On the Non-Hypercyclicity of Normal Operators, Their Exponentials, and Symmetric Operators

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.

1. Introduction

In [1], 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 ${\left\{e^{tA}\right\}}_{t\ge 0}$ of its exponentials (see, e.g., [2]), the important particular case of a normal operator A in a complex Hilbert space (see, e.g., [3,4]) 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 [5] [Corollary $5.31$] for bounded normal operators, and further establish non-hypercyclicity for symmetric operators (see, e.g., [6]).

Definition 1

(Hypercyclicity). Let

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

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

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

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

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

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

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

is dense in X.

Cf. [5,7,8,9,10,11,12].

Remark 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 $(X,\parallel ·\parallel )$ is the subspace $C^{\infty }\left(A\right)$ (cf., e.g., [1]), which, in particular, implies that any hypercyclic linear operator is densely defined (i.e., $\overline{D\left(A\right)}=X$).
• Bounded normal operators on a complex Hilbert space are known to be non-hypercyclic [5] [Corollary $5.31$].

2. Preliminaries

Here, we briefly outline certain preliminaries essential for the subsequent discourse (for more, see, e.g., [13,14,15]).

Henceforth, unless specified otherwise, A is a normal operator in a complex Hilbert space $(X,(·,·),\parallel ·\parallel )$ with strongly $\sigma$-additive spectral measure (the resolution of the identity) $E_A(·)$ assigning to Borel sets of the complex plane $\mathbb{C}$ orthogonal projection operators on X and having the operator’s spectrum $\sigma \left(A\right)$ as its support [3,4].

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

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

with

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

where $(E_A(·)f,f)$ is a Borel measure, in which case [3,4]

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

(1)

In particular,

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

Provided

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

with some $\omega \in \mathbb{R}$, the collection of exponentials ${\left\{e^{tA}\right\}}_{t\ge 0}$ is the $C_0$-semigroup generated by A [4,16].

Remark 2.

• By [13] [Theorem 1], the orbits

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

  (2)
  describe all weak/mild solutions of the abstract evolution equation

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

  (3)
  (see [17], cf. also [16] [Ch. II, Definition 6.3]).
• The subspaces

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

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

  $$\bigcup _{\alpha >0}{E}_A\left({\Delta }_{\alpha }\right)X,where{\Delta }_{\alpha }:=\left\{\lambda \in \mathbb{C}\left|\right|\lambda |\le \alpha \right\},\alpha >0,$$
  which is dense in X and coincides with the class ${\mathcal{E}}^{\left\{0\right\}}$(A) of the entire vectors of A of exponential type (see, e.g., [18,19], cf. also [20]).

3. Normal Operators and Their Exponentials

We are to prove [1] [Corollary $4.1$] directly generalizing in part [5] [Corollary $5.31$].

Theorem 1

([1] (Corollary $4.1$)). An arbitrary normal, in particular self-adjoint, operator A in a nonzero complex Hilbert space $(X,(·,·),\parallel ·\parallel )$ with spectral measure $E_A(·)$ is not hypercyclic and neither is the collection ${\left\{e^{tA}\right\}}_{t\ge 0}$ of its exponentials, which, provided the spectrum of A is located in a left half-plane

$$\left\{\lambda \in \mathbb{C}|Re\lambda \le \omega \right\},$$

with some $\omega \in \mathbb{R}$, is the $C_0$-semigroup generated by A.

Proof. 

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

There are two possibilities: either

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

or

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

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

$$\begin{array}{c}\parallel A^n{f\parallel }^2\mathrm{by}(1);\hfill \\ \multicolumn{1}{c}{=\underset{\sigma \left(A\right)}{\int }{\left|\lambda \right|}^{2n}d(E_A\left(\lambda \right)f,f)\ge \underset{\{\lambda \in \sigma (A\left)\right|\left|\lambda \right|>1\}}{\int }{\left|\lambda \right|}^{2n}d(E_A\left(\lambda \right)f,f)}\\ \multicolumn{1}{c}{\ge \underset{\{\lambda \in \sigma (A\left)\right|\left|\lambda \right|>1\}}{\int }1d(E_A\left(\lambda \right)f,f)=\left(E_A\left(\right\{\lambda \in \sigma \left(A\right)\left|\right|\lambda |>1\})f,f\right)}\\ ={∥E_A\left(\right\{\lambda \in \sigma \left(A\right)\left|\right|\lambda |>1\})f∥}^2>0,\hfill \end{array}$$

