О λ-коммутировании, левом (правом) псевдоспектре и левом (правом) условном псевдоспектре линейных непрерывных операторов на ультраметрических банаховых пространствах

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1. Introduction and preliminaries

The classical theory of commutators was studied by H. Weyl [1] and J. von Neumann [2] and it played an important role in quantum mechanics [3–5]. In [6], C. R. Putnam collected some properties of the commutation of continuous linear operators in a Hilbert space over the field of complex numbers $ℂ\mathrm{.}$ Recently, many researchers studied and explored the operator equation $AS=\lambda SA$ where $\lambda \in ℂ\backslash \left\{0\right\},$ A and S are continuous linear operators on complex Hilbert spaces, see [7–9].

In ultrametric operator theory, the author [10] extended and studied the operator equation of the form $AS=\lambda SA$ where $\lambda \in \mathbb{K}\backslash \left\{0\right\},$ A and S are continuous linear operators on ultrametric Banach spaces over $\mathbb{K}\mathrm{.}$ He presented some spectral properties of $\lambda$-commuting operators on ultrametric Banach spaces over  and he gave an illustrative examples, see [10].

Recently, A. Ammar et al. [11] introduced and studied the pseudospectra of closed linear operators on ultrametric Banach spaces. On the other hand, A. Ammar et al. [12] introduced and studied the condition pseudospectra of continuous linear operators on ultrametric Banach spaces and gave some of its properties.

In [13], the author presented and studied the determinant spectrum, the -determinant spectrum, and the C-trace pseudospectrum of ultrametric matrix pencils.

There are many studies on pseudospectra and condition pseudospectra of continuous linear operator pencils and $\lambda$-commuting of operators in ultrametric operator theory, see [14–17]. In Section 5, we consider the problem of finding the eigenvalues of the generalized eigenvalue problem of the form

$P\left(\lambda \right)x=\mathrm{0,}$

where $P\left(\lambda \right)={\sum }_{k=0}^m{\lambda }^k{A}_k,$ $A_k\in {\mathcal{M}}_n\left(\mathbb{K}\right),$ $\lambda \in \mathbb{K}\mathrm{,}$ $x\in {\mathbb{K}}^n$ and ${\mathcal{M}}_n\left(\mathbb{K}\right)$ is the space of all $n\times n$ matrices over $\mathbb{K}\mathrm{.}$ I is the identity matrix of ${\mathcal{M}}_n\left(\mathbb{K}\right).$ If $C\in {\mathcal{M}}_n\left(\mathbb{K}\right),$ the determinant of C is denoted by $\det \left(C\right)$ (for details on the space ${\mathcal{M}}_n\left(\mathbb{K}\right)$ see [18] and [19]).

Throughout this paper, ${\mathbb{Q}}_p$ is the field of p-adic numbers, $\mathcal{E}$ is an ultrametric infinite-dimensional Banach space over a complete ultrametric valued field $\mathbb{K}$ with a non-trivial valuation $|\cdot |$ and $\mathcal{L}\left({\rm E}\right)$ denotes the set of all continuous linear operators on $\mathcal{E}\mathrm{.}$ Recall that $\mathbb{K}$ is called spherically complete if each decreasing sequence of balls in $\mathbb{K}$ has a non-empty intersection. For more details, see [20]. Let $S\in \mathcal{L}\left({\rm E}\right),$ $R\left(S\right),$ $N\left(S\right),$ $S^{\ast },$ ${\sigma }_p\left(S\right),$ $\sigma \left(S\right)$ and $\rho \left(S\right)$ denote the range, the kernel, the adjoint, the point spectrum, the spectrum and the resolvent set of S respectively [20].

The aim of this paper is to demonstrate some spectral properties of $\lambda$-commuting of continuous linear operators on ultrametric Banach spaces and we introduce and study the operator equations $AS=SB$ and $ASB=S$ for some $S\in \mathcal{L}\left({\rm E}\right).$ Moreover, some illustrative examples are provided. On the other hand, we introduce and study the left (right) pseudospectrum and the left (right) condition pseudospectrum of continuous linear operators on ultrametric Banach spaces. We obtain some results related to them. We continue by recalling some preliminaries.

Definition 1.1. [20] A field $\mathbb{K}$ is said to be ultrametric if it is endowed with an absolute value $|\cdot |:\mathbb{K}\to {ℝ}_{+}$ such that 

(i) $\left|\alpha \right|=0$ if, and only if, $\alpha =0;$ 

(ii) For all $\lambda ,\alpha \in \mathbb{K},\left|\lambda \alpha \right|=\left|\lambda \right|\left|\alpha \right|;$ 

(iii) For each $\lambda ,\alpha \in \mathbb{K},|\lambda +\alpha |\le \max \left\{\right|\lambda |,|\alpha \left|\right\}.$

Definition 1.2. [20] Let $\mathcal{E}$ be a vector space over $\mathbb{K}\mathrm{.}$ A mapping $\parallel \cdot \parallel :{\rm E}\to {ℝ}_{+}$ is said to be an ultrametric norm if: 

(i) For all $x\in \mathcal{E}\mathrm{,}$ $\parallel x\parallel =0$ if and only if $x=\mathrm{0,}$ 

(ii) For any $x\in \mathcal{E}$ and $\lambda \in \mathbb{K}\mathrm{,}$ $\parallel \lambda x\parallel =\left|\lambda \right|\parallel x\parallel ,$ 

(iii) For each $x\mathrm{,}y\in \mathcal{E}\mathrm{,}$ $\parallel x+y\parallel \le \max (\parallel x\parallel ,\parallel y\parallel ).$ 

Definition 1.3. [20] An ultrametric Banach space is a complete ultrametric normed space.

Example 1.1. [20] Let $c_0\left(\mathbb{K}\right)$ be the space of all sequences ${\left(x_i\right)}_{i\in \mathbb{N}}$ in $\mathbb{K}$ such that ${\lim }_{i\to \infty }{x}_i=0.$ Then $c_0\left(\mathbb{K}\right)$ is a vector space over $\mathbb{K}$ and

$\parallel {\left(x_i\right)}_{i\in \mathbb{N}}\parallel =\underset{i\in \mathbb{N}}{sup}\left|x_i\right|$

is an ultrametric norm for which $\left(c_0\left(\mathbb{K}\right),\parallel \cdot \parallel \right)$ is an ultrametric Banach space.

