Maximal operators on BMO and slices

TL;DR

This paper investigates the behavior of maximal operators on BMO spaces, showing discontinuity of the Hardy-Littlewood operator on certain BMO variants, providing counterexamples for others, and exploring properties of slices of BMO functions.

Contribution

It demonstrates the discontinuity of the Hardy-Littlewood maximal operator on morn, shows that some maximal operators are unbounded on morn, and analyzes slices of BMO functions.

Findings

Hardy-Littlewood maximal operator is discontinuous on morn

Counterexamples show strong and directional maximal operators are unbounded on morn

Properties of slices of morn functions are characterized

Abstract

We prove that the Hardy-Littlewood maximal operator is discontinuous on $\bmorn$ and maps $\vmorn$ to itself. A counterexample to boundedness of the strong and directional maximal operators on $\bmorn$ is given, and properties of slices of $\bmorn$ functions are discussed.

Equations

$$O(f,A):=\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_A|f-f_A|,$$

$$\Vert f{\Vert }_{BMO({\mathbb{R}}^n)}=\underset{Q}{\sup }O(f,Q).$$

$$Mf(x)=\underset{x\in Q}{\sup }\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_Q|f|,f\in L_{loc}^1({\mathbb{R}}^n),$$

$$\omega (f,\delta ):=\underset{l(Q)\le \delta }{\sup }O(f,Q),$$

$$M_{e_1}f(x_1,x^{\prime }):=\underset{x_1\in I}{\sup }\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_I|f_{x^{\prime }}|,M_{s}f(x):=\underset{x\in R}{\sup }\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_R|f|.$$

$$H(-x)=H(x),H(x+T)=H(x),x\in \mathbb{R},$$

$$O(H,I)\le 2\frac{|J|}{|I|}O(H,J)\le 4O(H,J^{+})\le 4\Vert h{\Vert }_{BMO[0,\frac{T}{2}]}.$$

$$O(H,I)\le 2\frac{|J|}{|I|}O(H,J)\le 10O(h,[0,\frac{T}{2}])\le 10\Vert h{\Vert }_{BMO[0,\frac{T}{2}]}.$$

$$\Vert g_n{\Vert }_{BMO(\mathbb{R})}\le \frac{1}{\log n}(\Vert H_n{\Vert }_{BMO(\mathbb{R})}+\Vert H_n{\chi }_{[c,0]}{\Vert }_{L^{\infty }})\lesssim \frac{\log |c|}{\log n}.$$

$$\underset{n\to \infty }{\underline{\lim }}\Vert M(f+\frac{a}{1-c}{g}_n)-Mf{\Vert }_{BMO(\mathbb{R})}>0.$$

$$\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_I(f+\frac{a}{1-c}{g}_n)=\frac{a}{1-c}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_I{g}_n\le \frac{a}{1-c}\Vert g_n{\Vert }_{L^{\infty }}\le \frac{a}{1-c}\le Mf(x),$$

$$\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_I(f+\frac{a}{1-c}{g}_n)=\frac{|I\cap [x,1]|}{|I|}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{I\cap [x,1]}f+\frac{|I\backslash [x,1]|}{|I|}\frac{a}{1-c}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{I\backslash [x,1]}{g}_n$$

$$\le \frac{|I\cap [x,1]|}{|I|}Mf(x)+\frac{|I\backslash [x,1]|}{|I|}\frac{a}{1-c}\le Mf(x).$$

$$O(M(f+\frac{a}{1-c}{g}_n)-Mf,[2c,0])\ge \frac{1}{4}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[2c,c]}(M(f+\frac{a}{1-c}{g}_n)-Mf).$$

$$M(f+\frac{a}{1-c}{g}_n)(x)\ge \frac{a}{1-c}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[x,n]}{g}_n\ge \frac{a}{1-c}.\frac{n}{n-x}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[0,n]}{g}_n.$$

$$\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[2c,c]}(M(f+\frac{a}{1-c}{g}_n)-Mf)\ge \frac{a}{1-c}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[0,n]}{g}_n\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[2c,c]}\frac{n}{n-x}dx-\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[2c,c]}\frac{1}{1-x}dx.$$

$$\Vert M(f+\frac{a}{1-c}{g}_n)-Mf{\Vert }_{BMO(\mathbb{R})}\ge \frac{1}{4}(\frac{a}{1-c}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[0,n]}{g}_n\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{[2c,c]}\frac{n}{n-x}dx+\frac{1}{c}\log (1+\frac{c}{c-1})).$$