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

In the second case, since

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

we infer that

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

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

$$\begin{array}{c}{∥A^{n}f∥}^2={∥A^n{E}_A\left(\left\{\lambda \in \sigma \left(A\right)\left|\right|\lambda |\le 1\right\}\right)f∥}^2\hfill \\ \hfill \mathrm{by}\left(1\right)\mathrm{and}\mathrm{the}\mathrm{properties}\mathrm{of}\mathrm{the}operationalcalculus;\\ \multicolumn{1}{c}{=\underset{\{\lambda \in \sigma (A\left)\right|\left|\lambda \right|\le 1\}}{\int }{\left|\lambda \right|}^{2n}d(E_A\left(\lambda \right)f,f)\le \underset{\{\lambda \in \sigma (A\left)\right|\left|\lambda \right|\le 1\}}{\int }1d(E_A\left(\lambda \right)f,f)}\\ =\left(E_A\left(\right\{\lambda \in \sigma \left(A\right)\left|\right|\lambda |\le 1\left\}\right)f,f\right)={∥E_A\left(\right\{\lambda \in \sigma \left(A\right)\left|\right|\lambda |\le 1\left\}\right)f∥}^2={\parallel f\parallel }^2,\hfill \end{array}$$

which also implies that the orbit $orb(f,A)$ 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 ${\left\{e^{tA}\right\}}_{t\ge 0}$ assuming that ${\displaystyle f\in \bigcap _{t\ge 0}D\left(e^{tA}\right)\backslash \left\{0\right\}}$ is arbitrary.

There are two possibilities: either

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

or

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

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

$$\begin{array}{c}\parallel e^{tA}{f\parallel }^2\mathrm{by}(1);\hfill \\ \multicolumn{1}{c}{=\underset{\sigma \left(A\right)}{\int }{\left|e^{t\lambda }\right|}^{2}d(E_A\left(\lambda \right)f,f)=\underset{\sigma \left(A\right)}{\int }{e}^{2tRe\lambda }d(E_A\left(\lambda \right)f,f)}\\ \multicolumn{1}{c}{\ge \underset{\{\lambda \in \sigma (A\left)\right|Re\lambda >0\}}{\int }{e}^{2tRe\lambda }d(E_A\left(\lambda \right)f,f)\ge \underset{\{\lambda \in \sigma (A\left)\right|Re\lambda >0\}}{\int }1d(E_A\left(\lambda \right)f,f)}\\ =\left(E_A\left(\{\lambda \in \sigma \left(A\right)|Re\lambda >0\}\right)f,f\right)={∥E_A\left(\{\lambda \in \sigma \left(A\right)|Re\lambda >0\}\right)f∥}^2>0,\hfill \end{array}$$

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

In the second case, since

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

we infer that

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

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

$$\begin{array}{c}{∥e^{tA}f∥}^2={∥e^{tA}{E}_A\left(\left\{\lambda \in \sigma \left(A\right)|Re\lambda \le 0\right\}\right)f∥}^2\hfill \\ \hfill \mathrm{by}\left(1\right)\mathrm{and}\mathrm{the}\mathrm{properties}\mathrm{of}\mathrm{the}operationalcalculus;\\ \multicolumn{1}{c}{=\underset{\{\lambda \in \sigma (A\left)\right|Re\lambda \le 0\}}{\int }{\left|e^{t\lambda }\right|}^{2}d(E_A\left(\lambda \right)f,f)=\underset{\{\lambda \in \sigma (A\left)\right|Re\lambda \le 0\}}{\int }{e}^{2tRe\lambda }d(E_A\left(\lambda \right)f,f)}\\ \multicolumn{1}{c}{\le \underset{\{\lambda \in \sigma (A\left)\right|Re\lambda \le 0\}}{\int }1d(E_A\left(\lambda \right)f,f)=\left(E_A\left(\{\lambda \in \sigma \left(A\right)|Re\lambda \le 0\}\right)f,f\right)}\\ ={∥E_A\left(\{\lambda \in \sigma \left(A\right)|Re\lambda \le 0\}\right)f∥}^2={\parallel f\parallel }^2,\hfill \end{array}$$

which also implies that the orbit ${\left\{e^{tA}f\right\}}_{t\ge 0}$ 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 [5] [Lemma $2.53$ (a)] to the case of a densely defined unbounded linear operator in a Hilbert space.