Theorem 1.1. [21] Let $\mathcal{E}$ be an ultrametric Banach space over a spherically complete field $\mathbb{K}\mathrm{.}$ For each $x\in {{\rm E}}^{\ast }={\rm E}\backslash \left\{0\right\},$ there exists $x^{\ast }\in {\mathcal{E}}^{\ast }$ such that $x^{\ast }\left(x\right)=1$ and $\parallel x^{\ast }\parallel =\parallel x{\parallel }^{-1}.$ 

Definition 1.4. [20] An ultrametric Banach space $\mathcal{E}$ over $\mathbb{K}$ is called a free Banach space if there is a family ${\left(x_i\right)}_{i\in \mathcal{I}}\in {\rm E}$ indexed by a set $\mathcal{I}$ such that all $x\in \mathcal{E}$ is written in a unique fashion as $x={\sum }_{i\in \mathcal{I}}{\lambda }_i{x}_i$ and $\parallel x\parallel =su{p}_{i\in \mathcal{I}}\left|{\lambda }_i\right|\parallel x_i\parallel .$ The family ${\left(x_i\right)}_{i\in \mathcal{I}}$ is called an orthogonal basis for $\mathcal{E}\mathrm{.}$ If, for each $i\in \mathcal{I}\mathrm{,}$ $\parallel x_i\parallel =\mathrm{1,}$ hence ${\left(x_i\right)}_{i\in \mathcal{I}}$ is called an orthonormal basis of $\mathcal{E}\mathrm{.}$ 

Definition 1.5. [20] Let $\omega ={\left({\omega }_i\right)}_i$ be a sequence of ${\mathbb{K}}^{\ast }=\mathbb{K}\backslash \left\{0\right\}.$ We define ${\mathcal{E}}_{\omega }$ by

${{\rm E}}_{\omega }=\{x={\left(x_i\right)}_i:\forall i\in \mathbb{N}{x}_i\in \mathbb{K},and\underset{i\to \infty }{lim}|{\omega }_i{|}^{\frac{1}{2}}\left|x_i\right|=0\},$

and it is equipped with the norm

$\forall x\in {{\rm E}}_{\omega }:x={\left(x_i\right)}_i\parallel x\parallel =\underset{i\in \mathbb{N}}{sup}\left(\right|{\omega }_i{|}^{\frac{1}{2}}\left|x_i\right|).$

Remark 1.1. [20]

(i) The space $({{\rm E}}_{\omega },\parallel \cdot \parallel )$ is an ultrametric Banach space.

(ii) If

$⟨\cdot ,\cdot ⟩:{{\rm E}}_{\omega }\times {{\rm E}}_{\omega }\to \mathbb{K},(x,y)\mapsto \sum _{i=0}^{\infty }{x}_i{y}_i{\omega }_i,$

where $x={\left(x_i\right)}_i$ and $y={\left(y_i\right)}_i.$ Then the space $({{\rm E}}_{\omega },\parallel \cdot \parallel ,⟨\cdot ,\cdot ⟩)$ is called an ultrametric Hilbert space.

(iii) The orthogonal basis $\{e_i,i\in \mathbb{N}\}$ is called the canonical basis of ${\mathbb{E}}_{\omega }$ where for all $i\in \mathbb{N}\mathrm{,}$ $\parallel e_i\parallel =|{\omega }_i{|}^{\frac{1}{2}}.$ 

Remark 1.2. [20] Let $\mathbb{K}={\mathbb{Q}}_p,$ if $p\equiv 1\left(\mathrm{mod}4\right),$ then $i=\sqrt{-1}\in {\mathbb{Q}}_p$ and $i^2=-1.$ 

Definition 1.6. [20] Let $S\in {\mathcal{M}}_n\left(\mathbb{K}\right).$ The spectrum $\sigma \left(S\right)$ of S is defined by

$\sigma \left(S\right)=\{\lambda \in \mathbb{K}:S-\lambda Iis not invertible\}.$

By Definition 6 of [13] (where $B=I$), we have the following:

Definition 1.7. If $S\in {\mathcal{M}}_n\left(\mathbb{K}\right)$ and $\epsilon >0.$ Then the $\epsilon$-determinant spectrum $d_{\epsilon }\left(S\right)$ of S is the following set:

From Remark 2 of [13] (where $B=I$), we get

Remark 1.3. Note that for each $S\in {\mathcal{M}}_n\left(\mathbb{K}\right)$ and $\epsilon >\mathrm{0,}$ $\sigma \left(S\right)\subseteq d_{\epsilon }\left(S\right)$ and $d_0\left(S\right)=\sigma \left(S\right).$ 

The $\lambda$-commuting of operators is defined as follows.

Definition 1.8. [10] Let $A,B\in \mathcal{L}\left({\rm E}\right),$ A and B are called $\lambda$-commuting operators if $AB=\lambda BA$ for some $\lambda \in {\mathbb{K}}^{\ast }.$ 

Example 1.2. [10] Let $\mathbb{K}={\mathbb{Q}}_p$ with $p\equiv 1\left(\mathrm{mod}4\right),$ let A and B be defined on ${\mathbb{Q}}_p\times {\mathbb{Q}}_p$ respectively by

$A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)andB=\left(\begin{array}{cc}0& i\\ -i& 0\end{array}\right).$

Then $AB=-BA.$

Example 1.3. [10] Let $\lambda \in {\mathbb{K}}^{\ast }\mathrm{,}$ let A and B be defined on ${\mathbb{K}}^3$ by

$A=\left(\begin{array}{ccc}1& 0& 0\\ 1& \lambda & 0\\ 1& \lambda & {\lambda }^2\end{array}\right)andB=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\end{array}\right).$

Then $AB=\lambda BA.$ 

Example 1.3. [10] Let $\lambda \in \mathbb{K}$ such that $\left|\lambda \right|>1$ and let $A,B\in \mathcal{L}\left(c_0\right(\mathbb{K}\left)\right)$ be given respectively by

$B(x_1,x_2,x_3,\dots )=(\frac{x_1}{\lambda },\frac{x_2}{{\lambda }^2},\frac{x_3}{{\lambda }^3},\dots )forall(x_1,x_2,x_3,\dots )\in c_0\left(\mathbb{K}\right)$

and

$foralln\ge \mathrm{1,}A{e}_n=e_{n+1},$

where ${\left(e_n\right)}_{n\ge 1}$ is a base of $c_0\left(\mathbb{K}\right).$ Hence $AB=\lambda BA.$ 

Let $A\in \mathcal{L}\left({\rm E}\right)$ be given, set $S_{\lambda }\left(A\right)=\{B\in \mathcal{L}({\rm E}):AB=\lambda BA\}.$ We collect some properties of $\lambda$-commuting operators.

