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Maximal orthogonal family of cyclic reducing subspaces

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ . Then:

there is a family $(H_{j})_{j \in J}$ of pairwise orthogonal nonzero closed subspaces of $H$ , each reducing $T$ and cyclic for $T ∣_{H_{j}}$ , whose Hilbert sum is all of $H$ : the closed span of $⋃_{j} H_{j}$ equals $H$ ;
if in addition $H$ is separable and $(v_{n})_{n \geq 0}$ is a dense sequence in $H$ , put $S_{n} = H_{0} \oplus \dots \oplus H_{n - 1}$ with $S_{0} = {0}$ . Project $v_{n}$ onto $S_{n}^{⊥}$ to obtain $x_{n}$ , and take $H_{n}$ to be its cyclic subspace for the original operator $T$ on $H$ . The $H_{n}$ are closed and reducing, zero summands are allowed, and the nonzero summands are cyclic for their restrictions. Their closed span is $H$ , denoted $H = ⨁_{n \geq 0} H_{n}$ . After discarding zero summands this is a finite or countable family with the properties in claim 1.

Facts & Assumptions

[A1]

A closed subspace reduces $T$ if it is invariant under $T$ and $T^{*}$ . For every $y \in H$ the ambient cyclic subspace $H_{y}$ is closed, reducing and contains $y$ ; it is the closed span of the unital polynomial orbit in $T, T^{*}$ (Cyclic vector and cyclic normal operator). If $y = 0$ it is ${0}$ . On a nonzero reducing subspace $M$ , the restriction of $T^{*}$ is the adjoint of $T ∣_{M}$ by the defining pairing, so $T ∣_{M}$ is normal. A closed subspace is complete because Cauchy sequences converge in $H$ and their limits stay in the subspace. For nonzero $y$ , restricting the same polynomial orbit to $H_{y}$ shows $y$ is cyclic for $T ∣_{H_{y}}$ . No spectrum or calculus on a zero space is used. The adjoint pairing and involution are supplied by The Hilbert-space adjoint of a bounded operator and Hilbert-adjoint identities, and bounded operators are continuous by For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent.

[A2]

If $M$ reduces $T$ , then $M^{⊥}$ reduces $T$ : for $y \in M^{⊥}$ and $m \in M$ one has $⟨ T y, m ⟩ = ⟨ y, T^{*} m ⟩ = 0$ and $⟨ T^{*} y, m ⟩ = ⟨ y, T m ⟩ = 0$ , since $T^{*} M \subseteq M$ and $T M \subseteq M$ (Orthogonality and the orthogonal complement, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[A3]

Orthogonal decompositions: for a closed subspace $M$ one has $H = M \oplus M^{⊥}$ , and the orthogonal complement of a closed subspace is closed; a vector orthogonal to a closed subspace $N$ lies in $N^{⊥}$ (Orthogonal decomposition by a closed subspace, Orthogonal complements are closed, Orthogonality and the orthogonal complement).

[A4]

Zorn's lemma: a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma, The Axiom of Choice).

[A5]

If a closed linear subspace $L$ has $L^{⊥} = {0}$ , then $H = L \oplus L^{⊥} = L$ . Equivalently, double orthogonal complementation of any linear subspace gives its closure (Orthogonal decomposition by a closed subspace, The double orthogonal complement of a subspace is its closure). A closed set containing a dense subset is the whole space (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

[A6]

Separability means existence of an at most countable dense subset (Separability: the existence of an at most countable dense subset). Finite sums of pairwise orthogonal closed subspaces are closed: their orthogonal projections $P_{j}$ are bounded, are the identity on their own subspace and vanish on the others (The Hilbert orthogonal projection onto a closed subspace, Hilbert projections are linear, self-adjoint and contractive). Thus $P = \sum_{j < n} P_{j}$ satisfies $P^{2} = P$ and has range $S = \sum_{j < n} H_{j}$ , so $S = \ker (I - P)$ . This kernel is closed: for $x \notin S$ , a ball of radius $∥ x - P x ∥ / (2 (1 + ∥ P ∥))$ misses it, by $∥ (I - P) (y - x) ∥ \leq (1 + ∥ P ∥) ∥ y - x ∥$ . For $n = 0$ , $P = 0$ and $S = {0}$ . If each summand reduces $T$ , their finite sum does too by linearity. No closedness of an infinite algebraic sum is asserted.

