Some Geometric Properties of the Domain of the Double Sequential Band Matrix $$ in the Sequence Space ℓ(p)

Definition 1. Let S(X) be the unit sphere of a Banach space X. Then, a point x ∈ S(X) is called an extreme point if 2x = y + z implies y = z for every y, z ∈ S(X). A Banach space X is said to be rotund (strictly convex) if every point of S(X) is an extreme point.

Definition 2. A Banach space X is said to have Kadec-Klee property (or property (H)) if every weakly convergent sequence on the unit sphere is convergent in norm.

Definition 3. A Banach space X is said to have

• (i)

  the Opial property if every sequence (x_n) weakly convergent to x₀ ∈ X satisfies

  $$ ()
  for every x ∈ X with x ≠ x₀;

• (ii)

  the uniform Opial property if for each ϵ > 0, there exists an r > 0 such that

  $$ ()
  for each x ∈ X with ∥x∥≥ϵ and each sequence (x_n) in X such that x_n → 0 and liminf _n→∞∥x_n∥≥1.

Definition 4. Let X be a real vector space. A functional σ : X → [0, ∞) is called a modular if

• (i)

  σ(x) = 0 if and only if x = θ;

• (ii)

  σ(αx) = σ(x) for all scalars α with |α | = 1;

• (iii)

  σ(αx + βy) ≤ σ(x) + σ(y) for all x, y ∈ X and α, β ≥ 0 with α + β = 1;

• (iv)

  the modular σ is called convex if σ(αx + βy) ≤ ασ(x) + βσ(y) for all x, y ∈ X and α, β > 0 with α + β = 1.

Proposition 5. The modular σ_p on $$ satisfies the following properties with p_k ≥ 1 for all k ∈ ℕ:

• (i)

  if 0 < α ≤ 1, then α^Mσ_p(x/α) ≤ σ_p(x)    and σ_p(αx) ≤ ασ_p(x).

• (ii)

  If α ≥ 1, then σ_p(x) ≤ α^Mσ_p(x/α).

• (iii)

  If α ≥ 1, then σ_p(x) ≤ ασ_p(x/α).

• (iv)

  The modular σ_p   is continuous on the space $$.

Proof. Consider the modular σ_p on $$.

• (i)

  Let 0 < α ≤ 1, then $$. So, we have

  $$ ()

• (ii)

  Let α ≥ 1. Then, $$ for all p_k ≥ 1. So, we have

  $$ ()

• (iii)

  Let α ≥ 1. Then, $$ for all p_k ≥ 1. So, we have

  $$ ()

• (iv)

  By (ii) and (iii), one can immediately see for α > 1 that

  $$ ()
  By passing to limit as α  →  1⁺ in (11), we have $$. Hence, σ_p is right continuous. If 0 < α < 1, by (i) we have
  $$ ()
  By letting α  →  1⁻ in (12), we observe that $$. Hence, σ_p is also left continuous, and so, it is continuous.

Proposition 6. For any $$, the following statements hold:

• (i)

  if ∥x∥<1, then σ_p(x)≤∥x∥.

• (ii)

  If ∥x∥>1, then σ_p(x)≥∥x∥.

• (iii)

  ∥x∥ = 1   if and only if σ_p(x) = 1.

• (iv)

  ∥x∥<1   if and only if σ_p(x) < 1.

• (v)

  ∥x∥>1   if and only if σ_p(x) > 1.

Proof. Let $$.

• (i)

  Let ϵ > 0 be such that 0 < ϵ < 1 − ∥x∥. By the definition of ∥·∥, there exists an α > 0 such that ∥x∥+ϵ > α and σ_p(x) ≤ 1. From Parts (i) and (ii) of Proposition 5, we obtain

  $$ ()
  Since ϵ is arbitrary, we have (i).

• (ii)

  If we choose ϵ > 0 such that 0 < ϵ < 1 − (1/∥x∥), then 1 < (1 − ϵ)∥x∥<∥x∥. By the definition of ∥·∥ and Part (i) of Proposition 5, we have

  $$ ()
  So, (1 − ϵ)∥x∥<σ_p(x) for all ϵ ∈ (0,1 − (1/∥x∥)). This implies that ∥x∥<σ_p(x).

• (iii)

  Since σ_p is continuous, by Theorem 1.4 of [12] we directly have (iii).

• (iv)

  This follows from Parts (i) and (iii).

• (v)

  This follows from Parts (ii) and (iii).

Theorem 7. $$ is a Banach space with Luxemburg norm.

