On the Generalized Krätzel Transform and Its Extension to Bohemian Spaces

Theorem 1. Let 1 ≤ ρ ≤ ∞, μ, v ∈ ℂ, ρ > 0, (1/ρ)+(1/ρ^′) = 1 and Reμ > −(1/ρ^′) − min {0, ρRev} and then $$ is a continuous linear mapping from l_ρ,μ into l_{ρ,2/(ρ−μ−1)}.

Definition 2. Let f, g ∈ l_ρ,μ. For f and g, we define the operation ⊗ given by

Theorem 3. Let f ∈ l_ρ,μ and φ ∈ 𝒟, then $$.

Proof. Let x > 0. For f ∈ l_ρ,μ and φ ∈ 𝒟 we, by (3) and (8), get that

A change of variables y = tz in (10) implies Hence, by (9), we have This completes the proof of theorem.

Theorem 4. $$ as n → ∞ if and only if there are f_n,k, f_k ∈ X and υ_k ∈ Δ such that h_n = [f_n,k/υ_k], h = [f_k/υ_k] and, for all k ∈ ℕ, f_n,k → f_k ∈ X as n → ∞. See [6–17] for more details.

Theorem 5. Let f ∈ l_ρ,μ and φ, ψ ∈ 𝒟; then, f ⊗ (φ⋎ψ) = (f ⊗ φ) ⊗ ψ.

Proof. Let f ∈ l_ρ,μ and φ, ψ ∈ 𝒟; then, applying the Fubinitz theorem yields

The change of variables t = yz implies dt = ydz and further Therefore, Hence, (15) implies This completes the proof of the theorem.

Theorem 6. Let f ∈ l_ρ,μ and φ ∈ 𝒟 and then f⋎φ ∈ l_ρ,μ.

Proof. For each f ∈ l_ρ,μ and φ ∈ 𝒟, we have

By Jensen’s theorem, we write where K = [a, b], b > a > 0 is a compact subset containing the support of φ. Hence, from (18), we get where M is certain positive real number. This completes the proof of the theorem.

Theorem 7. Let f ∈ l_ρ,μ and φ, ψ ∈ 𝒟 and then

• (i)

  f⋎(φ + ψ) = f⋎φ + f⋎ψ;

• (ii)

  (αf)⋎φ = α(f⋎φ);

• (iii)

  let (f_n) ∈ l_ρ,μ and f_n → f as n → ∞ then for φ ∈ 𝒟, and we have f_n⋎φ → f⋎φ as n → ∞;

• (iv)

  f⋎(φ⋎ψ) = (f⋎φ)⋎ψ.

Theorem 8. Let f ∈ l_ρ,μ and (μ_n) ∈ Δ then f⋎μ_n → f as n → ∞.

Proof. Since 𝒟 is a dense subspace of l_ρ then it is a dense subspace of l_ρ,μ. Hence, there can be found ψ ∈ 𝒟 such that

for ε > 0. Also, from (18), we have Let g(t) = y^μψ(yt⁻¹)t⁻¹ then; g(t) ∈ 𝒟 and hence uniformly continuous on ℝ₊.

Therefore, there is δ > 0 such that

Thus, using (22), we get By (22), supp μ_n(t) → 0 as n → ∞ implies that there can be found N ∈ ℕ, such that supp μ_n⊆[0, δ], for all n ≥ N. Further, the fact that the function ψ is of compact support thus this implies that supp ψ(y)⊆K = [a, b], where K is a compact subset of ℝ₊. Thus, from (23), we write Now, we have On using (20), (21) and (24) prove that Thus the theorem is proved. Then the Bohemian space 𝔹ℍ₁(l_ρ,μ) is therefore constructed. The operations such as sum and multiplication by a scalar of two Bohemian in 𝔹ℍ₁(l_ρ,μ) are defined in a natural way where α  is a complex number.

Similarly, the operation ⋎ and differentiation are defined by

Now constructing the space 𝔹ℍ₂(l_{ρ,2/(ρ−μ−1)}) follows from theorems which were used for constructing the space 𝔹ℍ₁(l_ρ,μ). Therefore, the corresponding proofs of Theorems 10 and 11 are omitted.

Theorem 9. Let f ∈ l_{ρ,2/(ρ−μ−1)} and φ ∈ 𝒟 and then f ⊗ φ ∈ l_{ρ,2/(ρ−μ−1)}.

Theorem 10. Let f ∈ l_{ρ,2/(ρ−μ−1)}, φ, ψ ∈ 𝒟 and then the following hold:

• (i)

  f ⊗ (φ + ψ) = f ⊗ φ + f ⊗ ψ;

• (ii)

  (αf) ⊗ φ = α(f ⊗ φ);

• (iii)

  if f_n → f in l_{ρ,2/(ρ−μ−1)}, as n → ∞ then f_n ⊗ φ → f ⊗ φ as n → ∞;

• (iv)

  if (μ_n) ∈ Δ, then f ⊗ μ_n → f as n → ∞.

Theorem 11. Let f ∈ l_{ρ,2/(ρ−μ−1)} and φ, ψ ∈ 𝒟 and then f ⊗ (φ⋎ψ) = (f ⊗ φ) ⊗ ψ.

Theorem 12. The mapping $$ is

• (i)

  well defined,

• (ii)

  linear.

