On the Fine Spectrum of the Operator Defined by the Lambda Matrix over the Spaces of Null and Convergent Sequences

Theorem 6. σ(Λ, c₀)⊆{α ∈ ℂ : |2α − 1| ≤ 1}.

Proof. Let |2α − 1| > 1. Since Λ − αI is triangle, (Λ−αI)⁻¹ exists, and solving the matrix equation (Λ − αI)x = y for x in terms of y gives the matrix (Λ−αI)⁻¹ = B = (b_nk), where

for all k, n ∈ ℕ. Thus, we observe that The inequality |2α − 1| > 1 is equivalent to γ > −1, where −(1/α) = γ + iβ. For all α ∈ ℂ, holds for all j ∈ ℕ. So, 1/|1 − (c_j/α)| ≤ 1/(1 + γc_j).

Firstly we take −1 < γ < 0. Since 0 < c_j ≤ 1, we have 1 + γ ≤ 1 + γc_j < 1. Therefore 1/(1 + γc_j) < 1/(1 + γ) and 1 < 1/(1 + γ) < ∞ for 0 < 1 + γ < 1.

Secondly we get 0 ≤ γ. Since 1 < 1 + γc_j ≤ 1 + γ, 1/(1 + γc_j) < 1. So,

Therefore, we have that is, (Λ−αI)⁻¹ ∈ (c₀ : c₀). But for |2α − 1| ≤ 1, that is, (Λ−αI)⁻¹ is not in B(c₀). This completes the proof.

Theorem 7. Define μ and η by μ = limsup _j→∞c_j and η = liminf _j→∞c_j. Then,

Proof. Let |α − 1/(2 − μ)| < (1 − μ)/(2 − μ) and α ≠ c_j for any j ∈ ℕ. Then,

So we have, Note that |1 + (1 − α⁻¹)(λ_j − λ_j−1)/λ_j−1| ≤ 1 if and only if where −α⁻¹ = γ + iβ. So, one can see that which is equivalent to the inequality For inequality (26) to be true for all sufficiently large j, it is sufficient to have We can write (λ_j − λ_j−1)/λ_j−1 = λ_j(λ_j − λ_j−1)/(λ_jλ_j−1) and λ_j/λ_j−1 = 1/(1 − c_j). Therefore, since the function g defined by g(x) = x/(1 − x) is monotone increasing in x for 0 < x < 1.

For (27) to be true for all sufficiently large j, it is sufficient to have μ satisfying

which is equivalent to Therefore, for all n ≥ N, for some fixed N, which diverges in the light of (29).

If α = c_j for any j ∈ ℕ, then clearly α lies in the spectrum of Λ. This completes the proof.

Theorem 8. σ(Λ, c₀)⊆{α ∈ ℂ : |α − 1/(2 − η)| ≤ (1 − η)/(2 − η)} ∪ S.

Proof. Let α be fixed and satisfy the inequality

and α ≠ c_j for any j ∈ ℕ. We will show that α ∈ ρ(Λ, c₀). From Theorem 6, we need consider only those values of α satisfying |2α − 1| > 1; that is, γ > −1. Under the assumption on α, we wish to verify that for all sufficiently large j. It will be sufficient to show that that is, which is equivalent to (33).

Define the function f by f(t) = 1 + 2(1 + γ)t + [(1 + γ) ² + β²]t². f has a minumum at t₀ = −(1 + γ)/[(1+γ)² + β²]. The above inequality is equivalent to η(γ² + β²) + 2γ > η − 2 and is also equivalent to

Therefore, for those values of η satisfying (37), f is monotone increasing. Let ϵ > 0 and small. Then f(η/(1 − η) − ϵ) = f(η/(1 − η)) − (2ϵ)g(ϵ), where g(ϵ) = 1 + γ + [(1 + γ) ² + β²][η/(1 − η) − ϵ/2]. Note that g(ϵ) > 0 for small ϵ, since f is monotone increasing for t > η/[2(1 − η)], we will now show that f(η/(1 − η)) > 1. From (37), which is equivalent to But 1/(1 − η) = 1 + η/(1 − η), so we have f(η/(1 − η)) = $$. Now choose ϵ > 0 and so small that f(η/(1 − η) − ϵ) = f(η/(1 − η)) − 2ϵg(ϵ) = m² > 1. Then, by the definition of η, there exists an N such that n > N implies (λ_n+1 − λ_n)/λ_n > η/(1 − η) − ϵ, so that f((λ_n+1 − λ_n)/λ_n) > f(η/(1 − η) − ϵ) = m². Using (23), for all n ≥ N. Therefore {|b_nk|} is monotone decreasing in n for each k, n > N, so that B has bounded columns. It remains to show that B has finite norm.

