Consider an unbounded normal operator $A$ with domain $\mathcal{D}(A)$ on a Hilbert space $\mathcal{H}$, i.e. its adjoint $A^*$ satisfies $\mathcal{D}(A^*)=\mathcal{D}(A)$ and, for all $x\in\mathcal{D}(A)$, $\|Ax\|=\|A^*x\|$. We can thus define its self-adjoint and skew-adjoint parts:
\begin{equation}
\mathrm{Re}\,A=\frac{A+A^*}{2},\qquad\mathrm{Im}\,A=\frac{A-A^*}{2\mathrm{i}},
\end{equation}
both being self-adjoint operators with domain $\mathcal{D}(A)$.
Now, suppose that one of them (say, $\mathrm{Re}\,A$) is a strictly positive operator, thus implying that its spectrum is contained in $[\epsilon,\infty)$ for some $\epsilon>0$ and, consequently, it admits a bounded inverse $(\mathrm{Re}\,A)^{-1}\in\mathcal{B}(\mathcal{H})$. Can we state that $A$ admits a bounded inverse $A^{-1}$ as well? (Equivalently, is $0$ an element of the resolvent set of $A$?)
Best Answer
This should be the answer. Since $A$ is normal, it is closed. A closed operator admits a bounded inverse if and only if there is $c>0$ such that, for all $x,y\in\mathcal{D}(A)$, $$ \|Ax\|\geq c\|x\|,\qquad \|A^*y\|\geq c\|y\|; $$ see e.g. J. Derezinski, Unbounded linear operators, Theorem 3.3.2. Now, if $\mathrm{Re}\,A$ is strictly positive, i.e. $\langle x,\mathrm{Re}\,Ax\rangle\geq c\|x\|^2$, then $$ |\langle x,Ax\rangle|\geq c\|x\|^2,\qquad|\langle x,A^*x\rangle|\geq c\|x\|^2 $$ and therefore both conditions hold, since e.g. $$ \|x\|\|Ax\|\geq|\langle x,Ax\rangle|\geq c\|x\|^2. $$