On the non-hypercyclicity of normal operators, their exponentials, and symmetric operators

TL;DR

This paper provides a simple proof that normal and symmetric operators, as well as their exponentials under certain spectral conditions, are not hypercyclic in complex Hilbert spaces.

Contribution

It offers a straightforward proof of non-hypercyclicity for normal and symmetric operators and their exponentials, extending known results with minimal complexity.

Findings

Normal operators are not hypercyclic.

Exponentials of normal operators are not hypercyclic under certain spectral conditions.

Symmetric operators are not hypercyclic.

Abstract

We give a simple, straightforward proof of the non-hypercyclicity of an arbitrary (bounded or not) normal operator $A$ in a complex Hilbert space as well as of the collection $\left\{e^{tA}\right\}_{t\ge 0}$ of its exponentials, which, under a certain condition on the spectrum of $A$, coincides with the $C_0$-semigroup generated by it. We also establish non-hypercyclicity for symmetric operators.

Equations

$$A:X\supseteq D(A)\to X$$

$$f\in C^{\infty }(A):=\bigcap _{n=0}^{\infty }D(A^n)$$

$$\mathrm{orb}(f,A):={\left\{A^{n}f\right\}}_{n\in {\mathbb{Z}}_{+}}$$

$${\left\{T(t)f\right\}}_{t\in J}$$

$$F(A):=\underset{\sigma (A)}{\int }F(\lambda )d{E}_A(\lambda )$$

$$f\in D(F(A))\iff \underset{\sigma (A)}{\int }|F(\lambda ){|}^{2}d(E_A(\lambda )f,f)<\infty ,$$

$$\Vert F(A)f{\Vert }^2=\underset{\sigma (A)}{\int }|F(\lambda ){|}^{2}d(E_A(\lambda )f,f)$$

$$A^n=\underset{\sigma (A)}{\int }{\lambda }^{n}d{E}_A(\lambda ),n\in {\mathbb{Z}}_{+},\text{and}{e}^{tA}:=\underset{\sigma (A)}{\int }{e}^{t\lambda }d{E}_A(\lambda ),t\in \mathbb{R}.$$

$$\sigma (A)\subseteq \left\{\lambda \in \mathbb{C}|\mathrm{Re}\text{ignorespaces}\lambda \le \omega \right\}$$

$$y(t)=e^{tA}f,t\ge 0,f\in \bigcap _{t\ge 0}D(e^{tA}),$$

$$y^{\prime }(t)=Ay(t),t\ge 0,$$

$$C^{\infty }(A)\text{and}\bigcap _{t\ge 0}D(e^{tA})$$

$$\bigcup _{\alpha >0}{E}_A({\Delta }_{\alpha })X,\text{where}{\Delta }_{\alpha }:=\left\{\lambda \in \mathbb{C}||\lambda |\le \alpha \right\},\alpha >0,$$

$$\left\{\lambda \in \mathbb{C}|\mathrm{Re}\text{ignorespaces}\lambda \le \omega \right\}$$

$$E_A\left(\left\{\lambda \in \sigma (A)||\lambda |>1\right\}\right)f\ne 0$$

$$E_A\left(\left\{\lambda \in \sigma (A)||\lambda |>1\right\}\right)f=0.$$

$$\begin{gathered} \Vert A^{n}f{\Vert }^2\text{hfill}\text{by eqrefn2}; \\ =\underset{\sigma (A)}{\int }|\lambda {|}^{2n}d(E_A(\lambda )f,f)\ge \underset{\{\lambda \in \sigma (A)||\lambda |>1\}}{\int }|\lambda {|}^{2n}d(E_A(\lambda )f,f) \\ \ge \underset{\{\lambda \in \sigma (A)||\lambda |>1\}}{\int }1d(E_A(\lambda )f,f)=\left(E_A(\{\lambda \in \sigma (A)||\lambda |>1\})f,f\right) \\ ={\Vert E_A(\{\lambda \in \sigma (A)||\lambda |>1\})f\Vert }^2>0,\text{hfill} \end{gathered}$$

$$f=E_A\left(\left\{\lambda \in \sigma (A)||\lambda |>1\right\}\right)f+E_A\left(\left\{\lambda \in \sigma (A)||\lambda |\le 1\right\}\right)f,$$

$$f=E_A\left(\left\{\lambda \in \sigma (A)||\lambda |\le 1\right\}\right)f\ne 0$$

