Lemma 1. If Δ_α,β is nonempty, β⋡k + l, then Δ_α,β is a finite set.

Proof. Since ω_j > 0 for j ∈ ℤ₊, we have λ_α, λ_β ∈ ℝ. For every $$, it is easy to check that

So, Δ_α,β∖{0}⊆{p ∈ ℕ : Q_β(p) = λ_α}. If Δ_α,β is infinite, there exists $$ such that p_j ∈ Δ_α,β and $$. Then, Q_β(p_j) = λ_α,  ∀ j ∈ ℕ. (A1) shows that $$, which contradicts with (A2).

For s ∈ ℤ₊, if Δ_{α+s(k + l),β} is nonempty, β⋡k + l, by the above lemma, we can set

Lemma 2. Let α, β⋡k + l and s ∈ ℤ₊. Then, M_α,β(s + 1) = M_α,β(s) + 1.

Proof. Let

If s = 0, then η = 0 in equation (20). Since (F + G)⊥z^{β+(t + 1)(k + l)} and Az^α⊥z^{β+(t + 1)(k + l)}, we have

By equation (20), we get M_α,β(1) = t + 1 = M_α,β(0) + 1.

Suppose the statement holds when s = N − 1(N ≥ 1). If s = N, η > 0 in equation (20). By hypothesis, M_α,β(N − 1) = M_α,β(N) − 1 = t − 1, which implies that Az^{α+(N − 1)(k + l)}⊥z^{β+(t + 1)(k + l)}. Together with (F + G)⊥z^{β+(t + 1)(k + l)} and Az^{α+N(k + l)}⊥z^{β+(t + 1)(k + l)}, equation (20) shows that

So, M_α,β(N + 1) = t + 1 = M_α,β(N) + 1.

Lemma 3. Suppose α⋡k + l and $$. If Δ_α,β is nonempty, then CardΔ_α,β = 1. Moreover, if Δ_α,β = {p₀}, then Δ_{α+h(k + l),β} = {p₀ + h} for any h ∈ ℤ₊.

Proof. Without loss of generality, we may assume β⋡k + l. By Lemma 1, let p₀ = M_α,β(0). For every p ∈ ℤ₊, as in Lemma 1, we can prove that {h ∈ ℤ₊ : 〈Az^{α+h(k + l)}, z^{β+p(k + l)}〉 ≠ 0} is a finite set at most. Set

Then, $$, and $$ is finite. Let $$; we prove the following statements hold by induction:

• (i)

  For every h ∈ ℤ₊, m_α,β(h₀ + h + q) > p₀ + h, ∀q ∈ ℕ.

• (ii)

  For every h ∈ ℤ₊, m_α,β(h₀ + h + 1) = p₀ + h + 1.

Step 1. It is clear that (i) holds for h = 0 from the definition of h₀. And we can set

Step 2. Suppose (i) and (ii) hold for h ≤ N; then, m_α,β(h₀ + N + q) > p₀ + N, ∀q ∈ ℕ. So, m_α,β(h₀ + N + 1 + q) > p₀ + N. We can set

Equation (27) and (ii) show that m_α,β(h₀ + h) = p₀ + h, ∀h ∈ ℤ₊. Together with Lemma 2, we get

So,

If h₀ ≠ 0, then $$ from (A1), which contradicts with (A2). Therefore, equation (30) implies that m_α,β(h) = p₀ + h = M_α,β(h), ∀h ∈ ℤ₊. So, we get the desired results.

Lemma 4. Let α ∈ Ω₁ and $$. Then, $$.

Proof. Suppose there exists some β ∈ Ω₂ ∪ Ω₃ ∪ Ω₄ such that 〈Az^α, z^β〉 ≠ 0. Lemma 3 shows that Az^α = c_βz^β + h(z), where c_β ≠ 0 and $$. Now, we discuss in two cases.

Case one: β ∈ Ω₂ ∪ Ω₄. By $$, we have A(az^α+l) = ac_βz^β+l + c_β(ω_β/ω_β−k)z^β−k + H₁(z), where $$. Hence, Δ_α+l,β−k = {p ∈ ℤ : 〈Az^α+l, z^{β−k+p(k + l)}〉 ≠ 0} = {0,1}. This is a contradiction to Lemma 3.

Case two: β ∈ Ω₃. By AT_φ(z^α) = T_φA(z^α), we have A(az^α+k) = c_βz^β+k + ac_β(ω_β/ω_β−l)z^β−l + H₂(z), where $$. Hence, Δ_α+k,β−l = {p ∈ ℤ : 〈Az^α+k, z^{β−l+p(k + l)}〉 ≠ 0} = {0,1}. This is also a contradiction.

So, $$; then, $$.

Theorem 1. Let $$ be a weighted Hardy space satisfying assumptions (A1) and (A2). Then, L_m = [z^m] for every m ∈ Δ.

Proof. To show L_m⊆[z^m], we only need to prove $$. Obviously, $$. Since {c_n} and {d_n} are strictly monotonically increasing series (see Theorem 3.1 in [15]),

To show $$, we will prove the following statements hold for every n ∈ ℕ by induction.