For ϵ being used, from (29), we can enlarge N, if necessary, to ensure that (λ_n − λ_n−1)/λ_n−1 < μ/(1 − μ) + 1 for n ≥ N. From (23),

where H is a constant independent of n. Further Hence, B has a finite norm.

Corollary 9. Let δ = lim _j→∞c_j exist. Then,

Theorem 10. Let δ be defined as in Corollary 9. Then, $$.

Proof. Suppose that Λ^*x = αx for x ≠ θ in $$. Then, by solving the system of linear equations

we can write x_n = (λ_n − λ_n−1)/λ_n−1(1 − α⁻¹)$$[1+(1 − α⁻¹)(λ_j − λ_j−1)/λ_j−1]. Let |α − 1/(2 − δ)| < (1 − δ)/(2 − δ) or α ∈ S and $$[1 + (1 − α⁻¹)(λ_j − λ_j−1)/λ_j−1]. One can see that |1 + (1 − α⁻¹)(λ_j − λ_j−1)/λ_j−1| < 1 for all sufficiently large j if and only if Then, we have, from the discussion in Theorem 7 and the hypothesis on α, for all sufficiently large n, so $$ is convergent. Since |(1 − α⁻¹)(λ_n − λ_n−1)/λ_n−1x₀| is bounded, it follows that $$ is convergent, so that Λ^*x = αx has nonzero solutions. Therefore, the proof is completed.

Theorem 11. Let δ be defined as in Corollary 9. Then

Proof. Let c_k be any diagonal entry satisfying 0 < c_k ≤ δ/(2 − δ). Let j be the smallest integer such that c_j = c_k. By setting x_n = 0 for n > j + 1, x₀ = 0, the system (Λ^* − c_jI)x = θ reduces to a homogeneous linear system of j equations in j + 1 unknowns, so that nontrivial solutions exist. Therefore Λ − c_jI ∈ 3.

Λ − αI is not one to one for α = 0,1 and so Λ − αI ∈ 3. This step concludes the proof.

Lemma 12 (see [3], page 59.)T has a dense range if and only if T^* is one to one.

Theorem 13. $$.

Proof. For $$, the operator Λ − αI is triangle, so has an inverse. But Λ^* − αI is not one to one by Theorem 10. Therefore by Lemma 12, $$, and this step concludes the proof.

Theorem 14. Let δ be defined as in Corollary 9 and c_n ≥ δ for all sufficiently large n. Then,

Proof. Fix α ≠ 1, δ/(2 − δ), and satisfying |α − 1/(2 − δ)| = (1 − δ)/(2 − δ). Since the operator Λ − αI is triangle, it has an inverse. Consider the adjoint operator Λ^* − αI. As in Theorem 11, x₀ is arbitrary and

for all positive n. From the hypothesis, there exists a positive integer N such that n ≥ N implies c_n ≥ δ. This fact, together with the condition on α, implies that |1 + (1 − α⁻¹)(λ_n − λ_n−1)/λ_n−1| ≥ 1 for n ≥ N. Thus, |x_n | = C(λ_n − λ_n−1)/λ_n−1 for n ≥ N, where C is a positive constant independent of n. We can write Therefore (x_n) ∈ ℓ₁⇔x₀ = 0; that is, Λ^* − αI is one to one. From Lemma 12, the range of Λ − αI is dense in c₀. This completes the proof.

Lemma 15 (see [3], page 60.)T has a bounded inverse if and only if T^* is onto.

Theorem 16. Let δ be defined as in Corollary 9 and less than 1. If α satisfies |α − 1/(2 − δ)| < (1 − δ)/(2 − δ) and α ∉ S, then α ∈ C₁σ(Λ, c₀).

Proof . First of all Λ − αI is a triangle; hence 1 − 1. Therefore Λ − αI ∈ 1 ∪ 2. To verify that Λ − αI ∈ C₁σ(Λ, c₀) it is sufficient to show that Λ^* − αI is onto by Lemma 15.

