Proposition 13

Tag 033M, permanent. Cite it as [Bourbaki, Tag 033M].

Théories spectrales, chapter IV, § 4. Opérateurs partiels

Let $u$ be a partial operator on $L^2(U)$ . If $D_-\subset u\subset D_+$ , then $u$ is closable and $(^tD)_-\subset u^*\subset (^tD)_+$ .

Let $\varphi$ and $\psi$ belong to $\mathscr{D}(U)$ . Then one has

\langle \varphi |D(\psi )\rangle =\sum_{|\alpha|\leqslant}\int_{kU}n_{\alpha}\varphi \partial^{\alpha}\psi d\mu

=\sum_{|\alpha|\leqslant k}(-1)^{|\alpha|}\langle \partial^{\alpha}(\overline{n}_{\alpha}\varphi )|\psi \rangle =\langle^tD(\varphi )|\psi \rangle

(cf. VAR, R2, 14.3.8). Since $\mathscr{D}(U)$ is dense in $L^2(U)$ (prop. 4 of IV, p. 202), this implies that $\varphi \in$ dom( $u^*$ ) and $u^*(\varphi ) =^tD(\varphi )$ . Hence one has $(^tD)_-\subset u^*$ ; in particular, $u^*$ has dense domain and $u$ is closable (prop. 8 of IV, p. 237).

Let $f\in$ dom( $u^*$ ) and $\varphi \in \mathscr{D}(U)$ . Since $\mathscr{D}(U)\subset$ dom( $u$ ), the distribution associated with $u^*(f)$ satisfies

\langle u^*(f), \varphi \rangle =\langle \overline{\varphi}|u^*(f)\rangle =\langle u(\overline{\varphi})|f\rangle

Since $D_-\subset u$ , $u(\overline{\varphi})$ is calculated by formula (5), whence

\langle u^*(f), \varphi \rangle =\sum_{\alpha}\langle n_{\alpha}\partial^{\alpha}\overline{\varphi}|f\rangle =\sum_{\alpha}\langle \partial^{\alpha}\overline{\varphi}|\overline{n}_{\alpha}f\rangle

=\sum_{\alpha}\langle \overline{n}_{\alpha}f, \partial^{\alpha}\varphi \rangle =\sum_{\alpha}(-1)^{|\alpha|}\langle \partial^{\alpha}(\overline{n}_{\alpha}f), \varphi \rangle =\langle^tD(f), \varphi \rangle

The distributions $u^*(f)$ and $^tD(f)$ are therefore equal; the distribution $f$ therefore belongs to $H_{_tD}$ and $u^*(f) =^tD(f)$ , whence $u^*\subset (^tD)_+$ .

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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