Proof. Let S_x = {α > 0 : σ_p(x/α) ≤ 1} and ∥x∥ = inf S_x for all $$. Then, S_x ⊂ (0, ∞). Therefore, ∥x∥≥0 for all $$.

For x = θ, σ_p(θ) = 0 for all α > 0. Hence, S₀ = (0, ∞) and ∥θ∥ = inf S₀ = inf (0, ∞) = 0.

Let x ≠ θ and Y = {kx : k ∈ ℂ and $$ be a nonempty subset of $$. Since $$, there exists k₁ ∈ ℂ such that $$. Obviously, k₁ ≠ 0. We assume that 0 < α < 1/k₁ and α ∈ S_x. Then, $$. Since |k₁α | < 1, we get

which contradicts the assumption. Hence, we obtain that if α ∈ S_x, then α > 1/|k₁|. This means that ∥x∥≥1/|k₁ | > 0. Thus, we conclude that ∥x∥ = 0 if and only if x = θ.

Now, let k ≠ 0 and α ∈ S_kx. Then, we have

Therefore, we obtain That is, ∥x∥≤α/|k| and |k | ∥x∥≤α for all α ∈ S_kx. So, |k | ∥x∥≤∥kx∥.

If we take 1/k and kx instead of k and x, respectively, then we obtain that

Hence, we get ∥kx∥ = |k | ∥x∥. This also holds when k = 0.

To prove the triangle inequality, let $$ and ϵ > 0 be given. Then, there exist α ∈ S_x and β ∈ S_y such that α < ∥x∥+ϵ and β < ∥y∥+ϵ. Since $$ is convex,

Therefore, α + β ∈ S_x+y. Then, we have ∥x + y∥≤α + β < ∥x∥+∥y∥+2ϵ. Since ϵ > 0 was arbitrary, we obtain ∥x + y∥≤∥x∥+∥y∥. Hence, ∥x∥ = inf {α > 0 : σ_p(x/α) ≤ 1} is a norm on $$.

Now, we need to show that every Cauchy sequence in $$ is convergent according to the Luxemburg norm. Let $$ be a Cauchy sequence in $$ and ϵ ∈ (0,1). Thus, there exists n₀ such that ∥x⁽ⁿ⁾ − x^(m)∥<ϵ for all n, m ≥ n₀. By Part (i) of Proposition 6, we have

for all n, m ≥ n₀. This implies that Then, for each fixed k and for all n, m ≥ n₀, Hence, the sequence $$ is a Cauchy sequence in ℝ. Since ℝ is complete, there is a $$ such that $$ as m → ∞. Therefore, as m → ∞ by (22), we have for all n ≥ n₀.

Now, we have to show that (x_k) is an element of $$. Since $$ as m → ∞, we have

Then, we see by (21) that σ_p(x⁽ⁿ⁾ − x)≤∥x⁽ⁿ⁾ − x∥<ϵ for all n ≥ n₀. This implies that xⁿ → x as n → ∞. So, we have $$. Therefore, the sequence space $$ is complete with respect to Luxemburg norm. This completes the proof.

Theorem 8. The space $$ is rotund if and only if p_k > 1 for all k ∈ ℕ.

Proof. Let $$ be rotund and choose k ∈ ℕ such that p_k = 1 for k < 3. Consider the following sequences given by

Then, obviously x ≠ y and By Part (iii) of Proposition 6, $$ which leads us to the contradiction that the sequence space $$ is not rotund. Hence, p_k > 1 for all k ∈ ℕ.

Conversely, let $$ and $$ with x = (v + z)/2. By convexity of σ_p and Part (iii) of Proposition 6, we have

which gives that σ_p(v) = σ_p(z) = 1, and Also, we obtain from (29) that Since x = (v + z)/2, we have This implies that for all k ∈ ℕ. Since the function $$ is strictly convex for all k ∈ ℕ, it follows by (32) that v_k = z_k for all k ∈ ℕ. Hence, v = z. That is, the sequence space $$ is rotund.

Theorem 9. Let $$. Then, the following statements hold:

• (i)

  0 < α < 1 and ∥x∥>α imply σ_p(x) > α^M.

• (ii)

  α ≥ 1 and ∥x∥<α imply σ_p(x) < α^M.

Proof. Let $$.

• (i)

  Suppose that ∥x∥>α with 0 < α < 1. Then, ∥x/α∥>1. By Part (ii) of Proposition 6, ∥x/α∥>1 implies σ_p(x/α)≥∥x/α∥>1. That is, σ_p(x/α) > 1. Since 0 < α < 1, by Part (i) of Proposition 5, we get α^Mσ_p(x/α) ≤ σ_p(x). Thus, we have α^M < σ_p(x).

