Do you know an example of an unbounded closed operator $T:D(T)\subseteq X \to X$, defined in a complex Banach space $X$, such that $\mathbb{C} \setminus\sigma(T)$ is unbounded and the equality \begin{equation}\tag{1} \lim_{z \to \infty} \ R(z,T)\=0 \end{equation} holds?, where $R(z,T):=(TzI)^{1}$ for $z \notin \sigma(T)$. I know that if $T$ is a bounded operator then $(1)$ is true. But, does $(1)$ imply that $T$ is bounded?

$\begingroup$ Say, how should this be understood if the spectrum is the whole plane? $\endgroup$ – Fedor Petrov Sep 14 '19 at 23:38

$\begingroup$ @FedorPetrov thanks for your observation. I want an example where the spectrum is not the whole plane $\endgroup$ – Mainkit Sep 14 '19 at 23:43

$\begingroup$ Also, isn't it true in general that $\R(z)\\to\infty$ as $z\to w\in\sigma$ ? That would give a stronger statement along these lines. $\endgroup$ – Christian Remling Sep 15 '19 at 2:24

$\begingroup$ @ChristianRemling I have edited my question. You are right, in general it is true that $\ R(z_n,T)\ \to \infty$ if $z_n \to z \in \sigma(T)$. But, that doesn't help me because I'm interested in the behavior of $\ R(z_n,T)\$ far from the spectrum. $\endgroup$ – Mainkit Sep 15 '19 at 12:53

$\begingroup$ @Mainkit so the limit is for $z\to\infty$ but with $\text{dist}(z, \sigma(T))\ge \delta>0$? $\endgroup$ – Pietro Majer Sep 15 '19 at 13:10
As already noted by several users in this comments, something seems to be a bit odd with the question due to the behaviour of the resolvent close to the spectrum. Here are a few details about what is true and what is not:
1) If $(z_n)$ is a sequence in $\mathbb{C} \setminus \sigma(T)$ such that $\R(z_n,T)\ \to 0$, then we necessarily have $\operatorname{dist}(z_n, \sigma(C)) \to \infty$; this follows from the inequality \begin{align*} \R(z,T)\ \ge \frac{1}{\operatorname{dist}(z, T)} \qquad (*) \end{align*} which is true for all $z\in \mathbb{C} \setminus \sigma(T)$. [To also cover the case $\sigma(T) = \emptyset$ in this formula, we define the distance of any number from the empty set to be $\infty$.]
Thus, it does not suffice to consider only numbers $z$ which are bounded away from the spectrum and tend to infinity; one has to assume that the distance of $z$ to the spectrum also tends to infinity in order to have any chance for the resolvent to tend to $0$.
2) If, on the other hand, the question is whether there exists a Banach space $X$ and a closed nonbounded operator $T$ such that $\R(z_n,T)\ \to 0$ for any sequence $(z_n) \subseteq \mathbb{C} \setminus \sigma(T)$ which satisfies $z_n \to \infty$ and $\operatorname{dist}(z_n,\sigma(T)) \to \infty$, then the answer is yes:
Just choose $X$ as an infinite dimensional Hilbert space and let $H$ be your favourite nonbounded selfadjoint operator on $X$. (It follows from the spectral theorem that we have equality in $(*)$ for selfadjoint operators).
3) The question probably becomes a bit more interesting if we consider the following variant: If $T$ is a closed operator on a Banach space $X$ such that $\sigma(T)$ is bounded and such that $\R(z,T)\ \to 0$ as $z \to \infty$, does it follow that $T$ is bounded?
The answer is yes (here, by ``bounded'' I mean not only that $T$ is continuous, but also that $D(T) = X$).
Proof. Since $\sigma(T)$ is bounded, we can compute the spectral projection corresponding to the entire spectrum by integrating that resolvent over a path that encloses $\sigma(T)$. This yields a decomposition of $X$ into two closed subspaces, $X = V \oplus W$, such that $T$ splits along this decomposition into a bounded operator $T_V$ on $V$ and a closed operator $T_W$, with empty spectrum, on $W$.
The resolvent of $T_W$ is an analytic function defined everywhere on $\mathbb{C}$ and by assumption the resolvent of $T_W$ decays to $0$ as $z \to \infty$. Hence, the resolvent of $T_V$ is constant by Liouville's theorem and thus, it is constantly $0$. But each resolvent operator has to be injective, so $W = 0$. Thus, $X = V$ and $T$ is equal to the bounded operator $T_V$.
EDIT. Maybe it is worthwhile to mention the following generalization of 3) (which is inspired by a comment of Christian Remling below the question):
In fact, the same assertion remains true if we only assume that the resolvent is bounded outside a ball of sufficiently large radius. To see this, one argues similarly as in the proof of 3), but now Liouville's theorem only yields that the resolvent of $T_W$ is constant; so say $R(z,T_W) = R$ for a bounded operator $R$ on $W$. Now the resolvent identity implies that $R^2 = 0$. But $R^2$ is an injective operator, so we again obtain $W = 0$.