Proportion 1.1. [10] Let $A\in \mathcal{L}\left({\rm E}\right)$ and $\lambda \in \mathbb{K}\mathrm{.}$  

(i) If $B_1,B_2\in S_{\lambda }\left(A\right),$ hence $B_1+B_2\in S_{\lambda }\left(A\right)$ and $B_1{B}_2\in S_{{\lambda }^2}\left(A\right);$ 

(ii) If B is invertible and $\lambda \ne \mathrm{0,}$ then $B^{-1}\in S_{\frac{1}{\lambda }}\left(A\right);$ 

(iii) $S_{\lambda }\left(A\right)$ is closed in the uniform operator topology 

(iv) If $AB=\lambda BA$ and $AB\ne \mathrm{0,}$ then $Ap\left(B\right)=p\left(\lambda B\right)A$ where p is a non-constant polynomial.

Proposition 1.2. [10] Let $A,B\in \mathcal{L}\left({\rm E}\right)$ and $\lambda \in {\mathbb{Z}}_p$ such that $AB=\lambda BA,$ $AB\ne 0$ and $\parallel B\parallel <p^{\frac{1}{1-p}}.$ Then

$A{e}^B=e^{\lambda B}A.$

Proposition 1.3. [10] If $A,B\in \mathcal{L}\left({\rm E}\right),$ $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA,$ hence ${\sigma }_p\left(AB\right)=\lambda {\sigma }_p\left(BA\right).$ 

Proposition 1.4. [10] Let $A,B\in \mathcal{L}\left({\rm E}\right),$ $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA.$ Then $\rho \left(AB\right)\subset \lambda \rho \left(BA\right).$ Furthermore, for any $\mu \in \rho \left(AB\right),$  

Proposition 1.5. [10] Let $A,B\in \mathcal{L}\left({\rm E}\right)$ and $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA.$ Then 

(i) $N\left(AB\right)=N\left(BA\right);$ 

(ii) $R\left(AB\right)=R\left(BA\right);$ 

(iii) For all $\mu \in \mathbb{K},N(AB-\mu )=N(BA-{\lambda }^{-1}\mu );$ 

(iv) For any $\mu \in \mathbb{K},R(AB-\mu )=R(BA-{\lambda }^{-1}\mu ).$ 

From Proposition 1.5, we conclude:

Theorem 1.2. [10] If $A,B\in \mathcal{L}\left({\rm E}\right)$ and $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA,$ then

${\sigma }_e\left(AB\right)=\lambda {\sigma }_e\left(BA\right).$

For $A\in \mathcal{L}\left({\rm E}\right),$ set $r\left(A\right)=li{m}_{n\to \infty }\parallel A^n{\parallel }^{\frac{1}{n}}.$ We have the following proposition.

Proposition 1.6. [10] Let $A,B\in \mathcal{L}\left({\rm E}\right)$ and $\lambda \in \mathbb{K}$ with $\left|\lambda \right|=1$ and $AB=\lambda BA.$ Then $r\left(AB\right)\le r\left(A\right)r\left(B\right).$ 

We continue with the following definitions.

Definition 1.9. [20] Let $\mathcal{E}$ be a non-Archimedean Banach space over $\mathbb{K}$ and let $B\in \mathcal{L}\left({\rm E}\right),$ the spectrum $\sigma \left(B\right)$ of B is defined by

$\sigma \left(B\right)=\{\mu \in \mathbb{K}:B-\mu Iisnotinvertible\},$

the resolvent set of B is defined by $\rho \left(B\right)=\mathbb{K}\backslash \sigma \left(B\right).$ 

Definition 1.10. [13] Let $B\in {\mathcal{M}}_n\left(\mathbb{K}\right),$ the trace $Tr\left(B\right)$ of B is defined by ${\sum }_i^n{b}_{i\mathrm{,}i}$ where for each $i\in \{\mathrm{1,}\dots ,n\},$ $b_{i\mathrm{,}i}\in \mathbb{K}$ are diagonal coefficients of B. 

Proposition 1.7. [13] Let $B,C\in {\mathcal{M}}_n\left(\mathbb{K}\right).$ Then 

(i) For any $\lambda \in \mathbb{K}\mathrm{,}$ $Tr(B+\lambda C)=Tr\left(B\right)+\lambda Tr\left(C\right),$ 

(ii) $Tr\left(BC\right)=Tr\left(CB\right).$ 

We have:

Definition 1.11. [13] Let $B\in {\mathcal{M}}_n\left(\mathbb{K}\right),$ $\epsilon >\mathrm{0,}$ the trace pseudospectrum $T{r}_{\epsilon }\left(B\right)$ of B is given by

$T{r}_{\epsilon }\left(B\right)=\sigma \left(B\right)\cup \{\lambda \in \mathbb{K}:|Tr(B-\lambda I)|\le \epsilon \}.$

The trace pseudoresolvent $Tr{\rho }_{\epsilon }\left(B\right)$ of B is defined by

$Tr{\rho }_{\epsilon }\left(B\right)=\rho \left(B\right)\cap \{\mu \in \mathbb{K}:|Tr(B-\mu I)|>\epsilon \}.$

Lemma 1.1. [20] Let $S\in \mathcal{L}\left({\rm E}\right)$ with $\parallel S\parallel <\mathrm{1,}$ then $\parallel {(I-S)}^{-1}\parallel \le 1.$ 

2. $\lambda$-commuting of ultrametric operators

Similar to the proof of Proposition 1, we conclude:

Proposition 2.1. Let $A,B\in \mathcal{L}\left({\rm E}\right)$ and $\lambda \in \mathbb{K}$ such that $\left|\lambda \right|=1$ and $AB=\lambda BA.$ Then $r\left(BA\right)\le r\left(A\right)r\left(B\right).$ 

Question: In Proposition 2.1, if $\left|\lambda \right|\ne \mathrm{1,}$ does $r\left(BA\right)\le r\left(A\right)r\left(B\right)$ hold?

