A New Approach For Weighted Hardy’s Operator In VELS

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 L^p(.) 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

$\begin{matrix} {∥\frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t∥}_{L^{p} (0, \infty)} \leq C {∥f∥}_{L^{p} (0, \infty)} \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert \frac{1}{x}v(x)\int\limits_{0}^{x}f(t)w(t)dt\right\Vert _{L^{p}(0,\infty )}\leq C\left\Vert f\right \Vert _{L^{p}(0,\infty )} \end{array}$$(1)

of Hardy’s operator $\begin{matrix} \frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t \end{matrix}$ $\begin{array}{} \frac{1}{x}v(x)\int_{0}^{x}f(t)w(t)dt \end{array}$ 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

$\begin{matrix} {∥\frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t∥}_{L^{p} (0, \infty)} \leq C_{1} {∥f∥}_{L^{p} (0, \infty)} \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert \frac{1}{x}v(x)\int_0^x f(t)w(t)dt \right \Vert _{L^{p}(0,\infty )}\leq C_1\left\Vert f\right\Vert _{L^{p}(0,\infty )} \end{array}$$

for any positive measurable function f(x) on (0, ∞) with a positive constant C₁ depending on constant C₂ below and p⁺, p(∞) if and only if the condition

$\begin{matrix} \int_{b}^{\infty} x^{- \frac{1}{p^{'} (x)}} \frac{d x}{x} \leq C_{2} b^{- \frac{1}{p^{'} (b)}}, b > 0 . \end{matrix}$ $$\begin{array}{} \displaystyle \int_b^\infty x^{- \frac{1}{p^\prime (x)}}\, \frac{dx}{x}\leq C_{2}b^{- \frac{1}{p^\prime (b)}},\quad b \gt 0\text{.} \end{array}$$(2)

is satisfied.

We use the following notation.By C, C_i 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 L^p(.)(0, ∞) given as ∥f∥ = inf {λ > 0 : I_p(.) $\begin{matrix} (\frac{f}{λ}) \end{matrix}$ $\begin{array}{} \displaystyle \big(\frac{f}{\lambda} \big) \end{array}$ ≤ 1} makes it as a Banach space, where the modular I_p(.)(f) = $\begin{matrix} \int_{0}^{\infty} \end{matrix}$ $\begin{array}{} \int_0^\infty \end{array}$ | f(t)|^p(t)dt. Denote ∥f∥_p(.) or ∥f∥_{L^p(.)(0,∞)} for the L^p(.)(0, ∞) variable exponent Lebesgue norm of function f. For a function p : [0, ∞) → [1, ∞) denote p′(x) the function satisfying $\begin{matrix} \frac{1}{p (x)} + \frac{1}{p^{'} (x)} = 1 \end{matrix}$ $\begin{array}{} \displaystyle \frac{1}{p(x)}+\frac{1}{p'(x)} = 1 \end{array}$ and p′ = ∞ if p = 1.

Denote by χ_E 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 that p⁺ < ∞. Then there exists a positive constant C₃ depending on C₂, p⁺ such that

$\begin{matrix} [p (2 b) - p (b)] \ln \frac{1}{b} \leq C_{3} f o r b \in (0, δ) \end{matrix}$ $$\begin{array}{} \displaystyle \lbrack p(2b)-p(b)]\ln \frac{1}{b}\leq C_{3}{ \,\, for \,\, }b\in (0,\delta ) \end{array}$$(3)

and

$\begin{matrix} [p (b) - p (2 b)] \ln b \leq C_{4} f o r b \in (N, \infty) . \end{matrix}$ $$\begin{array}{} \displaystyle [ p(b)-p(2b)] \ln b \leq C_{4} ~~ for ~~b\in (N,\infty ). \end{array}$$(4)

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):

$\begin{matrix} \int_{b}^{\infty} x^{- \frac{1}{p^{^{'}} (x)}} \frac{d x}{x} \leq C_{2} b^{- \frac{1}{p^{^{'}} (b)}}, 0 < b < \frac{1}{4} . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{b}^{\infty }x^{-\frac{1}{p^{^{\prime }}(x)}}\frac{dx}{x}\leq C_{2}b^{-\frac{1}{p^{^{\prime }}(b)}},\quad 0 \lt b \lt \frac{1}{4}. \end{array}$$

By monotony increasing of p, $\begin{matrix} x^{- \frac{1}{p^{^{'}} (x)}} \geq (4 b)^{- \frac{1}{p^{^{'}} (2 b)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1}{p^{^{\prime }}(x)}}\geq (4b)^{-\frac{1}{ p^{^{\prime }}(2b)}} \end{array}$ for x ∈ (2b, 4b), where 0 < b < $\begin{matrix} \frac{1}{4} \end{matrix}$ $\begin{array}{} \displaystyle \frac{1}{4} \end{array}$. From this, it follows that

$\begin{matrix} C_{2} b^{- \frac{1}{p^{^{'}} (b)}} \geq \int_{b}^{2 b} x^{- \frac{1}{p^{^{'}} (x)}} \frac{d x}{x} \geq (4 b)^{- \frac{1}{p^{^{'}} (2 b)}} \ln 2. \end{matrix}$ $$\begin{array}{} \displaystyle C_{2}b^{-\frac{1}{p^{^{\prime }}(b)}}\geq \int\limits_{b}^{2b}x^{-\frac{1}{ p^{^{\prime }}(x)}}\frac{dx}{x}\geq (4b)^{-\frac{1}{p^{^{\prime }}(2b)}}\ln 2. \end{array}$$

Therefore, for 0 < b < $\begin{matrix} \frac{1}{4} \end{matrix}$ $\begin{array}{} \displaystyle \frac{1}{4} \end{array}$, we have

$\begin{matrix} {(\frac{1}{b})}^{\frac{1}{p (b)} - \frac{1}{p (2 b)}} \leq \frac{C_{2}}{\ln 2} 4^{^{\frac{1}{p^{^{'}} (2 b)}}} \leq \frac{4 C_{2}}{\ln 2} = C_{3} . \end{matrix}$ $$\begin{array}{} \displaystyle \left( \frac{1}{b}\right) ^{\frac{1}{p(b)}-\frac{1}{p(2b)}}\leq \frac{C_{2}}{ \ln 2}4^{^{\frac{1}{p^{^{\prime }}(2b)}}}\leq \frac{4C_{2}}{\ln 2}=C_{3} \text{.} \end{array}$$

This proves (3).

Let b > N and p(x) decreases near infinity, say for x > M. Then $\begin{matrix} x^{- \frac{1}{p^{^{'}} (x)}} \geq x^{- \frac{1}{p^{^{'}} (2 b)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1}{p^{^{\prime }}(x)}}\geq x^{-\frac{1}{ p^{^{\prime }}(2b)}} \end{array}$ for x ∈ (b, 2b) and b > N. Therefore, from condition (2) over (N, ∞), it follows that,

$\begin{matrix} \int_{b}^{\infty} x^{- \frac{1}{p^{^{'}} (x)}} \frac{d x}{x} \leq C_{2} b^{- \frac{1}{p^{^{'}} (b)}}, b > N . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{b}^{\infty }x^{-\frac{1}{p^{^{\prime }}(x)}}\frac{dx}{x}\leq C_{2}b^{-\frac{1}{p^{^{\prime }}(b)}},\quad b \gt N. \end{array}$$

Then

$\begin{matrix} C_{2} b^{- \frac{1}{p^{^{'}} (b)}} \geq \int_{b}^{2 b} x^{- \frac{1}{p^{^{'}} (x)}} \frac{d x}{x} \geq \int_{b}^{2 b} x^{- \frac{1}{p^{^{'}} (2 b)}} \frac{d x}{x} \geq a^{- \frac{1}{p^{^{'}} (2 b)}} \ln 2 . \end{matrix}$ $$\begin{array}{} \displaystyle C_{2}b^{-\frac{1}{p^{^{\prime }}(b)}}\geq \int\limits_{b}^{2b}x^{-\frac{1}{ p^{^{\prime }}(x)}}\frac{dx}{x} \geq \int\limits_{b}^{2b}x^{-\frac{1}{ p^{^{\prime }}(2b)}}\frac{dx}{x} \geq a^{-\frac{1}{p^{^{\prime }}(2b)}}\ln 2 \text{.} \end{array}$$

and a decreasing of p over (b, 2b) provides

$\begin{matrix} x^{- \frac{1}{p^{^{'}} (2 b)}} \geq 2^{\frac{1}{p^{^{'}} (2 b)}} b^{- \frac{1}{p^{^{'}} (2 b)}} \geq b^{- \frac{1}{p^{^{'}} (2 b)}} \ln 2. \end{matrix}$ $$\begin{array}{} \displaystyle x^{-\frac{1}{ p^{^{\prime }}(2b)}} \ge 2^{\frac{1}{ p^{^{\prime }}(2b)}} b^{-\frac{1}{ p^{^{\prime }}(2b)}} \ge b^{-\frac{1}{ p^{^{\prime }}(2b)}} \ln 2. \end{array}$$

Therefore, for b > N, we have

$\begin{matrix} b^{\frac{1}{p (2 b)} - \frac{1}{p (b)}} \leq \frac{C_{2}}{\ln 2} = C_{5} or b^{p (b) - p (2 b)} \leq C_{6} . \end{matrix}$ $$\begin{array}{} \displaystyle b^{\frac{1}{p(2b)}-\frac{1}{p(b)}} \leq \frac{C_{2}}{\ln 2}=C_5 ~~~ \text{ or} ~ ~~ b ^{p(b)-p(2b)}\leq C_{6}. \end{array}$$

Whence,

$\begin{matrix} b^{p (b) - p (2 b)} \leq C_{6}, b > N \end{matrix}$ $$\begin{array}{} \displaystyle b^{p(b)-p(2b)}\leq C_{6}, \, \,~b \gt N \end{array}$$

where $\begin{matrix} C_{6} = C_{5}^{p (N)^{2}} . \end{matrix}$ $\begin{array}{} \displaystyle C_6=C_5 ^{p(N)^2}. \end{array}$ 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, ∞) function p : (0, ∞) → (1, ∞) such that p⁺ < ∞. Then there exists an δ > 0 depending on C₂, p⁺ such that the function $\begin{matrix} x^{δ - \frac{1}{p^{'} (x)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{\delta - \frac{1}{p^\prime (x)}} \end{array}$ is almost decreasing near origin (0, ε) and infinity (N, ∞).