$$\begin{gathered} {\Vert A^{n}f\Vert }^2={\Vert A^n{E}_A\left(\left\{\lambda \in \sigma (A)||\lambda |\le 1\right\}\right)f\Vert }^2 \\ \text{hfill}\text{by eqrefn2 and the properties of the operational calculus}; \\ =\underset{\{\lambda \in \sigma (A)||\lambda |\le 1\}}{\int }|\lambda {|}^{2n}d(E_A(\lambda )f,f)\le \underset{\{\lambda \in \sigma (A)||\lambda |\le 1\}}{\int }1d(E_A(\lambda )f,f) \\ =\left(E_A(\{\lambda \in \sigma (A)||\lambda |\le 1\})f,f\right)={\Vert E_A(\{\lambda \in \sigma (A)||\lambda |\le 1\})f\Vert }^2=\Vert f{\Vert }^2,\text{hfill} \end{gathered}$$

$$E_A\left(\left\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda >0\right\}\right)f\ne 0$$

$$E_A\left(\left\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda >0\right\}\right)f=0.$$

$$\begin{gathered} \Vert e^{tA}f{\Vert }^2\text{hfill}\text{by eqrefn2}; \\ =\underset{\sigma (A)}{\int }{\left|e^{t\lambda }\right|}^{2}d(E_A(\lambda )f,f)=\underset{\sigma (A)}{\int }{e}^{2t\mathrm{Re}\text{ignorespaces}\lambda }d(E_A(\lambda )f,f) \\ \ge \underset{\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda >0\}}{\int }{e}^{2t\mathrm{Re}\text{ignorespaces}\lambda }d(E_A(\lambda )f,f)\ge \underset{\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda >0\}}{\int }1d(E_A(\lambda )f,f) \\ =\left(E_A(\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda >0\})f,f\right)={\Vert E_A(\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda >0\})f\Vert }^2>0,\text{hfill} \end{gathered}$$

$$f=E_A\left(\left\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda >0\right\}\right)f+E_A\left(\left\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\right\}\right)f,$$

$$f=E_A\left(\left\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\right\}\right)f\ne 0$$

$$\begin{gathered} {\Vert e^{tA}f\Vert }^2={\Vert e^{tA}{E}_A\left(\left\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\right\}\right)f\Vert }^2 \\ \text{hfill}\text{by eqrefn2 and the properties of the operational calculus}; \\ =\underset{\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\}}{\int }{\left|e^{t\lambda }\right|}^{2}d(E_A(\lambda )f,f)=\underset{\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\}}{\int }{e}^{2t\mathrm{Re}\text{ignorespaces}\lambda }d(E_A(\lambda )f,f) \\ \le \underset{\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\}}{\int }1d(E_A(\lambda )f,f)=\left(E_A(\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\})f,f\right) \\ ={\Vert E_A(\{\lambda \in \sigma (A)|\mathrm{Re}\text{ignorespaces}\lambda \le 0\})f\Vert }^2=\Vert f{\Vert }^2,\text{hfill} \end{gathered}$$

$$\overline{R(A-\lambda I)}=X$$

$${\sigma }_r(A)=\varnothing .$$

$$\exists g\in X\setminus \{0\}:A^{\ast }g=\lambda g,$$

$$\forall n\in \mathbb{N}:{(A^{\ast })}^{n}g={\lambda }^{n}g.$$

Full text

1. Introduction

In [19], furnished is a straightforward proof of the non-hypercyclicity of an arbitrary (bounded or not) scalar type spectral operator $A$ in a complex Banach space as well as of the collection $\left\{e^{tA}\right\}_{t\geq 0}$ of its exponentials (see, e.g., [7]), the important particular case of a normal operator $A$ in a complex Hilbert space (see, e.g., [6, 23]) following immediately.

Without the need to resort to the machinery of dual space, we provide a shorter, simpler, and more transparent direct proof for the normal operator case, in particular, generalizing the known result [10, Corollary $5.31$] for bounded normal operators, and further establish non-hypercyclicity for symmetric operators (see, e.g., [1]).

Definition 1.1** (Hypercyclicity).**

Let

[TABLE]

($D(\cdot)$ is the domain of an operator) be a (bounded or unbounded) linear operator in a (real or complex) Banach space $(X,\|\cdot\|)$. A vector

[TABLE]

($A^{0}:=I$, $I$ is the identity operator on $X$) is called hypercyclic if its orbit

[TABLE]

under $A$ (${\mathbb{Z}}_{+}:=\left\{0,1,2,\dots\right\}$ is the set of nonnegative integers) is dense in $X$.

Linear operators possessing hypercyclic vectors are said to be hypercyclic.

More generally, a collection $\left\{T(t)\right\}_{t\in J}$ ($J$ is a nonempty indexing set) of linear operators in $X$ is called hypercyclic if it possesses hypercyclic vectors, i.e., such vectors $\displaystyle f\in\bigcap_{t\in J}D(T(t))$, whose orbit

[TABLE]

is dense in $X$.