• (ii)

  Let ∥x∥<α and α ≥ 1. Then, ∥x/α∥<1. By Part (i) of Proposition 6, ∥x/α∥<1 implies σ_p(x/α)≤∥x/α∥<1. That is, σ_p(x/α) < 1. If α = 1, then σ_p(x/α) = σ_p(x) < 1 = α^M. If α > 1, then by Part (ii) of Proposition 5, we have σ_p(x) ≤ α^Mσ_p(x/α). This means that σ_p(x) < α^M.

Theorem 10. Let (x_n) be a sequence in $$. Then, the following statements hold:

• (i)

  lim _n→∞∥x_n∥ = 1 implies lim _n→∞σ_p(x_n) = 1.

• (ii)

  lim _n→∞σ_p(x_n) = 0 implies lim _n→∞∥x_n∥ = 0.

Proof. Let (x_n) be a sequence in $$.

• (i)

  Let lim _n→∞∥x_n∥ = 1 and ϵ ∈ (0,1). Then, there exists n₀ ∈ ℕ such that 1 − ϵ < ∥x_n∥<ϵ + 1 for all n ≥ n₀. By Parts (i) and (ii) of Theorem 9, 1 − ϵ < ∥x_n∥ implies σ_p(x_n)>(1 − ϵ) ^M and ∥x_n∥<ϵ + 1 implies σ_p(x_n)<(1 + ϵ) ^M for all n ≥ n₀. This means ϵ ∈ (0,1) and for all n ≥ n₀ there exists n₀ ∈ ℕ such that (1 − ϵ) ^M < σ_p(x_n)<(1 + ϵ) ^M. That is, lim _n→∞σ_p(x_n) = 1.

• (ii)

  We assume that lim _n→∞∥x_n∥≠0 and ϵ ∈ (0,1). Then, there exists a subsequence $$ of (x_n) such that $$ for all k ∈ ℕ. By Part (i) of Theorem 9, 0 < ϵ < 1 and $$ imply $$. Thus, lim _n→∞σ_p(x_n) ≠ 0 for all k ∈ ℕ. Hence, we obtain that lim _n→∞σ_p(x_n) = 0 implies lim _n→∞∥x_n∥ = 0.

Theorem 11. Let $$ and $$. If σ_p(x⁽ⁿ⁾) → σ_p(x) as n → ∞ and $$ as n → ∞ for all k ∈ ℕ, then x⁽ⁿ⁾ → x as n → ∞.

Proof. Let ϵ > 0 be given. Since $$, there exists k₀ ∈ ℕ such that

It follows from the fact that there exists n₀ ∈ ℕ such that for all n ≥ n₀ and for all k ∈ ℕ, and for all n ≥ n₀, Therefore, we obtain from (33), (35), and (36) that This means that σ_p(x⁽ⁿ⁾ − x) → 0 as n → ∞. By Part (ii) of Theorem 10, σ_p(x⁽ⁿ⁾ − x) → 0 as n → ∞ implies ∥x_n − x∥→0 as n → ∞. Hence, x_n → x as n → ∞.

Theorem 12. The sequence space $$ has the Kadec-Klee property.

Proof. Let $$ and $$ such that ∥x⁽ⁿ⁾∥→1 and $$ are given. By Part (ii) of Theorem 10, we have σ_p(x⁽ⁿ⁾) → 1 as n → ∞. Also $$ implies ∥x∥ = 1. By Part (iii) of Proposition 6, we obtain σ_p(x) = 1. Therefore, we have σ_p(x⁽ⁿ⁾) → σ_p(x) as n → ∞.

Since $$ and $$ defined by q_k(x) = x_k is continuous, $$ as n → ∞ for all k ∈ ℕ. Therefore, x⁽ⁿ⁾ → x as n → ∞.

Since any weakly convergent sequence in $$ is convergent, the sequence space $$ has the Kadec-Klee property.

Theorem 13. For any 1 < p < ∞, the space $$ has the uniform Opial property.

Proof. Let ϵ > 0 and ϵ₀ ∈ (0, ϵ) be given such that 1 + (ϵ^p/2)>(1 + ϵ₀) ^p. Also let $$ and ∥x∥≥ϵ. There exists k₁ ∈ ℕ such that

Hence, we have Furthermore, we have which yields that For any weakly null sequence $$, since $$ as m → ∞ for each k ∈ ℕ, there exists m₀ ∈ ℕ such that for all m > m₀, Therefore, for all m > m₀, Moreover, Then, we have This means that $$ has the uniform Opial property.