Find an infinite family of operators that are not pairwise unitarily equivalent

Question

$H$ is a separable infinite-dimensional Hilbert space.

1- Find two self-adjoint operators on $H$ with spectrum $[0, 1]$ and empty point-spectrum that are not unitarily equivalent.

2- Find uncountably many self-adjoint operators on H with spectrum $[0, 1]$ and empty point-spectrum that are pairwise unitarily inequivalent.

I can refer to Is there an injective, self-adjoint operator on $\ell_2(\mathbb{Z})$ with spectrum $[0,1]$?

And to "Conway's A Course in Functional Analysis" section IX. 10 on 'Multiplicity Theory for Normal Operators: A Complete Set of Unitary Invariants' as mentioned in the comments by @Martin .

I am thinking toward using spectral measures, though I think that producing an uncountable family would be challenging, or obvious from the cardinality of the space of measures being larger than (or equal to) the uncountable $c$.

Answer 1

Concerning the first question, consider $L^2(0,1)$ and $L^2(0,1)\times L^2(0,1).$ These spaces are isometrically isomorphic, as both are separable Hilbert spaces. Let $T$ acts on $L^2(0,1)$ as $(Tf)(x)=xf(x)$ and $S$ acts on the second space as $S(f,g)(x)=(xf(x),xg(x)).$ Then $\sigma(T)=\sigma(S)=[0,1].$ The point spectra of both operators are empty.

The operator $T$ has a cyclic vector, namely $v(x)=1.$ By definition for a cyclic vector $v$ the linear span of all vectors $T^nv,$ $n\ge 0,$ is dense in the space. For our $T$ it follows from the Weierstrass theorem. On the other hand $S$ does not admit a cyclic vector. Indeed, assume $(f,g)$ is a cyclic vector for $S.$ Consider $(1,0)\in L^2(0,1)\times L^2(0,1).$ Therefore, there exists a sequence of polynomials $p_n(x),$ such that $p_n(S)(f,g)=(p_n(x)f(x), p_n(x)g(x))$ tends to $(1,0)$ in $L^2(0,1)\times L^2(0,1).$ Thus $p_n(x)f(x)$ tends to $1$ and $p_n(x)g(x)$ tends to $0$ in $L^2(0,1).$ There exists $a>0$ such that the set $A=\{x\,:\, |g(x)|>a\}$ has positive Lebesgue measure. Then $p_n(x)\chi_A(x)$ tends to $0$ and $p_n(x)f(x)\chi_A(x)$ tends to $\chi_A(x).$ This leads to a contradiction.

The operators $T$ and $S$ cannot be unitarily equivalent as the equivalence relation preserves the property of having a cyclic vector.

Answer 2

We will use spectral measures, two different spectral measures give two non-equivalent operators. As we can find in Conway's 'A course in functional analysis' the following theorem:
"Two normal operators are unitarily equivalent if and only if they have the same spectral measure E and their multiplicity functions are equal almost everywhere w.r. to E".

Recall that every normal operator $N$ can be determined by a spectral measure, which is a map $P: B(\mathbb{C})\rightarrow \mathcal{B}(H)$ sending a Borel set $E$ to a projection operator $P(E)$ and such that $P(\emptyset)= 0,\ P(\mathbb{C})= 1$ and whenever $(E_n)$ is a sequence of mutually disjoint sets: $P(\bigcup_n E_n)= \sum_n P(E_n)$ where the limit is in the SOT.
A bounded normal operator $N$ can be regarded as the integral, (invoking Borel calculus) : $$N= \int_{\mathbb{C}} z\,dP(z)= \int_{K} z\,dP(z)\ ,\quad K\subset \mathbb{C} \text{ a compact}$$

Let $\{e_n\}$ be an orthonormal basis of $H$, $H$ is separable so $\{e_n\}$ is countable, consider all of the rational numbers in $[0,1]$ and assign each one of them to one of the $\{e_n\}$'s bijectively, say this bijection is $\alpha_1: n\mapsto r_n\in [0,1]\cap \mathbb{Q}$

Now consider the spectral measure $P_1$ to be mapping that sends the empty set, and each Borel set to the zero projection, except for the borel sets intersecting $[0,1]$ : for subsets $I$ of $[0,1]$ $$P_1(I)= P_{<\{e_n|\ r_n\in I\}>} $$ the projection onto the span of all the $e_i$'s which are assigned to the rationals in this subset, and extending to the rest of Borel sets by the rule :
$P_1(E)= P_1((E\cap [0,1])\cup (E\cap [0,1]^c))= P_1(E\cap [0,1])= P_{<\{e_n|\ r_n\in E\cap [0,1]\}>}$ , but with $P_1([0,1]\cap \mathbb{Q})= 0$ , to avoid having eigenvectors, that is, in some sense, the information is carried by the rationals but only if they are surrounded by some subset of non-zero measure (essentially) , we can check that this is a well defined spectral measure, it satisfies :

• $P_1(\emptyset)= 0$ and $P_1(\mathbb{C}) = 1$
• $P_1(E\cap F) = P_1(E)P_1(F)$ for $E$ and $F$ in $B(\mathbb{C})$
• Whenever $(E_n)$ is a sequence of mutually disjoint sets: $P_1(\bigcup_n E_n)= \sum_n P_1(E_n)$
• if $O$ is a non-empty relatively open subset of $[0,1]$ then $P_1(O)\neq 0$
  So our normal operator is $$ N_1= \int_{[0,1]} z\,dP_1(z) \quad \text{ (self-adjoint, and has no eigenvectors)}$$

We can now construct further spectral measures that are mutually different, consider projections, say for subsets of $[0,1]:\ P_2(I)= P_{<\{e_{n}|\ r_{2n}\in I\}>}$ , $P_3(I)= P_{<\{e_{n}|\ r_{3n}\in I\}>}$ ... $$\text{ we can consider then the countably many }\ \Big\{N_k= \int_{[0,1]} z\,dP_k(z)\Big\}$$ But why stop here, how many proper injections $\phi:\mathbb{N}\to \mathbb{N}$ (non-bijective) such that $\phi_i(\mathbb{N})\neq\phi_j(\mathbb{N})\ \forall i\neq j$ are there ? $2^{\aleph_o}- \aleph_o= \aleph_1= c$ , so we can also consider $$\Big\{N_\phi= \int_{[0,1]} z\,dP_\phi(z)\ |\ \phi\in \mathcal{\mathbb{N}}^\mathbb{N} \text{ are proper injections with mutually different ranges } \Big\}$$ where $P_\phi$ is the spectral measure given as for $P_1$ by considering $P_\phi(I)= P_{<\{e_{n}|\ r_{\phi(n)}\in I\}>}$, the fact that the $\phi$'s are embeddings ensure that the $N_\phi$ are not projections, so they have no eigenvectors.