Proposition 2

Let $K =\mathbf{R}$ and let E and F be topological vector spaces over $\mathbf{R}$. Let $u$ be a partial operator of E into F. Then $u$ has dense domain (resp. is closed, is closable) if and only if $u_{(\mathbf{C})}$ has dense domain (resp. is closed, is closable).

Suppose that the domain of $u$ is dense in E. Every neighbourhood of 0 in $E_{(\mathbf{C})}$ contains a neighbourhood of the form $V +iV$ (TVS, II, p. 65), where V is a neighbourhood of 0 in E, and therefore contains an element of the domain of $u_{(\mathbf{C})}$; this partial operator therefore has dense domain. The converse is also true since the mapping of $E_{(\mathbf{C})}$ into E which associates $x$ to $x+iy$ for every $(x, y)\in E\times E$ is continuous and surjective.

The graph of $u_{(\mathbf{C})}$ identifies with the complexified topological vector space of the graph of $u$. It is therefore closed in $E_{(\mathbf{C})}\times E_{(\mathbf{C})}$ if the graph of $u$ is closed in $E\times E$. Conversely, one has $\Gamma_u= \Gamma_{u_{(\mathbf{C})}}\cap (E\times E)$ in $E_{(\mathbf{C})}\times E_{(\mathbf{C})}$; since $E\times E$ is closed in $E_{(\mathbf{C})}\times E_{(\mathbf{C})}$, the partial operator $u$ is closed when $u_{(\mathbf{C})}$ is.

A partial operator $v$ from E into F is an extension of $u$ if and only if $v_{(\mathbf{C})}$ is an extension of $u_{(\mathbf{C})}$, hence the partial operator $u_{(\mathbf{C})}$ is closable if $u$ is closable. Conversely, if $u_{(\mathbf{C})}$ is closable, the partial operator $u$ is also, since $\Gamma_u= \Gamma_{u_{(\mathbf{C})}}\cap (E\times E)$, which is then a functional graph (prop. 1).