A second alternative approach for the study of the Muckenhoupt class $A_1({\mathbb {R}})$

Abstract

We find the exact best possible range of those $p >1$ for which any $\varphi \in A_1({\mathbb {R}})$, with $A_1$ constant equal to c, must also belong to $L^p$. In this way, we provide an alternative proof of the corresponding result in Bojarski and Sbordone (Studia Math 101(2):155–163, 1992) and Nikolidakis (Ann Acad Scient Fenn Math 40:949–955, 2015).

1 Introduction

The study of Muckenhoupt weights has been proved to be important in analysis. One of the most important facts about these is their self improving property. A way to express this is through the so-called reverse Hölder inequalities (see [3, 4, 6]).

For an interval $\mathcal {J}$ on ${\mathbb {R}}$, we define the class $A_1(\mathcal {J})$ to be the set of all those $\varphi : \mathcal {J} \rightarrow {\mathbb {R}}^+$ for which there exists a constant $c\ge 1$, such that the following inequality is satisfied:

$$\begin{aligned} \frac{1}{|\mathcal {I}|}\int \limits _{\mathcal {I}} \varphi (x)dx \le c\cdot \mathop {\text {essinf}}\limits _{\mathcal {I}}(\varphi ) \end{aligned}$$

(1.1)

for every subinterval $\mathcal {I}$ of $\mathcal {J}$, where $|\cdot |$ is the Lebesque measure on ${\mathbb {R}}$. The least constant c for which (1.1) holds, is called the $A_1$-constant of $\varphi$ and is denoted by $[\varphi ]_1$. We will then say that $\varphi$ belongs to the class $A_1(\mathcal {J})$ with constant c, and we will write $\varphi \in A_1(\mathcal {J},c)$.

The class $A_1(\mathcal {J},c)$ has been studied for the first time in [2]. In the present paper, we work on such weights by using the notion of the non-increasing rearrangement of $\varphi$, denoted by $\varphi ^*$, which is a non-negative and non-increasing function defined on $(0,|\mathcal {J}|]$. It is characterized by the following two additional properties. It is equimeasurable to $\varphi$ (in the sense that $|\{\varphi> \lambda \}|=|\{\varphi ^* >\lambda \}|$ for every $\lambda >0$) and is also left continuous. All these properties uniquely define $\varphi ^*$ as can be seen in [1, 5], or [8]. Nevertheless, an equivalent definition of $\varphi ^*$ can be given by the following formula

$$\begin{aligned} \varphi ^*(t)=\sup _{\begin{array}{c} E\subseteq \mathcal {J} \\ |E|=t \end{array}}[\inf _{x\in E}\varphi (x)], \text {for} \; t \in (0,|\mathcal {J}|], \end{aligned}$$

as can be seen in [8].

In [2], it is proved the following

Theorem 1

Let $\varphi \in A_1(\mathcal {J},c)$. $\varphi ^*$ satisfies

$$\begin{aligned} \frac{1}{t} \int \limits _{0}^{t} \varphi ^*(y)dy \le c \varphi ^*(t), \text {for t} \in (0,|\mathcal {J}|]. \end{aligned}$$

(1.2)

That is $\varphi ^*$ belongs to the class $A_1(\mathcal {J})$, with $A_1$-constant not more than c.

The above theorem describes the $A_1$-properties of $\varphi ^*$, in terms of those of $\varphi$. It was used effectively by the authors in [2] in order to prove the following:

Theorem 2

Let $\varphi \in A_1(\mathcal {J},c)$. Then $\varphi \in L^p$ for every $p\in [1 ,\frac{c}{c-1})$. Moreover, the following inequality must hold for every subinterval $\mathcal {I}$ of $\mathcal {J}$, and every p in the above range,

$$\begin{aligned} \frac{1}{|\mathcal {I}|}\int \limits _{\mathcal {I}} \varphi ^p(x)dx \le \frac{1}{c^{p-1}(c+p-pc)}\Big (\frac{1}{|\mathcal {I}|}\int \limits _{\mathcal {I}} \varphi (x)dx \Big )^p. \end{aligned}$$

(1.3)

Additionally, the above inequality is sharp, that is the constant appearing in the right side of (1.3) cannot be decreased.