Proof. Let [(f_n)/(μ_n)], [(g_n)/(ψ_n)] ∈ 𝔹ℍ₁(l_ρ,μ) be such that [(f_n)/(μ_n)] = [(g_n)/(ψ_n)], and then f_n⋎ψ_m = g_m⋎μ_n = g_n⋎μ_m. Using Theorem 4 implies $$, for all n, m. The idea of quotient of sequences in 𝔹ℍ₂(l_{ρ,2/(ρ−μ−1)}) implies that

That is, To prove part (ii) of the theorem, if [(f_n)/(μ_n)], [(g_n)/(ψ_n)] ∈ 𝔹ℍ₁(l_ρ,μ), then Hence Also, if α ∈ ℂ, then Hence This completes the proof.

Definition 13. Let $$. We define the inverse of $$ transform of the Bohemian $$ as

for each (μ_n) ∈ Δ.

Theorem 14. $$ is an isomorphism.

Proof. Let $$ and then by (29) we get $$. Therefore, Theorem 4 implies

Thus f_n⋎ψ_m = g_m⋎μ_n, for all m, n ∈ ℕ. The concept of quotients of equivalent classes of 𝔹ℍ₁(l_ρ,μ) then gives This proves that $$ is injective.

To show that $$ is surjective, let $$. Then $$ is a quotient of sequences in 𝔹ℍ₂(l_{ρ,2/(ρ−μ−1)}).  Hence, $$, for all m, n ∈ ℕ. Once again, Theorem 4 implies that $$. Hence [(f_n)/(μ_n)] ∈ 𝔹ℍ₁(l_ρ,μ) satisfies

This completes the proof of the theorem.

Theorem 15. Let ψ ∈ 𝒟 and$$ then one has

• (i)

  $$;

• (ii)

  $$.

Proof. By using (29), we have

Theorem 4 then gives Hence the part (i) of the theorem is proved. The proof of part (ii) is similar thus we omit the details. This completes the proof of the theorem.

Theorem 16. The mappings

• (i)

  $$ are continuous with respect to δ and Δ-convergence.

• (ii)

  $$ are continuous with respect to δ and Δ-convergence.

Proof. At first, let us show that $$ and $$ are continuous with respect to δ-convergence.

Let $$ in 𝔹ℍ₁(l_ρ,μ) as n → ∞ and then we show that $$ as n → ∞. By virtue of Theorem 5, we can find f_n,k and f_k in l_ρ,μ such that

such that f_n,k → f_k as n → ∞ for every k ∈ ℕ. Hence, $$ as n → ∞. Thus, as n → ∞.

To prove the Part (ii), Let g_n, g ∈ 𝔹ℍ₂(l_{ρ,2/(ρ−μ−1)}) be such that $$ as n → ∞. Then, once again, by Theorem 5, $$ and $$ for some f_n,k, f_k ∈ l_ρ,μ and $$ as n → ∞. Hence [f_n,k/μ_k] → [f_k/μ_k] as n → ∞.

Using (36), we get

Now, we establish that $$ and $$ are continuous with respect to Δ-convergence.

Let $$ in 𝔹ℍ₁(l_ρ,μ) as n → ∞. Then, there exist f_n ∈ l_ρ,μ and (μ_n) ∈ Δ such that (β_n − β)⋎μ_n = [((f_n)⋎μ_k)/μ_k] and f_n → 0 as n → ∞. By applying (29) we get

Hence, we have $$ as n → ∞ in l_{ρ,2/(ρ−μ−1)}.

Therefore

Hence, $$ as n → ∞.

Finally, let $$ as n → ∞ and then we find $$ such that $$ and $$ as n → ∞ for some (μ_n) ∈ Δ.

Now, using (36), we obtain

Theorem 4 implies Thus From this, we find that $$ as n → ∞ in 𝔹ℍ₁(l_ρ,μ).

This completes the proof of the theorem.

Theorem 17. The $$ transform is consistent with the classical transform $$.

Proof. For every f ∈ l_ρ,μ, let β be its representative in 𝔹ℍ₁(l_ρ,μ) and then β = [(f⋎(μ_n))/(μ_n)] where (μ_n) ∈ Δ, for all n ∈ ℕ. Since (μ_n) is independent from the representative, for all n ∈ ℕ, therefore

which is the representative of $$. Thus this completes the proof.

Theorem 18. Let [(g_n)/(ψ_n)] ∈ 𝔹ℍ₂(l_{ρ,2/(ρ−μ−1)}) and then a necessary and sufficient condition for [(g_n)/(ψ_n)] to be in the range of $$ is that g_n belongs to range of $$, for all n ∈ ℕ.

Proof. Let [(g_n)/(ψ_n)] be in the range of $$ and then it is clear that g_n belongs to the range of $$, for all n ∈ ℕ.

To establish the converse, let g_n be in the range of $$, for all n ∈ ℕ. Then there is f_n ∈ l_ρ,μ such that $$, n ∈ ℕ. Since [(g_n)/(ψ_n)] ∈ 𝔹ℍ₂(l_{ρ,2/(ρ−μ−1)}),

for all m, n ∈ ℕ, therefore, where f_n ∈ l_ρ,μ and φ_n ∈ Δ, for all n ∈ ℕ. Then it follows that we get f_n⋎φ_m = f_m⋎φ_n, for all m, n ∈ ℕ. Thus f_n/φ_n is a quotient of sequences in 𝔹ℍ₁(l_ρ,μ). Therefore, we have [(f_n)/(φ_n)] ∈ 𝔹ℍ₁(l_ρ,μ) and Hence the theorem is proved.

Theorem 19. If β = [(f_n)/(φ_n)] ∈ 𝔹ℍ₁(l_ρ,μ) and γ = [(κ_n)/(μ_n)] ∈ 𝔹ℍ₁(l_ρ,μ), then one has

Proof. Let the hypothesis of the theorem be satisfied for some [(f_n)/(φ_n)] and [(κ_n)/(μ_n)]. Therefore,

This completes the proof.