On the generalized spectrum of bounded linear operators in Banach spaces

1.   Introduction and preliminaries

Let $X$ be a Banach space and $B(X)$ denote the Banach space of all bounded linear operators from $X$ into itself. The identity operator will be denoted by $I$. For any $T\in B(X)$, we denote by $N(T)$ and $R(T)$ the null space and the range of $T$, respectively.

The resolvent set $\rho(T)$ of $T\in B(X)$ is, by definition,

And, its resolvent $R(\lambda) = (T-\lambda I)^{-1}$ is an analytic function on $\rho(T)$ since it satisfies the resolvent identity:

The spectrum $\sigma(T)$ is the complement of $\rho(T)$ in $\mathbb{C}$. As we all know, the spectral theory plays a fundamental role in functional analysis. If $T_\lambda$ is not invertible in $B(X)$, we can consider its generalized inverse. Recall that $T\in B(X)$ is generalized invertible if there exists an operator $S\in B(X)$ such that $TST = T$ and $STS = S$. We also say that such $S$ is a generalized inverse of $T$, which is always denoted by $T^+$. If $T$ has a bounded generalized inverse $T^+$, then, from ^[1], we know that both $TT^+$ and $T^+T$ are projectors on $X$ and

If $X$ is a Hilbert space and the direct sum decompositions in (1.1) are orthogonal, the corresponding generalized inverse is the Moore-Penrose inverse. Recall that the operator $T^\dagger\in B(X)$ is said to be the Moore-Penrose inverse of $T$ if $T^\dagger$ satisfies

where $T^*$ denotes the adjoint operator of $T$.

If the operator $T^\sharp\in B(X)$ satisfies

then $T^\sharp$ is called the group inverse of $T$. If $T^\sharp$ is the group inverse of $T$, then $N(T^\sharp) = N(T)$, $R(T^\sharp) = R(T)$ and $X = N(T^\sharp)\oplus R(T^\sharp)$ ^[1].

If, as the definition of $\rho(T)$, the generalized resolvent set is defined by

we can find that such $\rho_g(T)$ is meaningless in the case of matrices, since every matrix is generalized invertible and $\rho_g(T) = \mathbb{C}$. To define reasonably the generalized resolvent set, we should add some additional conditions.

Definition 1.1. Let $U$ be an open set in the complex plane $\mathbb{C}$; the function

is said to be a generalized resolvent of $T_\lambda = T-\lambda I$ on $U$ if

$(1)$ for all $\lambda\in U$,

$(2)$ for all $\lambda\in U$,

$(3)$ for all $\lambda$ and $\mu$ in $U$,

The conditions (1) and (2) say that $R_g(\lambda)$ is a generalized inverse of $T_\lambda$. While the equality in (3) is an analogue of the classical resolvent identity, we refer to it as the generalized resolvent identity, which assures that $R_g(\lambda)$ is locally analytic. In ^[2], Shubin points out that there exists a continuous generalized inverse function (satisfying $(1)$ and $(2)$ but not possibly $(3)$) meromorphic in the Fredholm domain $\rho_\phi(T) = \{\lambda\in C: T-\lambda I$ is Fredholm$\}$. And, it remains an open problem whether or not this can be done while also satisfying $(3)$, i.e., it is not known whether generalized resolvents always exist. Many authors have been interested in the existence of the generalized resolvents and the property of the corresponding spectrum in ^{[3,4,5,6,7,8,9,10,11,12,13]}.

Definition 1.2. The generalized resolvent set is

and the generalized spectrum $\sigma_g(T)$ is the complement of $\rho_g(T)$ in $\mathbb{C};$ the generalized spectral radius is

In this paper, we utilize the stability characterization of generalized inverses to investigate the properties of the generalized resolvent set in Banach spaces. We also introduce two sets

and

and prove that they are identical to $\rho_g(T)$. Based on this result, we discuss the relationship between the resolvent set and the generalized resolvent set, as well as the spectrum and the generalized spectrum. We also prove that several properties of the classical spectrum remain true in the case of the generalized one. Finally, we explain why we use the generalized inverse rather than the Moore-Penrose inverse or the group inverse to define the generalized resolvent.

2.   Main results

We start with the following lemma, which is preparation for the proofs of our main results.