The above two theorems have been proved in [2] for the first time and in [10] alternatively. Our aim in this paper is to give a second alternative proof of Theorem 2 by using Theorem 1 and certain techniques involving the well known Hardy operator on ${\mathbb {R}}$. Additionally, we need to mention that in [7] and [9] related problems for estimates for the respective range of p in higher dimensions have been treated. At last one can consult [11] for further reading.

The paper is organized as follows: In Section 2, we give a brief discussion of the proof of the Theorem 1, as is presented in [2], and in Section 3, we provide the proof of Theorem 2.

2 $\varphi ^*$ as an $A_1$ weight on ${\mathbb {R}}$.

A similar lemma as the one that is presented below is proved in [2]. It’s proof is essentially the same and for this reason we omit it.

Lemma 2.1

Let E be a measurable bounded subset of ${\mathbb {R}}$ and $\epsilon >0$. More precisely, suppose that $E\subseteq I$ for a certain bounded interval I of ${\mathbb {R}}$ for which $|I-E|>0$. Then there exists a sequence $(I_\nu )_{\nu =1}^{\infty }$ of subintervals of I with disjoint interiors and a subset $E_1$ of E with the properties that $|E_1|=|E|$ and

• (i) $E_1\subseteq \bigcup \nolimits _{\nu =1}^{\infty }I_{\nu }$,

• (ii) $(1-\epsilon ) |I_{\nu }| \le |I_{\nu }\cap E|<|I_{\nu }|$ for every $\nu$.

We now proceed to the

Proof of Theorem 1

Suppose without loss of generality that $\mathcal {J}=(0,1)$ and that $\varphi$ satisfies (1.1) for every subinterval $\mathcal {I}$ of $\mathcal {J}$. Let $t\in (0,1)$ and $\epsilon >0$. Let $E_t$ be a subset of (0, 1) such that $|E_t|=t$ and $\varphi (x)\le \varphi ^*(t)$ for any $x\notin E_t$. Obviously $|J-E_t|>0$. Using Lemma 2.1, we produce a subset $E_{t,1}$ of $E_t$ such that $|E_{t,1}|=t$ and $E_{t,1}\subseteq \bigcup \nolimits _{\nu =1}^{\infty }I_{\nu }$, where for every $\nu =1,2,\ldots$, the following holds:

$$\begin{aligned} (1-\epsilon )|I_\nu |\le |I_\nu \cap E_t|<|I_\nu | \end{aligned}$$

(2.1)

for a suitable family $(I_\nu )_{\nu =1}^{\infty }$ of subintervals of (0, 1) with disjoint interiors. By the strict inequality in (2.1), we conclude that $I_\nu$ contains a set of positive measure in the complement of $E_t$, therefore we must have that

$$\begin{aligned} \mathop {\text {essinf}}\limits _{x\in I_\nu } \varphi (x)\le \varphi ^*(t), \end{aligned}$$

so using (1.1) and (2.1), we have as a consequence that

$$\begin{aligned} \int \limits _{0}^{t} \varphi ^*(y)dy= & {} \int \limits _{E_t} \varphi (x)dx=\int \limits _{E_t,1} \varphi (x)dx\le \sum _{\nu =1}^{\infty } \int \limits _{I_\nu }\varphi (x)dx\le c \sum _{\nu =1}^{\infty }|I_\nu |\cdot \varphi ^*(t) \\\le & {} \frac{c}{1-\epsilon }\Big (\sum _{\nu =1}^{\infty }|I_\nu \cap E_t|\Big )\cdot \varphi ^*(t)=\frac{c}{1-\epsilon }\cdot t\cdot \varphi ^*(t) \\\Rightarrow & {} \frac{1}{t} \int \limits _{0}^{t}\varphi ^*(y)dy\le \frac{c}{1-\epsilon }\varphi ^*(t) \end{aligned}$$

for every $\epsilon >0$. Letting $\epsilon \rightarrow 0^+$, we conclude (1.2) for any $t\in (0,1)$. The case $t=1$ is handled by letting $t\rightarrow 1^{-}$ in (1.2) and noting that $\varphi ^*$ is left continuous on (0, 1]. $\square$