$$\underset{n\to \infty }{\underline{\lim }}\Vert M(f+\frac{a}{1-c}{g}_n)-Mf{\Vert }_{BMO(\mathbb{R})}\ge \frac{1}{4}(\frac{a}{1-c}+\frac{1}{c}\log (1+\frac{c}{c-1})).$$

$$a>\frac{c-1}{c}\log (1+\frac{c}{c-1})$$

$$f(x_1,..,x_n)=f(x_1),g_m(x_1,..,x_n)=g_m(x_1),$$

$$\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_A|f-f_Q|\lesssim 1+\log \frac{|Q|}{|A|}.$$

$$\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_A{e}^{c|f-f_Q|}\le \frac{|Q|}{|A|}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_Q{e}^{c|f-f_Q|}\lesssim \frac{|Q|}{|A|}.$$

$$\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_A|f-f_Q|=\frac{1}{c}\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_A\log e^{c|f-f_Q|}\le \frac{1}{c}\log \text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_A{e}^{c|f-f_Q|}\lesssim 1+\log \frac{|Q|}{|A|}.$$

$$O(f,A)\lesssim 1+\log e(A),$$

$$M_{1}f(x)=\underset{l(Q)\le cl(Q_0)}{\underset{x\in Q}{\sup }{f}_Q},M_{2}f(x)=\underset{l(Q)\ge cl(Q_0)}{\underset{x\in Q}{\sup }{f}_Q}.$$

$$O(Mf,Q_0)\lesssim O(M_{1}f,Q_0)+O(M_{2}f,Q_0).$$

$$O(M_{1}f,Q_0)\le 2\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{Q_0}|M_{1}f-f_{Q_0^{\ast }}|\le 2\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{Q_0}{M}_1|f-f_{Q_0^{\ast }}|$$

$$\le 2(\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{Q_0}(M_1|f-f_{Q_0^{\ast }}|{)}^2{)}^{\frac{1}{2}}\le 2(\frac{1}{|Q_0|}\int (M|f-f_{Q_0^{\ast }}|{\chi }_{Q_0^{\ast }}{)}^2{)}^{\frac{1}{2}}.$$

$$O(M_{1}f,Q_0)\lesssim c^{\frac{n}{2}}(\text{ThisStyle}\text{ensurestackMath}\text{stackinset}c.2\text{LMpt}c.5\text{LMpt}\text{SavedStyle}-\text{SavedStyle}{\int }_{Q_0^{\ast }}|f-f_{Q_0^{\ast }}{|}^2{)}^{\frac{1}{2}},$$

$$O(M_{1}f,Q_0)\lesssim c^{\frac{n}{2}}\Vert f{\Vert }_{BMO(Q_0^{\ast })}.$$

$$f_Q-M_{2}f(y)\le f_Q-f_{Q^{\prime }}=f_Q-(\frac{|A|}{|Q^{\prime }|}{f}_A+\frac{|Q|}{|Q^{\prime }|}{f}_Q)=\frac{|A|}{|Q^{\prime }|}(f_Q-f_A)$$

Full text

1. Introduction

Let $A\subset\mathbb{R}^{n}$ be a measurable set with positive finite measure and $f\in{L}_{loc}^{1}(\mathbb{R}^{n})$. By the mean oscillation of $f$ on $A$ we mean the quantity

[TABLE]

where $\,\ThisStyle{\ensurestackMath{\stackinset{c}{.2\LMpt}{c}{.5\LMpt}{\SavedStyle-}{\SavedStyle\phantom{\int}}}\kern-16.56937pt}\int_{A}{f}$ and $f_{A}$ mean the average of $f$ over $A$ i.e. $\frac{1}{|A|}\int_{A}f$. Then it is said that $f$ is of bounded mean oscillation if $O(f,Q)$ is uniformly bounded on all cubes $Q$ (by a cube we mean one with sides parallel to the axes). The space of such functions is denoted by ${BMO}(\mathbb{R}^{n})$, and modulo constants the following quantity defines a norm on this space.

[TABLE]

(Sometimes we use $\|f\|_{BMO(Q_{0})}$ which means that we take the above supremum over all cubes contained in $Q_{0}$.) ${BMO}(\mathbb{R}^{n})$ is a Banach space, and since its introduction has played an important role in harmonic analysis. It appears in many problems and somehow serves as a substitute for ${L}^{\infty}(\mathbb{R}^{n})$. For instance, it’s the dual of the Hardy space ${H}^{1}(\mathbb{R}^{n})$, and singular integral operators map ${L}^{\infty}(\mathbb{R}^{n})$ to ${BMO}(\mathbb{R}^{n})$ and not to itself [12],[6].