Definition 2.1. Suppose that $\parallel {\rm E}\parallel \subseteq \left|\mathbb{K}\right|.$ Let $A\in \mathcal{L}\left({\rm E}\right),$ the approximate spectrum ${\sigma }_{ap}\left(A\right)$ of A is defined by

${\sigma }_{ap}\left(A\right)=\{\mu \in \mathbb{K}:\exists {\left(x_n\right)}_{n\in \mathbb{N}}\in {\rm E}\forall n\in \mathbb{N}\parallel x_n\parallel =1and\underset{n\to \infty }{lim}\parallel (A-\mu I)x_n\parallel =0\}.$

Proposition 2.2. Suppose that $\parallel {\rm E}\parallel \subseteq \left|\mathbb{K}\right|.$ If $A,B\in \mathcal{L}\left({\rm E}\right),$ $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA\ne \mathrm{0,}$ then ${\sigma }_{ap}\left(AB\right)=\lambda {\sigma }_{ap}\left(BA\right).$ 

Proof. Let $\mu \in {\sigma }_{ap}\left(AB\right),$ then there is ${\left(x_n\right)}_{n\in \mathbb{N}}$ in $\mathcal{E}$ such that for each $n\in \mathbb{N},$ $\parallel x_n\parallel =1$ and ${\lim }_{n\to \infty }\parallel (AB-\mu I)x_n\parallel =0.$ Since

$\parallel (BA-\frac{\mu }{\lambda }I)x_n\parallel =\frac{\parallel (AB-\mu I)x_n\parallel }{\left|\lambda \right|}.$ (2.1)

Then $\frac{\mu }{\lambda }\in {\sigma }_{ap}\left(BA\right)$ that is, $\mu \in \lambda {\sigma }_{ap}\left(BA\right).$ Similarly, if $\frac{\mu }{\lambda }\in {\sigma }_{ap}\left(BA\right)$ and using (1), we get $\mu \in {\sigma }_{ap}\left(AB\right).$ 

Lemma 2.1. Let $A,B\in \mathcal{L}\left({\rm E}\right),$ $\lambda \in \mathbb{K}$ such that $AB=\lambda BA\ne 0.$ Then for any $n\in \mathbb{N}\mathrm{,}$ $A^{n}B={\lambda }^{n}B{A}^n.$ 

Proof. Since $AB=\lambda BA\ne 0.$ Then $A^{2}B=\lambda ABA={\lambda }^{2}B{A}^2.$ One can see that for all $n\in \mathbb{N}\mathrm{,}$ $A^{n}B={\lambda }^{n}B{A}^n.$ 

Proposition 2.3. Let $A,B\in \mathcal{L}\left({\rm E}\right),$ $\lambda \in {\mathbb{Z}}_p$ such that $AB=\lambda BA,$ $AB\ne 0$ and $\parallel A\parallel <p^{\frac{1}{1-p}}.$ Then

$e^{A}B=B{e}^{\lambda A}.$

Proof. By $\parallel A\parallel <p^{\frac{1}{1-p}},$ we get $e^A$ and $e^{\lambda A}$ exist. Since $AB=\lambda BA\ne 0.$ Using Lemma 2, we conclude that

$e^{A}B=B{e}^{\lambda A}.$  

In the finite-dimensional ultrametric Banach space, we obtain.

Proposition 2.4. If $A,B\in {\mathcal{M}}_n\left(\mathbb{K}\right)$ are invertible matrices and $\lambda \in {\mathbb{K}}^{\ast }$ such that $AB=\lambda BA,$ then ${\lambda }^n=1.$ 

Proof. From $\det \left(AB\right)={\lambda }^n\det \left(BA\right)$ and $\det \left(BA\right)=\det \left(AB\right).$ We get ${\lambda }^n=1.$ 

From Proposition 4.2, we have the following:

Corollary 2.1. If $A,B\in {\mathcal{M}}_n\left(\mathbb{K}\right)$ are invertible matrices and $\lambda \in {\mathbb{K}}^{\ast }$ such that $AB=\lambda BA,$ then $\left|\lambda \right|=1.$ 

Proposition 2.5. Let $A,B\in {\mathcal{M}}_n\left(\mathbb{K}\right)$ and $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA.$ Then $\mu \in d_{\epsilon }\left(AB\right)$ if and only if $\frac{\mu }{\lambda }\in d_{\frac{\epsilon }{|\lambda {|}^n}}\left(BA\right).$ 

Proof. From $\det (AB-\mu I)=\det (\lambda BA-\mu I)={\lambda }^n\det (BA-\frac{\mu }{\lambda }I)$ for $\mu \in \mathbb{K}$ and $\lambda \in {\mathbb{K}}^{\ast }.$ Then $\mu \in d_{\epsilon }\left(AB\right)$ if and only if $\frac{\mu }{\lambda }\in d_{\frac{\epsilon }{|\lambda {|}^n}}\left(BA\right).$ 

Proposition 2.6. Let $A,B\in {\mathcal{M}}_n\left(\mathbb{K}\right)$ and $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA$ and $AB\ne 0.$ If $tr\left(AB\right)\ne 0$ or $tr\left(BA\right)\ne \mathrm{0,}$ then $\lambda =1.$ 

Proof. Since $tr\left(AB\right)=\lambda tr\left(BA\right)=\lambda tr\left(AB\right)$ and ($tr\left(AB\right)\ne 0$ or $tr\left(BA\right)\ne 0$), we get $\lambda =1.$ 

Let $\mathcal{E}$ is a free Banach space over $\mathbb{K}\mathrm{,}$ we set ${\mathcal{L}}_0\left({\rm E}\right)=\{A\in \mathcal{L}({\rm E}):A^{\ast }exists\}.$

Proposition 2.7. [20] If $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ and $\lambda \in \mathbb{K}\mathrm{,}$ then 

(i) ${(A+\lambda B)}^{\ast }=A^{\ast }+\lambda B^{\ast }.$ 

(ii)  ${\left(AB\right)}^{\ast }=B^{\ast }{A}^{\ast }.$

Definition 2.2. [20] Let $A\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right).$ We have 

(i) A is said to be selfadjoint if $A^{\ast }=A;$ 

(ii) A is said to be normal if $A^{\ast }A=A{A}^{\ast };$ 

(iii) A is said to be unitary if $A^{\ast }A=A{A}^{\ast }=I.$ 

The following proposition describes some spectral properties of $\lambda$-commuting operators.

Proposition 2.8. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ and $\lambda \in \mathbb{K}$ with $AB\lambda BA\ne 0.$ If A is a selfadjoint, then $AB{B}^{\ast }=B{B}^{\ast }A$ and $AB{B}^{\ast }$is selfadjoint.