Cf. [10, 11, 25, 3, 4, 20, 21].

Remarks 1.1**.**

• •

Clearly, hypercyclicity for a linear operator can only be discussed in a separable Banach space setting. Generally, for a collection of operators, this need not be the case.

• •

For a hypercyclic linear operator $A$, dense in $(X,\|\cdot\|)$ is the subspace $C^{\infty}(A)$ (cf., e.g., [19]), which, in particular, implies that any hypercyclic linear operator is densely defined (i.e., $\overline{D(A)}=X$).

• •

Bounded normal operators on a complex Hilbert space are known to be non-hypercyclic [10, Corollary $5.31$].

2. Preliminaries

Here, we briefly outline certain preliminaries essential for the subsequent discourse (for more, see, e.g., [12, 13, 14]).

Henceforth, unless specified otherwise, $A$ is a normal operator in a complex Hilbert space $(X,(\cdot,\cdot),\|\cdot\|)$ with strongly $\sigma$-additive spectral measure (the resolution of the identity) $E_{A}(\cdot)$ assigning to Borel sets of the complex plane ${\mathbb{C}}$ orthogonal projection operators on $X$ and having the operator’s spectrum $\sigma(A)$ as its support [6, 23].

Associated with a normal operator $A$ is the Borel operational calculus assigning to any Borel measurable function $F:\sigma(A)\to{\mathbb{C}}$ a normal operator

[TABLE]

with

[TABLE]

where $(E_{A}(\cdot)f,f)$ is a Borel measure, in which case

[TABLE]

[6, 23].

In particular,

[TABLE]

Provided

[TABLE]

with some $\omega\in{\mathbb{R}}$, the collection of exponentials $\left\{e^{tA}\right\}_{t\geq 0}$ is the $C_{0}$-semigroup generated by $A$ [8, 23].

Remarks 2.1**.**

• •

By [12, Theorem $3.1$], the orbits

[TABLE]

describe all weak/mild solutions of the abstract evolution equation

[TABLE]

(see [2], cf. also [8, Ch. II, Definition 6.3]).

• •

The subspaces

[TABLE]

of all possible initial values for the corresponding orbits are dense in $X$ since they contain the subspace

[TABLE]

which is dense in $X$ and coincides with the class ${\mathscr{E}}^{\{0\}}(A)$ of the entire vectors of $A$ of exponential type (see, e.g., [9, 24], cf. also [15]).

3. Normal Operators and Their Exponentials

We are to prove [19, Corollary $4.1$] directly generalizing in part [10, Corollary $5.31$].

Theorem 3.1** ([19, Corollary $4.1$]).**

An arbitrary normal, in particular self-adjoint, operator $A$ in a nonzero complex Hilbert space $(X,(\cdot,\cdot),\|\cdot\|)$ with spectral measure $E_{A}(\cdot)$ is not hypercyclic and neither is the collection $\left\{e^{tA}\right\}_{t\geq 0}$ of its exponentials, which, provided the spectrum of $A$ is located in a left half-plane

[TABLE]

with some $\omega\in{\mathbb{R}}$, is the $C_{0}$-semigroup generated by $A$.

Proof.

Let $f\in C^{\infty}(A)\setminus\{0\}$ be arbitrary.

There are two possibilities: either

[TABLE]

or

[TABLE]

In the first case, for any $n\in{\mathbb{Z}}_{+}$,

[TABLE]

which implies that the orbit $\operatorname{orb}(f,A)$ of $f$ under $A$ cannot approximate the zero vector, and hence, is not dense in $X$.

In the second case, since

[TABLE]

we infer that

[TABLE]

and hence, for any $n\in{\mathbb{Z}}_{+}$,

[TABLE]

which also implies that the orbit $\operatorname{orb}(f,A)$ of $f$ under $A$, being bounded, is not dense in $X$ and completes the proof for the operator case.

Now, let us consider the case of the exponential collection $\left\{e^{tA}\right\}_{t\geq 0}$ assuming that $\displaystyle f\in\bigcap_{t\geq 0}D(e^{tA})\setminus\{0\}$ is arbitrary.

There are two possibilities: either

[TABLE]

or

[TABLE]

In the first case, for any $t\geq 0$,

[TABLE]

which implies that the orbit $\left\{e^{tA}f\right\}_{t\geq 0}$ of $f$ cannot approximate the zero vector, and hence, is not dense in $X$.

