Remark

Thus $y\in D^*$ if and only if there exists $c\in \mathbf{R}_+$ such that $|\langle y|u(x)\rangle |\leqslant c\|x\|$ for every $x\in D$. The element $u^*(y)$ is then characterised by the relation

for every $x\in D$. We then have $|\langle y|u(x)\rangle |\leqslant \|u^*(y)\| \|x\|$ for every $x\in D$.

In the case where $u$ is a continuous linear mapping from E into F, its adjoint in the sense of the preceding definition coincides with the adjoint defined in EVT, V, p. 38, Definition 1, since $D^*$ is equal to F in this case.

Let $v\in \mathscr{P}(E; F)$ be such that $u\subset v$. Then $v^*\subset u^*$.

Let $v\in \mathscr{P}(E; F)$ be such that $u+v$ has dense domain. Then the partial operator $v$ has dense domain and we have $u^*+v^*\subset (u+v)^*$. In general, there is no equality (exercise 9 of IV, p. 347). If $v\in \mathscr{L}(E; F)$, then $u+v$ has dense domain and $(u+v)^*=u^*+v^*$. This is so, for example, if F = E and if $v=\lambda 1_E$ where $\lambda \in K$.

Let G be a Hilbert space over K and let $v\in \mathscr{P}(F; G)$ be an operator with dense domain. If $v\circ u$ has dense domain, then $u^*\circ v^*\subset (v\circ u)^*$. In general, there is no equality (loc. cit.). If $u$ (resp. $v$) is an isomorphism, then $v\circ u$ has dense domain and we have $(v\circ u)^*=u^*\circ v^*$. This is so, for example, if E = F (resp. F = G) and $u=\lambda 1_E$ (resp. $v=\lambda 1_F$) where $\lambda \in K^*$.

Let $s$ denote the isometric isomorphism of Hilbert spaces from $E\oplus F$ into $F\oplus E$ defined by $s(x, y) = (-y, x)$ for every $(x, y)\in E\oplus F$.