If I know $\;T\;$ is a self-adjoint operator on a Hilbert space $\;\mathcal H\;$ and it also holds $\; \vert \vert (z-T)x \vert \vert \ge \vert Im(z) \vert\;\vert \vert x \vert \vert\;\;\forall x \in \mathcal H\;$ where $\;z\in \mathbb C\;$, can I conclude $\; (z-T)\;$ is a bounded operator?
I'm really confused here. Thanks in advance!!!
If $T : \mathcal{H}\rightarrow\mathcal{H}$ is a symmetric operator, then it is bounded. This is because the graph of $T$ is closed, which you can show by assuming $\{ x_n \}$ converges to $x$ and $\{ Tx_n \}$ converges to $y$, and showing that $Tx=y$. This follows because $$ (Tx_n,z) = (x_n,Tz), \;\;\; z\in\mathcal{H} \\ (y,z) = \lim_n (Tx_n,z) = \lim_n (x_n,Tz) = (x,Tz) \\ (y,z) = (x,Tz) = (Tx,z), \;\;\; z\in\mathcal{H}. $$ A bounded symmetric operator on a Hilbert space is selfadjoint.
The inequality you gave implies that $T-\lambda I$ is injective for $\lambda\notin\mathbb{R}$, and it implies that the inverse $(T-\lambda I)^{-1} : \mathcal{R}(T-\lambda I)\rightarrow\mathcal{H}$ is bounded for $\lambda\notin\mathbb{R}$.