3 $L^p$ integrability of $A_1$ weights on ${\mathbb {R}}$

We will now prove the following

Lemma 3.1

Let $g:(0,1]\rightarrow {\mathbb {R}}^+$ be a non-increasing, left continuous function which satisfies the following inequality:

$$\begin{aligned} \frac{1}{t} \int \limits _{0}^{t}g(y)dy\le c\cdot g(t), \; \; \forall t\in (0,1], \end{aligned}$$

(3.1)

for a fixed $c>1$. Then for any $p\in [1,\frac{c}{c-1})$, the following is true:

$$\begin{aligned} \int \limits _{0}^{1}g^p(y)dy\le \frac{1}{c^{p-1}(c+p-pc)}\Big ( \int \limits _{0}^{1}g(y)dy\Big )^p. \end{aligned}$$

(3.2)

Moreover, inequality (3.2) is best possible.

Proof

Fix a p such that $1\le p< \frac{c}{c-1}$ and let $F=\int _{0}^{1}g^p(y)dy$ and $f=\int _{0}^{1}g(y)dy$. Then by Hölder’s inequality, $f^p\le F$. We need to prove that

$$\begin{aligned} F\le \frac{1}{c^{p-1}(c+p-pc)}\cdot f^p. \end{aligned}$$

(3.3)

We define the function

$$\begin{aligned} H_p:\Big [ 1,\frac{p}{p-1}\Big ] \rightarrow [0,1] \end{aligned}$$

by $H_p(z)=pz^{p-1}-(p-1)z^p$. Then we easily see that $H_p$ is one to one and onto. We denote it’s inverse function by $\omega _p$ defined on [0, 1], which is decreasing as $H_p$ also is. We shall prove that (3.3) holds, equivalently, $H_p(c)\le \frac{f^p}{F}$$\Leftrightarrow c\ge \omega _p \Big (\frac{f^p}{F}\Big )=: \tau$.

Suppose on the contrary that $c<\tau$. We are going to reach a contradiction.

Define the function $g_1$ on (0, 1] by $g_1(t)=\frac{f}{\tau }t^{-1+\frac{1}{\tau }}$. This is obviously non-increasing and continuous (0, 1]. Additionally, it satisfies for any $t\in (0,1]$, the following equality:

$$\begin{aligned} \frac{1}{t}\int \limits _{0}^{t}g_1(y)dy=\tau \cdot g_1(t). \end{aligned}$$

(3.4)

Indeed: $\frac{1}{t} \int _{0}^{t}g_1(y)dy=\frac{1}{t} \frac{f}{\tau }\int _{0}^{t}y^{-1+\frac{1}{\tau }}dy=\frac{f}{t}\big [y^{\frac{1}{\tau }}\big ]_{y=0}^t=\frac{f}{t}\cdot t^{\frac{1}{\tau }}=\tau \cdot \Big (\frac{f}{\tau }t^{-1+\frac{1}{\tau }}\Big )=\tau g_1(t)$. Moreover, it satisfies $\int _{0}^{1}g_1(y)dy=f$ and $\int _{0}^{1}g_1^p(y)dy=F$. The first equation is obvious, in view of (3.4). As for the second, it is equivalent to $\frac{f^p}{\tau ^p}\int _{0}^{1}y^{-p+\frac{p}{\tau }}dy=F$$\Leftrightarrow \frac{f^p}{\tau ^p(1+\frac{p}{\tau }-p)}=F$$\Leftrightarrow p\tau ^{p-1}-(p-1)\tau ^p=\frac{f^p}{F}$$\Leftrightarrow H_p(\tau )=\frac{f^p}{F}$$\Leftrightarrow \tau =\omega _p(\frac{f^p}{F})$, which is true by the definition of $\tau$.