In [2] the authors proved that the uncentered Hardy-Littlewood maximal operator, defined by

[TABLE]

is bounded on ${BMO}(\mathbb{R}^{n})$. (The above supremum is taken over all cubes containing $x$.) Another proof of this was given in [1], and a third one in [10], where the author proved that $M$ preserves Poincare inequalities. In this paper we want to give an answer to the following related questions:

Q1. Is the Hardy-Littlewood maximal operator $M$ continuous on ${BMO}(\mathbb{R}^{n})$?

The operator $M$ is nonlinear and for such operators continuity doesn’t follow from boundedness. However, it is sublinear and this makes it continuous on ${L}^{p}(\mathbb{R}^{n})$ for $1<p\leq\infty$. In [9], a similar question has been studied for Sobolev spaces, where the author proved that $M$ is continuous on $W^{1,p}(\mathbb{R}^{n})$ for $1<p<\infty$. However, in Section 2 we give a negative answer to question Q1.

${BMO}(\mathbb{R}^{n})$ has an important subspace, namely ${VMO}(\mathbb{R}^{n})$ or functions of vanishing mean oscillation. ${VMO}(\mathbb{R}^{n})$ is the closure of the uniformly continuous functions in ${BMO}(\mathbb{R}^{n})$ [11],[butaev]. Another characterization of ${VMO}(\mathbb{R}^{n})$ is given in terms of the moduli of mean oscillation which is defined by

[TABLE]

and $f\in{VMO}(\mathbb{R}^{n})$ exactly when $\lim\limits_{\delta\to 0}\omega(f,\delta)=0$. (In the above by $l(Q)$ we mean the side length of $Q$). Regarding this subspace we ask:

Q2. Dose $M$ map ${VMO}(\mathbb{R}^{n})$ to itself?

In Section 3 we provide a positive answer to this question.

In the last section, we consider the action of other maximal operators on ${BMO}(\mathbb{R}^{n})$. More specifically, the directional maximal operator $M_{e_{1}}$, and the strong maximal operator $M_{s}$ which are defined as the following:

[TABLE]

In the above, $f_{x^{\prime}}(t)=f(t,x^{\prime})$ and the left supremum is taken over all intervals containing $x_{1}$. The right one is taken over all rectangles (here and afterwards by a rectangle we mean one with sides parallel to the axes) containing $x$. These are the most important maximal operators in multi-parameter harmonic analysis and are bounded and continuous on ${L}^{p}(\mathbb{R}^{n})$ for $1<p<\infty$[4]. Regarding this we ask:

Q3. Do $M_{e_{1}}$ and $M_{s}$ map ${BMO}(\mathbb{R}^{n})$ to itself boundedly?

We answer this question negatively in Section 4, where some properties of slices of functions in ${BMO}(\mathbb{R}^{n})$ are also investigated.

2. Discontinuity of $M$ on ${BMO}(\mathbb{R}^{n})$

Our theorem in this section is the following:

Theorem 2.1**.**

Let $f$ be a nonnegative function supported in $[0,1]$, $\|f\|_{{L}^{\infty}}\leq 1$ and $\|f\|_{{L}^{1}}>\log 2$. Then the Hardy-Littlewood maximal operator $M:{BMO}(\mathbb{R})\rightarrow{BMO}(\mathbb{R})$ is discontinuous at $f$.

To prove this we need a couple of simple lemmas which we give below.

Lemma 2.2**.**

Let $T>0$ and $h\in BMO[0,\frac{T}{2}]$. Then the even periodic extension of $h$, which is defined by

[TABLE]

is in ${BMO}(\mathbb{R})$ and $\|H\|_{{{BMO}(\mathbb{R})}}\leq 10\|h\|_{BMO[0,\frac{T}{2}]}$.

Proof.

For an arbitrary interval $I$, there are two possibilities:

• •

$|I|\leq\frac{T}{2}$.