Proof

Set $\begin{matrix} g (x) = \int_{x}^{\infty} t^{- \frac{1}{p^{^{'}} (t)}} \frac{d t}{t}, x > 0. \end{matrix}$ $\begin{array}{} g(x)=\int\limits_{x}^{\infty }t^{-\frac{1}{p^{^{\prime }}(t)}}\frac{ dt}{t}, \quad x \gt 0. \end{array}$ From (2), it follows that

$\begin{matrix} g (x) \leq - C_{2} x g^{'} (x), 0 < x < ε . \end{matrix}$ $$\begin{array}{} \displaystyle g(x)\leq -C_{2}xg^{\prime }(x),~~~~~~0 \lt x \lt \varepsilon \text{.} \end{array}$$

First, integrating this inequality over (t₁, t₂) with 0 < t₁ < t₂ < ε, we get

$\begin{matrix} t_{2}^{\frac{1}{C_{2}}} g (t_{2}) \leq C_{2} t_{1}^{- \frac{1}{p^{'} (t_{1})} + \frac{1}{C_{2}}} . \end{matrix}$ $$\begin{array}{} \displaystyle t_{2}^{\frac{1}{C_{2}}}g(t_{2})\leq C_{2}t_{1}^{-\frac{1}{p^{\prime }(t_{1})} +\frac{1}{C_{2}}} \text{.} \end{array}$$(5)

On the other hand, using the conditions (2) and (3) for 0 < t₂ < ε, we get

$\begin{matrix} g (t_{2}) \geq \int_{t_{2}}^{2 t_{2}} x^{- \frac{1}{p^{'} (x)}} \frac{d x}{x} \geq {(\frac{1}{2})}^{\frac{1}{p^{'} (2 t_{2})}} \ln 2 {(\frac{1}{t_{2}})}^{\frac{1}{p^{'} (t_{2})}} \geq \frac{\ln 2}{2} t_{2}^{- \frac{1}{p^{'} (t_{2})}} . \end{matrix}$ $$\begin{array}{} \displaystyle g(t_{2})\geq \int\limits_{t_{2}}^{2t_{2}}x^{-\frac{1}{p^{\prime }(x)}}\frac{ dx}{x}\geq \left( \frac{1}{2}\right) ^{\frac{1}{p^{\prime }(2t_{2})}}\ln 2\left( \frac{1}{t_{2}}\right) ^{\frac{1}{p^{\prime }(t_{2})}}\geq \frac{\ln 2}{2}t_{2}^{-\frac{1}{p^{\prime }(t_{2})}}. \end{array}$$

Inserting this in (5) for all 0 < t₁ < t₂ < ε we get

$\begin{matrix} t_{2}^{- \frac{1}{p^{'} (t_{2})} + \frac{1}{C_{2}}} \leq \frac{2 C_{2}}{\ln 2} t_{1}^{- \frac{1}{p^{'} (t_{1})} + \frac{1}{C_{2}}} . \end{matrix}$ $$\begin{array}{} \displaystyle t_2^{-\frac{1}{p^{\prime }(t_{2})}+\frac{1}{C_{2}}}\le \frac{2C_2 }{\ln 2 }t_{1}^{-\frac{1}{p^{\prime }(t_{1})}+\frac{1}{C_{2}}} . \end{array}$$

Now, let N < t₁ < t₂ < ∞. We will show that again (5) holds for all N < t₁ < t₂ < ∞. From (3), it follows that

$\begin{matrix} g (x) \leq - C_{2} x g^{'} (x), N < x < \infty . \end{matrix}$ $$\begin{array}{} \displaystyle g(x)\le -C_2 x g'(x), \quad N \lt x \lt \infty. \end{array}$$

Integrating this inequality over (t₁, t₂) for N < t₁ < t₂ < ∞ from (2), it follows that

$\begin{matrix} g (x) \leq - C_{2} x g^{'} (x), N < x < \infty . \end{matrix}$ $$\begin{array}{} \displaystyle g(x)\leq -C_{2}xg^{\prime }(x),~~~~~N \lt x \lt \infty \text{.} \end{array}$$

Integrating this inequality over (t₁, t₂), for N < t₁ < t₂ < ∞, we get

$\begin{matrix} t_{2}^{\frac{1}{C_{2}}} g (t_{2}) \leq C_{2} t_{1}^{- \frac{1}{p^{'} (t_{1})} + \frac{1}{C_{2}}} . \end{matrix}$ $$\begin{array}{} \displaystyle t_2^{\frac{1}{C_2}}g(t_2)\le C_2 t_{1}^{-\frac{1}{ p^\prime (t_{1})}+ \frac{1}{C_2}}. \end{array}$$(6)

On the other hand, using conditions (2) and (4), we get

$\begin{matrix} g (t_{2}) \geq \int_{t_{2}}^{2 t_{2}} x^{- \frac{1}{p^{'} (x)}} \frac{d x}{x} \geq \ln 2 {(\frac{1}{2 t_{2}})}^{\frac{1}{p^{'} (2 t_{2})}} \geq 2^{- \frac{1}{p^{'} (N)}} \ln 2 t_{2}^{- \frac{1}{p^{'} (t_{2})}}, t_{2} \geq N . \end{matrix}$ $$\begin{array}{} \displaystyle g(t_{2})\geq \int\limits_{t_{2}}^{2t_{2}}x^{-\frac{1}{p^{\prime }(x)}}\frac{ dx}{x}\geq \ln 2 \left( \frac{1}{2t_{2}}\right) ^{\frac{1}{p^{\prime}(2t_{2})}} \geq ~ 2^{-\frac{1}{p'(N)}} \ln 2 ~ t_{2}^{-\frac{1}{p^{\prime }( t_{2})}}, \quad t_2 \geq N. \end{array}$$

Inserting this in (6), we obtain,

$\begin{matrix} t_{2}^{- \frac{1}{p^{'} (t_{2})} + \frac{1}{C_{2}}} \leq \frac{2 C_{2}}{\ln 2} t_{1}^{- \frac{1}{p^{'} (t_{1})} + \frac{1}{C_{2}}} \end{matrix}$ $$\begin{array}{} \displaystyle t_2^{-\frac{1}{p^{\prime }(t_{2})}+\frac{1}{C_{2}}}\le \frac{2C_2}{\ln 2}t_{1}^{-\frac{ 1}{p^{\prime }(t_{1})}+\frac{1}{C_{2}}} \end{array}$$

for all N < t₁ < t₂ < ∞. Whence, the function $\begin{matrix} x^{δ - \frac{1}{p^{^{'}} (x)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{\delta -\frac{1}{p^{^{\prime }}(x)}} \end{array}$ is almost decreasing on (N, ∞). Therefore, we have proved that

$\begin{matrix} t_{2}^{- \frac{1}{p^{'} (t_{2})} + δ} \leq C_{7} t_{1}^{- \frac{1}{p^{'} (t_{1})} + δ} \end{matrix}$ $$\begin{array}{} \displaystyle t_{2}^{-\frac{1}{p^{\prime }(t_{2})}+\delta }\leq C_{7}t_{1}^{-\frac{1}{ p^{\prime }(t_{1})}+\delta } \end{array}$$

for all 0 < t₁ < t₂ < ε and N < t₁ < t₂ < ∞. The constants $\begin{matrix} C_{7} = \frac{2 C_{2}}{\ln 2}, δ = \frac{1}{C_{2}} . \end{matrix}$ $\begin{array}{} \displaystyle C_{7}= \frac{2C_{2}}{\ln 2}, ~ \delta =\frac{1}{C_{2}}. \end{array}$

This proves Lemma 2.2

Lemma 2.3

Let the condition (2) be satisfied for a monotone increasing near origin on (0, ε) and decreasing near infinity on (N, ∞) function p : (0, ∞) → (1, ∞) such that p⁺ < ∞. Then for 0 < x < ε or for x > N and t ∈ (2^–n–1 x, 2^–n x), the following inequality holds:

$\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \leq C_{7} 2^{- n δ} t^{- \frac{1}{p^{'} (t)}} . \end{matrix}$ $$\begin{array}{} \displaystyle x^{-\frac{1}{p^{\prime }(x)}}\leq C_{7}2^{-n\delta } t^{-\frac{1}{ p^\prime (t)} } \text{ .} \end{array}$$

Proof

By applying Lemma 2.2 for 2^–n–1x < t < 2^–nx, 0 < x < ε or x > N, we have

$\begin{matrix} x^{δ - \frac{1}{p^{'} (x)}} \leq C_{7} t^{δ - \frac{1}{p^{'} (t)}}, \end{matrix}$ $$\begin{array}{} \displaystyle x^{\delta -\frac{1}{p^{\prime }(x)}}\leq C_{7}t^{\delta -\frac{1}{ p^{\prime }(t)}}, \end{array}$$

since in both cases t < x. Indeed, using that 2^–n–1x < t < 2^–nx and Lemma 2.2, we get

$\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \leq C_{7} {(\frac{t}{x})}^{δ} t^{- \frac{1}{p^{'} (t)}} \leq C_{7} 2^{- n δ} t^{- \frac{1}{p^{'} (t)}} \end{matrix}$ $$\begin{array}{} \displaystyle x^{-\frac{1}{p^{\prime }(x)}}\leq C_{7}\left( \frac{t}{x}\right) ^{\delta }t^{-\frac{1}{p^{\prime }(t)}}\leq C_{7}2^{-n\delta }t^{- \frac{1}{p^{\prime }(t)}} \end{array}$$

$\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \leq C_{7} 2^{- n δ} t^{- \frac{1}{p^{'} (t)}} for 0 < x < ε or x > N . \end{matrix}$ $$\begin{array}{} \displaystyle x^{-\frac{1}{p^{\prime }(x)}}\leq C_{7}2^{-n\delta }t^{-\frac{1}{ p^{\prime }(t)}} \quad \text{for}\quad 0 \lt x \lt \varepsilon \quad \text{or} \quad x \gt N. \end{array}$$

This completes the proof of Lemma 2.3

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, ∞) function p : (0, ∞) → (1, ∞) such that p⁺ < ∞. Then for 0 < x < ε or x > N and t ∈ (2^–n–1x, 2^–nx), the following estimate holds

$\begin{matrix} {(2^{- n} x)}^{\frac{1}{{(p_{x, n}^{-})}^{'}}} \leq C_{8} t^{\frac{1}{p^{'} (t)}}, \end{matrix}$ $$\begin{array}{} \displaystyle \left( 2^{-n}x\right) ^{\frac{1}{\left( p_{x,n}^{-}\right) ^{\prime }}}\leq C_{8}t^{\frac{1}{p^{\prime }(t)}}\text{,} \end{array}$$(7)

where $\begin{matrix} p_{x, n}^{-} \end{matrix}$ $\begin{array}{} \displaystyle p_{x,n}^- \end{array}$ = inf {p(s) : s ∈ (2^–n–1x, 2^–nx)}.

