Multiplication-operator models, unitary equivalence and multiplicity

Under what conditions on the complex numbers $\left\{ \lambda_{k}\right\}_{k = 1}^{n}$ does the diagonal operator $A\left(x_{1}, \ldots , x_{n}\right) = \left(\lambda_{1} x_{1}, \ldots , \lambda_{n} x_{n}\right)$ in $l_{2}(n)$ have a cyclic vector? State the condition for the existence of a cyclic vector in terms of the characteristic polynomial of an arbitrary operator $A: \mathbb {C}^{n} \rightarrow \mathbb {C}^{n}$.

Under what conditions on the bounded sequence of complex numbers $\left\{ \lambda_{k}\right\}_{k = 1}^{\infty }$ does the diagonal operator

in $l_{2}$ have a cyclic vector?

Prove that the operator $(A x)(t) = a(t) x(t)$ of multiplication by a continuous strictly monotone function $a(t)$ has a cyclic vector:

Prove that the operator $(A x)(t) = a(t) x(t)$ of multiplication by a continuous non-monotone function $a(t)$, acting in the space $C[0,1]$, does not have a cyclic vector.

Reduce the operator

acting in the space $L_{2}[0,1]$, to the form of multiplication by the independent variable (find an isometric isomorphism and measure $\mu$).

Let $A$ be a linear self-adjoint operator in a finite-dimensional space. Find a unitary operator $U$, a measure $\mu$, and a function $a$ such that $U A U^{-1} = B$, where $B \in \mathcal{B}\left(L_{2}([0,1], \mu )\right),(B x)(t) = a(t) x(t)$.

In the space $l_{2}(\mathbb {Z})$, consider the operator $A = T_{r}+T_{l}$, where $T_{r}, T_{l}$ are the operators of right and left shift, respectively. Find a unitary operator $U$ and a function $a$ such that $U A U^{-1} = B$, where $B \in \mathcal{B}\left(L_{2}[-\pi , \pi ]\right), B x(t) = a(t) x(t)$.

Prove that every normal operator $A$ in a Hilbert space $H$ is unitarily equivalent to an operator of multiplication by some function $a$ in the space $L_{2}\left([0,1]^{2}, \mu \right)$ with some measure $\mu$.

Let $\mu$ and $v$ be two nonnegative Borel measures on the segment $[0,1]$, and let $T_{\mu }$ and $T_{v}$ be the operators of multiplication $x(t) \mapsto t x(t)$, acting in the spaces $L_{2}([0,1], \mu )$ and $L_{2}([0,1], v)$ respectively. These operators are unitarily equivalent if and only if the measures $\mu$ and $v$ are equivalent. ¹

Using Theorem 13.8, prove Theorem 13.9.

Theorem 13.8: For every self-adjoint operator $A$ in a separable Hilbert space $H$ there exist a decomposition $H = \bigcup_{k \in \mathcal{K}} H_{k}$, where $\mathcal{K} = \mathbb {N} \cup \left\{ \infty \right\}$ (some of the subspaces may be zero), a family of pairwise singular Borel measures $\left\{ \mu_{k}\right\}_{k \in \mathcal{K}}$ with supports $\operatorname {supp} \mu_{k} =: X_{k} \subset \sigma (A)$ (some of the sets $X_{k}$ may be empty), and a unitary isomorphism

such that the restriction of $A$ to $H_{k}$ is unitarily equivalent to the orthogonal direct sum of $k$ copies of the operator of multiplication by the independent variable in $L_{2}\left(X_{k}, \mu_{k}\right)$:

where $M_{k} \in \mathcal{B}\left(L_{2}\left(X_{k}, \mu_{k}\right)\right)$, $\left(M_{k} x\right)(t) = t x(t)$.

