Between alternative sources, there appear to be subtle differences in the definitions of the resolvent set for an unbounded operator on a Hilbert space. I am not able to reconcile the differences between two of the more precise definitions that I have found and would appreciate any insights that would help resolve and clarify these.
Suppose $H$ is a Hilbert space and $T: H \supset \mathfrak{D}(T)\rightarrow H$ is an unbounded linear operator. Then, I have the following two definitions of the resolvent set (adapted to the current setting, notation harmonized for consistency, and augmented for my questions)
From A.W. Naylor and G.R. Sell, Linear Operator Theory in Engineering and Science, Springer-Verlag, 1982, pp. 412:
6.5.2 Definition. The set of all [complex numbers] $\lambda$ such that the range of $(\lambda I - T)$ is dense in $H$ and $(\lambda I - T)$ has a continuous inverse defined on its range is said to be the resolvent set $\rho_1(T)$ of $T$.
From B.C. Hall., Quantum Theory for Mathematicians, Springer,2013, pp. 177:
Definition 9.16. A number $\lambda \in \mathbb{C}$ belongs to the resolvent set $\rho_2(T)$ of $T$ it there exists a bounded operator $B$ with the following properties: (1) For all $x \in H$, $B x$ belongs to $\mathfrak{D}(T)$ and $(T- \lambda I) B x=x$, and (2) for $x \in \mathfrak{D}(T)$, we have $B (T- \lambda I) x=x$.
Because a densely defined bounded operator in $H$ has a unique extension to the entire Hilbert space $H$, the required operator $B$ in the second definition appears to be the extension, to all of $H$, of the bounded inverse for $(T - \lambda I)$ defined over its dense range in the first definition. However, the two definitions do not appear to be equivalent. Because the resolvent set would be trivially empty for operators that are not closable, it would also be okay if I could show the second definition is equivalent to the first for closed operators. Specific questions and my line of reasoning associated with those are below:
The first property in the second definition seems to imply that the range of $(T- \lambda I)$ is $H$, which is a more stringent condition than the requirement from the first definition that the range of $(T- \lambda I)$ is dense in $H$. Based on this observation, it seems to follow that $\rho_2(T) \subset \rho_1(T)$ for any unbounded operator $T$.
Question 1. Is there an example of an unbounded operator for which $\rho_2(T)$ is a strict subset of $\rho_1(T)$?
Question 2. Does the second definition coincide with the first in the more meaningful setting of closed operators, i.e., is $\rho_1(T) = \rho_2(T)$ for any closed operator $T$?
Based on the preceding discussion, to answer the second question affirmatively, we only need to show that for any closed operator $T$, $\rho_1(T) \subset \rho_2(T)$. I outline below my attempted sketch at a proof till the point I hit a wall. I would really appreciate answers to the above questions and guidelines for completing the proof sketch if the answer to the second question is a yes.
Suppose $T$ is a closed operator and $\lambda \in \rho_1(T)$. Define the bounded operator $B$ as the unique extension of the bounded inverse of $(T - \lambda I)$ from its dense range to all of $H$. $B$ is bounded and therefore closed. However, I cannot show that for all $x \in H$, $B x$ belongs to $\mathfrak{D}(T)$, which seems like a non-intuitive a posteriori imposition of a constraint on the operator $T$. If I could establish that $B$ is injective (which it needs to be), it seems like I could use the fact that $T$ is closed. However, the fact that the restriction of $B$ to the (dense in $H$) range of $(T - \lambda I)$ is injective does not seem to readily imply that $B$ is also injective. My guess is that the Closed Graph Theorem should play a role here and, if so, would also like a pointer to the specific statement of the Closed Graph Theorem applicable for this setting.
Responses are much appreciated; thanks in advance for your help!
For question 2, you're right that an operator $T$ that is not closed has an empty resolvent by the second definition (which is the more common one.) The graph of $B$ is closed because it's bounded. The graph of $T-\lambda I$ is obtained from the graph of $B$ by swapping the coordinates. So the graph of $T-\lambda I$ is closed, so $T-\lambda I$ is closed, and it's simple to show that $T$ is closed iff $T-\lambda I$ is closed for any $\lambda\in\Bbb{C}$.
To complete your proof - the extended $B$ maps into $\mathfrak{D}(T)$. Let $v\in H$. The range of $T-\lambda I$ is dense, so there is a sequence $x_n\in\mathfrak{D}(T)$ such that $(T-\lambda I)x_n\to v$. Since $B$ is bounded, $x_n=B(T-\lambda I)x_n\to Bv$. So $x_n\to Bv$ and $(T-\lambda I)x_n\to v$. Since $T$ is closed, so is $T-\lambda I$ and it follows that $Bv\in\mathfrak{D}(T)$ and $(T-\lambda I)Bv=v$. We are already assuming that for $v\in\mathfrak{D}(T), B(T-\lambda I)v=v$ so nothing further to prove. It also follows that $B$ is injective because if $Bv=0$ then $v=(T-\lambda I)Bv=0$.
For your first question, let $T$ be densely defined, closable but not closed. Let $\lambda$ be in the resolvent of $\overline{T}$ and $B$ a bounded inverse to $(\overline{T}-\lambda I)$. Then $B':=B\rvert_{\operatorname{range(T-\lambda I)}}$ gives an inverse for $T-\lambda I$ by the first definition. OTOH, since $T$ is not closed, its resolvent by the second definition is empty.