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Introduction
Variable exponent studies have been stimulated by problems of elasticity, fluid dynamics, calculus of variations and differential equations with a non-standard growth condition, we refer to monographs [7, 11, 12, 13, 15, 18, 21, 23, 29, 30, 31]. The boundedness problems for weighted Hardy’s operator in variable exponent Lebesgue spaces Lp(.) are well studied, we refer to monographs [4, 9, 14, 16, 17, 19, 20, 22, 24, 26, 27, 28]. In this connection, a necessary and sufficient condition that assumes a log-regularity of exponent function near origin and infinity has been proved in [1, 5, 6, 8, 10, 17, 23]. For a compactness problem of main integral operators in variable exponent Lebesgue space, we refer to [2, 3, 25] and especially for Hardy’s operator [4].
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 $\begin{array}{}
\frac{1}{x}v(x)\int_{0}^{x}f(t)w(t)dt
\end{array}$ in Lp(x)(0, ∞).
The following main result has been obtained in this study.
Theorem 1.1
Letp : (0, ∞) → (1, ∞) be a measurable function monotony increasing on some neighborhood of origin (0, ε) and decreasing on some neighborhood of infinity (N, ∞) such thatp+ < ∞ andp(∞) > 1. Then it holds the inequality
We use the following notation.By C, Ci we denote a positive constant depending on p(∞), p+ and C2 from the conditions (2). We use also notation p+ = sup {p(x) : x ∈ (0, ∞)} and p– = inf{p(x) : x ∈ (0, ∞)}. Recall the norm in variable exponent Lebesgue space Lp(.)(0, ∞) given as ∥f∥ = inf {λ > 0 : Ip(.)$\begin{array}{}
\displaystyle
\big(\frac{f}{\lambda} \big)
\end{array}$ ≤ 1} makes it as a Banach space, where the modular Ip(.)(f) = $\begin{array}{}
\int_0^\infty
\end{array}$ | f(t)|p(t)dt. Denote ∥f∥p(.) or ∥f∥Lp(.)(0,∞) for the Lp(.)(0, ∞) variable exponent Lebesgue norm of function f. For a function p : [0, ∞) → [1, ∞) denote p′(x) the function satisfying $\begin{array}{}
\displaystyle
\frac{1}{p(x)}+\frac{1}{p'(x)} = 1
\end{array}$ and p′ = ∞ if p = 1.
Denote by χE the characteristic function of set E ⊂ R. 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 thatp+ < ∞. Then there exists a positive constantC3depending onC2, p+such that
$$\begin{array}{}
\displaystyle
[ p(b)-p(2b)] \ln b \leq C_{4} ~~ for ~~b\in (N,\infty ).
\end{array}$$
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, ∞).
Let the case b ∈ (0, 1) be considered. Write the condition (2) over interval (0, 1):
By monotony increasing of p, $\begin{array}{}
\displaystyle
x^{-\frac{1}{p^{^{\prime }}(x)}}\geq (4b)^{-\frac{1}{
p^{^{\prime }}(2b)}}
\end{array}$ for x ∈ (2b, 4b), where 0 < b < $\begin{array}{}
\displaystyle
\frac{1}{4}
\end{array}$. From this, it follows that
Let b > N and p(x) decreases near infinity, say for x > M. Then $\begin{array}{}
\displaystyle
x^{-\frac{1}{p^{^{\prime }}(x)}}\geq x^{-\frac{1}{
p^{^{\prime }}(2b)}}
\end{array}$ for x ∈ (b, 2b) and b > N. Therefore, from condition (2) over (N, ∞), it follows that,
$$\begin{array}{}
\displaystyle
\int\limits_{b}^{\infty }x^{-\frac{1}{p^{^{\prime }}(x)}}\frac{dx}{x}\leq
C_{2}b^{-\frac{1}{p^{^{\prime }}(b)}},\quad b \gt N.
\end{array}$$
$$\begin{array}{}
\displaystyle
b^{p(b)-p(2b)}\leq C_{6}, \, \,~b \gt N
\end{array}$$
where $\begin{array}{}
\displaystyle
C_6=C_5 ^{p(N)^2}.