Suppose y = (Λ^* − αI)x, where x, y ∈ ℓ₁. Then, x₀ = 1/(1 − α)y₀ − λ₀/[(λ₁ − λ₀)(1 − α)]y₁ and

Choose x₁ = 0 and solve (51) for x in terms of y to get For example, substituting (52) into (53), with n = 2, yields so that x₂ = (λ₂ − λ₁)/[α(λ₁ − λ₀)]y₁ − (1/α)y₂. For n = 3,

Continuing this process, the entries of the matrix B = (b_nk) such that By = x are calculated as

and b_nk = 0 otherwise.

To show that B ∈ B(ℓ₁), it is sufficient to establish that $$ is finite independent of k. $$. We may write 1 − (c_j/α) = (λ_j−1/λ_j)[1 + (1 − α⁻¹)(λ_j − λ_j−1)/λ_j−1]. Also, sup _n∈ℕ|(λ_n − λ_n−1)/λ_n−1| ≤ M < ∞. Therefore,

and, for k > 1,

Since k > 1, the series in inequality (24) is absolutely convergent from Theorem 7. Therefore, $$ is finite.

Because of (Λ−αI)⁻¹ is bounded, it is continuous, and α ∈ C₁σ(Λ, c₀). This completes the proof.

Theorem 17. Let δ be defined as in Corollary 9 and δ < 1. If α = δ or α = c_n for all n ∈ ℕ and δ/(2 − δ) < α < 1, then α ∈ C₁σ(Λ, c₀).

Proof. First assume that Λ has distinct diagonal entries and fix j ≥ 1. Then the system (Λ − c_jI)x = θ implies that x_n = 0 for n = 0,1, …, j − 1, and for n ≥ j

The system (59) yields the following recursion relation: which can be solved for x_n to yield

Since 0 < c_j < 1, the argument of Theorem 7 applies and (24) is true. Therefore x ∈ c₀ implies x = θ and Λ − c_jI is 1 − 1, so that Λ − c_jI ∈ 1 ∪ 2.

Clearly Λ − c_jI ∈ C. It remains to show that Λ^* − c_jI is onto.

Suppose that (Λ^* − c_jI)x = y, where x, y ∈ ℓ₁. By choosing x_j+1 = 0 we can solve for x₀, x₁, …, x_j in terms of y₀, y₁, …, y_j+1. As in Theorem 16, the remaining equations can be written in the form x = By, where the nonzero entries of B = (b_nk) are as follows

From (62), we have For m > 1, Using 1 − (c_j/α) = (λ_j−1/λ_j)[1 + (1 − α⁻¹)(λ_j − λ_j−1)/λ_j−1], one can convert (63) and (64) the similar expressions to (57) and (58), and therefore $$ is finite.

Suppose that Λ does not have distinct diagonal entries. The restriction on α guarantees that no zero diagonal entries are being considered. Let c_j ≠ 0 be any diagonal entry which occurs more than once, and let k, r denote, respectively, the smallest and largest integers for which c_j = c_k = c_r. From (61) it follows that x_n = 0 for n ≥ r. Also, x_n = 0 for 0 ≤ n < k. Therefore the system (Λ − c_jI)x = θ becomes

Case 1. Let r = k + 1. Then (65) reduces to the single equation which implies that x_k = 0, since c_j = c_r = c_k+1 and c_j ≠ 0. Therefore x = θ.

Case 2. Let r > k + 1. From (65) one can obtain the recursion formula x_n = λ_n+1(c_j − c_n+1)x_n+1/(λ_nc_j) with k < n < r. Since x_r = 0 it then follows that x_n = 0 for k < n < r. Using (65) with n = k + 1 yields x_k = 0 and so again x = θ.

To show that Λ^* − c_jI is onto, suppose (Λ^* − c_jI)x = y, where x, y ∈ ℓ₁. By choosing x_j+1 = 0 one can solve for x₀, x₁, …, x_j in terms of y₀, y₁, …, y_j+1. As in Theorem 16, the remaining equations can be written in the form x = By, where the nonzero entries of B are as in (62) with the other entries of B clearly zero.

Since k ≤ j ≤ r, there are two cases to consider.