Theorem 13.9: Two self-adjoint operators $A$ and $B$ in a Hilbert space are unitarily equivalent if and only if their multiplicity functions coincide and the spectral measures defined in Theorem 13.8 are equivalent: $\mu_{k, A} \sim \mu_{k, B}$ for every $k \in \mathcal{K}$. (Here the multiplicity function of $A$ is $n_{A}(\lambda ) = 0$ outside $\sigma (A)$, $n_{A}(\lambda ) = k$ for $\lambda \in X_{k}$, $k \in \mathcal{K}$.)

Determine whether among the following operators there are unitarily equivalent ones: $A_{1} \in \mathcal{B}\left(L_{2}[0,1]\right),\left(A_{1} x\right)(t) = t x(t) ;$ $A_{2} \in \mathcal{B}\left(L_{2}[-1,1]\right),\left(A_{2} x\right)(t) = \left|t\right| x(t) ;$ $A_{3} \in \mathcal{B}\left(L_{2}[0,1]\right),\left(A_{3} x\right)(t) = t^{2} x(t) ;$ $A_{4} \in \mathcal{B}\left(L_{2}[0,1]\right),\left(A_{4} x\right)(t) = t^{3} x(t) ;$ $A_{5} \in \mathcal{B}\left(L_{2}[0,1]\right),\left(A_{5} x\right)(t) = \sqrt{t} x(t) ;$ $A_{6} \in \mathcal{B}\left(L_{2}[0, \pi / 2]\right),\left(A_{6} x\right)(t) = \sin t x(t) ;$ $A_{7} \in \mathcal{B}\left(L_{2}[0, \pi ]\right),\left(A_{7} x\right)(t) = \sin t x(t) ;$ $A_{8} \in \mathcal{B}\left(L_{2}[0,1]\right),\left(A_{8} x\right)(t) = (1-t) x(t) ;$ $A_{9} \in \mathcal{B}\left(L_{2}[0,1]\right),\left(A_{9} x\right)(t) = L K(t) x(t)$.

Construct the canonical functional model for the following operators in the space $l_{2}$:

Prove that the convolution operator $S_{f}$ in $L_{2}(\mathbb {R})$ is unitarily equivalent to the operator of multiplication by the function $\sqrt{2 \pi } \hat{f}: \hat{F} S_{f} \hat{F}^{-1} h = \sqrt{2 \pi } \hat{f} h$ for every function $h \in L_{2}(\mathbb {R})$.

Let $\mu =$ Area measure on $\mathbb {D} \equiv \left\{ z \in \mathbb {C}: \left|z\right|<1\right\}$ and define $A: L^{2}(\mu ) \rightarrow L^{2}(\mu )$ by $(Af)(z) = zf(z)$ for $\left|z\right|<1$ and $f$ in $L^{2}(\mu )$. Find a nontrivial reducing subspace for $A$ and an invariant subspace that does not reduce $A$.

If $\mathcal{H}$ is separable, show that the definitions of a diagonalizable operator in (4.6) and (7.3) are equivalent.

Proposition 4.6 (context): Let $\mathcal{H}$ be a separable Hilbert space with basis $\left\{ e_{n}\right\}$; let $\left\{ \alpha_{n}\right\} \subseteq \mathbb {F}$ with $M = \sup \left\{ \left|\alpha_{n}\right|: n \geq 1\right\} < \infty$. If $Ae_{n} = \alpha_{n} e_{n}$ for all $n$, then $A$ extends by linearity to a bounded operator on $\mathcal{H}$ with $\left\| A\right\| = M$. The operator $A$ is compact if and only if $\alpha_{n} \rightarrow 0$ as $n \rightarrow \infty$. (Such an operator $A$ is said to be diagonalizable.)

(7.3) (context): An operator $A$ on $\mathcal{H}$ is diagonalizable if there is a partition of the identity on $\mathcal{H}$, $\left\{ P_{i}: i \in I\right\}$, and a family of scalars $\left\{ \alpha_{i}: i \in I\right\}$ such that $\sup_{i} \left|\alpha_{i}\right| < \infty$ and $Ah = \alpha_{i} h$ whenever $h \in \operatorname {ran} P_{i}$.

