13.6 Unbounded operators

中文

A linear operator $A$ on a Hilbert space is unbounded if for any $M>0$ there exists a vector $u$ such that $\Vert Au\Vert \ge M\Vert u\Vert$ . Very few interesting examples of unbounded operators are defined on all of $\mathcal{H}$ – for self-adjoint operators, there are none at all. It is therefore usual to consider an unbounded operator $A$ as not being necessarily defined over all of $\mathcal{H}$ but only on some vector subspace $D_{\mathcal{A}}\subseteq \mathcal{H}$ called the domain of $A$. Its range is defined as the set of vectors that are mapped onto, $R_A=A(D_A)$ . In general we will refer to a pair $(A,D_A)$ , where $D_A$ is a vector subspace of $\mathcal{H}$ and $A:D_A\to R_A\subseteq \mathcal{H}$ is a linear map, as being an operator in $\mathcal{H}$. Often we will simply refer to the operator $A$ when the domain $D_A$ is understood.

We say the domain $D_A$ is a dense subspace of $\mathcal{H}$ if for every vector $u\in \mathcal{H}$ and any $ϵ>0$ there exists a vector $v\in D_A$ such that $\Vert u-v\Vert <ϵ$ . The operator $A$ is then said to be densely defined.

We say $A$ is an extension of $B$, written $B\subseteq A,\mathrm{if}{D}_B\subseteq D_A$ and $A{|}_{D_B}=B$ . Two operators $(A,D_A)$ and $(B,D_B)$ in $\mathcal{H}$ are called equal if and only if they are extensions of each other – their domains are equal, $D_A=D_B$ and $Au=Bu$ for all $u\in D_A$

For any two operators in we must be careful about simple operations such as addition $A+B$ and multiplication $AB$. The former only exists on the domain $D_{A+B}=D_A\cap D_B$ , while the latter only exists on the set $B^{-1}(R_B\cap D_A)$ ). Thus operators in do not form a vector space or algebra in any natural sense.

Example 13.16

In $\mathcal{H}={\ell }^2$ let $A:\mathcal{H}\to \mathcal{H}$ be the operator defined by

$$(Ax{)}_n=\frac{1}{n}{x}_n.$$

This operator is bounded, hermitian and has domain $D_A=\mathcal{H}$ since

$$\sum _{n=1}^{\infty }{\left|x_n\right|}^2<\infty ⟹\sum _{n=1}^{\infty }{\left|\frac{x_n}{n}\right|}^2<\infty .$$

The range of this operator is

$$R_A=\left\{y\mid \sum _{n=1}^{\infty }{n}^2|y_n{|}^2<\infty \right\},$$

which is dense in ${\ell }^2-$ since every $x\in {\ell }^2$ can be approximated arbitrarily closely by, for example, a finite sum ${\sum }_{n=1}^N{x}_n{e}_n$ where $e_n$ are the standard basis vectors having components $(e_n{)}_m={\delta }_{nm}$ . The inverse operator $A^{-1}$ , defined on the dense domain $D_{A^{-1}}=R_A$ , is unbounded since

$$\Vert A^{-1}{e}_n\Vert =\Vert n{e}_n\Vert =n\to \infty .$$

Example 13.17

In the Hilbert space $L^2(\mathbb{R})$ of equivalence classes of square integrable functions (see Example 13.4), set $D$ to be the vector subspace of elements $\tilde{\phi }$ having a representative $\phi$ from the $C^{\infty }$ functions on $\mathbb{R}$ of compact support. This is essentially the space of test functions ${\mathcal{D}}^{\infty }(\mathbb{R})$ defined in Chapter 12. An argument similar to that outlined in Example 13.5 shows that $D$ is a dense subspace of $L^2(\mathbb{R})$ . We define the position operator $Q:D\to D\subset L^2(\mathbb{R})$ by $Q\tilde{\phi }=\widetilde{x\phi }$ . We may write this more informally as

$$(Q\phi )(x)=x\phi (x).$$

Similarly the momentum operator $P:D\to D$ is defined by

$$P\phi (x)=-i\frac{\mathrm{d}}{\mathrm{d}x}\phi (x).$$

Both these operators are evidently linear on their domains.