Case 1. If j = r, then the proof proceeds exactly as in the argument following (62).

Case 2. If j < r, then from (62), b_j+m,j+k = b_j+m,j+1 = 0 at least for m ≥ r − j + 2. If there are other values of n with j < n < r for which c_n − c_j, then additional entries of B will be zero. These zero entries do not affect the validity of the argument showing that (63) converges.

If δ = 0, then 0 does not lie inside the disc, and so it is not considered in this theorem.

Let α = δ > 0. If λ_nn ≤ δ for each n ≥ 1, all i sufficiently large, then the argument of Theorem 16 applies and Λ − δI ∈ C₁. If λ_nn = δ for some n, then the proof of Theorem 17 applies with c_j replaced by δ and again, Λ − δI ∈ C₁.

Therefore, in all cases, Λ − c_jI ∈ 1 ∪ 2.

Theorem 18. If α ∈ σ_p(Λ, c₀), α ∈ C₃σ(Λ, c₀).

Proof . For α ∈ σ_p(Λ, c₀), Λ − αI ∈ 3 and Λ^* − αI is not one to one. Therefore $$ by Lemma 12. This concludes the proof.

Theorem 19. The statement A₃σ(Λ, c₀) = C₂σ(Λ, c₀) = ∅ holds.

Proof. Let δ be defined as in Corollary 9 and c_n ≥ δ for all sufficiently large n, then A₃σ(Λ, c₀) = ∅ and C₂σ(Λ, c₀) = ∅ follow from Corollary 9, Theorems 14, and 16–18.

Theorem 20. The following results hold:

• (a)

  σ_ap(Λ, c₀) = {α : |α − (2 − δ) ⁻¹| = (1 − δ)/(2 − δ)} ∪ E,

• (b)

  σ_δ(Λ, c₀) = σ(Λ, c₀),

• (c)

  σ_co(Λ, c₀) = {α ∈ ℂ : |α − (2 − δ) ⁻¹| < (1 − δ)/(2 − δ)} ∪ S.

Proof . (a) Since the relation

holds by Theorems 16 and 17 and from Table 1, σ_ap(Λ, c₀) = σ(Λ, c₀)∖C₁σ(Λ, c₀). Therefore, we have σ_ap(Λ, c₀) = {α : |α − (2−δ)⁻¹| = (1 − δ)/(2 − δ)} ∪ E.

(b) Since σ_δ(Λ, c₀) = σ(Λ, c₀)∖A₃σ(Λ, c₀) from Table 1 and A₃σ(Λ, c₀) = ∅ by Theorem 19, we have σ_δ(Λ, c₀) = σ(Λ, c₀).

(c) Since the equality σ_co(Λ, c₀) = C₁σ(Λ, c₀) ∪ C₂σ(Λ, c₀) ∪ C₃σ(Λ, c₀) holds from Table 1, we have σ_co(Λ, c₀) = {α ∈ ℂ : |α − (2 − δ) ⁻¹| < (1 − δ)/(2 − δ)} ∪ S by Theorems 16–19.

Corollary 21. The following results hold:

• (a)

  σ_ap(Λ^*, ℓ₁) = σ(Λ, c₀),

• (b)

  σ_δ(Λ^*, ℓ₁) = {α : |α − (2−δ)⁻¹| = (1 − δ)(2 − δ)} ∪ E,

• (c)

  σ_p(Λ^*, ℓ₁) = {α ∈ ℂ : |α − (2−δ)⁻¹| < (1 − δ)/(2 − δ)} ∪ S.

Theorem 22. σ(Λ, c)⊆{α ∈ ℂ : |2α − 1| ≤ 1}.

Proof. This is obtained in a similar way to that used in the proof of Theorem 6.

Theorem 23. Suppose that μ, η and S be defined as in Theorem 7. Then,

Proof. This is similar to the proof of Theorems 7 and 8. To avoid the repetition of the similar statements, we omit the detail.

Corollary 24. Let δ be defined as in Corollary 9. Then,

Theorem 25. Let δ be defined as in Corollary 9. Then,

Proof. Suppose that Λ^*x = αx for x ≠ θ in c^*≅ℓ₁. Then, by solving the system of linear equations

we get by assumption (1 − α)x₀ = 0 with α = 1 that x = (x₀, x₁, 0,0, …) ∈ c. If α ≠ 1, we have x₀ = 0 and (x_n) ∈ ℓ₁ if and only if |1 + (1 − α⁻¹)(λ_j − λ_j−1)/λ_j−1| < 1, by Theorem 11. This completes the proof.