If $\lambda \in \mathbb {C}$, let $\delta_{\lambda }$ be the unit point mass at $\lambda$; that is, $\delta_{\lambda }$ is the measure on $\mathbb {C}$ such that $\delta_{\lambda }(\Delta ) = 1$ if $\lambda \in \Delta$ and $\delta_{\lambda }(\Delta ) = 0$ if $\lambda \notin \Delta$. If $\left\{ \lambda_{1}, \lambda_{2}, \ldots \right\}$ is a bounded sequence of distinct complex numbers and $\left\{ \alpha_{n}\right\}$ is a sequence of real numbers with $\alpha_{n} > 0$ and $\sum_{n} \alpha_{n} < \infty$, let $\mu = \sum_{n=1}^{\infty } \alpha_{n} \delta_{\lambda_{n}}$; so $\mu$ is a finite measure. If $\phi \in l^{\infty }(\mathbb {C})$, let $M_{\phi }$ be the multiplication operator on $L^{2}(\mu )$. Define $T: L^{2}(\mu ) \rightarrow L^{2}(\mu )$ by $(Tf)(\lambda_{n}) = \lambda_{n} f(\lambda_{n})$. Prove:

Let $A: L^{2}(0,1) \rightarrow L^{2}(0,1)$ be defined by $(Af)(x) = xf(x)$ for $f$ in $L^{2}(0,1)$ and $x$ in $(0,1)$. Show that $A \cong A^{2}$.

Show that the following representation is a cyclic representation and find all of the cyclic vectors: if $(X, \Omega , \mu )$ is a $\sigma$-finite measure space and $\mathcal{H} = L^2(\mu )$, then $\pi : L^\infty (\mu ) \to \mathcal{B}(\mathcal{H})$ defined by $\pi (\phi ) = M_\phi$ (the multiplication operator) is a representation.

Show that the following representation is a cyclic representation and find all the cyclic vectors: if $X$ is a compact space and $\mu$ is a positive Borel measure on $X$, then $\pi : C(X) \to \mathcal{B}(L^2(\mu ))$ defined by $\pi (f) = M_f$ is a representation.

Verify the statements in the following example (Example 2.5). If $\mu$ is a regular Borel measure on $\mathbb {C}$ with compact support $K$, define $N_\mu$ on $L^2(\mu )$ by $N_\mu f = zf$ for each $f$ in $L^2(\mu )$. It is easy to check that $N_\mu^* f = \bar{z} f$, and hence $N_\mu$ is normal.

Let $(X, \Omega , \mu )$ be any $\sigma$-finite measure space and put $\mathcal{H} = L^2(X, \Omega , \mu )$. For $\phi$ in $L^\infty (\mu ) \equiv L^\infty (X, \Omega , \mu )$, define $M_\phi$ on $\mathcal{H}$ by $M_\phi f = \phi f$; then $M_\phi$ is normal with $M_\phi^* = M_{\bar\phi }$, and $\sigma (M_\phi ) = \operatorname {ess-ran}(\phi )$, the essential range of $\phi$, defined by

Verify the following statement (Example 2.6(e)). If $E$ is the spectral measure for $M_\phi$ [so $E$ is defined on the Borel subsets of $\sigma (M_\phi ) = \operatorname {ess-ran}(\phi ) \subseteq \mathbb {C}$], then for every Borel subset $\Delta$ of $\sigma (M_\phi )$,

If $\mu$ is a compactly supported measure on $\mathbb {C}$ and $f \in L^2(\mu )$, $f$ is a star-cyclic vector for $N_\mu$ if and only if $\mu (\{ x : f(x) = 0\} ) = 0$.