Proof. If $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ with $AB=\lambda BA,$ then ${\left(AB\right)}^{\ast }={\left(\lambda BA\right)}^{\ast }.$ Hence

$B^{\ast }{A}^{\ast }=\lambda A^{\ast }{B}^{\ast }.$ (2.2)

Since A is a selfadjoint and by (2), we get $B^{\ast }A=\lambda A{B}^{\ast }.$ On the other hand

$AB{B}^{\ast }=\lambda BA{B}^{\ast }=\lambda {\lambda }^{-1}B{B}^{\ast }A=B{B}^{\ast }A,$

and

${\left(AB{B}^{\ast }\right)}^{\ast }={\left(B{B}^{\ast }\right)}^{\ast }{A}^{\ast }=B{B}^{\ast }A=AB{B}^{\ast }.$

As the proof of Proposition 2.8, we get the following:

Proposition 2.9. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ and $\lambda \in \mathbb{K}$ with $AB=\lambda BA\ne 0.$ If B is a selfadjoint, then $BA{A}^{\ast }=A{A}^{\ast }B$ and $A{A}^{\ast }B$ is selfadjoint. 

By Proposition 2.8 and Proposition 2.9, we conclude that: 

Lemma 2.2. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ and $\lambda \in \mathbb{K}$ with $AB=\lambda BA\ne 0.$ If A and B are selfadjoint operators, then $A{B}^2=B^{2}A$ and $B{A}^2=A^{2}B.$   

Proposition 2.10. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ and $\lambda \in \mathbb{K}$ with $AB=\lambda BA\ne 0$ and $B{A}^2\ne 0.$ If A and B are selfadjoint operators, then $\lambda \in \{-\mathrm{1,1}\}.$  

Proof. From $AB=\lambda BA,$ we get $A^{2}B={\lambda }^{2}B{A}^2.$ By Lemma 2.2, we have $A^{2}B=B^{2}A={\lambda }^{2}B{A}^2.$ Since $B{A}^2\ne \mathrm{0,}$ we get ${\lambda }^2=1.$ Then $\lambda \in \{-\mathrm{1,1}\}.$ 

We give another proof of Proposition 2 without the condition $B{A}^2\ne 0.$  

Proposition 2.11. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ and $\lambda \in \mathbb{K}$ with $AB=\lambda BA\ne 0.$ If A and B are selfadjoint operators, then $\lambda \in \{-\mathrm{1,1}\}.$ 

Proof. From A and B are selfadjoint operators, we get ${\left(AB\right)}^{\ast }={\left(\lambda BA\right)}^{\ast },$ hence

$BA=\lambda AB.$ (2.3)

Using $AB=\lambda BA$ and (3), we get $AB={\lambda }^{2}AB.$ Hence ${\lambda }^2=1.$ 

Thus $\lambda \in \{-\mathrm{1,1}\}.$ 

Proposition 2.12. Let $A,B\in \mathcal{L}\left({{\rm E}}_{\omega }\right).$ If there is an unitary operator $U\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ with $AB=UBA=BAU,$ then $A{B}^{2}A=B{A}^{2}B.$ 

Proof. Since $AB=UBA=BAU,$ we have

$A{B}^{2}A=UBABA=BAUBA=BAAB=B{A}^{2}B.$

Lemma 2.3. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ be selfadjoint operators. If there is an unitary operator $U\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ with $AB=UBA.$ Then 

(i) U and U^* commute with $AB$; 

(ii) U and U^* commute with BA. 

Proof. (i)  From $AB=UBA,$ we have $BA=AB{U}^{\ast }.$ Hence $AB=UBA=UAB{U}^{\ast }.$ Thus $ABU=UAB$ and $U^{\ast }AB=AB{U}^{\ast }.$

(ii)  From (i), we get $U^{\ast }BA={\left(ABU\right)}^{\ast }={\left(UAB\right)}^{\ast }=BA{U}^{\ast }.$ On the other hand $BAU=U{U}^{\ast }BAU=UBA{U}^{\ast }U=UBA.$ 

Proposition 2.13. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ be selfadjoint operators. If there is an unitary operator $U\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ with $AB=UBA.$ Then $A{B}^{2}A=B{A}^{2}B.$ 

Proof. From Lemma 2.3, $UBA=BAU.$ Therefore, $B{A}^{2}B=BAAB=BAUBA=UBABA=ABBA=A{B}^{2}A.$ 

Lemma 2.4. Let $A,B\in \mathcal{L}\left({\rm E}\right).$ If $A{B}^2=B^{2}A$ and $B{A}^2=A^{2}B,$ then $A{B}^{2}A=B{A}^{2}B.$ 

Proof. One can see that $A{B}^{2}A=B^2{A}^2$ and $B{A}^{2}B=B^2{A}^2.$ Thus $A{B}^{2}A=B{A}^{2}B.$ 

Question: Let $A,B\in {\mathcal{L}}_0\left({\rm E}\right)$ be selfadjoint operators. Is the converse of Lemma 2 hold?

Example 2.1. Let $\mathbb{K}={\mathbb{Q}}_p$ and let A and B be defined on ${\mathbb{Q}}_p^2$ by

$A=\left(\begin{array}{cc}a& 0\\ b& \lambda a\end{array}\right)andB=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),$

where $\lambda a\in {\mathbb{Q}}_p\backslash \left\{0\right\}.$ Then $AB=\lambda BA.$ Let $\lambda \in {\mathbb{K}}^{\ast }\mathrm{,}$ let A and B be defined on ${\mathbb{K}}^3$ by

$A=\left(\begin{array}{ccc}a& 0& 0\\ b& \lambda a& 0\\ c& \lambda b& {\lambda }^{2}a\end{array}\right)andB=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\end{array}\right).$

Then $AB=\lambda BA.$

Lemma 2.5. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right).$ If there is unitary operators $U,V\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ with $AB=U{B}^{\ast }{A}^{\ast }$ and $BA=V{A}^{\ast }{B}^{\ast },$ then the following statements hold: 

(i) AB commutes with U and $U^{\ast };$ 

(ii) BA commutes with V and $V^{\ast }.$ 

Proof. (i) From $AB=U{B}^{\ast }{A}^{\ast },$ we have $B^{\ast }{A}^{\ast }=AB{U}^{\ast }.$ Hence $AB=U{B}^{\ast }{A}^{\ast }=UAB{U}^{\ast }.$ Thus $ABU=UAB$ and $U^{\ast }AB=AB{U}^{\ast }.$

(ii) By $BA=V{A}^{\ast }{B}^{\ast },$ we get $A^{\ast }{B}^{\ast }=BA{V}^{\ast }.$ Thus $BA=VBA{V}^{\ast }.$ Hence $BAV=VBA$ and $V^{\ast }BA=BA{V}^{\ast }.$ 

Theorem 2.1. If $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ are selfadjoint operators and $\lambda \in {\mathbb{K}}^{\ast }\text{}$ with $AB=\lambda BA\ne \mathrm{0,}$ then AB and BA are normal commuting operators.