Lemma 1.

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

• 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.,

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

  ($R(·)$ is the range of an operator);
• 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.,

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

Proof. 

• 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 Remark 1), has an eigenvalue $\lambda \in \mathbb{F}$, and hence,

  $$\exists g\in X\backslash \left\{0\right\}:A^{*}g=\lambda g,$$

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

  $$\forall n\in \mathbb{N}:{\left(A^{*}\right)}^{n}g={\lambda }^{n}g.$$
  In view of the above, we have inductively:

  $$\forall n\in \mathbb{N}:(A^{n}f,g)=(A^{n-1}f,A^{*}g)=(f,{\left(A^{*}\right)}^{n}g)=(f,{\lambda }^{n}g)={\overline{\lambda }}^n(f,g),$$

  the conjugation being superfluous when the space is real.
  Since $g\ne 0$, by the Riesz representation theorem (see, e.g., [21,22]), the hypercyclicity of f implies that the set

  $${\left\{(A^{n}f,g)\right\}}_{n\in \mathbb{N}}$$

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

  $${\left\{{\overline{\lambda }}^n(f,g)\right\}}_{n\in \mathbb{N}}$$

  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

  $$X=ker(A^{*}-\overline{\lambda }I)\oplus \overline{R(A-\lambda I)},\lambda \in \mathbb{F},$$

  the conjugation being superfluous when the space is real, (see, e.g., [21]).
• Suppose that the space X is complex (i.e., $\mathbb{F}=\mathbb{C}$) and the operator A is closed. Recalling that

  $${\sigma }_r\left(A\right)=\left\{\lambda \in \mathbb{C}|A-\lambda I\mathrm{is}\mathrm{one}-\mathrm{to}-\mathrm{one}\mathrm{and}\overline{R(A-\lambda I)}\ne X\right\}$$

  (see, e.g., [21,23]), we infer from part (1) that

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

□

We immediately arrive at the following

Proposition 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., ${\sigma }_r\left(A\right)\ne \varnothing$) is not hypercyclic.

Now, we are ready to prove the subsequent.

Theorem 2.

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

Proof. 

Since

$$A\subseteq A^{*},$$

without loss of generality, we can regard the symmetric operator A to be closed (see, e.g., [24]).

If both deficiency indices of the operator A are equal to zero, A is self-adjoint ($A=A^{*}$) (see, e.g., [6]), and hence, by Theorem 1, is not hypercyclic.

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

$${\sigma }_r\left(A\right)\ne \varnothing$$

(see, e.g., [6,22]), and hence, by Proposition 1, A is not hypercyclic. □

5. Some Examples

Example 1.

• In the complex Hilbert space $L_2\left(\mathbb{R}\right)$, the self-adjoint differential operator $A:=i{\displaystyle \frac{d}{dx}}$ (i is the imaginary unit) with the domain

  $$D\left(A\right):=W_2^1\left(\mathbb{R}\right):=\left\{f\in L_2\left(\mathbb{R}\right)|f(·)\in AC\left(\mathbb{R}\right),f^{\prime }\in L_2\left(\mathbb{R}\right)\right\}$$
  ($AC(·)$ is the set of absolutely continuous functions on an interval) is non-hypercyclic by Theorem 1 (cf. [1] [Corollary $5.1$]).
• In the complex Hilbert space $L_2(0,\infty )$, the symmetric differential operator $A:=i{\displaystyle \frac{d}{dx}}$ with the domain

  $$D\left(A\right):=\left\{f\in L_2(0,\infty )|f(·)\in AC[0,\infty ),f^{\prime }\in L_2(0,\infty ),f\left(0\right)=0\right\}$$

  and deficiency indices $(0,1)$ is non-hypercyclic by Theorem 2.
• In the complex Hilbert space $L_2(0,2\pi )$, the symmetric differential operator $A:=i{\displaystyle \frac{d}{dx}}$ with the domain

  $$D\left(A\right):=\left\{f\in L_2(0,2\pi )|f(·)\in AC[0,2\pi ],f^{\prime }\in L_2(0,2\pi ),f\left(0\right)=f\left(2\pi \right)=0\right\}$$

  and deficiency indices $(1,1)$ is non-hypercyclic by Theorem 2.

Cf. [6] (Sections 49 and 80).