therefore equal $\mu$-almost everywhere (loc. cit.). Thus $\widetilde{f}$ belong to the domain of $\widetilde{m}_g$ and $\widetilde{h}=\widetilde{m}_g(\widetilde{f})$. This proves that $\widetilde{m}_g$ is closed.

Let us prove that the domain of $\widetilde{m}_g$ is dense in $L^{p_1}_K(X, \mu)$. It is enough to verify that the classes of functions $f\in \mathscr{K}(X; K)$ belong to the closure of the domain of $\widetilde{m}_g$ in $L^{p_1}_K(X, \mu)$. Let $f\in \mathscr{K}(X; K)$ and let $\widetilde{f}$ be its class in $L^{p_1}_K(X, \mu)$. For every integer $n\in \mathbf{N}$, let $\varphi_n$ denote the characteristic function of the set of elements $x\in X$ such that $|g(x)|\leqslant n$, and put $f_n=f \varphi_n$. Then $|gf_n|\leqslant n|f|$, which belong to $\mathscr{L}_K^{p_2}(X, \mu)$, hence $f_n$ belong to the domain of $\widetilde{m}_g$. For every element $x$ of X, the sequence $(f_n(x))_{n\in\mathbf{N}}$ converges to $f(x)$ when $n\rightarrow +\infty$; moreover, one has $|f_n|\leqslant |f|$, which belong to $\mathscr{L}_K^{p_1}(X, \mu)$. By Lebesgue's theorem (INT, IV, p. 137, § 3, n$^o7$, th. 6), the sequence of classes of $f_n$ converges to $\widetilde{f}$ in $L^{p_1}_K(X, \mu)$. Thus, the class of $f$ belong to the closure of the domain of $\widetilde{m}_g$.

In the following proposition, it is supposed that $p_1=p_2= 2$.