Proof. Set $S=AB,$ hence $S^{\ast }=BA.$ From $S=AB=\lambda BA,$ we get $S=\lambda S^{\ast }.$ Then $S{S}^{\ast }=\lambda {\left(S^{\ast }\right)}^2=S^{\ast }S.$ Thus AB is normal and $S{S}^{\ast }=S^{\ast }S.$ Hence AB and BA are normal commuting operators.

Theorem 2.2. Let $A\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ and let $B\in \mathcal{L}\left({{\rm E}}_{\omega }\right)$ and $\lambda \in {\mathbb{K}}^{\ast }$ with $AB=\lambda BA\ne \mathrm{0,}$ $A^{\ast }B=B{A}^{\ast }$ and $A^{\ast }A=I$ with $B\ne 0.$ Then $\lambda =1.$ 

Proof. From $AB=\lambda BA,$ hence $A^{\ast }AB=\lambda A^{\ast }BA.$ Since $A^{\ast }B=B{A}^{\ast }$ and $A^{\ast }A=I,$ we get $B=\lambda A^{\ast }BA=\lambda B{A}^{\ast }A=\lambda B.$ Hence $(\lambda -1)B=0.$ Since $B\ne \mathrm{0,}$ we get $\lambda =1.$ 

Proposition 2.14. Let $A,B\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right).$ If there is unitary operators $U,V\in {\mathcal{L}}_0\left({{\rm E}}_{\omega }\right)$ with $AB=U{B}^{\ast }{A}^{\ast }$ and $BA=V{A}^{\ast }{B}^{\ast }.$ Then AB and BA are normal.

Proof. By Lemma 2, we get $U{B}^{\ast }{A}^{\ast }=B^{\ast }{A}^{\ast }U.$ Then

$AB{\left(AB\right)}^{\ast }=AB{B}^{\ast }{A}^{\ast }=U{B}^{\ast }{A}^{\ast }{B}^{\ast }{A}^{\ast }=B^{\ast }{A}^{\ast }U{B}^{\ast }{A}^{\ast }=B^{\ast }{A}^{\ast }AB={\left(AB\right)}^{\ast }AB$

and

${\left(BA\right)}^{\ast }BA=A^{\ast }{B}^{\ast }BA=BA{V}^{\ast }BA=BABA{V}^{\ast }=BA{\left(BA\right)}^{\ast }.$

Definition 2.3. [22] Let $A\in \mathcal{L}\left({\rm E}\right),$ A is said to be bounded below if for each $x\in \mathcal{E}\mathrm{,}$ $M\parallel x\parallel \le \parallel Ax\parallel$ for some $M>0.$ 

We have the following statement.

Theorem 2.3. Let $A,B\in \mathcal{L}\left({\rm E}\right)$ and $\lambda \in {\mathbb{K}}^{\ast }$ such that $AB=\lambda BA\ne 0.$ Then AB is bounded below if and only if A and B are bounded below. 

Proof. Suppose that AB is bounded below and $AB=\lambda BA\ne 0.$ Then there is $M>0$ with for each $x\in \mathcal{E}\mathrm{,}$ $M\parallel x\parallel \le \parallel ABx\parallel \le \parallel A\parallel \parallel Bx\parallel .$

Hence B is bounded below. Since $M\parallel x\parallel \le \parallel ABx\parallel$ for any $x\in \mathcal{E}\mathrm{,}$ and $AB=\lambda BA\ne \mathrm{0,}$ it follows that for every $x\in \mathcal{E}\mathrm{,}$ $\frac{M}{\left|\lambda \right|\parallel B\parallel }\parallel x\parallel \le \parallel Ax\parallel .$ Consequently, A is bounded below.

Conversely, it is easy to see that if A and B are bounded below, then AB is bounded below.

3. Some properties of ultrametric operator equations

In this section, let $A,B\in \mathcal{L}\left({\rm E}\right).$ We shall study the operator equations $AS=SB$ and $ASB=S$ for some $S\in \mathcal{L}\left({\rm E}\right).$ We continue with the following results.

Lemma 3.1. Let $A,B,S\in \mathcal{L}\left({\rm E}\right)$ such that $AS=SB$ and $ASB=S.$ If $A^2-I$ or $I-B^2$ is invertible, then $S=0.$ 

Proof. From $AS=SB$ and $ASB=S,$ we have $S=S{B}^2.$ Then $S(I-B^2)=0.$ Since $I-B^2$ is invertible, we conclude that $S=0.$ Similarly, one can see that $(A^2-I)S=0.$ From $A^2-I$ is invertible, then $S=0.$ 

Further, $R\left(S\right)$ denotes the range of S dense in $\mathcal{E}\mathrm{,}$ i.e. $\overline{R\left(S\right)}={\rm E}.$

Proposition 3.1. Let $A,B,S\in \mathcal{L}\left({\rm E}\right)$ and $\overline{R\left(S\right)}={\rm E}$ such that $AS=SB$ and $ASB=S.$ Then $A^2=I.$  

Proof. From $AS=SB$ and $ASB=S,$ then $(A^2-I)S=0.$ Hence $R\left(S\right)\subseteq N(A^2-I).$ Since $\overline{R\left(S\right)}={\rm E},$ we get $A^2=I.$ 

One can see the following: 

Lemma 3.2. Let $A,B,S\in \mathcal{L}\left({\rm E}\right)$ with $ASB=S.$ If S is one to one, then B is one to one. 

Proof. It follows by S is one to one and $N\left(ASB\right)=N\left(S\right).$ 

Theorem 3.1. Let $A,B\in \mathcal{L}\left({\rm E}\right)$ such that A is injective and $R\left(B\right)$ is dense. If $A^{2}S=S{B}^2$ and $A^{3}S=S{B}^3,$ then $AS=SB$ for some $S\in \mathcal{L}\left({\rm E}\right).$  

Proof. Set $U=AS$ and $V=SB.$ Using $A^{2}S=S{B}^2$ and $A^{3}S=S{B}^3.$ We get $AU=VB$ and $A^{2}U=V{B}^2.$ Then $A\left(AU\right)=AVB=\left(UB\right)B,$ thus $(AV-VB)B=0.$ By $R\left(B\right)$ is dense, we get $B\ne 0$ then $AV=VB.$ From $AU=VB=AV,$ we get $A(U-V)=0.$ From A is injective, then $U=V.$ Hence $AS=SB.$ 

Theorem 3.1. Let $A,B\in \mathcal{L}\left({\rm E}\right)$ with A is injective and $R\left(B\right)$ is dense. If $A^{2}S{B}^2=S$ and $A^{3}S{B}^3=S,$ then $ASB=S$ for some $S\in \mathcal{L}\left({\rm E}\right).$  

Proof. From $A^{2}S{B}^2=S$ and $A^{3}S{B}^3=S,$ then $A^{2}S{B}^2=A^{3}S{B}^3.$ Hence $A^{2}S{B}^2-A^{3}S{B}^3=0.$ Thus $A(A^{2}S{B}^2-ASB)B=0.$ From A is injective and $R\left(B\right)$ is dense, we obtain that $A^{2}S{B}^2-ASB=0.$ Thus $ASB=S.$

4. Left (right) pseudospectum and left (right) condition pseudospectum of bounded linear operators on ultrametric Banach spaces

We introduce the following definitions.