\end{array}$ Therefore (4) satisfied by a constant C4 = (p+)2 ln C5. This proves Lemma 2.1
Lemma 2.2
Let the condition(2)be satisfied for a monotone increasing near origin on (0, ε) and decreasing near infinity on (N, ∞) functionp : (0, ∞) → (1, ∞) such thatp+ < ∞. Then there exists anδ > 0 depending onC2, p+such that the function$\begin{array}{}
\displaystyle
x^{\delta - \frac{1}{p^\prime (x)}}
\end{array}$is almost decreasing near origin (0, ε) and infinity (N, ∞).
Proof
Set $\begin{array}{}
g(x)=\int\limits_{x}^{\infty }t^{-\frac{1}{p^{^{\prime }}(t)}}\frac{
dt}{t}, \quad x \gt 0.
\end{array}$ From (2), it follows that
for all N < t1 < t2 < ∞. Whence, the function $\begin{array}{}
\displaystyle
x^{\delta -\frac{1}{p^{^{\prime }}(x)}}
\end{array}$ is almost decreasing on (N, ∞). Therefore, we have proved that
Let the condition(2)be satisfied for a monotone increasing near origin on (0, ε) and decreasing near infinity on (N, ∞) functionp : (0, ∞) → (1, ∞) such thatp+ < ∞. Then for 0 < x < εor forx > Nandt ∈ (2–n–1x, 2–nx), the following inequality holds:
By applying preceding lemmas, we easily get the following assertion.
Lemma 2.4
Let the condition(2)be satisfied for a monotone increasing near origin on (0, ε) and decreasing near infinity on (N, ∞) functionp : (0, ∞) → (1, ∞) such thatp+ < ∞. Thenfor 0 < x < εorx > Nandt ∈ (2–n–1x, 2–nx), the following estimate holds
Let the condition (2) be satisfied. we will show that inequality (1) holds. Take a measurable positive function f with
$$\begin{array}{}
\displaystyle
\Vert f \Vert_{p(.)}\leq 1.
\end{array}$$
In order to prove sufficiency, we have to show that $\begin{array}{}
\Vert \frac{1}{x}v(x)\int_0^x f(t)w(t)dt) \Vert_{p(.)} \leq 1
\end{array}$ or the same is to show that $\begin{array}{}
I_{p(.)}\left(\frac{1}{x}v(x)\int_0^x f(t)w(t)dt) \right)\leq 1
\end{array}$ (see, e.g. in [10]).
Using Minkowski’s inequality for p(.)-norms we have
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 = $\begin{array}{}
\left\Vert x^{-1}\int\limits_{0}^{x}f(t)dt\right\Vert_{L^{p(.)}(N,\infty )}.
\end{array}$
By using Minkowski’s inequality for p(.)-norms, it follows that,
We shall estimate every summand on the right hand side. By using Lemma 2.3 for N < t < x < ∞, it follows that $\begin{array}{}
\displaystyle
x^{-\frac{1
}{p^{\prime }(x)}}\leq C_{7}2^{-n\delta }t^{-\frac{1}{p^{\prime }(t)}}.
\end{array}$ Therefore
We set $\begin{array}{}
\displaystyle
\delta =\frac{1}{2}\left\Vert \chi_{\left \{ t \gt N \right \} } (.) \chi_{(2^{-n-1}x,2^{-n}x)}(.)\right\Vert _{L^{p^\prime(.)}}
\end{array}$ in (15), to get
For any points t, y lying in (2–n–1x, 2–nx) it follows from Lemma 2.1 that $\begin{array}{}
\displaystyle
y^{\frac{1}{p^{\prime }(y)}}
\end{array}$ is comparable with $\begin{array}{}
\displaystyle
t^{\frac{1}{p^{\prime }(t)}}.
\end{array}$ This means, there exist two positive constants C10, C11 depending on C2, p+ such that
By applying here Lemma 2.1, we get that, the term $\begin{array}{}
\displaystyle
y^{-\frac{p(x)
}{p^{\prime }(y)}}
\end{array}$ is comparable with $\begin{array}{}
\displaystyle
t^{-\frac{p(x)
}{p^{\prime }(t)}}
\end{array}$ 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 $\begin{array}{}
\displaystyle
p_{x,n}^{-},
\end{array}$ we will increase the fraction. Then the last expression is less then
Evidently, $\begin{array}{}
\displaystyle
\left(
\frac{1}{1+t^{2}}\right) ^{p_{x,n}^{-}}\leq \frac{1}{1+t^{2}}.