Proof

Using the assertions of Lemma 2.1 and condition (2), we get

$\begin{matrix} {(2^{- n} x)}^{\frac{1}{{(p_{x, n}^{-})}^{'}}} \leq {(2 t)}^{\frac{1}{{(p_{x, n}^{-})}^{'}}} \leq 2 t^{\frac{1}{{(p_{x, n}^{-})}^{'}}} \\ \leq 2 t^{\frac{1}{p^{'} (t)}} t^{\frac{1}{{(p_{x, n}^{-})}^{'}} - \frac{1}{p^{'} (t)}} \leq 2 t^{\frac{1}{p^{'} (t)}} t^{\frac{p_{x, n}^{-} - p (t)}{p_{x, n}^{-} p (t)}} \leq 2 t^{\frac{1}{p^{'} (t)}} {(\frac{1}{t})}^{\frac{p (t) - p_{x, n}^{-}}{p_{x, n}^{-} p (t)}} \\ \leq C_{8} t^{\frac{1}{p^{'} (t)}} . \end{matrix}$ $$\begin{array}{c} \displaystyle \left( 2^{-n}x\right) ^{\frac{1}{\left( p_{x,n}^{-}\right) ^{\prime }}}\leq \left( 2t\right) ^{\frac{1}{\left( p_{x,n}^{-}\right) ^{\prime }} }\leq 2t^{\frac{1}{\left( p_{x,n}^{-}\right) ^{\prime }}}\\ \displaystyle\leq 2t^{\frac{1}{p^{\prime }(t)}}t^{\frac{1}{\left( p_{x,n}^{-}\right) ^{\prime }}-\frac{1}{p^{\prime }(t)}}\leq 2t^{\frac{1}{ p^{\prime }(t)}}t^{\frac{p_{x,n}^{-}-p(t)}{p_{x,n}^{-}p(t)}}\leq 2t^{\frac{1 }{p^{\prime }(t)}}\left( \frac{1}{t}\right) ^{\frac{p(t)-p_{x,n}^{-}}{ p_{x,n}^{-}p(t)}}\\ \displaystyle\leq C_{8} \, t^{\frac{1}{p^{\prime }(t)}}\text{.} \end{array}$$

Proof of Theorem 1.1

3.1

Sufficiency

Let the condition (2) be satisfied. we will show that inequality (1) holds. Take a measurable positive function f with

$\begin{matrix} ∥ f ∥_{p (.)} \leq 1. \end{matrix}$ $$\begin{array}{} \displaystyle \Vert f \Vert_{p(.)}\leq 1. \end{array}$$(8)

In order to prove sufficiency, we have to show that $\begin{matrix} ∥ \frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t) ∥_{p (.)} \leq 1 \end{matrix}$ $\begin{array}{} \Vert \frac{1}{x}v(x)\int_0^x f(t)w(t)dt) \Vert_{p(.)} \leq 1 \end{array}$ or the same is to show that $\begin{matrix} I_{p (.)} (\frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t)) \leq 1 \end{matrix}$ $\begin{array}{} I_{p(.)}\left(\frac{1}{x}v(x)\int_0^x f(t)w(t)dt) \right)\leq 1 \end{array}$ (see, e.g. in [10]).

Using Minkowski’s inequality for p(.)-norms we have

$\begin{matrix} ∥ \frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t ∥_{p (.); (0, \infty)} \\ \leq ∥ \frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t ∥_{p (.); (0, N)} + ∥ \frac{1}{x} v (t) \int_{0}^{x} f (t) w (t) d t ∥_{p (.); (N, \infty)} \end{matrix}$ $$\begin{array}{c} \displaystyle \Big \Vert \frac{1}{x} v(x)\int_0^x f(t)w(t) dt \Big \Vert_{p(.); (0,\infty)}\\ \displaystyle\leq \Big \Vert \frac{1}{x}v(x) \int_0^x f(t)w(t) dt \Big \Vert_{p(.); (0,N)}+\Big \Vert \frac{1}{x}v(t) \int_0^x f(t)w(t) dt \Big \Vert_{p(.); (N,\infty)} \end{array}$$(9)

For the first summand it follows from the results of the paper [10] (for α = 0 therein) that it may be estimated as

$\begin{matrix} ∥ \frac{1}{x} v (x) \int_{0}^{x} f (t) w (t) d t ∥_{p (.); (0, N)} \leq C_{1} ∥ f ∥_{p; (0, N)} \leq 1 \end{matrix}$ $$\begin{array}{} \displaystyle \Big \Vert \frac{1}{x}v(x)\int_0^x f(t)w(t) dt \Big \Vert_{p(.); (0,N)}\leq C_1 \big \Vert f \big \Vert_{p; (0,N)} \leq 1 \end{array}$$

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 = $\begin{matrix} {∥x^{- 1} \int_{0}^{x} f (t) d t∥}_{L^{p (.)} (N, \infty)} . \end{matrix}$ $\begin{array}{} \left\Vert x^{-1}\int\limits_{0}^{x}f(t)dt\right\Vert_{L^{p(.)}(N,\infty )}. \end{array}$ By using Minkowski’s inequality for p(.)-norms, it follows that,

$\begin{matrix} \begin{aligned} {∥x^{- 1} v (x) \int_{0}^{x} f (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} & = & {∥x^{- 1} v (x) \int_{0}^{x} f (t) χ_{\{t > N\}} (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} \\ + & {∥x^{- 1} v (x) \int_{0}^{N} f (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle \left\Vert x^{-1}v(x)\int\limits_{0}^{x}f(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty )}&=&\left\Vert x^{-1}v(x)\int\limits_{0}^{x}f(t)\chi _{\left\{ t \gt N\right\}}(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty )}\notag \\ & +&\left\Vert x^{-1}v(x)\int\limits_{0}^{N}f(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty )}. \end{split} \end{array}$$(10)

Since p^– > 1 and ∥f∥_p(.) ≤ 1 it follows that

$\begin{matrix} \int_{0}^{N} f (t) d t \leq {∥f∥}_{p (.)} {∥χ_{\{0 < t < N\}}∥}_{p^{'} (.)} \leq {∥χ_{\{0 < t < N\}}∥}_{p^{'} (.)} \leq N^{\frac{1}{(p^{+})^{'}}} . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{0}^{N}f(t)dt\leq \left\Vert f\right\Vert _{p(.)}\left\Vert \chi _{\left\{ 0 \lt t \lt N\right\} }\right\Vert _{p^{\prime }(.)}\leq \left\Vert \chi _{\left\{ 0 \lt t \lt N\right\} }\right\Vert _{p^{\prime }(.)}\leq N^{\frac{1}{(p^+)^\prime}}. \end{array}$$

Whence

$\begin{matrix} {∥x^{- 1} v (x) \int_{0}^{N} f (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} \\ \leq {∥χ_{\{0 < t < N\}}∥}_{p^{'} (.)} {∥x^{- 1}∥}_{L^{p (.)} (N, \infty)} \leq C_{9} (N, p^{+}, p (\infty)) \end{matrix}$ $$\begin{array}{c} \displaystyle \left\Vert x^{-1}v(x)\int\limits_{0}^{N}f(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty)}\\ \displaystyle\leq \left\Vert \chi _{\left\{ 0 \lt t \lt N\right\} }\right\Vert _{p^{\prime}(.)}\left\Vert x^{-1}\right\Vert _{L^{p(.)}(N,\infty )}\leq C_{9}(N,p^{+},p(\infty)) \end{array}$$(11)

since p(∞) > 1,

$\begin{matrix} I_{p (.)} (t^{- 1} χ_{(N, \infty)} (t)) = \int_{N}^{\infty} \frac{d t}{t^{p (t)}} \leq \int_{N}^{\infty} \frac{d t}{t^{p (\infty)}} = \frac{N^{1 - p (\infty)}}{p (\infty) - 1} \end{matrix}$ $$\begin{array}{} \displaystyle I_{p(.)} \left( t^{-1} \chi_{(N, \infty)} (t) \right ) =\int_N^\infty \frac{dt}{t^{p(t)}}\leq \int_N^\infty \frac{dt}{t^{p(\infty)}} =\frac{N^{1-p(\infty)}}{p(\infty)-1} \end{array}$$

and therefore,

$\begin{matrix} {∥x^{- 1}∥}_{L^{p (.)} (n, \infty)} \leq N^{\frac{1 - p (\infty)}{p (\infty)}} (p (\infty) - 1)^{- \frac{1}{p (\infty)}} . \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert x^{-1}\right\Vert _{L^{p(.)}(n,\infty )}\leq N^{\frac{1-p(\infty)}{p(\infty)}}(p(\infty)-1)^{-\frac{1}{p(\infty)}}. \end{array}$$

Thus, in order to get an estimation for i it suffices to estimate

$\begin{matrix} {∥x^{- 1} v (t) \int_{0}^{x} f (u) χ_{\{t > N\}} (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} . \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert x^{-1}v(t)\int\limits_{0}^{x}f(u)\chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty )}\text{.} \end{array}$$

We have

$\begin{matrix} {∥x^{- 1} v (x) \int_{N}^{x} f (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} \leq {∥x^{- 1} \sum_{n = 0}^{\infty} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} \\ \leq \sum_{n = 0}^{\infty} {∥x^{- 1} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t∥}_{L^{p (.)} (N, \infty)} . \end{matrix}$ $$\begin{array}{c} \displaystyle \left\Vert x^{-1}v(x)\int\limits_{N}^{x}f(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty )}\leq \left\Vert x^{-1}\sum\limits_{n=0}^{\infty }v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f(t)\chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty )}\\ \displaystyle\leq \sum\limits_{n=0}^{\infty }\left\Vert x^{-1}v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f(t)\chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right\Vert _{L^{p(.)}(N,\infty )}. \end{array}$$(12)

We shall estimate every summand on the right hand side. By using Lemma 2.3 for N < t < x < ∞, it follows that $\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \leq C_{7} 2^{- n δ} t^{- \frac{1}{p^{'} (t)}} . \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1 }{p^{\prime }(x)}}\leq C_{7}2^{-n\delta }t^{-\frac{1}{p^{\prime }(t)}}. \end{array}$ Therefore