Definition 4.1. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}$ and let $A\in \mathcal{L}\left({\rm E}\right).$  

(i) A is said to be left invertible if there exists $B\in \mathcal{L}\left({\rm E}\right)$ such that $BA=I.$ 

(ii) A is said to be right invertible if there exists $C\in \mathcal{L}\left({\rm E}\right)$ such that $AC=I.$ 

Definition 4.2. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{.}$ Let $A\in \mathcal{L}\left({\rm E}\right),$ the left spectrum ${\sigma }^l\left(A\right)$ of A is defined by

${\sigma }^l\left(A\right)=\{\lambda \in \mathbb{K}:A-\lambda Iis not left invertible in\mathcal{L}({\rm E}\left)\right\}.$

Definition 4.3. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{.}$ Let $A\in \mathcal{L}\left({\rm E}\right),$ the right spectrum ${\sigma }^r\left(A\right)$ of A is defined by

${\sigma }^r\left(A\right)=\{\lambda \in \mathbb{K}:A-\lambda Iis not right invertible in\mathcal{L}({\rm E}\left)\right\}.$

Definition 4.4. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{,}$ let $A\in \mathcal{L}\left({\rm E}\right)$ and $\epsilon >\mathrm{0,}$ the left spectrum ${\sigma }_{\epsilon }^l\left(A\right)$ of A is defined by

${\sigma }_{\epsilon }^l\left(A\right)={\sigma }^l\left(A\right)\cup \{\lambda \in \mathbb{K}:\inf \{\parallel C_l\parallel :C_{l}a left inverse ofA-\lambda I\}>{\epsilon }^{-1}\},$

with the convention $\inf \{\parallel C_l\parallel :C_{l}a left inverse ofA-\lambda I\}=\infty$ if $A-\lambda I$ is not left invertible.

 Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{,}$ let $A\in \mathcal{L}\left({\rm E}\right)$ and $\epsilon >\mathrm{0,}$ the right spectrum ${\sigma }_{\epsilon }^r\left(A\right)$ of A is defined by

${\sigma }_{\epsilon }^r\left(A\right)={\sigma }^r\left(A\right)\cup \{\lambda \in \mathbb{K}:\inf \{\parallel C_r\parallel :C_{r}a right inverse ofA-\lambda I\}>{\epsilon }^{-1}\},$

with the convention $\inf \{\parallel C_r\parallel :C_{r}a right inverse ofA-\lambda I\}=\infty$ if $A-\lambda I$ is not right invertible.

We obtain the following results.

Remark 4.1. From Definition 4.4 and Definition 4.5, we get

${\sigma }^l\left(A\right)\subset {\sigma }_{\epsilon }^l\left(A\right)\subset {\sigma }_{\epsilon }\left(A\right)$

and

${\sigma }^r\left(A\right)\subset {\sigma }_{\epsilon }^r\left(A\right)\subset {\sigma }_{\epsilon }\left(A\right).$

Proposition 4.1. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{,}$ let $A\in \mathcal{L}\left({\rm E}\right)$ and $\epsilon >\mathrm{0,}$ we have 

(i) ${\sigma }^l\left(A\right)={\cap }_{\epsilon >0}{\sigma }_{\epsilon }^l\left(A\right)$ and ${\sigma }^r\left(A\right)={\cap }_{\epsilon >0}{\sigma }_{\epsilon }^r\left(A\right).$ 

(ii) For all ${\epsilon }_1$ and ${\epsilon }_2$ such that $0<{\epsilon }_1<{\epsilon }_2,$ ${\sigma }^l\left(A\right)\subset {\sigma }_{{\epsilon }_1}^l\left(A\right)\subset {\sigma }_{{\epsilon }_2}^l\left(A\right)$ and ${\sigma }^r\left(A\right)\subset {\sigma }_{{\epsilon }_1}^r\left(A\right)\subset {\sigma }_{{\epsilon }_2}^r\left(A\right).$ 

Proof. (i) From Definition 4.4, for any $\epsilon >\mathrm{0,}$ ${\sigma }^l\left(A\right)\subset {\sigma }_{\epsilon }^l\left(A\right).$ Conversely, if $\lambda \in {\cap }_{\epsilon >0}{\sigma }_{\epsilon }^l\left(A\right),$ hence for all $\epsilon >\mathrm{0,}\lambda \in {\sigma }_{\epsilon }^l\left(A\right).$ If $\lambda \cancel{\in }{\sigma }^l\left(A\right),$ then

$\lambda \in \{\lambda \in \mathbb{K}:\inf \{\parallel C_l\parallel :C_{l}a left inverse ofA-\lambda I\}>{\epsilon }^{-1}\},$

taking limits as $\epsilon \to 0^{+}\mathrm{,}$ we get $\inf \{\parallel C_l\parallel :C_{l}a left inverse ofA-\lambda I\}=\infty .$ Thus $\lambda \in {\sigma }^l\left(A\right).$ Similarly, we obtain ${\sigma }^r\left(A\right)={\cap }_{\epsilon >0}{\sigma }_{\epsilon }^r\left(A\right).$

(ii)  For ${\epsilon }_1$ and ${\epsilon }_2$ such that $0<{\epsilon }_1<{\epsilon }_2.$ Let $\lambda \in {\sigma }_{{\epsilon }_1}^l\left(A\right),$ then

$\inf \{\parallel C_l\parallel :C_{l}a left inverse ofA-\lambda I\}>{\epsilon }_1^{-1}>{\epsilon }_2^{-1},$

hence $\lambda \in {\sigma }_{{\epsilon }_2}^l\left(A\right).$ Similarly, we have ${\sigma }^r\left(A\right)\subset {\sigma }_{{\epsilon }_1}^r\left(A\right)\subset {\sigma }_{{\epsilon }_2}^r\left(A\right).$ 

Proposition 4.2. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{,}$ let $A\in \mathcal{L}\left({\rm E}\right)$ and  $\epsilon >0.$Then