Exercise

Show that the position and momentum operators in $L^2(\mathbb{R})$ are unbounded.

If $A$ is a bounded operator defined on a dense domain $D_A$ , it has a unique extension to all of $\mathcal{H}$ (see Problem 13.30). We may always assume then that a bounded operator is defined on all of $\mathcal{H},$ , and when we refer to a densely defined operator whose domain is a proper subspace of $\mathcal{H}$ we implicitly assume it to be an unbounded operator.

Self-adjoint and symmetric operators

Lemma 13.20 · Orthogonality criterion on a dense domain

If $D_A$ is a dense domain and $u$ a vector in $\mathcal{H}$ such that $⟨u|v⟩=0$ for all $v\in D_A$ , then $u=0$

Proof

Let $w$ be any vector in $\mathcal{H}$ and $ϵ>0$ . Since $D_A$ is dense there exists a vector $v\in D_A$ such that $\Vert w-v\Vert <ϵ$ . By the Cauchy–Schwarz inequality

$$|⟨u|w⟩|=|⟨u|w-v⟩|\le \Vert u\Vert \Vert w-v\Vert <ϵ\Vert u\Vert .$$

Since $ϵ$ is an arbitrary positive number, $⟨u|w⟩=0$ for all $w\in \mathcal{H};$ hence $u=0,$

$\mathrm{If}\left(A,D_A\right)$ is an operator in $\mathcal{H}$ with dense domain $D_A$ , then let $D_{A^{\ast }}$ be defined by

$$u\in D_{A^{\ast }}⟺\exists u^{\ast }\text{ such that }⟨u|Av⟩=⟨u^{\ast }|v⟩,\forall v\in D_A.$$

If $u\in D_A$ we set $A^{\ast }u=u^{\ast }$ . This is uniquely defined, for if $⟨u_1^{\ast }-u_2^{\ast }|\mathbf{v}⟩=0$ for all $v\in D_A$ then $u_1^{\ast }=u_2^{\ast }$ by Lemma 13.20. The operator $\left(A^{\ast },D_{A^{\ast }}\right)$ is called the adjoint of $(A,D_A)$

We say a densely defined operator $(A,D_A)$ in $\mathcal{H}$ is closed if for every sequence $u_n\in$ $D_A$ such that $u_n\to u$ and $A{u}_n\to v$ it follows that $u\in D_A$ and $Au=v$ . Another way of expressing this is to say that an operator is closed if and only if its graph $G_A=\{(x,Ax)|x\in$ $D_A\}$ is a closed subset of the product set $\mathcal{H}\times \mathcal{H}$ . The notion of closedness is similar to continuity, but differs in that we must assert the limit $A{u}_n\to v$ , while for continuity it is deduced. Clearly every continuous operator is closed, but the converse does not hold in general.

Theorem 13.21 · The adjoint of a densely defined operator is closed

If $A$ is a densely defined operator then its adjoint $A^{\ast }$ is closed.

Proof

Let $y_n$ be any sequence of vectors in $D_{A^{\ast }}$ such that $y_n\to y$ and $A^{\ast }{y}_n\to z$ . Then for all $x\in D_A$

$$⟨y|Ax⟩=\underset{n\to \infty }{\lim }⟨y_n|Ax⟩=\underset{n\to \infty }{\lim }⟨A^{\ast }{y}_n|x⟩=⟨z|x⟩.$$

Since $D_A$ is a dense domain, it follows from Lemma 13.20 that $y\in D_A,$ and $A^{\ast }y=z$ -