•  

  (T1) $$

•  

  (T2) $$

•  

  (T3) $$

We mainly use the following two important equalities:

Step 1. We prove (T1), (T2), and (T3) hold when n = 1. By calculating, we know

Thus, $$ and $$.

If d₁ = 0, then (T1) holds. If d₁ ≥ 1, equation (35) implies that $$ since $$ and $$. Similarly, it is easy to check that $$. Thus, $$. Together with $$, we obtain $$. (T1) holds for n = 1.

Let P be the orthogonal projection from $$ into [z^m]. Since $$, we have

So,

By the definition of c_n, we know m + c₁k − l ∈ Ω₁ and m + (c₁ + 1)k ∈ Ω₄. Lemma 4 implies that $$. Therefore, $$ from equation (39). So, (T2) holds for n = 1. The proof of (T3) is similar to that of (T2). We can prove $$, which implies (T3) holds for n = 1.

Step 2. Suppose (T1), (T2), and (T3) hold for 1 ≤ n ≤ p (p ∈ ℕ), and then we prove that (T1), (T2), and (T3) also hold for n = p + 1. Since $$,

By equation (36), $$ and $$ imply that

Also, $$ for 1 ≤ j ≤ d_p + p. Therefore, $$.

By equation (35), $$ and $$ imply that

Using the same technique, we finally get $$. By hypothesis to (T1), we obtain $$. So, (T1) holds for n = p + 1.

By (T1), we have $$. Notice that m + c_p+1k − (p + 1)l ∈ Ω₁ and m + (c_p+1 + 1)k − pl ∈ Ω₄. Put u = c_p+1 and v = −p in equation (35). As in Step 1, it is easy to prove that

That is, (T2) holds for n = p + 1. Similarly, by $$ and equation (36), we can prove that (T3) holds for n = p + 1. So, we finish the proof.

Lemma 5. Let m, n ∈ Ω₁ and n ≠ m. If 〈Az^m, zⁿ〉 ≠ 0, then ω_m+k = ω_n+k and ω_{m+h(k + l)} = ω_{n+h(k + l)} for every h ∈ ℤ₊.

Proof. Since 〈Az^m, zⁿ〉 ≠ 0, Lemma 3 shows that Δ_{m+p(k + l),n} = {p},  ∀ p ∈ ℤ₊. Set

For every N ∈ ℕ, $$ deduces that

Furthermore, equation (46) implies that

Let p⟶+∞; then,

That is, ω_m = ω_n and ω_{m+p(k + l)} = ω_{n+p(k + l)} for every p ∈ ℕ.

When p = 0, Az^m = a₀zⁿ + f₀(z). By $$, we have

By the expression of Az^m+k+l and Az^m, we get (ω_m+k/ω_m) = (ω_n+k/ω_n). Therefore, ω_m+k = ω_n+k.

Theorem 2. Let $$ be a weighted Hardy space satisfying assumptions (A1), (A2), and (A3). Then, for every m ∈ Δ, L_m = [z^m] is a minimal reducing subspace for T_φ.

Proof. Assume a nonzero subspace ℳ⊆L_m is a reducing subspace for T_φ. Let P_ℳ be the orthogonal projection from $$ into ℳ; then, $$. By Lemma 4, set

Lemma 5 and (A3) show that if n ≠ m, then zⁿ⊥L_m. So, P_ℳz^m = a_mz^m for some a_m ∈ ℂ. If a_m = 0, then L_m⊥ℳ, which contradicts with ℳ ≠ {0}. Hence, a_m ≠ 0 and L_m = [z^m]⊆ℳ.

Lemma 6. For $$, {p ∈ ℕ : Q_n(p) = 0} is a finite set at most.

Proof. Suppose {p ∈ ℕ : Q_n(p) = 0} is an infinite set; then, there exists $$ such that Q_n(p_j) = 0 and $$. By a direct computation, we get Q_n(p) = 0 if and only if

Let

Thus, for t_j = (1/p_j),

Set

A careful computation gives

It follows that $$. By L’Hospital’s rule, we have

That is,

Moreover, equation (60) shows that $$. By L’Hospital’s rule again, we get

That is,

Similarly,

Together with equations (61), (64), and (66), we get $$. That is, y₁ + y₂ = x₁ + x₂. Notice that x₁ + y₁ = x₂ + y₂; we obtain x₁ = y₂ and x₂ = y₁. Combining with equation (63), we have a = 1, which contradicts with a ∈ (0,1).

Lemma 7. Let m,  n ∈ Δ and n ≠ m. If ω_m+k = ω_n+k and ω_{m+h(k + l)} = ω_{n+h(k + l)} for every h ∈ ℤ₊, then

Proof. By the definition of ω_n and the equalities ω_m = ω_n and ω_m+k = ω_n+k, we have

Therefore, k₁(m₂ − n₂) + k₂(m₁ − n₁) = 0. Thus, m₁ = n₁ if and only if m₂ = n₂. Similarly, by ω_m+k = ω_n+k and ω_m+k+l = ω_n+k+l, we can prove that l₁(m₂ − n₂) + l₂(m₁ − n₁) = 0. Since n ≠ m, we get n₁ ≠ m₁, n₂ ≠ m₂, and k₁l₂ = k₂l₁. Furthermore, by k₁((m₂ + 1) − (n₂ + 1)) = k₂((n₁ + 1) − (m₁ + 1)) and equation (69), we get the desired results.