In the second case, since

[TABLE]

we infer that

[TABLE]

and hence, for any $t\geq 0$,

[TABLE]

which also implies that the orbit $\left\{e^{tA}f\right\}_{t\geq 0}$ of $f$, being bounded, is not dense on $X$ and completes the proof of the exponential case and the entire statement. ∎

4. Symmetric Operators

The following generalizes in part [10, Lemma $2.53$ (a)] to the case of a densely defined unbounded linear operator in a Hilbert space.

Lemma 4.1**.**

Let $A$ be a hypercyclic linear operator in a nonzero Hilbert space $(X,(\cdot,\cdot),\|\cdot\|)$ over the scalar field ${\mathbb{F}}$ of real or complex numbers (i.e., ${\mathbb{F}}={\mathbb{R}}$ or ${\mathbb{F}}={\mathbb{C}}$). Then

1. (1)

the adjoint operator $A^{*}$ has no eigenvalues, or equivalently, for any $\lambda\in{\mathbb{F}}$, the range of the operator $A-\lambda I$ is dense in $X$, i.e.,

[TABLE]

($R(\cdot)$* is the range of an operator);* 2. (2)

provided the space $X$ is complex (i.e., ${\mathbb{F}}={\mathbb{C}}$) and the operator $A$ is closed, the residual spectrum of $A$ is empty, i.e.,

[TABLE]

Proof.

1. (1)

Let $f\in X$ be a hypercyclic vector for $A$.

We proceed by contradiction, assuming that the adjoint operator $A^{*}$, which exists since $A$ is densely defined (see Remarks 1.1), has an eigenvalue $\lambda\in{\mathbb{F}}$, and hence,

[TABLE]

which, in particular, implies that $g\in C^{\infty}(A^{*}):=\bigcap_{n=0}^{\infty}D\left({(A^{*})}^{n}\right)$ and

[TABLE]

In view of the above, we have inductively:

[TABLE]

the conjugation being superfluous when the space is real.

Since $g\neq 0$, by the Riesz representation theorem (see, e.g., [17, 18]), the hypercyclicity of $f$ implies that the set

[TABLE]

is dense in ${\mathbb{F}}$, which contradicts the fact that the same set

[TABLE]

is clearly not.

Thus, the adjoint operator $A^{*}$ has no eigenvalues.

The rest of the statement of part (1) immediately follows from the orthogonal sum decomposition

[TABLE]

the conjugation being superfluous when the space is real, (see, e.g., [18]). 2. (2)

Suppose that the space $X$ is complex (i.e., ${\mathbb{F}}={\mathbb{C}}$) and the operator $A$ is closed. Recalling that

[TABLE]

(see, e.g., [16, 18]), we infer from part (1) that

[TABLE]

∎

We immediately arrive at the following

Proposition 4.1** (Non-Hypercyclicity Test).**

Any densely defined closed linear operator $A$ in a nonzero complex Hilbert space $X$ with a nonempty residual spectrum (i.e., $\sigma_{r}(A)\neq\emptyset$) is not hypercyclic.

Now, we are ready to prove the subsequent

Theorem 4.1**.**

An arbitrary symmetric operator $A$ in a complex Hilbert space $X$ is not hypercyclic.

Proof.

Since

[TABLE]

without loss of generality, we can regard the symmetric operator $A$ to be closed (see, e.g., [5]).

If both deficiency indices of the operator $A$ are equal to zero, $A$ is self-adjoint ($A=A^{*}$) (see, e.g., [1]), and hence, by Theorem 3.1, is not hypercyclic.

If at least one of the deficiency indices of the operator $A$ is nonzero, then

[TABLE]

(see, e.g., [1, 17]), and hence, by Proposition 4.1, $A$ is not hypercyclic. ∎

5. Some Examples

Examples 5.1**.**

In the complex Hilbert space $L_{2}({\mathbb{R}})$, the self-adjoint differential operator $A:=i\dfrac{d}{dx}$ ($i$ is the imaginary unit) with the domain

[TABLE]

($AC(\cdot)$ is the set of absolutely continuous functions on an interval) is non-hypercyclic by Theorem 3.1 (cf. [19, Corollary $5.1$]). 2. 2.

In the complex Hilbert space $L_{2}(0,\infty)$, the symmetric differential operator $A:=i\dfrac{d}{dx}$ with the domain

[TABLE]

and deficiency indices $(0,1)$ is non-hypercyclic by Theorem 4.1. 3. 3.

In the complex Hilbert space $L_{2}(0,2\pi)$, the symmetric differential operator $A:=i\dfrac{d}{dx}$ with the domain

[TABLE]

and deficiency indices $(1,1)$ is non-hypercyclic by Theorem 4.1.

Cf. [1, Sections $49$ and $80$].