$\begin{matrix} x^{- \frac{1}{p (x)}} x^{- \frac{1}{p^{'} (x)}} \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) d t \\ \leq C_{7} 2^{- n δ} x^{- \frac{1}{p (x)}} t^{- \frac{1}{p^{'} (x)}} \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) d t . \end{matrix}$ $$\begin{array}{c} \displaystyle x^{-\frac{1}{p(x)}}x^{-\frac{1}{p^{\prime }(x)}}\int \limits_{2^{-n-1}x}^{2^{-n}x}f(t)\chi _{\left\{ t \gt N\right\} }(t)dt \\ \displaystyle\leq C_{7}2^{-n\delta }x^{-\frac{1}{p(x)}}t^{-\frac{1}{p^{\prime }(x)} }\int\limits_{2^{-n-1}x}^{2^{-n}x}f(t)\chi _{\left\{ t \gt N\right\} }(t)dt \text{ .} \end{array}$$

By using Hölder’s inequality for p(x) -norms and assumption (8), we get

$\begin{matrix} \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) d t \\ \leq C_{0} {∥f (t) χ_{\{t > N\}} (t)∥}_{L^{p (.)} (2^{- n - 1} x, 2^{- n} x)} \times {∥χ_{\{t > N\}} (.)∥}_{L^{p^{'} (.)} (2^{- n - 1} x, 2^{- n} x)} \\ \leq {∥χ_{\{t > N\}} (.)∥}_{L^{p^{'} (.)} (2^{- n - 1} x, 2^{- n} x)} . \end{matrix}$ $$\begin{array}{c} \displaystyle \int\limits_{2^{-n-1}x}^{2^{-n}x}f(t)\chi _{\left\{ t \gt N\right\} }(t)dt\\ \displaystyle\leq C_{0}\left\Vert f(t)\chi _{\left\{ t \gt N\right\} }(t)\right\Vert _{L^{p(.)}(2^{-n-1}x,2^{-n}x)} \times \left\Vert \chi _{\left\{ t \gt N\right\} }(.)\right\Vert _{L^{p^{\prime }(.)}(2^{-n-1}x,2^{-n}x)} \\ \displaystyle\leq \left\Vert \chi _{\left\{ t \gt N\right\} }(.)\right\Vert _{L^{p^{\prime }(.)}(2^{-n-1}x,2^{-n}x)}. \end{array}$$

Lemma 3.1

There exists a positive constant C₉ > 1 depending on C₂, p⁺, p(∞), such that

$\begin{matrix} {∥χ_{\{t > N (.)\}} χ_{(2^{- n - 1} x, 2^{- n} x)} (.)∥}_{L^{p^{'} (.)}} \leq C_{9} y^{\frac{1}{p^{'} (y)}}, \forall y \in (2^{- n - 1} x, 2^{- n} x) \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert \chi _{\left\{ t \gt N(.)\right\} }\chi_{(2^{-n-1}x,2^{-n}x)}(.)\right\Vert _{L^{p^{\prime }(.)}}\leq C_{9}y^{\frac{1}{p^{\prime }(y)}}, \quad \forall y\in (2^{-n-1}x, 2^{-n}x) \end{array}$$(13)

for x > N.

Proof

We prove estimation (13) from the opposite. Let inequality (13) be violated. We show that a contradiction occurs. Let

$\begin{matrix} {∥χ_{\{t > N\}} (.) χ_{(2^{- n - 1} x, 2^{- n} x)} (.)∥}_{L^{p^{'} (.)}} > C_{9} y^{\frac{1}{p^{'} (y)}} . \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert \chi_{\left \{ t \gt N \right \} } (.) \chi_{(2^{-n-1}x,2^{-n}x)}(.)\right\Vert _{L^{p^\prime(.)}} \gt C_{9}\, y^{\frac{1}{p^{\prime }(y)}}. \end{array}$$(14)

From the definition of p^′(.)-norms, it follows that for any sufficiently small δ > 0 the inequality holds

$\begin{matrix} \int_{2^{- n - 1} x}^{2^{- n} x} {(\frac{1}{{∥χ_{\{t > N\}} (.) χ_{(2^{- n - 1} x, 2^{- n} x)} (.)∥}_{L^{p^{'} (.)}} - δ})}^{p^{'} (t)} d t > 1 . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{2^{-n-1}x}^{2^{-n}x}\left( \frac{1}{\left\Vert \chi_{\left \{ t \gt N \right \} } (.) \chi_{(2^{-n-1}x,2^{-n}x)}(.)\right\Vert _{L^{p^\prime(.)}}-\delta }\right) ^{p^{\prime }(t)}dt \gt 1 \text{ .} \end{array}$$(15)

Indeed, if (15) does not hold, we get a contradiction, that is,

$\begin{matrix} {∥χ_{\{t > N\}} (.) χ_{(2^{- n - 1} x, 2^{- n} x)} (.)∥}_{L^{p^{'} (.)}} \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert \chi_{\left \{ t \gt N \right \} } (.) \chi_{(2^{-n-1}x,2^{-n}x)}(.)\right\Vert _{L^{p^\prime(.)}} \end{array}$$

is a list number satisfying

$\begin{matrix} \int_{2^{- n - 1} x}^{2^{- n} x} {(\frac{1}{λ})}^{p^{'} (t)} d t \leq 1 . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{2^{-n-1}x}^{2^{-n}x}\left( \frac{1}{\lambda }\right) ^{p^{\prime }(t)}dt\leq 1\text{.} \end{array}$$

We set $\begin{matrix} δ = \frac{1}{2} {∥χ_{\{t > N\}} (.) χ_{(2^{- n - 1} x, 2^{- n} x)} (.)∥}_{L^{p^{'} (.)}} \end{matrix}$ $\begin{array}{} \displaystyle \delta =\frac{1}{2}\left\Vert \chi_{\left \{ t \gt N \right \} } (.) \chi_{(2^{-n-1}x,2^{-n}x)}(.)\right\Vert _{L^{p^\prime(.)}} \end{array}$ in (15), to get

$\begin{matrix} \int_{2^{- n - 1} x}^{2^{- n} x} {(\frac{2}{{∥χ_{\{t > N\}} (.) χ_{(2^{- n - 1} x, 2^{- n} x)} (.)∥}_{L^{p^{'} (.)}}})}^{p^{'} (t)} d t > 1 . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{2^{-n-1}x}^{2^{-n}x}\left( \frac{2}{\left\Vert \chi_{\left \{ t \gt N \right \} } (.) \chi_{(2^{-n-1}x,2^{-n}x)}(.)\right\Vert _{L^{p^\prime(.)}}}\right) ^{p^{\prime }(t)}dt \gt 1 \text{ .} \end{array}$$

By applying here (14), it follows that

$\begin{matrix} \int_{2^{- n - 1} x}^{2^{- n} x} {(\frac{2}{C_{9} y^{\frac{1}{p^{'} (y)}}})}^{p^{'} (t)} d t > 1, \forall y \in (2^{- n - 1} x, 2^{- n} x), x > N . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{2^{-n-1}x}^{2^{-n}x}\left( \frac{2}{C_{9}\, y^{\frac{1}{p^{\prime }(y)}}} \right) ^{p^{\prime }(t)}dt \gt 1, \quad \forall y\in (2^{-n-1}x, 2^{-n}x), \quad x \gt N. \end{array}$$

For any points t, y lying in (2^–n–1x, 2^–nx) it follows from Lemma 2.1 that $\begin{matrix} y^{\frac{1}{p^{'} (y)}} \end{matrix}$ $\begin{array}{} \displaystyle y^{\frac{1}{p^{\prime }(y)}} \end{array}$ is comparable with $\begin{matrix} t^{\frac{1}{p^{'} (t)}} . \end{matrix}$ $\begin{array}{} \displaystyle t^{\frac{1}{p^{\prime }(t)}}. \end{array}$ This means, there exist two positive constants C₁₀, C₁₁ depending on C₂, p⁺ such that

$\begin{matrix} C_{10} t^{\frac{1}{p^{'} (t)}} \leq y^{\frac{1}{p^{'} (y)}} \leq C_{11} t^{\frac{1}{p^{'} (t)}} . \end{matrix}$ $$\begin{array}{} \displaystyle C_{10} t^{\frac{1}{p^{\prime }(t)}}\leq y^{\frac{1}{p^{\prime }(y)}}\leq C_{11}t^{\frac{1}{p^{\prime }(t)}}. \end{array}$$

Therefore,

$\begin{matrix} \int_{2^{- n - 1} x}^{2^{- n} x} {(\frac{2}{C_{10} C_{9} t^{\frac{1}{p^{'} (t)}}})}^{p^{'} (t)} d t > 1, x > N . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{2^{-n-1}x}^{2^{-n}x}\left( \frac{2}{C_{10}C_{9}t^{\frac{1}{p^{\prime }(t)}}} \right) ^{p^{\prime }(t)}dt \gt 1, \quad x \gt N\text{ .} \end{array}$$

Whence

$\begin{matrix} \frac{2 (p^{-})^{'} \ln 2}{(C_{10} C_{9})^{(p^{+})^{'}}} \geq 1 . \end{matrix}$ $$\begin{array}{} \displaystyle \frac{2(p^{-})^{\prime }\ln 2}{(C_{10}C_{9})^{(p^{+})^{\prime }}}\geq 1\text{.} \end{array}$$(16)

By choosing constant C₉ sufficiently large, we get a contradiction with (16). This proves estimate (13). Thus we have proved that

$\begin{matrix} \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) d t \leq C_{9} y^{\frac{1}{p^{'} (y)}}, \forall y \in (2^{- n - 1} x, 2^{- n} x), x > N . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{2^{-n-1}x}^{2^{-n}x} f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)dt\leq C_{9}\, y^{\frac{1}{p^{\prime }(y)}},\quad \forall y\in ( 2^{-n-1}x, 2^{-n}x), \quad x \gt N\text{ .} \end{array}$$(17)

This proves Lemma 3.1.