\end{array}$ Therefore, we get an estimate
$$\begin{array}{}
\displaystyle
\int\limits_{B_{x,n}}\left( f\left( t\right) \right) ^{p_{x,n}^{-}}\chi
_{\left\{ t \gt N\right\} }(t)dt\leq C_{15}\int\limits_{B_{x,n}}f\left( t\right)
^{p(t)}\chi _{\left\{ t \gt N\right\} }(t)dt+\int\limits_{B_{x,n}}\left( \frac{
\chi _{\left\{ t \gt N\right\} }(t)}{1+t^{2}}\right) dt .
\end{array}$$
Inserting this inequality in (20) and applying Fubini’s therom, it follows that
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
$$\begin{array}{}
\displaystyle
f_0(x)=x^{-\frac{1}{p(x)}} \chi_{(b,2b)}(x), \quad x \gt N
\end{array}$$
into inequality (1.2) with a parameter b > N be fixed. It is clear that, Ip(.) (f0) = ln 2. It follows from the inequality (1) that
By monotony of p in (N, ∞) it follows the functions $\begin{array}{}
\displaystyle
x^{-\frac{1}{p^\prime(x)}}
\end{array}$ and $\begin{array}{}
\displaystyle
t^{-\frac{1}{p(t)}}
\end{array}$ are decreasing therein. This yields
Now, having property (25) and using inequality (1), we shall derive condition (2) for b > N, too. In connection, we need some assertions below, e.g. following lemma is similar to those in [10].
Lemma 3.2
For any$\begin{array}{}
\displaystyle
\frac{1}{2}x
\end{array}$ < y < 2xwithx > Nthe estimates
holds, and $\begin{array}{}
\displaystyle
\sum \limits_{k \in N^{\prime\prime}}
\end{array}$ is a summation of the opposite case. Therefore, for any k ∈ N one gets
i.e. function $\begin{array}{}
\displaystyle
x^{-\frac{1}{p\prime(x)}}
\end{array}$ is almost decreasing. Since $\begin{array}{}
\displaystyle
x^{-\frac{1}{p\prime(x)}}
\end{array}$ is almost decreasing on [N, ∞), we get
$$\begin{array}{}
\displaystyle
\int_b^\infty x^{-\frac{p^+}{p^\prime(x)}}\frac{dx}{x}\leq C_{27}b^{\frac {p^+} {p^\prime(b)}}, \quad b \gt N.
\end{array}$$
Now, let N < t1 < t2 < ∞. We show that condition (26) entails (2). We take any N < t1 < t2 < ∞ and set $\begin{array}{}
K(x)=\int_x^\infty t^{-\frac{p^+}{p^\prime(t)}}\frac{dt}{t},\quad x \gt N.
\end{array}$ From (26) it follows that
$$\begin{array}{}
\displaystyle
K(x)\le -C_{27} x K'(x), \quad N \lt x \lt \infty.
\end{array}$$
Integrating this inequality over (t1, t2), for N < t1 < t2 < ∞ and using (26), we get
On the other hand, by using monotone decreasing of $\begin{array}{}
\displaystyle
x^{-\frac{p^+}{p^\prime(x)}},
\end{array}$ the conditions (25) and (26), it follows that
for all N < t1 < t2 < ∞. Whence, the function $\begin{array}{}
\displaystyle
x^{\delta -\frac{p^+}{p^{^{\prime }}(x)}}
\end{array}$ is almost decreasing on (N, ∞). Therefore, we have proved that
for all N < t1 < t2 < ∞; a constant $\begin{array}{}
\displaystyle
\delta_1 =\frac{\delta}{p^+}.
\end{array}$
Therefore $\begin{array}{}
\displaystyle
x^{-\frac{1}{p^\prime(x)}+\delta_1}
\end{array}$ is almost decreasing with $\begin{array}{}
\displaystyle
\delta_1=\frac{\delta }{ p^+}, \, \delta=\frac{1}{C_{27}}.
\end{array}$ From this it follows
$$\begin{array}{c}
\displaystyle
\int_b^\infty x^{-\frac{1}{p^\prime(x)}}\frac{dx}{x}\leq C_{28}\int_a^\infty b^{\delta-\frac{1}{p^\prime(x)}}\frac{dx}{x^{1+\delta}}
\\
\displaystyle\leq C_{28}b^{\delta-\frac{1}{p^\prime(b)}}\int_b^\infty\frac{dx}{x^{1+\delta}}=\frac{C_{28}}{\delta}b^{-\frac{1}{p^\prime(a)}}, \quad b \gt N
\end{array}$$
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.