Lemma 2.1. $\rm (1)$ If $R_{g}(\lambda)$ and $R_{g}(\mu)$ satisfy the generalized resolvent identity:

then

and

$\rm (2)$ Let $P_\lambda = T_\lambda R_{g}(\lambda)$ and $Q_\lambda = R_{g}(\lambda)T_\lambda;$ then, $P_\lambda$ and $Q_\lambda$ are projectors with

$\rm (3)$ The resolvent set is included in the generalized resolvent set, i.e. $\rho(T)\subset \rho_{g}(T)$, the generalized resolvent set $\rho_{g}(T)$ is open in $\mathbb{C}$ and the generalized resolvent $R_{g}(\lambda)$ is locally analytic on $\rho_{g}(T)$.

Proof. $\rm (1)$ We exchange $\lambda$ with $\mu$ in the generalized resolvent identity and obtain

and so

Then, $N\left(R_{g}(\mu)\right)\subset N\left(R_{g}(\lambda)\right)$ and $R\left(R_{g}(\lambda)\right)\subset R\left(R_{g}(\mu)\right)$. Thus, exchanging $\lambda$ with $\mu$ again, we can get

$\rm (2)$ Obviously, $P_\lambda$ and $Q_\lambda$ are projectors on $X$. Noting that

and

we have $P_\lambda P_\mu = P_\lambda$ and $Q_\lambda Q_\mu = Q_\mu.$

$\rm (3)$ Obviously, $\rho(T)\subset \rho_{g}(T)$. It follows from the definition of the generalized resolvent that the set $\rho_{g}(T)$ is open. Since

we can see that the operator $I+(\lambda-\mu)R_{g}(\lambda)$ is invertible for all $\mu$ satisfying $|\mu-\lambda| \|R_{g}(\lambda)\| < 1$. So,

Hence, $\lim\limits_{ \mu\rightarrow \lambda} R_{g}(\mu) = R_{g}(\lambda)$ and

Therefore, $R_{g}(\lambda)$ is locally analytic on $\rho_{g}(T)$ and $\left[R_{g}(\lambda)\right]^{\prime} = R_{g}^{2}(\lambda)$. □

Theorem 2.2. Let $X$ be a Banach space and $T\in B(X);$ then,

Proof. From Lemma 2.1, we can easily see that $\rho_{g}(T)\subset\rho_{g}^{1}(T)\subset\rho_{g}^{2}(T).$ To complete the proof, we need show that $\rho_{g}^{2}(T)\subset\rho_{g}(T).$ In fact, for any $\lambda\in \rho^2_g(T)$, we can find $M > 0$ and $\delta > 0$, such that, for all $\mu$ satisfying $|\mu-\lambda| < \delta$, $T_\mu^+$ exists and $\|T_\mu^+\|\leq M$.

Step 1. We first prove that there exists $\delta_1 < \delta$,

for all $\mu\in \{\mu\in \mathbb{C}: |\mu-\lambda| < \delta_1\}$. In fact, if $N(T_\lambda^+) = \{0\}$, obviously, $R(T_\mu)\cap N(T_\lambda^+) = \{0\}$. We can assume $N(T_\lambda^+)\neq\{0\}$; then, $I-T_\lambda T_\lambda^{+}\neq 0$. Let

and consider $\mu\in \mathbb{C}$ such that $|\mu-\lambda| < \delta_1$, Then, for any $y_\mu\in R(T_\mu)\cap N(T_\lambda^+)$, we can get

Hence $y_\mu = 0$. This implies $R(T_\mu)\cap N(T_\lambda^+) = \{0\}$.

Step 2. We shall prove that

is the generalized resolvent of $T_\lambda$ on $U = \{\mu\in \mathbb{C}: |\mu-\lambda| < \delta_1\}$. First, by $\|(\mu-\lambda)T_\lambda^+\| < 1$ and the Banach theorem, we can see that $I+(\mu-\lambda)T_\lambda^+$ is invertible and so $B_\mu$ is well defined. Second, from the equivalences between (1) and (3) in [14, Theorem 1.1], it follows that $B_\mu$ is a generalized inverse of $T_\lambda$ with $N(B_\mu) = N(T_\lambda^+)$ and $R(B_\mu) = R(T_\lambda^+)$. Third, we shall show that

Define $P_\mu = T_\mu B_\mu$ and $Q_\mu = B_\mu T_\mu;$ then, $P_\mu$ and $Q_\mu$ are projectors from $X$ onto $R(T_\mu)$ and $R(B_\mu) = B(T_\lambda^+)$, respectively. Hence

and

Thus, we can conclude

Therefore,

So, $B_\mu$ is the generalized resolvent of $T_\lambda$ on $U$, which means $\lambda\in \rho_g(T)$. □

Remark 2.3. According to Shubin, there exists a continuous generalized inverse function but not an analytic generalized resolvent ^[2]. From Theorem 2.1, we can see that if there exists a continuous or locally bounded generalized inverse function, then we can find a relevant analytic generalized resolvent.

