Let $T:H\to H$ a densely defined operator, with $H$ a Hilbert space such that:
$$Re(x,Tx)\geq 0, \forall x\in Dom(T) $$
I want to prove that $T$ is a closable operator, that means… that there exists a closed extension $S:H\to H$ where, $Dom(T)\subset Dom(S)$, alternatively $T$ is closed if its graph $\Gamma(T)$ is closed in the direct sum $H⊕H$.
First of all, If I prove that $T$ is bounded (equivalent to be continuous)I would finish because every linear continuous opearator is closable, neverless I don't know if $Re(x,Tx)\geq 0$ let me what I want. All I can imagine with that hypothesis is that we're working in a half-plane, and see what happens with the adjoint of $T%$ if it's self-adjoint i.e. if $(x,Tx)=(Tx,x) \forall x\in H$, because of Hellinger-Toeplitz theorem tell me that every symmetric operator is bounded, but symmetric implies self-adjointness…
Any idea to undertand what $Re(x,Tx)\geq 0$ is telling me would be appreciated.
Best Answer
First of all, the condition you state above does NOT imply that $T$ is bounded. Moreover, Heilinger Toeplitz can only be applied if $T$ is defined on all of $H$, i.e. $\mathcal{D}(T)=H$, which is not one of your assumptions.
However, your assumption does indeed imply that $T$ is a (possibly unbounded) densely defined closable operator. To see this, your assumption implies that the numerical range $W(T)$ of $T$, defined as $$ W(T):=\{(Tx, x) \in \mathbb{C} \mid x \in \mathcal{D}(T),\,\|x\|=1\}, $$ (I use the convention that the inner product is linear in the first argument) is contained in the closed right half-plane. That means, in particular, $W(T)\neq \mathbb{C}$. All densely defined operators satisfying this must be closable. This is a classical result which can be seen for instance in Chapter V Thm. 3.4. of Kato, T. Perturbation theory for linear operators.
By the way, the condition you state above is called accretive.