We are now aiming to prove that the following inequality is satisfied:

$$\begin{aligned} \int \limits _{0}^{t}g(y)dy\le \int \limits _{0}^{t} g_1(y)dy, \; \text {for any} \; t \in (0,1]. \end{aligned}$$

(3.5)

For this reason, we define the following subset of (0, 1):

$G=\Big \{t\in (0,1): \int _{0}^{t}g(y)dy> \int _{0}^{t}g_1(y)dy\Big \}$, and we suppose that G is non-empty. By the continuity of the involving integral functions on t, we have as a consequence that G is an open subset of (0, 1). Since $G\ne \emptyset$$\Rightarrow G=\bigcup \nolimits _\nu I_\nu$, where $(I_\nu )_\nu$ is a (possibly finite) sequence of pairwise disjoint open intervals on (0, 1). Let us choose one of them, $I_\nu =(\alpha _\nu ,b_\nu )$. Since $\alpha _\nu \notin G$,

$$\begin{aligned} \int \limits _{0}^{\alpha _\nu } g(y)dy \le \int \limits _{0}^{\alpha _\nu } g_1(y)dy. \end{aligned}$$

(3.6)

Let now $(x_n)_n\subseteq I_\nu$ be a sequence such that $x_n\rightarrow \alpha _\nu$, as $n\rightarrow \infty$. Since $x_n\in G, \forall n=1,2,\ldots$, we must have that $\int _{0}^{x_n}g(y)dy> \int _{0}^{x_n}g_1(y)dy$, so letting $n\rightarrow \infty$, we conclude that

$$\begin{aligned} \int \limits _{0}^{\alpha _\nu }g(y)dy\ge \int \limits _{0}^{\alpha _\nu }g_1(y)dy. \end{aligned}$$

(3.7)

By (3.6) and (3.7), we see that $\int _{0}^{\alpha _\nu }g(y)dy=\int _{0}^{\alpha _\nu }g_1(y)dy$. In the same way, we prove that $\int _{0}^{b_\nu }g(y)dy=\int _{0}^{b_\nu }g_1(y)dy$. As a consequence, we must have that

$$\begin{aligned} \int \limits _{\alpha _\nu }^{b_\nu }g(y)dy=\int \limits _{\alpha _\nu }^{b_\nu }g_1(y)dy. \end{aligned}$$

(3.8)

Let now $t\in I_\nu =(\alpha _\nu ,b_\nu )$. Since $t\in G$ and because of (3.1) and (3.4) and the assumption on $\tau$, we must have the following: $cg(t)\ge \frac{1}{t}\int _{0}^{t}g(y)dy>\frac{1}{t}\int _{0}^{t}g_1(y)dy=\tau \cdot g_1(t)>c g_1(t)$ thus $g(t)>g_1(t)$ for every $t\in I_\nu$. This is impossible in view of (3.8). Thus we have proved (3.5).

For the following, consult [5, page 88].

Lemma 3.2

Let $\varphi _1, \varphi _2: (0,1]\rightarrow {\mathbb {R}}^+$ be integrable functions. Then the following are equivalent

• (i) $\int _{0}^{t}\varphi _1^*(y)dy \le \int _{0}^{t}\varphi _2^*(y)dy$ for every $t\in (0,1]$.

• (ii) $\int _{0}^{1}G(\varphi _1(x))dx \le \int _{0}^{1}G(\varphi _2(x))dx$

for any convex, non-negative, increasing, and left continuous function G on $[0,+\infty )$.

We consider now two cases:

• (A) We have equality in (3.5) for every $t\in (0,1]$. That is $\int _{0}^{t}g(y)dy =\int _{0}^{t}g_1(y)dy$ for every $t\in (0,1]$. This immediately gives as a consequence that $g(t)=g_1(t)$ almost everywhere on (0, 1], and since $g_1$ is continuous on (0, 1], we must have that $g(t)=g_1(t) \,\,\forall t\in (0,1]$$\Rightarrow g(t)=\frac{f}{\tau }t^{-1+\frac{1}{\tau }}\, \,\forall t\in (0,1]$$\Rightarrow \frac{1}{t}\int _{0}^{t}g(y)dy=\tau g(t)\,\, \forall t\in (0,1]$. Then in view of (3.1), we conclude that $c\ge \tau$ which is a contradiction since we have supposed the opposite inequality.