Lemma 2.4. $\rm (1)$ Let $U$ and $V$ be two open sets in $\mathbb{C}$ such that the generalized resolvent identity holds on $U$ and $V$. If $U \cap V \neq \emptyset$, then the generalized resolvent identity holds on $U \cup V$, i.e.,

$\rm (2)$ Let $U$ be a convex open set in $\rho_{g}(T);$ then,

$\rm (3)$ If $U$ is a convex open set in $\rho_{g}(T)$ and $\rho(T) \cap U\neq\emptyset$, then

Proof. $\rm (1)$ For all $\lambda, \mu\in U\cup V$, if $\lambda, \mu\in U$ or $\lambda, \mu\in V$, then the generalized resolvent identity holds. It is sufficient to prove that

holds for $\lambda \in U$ and $\mu\in V$. Let $\nu\in U\cap V$; then,

and

By Lemma 2.1, $P_\lambda P_\nu = P_\lambda$, $Q_\lambda Q_\nu = Q_\nu$, $P_\nu P_\mu = P_\nu$ and $Q_\nu Q_\mu = Q_\mu$. Hence,

$\rm (2)$ For all $\lambda, \mu \in U$, the segment $[\lambda, \mu] \subset U$. Then for any $\omega \in[\lambda, \mu]$, there exists a neighborhood $U(\omega) \subset \rho_{g}(T)$ such that the generalized resolvent identity holds on $U(\omega)$. It follows from the finite covering theorem that we can find $\omega_{1}, \omega_{2}, \cdots, \omega_{n} \in[\lambda, \mu]$, $n\in N,$ such that $[\lambda, \mu] \subset \bigcup\limits_{i = 1}^{n} U\left(\omega_{i}\right)$. Hence, by (1), we have

$\rm (3)$ Let $\mu\in\rho(T) \cap U;$ then, for all $\lambda\in U$, by Lemma 2.1,

This implies that $R_{g}(\lambda)$ is invertible, and so $\lambda\in \rho(T)$. □

Theorem 2.5. Let $X$ be a Banach space and $T\in B(X);$ then, the generalized spectrum $\sigma_{g}(T)$ is a nonempty bounded closed subset in ${\mathbb{C}}$.

Proof. Since $\rho_{g}(T)$ is open, $\sigma_{g}(T) = {\mathbb{C}}\backslash\rho_{g}(T)$ is closed. If $|\lambda| > \|T\|$, then, by the Banach's theorem, $T-\lambda I = \lambda\left(\frac{1}{\lambda} T-I\right)$ is invertible and its inverse $(T-\lambda I)^{-1}$ is bounded. Hence

So $\sigma_{g}(T) \subset\{\lambda\in {\mathbb{C}}: |\lambda|\leq\|T\|\}$ and $\sigma_{g}(T)$ is bounded. Finally, we prove that $\sigma_{g}(T)$ is nonempty. In fact, if $\sigma_{g}(T) = \emptyset$, then $\rho_{g}(T) = \mathbb{C}$. By (3) in Lemma 2.2 and $\{\lambda\in{\mathbb{C}}: |\lambda| > \|T\|\}\subset \rho(T)$, we can get $\rho(T) = \mathbb{C}$. This is a contradiction with $\sigma(T)\neq\emptyset$. □

Proposition 2.6. Let $X$ be a Banach space and $T\in B(X);$ then,

$\rm (1)$ $\partial\sigma(T)\subset\sigma_{g}(T)\subset\sigma(T);$

$\rm (2)$ $\sigma(T) \backslash \sigma_{g}(T) = \sigma(T)\cap\rho_{g}(T)$ is open in $\mathbb{C}$;

$\rm (3)$ $\rho_{g}(T) = \rho(T) \cup\left[\sigma(T) \backslash \sigma_{g}(T)\right].$

Proof. (1) It follows from $\rho(T) \subset \rho_{g}(T)$ that $\sigma_{g}(T) \subset \sigma(T)$. Now we shall show that $\partial \sigma(T) \subset \sigma_{g}(T)$. If there is a $\lambda \in \partial \sigma(T)$ and $\lambda \notin \sigma_{g}(T)$, then $\lambda \in \rho_{g}(T)$ and we can find a neighborhood $U(\lambda) \subset \rho_{g}(T)$. Noting that $\lambda \in \partial \sigma(T)$, we can see that $U(\lambda) \cap \rho(T) \neq \emptyset$. It follows from Lemma 2.2 that $U(\lambda) \subset \rho(T)$, which is contradictory with $\lambda \in \partial \sigma(T)$.

