A New Approach For Weighted Hardy’s Operator In VELS

A considerable number of research has been carried out on the generalized Lebesgue spaces Lp(x) and boundedness of different integral operators therein. In this study, a new approach for weighted increasing near the origin and decreasing near infinity exponent function that provides a boundedness of the Hardy’s operator in variable exponent space is given.

In this paper, we establish an integral-type necessary and sufficiency condition on a monotone weighted exponent function p : (0, ∞) → (1, ∞) governing the boundedness of Hardy's operator 1 x v(x)´x 0 f (t)w(t)dt in L p(x) (0, ∞).
The following main result has been obtained in this study.
Theorem 1.1. Let p : (0, ∞) → (1, ∞) be a measurable function monotony increasing on some neighborhood of origin (0, ε) and decreasing on some neighborhood of infinity (N, ∞) such that p + < ∞ and p(∞) > 1. Then it holds the inequality 1 x for any positive measurable function f (x) on (0, ∞) with a positive constant C 1 depending on constant C 2 below and p + , p(∞) if and only if the condition is satisfied.
We use the following notation.By C,C i we denote a positive constant depending on p(∞), p + and C 2 from the conditions (2). We use also notation p The weight functions v and w are assumed to be measurable and having non-negative finite values almost everywhere in (0, ∞).

Auxiliary Statements
In this section, we prove some auxiliary assertions in order to prove the main result of this paper. Lemma 2.1. Let the condition (2) be satisfied for a monotone increasing near the origin on (0, ε) and decreasing near infinity on (N, ∞) function p : (0, ∞) → (1, ∞) such that p + < ∞. Then there exists a positive constant C 3 depending on C 2 , p + such that and Proof. The proof for b ∈ (0, 1) similar to those in (see [10] the Lemma 4.1 therein) . For the completeness of presentation, we consider here the both cases b ∈ (0, ε) and b ∈ (N, ∞).
since in both cases t < x. Indeed, using that 2 −n−1 x < t < 2 −n x and Lemma 2.2, we get This completes the proof of Lemma 2.3 By applying preceding lemmas, we easily get the following assertion.

Sufficiency
Let the condition (2) be satisfied. we will show that inequality (1) holds. Take a measurable positive function f with f p(.) ≤ 1.
In order to prove sufficiency, we have to show that 1 x v(x)´x 0 f (t)w(t)dt) ≤ 1 (see, e.g. in [10]). Using Minkowski's inequality for p(.) -norms we have For the first summand it follows from the results of the paper [10] (for α = 0 therein) that it may be estimated as since an integration interval (0, N) is finite in this part and condition (2) is satisfied. Now, we pass to an estimation for a second summand in (9), i.e. we get an estimate for the term i = . By using Minkowski's inequality for p(.)-norms, it follows that, Since p − > 1 and f p(.) ≤ 1 it follows that since p(∞) > 1, Thus, in order to get an estimation for i it suffices to estimate .
We have We shall estimate every summand on the right hand side. By using Lemma 2.3 for N < t < x < ∞, it follows that . Therefore By using Hölder's inequality for p(x) -norms and assumption (8), we get Lemma 3.1. There exists a positive constant C 9 > 1 depending on C 2 , p + , p(∞), such that for x > N.
Proof. We prove estimation (13) from the opposite. Let inequality (13) be violated. We show that a contradiction occurs. Let From the definition of p (.)-norms, it follows that for any sufficiently small δ > 0 the inequality holds Indeed, if (15) does not hold, we get a contradiction, that is, is a list number satisfying By applying here (14), it follows that For any points t, y lying in (2 −n−1 x, 2 −n x) it follows from Lemma 2.1 that y 1 p (y) is comparable with t 1 p (t) . This means, there exist two positive constants C 10 ,C 11 depending on C 2 , p + such that Therefore, By choosing constant C 9 sufficiently large, we get a contradiction with (16). This proves estimate (13). Thus we have proved that This proves Lemma 3.1. Now, we shall derive an estimate for .
In order to carry out it, we get an estimation for the proper modular for ∀y,t ∈ (2 −n−1 x, 2 −n x) and x > N. By using the last estimate the expression (18) is exceeded x .
By applying here Lemma 2.1, we get that, the term y p (t) in the integral terms, i.e.
By a use of (13) a parentheses term in the preceding integral is less than 1. By decreasing the power p(x) on the power of parentheses to p − x,n , we will increase the fraction. Then the last expression is less then Using Holder's inequality, it follows that v(x) with x ≥ N2 n+1 . By inserting this in the interior integral (19), that is exceeded by Since 2 −n−1 x ≤ t and by using Lemma 2.4 and Lemma 2.1 , it follows that, the last expression is exceeded by Now, we estimate the interior integral through I p(.) ( f ). The interior integral here is exceeded since the assumption (2) is made using (4) from Lemma 2.1 and p − x,n ≥ 1 , it follows that Evidently, Inserting this inequality in (20) and applying Fubini's therom, it follows that Whence, From this it follows that Inserting (23) in (12), we get Further, substitute this estimate and (11) into (10), we complete the proof of the sufficiency part of Theorem 1.1.

Necessity.
Let us note, for an increasing near origin exponent functions p(.) it was proved in [19] that a necessity condition for inequality (1) to hold in the class of measurable positive functions f with support in finite interval (0, N) is the same condition (2) over the points b ∈ (0, N)(observe not over all axes (0, ∞)).To finish the necessity in Theorem 1.1 it remains to get this condition over the points b > N, where we shall essentially use the decreasing of exponent near infinity.
Below, we shall prove the necessity of condition (2) over points x > N for decreasing near infinity exponent functions. We insert a function into inequality (1.2) with a parameter b > N be fixed. It is clear that, I p(.) ( f 0 ) = ln 2. It follows from the inequality (1) that By monotony of p in (N, ∞) it follows the functions x − 1 p (x) and t − 1 p(t) are decreasing therein. This yields By taking into account this inequalities, it follows from (24)  for all N < t 1 < t 2 < ∞. Whence, the function x δ − p + p (x) is almost decreasing on (N, ∞). Therefore, we have proved that t − 1 p (t 2 ) +δ 1 2 ≤ C 28 t − 1 p (t 1 ) +δ 1 1 for all N < t 1 < t 2 < ∞; a constant δ 1 = δ p + . Therefore x − 1 p (x) +δ 1 is almost decreasing with δ 1 = δ p + , δ = 1 C 27 . From this it followŝ This proves the necessity of condition (2) near infinity.

Conclusion
A new method of weighted increasing near origin and decreasing near infinity exponent function that provides a boundedness of the Hardy's operator in variable exponent Lebesgue space was obtained. The method we use here leads us to the most general sense. We don't have to work in a particular interval. This method can be applied to different operators. And this method brings important facilities in the study of operator theory.
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