Let $(X, \Omega , \mu )$ be the measure space obtained in the proof of the following theorem (Theorem 4.6: if $N$ is a normal operator on $\mathcal{H}$, then there is a measure space $(X, \Omega , \mu )$ and a function $\phi$ in $L^\infty (X, \Omega , \mu )$ such that $N$ is unitarily equivalent to $M_\phi$ on $L^2(X, \Omega , \mu )$) and show that $L^1(X, \Omega , \mu )^*$ is isometrically isomorphic to $L^\infty (X, \Omega , \mu )$.

If $(X, \Omega , \mu )$ is a measure space, then $(X, \Omega , \mu )$ is $\sigma$-finite or $L^2(\mu )$ is finite dimensional if and only if every collection of pairwise orthogonal projections in $\{ M_\phi \in L^\infty (\mu )\}$ is countable.

Let $\mathcal{H}$ be separable with basis $\{ e_n\}$. Let $A$ be the diagonal operator on $\mathcal{H}$ given by $Ae_n = \lambda_n e_n$, where $\sup_n \left|\lambda_n\right| < \infty$. Determine $\{ A\} '$ and $\{ A\} ''$. Give necessary and sufficient conditions on $\{ \lambda_n\}$ such that $\{ A\} ' = \{ A\} ''$.

If $\mu$ is Lebesgue measure on $[0, 1]$, show that $N_\mu \cong N_\mu^p$ for $0 < p < \infty$.

Let $\mu$ be Lebesgue measure on $\mathbb {R}$ and $N =$ multiplication by $\sin x$ on $L^2(\mu )$. Find the decompositions of $N$ obtained in Theorem 10.1 and Theorem 10.16, stated below.

Theorem 10.1: (a) If $N$ is a normal operator, then there is a sequence (possibly finite) of measures $\{ \mu_n\}$ on $\mathbb {C}$ such that $\mu_{n+1} \ll \mu_n$ for all $n$ and $N \cong N_{\mu_1} \oplus N_{\mu_2} \oplus \cdots$. (b) If $N$ and $\{ \mu_n\}$ are as in (a) and $M \cong N_{\nu_1} \oplus N_{\nu_2} \oplus \cdots$, where $\nu_{n+1} \ll \nu_n$ for all $n$, then $N \cong M$ if and only if $[\mu_n] = [\nu_n]$ for all $n$.

Theorem 10.16: If $N$ is a normal operator, then there are mutually singular measures $\mu_\infty , \mu_1, \mu_2, \dots$ (some of which may be zero) such that

If $M$ is another normal operator with corresponding measures $\nu_\infty , \nu_1, \nu_2, \dots$, then $N \cong M$ if and only if $[\mu_n] = [\nu_n]$ for $1 \leq n \leq \infty$.

Let $(X, \Omega , \mu )$ be a measure space such that $L^2(\mu )$ is separable, let $\phi \in L^\infty (\mu )$, and let $N = M_\phi$ on $L^2(\mu )$. Find the decompositions of $N$ obtained in Theorem 10.1 and Theorem 10.16, stated below.

Theorem 10.1: (a) If $N$ is a normal operator, then there is a sequence (possibly finite) of measures $\{ \mu_n\}$ on $\mathbb {C}$ such that $\mu_{n+1} \ll \mu_n$ for all $n$ and $N \cong N_{\mu_1} \oplus N_{\mu_2} \oplus \cdots$. (b) If $N$ and $\{ \mu_n\}$ are as in (a) and $M \cong N_{\nu_1} \oplus N_{\nu_2} \oplus \cdots$, where $\nu_{n+1} \ll \nu_n$ for all $n$, then $N \cong M$ if and only if $[\mu_n] = [\nu_n]$ for all $n$.

Theorem 10.16: If $N$ is a normal operator, then there are mutually singular measures $\mu_\infty , \mu_1, \mu_2, \dots$ (some of which may be zero) such that

If $M$ is another normal operator with corresponding measures $\nu_\infty , \nu_1, \nu_2, \dots$, then $N \cong M$ if and only if $[\mu_n] = [\nu_n]$ for $1 \leq n \leq \infty$.