Example 13.18

Let $\mathcal{H}$ be a separable Hilbert space with complete orthonormal basis $e_n$ $(n=0,1,2,\dots )$ . Let the operators $a$ and $a^{\ast }$ be defined by

$$a{e}_n=\sqrt{n}{e}_{n-1},a^{\ast }{e}_n=\sqrt{n+1}{e}_{n+1}.$$

The effect on a typical vector $x={\sum }_{n=0}^{\infty }{x}_n{e}_n$ , where ${\mathbf{x}}_n=⟨\mathbf{x}|e_n⟩$ , is

$$ax=\sum _{n=0}^{\infty }{x}_{n+1}\sqrt{n+1}{e}_n,a^{\ast }x=\sum _{n=1}^{\infty }{x}_{n-1}\sqrt{n}{e}_n.$$

The operator $a^{\ast }$ is the adjoint of $a$ since

$$⟨a^{\ast }y|x⟩=⟨y|ax⟩=\sum _{n=1}^{\infty }\overline{y_n}\sqrt{n+1}{x}_{n+1}$$

and both operators have domain of definition

$$D=D_a=D_{a\ast }=\left\{y|\sum _{n=1}|y_n{|}^{2}n<\infty \right\},$$

which is dense in $\mathcal{H}$ (see Example 13.16). In physics, $\mathcal{H}$ is the symmetric Fock space, in which $e_n$ represents $n$ identical (bosonic) particles in a given state, and $a^{\ast }$ and $a$ are interpreted as creation operator and annihilation operator, respectively.

Exercise

Show that $N=a^{\ast }a$ is the particle number operator, $N{e}_n=n{e}_n$ , and the commutator $i$ $[a,a^{\ast }]=a{a}^{\ast }-a^{\ast }a=I$ . What are the domains of validity of these equations?

Theorem 13.22 · Operator extensions and reversed inclusion of adjoints

$If(A,D_A)$ and $(B,D_B)$ are densely defined operators in $\mathcal{H}$ then $A\subseteq$ $B⟹B^{\ast }\subseteq A^{\ast }$

Proof

If $A\subseteq B$ then for any vectors $u\in D_A$ and $v\in D_{B^{\ast }}$

$$⟨v\mid Au⟩=⟨v\mid Bu⟩=⟨B^{\ast }v\mid u⟩.$$

Hence $v\in D_{A^{\ast }}$ , so that $D_{B^{\ast }}\subseteq D_{A^{\ast }}$ and

$$⟨v\mid Au⟩=⟨A^{\ast }v\mid u⟩=⟨B^{\ast }v\mid u⟩.$$

By Lemma 13.20 $A^{\ast }v=B^{\ast }v$ , hence $B^{\ast }\subseteq A^{\ast }$

An operator $(A,D_A)$ on a dense domain in $\mathcal{H}$ is said to be self-adjoint if $A=A^{\ast }$ . This means that not only is $Au=A^{\ast }u$ wherever both sides are defined, but also that the domains are equal, $D_A=D_{A^{\ast }}$ . By Theorem 13.21 every self-adjoint operator is closed. This is not the only definition that generalizes the concept of a hermitian operator to unbounded operators. The following related definition is also useful. A densely defined operator $(A,D_A)$ in $\mathcal{H}$ is called a symmetric operator if $⟨Au\mid v⟩=⟨u\mid Av⟩$ for all $u,v\in D_A$

Theorem 13.23 · Symmetric operators and extension by the adjoint

An operator $(A,D_A)$ on a dense domain in $\mathcal{H}$ is symmetric if and only if $A^{\ast }$ is an extension of $A$, $A\subseteq A^{\ast }$

Proof

If $A\subseteq A^{\ast }$ then for all $u$, $v\in D_A\subseteq D_{A^{\ast }}$

$$⟨u\mid Av⟩=⟨A^{\ast }u\mid v⟩.$$

Furthermore, since $A^{\ast }u=Au$ for all $u\in D_A$ , we have the symmetry condition $⟨u\mid Av⟩=$ $⟨Au\mid v⟩$