(2) Since $\partial\sigma(T)\subset\sigma_{g}(T)$, we have

and it is an open set.

(3)

□

Example 2.7. Let $T$ be the right translation operator on $l^2$, i.e.,

Then $T$ is a Fredholm operator with

Noting that the nullity $n(T_\lambda) = \dim N(T_\lambda)\equiv 0$ and the defect $d(T_\lambda) = {\rm codim}R(T_\lambda)\equiv 1$ on $\{\lambda\in {\mathbb{C}}: |\lambda| < 1\}$, by Theorem 1.2 in ^[14] and the proof of Theorem 2.1, we know that $\{\lambda\in {\mathbb{C}}: |\lambda| < 1\}\subset \rho_g(T)$. Since $R(T_\lambda)$ is not closed for $\lambda$ satisfying $|\lambda| = 1$, $T_\lambda$ is not generalized invertible and so

Thus

Corollary 2.8. Let $X$ be a Banach space and $T\in B(X);$ then, the generalized spectral radius is just equal to the spectral radius, i.e.,

Proof. By $\sigma_{g}(T) \subset \sigma(T)$, we have $r_{\sigma_{g}}(T) \leq r_{\sigma}(T)$. Since $\sigma(T)$ is bounded and closed, we can find $\lambda_{0} \in \partial \sigma(T)$ such that $\left|\lambda_{0}\right| = r_{\sigma}(T)$. By Proposition 2.1, $\lambda_{0} \in \sigma_{g}(T)$ and then

Hence $r_{\sigma_{g}}(T) \geq r_{\sigma}(T)$ and so $r_{\sigma_{g}}(T) = r_{\sigma}(T)$. □

At the end, we shall explain why we use the generalized inverse rather than two of the most important unique generalized inverses (the Moore-Penrose inverse and the group inverse ^[1,15]) to define the generalized resolvent.

Theorem 2.9. Let $T\in B(X)$. Then, the Moore-Penrose inverse $T_\lambda^\dagger$ or the group inverse $T_\lambda^\sharp$ of $T_\lambda = T-\lambda I$ is the analytic generalized resolvent on $U$ if and only if

In this case, $T_\lambda$ is invertible, the Moore-Penrose inverse or the group inverse is the inverse and the generalized resolvent is exactly its classical resolvent.

Proof. It suffices to prove the necessity. We first claim that for all $\lambda, \mu\in U$, $N(T_\lambda) = N(T_\mu)$ and $R(T_\lambda) = R(T_\mu).$ In fact, if the Moore-Penrose inverse $T_\lambda^\dagger$ is the generalized resolvent on $U$, then, by Lemma 2.1, we have

Hence,

If the group inverse $T_\lambda^\sharp$ is the generalized resolvent on $U$, then

Hence,

Now, we prove that $N(T_\lambda) = \{0\}$ and $R(T_\lambda) = X.$ For all $x\in N(T_\lambda)$, then $T_\lambda x = T_\mu x = 0$, i.e., $Tx = \lambda x$ and $Tx = \mu x$. So, $x = 0$. This means that $N(T_\lambda) = \{0\}$. For any $y\in X$, $T_\mu y\in R(T_\lambda)$ and there is an $x\in X$, such that $T_\lambda x = T_\mu y$. Then,

We can conclude that $T_\lambda$ is invertible and the generalized resolvent is exactly its classical resolvent. □

3.   Conclusions

In this paper, we have proved that the existence of the analytic generalized resolvents of the linear operators in Banach spaces is equivalent to the continuity and local boundedness of generalized inverse functions. Based on the properties of generalized resolvents, we have shown that the generalized spectrum is a nonempty bounded closed subset. Moreover, the relationship between the resolvent set and the generalized resolvent set, as well as that between the spectrum and the generalized spectrum has been given. An interesting example is given to illustrate our results. Finally, we explain why we use the generalized inverse rather than the Moore-Penrose inverse or the group inverse to define the generalized resolvent.

Acknowledgments

This manuscript has benefited greatly from the constructive comments and helpful suggestions of the anonymous referees, the authors would like to express their deep gratitude to them. This research was supported by the National Natural Science Foundation of China (No. 11771378).

Conflict of interest

The authors declare that they have no conflicts of interest.