Two-weight inequalities for singular integral operators satisfying a variant of Hormander's condition

In this paper, we present some sufficient conditions for the boundedness of convolution operators that their kernel satisfies a certain version of Hormander's condition, in the weighted Lebesgue spaces Lp,ω(R n ).

We write f ∈ L loc p (R n ), 1 ≤ p < ∞, if f belongs to L p (F ) on any closed bounded set F ⊂ R n .
Let K : R n 0 → R, K ∈ L loc 1 (R n 0 ), R n 0 = R n \ {0} , be a function satisfying the following conditions: Let f ∈ L p (R n ), 1 < p < ∞, and consider the following singular integral (1) In the following theorem Calderon and Zygmund [5] proved the boundedness of the operator T . Theorem 1. Suppose that the kernel K of the singular integral (1) satisfies conditions 1) − 3) and f ∈ L p (R n ), 1 ≤ p < ∞. Then the singular integral exists for x ∈ R n almost everywhere and the following inequalities holds where C 1 , C 2 > 0 is independent of f.
Hörmander [13] imposed a weaker constraint on the kernel of the singular integral (1), namely, where K ∈ L loc 1 (R n 0 ) and C > 0 is a constant independent of y . By replacing condition 3) with condition (2), under conditions 1), 2) he proved Theorem 1 for singular integrals with kernels satisfying condition (2). This condition is related to condition 3), and under this condition, inequality (2) holds ( see [19]).
On the other hand, singular integrals whose kernels do not satisfy Hörmander's condition (2) are widely considered, for example oscillatory and some other singular integrals ( see [20]).
for any balls B ⊂ R n . Suppose that the function K satisfies conditions (K1) − (K4). For f ∈ L p (R n ), 1 ≤ p < ∞ define the following convolution operator generated by the kernel K as For the convolution operator (5), the following theorem holds.
Theorem 2. [20] Suppose that w ∈ A p (R n ), 1 ≤ p < ∞, and the kernel of the convolution operator (5) satisfies conditions (K1) − (K4). Then the following inequalities holds: Note that in the "nonweighted" case, when condition (K2) is not imposed and condition (3) is replaced by condition (4), Theorem 2 was proved in [10]. (i) For the validity of the inequality
holds, where the constant c > 0 does not depend on F.
(i) For the n-dimensional Hardy inequality with a constant C 5 , independent on f , to hold, it is necessary and sufficient that the following condition be satisfied: with a constant C 6 , independent on f , to hold, it is necessary and sufficient that the following condition be satisfied: , ω 1 (x)) satisfies the following condition: |x|>2r Then there exists a positive constant C depending only on p, n such that, for any t > 0 , the following inequality holds: In the case ϕ = 1 Lemma 4 was proved also in [11].

Main results
Theorem 3. Suppose that the kernel K of the convolution operator (5) and there exist b > 0 such that Then there exists a C 7 > 0 such that, for any f ∈ L p,ω (R n ), 1 < p < ∞ the following inequality holds Moreover, the condition (6) can be replaced by the condition : there exist b > 0 such that Proof.
and the multiplicity of the covering {E k,2 } k∈Z is equal to 3. where Hence by condition (K2) Since A p (ω, ω 1 ) < ∞, the Hardy inequality holds and C 9 ≤ c A p (ω, ω 1 ), where c depends only on n and p. In fact the condition A p (ω, ω 1 ) < ∞ is necessary and sufficient for the validity of this inequality (see [1], [6]). Hence, we obtain where C 9 is independent of f . Next we estimate A 3 f Lp,ω 1 . It is easy to verify, for x ∈ E k , y ∈ E k,3 we have |y| > 2|x| and |x − y| ≥ |y|/2 . Since E k ∩ suppf k,3 = ∅ , for x ∈ E k by condition (K2) we obtain holds and C 6 ≤ c B p (ω, ω 1 ), where c depends only on n and p. In fact the condition B p (ω, ω 1 ) < ∞ is necessary and sufficient for the validity of this inequality (see [1], [6]). Hence, we obtain for almost all x ∈ E k,2 . Therefore we get (11) since the multiplicity of covering {E k,2 } k∈Z is equal to 3, where C 10 = 3 A p φ b . Inequalities (8), (9), (10), (11) imply (7) which completes the proof.
Analogously proved the following theorem.
We have If u 1 (0+) = 0 , then J 1 = 0. If u 1 (0+) = 0 by the weak L 1 boundedness of A, φ ∈ A 1 (R n ) thanks to Lemma 4 After changing the order of integration in J 2 we have Using the weak L 1 boundeedness of A and Lemma 4 we have Let us estimate J 22 . For |x| > t and |y| ≤ t/2 we have |x|/2 ≤ |x − y| ≤ 3|x|/2, and so The Hardy inequality for p = 1 is characterized by the condition C ≤ c A 1 (see [4], [14]), where |x|>2r Hence, applying the Hardy inequality, we obtain Combining the estimates of J 1 and J 2 , we get (12) for ω 1 (t) = ω 1 (0+) + t 0 ψ(τ )dτ . By Fatou's theorem on passing to the limit under the Lebesgue integral sign, this implies (12). The theorem is proved.
Analogously proved the following theorem.
We have If u 1 (+∞) = 0, then I 1 = 0. If u 1 (+∞) = 0, by the weak L 1 boundedness of A, φ ∈ A 1 (R n ) thanks to Lemma 4 After changing the order of integration in J 2 we have Using the weak L 1 boundedness of A and Lemma 4 we obtain Let us estimate J 22 . For |x| < t and |y| ≥ 2t we have |y|/2 ≤ |x − y| ≤ 3|y|/2, and so Condition (c ) of the theorem guarantees that B ≤ B < ∞. Hence, applying the Hardy inequality, we obtain Combining the estimates of I 1 and I 2 , we get (12) for ω 1 (t) = ω 1 (+∞) + ∞ t ψ(t)dt. By Fatou's theorem on passing to the limit under the Lebesgue integral sign, this implies (12). The theorem is proved.

Remark 2.
Note that for the case in which u = u 1 = 1 , Theorem 3 was proved in [20] by using different methods. Further, in the case 1 < p < ∞ Theorems 6 and 8 was proved in [3].