Now, we shall derive an estimate for

$\begin{matrix} {∥\frac{1}{x} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t∥}_{L^{p} (N, \infty)} . \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert \frac{1}{x} v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right\Vert _{L^{p}(N,\infty )}. \end{array}$$

In order to carry out it, we get an estimation for the proper modular

$\begin{matrix} I_{p (.) : (N, \infty)} (\frac{1}{x} v (t) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t) \\ = I_{p (.) : (N, \infty)} (x^{- \frac{1}{p (x)}} x^{- \frac{1}{p^{'} (x)}} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > M\}} (t) w (t) d t) \end{matrix}$ $$\begin{array}{c} \displaystyle I_{p(.):(N,\infty )}\left( \frac{1}{x} v(t)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right)\\ \displaystyle=I_{p(.):(N,\infty )}\left( x^{- \frac{1}{p(x)}}x^{-\frac{1}{p^{\prime }(x)}}v(x)\int \limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt M\right\} }(t)w(t)dt\right) \end{array}$$(18)

It follows from Lemma 2.3 that

$\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \leq C_{7} 2^{- n ε} t^{- \frac{1}{p^{'} (t)}} \end{matrix}$ $$\begin{array}{} \displaystyle x^{-\frac{1}{p^{\prime }(x)}}\leq C_{7}2^{-n\varepsilon }t^{-\frac{1}{p^{\prime }(t)}} \end{array}$$

for ∀y, t ∈ (2^–n–1x, 2^–nx) and x > N. By using the last estimate the expression (18) is exceeded

$\begin{matrix} \int_{N}^{\infty} {(C_{7} 2^{- n δ} y^{- \frac{1}{p^{'} (y)}} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t)}^{p (x)} \frac{d x}{x} \\ = \int_{N}^{\infty} C_{7}^{p (x)} 2^{- n p (x) δ} y^{- \frac{p (x)}{p^{'} (y)}} {(\frac{v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t}{C_{7} t^{\frac{1}{p^{'} (t)}}})}^{p (x)} {(C_{11} t^{\frac{1}{p^{'} (t)}})}^{p (x)} \frac{d x}{x} . \end{matrix}$ $$\begin{array}{c} \displaystyle \int\limits_{N}^{\infty }\left( C_{7}2^{-n\delta }y^{-\frac{1}{ p^{\prime }(y)}}v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right) ^{p(x)}\frac{dx}{x} \\ \displaystyle=\int\limits_{N}^{\infty }C_{7}^{p(x)}2^{-np(x)\delta }y^{-\frac{p(x) }{p^{\prime }(y)}}\left( \frac{v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt}{C_{7}t^{\frac{1}{p^{\prime }(t)}} }\right) ^{p(x)}\left( C_{11}t^{\frac{1}{p^{\prime }(t)}}\right) ^{p(x)}\frac{ dx}{x}. \end{array}$$

By applying here Lemma 2.1, we get that, the term $\begin{matrix} y^{- \frac{p (x)}{p^{'} (y)}} \end{matrix}$ $\begin{array}{} \displaystyle y^{-\frac{p(x) }{p^{\prime }(y)}} \end{array}$ is comparable with $\begin{matrix} t^{- \frac{p (x)}{p^{'} (t)}} \end{matrix}$ $\begin{array}{} \displaystyle t^{-\frac{p(x) }{p^{\prime }(t)}} \end{array}$ in the integral terms, i.e.

$\begin{matrix} {(y^{- \frac{1}{p^{'} (y)}} \cdot t^{\frac{1}{p^{'} (t)}})}^{p^{+}} \leq C_{12} for all t, y lying in (2^{- n - 1} x, 2^{- n} x) . \end{matrix}$ $$\begin{array}{} \displaystyle \left (y^{-\frac{1 }{p^{\prime }(y)}}\cdot t^{\frac{1 }{p^{\prime }(t)}}\right) ^{p^+}\leq C_{12} \quad \text{for all}~ t,y ~ \text{lying in} ~(2^{-n-1}x, 2^{-n}x). \end{array}$$

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 $\begin{matrix} p_{x, n}^{-}, \end{matrix}$ $\begin{array}{} \displaystyle p_{x,n}^{-}, \end{array}$ we will increase the fraction. Then the last expression is less then

$\begin{matrix} \begin{aligned} C_{12} 2^{- n δ p^{-}} \int_{M}^{\infty} {(\frac{v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > M\}} (t) w (t) d t}{C_{7} t^{\frac{1}{p^{'} (t)}}})}^{p_{x, n}^{-}} \frac{d x}{x} \\ \leq & C_{12} 2^{- n δ p^{-}} \int_{N}^{\infty} {(\frac{v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t}{C_{9} t^{\frac{1}{p^{'} (t)}}})}^{p_{x, n}^{-}} \frac{d x}{x} . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle &&C_{12}2^{-n\delta p^{-}}\int\limits_{M}^{\infty }\left( \frac{v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt M\right\} }(t)w(t)dt}{C_{7}t^{\frac{1}{p^{\prime }(t)}}}\right) ^{p_{x,n}^{-}}\frac{dx}{x} \notag \\ &\leq &C_{12}2^{-n\delta p^{-}}\int\limits_{N}^{\infty }\left( \frac{v(x)\int \limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt}{C_{9}t^{\frac{1}{p^{\prime }(t)}}}\right) ^{p_{x,n}^{-}}\frac{dx}{x}. \end{split} \end{array}$$(19)

Using Holder’s inequality, it follows that

$\begin{matrix} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t \\ \leq {(v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f {(t)}^{p_{x, n}^{-}} χ_{\{t > N\}} (t) w (t) d t)}^{\frac{1}{p_{x, n}^{-}}} {(2^{- n - 1} x)}^{^{\frac{1}{{(p_{x, n}^{-})}^{'}}}} \end{matrix}$ $$\begin{array}{c} \displaystyle v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt \\ \displaystyle\leq \left( v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) ^{p_{x,n}^{-}}\chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right) ^{\frac{1}{ p_{x,n}^{-}}}\left( 2^{-n-1}x\right) ^{^{\frac{1}{\left( p_{x,n}^{-}\right) ^{\prime }}}} \end{array}$$

with x ≥ N2ⁿ⁺¹. By inserting this in the interior integral (19), that is exceeded by

$\begin{matrix} C_{12} 2^{- n δ p^{-}} \int_{N}^{\infty} (v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f {(t)}^{p_{x, n}^{-}} χ_{\{t > N\}} (t) w (t) d t) {(2^{- n - 1} x)}^{^{\frac{p_{x, n}^{-}}{{(p_{x, n}^{-})}^{'}}}} t^{- \frac{p_{x, n}^{-}}{p^{'} (t)}} \frac{d x}{x} . \end{matrix}$ $$\begin{array}{} \displaystyle C_{12}2^{-n\delta p^{-}}\int\limits_{N}^{\infty }\left( v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) ^{p_{x,n}^{-}}\chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right) \left( 2^{-n-1}x\right) ^{^{\frac{ p_{x,n}^{-}}{\left( p_{x,n}^{-}\right) ^{\prime }}}}t^{-\frac{p_{x,n}^{-}}{ p^{\prime }(t)}}\frac{dx}{x}. \end{array}$$

Since 2^–n–1x ≤ t and by using Lemma 2.4 and Lemma 2.1, it follows that, the last expression is exceeded by

$\begin{matrix} \begin{aligned} \int_{N}^{\infty} {(\frac{1}{x} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f (t) χ_{\{t > N\}} (t) w (t) d t)}^{p (x)} d x \\ \leq C_{13} 2^{- n δ p^{-}} \int_{N}^{\infty} (v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f {(t)}^{p_{x, n}^{-}} χ_{\{t > N\}} (t) w (t) d t) \frac{d x}{x} . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle \int\limits_N^\infty \left( \frac{1}{x} v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x} f \left( t\right) \chi _{\left\{ t \gt N\right\} }(t) w(t) dt \right)^{p(x)}dx \notag \\ \leq C_{13}2^{-n\delta p^{-}}\int\limits_{N}^{\infty }\left( v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) ^{p_{x,n}^{-}}\chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right) \frac{dx}{x}. \end{split} \end{array}$$(20)

Now, we estimate the interior integral through I_p(.)(f). The interior integral here is exceeded

$\begin{matrix} \int_{B_{x, n} \cap \{t : f (t) > \frac{1}{1 + t^{2}}\}} {(\frac{f (t) χ_{\{t > N\}} (t)}{\frac{1}{1 + t^{2}}})}^{p_{x, n}^{-}} {(\frac{1}{1 + t^{2}})}^{p_{x, n}^{-}} d t \\ + \int_{B_{x, n} \cap \{t : f (t) \leq \frac{1}{1 + t^{2}}\}} {(f (t))}^{p_{x, n}^{-}} d t \\ \leq \int_{B_{x, n}} {(\frac{f (t) χ_{\{t > N\}} (t)}{\frac{1}{1 + t^{2}}})}^{p (t)} {(\frac{1}{1 + t^{2}})}^{p_{x, n}^{-}} d t + \int_{B_{x, n}} {(\frac{χ_{\{t > N\}} (t)}{1 + t^{2}})}^{p_{x, n}^{-}} d t \\ \leq \int_{B_{x, n}} {(f (t))}^{p (t)} {(1 + t^{2})}^{p (t) - p_{x, n}^{-}} χ_{\{t > N\}} (t) d t + \int_{B_{x, n}} {(\frac{χ_{\{t > N\}} (t)}{1 + t^{2}})}^{p_{x, n}^{-}} d t \\ \leq C_{14} \int_{B_{x, n}} f {(t)}^{p (t)} χ_{\{t > N\}} (t) d t + \int_{B_{x, n}} (\frac{χ_{\{t > N\}} (t)}{1 + t^{2}}) d t; B_{x, n} = (2^{- n - 1} x, 2^{- n} x) \end{matrix}$ $$\begin{array}{c} \displaystyle \int\limits_{B_{x,n}\cap \left\{ t:f(t) \gt \frac{1}{1+t^{2}}\right\} }\left( \frac{f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)}{\frac{1}{1+t^{2}}} \right) ^{p_{x,n}^{-}}\left( \frac{1}{1+t^{2}}\right) ^{p_{x,n}^{-}}dt\\ \displaystyle+ \int\limits_{B_{x,n}\cap \left\{ t:f(t)\leq \frac{1}{1+t^{2}}\right\} }\left( f(t)\right) ^{p_{x,n}^{-}}dt \\ \displaystyle\leq \int\limits_{B_{x,n}}\left( \frac{f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)}{\frac{1}{1+t^{2}}}\right) ^{p(t)}\left( \frac{1}{1+t^{2}} \right) ^{p_{x,n}^{-}}dt+\int\limits_{B_{x,n}}\left( \frac{\chi _{\left\{ t \gt N\right\} }(t)}{1+t^{2}}\right) ^{p_{x,n}^{-}}dt\\ \displaystyle\leq \int\limits_{B_{x,n}}\left( f\left( t\right) \right) ^{p(t)}\left( 1+t^{2}\right) ^{p(t)-p_{x,n}^{-}}\chi _{\left\{ t \gt N\right\} }(t)dt+\int\limits_{B_{x,n}}\left( \frac{\chi _{\left\{ t \gt N\right\} }(t)}{ 1+t^{2}}\right) ^{p_{x,n}^{-}}dt\\ \displaystyle\leq C_{14}\int\limits_{B_{x,n}}f\left( t\right) ^{p(t)}\chi _{\left\{ t \gt N\right\} }(t)dt+\int\limits_{B_{x,n}}\left( \frac{\chi _{\left\{ t \gt N\right\} }(t)}{1+t^{2}}\right) dt; \quad B_{x,n}=(2^{-n-1}x, 2^{-n}x) \end{array}$$

since the assumption (2) is made using (4) from Lemma 2.1 and $\begin{matrix} p_{x, n}^{-} \geq 1, \end{matrix}$ $\begin{array}{} \displaystyle p_{x,n}^{-}\geq 1 , \end{array}$ it follows that

