# Example of an unbounded closed operator $T$ such that $\lim_{|z| \to \infty} \| R(z,T)\|=0$

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 $$$$\tag{1} \lim_{|z| \to \infty} \| R(z,T)\|=0$$$$ holds?, where $$R(z,T):=(T-zI)^{-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?

• Say, how should this be understood if the spectrum is the whole plane? – Fedor Petrov Sep 14 '19 at 23:38
• @FedorPetrov thanks for your observation. I want an example where the spectrum is not the whole plane – Mainkit Sep 14 '19 at 23:43
• 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. – Christian Remling Sep 15 '19 at 2:24
• @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. – Mainkit Sep 15 '19 at 12:53
• @Mainkit so the limit is for $|z|\to\infty$ but with $\text{dist}(z, \sigma(T))\ge \delta>0$? – 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 non-bounded 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 non-bounded self-adjoint operator on $$X$$. (It follows from the spectral theorem that we have equality in $$(*)$$ for self-adjoint 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 $$W = 0$$.