In this case by a translation with an integer multiple of $T$ and using periodicity of $H$, we may assume either $I\subset[-\frac{T}{2},\frac{T}{2}]$ or $I\subset[0,T]$. Suppose $I\subset[-\frac{T}{2},\frac{T}{2}]$ and note that if $0\notin I$, we have either $I\subset[0,\frac{T}{2}]$ or $I\subset[-\frac{T}{2},0]$ and from the symmetry $O(H,I)\leq\|h\|_{BMO[0,\frac{T}{2}]}$. If $0\in I$, then take the interval $J$ centered at zero with the right half $J^{+}$, which contains $I$ and $|J|\leq 2|I|.$ Again from symmetry we get

[TABLE]

The same argument works for $I\subset[0,T]$. This time we use the symmetry of $H$ around $\frac{T}{2}$.

• •

$|I|\geq\frac{T}{2}$.

This time take $J=[nT,mT]$ with $n,m\in\mathbb{Z}$ which contains $I$ and $|J|\leq|I|+2T\leq 5|I|$. And again like the previous cases, from the symmetry and periodicity of $H$, we get

[TABLE]

The proof is now complete.

∎

Remark 2.3**.**

There are much more general ways to extend $BMO$ functions to the outside of domains, but for the purpose of our paper the above simple lemma is enough. See [7] for more on extensions.

Before we proceed further, let us fix some notations. By $A\lesssim B$, $A\gtrsim B$ and $A\approx B$, we mean $A\leq CB$, $A\geq CB$ and $C^{-1}B\leq A\leq CB$ respectively, where $C$ is a constant independent of the important parameters.

Lemma 2.4**.**

For $c<-1$, there exists a sequence of functions $g_{n}$ with the following properties:

1. (1)

$g_{n}\geq 0$** 2. (2)

$g_{n}=0$* on $[c,1]$* 3. (3)

$\|g_{n}\|_{{L}^{\infty}}\leq 1$** 4. (4)

$\lim\limits_{n\to\infty}\,\ThisStyle{\ensurestackMath{\stackinset{c}{.2\LMpt}{c}{.5\LMpt}{\SavedStyle-}{\SavedStyle\phantom{\int}}}\kern-16.56937pt}\int_{[0,n]}{g_{n}}=1$** 5. (5)

$\lim\limits_{n\to\infty}\|g_{n}\|_{{{BMO}(\mathbb{R})}}=0$.

Proof.

Let $\log^{+}|x|=\max\{0,\log|x|\}$ be the positive part of the logarithm, and consider the function $h_{n}(x)=\log^{+}x$ on the interval $[0,n]$, which belongs to $BMO[0,n]$ with $\|h_{n}\|_{BMO[0,n]}\leq\|\log^{+}|\cdot|\|_{{{BMO}(\mathbb{R})}}$. Then an application of lemma 2.2 with $T=2n$ gives us a sequence of nonnegative functions $H_{n}$ with $\|H_{n}\|_{{{BMO}(\mathbb{R})}}\lesssim 1$. Now let $g_{n}=\frac{1}{\log n}H_{n}(x)(1-\chi_{[c,0]}(x))$. The first three properties are immediate from the definition, the forth one follows from integration, and the last one from

[TABLE]

This finishes the proof. ∎

Now we turn to the proof of the above theorem.

Proof of Theorem 2.1.

Let $f$ be as in the theorem, $a=\int_{0}^{1}f$, $c<0$ a constant with large magnitude to be determined later, and let $g_{n}$ be the sequence constructed in lemma 2.4.

We will show that

[TABLE]

This, together with $\lim\limits_{n\to\infty}\|g_{n}\|_{{{BMO}(\mathbb{R})}}=0$, proves the theorem.

To begin with, we claim that $M(f+\frac{a}{1-c}g_{n})=Mf$ on $[c,0]$. To see this, note that from the positivity of $f$ and $g_{n}$, $M(f+\frac{a}{1-c}g_{n})(x)\geq Mf(x)$ for all values of $x$, and it remains to show that the reverse inequality holds also. For $x\in[c,0]$, $Mf(x)\geq\frac{\int_{c}^{1}f}{1-c}=\frac{a}{1-c}$, and for any interval $I$ which contains $x$, we have two possibilities:

• •

either $I\subset(-\infty,0)$, in which case from the third property of $g_{n}$ we have

[TABLE]

• •

or $I\cap[0,1]\neq\emptyset$, in which case the second and the third property of $g_{n}$ gives us

[TABLE]

This proves our claim.