$\begin{matrix} {(1 + t^{2})}^{p (t) - p_{x, n}^{-}} \leq C_{15} . \end{matrix}$ $$\begin{array}{} \displaystyle \left( 1+t^{2}\right)^{p(t)-p_{x,n}^-}\leq C_{15}. \end{array}$$

Evidently, $\begin{matrix} {(\frac{1}{1 + t^{2}})}^{p_{x, n}^{-}} \leq \frac{1}{1 + t^{2}} . \end{matrix}$ $\begin{array}{} \displaystyle \left( \frac{1}{1+t^{2}}\right) ^{p_{x,n}^{-}}\leq \frac{1}{1+t^{2}}. \end{array}$ Therefore, we get an estimate

$\begin{matrix} \int_{B_{x, n}} {(f (t))}^{p_{x, n}^{-}} χ_{\{t > N\}} (t) d t \leq C_{15} \int_{B_{x, n}} f {(t)}^{p (t)} χ_{\{t > N\}} (t) d t + \int_{B_{x, n}} (\frac{χ_{\{t > N\}} (t)}{1 + t^{2}}) d t . \end{matrix}$ $$\begin{array}{} \displaystyle \int\limits_{B_{x,n}}\left( f\left( t\right) \right) ^{p_{x,n}^{-}}\chi _{\left\{ t \gt N\right\} }(t)dt\leq C_{15}\int\limits_{B_{x,n}}f\left( t\right) ^{p(t)}\chi _{\left\{ t \gt N\right\} }(t)dt+\int\limits_{B_{x,n}}\left( \frac{ \chi _{\left\{ t \gt N\right\} }(t)}{1+t^{2}}\right) dt . \end{array}$$(21)

Inserting this inequality in (20) and applying Fubini’s therom, it follows that

$\begin{matrix} I_{p (.), (N, \infty)} (\frac{1}{x} v (x) \int_{B_{n, x}} f (t) χ_{\{t > N\}} (t) w (t) d t) \\ \leq C_{15} 2^{- n δ p^{-}} \int_{N}^{\infty} (v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f {(t)}^{p (t)} χ_{\{t > N\}} (t) w (t) d t + v (x) \int_{B_{x, n}} (\frac{χ_{\{t > N\}} (t)}{1 + t^{2}}) w (t) d t) \frac{d x}{x} \\ = C_{15} 2^{- n δ p^{-}} v (x) \int_{2^{- n} N}^{\infty} (\int_{2^{n} t}^{2^{n + 1} t} \frac{d x}{x}) f {(t)}^{p (t)} χ_{\{t > N\}} (t) w (t) d t + v (x) \int_{2^{- n} M}^{\infty} (\int_{2^{n} t}^{2^{n + 1} t} \frac{d x}{x}) \frac{χ_{\{t > N\}} (t) w (t) d t}{1 + t^{2}} \\ \leq C_{15} 2^{- n δ p^{-}} \ln 2 (\int_{2^{- n} N}^{\infty} f {(t)}^{p (t)} χ_{\{t > N\}} (t) d t + \ln 2 \int_{2^{- n} M}^{\infty} \frac{χ_{\{t > N\}} (t) d t}{1 + t^{2}}) \\ = C_{15} 2^{- n δ p^{-}} \ln 2 (\int_{N}^{\infty} f {(t)}^{p (t)} d t + \int_{N}^{\infty} \frac{d t}{1 + t^{2}}) \leq C_{15} 2^{- n δ p^{-}} \ln 2 (1 + \frac{π}{2}) = C_{16} 2^{- n δ p^{-}} . \end{matrix}$ $$\begin{array}{c} \displaystyle I_{p(.),(N,\infty )}\left( \frac{1}{x}v(x)\int\limits_{B_{n,x}}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right)\\ \displaystyle\leq C_{15}2^{-n\delta p^{-}}\int\limits_{N}^{\infty }\left( v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) ^{p(t)}\chi _{\left\{ t \gt N\right\} }(t)w(t)dt+v(x)\int\limits_{B_{x,n}}\left( \frac{\chi _{\left\{ t \gt N\right\} }(t)}{1+t^{2}}\right)w(t) dt\right) \frac{dx}{x}\\ \displaystyle=C_{15}2^{-n\delta p^{-}}v(x)\int\limits_{2^{-n}N}^{\infty }\left( \int\limits_{2^{n}t}^{2^{n+1}t}\frac{dx}{x}\right) f\left( t\right) ^{p(t)}\chi _{\left\{ t \gt N\right\} }(t)w(t)dt+v(x)\int\limits_{2^{-n}M}^{\infty }\left( \int\limits_{2^{n}t}^{2^{n+1}t}\frac{dx}{x}\right) \frac{\chi _{\left\{ t \gt N\right\} }(t)w(t)dt}{1+t^{2}} \\ \displaystyle\leq C_{15}2^{-n\delta p^{-}}\ln 2\left( \int\limits_{2^{-n}N}^{\infty }f\left( t\right) ^{p(t)}\chi _{\left\{ t \gt N\right\} }(t)dt+\ln 2\int\limits_{2^{-n}M}^{\infty }\frac{\chi _{\left\{ t \gt N\right\} }(t)dt}{1+t^{2}}\right)\\ \displaystyle=C_{15}2^{-n\delta p^{-}}\ln 2\left( \int\limits_{N}^{\infty }f\left( t\right) ^{p(t)}dt+\int\limits_{N}^{\infty }\frac{dt}{1+t^{2}}\right) \leq C_{15}2^{-n\delta p^{-}}\ln 2\left( 1+\frac{\pi}{2 }\right) =C_{16}2^{-n\delta p^{-}}. \end{array}$$

Whence,

$\begin{matrix} I_{p (.), (N, \infty)} (\frac{1}{x} v (x) \int_{B_{n, x}} f (t) χ_{\{t > N\}} (t) w (t) d t) \leq C_{16} 2^{- n δ p^{-}} . \end{matrix}$ $$\begin{array}{} \displaystyle I_{p(.),(N,\infty )}\left( \frac{1}{x}v(x)\int\limits_{B_{n,x}}f\left( t\right) \chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right) \leq C_{16}2^{-n\delta p^{-}}. \end{array}$$(22)

From this it follows that

$\begin{matrix} {∥\frac{1}{x} v (x) \int_{2^{- n - 1} x}^{2^{- n} x} f {(t)}^{p (t)} χ_{\{t > N\}} (t) w (t) d t∥}_{L^{p} (N, \infty)} \leq C_{17} 2^{- \frac{n δ p^{-}}{p^{+}}} . \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert \frac{1}{x}v(x)\int\limits_{2^{-n-1}x}^{2^{-n}x}f\left( t\right) ^{p(t)}\chi _{\left\{ t \gt N\right\} }(t)w(t)dt\right\Vert _{L^{p}(N,\infty )}\leq C_{17}2^{-\frac{n\delta p^{-}}{p^{+}}}. \end{array}$$(23)

Inserting (23) in (12), we get

$\begin{matrix} {∥x^{- 1} v (x) \int_{N}^{x} f (x) d x∥}_{L^{p (.)} (N, \infty)} \leq \sum_{n = 1}^{\infty} C_{17} 2^{- \frac{n δ p^{-}}{p^{+}}} = C_{18} . \end{matrix}$ $$\begin{array}{} \displaystyle \left\Vert x^{-1}v(x)\int\limits_{N}^{x}f\left( x\right) dx\right\Vert _{L^{p(.)}(N,\infty )}\leq \sum\limits_{n=1}^{\infty }C_{17}2^{-\frac{ n\delta p^{-}}{p^{+}}}=C_{18}. \end{array}$$

Further, substitute this estimate and (11) into (10), we complete the proof of the sufficiency part of Theorem 1.1.

3.2

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

$\begin{matrix} f_{0} (x) = x^{- \frac{1}{p (x)}} χ_{(b, 2 b)} (x), x > N \end{matrix}$ $$\begin{array}{} \displaystyle f_0(x)=x^{-\frac{1}{p(x)}} \chi_{(b,2b)}(x), \quad x \gt N \end{array}$$

into inequality (1.2) with a parameter b > N be fixed. It is clear that, I_p(.) (f₀) = ln 2. It follows from the inequality (1) that

$\begin{matrix} I_{p (.)} (\frac{1}{x} \int_{b}^{x} f_{0} (x) d x) \leq C_{19} . \end{matrix}$ $$\begin{array}{} \displaystyle I_{p(.)} \left ( \frac{1}{x} \int_b^x f_0 (x)dx \right ) \leq C_{19}. \end{array}$$

From this it follows

$\begin{matrix} \int_{2 b}^{4 b} {(x^{- \frac{1}{p^{'} (x)}} \int_{b}^{2 b} t^{- \frac{1}{p (t)}} d t)}^{p (x)} \frac{d x}{x} \leq C_{19} . \end{matrix}$ $$\begin{array}{} \displaystyle \int_{2b}^{4b} \left(x^{ -\frac{1}{p^\prime(x)}} \int_b^{2b} t^{-\frac{1}{p(t)}}dt\right)^{p(x)}\frac{dx}{x}\leq C_{19}. \end{array}$$(24)

By monotony of p in (N, ∞) it follows the functions $\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1}{p^\prime(x)}} \end{array}$ and $\begin{matrix} t^{- \frac{1}{p (t)}} \end{matrix}$ $\begin{array}{} \displaystyle t^{-\frac{1}{p(t)}} \end{array}$ are decreasing therein. This yields

$\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \geq (4 b)^{- \frac{1}{p^{'} (4 b)}}, 2 b < x < 4 b \end{matrix}$ $$\begin{array}{} \displaystyle x^{-\frac{1}{p^\prime(x)}}\geq (4b)^{-\frac{1}{p^\prime(4b)}}, ~ 2b \lt x \lt 4b \end{array}$$

and

$\begin{matrix} t^{- \frac{1}{p (t)}} \geq (2 b)^{- \frac{1}{p (2 b)}}, b < t < 2 b . \end{matrix}$ $$\begin{array}{} \displaystyle t^{-\frac{1}{p(t)}}\geq (2b)^{-\frac{1}{p(2b)}}, ~ b \lt t \lt 2b. \end{array}$$