Conversely, if $A$ is symmetric then

$$⟨u\mid Av⟩=⟨Au\mid v⟩\text{ for all }u,v\in D_A.$$

On the other hand, the definition of adjoint gives

$$⟨u\mid Av⟩=⟨A^{\ast }u\mid v⟩\text{ for all }u\in D_{A^{\ast }},v\in D_A.$$

Hence if $u\in D_A$ then $u\in D_A$ and $Au=A^{\ast }u$ , which two conditions are equivalent to $A\subseteq A^{\ast }$ -

From this theorem it is immediate that every self-adjoint operator is symmetric, since $A=A^{\ast }⟹A\subseteq A^{\ast }$

Exercise

Show that the operators $A$ and $A^{-1}$ of Example 13.16 are both self-adjoint.

Example 13.19

In Example 13.17 we defined the position operator $(Q,D)$ having domain $D,$ the space of $\cdot C^{\infty }$ functions of compact support on $\mathbb{R}$. This operator is symmetric in $L^2(\mathbb{R})$ , since

$$⟨\phi |Q\psi ⟩={\int }_{-\infty }^{\infty }\overline{\phi (x)}x\psi (x)dx={\int }_{-\infty }^{\infty }\overline{x\phi (x)}\psi (x)dx=⟨Q\phi |\psi ⟩$$

for all functions $\phi ,\psi \in D$ . However it is not self-adjoint, since there are many functions $\phi \notin D$ for which there exists a function ${\phi }^{\ast }$ such that $\phi \mid Q\psi ={\phi }^{\ast }\mid \psi$ for all $\psi \in D$ . For example, the function

$$\phi (x)=\{\begin{array}{ll}1& \text{ for }-1\le x\le 1\\ 0& \text{ for }|x|>1\end{array}$$

is not in $D$ since it is not $C^{\infty }$ , yet

$$⟨\phi |Q\psi ⟩=⟨{\phi }^{\ast }|\psi ⟩,\forall \psi \in D\text{ where }{\phi }^{\ast }(x)=x\phi (x).$$

Similarly, the function $\phi =1/(1+x^2)$ does not have compact support, yet satisfies the same equation. Thus the domain $D_{Q^{\ast }}$ of the adjoint operator $Q^{\ast }$ is larger than the domain $D$, and $(\mathbf{Q},\mathbf{D})$ is not self-adjoint.

To rectify the situation, let $D_Q$ be the subspace of $L^2(\mathbb{R})$ of functions $\phi$ such that $x\phi \in L^2(\mathbb{R})$ ,

$${\int }_{-\infty }^{\infty }|x\phi (x){|}^{2}dx<\infty .$$

Functions $\phi$ and ${\phi }^{\prime }$ are always to be identified, ofcourse, if they are equal almost everywhere. The operator $(Q,D_Q)$ is symmetric since

$$⟨\phi |Q\psi ⟩={\int }_{-\infty }^{\infty }\overline{\phi (x)}x\psi (x)dx=⟨Q\phi |\psi ⟩$$

for all $\phi ,\psi \in D_Q$ . The domain $D_Q$ is dense in $L^2(\mathbb{R})$ , for if $\phi$ is any square integrable function then the sequence of functions

$${\phi }_n(x)=\{\begin{array}{ll}\phi (x)& \text{ for }-n\le x\le n\\ 0& \text{ for }|x|>n\end{array}$$

all belong to $D_Q$ and ${\phi }_n\to \phi$ as $n\infty$ since

$$\Vert \phi -{\phi }_n{\Vert }^2={\int }_{-\infty }^{-n}|\phi (x){|}^{2}dx+{\int }_n^{\infty }|\phi (x){|}^{2}dx\to 0.$$

