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.