Proposition 14

Let $q$ be a closed positive partial form with dense domain on E and $u$ the partial operator representing $q$.

a) The domain of $u$ is dense in the Hilbert space $E_q$;

b) The partial operator $u$ is self-adjoint and positive.

Since $q$ is closed, the prehilbertian space $E_q$ of Definition 9 is a Hilbert space.

Let us prove that the partial operator $u+ 1_E$ with domain dom($u$) is bijective. Since

for every $x\in$ dom($u$), this partial operator is injective.

Let $y\in E$. The linear form on $E_q$ defined by $x\mapsto \langle y|x\rangle$ is continuous since $\|x\|\leqslant \|x\|_q$. There therefore exists $y'\in E_q$ such that

for every $x\in E_q$ (EVT, V, p. 15, Theorem 3). By definition, this means that $y'$ belongs to the domain of the partial operator $\widetilde{u}$ representing the partial form $(x, y)\mapsto (x|y)_q$ with domain dom($q$), and that $\widetilde{u}(y') =y$. Since $\widetilde{u}=u+ 1_E$ by the above remark, it follows that the partial operator $u+ 1_E$ is also surjective, hence bijective.

Let us prove that the domain of $u$ is dense in $E_q$. Let $y\in E_q$ be orthogonal to dom($u$) in $E_q$. There exists $y'\in$ dom($u$) such that $u(y')+y'=y$. We then have

therefore $y= 0$. Assertion a) is therefore proved.

Since dom($q$) is dense in E and $\|x\|\leqslant \|x\|_q$ for every $x\in$ dom($q$), assertion a) implies that dom($u$) is dense in E. Since the form $q$ is Hermitian (resp. positive), for all $x$ and $y$ in dom($u$), we have

(resp. $\langle x|u(x)\rangle =q(x, x)\geqslant 0$), so that $u$ is symmetric (resp. positive). Finally, the partial operator $u+ 1_E$ is self-adjoint by the corollary to proposition 10 of IV, p. 240, and the same is true of $u$.