By Theorem 13.23, $Q^{\ast }$ is an extension of $\mathbf{Q}$ since the operator $(Q,D_Q)$ is symmetric. $I$ only remains to show that $D_{Q^{\ast }}\subseteq D_Q$ . The domain $D_{Q^{\ast }}$ is the set of functions $\phi \in L^2(\mathbb{R})$ such that there exists a function ${\phi }^{\ast }$ such that

$$⟨\phi |Q\psi ⟩=⟨{\phi }^{\ast }|\psi ⟩,\forall \psi \in D_Q.$$

The function ${\phi }^{\ast }$ has the property

$${\int }_{-\infty }^{\infty }(\overline{x\phi (x)}-\overline{{\phi }^{\ast }})\psi (x)\mathrm{d}x=0,\forall \psi \in D_Q.$$

Since $D_Q$ is a dense domain this is only possible if ${\phi }^{\ast }(x)=x\phi (x)$ ) a.e. Since ${\phi }^{\ast }\in L^2(\mathbb{R})$ it must be true that $x\phi (x)\in L^2(\mathbb{R})$ , whence $\phi (x)\in D_Q$ . This proves that $D_{Q^{\ast }}\subseteq D_Q$ . Hence $D_{Q^{\ast }}=D_Q$ , and since ${\phi }^{\ast }(x)=x\phi (x)$ a.e., we have ${\phi }^{\ast }=Q\phi$ . The position operator is therefore self-adjoint, $Q=Q^{\ast }$

Example 13.20

The momentum operator defined in Example 13.17 on the domain $D$ of differentiable functions of compact support is symmetric, for

$$\begin{array}{l}⟨\phi |P\psi ⟩={\int }_{-\infty }^{\infty }-i\overline{\phi (x)}\frac{\mathrm{d}\psi }{\mathrm{d}x}\mathrm{d}x\\ =[-i\overline{\phi (x)}\psi (x){]}_{-\infty }^{\infty }+{\int }_{-\infty }^{\infty }i\frac{\mathrm{d}\overline{\phi (x)}}{\mathrm{d}x}\psi (x)\mathrm{d}x\\ ={\int }_{-\infty }^{\infty }\overline{i\frac{\mathrm{d}\phi (x)}{\mathrm{d}x}}\psi (x)\mathrm{d}x\\ =⟨P\phi |\psi ⟩\end{array}$$

for all $\phi ,\psi \in D$ . Again, it is not hard to find functions $\phi$ outside $D$ that satisfy this relation for all $\psi$ , so this operator is not self-adjoint. Extending the domain so that the momentum operator becomes self-adjoint is rather trickier than for the position operator. We only give the result; details may be found in [3, 7]. Recall from the discussion following Theorem 13.19 that a function $\phi :\mathbb{RC}$ is said to be absolutely continuous if there exists a measurable function $\rho$ on $\mathbb{R}$ such that

$$\phi (x)=c+{\int }_0^x\rho (x)dx.$$

We may then set $D\phi ={\phi }^{\prime }=\rho$ . When $\rho$ is a continuous function, $\phi (x)$ is differentiable and $D\phi =\mathrm{d}\phi (x)/\mathrm{d}x$ . Let $D_P$ consist of those absolutely continuous functions such that $\phi$ and $D\phi$ are square integrable. It may be shown that $D_P$ is a dense vector subspace of $L^2(\mathbb{R})$ and that the operator $(P,D_P)$ where $P\phi =-iD\phi$ is a self-adjoint extension of the momentum operator $P$ defined in Example 13.17.