The above lemma shows that ω satisfies (A3). In fact, if zⁿ ∈ L_m, then there exist u,  v ∈ ℤ such that n₁ = (l₁/l₂)(m₂ + 1) − 1 = m₁ + uk₁ + vl₁ and n₂ = (l₂/l₁)(m₁ + 1) − 1 = m₂ + uk₂ + vl₂. It is easy to see that uk₁ + vl₁ = uk₂ + vl₂ = 0. Then, n₁ = m₁ and n₂ = m₂, which contradict with n ≠ m.

Since ω satisfies (A1), (A2), and (A3), Theorem 2 implies the following result.

Corollary 1. Let k,  l ∈ ℕ²,  k ≠ l. Then, for every m ∈ Δ, L_m = [z^m] is a minimal reducing subspace for T_φ on $$.

Corollary 2. Let $$ and m ∈ Δ. Then,

• (i)

  If k₁l₂ ≠ k₂l₁, then Az^m = cz^m, where c ∈ ℂ.

• (ii)

  If k₁l₂ = k₂l₁, then $$, where m^′ = ((l₁/l₂)(m₂ + 1) − 1, (l₂/l₁)(m₁ + 1) − 1), a, b ∈ ℂ. In particular, if $$, then b = 0.

Theorem 3. Let k, l ∈ ℕ²,  k ≠ l. If ℳ is a reducing subspace for T_φ on $$, then ℳ is the orthogonal sum of some minimal reducing subspaces. Moreover, ℳ is a minimal reducing subspace for T_φ if and only if ℳ has the form as follows:

• (i)

  If k₁l₂ ≠ l₁k₂, then ℳ = L_m for some m ∈ Δ.

• (ii)

  If k₁l₂ = l₁k₂, then there exist m ∈ Δ and s, t ∈ ℂ such that ℳ = ℳ_st where ℳ_st are defined by

  $$ (70)

•  

  with m^′ = ((l₁/l₂)(m₂ + 1) − 1, (l₂/l₁)(m₁ + 1) − 1). In particular, if $$, then t = 0.

Proof. For n ∈ Δ, if zⁿ⊥ℳ, Corollary 1 shows that [zⁿ] = L_n⊥ℳ. There is m ∈ Δ such that P_ℳz^m ≠ 0. If k₁l₂ ≠ l₁k₂, Corollary 2 (i) implies that P_ℳz^m = cz^m with c ≠ 0. And then, L_m⊆ℳ. If k₁l₂ = l₁k₂, Corollary 2 (ii) implies that there are s, t ∈ ℂ such that $$, where m^′ = ((l₁/l₂)(m₂ + 1) − 1, (l₂/l₁)(m₁ + 1) − 1). If $$, then t = 0. If $$, we claim that $$. Therefore, ℳ_st⊆ℳ. In fact, we only need to prove there is a linear operator U on $$ such that $$ is unitary from L_m onto ℳ_s,t, UT_φ = T_φU, and $$. That is, L_m and ℳ_st are unitarily equivalent.

Let $$ and

Similarly, m⪰l implies that m^′⪰l; then,

Since $$ and $$, we obtain UT_φ = T_φU. The rest of the proof is obvious.

Let m, m^′ be as in the above theorem, and let $$ and $$ be a linear operator defined by

It is not difficult to get the following theorem.

Theorem 4. Let k, l ∈ ℕ²,  k ≠ l. Suppose m, m^′ ∈ Δ; then, the following statements hold:

• (i)

  If k₁l₂ ≠ k₂l₁, then L_m and $$ are unitarily equivalent if and only if m^′ = m.

• (ii)

  If k₁l₂ = k₂l₁, then L_m and $$ are unitarily equivalent if and only if m^′ = m or m^′ = ((l₁/l₂)(m₂ + 1) − 1, (l₂/l₁)(m₁ + 1) − 1). In particular, if $$, then L_m and $$ are unitarily equivalent if and only if m^′ = m.

Theorem 5. Let k, l ∈ ℕ²,  k ≠ l. Then, $$ is a type I von Neumann algebra. Furthermore, the following statements hold:

• (i)

  If k₁l₂ ≠ k₂l₁, then $$ is abelian and is ∗−isomorphic to $$, where j = |k₁l₂ − k₂l₁|.

• (ii)

  If k₁l₂ = k₂l₁ and s = (s₁, s₂) with s_i = gcd{k_i, l_i} (i = 1,2), then $$ and $$ is never abelian. Moreover, if s₁ = s₂ = r, then $$ is ∗−isomorphic to

  $$ (76)

If s₁ ≠ s₂, then $$ is ∗−isomorphic to the direct sum of countably many M₂(ℂ) ⊕ ℂ.