Theorem 26. Let δ be defined as in Corollary 9. Then,

Proof. The proof is identical to the proof of Theorem 11.

Theorem 27. σ_r(Λ, c) = σ_p(Λ^*, c^*)∖σ_p(Λ, c).

Proof. For α ∈ σ_p(Λ^*, c^*)∖σ_p(Λ, c), the operator Λ − αI is triangle, so has an inverse. But Λ^* − αI is not one to one by Theorem 26. Therefore by Lemma 12, $$ and this concludes the proof.

Theorem 28. Let δ be defined as in Corollary 9 and c_n ≥ δ for all sufficiently large n. Then,

Theorem 29. Let δ be defined as in Corollary 9 and less than 1. If α satisfies |α − 1/(2 − δ)| < (1 − δ)/(2 − δ) and α ∉ S, then α ∈ C₁σ(Λ, c).

Theorem 30. Let δ be defined as in Corollary 9 and δ < 1. If α = δ or α = c_n for all n ∈ ℕ and δ/(2 − δ) < α < 1, then α ∈ C₁σ(Λ, c).

Theorem 31. If α ∈  σ_p(Λ, c), α ∈ C₃ σ(Λ, c).

Theorem 32. The following statement holds:A₃σ(Λ, c) = C₂σ(Λ, c) = ∅.

Proof. Let δ be defined as in Corollary 9 and c_n ≥ δ for all sufficiently large n, then A₃σ(Λ, c) = ∅ and C₂σ(Λ, c) = ∅ follow from Corollary 24 and Theorems 28–31.

Theorem 33. The following results hold:

• (a)

  σ_ap(Λ, c) = {α : |α − (2 − δ) ⁻¹| = (1 − δ)/(2 − δ)} ∪ E,

• (b)

  σ_δ(Λ, c) = σ(Λ, c),

• (c)

  σ_co(Λ, c) = {α ∈ ℂ : |α − (2 − δ) ⁻¹| < (1 − δ)/(2 − δ)} ∪ S.

Proof. (a) Since the relation C₁σ(Λ, c) = {{α : |α − (2 − δ) ⁻¹| < (1 − δ)/(2 − δ)}∖S}⋃ {α = λ_nn : δ/(2 − δ) < α < 1} holds by Theorems 29 and 30 and from Table 1, σ_ap(Λ, c) = σ(Λ, c)∖C₁σ(Λ, c). Therefore, we have σ_ap(Λ, c) = {α : |α − (2 − δ) ⁻¹| = (1 − δ)/(2 − δ)} ∪ E.

(b) Since σ_δ(Λ, c) = σ(Λ, c)∖A₃σ(Λ, c) from Table 1 and A₃σ(Λ, c) = ∅ by Theorem 32, we have σ_δ(Λ, c) = σ(Λ, c).

(c) Since the equality σ_co(Λ, c) = C₁σ(Λ, c) ∪ C₂σ(Λ, c) ∪ C₃σ(Λ, c) holds from Table 1, we have σ_co(Λ, c) = {α ∈ ℂ : |α − (2 − δ) ⁻¹| < (1 − δ)/(2 − δ)} ∪ S by Theorems 29–32.

Corollary 34. The following results hold:

• (a)

  σ_ap(Λ^*, ℓ₁) = σ(Λ, c),

• (b)

  σ_δ(Λ^*, ℓ₁) = {α : |α − (2 − δ) ⁻¹| = (1 − δ)/(2 − δ)} ∪ E,

• (c)

  σ_p(Λ^*, ℓ₁) = {α ∈ ℂ : |α − (2 − δ) ⁻¹| < (1 − δ)/(2 − δ)} ∪ S.

Theorem 35. Suppose that |α + 1 | >|α − 1|. Then the convergence field of A = αI + (1 − α)Λ is c.

Proof. By Theorem 22, Λ − [α/(α − 1)]I has an inverse in B(c). That is to say that

Since A is a triangle and is in B(c), A⁻¹ is also conservative which implies that c_A = c [22, page 12].