$\underset{C\in \mathcal{L}\left({\rm E}\right):\parallel C\parallel <\epsilon }{\cup }{\sigma }^l(A+C)\subset {\sigma }_{\epsilon }^l\left(A\right).$ (4.1)

Proof. If $\lambda \in {\cup }_{C\in \mathcal{L}\left({\rm E}\right):\parallel C\parallel <\epsilon }{\sigma }^l(A+C).$ We argue by contradiction. Suppose that $\lambda \cancel{\in }{\sigma }_{\epsilon }^l\left(A\right),$ hence $\lambda \cancel{\in }{\sigma }^l\left(A\right)$ and $\inf \{\parallel C_l\parallel :C_{l}a left inverse ofA-\lambda I\}\le {\epsilon }^{-1},$ thus $\parallel C{C}_l\parallel <1.$ Let D defined on $\mathcal{E}$ by

$D=\sum _{n=0}^{\infty }{C}_l{(-C{C}_l)}^n.$

One can see that D is well-defined and $D=C_l{(I+C{C}_l)}^{-1}.$ Hence for all $y\in \mathcal{E}\mathrm{,}$ $D(I+C{C}_l)y=C_{l}y.$ Set $y=(A-\lambda I)x,$ we have for all $x\in \mathcal{E}\mathrm{,}$ 

$x=D(I+C{C}_l)(A-\lambda I)x=D(A-\lambda I+C{C}_l(A-\lambda I\left)\right)x=D(A-\lambda I+C)x.$

Hence $A+C-\lambda I$ is left invertible which is contradiction with $\lambda \in {\cup }_{C\in \mathcal{L}\left({\rm E}\right):\parallel C\parallel <\epsilon }{\sigma }^l(A+C).$ Thus, (4) holds.

Theorem 4.1. Let $\mathcal{E}$ be an ultrametric Banach space over a spherically complete field $\mathbb{K}$ such that $\parallel {\rm E}\parallel \subseteq \left|\mathbb{K}\right|,$ let $A\in \mathcal{L}\left({\rm E}\right)$ and $\epsilon >0.$ Then,

${\sigma }_{\epsilon }^l\left(A\right)=\underset{C\in \mathcal{L}\left({\rm E}\right):\parallel C\parallel <\epsilon }{\cup }{\sigma }^l(A+C).$

Proof. According to the proposition 4 the embedding (inclusion) (4) is satisfied.

Conversely, suppose that $\lambda \in {\sigma }_{\epsilon }^l\left(A\right).$ We discuss two cases.

First case: If $\lambda \in {\sigma }^l\left(A\right),$ we may set $C=0.$

Second case: Assume that $\lambda \in {\sigma }_{\epsilon }^l\left(A\right)$ and $\lambda \cancel{\in }{\sigma }^l\left(A\right),$ then for all $C_l$ a left inverse of $A-\lambda I\mathrm{,}$ we have $\parallel C_l\parallel >\frac{1}{\epsilon }.$ Hence, there exists $y\in {\rm E}\backslash \left\{0\right\}$ such that

$\frac{\parallel C_{l}y\parallel }{\parallel y\parallel }>\frac{1}{\epsilon }.$ 4.2)

Set $y=(A-\lambda I)x,$ then $C_{l}y=x.$ From (4.2), we have $\parallel (A-\lambda I)x\parallel <\epsilon \parallel x\parallel .$ Since $\parallel {\rm E}\parallel \subseteq \left|\mathbb{K}\right|,$ then there exists $c\in \mathbb{K}\backslash \left\{0\right\}$ such that $\left|c\right|=\parallel x\parallel .$ Putting $z=c^{-1}x,$ then $\parallel z\parallel =\mathrm{1,}$ hence $\parallel (A-\lambda I)z\parallel \text{}<\text{}\epsilon .$ By Theorem 1.1, there exists $\varphi \in {\mathcal{E}}^{\ast }$ such that $\varphi \left(z\right)\text{}=\text{}1$ and $\parallel \varphi \parallel =\parallel z{\parallel }^{-1}=1.$ Define

for all $y\in {\rm E},Cy=-\varphi \left(y\right)(A-\lambda I)z.$ 

Then $C\in \mathcal{L}\left({\rm E}\right)$ and $\parallel C\parallel <\epsilon ,$ since for all $y\in \mathcal{E}\mathrm{,}$ 

$\parallel Cy\parallel =\parallel \varphi \left(y\right)\parallel \parallel (A-\lambda I)z\parallel <\epsilon \parallel y\parallel .$

Furthermore, we have $(A-\lambda I+C)z=0.$ Thus $A-\lambda I+C$ is not left invertible. Consequently, $\lambda \in {\cup }_{C\in \mathcal{L}\left({\rm E}\right):\parallel C\parallel <\epsilon }{\sigma }^l(A+C).$ 

We continue with the following definitions.

Definition 4.6. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{,}$ let $A\in \mathcal{L}\left({\rm E}\right)$ and $\epsilon >\mathrm{0,}$ the left condition pseudospectrum ${\Lambda }_{\epsilon }^l\left(A\right)$ of A is defined by

${\Lambda }_{\epsilon }^l\left(A\right)={\sigma }^l\left(A\right)\cup \{\lambda \in \mathbb{K}:\inf \{\parallel (A-\lambda I)\parallel \parallel D_l\parallel :D_{l}a left inverse ofA-\lambda I\}>{\epsilon }^{-1}\},$

with the convention $\inf \{\parallel (A-\lambda I)\parallel \parallel D_l\parallel :D_{l}a left inverse ofA-\lambda I\}=\infty$ if $A-\lambda I$ is not left invertible.

Definition 4.7. Let $\mathcal{E}$ be an ultrametric Banach space over $\mathbb{K}\mathrm{,}$ let $A\in \mathcal{L}\left({\rm E}\right)$ and $\epsilon >\mathrm{0,}$ the right condition pseudospectrum ${\Lambda }_{\epsilon }^r\left(A\right)$ of A is defined by

${\Lambda }_{\epsilon }^r\left(A\right)={\sigma }^r\left(A\right)\cup \{\lambda \in \mathbb{K}:\inf \{\parallel A-\lambda I\parallel \parallel D_r\parallel :D_{r}a right inverse ofA-\lambda I\}>{\epsilon }^{-1}\},$

with the convention $\inf \{\parallel A-\lambda I\parallel \parallel D_r\parallel :D_{r}a right inverse ofA-\lambda I\}=\infty$ if $A-\lambda I$ is not right invertible.