Let $\mu$ be a compactly supported measure on $\mathbb {C}$, $\phi$ a bounded Borel function on $\mathbb {C}$, and suppose $\{ \Delta_n\}$ are pairwise disjoint Borel sets such that $\phi$ is one-to-one on each $\Delta_n$ and $\mu (\mathbb {C} \setminus \bigcup_{n=1}^\infty \Delta_n) = 0$. Let $\phi_n = \phi \chi_{\Delta_n}$ and $\mu_n = \mu \circ \phi_n^{-1}$ for $n \geq 1$. Prove that $M_\phi$ on $L^2(\mu )$ is unitarily equivalent to $\bigoplus_{n=1}^\infty N_{\mu_n}$.

Recall the construction from the proof of the Spectral Theorem (4.11): given a normal operator $N$, one lets $B = (1 + N^*N)^{-1} = \int_0^1 t \, dP(t)$ be the spectral decomposition of $B$, puts $P_n = P\left(\left[1/(n+1), 1/n\right]\right)$ and $\mathcal{H}_n = P_n \mathcal{H}$ for $n \geq 1$, so that $\mathcal{H}_n$ reduces $N$ and $N_n \equiv N \rvert \mathcal{H}_n$ is a bounded normal operator with $\sigma (N_n) \subseteq \Delta_n \equiv \left\{ z \in \mathbb {C} : (n-1)^{1/2} \leq \left|z\right| \leq n^{1/2}\right\}$; letting $N_n = \int z \, dE_n(z)$ be the spectral decomposition of $N_n$, the spectral measure of $N$ is then defined by $E(\Delta ) = \sum_{n=1}^{\infty } E_n(\Delta \cap \Delta_n)$.

If $\Delta_n$ and $E_n$ are as in this construction, show that $E_n(\Delta_{n+1}) = E_n(\Delta_{n-1}) = 0$.

Use the Spectral Theorem to show that if $0 < a \leq b < \infty$, $\Delta = \left\{ z \in \mathbb {C} : a \leq \left|z\right| \leq b\right\}$, and $N = \int z \, dE(z)$ is the spectral decomposition of the normal operator $N$, then

Recall Theorem 4.19: if $N$ is a normal operator on the separable Hilbert space $\mathcal{H}$, then there is a $\sigma$-finite measure space $(X, \Omega , \mu )$ and an $\Omega$-measurable function $\phi$ such that $N$ is unitarily equivalent to $M_\phi$ on $L^2(\mu )$.

The proof is only sketched in the text as follows. Write $N$ as the (unbounded) direct sum of bounded normal operators $\left\{ N_n\right\}$. For each $n$ there is a $\sigma$-finite measure space $(X_n, \Omega_n, \mu_n)$ and a bounded $\Omega_n$-measurable function $\phi_n$ such that $N_n \cong M_{\phi_n}$. Let $X$ be the disjoint union of $\left\{ X_n\right\}$ and let $\Omega = \left\{ \Delta \subseteq X : \Delta \cap X_n \in \Omega_n \text{ for every } n\right\}$. If $\Delta \in \Omega$, let $\mu (\Delta ) = \sum_{n=1}^{\infty } \mu_n(\Delta \cap X_n)$. Let $\phi : X \to \mathbb {C}$ be defined by $\phi (x) = \phi_n(x)$ if $x \in X_n$. Then $\phi$ is $\Omega$-measurable and $N \cong M_\phi$ on $L^2(X, \Omega , \mu )$.

Fill in the details of this proof.

A necessary and sufficient condition that an operator $A$ on a Hilbert space $\mathbf{H}$ be representable as a multiplication on a functional Hilbert space is that the eigenvectors of $A^{*}$ span $\mathbf{H}$.

If $A$ is a normal operator on a separable Hilbert space, and if $1 + A + \cdots + A^{n}$ is diagonal for some positive integer $n$, does it follow that $A$ is diagonal?

What are all the cyclic vectors of the position operator on $L^{2}\left((0,1)\right)$?