By taking into account this inequalities, it follows from (24) that

$\begin{matrix} \int_{2 b}^{4 b} ((4 b)^{- \frac{1}{p^{'} (4 b)}} (2 b)^{\frac{1}{p^{'} (2 b)}})^{p (x)} \frac{d x}{x} \leq C_{19} . \end{matrix}$ $$\begin{array}{} \displaystyle \int_{2b}^{4b}\Big( (4b)^{-\frac{1}{p^\prime(4b)}}(2b) ^{\frac{1}{p^\prime(2b)}}\Big)^{p(x)}\frac{dx}{x}\leq C_{19}. \end{array}$$

This inequality yields

$\begin{matrix} (\frac{1}{2} (2 b)^{\frac{1}{p^{'} (2 b)} - \frac{1}{p^{'} (4 b)}})^{p^{-}} \ln 2 \leq C_{19} \end{matrix}$ $$\begin{array}{} \displaystyle \Big( \frac{1}{2}(2b)^{\frac{1}{p^\prime(2b)}-\frac{1}{p^\prime(4b)}}\Big)^{p^-}\ln2 \leq C_{19} \end{array}$$

if the parenthesis term is greater 1. If not, we have

$\begin{matrix} \frac{1}{2} (2 b)^{\frac{1}{p^{'} (2 b)} - \frac{1}{p^{'} (4 b)}} \leq 1, \end{matrix}$ $$\begin{array}{} \displaystyle \frac{1}{2}(2b)^{\frac{1}{p^\prime(2b)}-\frac{1}{p^\prime(4b)}}\leq 1, \end{array}$$

i.e. it follows that

$\begin{matrix} (2 b)^{\frac{1}{p^{'} (2 b)} - \frac{1}{p^{'} (4 b)}} \leq C_{20}, \end{matrix}$ $$\begin{array}{} \displaystyle (2b)^{\frac{1}{p^\prime(2b)}-\frac{1}{p^\prime(4b)}}\leq C_{20}, \end{array}$$

$\begin{matrix} (2 b)^{\frac{p (2 b) - p (4 b)}{p (2 b) p (4 b)}} \leq C_{20} . \end{matrix}$ $$\begin{array}{} \displaystyle (2b)^{\frac{p(2b)-p(4b)}{p(2b)p(4b)}}\leq C_{20}. \end{array}$$

Replacing 2b by b, we get the inequality

$\begin{matrix} b^{\frac{p (b) - p (2 b)}{p (b) p (2 b)}} \leq C_{20}, b > 2 N . \end{matrix}$ $$\begin{array}{} \displaystyle b^{\frac{p(b)-p(2b)}{p(b)p(2b)}}\leq C_{20}, \quad b \gt 2N. \end{array}$$

Whence,

$\begin{matrix} b^{p (b) - p (2 b)} \leq C_{21}, C_{21} = C_{20}^{(p^{+})^{2}}, b > 2 N, \end{matrix}$ $$\begin{array}{} \displaystyle b^{p(b)-p(2b)}\leq C_{21}, \quad C_{21}=C_{20}^{(p^+)^2}, \quad b \gt 2N, \end{array}$$

$\begin{matrix} [p (b) - p (2 b)] \ln b \leq C_{22}, C_{22} = (p^{+})^{2} \ln C_{21} . \end{matrix}$ $$\begin{array}{} \displaystyle \big [p(b)-p(2b)\big ] \ln b \leq C_{22}, \quad C_{22}=(p^+)^2\ln C_{21}. \end{array}$$(25)

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 $\begin{matrix} \frac{1}{2} x \end{matrix}$ $\begin{array}{} \displaystyle \frac{1}{2}x \end{array}$ < y < 2x with x > N the estimates

$\begin{matrix} C_{23} x^{- \frac{1}{p^{'} (x)}} \leq y^{- \frac{1}{p^{'} (y)}} \leq C_{24} x^{- \frac{1}{p^{'} (x)}} \end{matrix}$ $$\begin{array}{} \displaystyle C_{23}x^{-\frac{1}{p^\prime(x)} }\leq y^{-\frac{1}{p^\prime(y)} } \leq C_{24} x^{-\frac{1}{p^\prime(x)} } \end{array}$$

are satisfied with positive constants C₂₃, C₂₄ depending on C₂₂.

Proof

By using estimate (25) and decreasing of $\begin{matrix} \frac{1}{p^{'} (x)}, \end{matrix}$ $\begin{array}{} \displaystyle \frac{1}{p^\prime(x)}, \end{array}$ it follows that

$\begin{matrix} y^{- \frac{1}{p^{'} (y)}} \leq {(\frac{x}{2})}^{- \frac{1}{p^{'} (y)}} \leq x^{\frac{1}{p^{'} (x)} - \frac{1}{p^{'} (2 x)}} \cdot x^{- \frac{1}{p^{'} (x)}} 2^{\frac{1}{p^{'} (N)}} \\ \leq 2 x^{p (x) - p (2 x)} \cdot x^{- \frac{1}{p^{'} (x)}} \leq 2 C_{21} x^{- \frac{1}{p^{'} (x)}} . \end{matrix}$ $$\begin{array}{c} \displaystyle y^{-\frac{1}{p^\prime(y)} }\leq \left( \frac{x}{2}\right)^{-\frac{1}{p^\prime(y)}} \leq x^{\frac{1}{p^\prime(x)}-\frac{1}{p^\prime(2x)}}\cdot x^{-\frac{1}{p^\prime(x)}}2^{\frac{1}{p^\prime(N)}}\\ \displaystyle\leq 2x^{p(x)-p(2x)}\cdot x^{-\frac{1}{p^\prime(x)}}\leq 2C_{21} x^{-\frac{1}{p^\prime(x)}}. \end{array}$$

By the same way,

$\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \leq {(\frac{y}{2})}^{- \frac{1}{p^{'} (x)}} \leq y^{\frac{1}{p^{'} (y)} - \frac{1}{p^{'} (2 y)}} \cdot y^{- \frac{1}{p^{'} (y)}} 2^{\frac{1}{p^{'} (N)}} \\ \leq 2 y^{p (y) - p (2 y)} \cdot y^{- \frac{1}{p^{'} (y)}} \leq 2 C_{21} y^{- \frac{1}{p^{'} (y)}} . \end{matrix}$ $$\begin{array}{c} \displaystyle x^{-\frac{1}{p^\prime(x)} }\leq \left( \frac{y}{2}\right)^{-\frac{1}{p^\prime(x)}} \leq y^{\frac{1}{p^\prime(y)}-\frac{1}{p^\prime(2y)}}\cdot y^{-\frac{1}{p^\prime(y)}}2^{\frac{1}{p^\prime(N)}}\\ \displaystyle\leq 2y^{p(y)-p(2y)}\cdot y^{-\frac{1}{p^\prime(y)}}\leq 2C_{21} y^{-\frac{1}{p^\prime(y)}}. \end{array}$$

This proves Lemma 3.2.

Lemma 3.3

The function $\begin{matrix} x^{- \frac{1}{p^{'} (x)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1}{p^\prime(x)}} \end{array}$ almost decreases on (N, ∞).

Proof

Take any N < t₁ ≤ t₂ < ∞, we show that there exists a constant C₂₅ > 1 depending on the constant C₂ of inequality (1) and p⁺ such that

$\begin{matrix} t_{2}^{- \frac{1}{p^{'} (t_{2})}} \leq C_{25} t_{1}^{- \frac{1}{p^{'} (t_{1})}} . \end{matrix}$ $$\begin{array}{} \displaystyle t_2^{-\frac{1}{p^\prime(t_2)}}\leq C_{25} t_1^{-\frac{1}{p^\prime(t_1)}}. \end{array}$$

We fix any t₁ = b > 2N and choose n ∈ N such that 2^n–1a < t₂ ≤ 2^nb. Then

$\begin{matrix} C_{2} \geq \int_{2 b}^{\infty} (x^{- \frac{1}{p^{'} (x)}} \int_{b}^{2 b} t^{- \frac{1}{p (t)}} d t)^{p (x)} \frac{d x}{x} \end{matrix}$ $$\begin{array}{} \displaystyle C_{2}\geq \int_{2b}^\infty \Big( x^{-\frac{1}{p^\prime(x)}} \int_b^{2b} t^{-\frac{1}{p(t)}}dt \Big)^{p(x)}\frac{dx}{x} \end{array}$$

$\begin{matrix} \geq \sum_{k = 1}^{\infty} \int_{2^{k - 1} b}^{2^{k} b} (x^{- \frac{1}{p (x)}} \int_{b}^{2 b} t^{- \frac{1}{p (t)}} d t)^{p (x)} \frac{d x}{x}, \end{matrix}$ $$\begin{array}{} \displaystyle \geq \sum_{k=1}^\infty\int_{2^{k-1}b}^{2^kb}\Big(x^{-\frac{1}{p(x)}}\int_b^{2b}t^{-\frac{1}{p(t)}}dt\Big)^{p(x)}\frac{dx}{x}, \end{array}$$

by using Lemma 3.2,

$\begin{matrix} \geq \sum_{k = 1}^{\infty} \int_{2^{k - 1} b}^{2^{k} b} [C_{24}^{- 1} b^{\frac{1}{p^{'} (b)}} x^{- \frac{1}{p^{'} (x)}}]^{p (x)} C_{24}^{p^{-}} \frac{d x}{x}, \end{matrix}$ $$\begin{array}{} \displaystyle \geq\sum_{k=1}^\infty\int_{2^{k-1}b}^{2^kb}\Big [C_{24}^{-1}b^{\frac{1}{p^\prime(b)}}x^{-\frac{1}{p^\prime(x)}}\Big]^{p(x)}C_{24}^{p^-}\frac{dx}{x}, \end{array}$$

by using Lemma 3.2, we obtain

$\begin{matrix} \geq \sum_{k = 1}^{\infty} \int_{2^{k - 1} b}^{2^{k} b} [C_{24}^{- 1} b^{\frac{1}{p^{'} (b)}} (2^{k} b)^{- \frac{1}{p^{'} (2^{k} b)}}]^{p (x)} C_{24}^{p^{-}} \frac{d x}{x} \\ \geq \sum_{k \in N^{'}} [C_{24}^{- 1} b^{\frac{1}{p^{'} (b)}} (2^{k} b)^{- \frac{1}{p' (2^{k} b)}}]^{p^{+}} C_{24}^{p^{-}} \ln 2 \\ + \sum_{k \in N^{''}} [C_{24}^{- 1} b^{\frac{1}{p^{'} (b)}} (2^{k} b)^{- \frac{1}{p' (2^{k} b)}}]^{p^{-}} C_{24}^{p^{-}} \ln 2, \end{matrix}$ $$\begin{array}{c} \displaystyle \geq\sum_{k=1}^\infty\int_{2^{k-1}b}^{2^kb}\Big [C_{24}^{-1}b^{\frac{1}{p^\prime(b)}}(2^kb)^{-\frac{1}{p^\prime(2^kb)}}\Big]^{p(x)}C_{24}^{p^-}\frac{dx}{x} \\ \displaystyle\geq\sum_{k\in N^\prime}\Big [C_{24}^{-1}b^{\frac{1}{p^\prime(b)}}(2^kb)^{-\frac{1}{p\prime(2^kb)}}\Big]^{p^+}C_{24}^{p^-}\ln2 \\ \displaystyle+\sum_{k\in N^{\prime\prime}}\Big [C_{24}^{-1}b^{\frac{1}{p^\prime(b)}}(2^kb)^{-\frac{1}{p\prime(2^kb)}}\Big]^{p^-}C_{24}^{p^-}\ln2, \end{array}$$

where $\begin{matrix} \sum_{k \in N^{'}} \end{matrix}$ $\begin{array}{} \displaystyle \sum\limits_{k\in N^\prime} \end{array}$ is a summation over k ∈ N such that the inequality