• (B) There exists a $t_0 \in (0,1)$ such that

  $$\begin{aligned} \int \limits _{0}^{t_0}g(y)dy<\int \limits _{0}^{t_0}g_1(y)dy. \end{aligned}$$

  Then, by continuity reasons, we have as a consequence that there exists a $\delta >0$ such that

  $$\begin{aligned} \int \limits _{0}^{t}g(y)dy<\int \limits _{0}^{t}g_1(y)dy, \,\text {for any} \; t \in (t_0-\delta , t_0+\delta )=I_\delta . \end{aligned}$$

  (3.9)

We define now the quantities $d_1,d_2$ by the following equations:

$$\begin{aligned} \frac{1}{\delta }\int \limits _{t_0-\delta }^{t_0}g_1(y)dy=d_1 \quad \text {and} \quad \frac{1}{\delta }\int \limits _{t_0}^{t_0+\delta }g_1(y)dy=d_2. \end{aligned}$$

(3.10)

Then by Hölder’s inequality on the interval $(t_0-\delta ,t_0)$ for $g_1$, we conclude that

$$\begin{aligned} \frac{1}{\delta }\int \limits _{t_0-\delta }^{t_0}g_1^p(y)dy>d_1^p, \end{aligned}$$

(3.11)

which is a strict inequality since $g_1$ is strictly decreasing (therefore not constant) on the interval $(t_0-\delta ,t_0)$. In the same way, we have

$$\begin{aligned} \frac{1}{\delta }\int \limits _{t_0}^{t_0+\delta }g_1^p(y)dy>d_2^p. \end{aligned}$$

(3.12)

Then since $g_1$ is decreasing, we have that $d_2<d_1$. We define now the following non-increasing (as can easily be seen) function on (0, 1]:

$$\begin{aligned} g_2(t) = \left\{ \begin{array}{ll} g_1(t), &{} t\in (0,1]\setminus (t_0-\delta ,t_0+\delta ),\\ d_1, &{} t\in [t_0-\delta ,t_0),\\ d_2, &{} t\in [t_0,t_0+\delta ].\end{array} \right. \end{aligned}$$

(3.13)

By (3.9) and since $g_1$ is decreasing, we easily see that we can choose $\delta >0$ small enough, so that

$$\begin{aligned} \int \limits _{0}^{t}g(y)dy\le \int \limits _{0}^{t}g_2(y)dy, \,\text {for any}\, t\in (0,1]. \end{aligned}$$

(3.14)

Additionally, because of (3.11) and (3.12), we must have that

$$\begin{aligned} \int \limits _{0}^{1}g_2^p(y)dy< \int \limits _{0}^{1}g_1^p(y)dy=F. \end{aligned}$$

Since (3.14) holds for any $t\in (0,1]$ and because of Lemma 3.2, we conclude that $\int _{0}^{1}g^p(y)dy\le \int _{0}^{1}g_2^p(y)dy<F$ by considering the function $G(t)=t^p$. This is obviously a contradiction according to the way that F is defined. In this way, we derive the proof of our lemma. $\square$

We now proceed to the

Proof of Theorem 2

Without loss of generality, we suppose that $\mathcal {J}=(0,1)$. Let $p\in [1,\frac{c}{c-1})$ and $\mathcal {I}\subseteq (0,1)$ and let also $\varphi _{\mathcal {I}}=\varphi /_{\mathcal {I}}$ be the restriction of $\varphi$ to $\mathcal {I}$. Consider now the function $g: (0,|\mathcal {I}|] \rightarrow {\mathbb {R}}^+$, defined by $g=(\varphi _{\mathcal {I}})^*$. Then since $\varphi _{\mathcal {I}}\in A_1(\mathcal {I})$ with $A_1$ constant not more than c, we must have, by using Theorem 1, that $\frac{1}{t}\int _{0}^{t}g(y)dy\le cg(t)$ for any $t\in (0,|\mathcal {I}|]$. Thus by Lemma 3.1, it is easy to see that the following is true:

$$\begin{aligned} \frac{1}{|\mathcal {I}|}\int \limits _{0}^{|\mathcal {I}|}g^p(y)dy\le \frac{1}{c^{p-1}(c+p-pc)} \Big (\frac{1}{|\mathcal {I}|} \int \limits _{0}^{|\mathcal {I}|}g(y)dy \Big )^p, \end{aligned}$$

which is

$$\begin{aligned} \frac{1}{|\mathcal {I}|}\int \limits _{\mathcal {I}}\varphi ^p(x)dx \le \frac{1}{c^{p-1}(c+p-pc)} \Big (\frac{1}{|\mathcal {I}|} \int \limits _{\mathcal {I}}\varphi (x)dx \Big )^p. \end{aligned}$$

The relation (1.3) is proved. $\square$