Lemma 1

Let E, F and G be topological vector spaces over K. Let $u$ be a closed operator from E into F.

a) For every $v\in \mathscr{L}(E; F)$, the partial operator $u+v$ is closed.

b) For every $v\in \mathscr{L}(G; E)$, the partial operator $u\circ v$ is closed.

Let us prove a). Let $\gamma$ be the mapping $(x, y)\mapsto (x, y-v(x))$ of $E\times F$ into itself; it is continuous. For every $(x, y)\in E\times F$, one has $\gamma (x, y)\in \Gamma_u$ if and only if $x\in$ dom($u$) and $y=u(x) +v(x)$, that is to say that $\overset{-1}{\gamma}(\Gamma_u) = \Gamma_{u+v}$. The assertion follows.

Let us prove b). The mapping $\eta = (v,1_F)$ of $G\times F$ into $E\times F$ is continuous; for every $(z, y)\in G\times F$, one has $\eta (z, y) = (v(z), y)$, hence $\overset{-1}{\eta}(\Gamma_u) = \Gamma_{u\circ v}$. The assertion follows.

Let E and F be topological vector spaces over K, F being separated. Let $a\in K$. If $u\in \mathscr{P}(E; F)$ is closable, the same is true of $au$. If $a\not = 0$, one has $\overline{au}=au$, and $u$ is closed if and only if $au$ is. If $a= 0$, the closure of $au$ is $0_{\overline{dom(u)}}$, and $au$ is equal to $0_{dom(\overline{u})}$; it may therefore happen that $u$ is closed but that $au$ is not.