$\begin{matrix} b^{\frac{1}{p^{'} (b)}} (2^{k} b)^{- \frac{1}{p^{'} (2^{k} b)}} \leq 1 \end{matrix}$ $$\begin{array}{} \displaystyle b^{\frac{1}{p^\prime{(b)}}}(2^kb)^{-\frac{1}{p^\prime{(2^kb)}}}\leq1 \end{array}$$

holds, and $\begin{matrix} \sum_{k \in N^{''}} \end{matrix}$ $\begin{array}{} \displaystyle \sum \limits_{k \in N^{\prime\prime}} \end{array}$ is a summation of the opposite case. Therefore, for any k ∈ N one gets

$\begin{matrix} b^{\frac{1}{p' (b)}} (2^{k} b)^{- \frac{1}{p' (2^{k} b)}} \leq C_{25} . \end{matrix}$ $$\begin{array}{} \displaystyle b^{\frac{1}{p\prime{(b)}}}(2^kb)^{-\frac{1}{p\prime{(2^kb)}}}\leq C_{25}. \end{array}$$

This yields

$\begin{matrix} t_{2}^{- \frac{1}{p^{'} (t_{2})}} \cdot t_{1}^{\frac{1}{p^{'} (t_{1})}} \leq C_{26} \end{matrix}$ $$\begin{array}{} \displaystyle t_2^{-\frac{1}{p^\prime(t_2)}}\cdot t_1^{\frac{1}{p^\prime(t_1)}}\leq C_{26} \end{array}$$

i.e. function $\begin{matrix} x^{- \frac{1}{p' (x)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1}{p\prime(x)}} \end{array}$ is almost decreasing. Since $\begin{matrix} x^{- \frac{1}{p' (x)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1}{p\prime(x)}} \end{array}$ is almost decreasing on [N, ∞), we get

$\begin{matrix} C_{2} \geq \int_{b}^{\infty} C_{26}^{p^{-}} [\frac{1}{C_{26}} b^{\frac{1}{p^{'} (b)}} x^{- \frac{1}{p^{'} (x)}}]^{p^{+}} d x \end{matrix}$ $$\begin{array}{} \displaystyle C_{2} \geq\int_b^\infty C_{26}^{p^-}\Big[\frac{1}{C_{26}}b^{\frac{1}{p^\prime(b)}}x^{-\frac{1}{p^\prime(x)}}\Big]^{p^+}dx \end{array}$$

$\begin{matrix} \int_{b}^{\infty} x^{- \frac{p^{+}}{p^{'} (x)}} \frac{d x}{x} \leq C_{27} b^{\frac{p^{+}}{p^{'} (b)}}, b > N . \end{matrix}$ $$\begin{array}{} \displaystyle \int_b^\infty x^{-\frac{p^+}{p^\prime(x)}}\frac{dx}{x}\leq C_{27}b^{\frac {p^+} {p^\prime(b)}}, \quad b \gt N. \end{array}$$(26)

Now, let N < t₁ < t₂ < ∞. We show that condition (26) entails (2). We take any N < t₁ < t₂ < ∞ and set $\begin{matrix} K (x) = \int_{x}^{\infty} t^{- \frac{p^{+}}{p^{'} (t)}} \frac{d t}{t}, x > N . \end{matrix}$ $\begin{array}{} K(x)=\int_x^\infty t^{-\frac{p^+}{p^\prime(t)}}\frac{dt}{t},\quad x \gt N. \end{array}$ From (26) it follows that

$\begin{matrix} K (x) \leq - C_{27} x K^{'} (x), N < x < \infty . \end{matrix}$ $$\begin{array}{} \displaystyle K(x)\le -C_{27} x K'(x), \quad N \lt x \lt \infty. \end{array}$$

Integrating this inequality over (t₁, t₂), for N < t₁ < t₂ < ∞ and using (26), we get

$\begin{matrix} t_{2}^{\frac{1}{C_{27}}} K (t_{2}) \leq C_{27} t_{1}^{- \frac{p^{+}}{p^{'} (t_{1})} + \frac{1}{C_{27}}} \end{matrix}$ $$\begin{array}{} \displaystyle t_2^{\frac{1}{C_{27}}}K(t_2)\le C_{27} t_{1}^{-\frac{p^+}{ p^\prime (t_{1})}+ \frac{1}{C_{27}}} \end{array}$$(27)

On the other hand, by using monotone decreasing of $\begin{matrix} x^{- \frac{p^{+}}{p^{'} (x)}}, \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{p^+}{p^\prime(x)}}, \end{array}$ the conditions (25) and (26), it follows that

$\begin{matrix} k (t_{2}) \geq \int_{t_{2}}^{2 t_{2}} x^{- \frac{p^{+}}{p^{'} (x)}} \frac{d x}{x} \geq \ln 2 {(\frac{1}{2 t_{2}})}^{\frac{p^{+}}{p^{'} (2 t_{2})}} \geq 2^{- \frac{p^{+}}{p^{'} (N)}} \ln 2 t_{2}^{- \frac{p^{+}}{p^{'} (t_{2})}}, t_{2} \geq N . \end{matrix}$ $$\begin{array}{} \displaystyle k(t_{2})\geq \int\limits_{t_{2}}^{2t_{2}}x^{-\frac{p^+}{p^{\prime }(x)}}\frac{ dx}{x}\geq \ln 2 \left( \frac{1}{2t_{2}}\right) ^{\frac{p^+}{p^{\prime}(2t_{2})}}\geq \, 2^{-\frac{p^+}{p'(N)}} \ln 2 ~ t_{2}^{-\frac{p^+}{p^{\prime }( t_{2})}},\quad t_2 \geq N. \end{array}$$

By inserting this in (27) one gets,

$\begin{matrix} t_{2}^{- \frac{p^{+}}{p^{'} (t_{2})} + \frac{1}{C_{27}}} \leq C_{28} t_{1}^{- \frac{p^{+}}{p^{'} (t_{1})} + \frac{1}{C_{27}}} \end{matrix}$ $$\begin{array}{} \displaystyle t_2^{-\frac{p^+}{p^{\prime }(t_{2})}+\frac{1}{C_{27}}}\le C_{28}t_{1}^{-\frac{ p^+}{p^{\prime }(t_{1})}+\frac{1}{C_{27}}} \end{array}$$

for all N < t₁ < t₂ < ∞. Whence, the function $\begin{matrix} x^{δ - \frac{p^{+}}{p^{^{'}} (x)}} \end{matrix}$ $\begin{array}{} \displaystyle x^{\delta -\frac{p^+}{p^{^{\prime }}(x)}} \end{array}$ is almost decreasing on (N, ∞). Therefore, we have proved that

$\begin{matrix} t_{2}^{- \frac{1}{p^{'} (t_{2})} + δ_{1}} \leq C_{28} t_{1}^{- \frac{1}{p^{'} (t_{1})} + δ_{1}} \end{matrix}$ $$\begin{array}{} \displaystyle t_{2}^{-\frac{1}{p^{\prime }(t_{2})}+\delta_1 }\leq C_{28}t_{1}^{-\frac{1}{ p^{\prime }(t_{1})}+\delta_1 } \end{array}$$

for all N < t₁ < t₂ < ∞; a constant $\begin{matrix} δ_{1} = \frac{δ}{p^{+}} . \end{matrix}$ $\begin{array}{} \displaystyle \delta_1 =\frac{\delta}{p^+}. \end{array}$

Therefore $\begin{matrix} x^{- \frac{1}{p^{'} (x)} + δ_{1}} \end{matrix}$ $\begin{array}{} \displaystyle x^{-\frac{1}{p^\prime(x)}+\delta_1} \end{array}$ is almost decreasing with $\begin{matrix} δ_{1} = \frac{δ}{p^{+}}, δ = \frac{1}{C_{27}} . \end{matrix}$ $\begin{array}{} \displaystyle \delta_1=\frac{\delta }{ p^+}, \, \delta=\frac{1}{C_{27}}. \end{array}$ From this it follows

$\begin{matrix} \int_{b}^{\infty} x^{- \frac{1}{p^{'} (x)}} \frac{d x}{x} \leq C_{28} \int_{a}^{\infty} b^{δ - \frac{1}{p^{'} (x)}} \frac{d x}{x^{1 + δ}} \\ \leq C_{28} b^{δ - \frac{1}{p^{'} (b)}} \int_{b}^{\infty} \frac{d x}{x^{1 + δ}} = \frac{C_{28}}{δ} b^{- \frac{1}{p^{'} (a)}}, b > N \end{matrix}$ $$\begin{array}{c} \displaystyle \int_b^\infty x^{-\frac{1}{p^\prime(x)}}\frac{dx}{x}\leq C_{28}\int_a^\infty b^{\delta-\frac{1}{p^\prime(x)}}\frac{dx}{x^{1+\delta}} \\ \displaystyle\leq C_{28}b^{\delta-\frac{1}{p^\prime(b)}}\int_b^\infty\frac{dx}{x^{1+\delta}}=\frac{C_{28}}{\delta}b^{-\frac{1}{p^\prime(a)}}, \quad b \gt N \end{array}$$

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.

eISSN:: 2444-8656
Langue:: Anglais

Périodicité:: Volume Open
Sujets de la revue:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

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A New Approach For Weighted Hardy’s Operator In VELS

Publié en ligne: 28 oct. 2019

Pages: 417 - 432

Reçu: 07 mai 2019

Accepté: 24 juin 2019

DOI: https://doi.org/10.2478/AMNS.2019.2.00040

Mots clésHardy inequality, Variable exponent, Boundedness

© 2019 Lutfi Akin and Faruk Dusunceli, published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Mots clés
Hardy inequality, Variable exponent, Boundedness