[A7]

AC is the declared choice hypothesis of this page from the construction item onward (The Axiom of Choice).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ and a bounded normal operator $T \in B (H)$ ; in the separable case also a dense sequence $(v_{n})$ .

1.1

Every chain in the poset $P \subseteq P (P (H))$ of sets ${M_{j}}_{j \in J}$ of pairwise orthogonal nonzero closed $T$ -reducing subspaces, cyclic for $T ∣_{M_{j}}$ , ordered by inclusion, has an upper bound (the empty family belongs to this poset): the union of the chain is again such a family, because any two of its members lie in a common family of the chain and are therefore orthogonal, and each is reducing and cyclic by membership.

A1A4

1.2

Separable construction: start with $S_{0} = {0}$ , $K_{0} = H$ , $x_{0} = v_{0}$ and the ambient cyclic subspace $H_{0} = H_{x_{0}}$ . Recursively, once $H_{j}$ for $j < n$ are defined, their finite orthogonal sum $S_{n}$ is closed and reducing by [A6]. Its orthogonal complement $K_{n} = S_{n}^{⊥}$ is closed and reducing by [A2, A3]. Let $x_{n} = P_{K_{n}} v_{n}$ and define $H_{n} = H_{x_{n}}$ using the original $T$ on $H$ . Since $K_{n}$ is closed and invariant under $T, T^{*}$ , the whole polynomial orbit of $x_{n}$ and its closure lie in $K_{n}$ . Thus $H_{n}$ is closed, reducing and orthogonal to all earlier summands. If $x_{n} = 0$ , set $H_{n} = {0}$ with no restriction calculus.

A1A2A3A6

2.1

By Zorn's lemma $P$ has a maximal element $(H_{j})_{j \in J}$ ; its members are pairwise orthogonal, nonzero, $T$ -reducing and cyclic for the restrictions.

step 1.1A4

2.2

The family $(H_{n})$ is pairwise orthogonal and every $H_{n}$ reduces $T$ and each nonzero $H_{n}$ is cyclic for the restriction, by construction and by the preceding paragraph; the nonzero members form a pairwise orthogonal family of cyclic reducing subspaces.

step 1.2A1

2.3

Spanning in the separable case: each $v_{n}$ splits as $v_{n} = p_{n} + x_{n}$ with $p_{n} \in H_{0} \oplus \dots \oplus H_{n - 1}$ and $x_{n} \in K_{n}$ , and $x_{n} \in H_{n}$ by the ambient cyclic-subspace construction, including $x_{n} = 0$ ; hence $v_{n} \in H_{0} \oplus \dots \oplus H_{n}$ for every $n$ , so the closed sum $⨁_{n} H_{n}$ contains the dense sequence $(v_{n})$ and therefore equals $H$ .

step 1.2A5A6

3.1

Maximality forces $K : = (span ⋃_{j} H_{j})^{⊥} = {0}$ : $K$ is the orthogonal complement of the closed span of a family of cyclic reducing subspaces, hence closed and $T$ -reducing: the algebraic span is invariant under $T, T^{*}$ , its closure stays invariant by their continuity, and [A2] applies; if $y \in K$ is nonzero, its ambient cyclic subspace $M : = H_{y}$ is nonzero, closed and $T$ -reducing, and lies in $K$ by invariance of $K$ under the polynomial orbit and orthogonal to every $H_{j}$ , and it is cyclic for $T ∣_{M}$ , so $(H_{j})_{j \in J} \cup {M}$ is a strictly larger element of $P$ , contradicting maximality.

step 2.1A1A2A3

4.1

Hence no nonzero vector is orthogonal to the closed span of $⋃_{j} H_{j}$ , so [A5] gives that this closed span equals $H$ , which is the first assertion.

step 3.1A3A5

5.1

The separable construction therefore yields a finite or countable family of nonzero members of $(H_{n})_{n \geq 0}$ , consisting of pairwise orthogonal cyclic reducing subspaces with $H = ⨁_{n} H_{n}$ , as claimed.

step 2.2step 2.3A7∎

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, Theorem 5.82 and §5.7, printed pp.293–296 (standard reference, not scraped)
John B. Conway, A Course in Functional Analysis, 2nd ed., Chapter IX §10, printed pp.293–297 (standard reference, not scraped)