Spectral theory of unbounded operators

As for hermitian operators, the eigenvalues of a self-adjoint operator $(A,D_A)$ are real and eigenvectors corresponding to different eigenvalues are orthogonal. If $Au=\lambda u$ , then $\lambda$ is real since

$$\lambda =\frac{⟨u|Au⟩}{\Vert u{\Vert }^2}=\frac{⟨Au|u⟩}{\Vert u{\Vert }^2}=\frac{\overline{⟨u|Au⟩}}{\Vert u{\Vert }^2}=\overline{\lambda }.$$

If $Au=\lambda u$ and $Av=\mu v$ , then

$$0=⟨Au|v⟩-⟨u|Av⟩=(\lambda -\mu )⟨u|v⟩$$

whence $⟨u|v⟩=0$ whenever $\lambda \ne \mu$

For each complex number define ${\Delta }_{\lambda }$ to be the domain of the resolvent operator $(A-\lambda I{)}^{-1}$ ,

$${\Delta }_{\lambda }=D_{(A-\lambda I{)}^{-1}}=R_{A-\lambda I}.$$

The operator $(A-\lambda I{)}^{-1}$ is well-defined with domain ${\Delta }_{\lambda }$ provided $\lambda$ is not an eigenvalue. For, if $\lambda$ is not an eigenvalue then $\ker (A-\lambda I)=\{0\}$ and for every $y\in R_{A-\lambda I}$ there exists a unique $x\in D_A$ such that $y=(A-\lambda I)x$

Exercise

Show that for all complex numbers $\lambda$, the operator $A-\lambda I$ is closed.

As for bounded operators, a complex number $\lambda$ is said to be a regular value for $A$ if ${\Delta }_{\lambda }=\mathcal{H}$ . The resolvent operator $(A-\lambda I{)}^{-1}$ can then be shown to be a bounded (continuous) operator. The set of complex numbers that are not regular are again known as the spectrum of $A$.

Theorem 13.24 · Eigenvalues and density of the resolvent range

$\lambda$ is an eigenvalue of a self-adjoint operator $(A,D_A)$ ) if and only ifthe resolvent set ${\Delta }_{\lambda }$ is not dense in $\mathcal{H}$.

Proof

If $Ax=\lambda x$ where $x\ne 0$ , then

$$0=⟨(A-\lambda I)x|u⟩=⟨x|(A-\lambda I)u⟩$$

for all $u\in D_A$ . Hence $⟨x\mid v⟩=0$ for all $v\in {\Delta }_{\lambda }=R_{A-\lambda I}.\mathrm{If}{\Delta }_{\lambda }$ is dense in $\mathcal{H}$ then, by Lemma 13.20, this can only be true for $x=0$ , contrary to assumption

Conversely if ${\Delta }_{\lambda }$ is not dense then by Theorem 13.8 there exists a non-zero vecto $x\in (\overline{{\Delta }_{\lambda }}{)}^{\perp }$ . This vector has the property

$$0=⟨x|(A-\lambda I)u⟩=⟨(A-\lambda I)x|u⟩$$

for all $u\in D_A$ . Since $D_A$ is a dense domain, $x$ must be an eigenvector, $Ax=\lambda x$

It is natural to classify the spectrum into two parts – the point spectrum consisting of eigenvalues, where the resolvent set ${\Delta }_{\lambda }$ is not dense in $\mathcal{H},$ and the continuous spectrum consisting of those values $\lambda$ for which ${\Delta }_{\lambda }$ is not closed. Note that these are not mutually exclusive; it is possible to have eigenvalues $\lambda$ for which the resolvent set is neither closed nor dense. The entire spectrum of a self-adjoint operator can, however, be shown to consist of real numbers. The spectral theorem 13.19 generalizes for self-adjoint operators as follows:

Theorem 13.25 · Spectral theorem for self-adjoint operators

Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$. There exists an increasingfamily ofprojection operators $E_{\lambda }\left(\lambda \in \mathbb{R}\right)$ , with $E_{\lambda }\le P_{{\lambda }^{\prime }}\text{for}\lambda \le {\lambda }^{\prime }$ , such that

$$E_{-\infty }=0\text{and}{E}_{\infty }=I$$

such that

$$A={\int }_{-\infty }^{\infty }\lambda \mathrm{d}{E}_{\lambda },$$

where the integral is interpreted as the Lebesgue–Stieltjes integra

$$⟨u\mid Au⟩={\int }_{-\infty }^{\infty }\lambda d⟨u\mid E_{\lambda }u⟩$$

valid for all $u\in D_A$

The proof is difficult and can be found in [7]. Its main use is that it permits us to define functions $f(A)$ of a self-adjoint operator $A$ for a very wide class of functions. For example if $f:\mathbb{R}\to \mathbb{C}$ is a Lebesgue integrable function then we set

$$f(A)={\int }_{-\infty }^{\infty }f(\lambda )\mathrm{d}{E}_{\lambda }.$$

This is shorthand for

$$⟨u\mid f(A)v⟩={\int }_{-\infty }^{\infty }f(\lambda )\mathrm{d}⟨u\mid E_{\lambda }v⟩$$

for arbitrary vectors $u\in \mathcal{H},v\in D_A$ . One of the most useful of such functions is $f={\mathbf{e}}^{ix}$ , giving rise to a unitary transformation

$$U={\mathrm{e}}^{iA}={\int }_{-\infty }^{\infty }{\mathrm{e}}^{i\lambda }\mathrm{d}{E}_{\lambda }.$$

This relation between unitary and self-adjoint operators has its main expression in Stone’s theorem, which generalizes the result for finite dimensional vector spaces, discussed in Example 6.12 and Problem 6.12.

Theorem 13.26 · Stone’s theorem

Every one-parameter unitary group oftransformations $U_t$ on a Hilbert space $\mathcal{H}$, such that $U_t{U}_s=U_{t+s}$ , can be expressed in theform

$$U_t={\mathrm{e}}^{iAt}={\int }_{-\infty }^{\infty }{\mathrm{e}}^{i\lambda t}\mathrm{d}{E}_{\lambda }.$$

Problems

Problem 13.29

For unbounded operators, show that

(a) $(AB)C=A(BC)$

(b) $(A+B)C=AC+BC.$

(c) $AB+AC\subseteq A(B+C)$ . Give an example where $A(B+C)\ne AB+AC$

Problem 13.30

Show that a densely defined bounded operator $A$ in $\mathcal{H}$ has a unique extension to an operator $\hat{A}$ defined on all of $\mathcal{H}$. Show that $\Vert \hat{A}\Vert =\Vert A\Vert$

Problem 13.31

If $A$ is self-adjoint and $B$ a bounded operator, show that $B^{\ast }AB$ is self-adjoint.

Problem 13.32

Show that $\mathrm{if}(A,D_A)$ and $(B,D_B)$ ) are operators on dense domains in $\mathcal{H}$ then $B^{\ast }{A}^{\ast }\subseteq$ $(AB{)}^{\ast }$

Problem 13.33

For unbounded operators, show that $A^{\ast }+B^{\ast }\subseteq (A+B{)}^{\ast }$

Problem 13.34

$\mathrm{If}(A,D_A)$ is a densely defined operator and $D_{A^{\ast }}$ is dense in $\mathcal{H}$, show that $A\subseteq A^{\ast \ast }$

Problem 13.35

If $A$ is a symmetric operator, show that $A^{\ast }$ is symmetric if and only if it is self-adjoint, $A^{\ast }=A^{\ast \ast }$

Problem 13.36

If $A_1,A_2,\dots ,A_n$ are operators on a dense domain such that

$$\sum _{i=1}^n{A}_i^{\ast }{A}_i=0,$$

show that $A_1=A_2=\cdots =A_n=0$

Problem 13.37

If $A$ is a self-adjoint operator show that

$$\Vert (A+iI)u{\Vert }^2=\Vert Au{\Vert }^2+\Vert u{\Vert }^2$$

and that the operator $A+iI$ is invertible. Show that the operator $U=(A-iI)(A+iI{)}^{-1}$ is unitary (called the Cayley transform of $A$).