Next we look at the mean oscillation of $M(f+\frac{a}{1-c}g_{n})-Mf$ on $[2c,0]$. Because this function vanishes on $[c,0]$, we have

[TABLE]

To bound the right hand side of the above inequality from below, we note that $f\leq\chi_{[0,1]}$ so $Mf(x)\leq M(\chi_{[0,1]})(x)=\frac{1}{1-x}$ for $x\leq 0$. Also for $x\leq 0$, we have

[TABLE]

So we get the following estimate for the right hand side in (3):

[TABLE]

Combining (3) and (4), gives us

[TABLE]

Now by taking the $\varliminf$ and using the forth property of $g_{n}$ we get

[TABLE]

This shows that if we have

[TABLE]

then (2) holds. Here we note that the function on the right hand side of (5), attains its minimum, which is $\log 2$, at infinity. So if we choose $|c|$ sufficiently large (5) holds, and this completes the proof. ∎

By lifting the above functions to higher dimensions with

[TABLE]

we obtain a counterexample for continuity of the $n$-dimensional Hardy-Littlewood maximal operator on ${BMO}(\mathbb{R}^{n})$, simply because the ${BMO}(\mathbb{R}^{n})$ norms and the maximal operator become one dimensional.

Corollary 2.5**.**

The Hardy-Littlewood maximal operator is discontinuous on ${BMO}(\mathbb{R}^{n})$ for $n\geq 1$.

3. Maximal function on ${VMO}(\mathbb{R}^{n})$

As it is mentioned before, ${VMO}(\mathbb{R}^{n})$ is the ${BMO}(\mathbb{R}^{n})$-closure of uniformly continuous functions which belong to ${BMO}(\mathbb{R}^{n})$. The operator $M$ reduces moduli of continuity, because it’s sublinear, so it preserves uniformly continuous functions. But from our previous result, one cannot deduce boundedness of $M$ on ${VMO}(\mathbb{R}^{n})$ by a limiting argument. Nevertheless, we have the following theorem:

Theorem 3.1**.**

The operator $M$ maps ${VMO}(\mathbb{R}^{n})$ to itself for $n\geq 1$.

Before we prove this, we bring the following lemma which is needed later.

Lemma 3.2**.**

Let $A$ be a measurable subset of a cube $Q$ and $f\in{BMO}(\mathbb{R}^{n})$ with $\|f\|_{{BMO}(\mathbb{R}^{n})}=1$; then we have

[TABLE]

Proof.

From the John-Nirenberg inequality, there’s a dimensional constant $c>0$ such that

[TABLE]

Now Jensen’s inequality gives us 7, as follows:

[TABLE]

∎

Remark 3.3**.**

In the above lemma, let $A$ be a rectangle and take $Q$ to be the smallest cube which contains it. Then

[TABLE]

where $e(A)$ is the eccentricity of $A$, or the ratio of the largest side to the smallest one.

We now turn to the proof of Theorem 3.1, which is a simpler variant of the local-nonlocal argument in [10].

Proof of Theorm 3.1.

We have to show that $\varlimsup\limits_{\delta\to 0}\omega(Mf,\delta)=0$ when $f\in{VMO}(\mathbb{R}^{n})$. Now for every cube Q, we have $O(|f|,Q)\leq 2O(f,Q)$, which means that $|f|\in{VMO}(\mathbb{R}^{n})$ too. From this together with $M(|f|)=M(f)$, it’s enough to prove the theorem for nonnegative functions. Also from the homogeneity of $M$ we may assume $\|f\|_{{BMO}(\mathbb{R}^{n})}=1$.

Let $Q_{0}$ be a cube and $c$ a constant with $c>e$. We decompose $M$ into the local part $M_{1}$, and the nonlocal part $M_{2}$ as follows:

[TABLE]

We have $Mf(x)=\max\{M_{1}f(x),M_{2}f(x)\}$ and so

[TABLE]

To estimate the first term in the right hand side of (8), let $Q_{0}^{*}$ be the concentric dilation of $Q_{0}$ with $l(Q_{0}^{*})=2cl(Q_{0})$. Then for the local part we have

[TABLE]

By using the boundedness of $M$ on $L^{2}(\mathbb{R}^{n})$ we get

[TABLE]

and an application of the John-Nirenberg inequality gives us

[TABLE]

To estimate the mean oscillation of the nonlocal part, suppose $x,y\in Q_{0}$, $M_{2}f(x)>M_{2}f(y)$ and let $Q$ be a cube with $l(Q)\geq cl(Q_{0})$, which contains $x$ and such that $M_{2}f(y)<f_{Q}$. Now let $Q^{\prime}$ be a cube such that $Q_{0}\cup Q\subset Q^{\prime}$, $l(Q^{\prime})=l(Q)+l(Q_{0})$, and let $A=Q^{\prime}\backslash Q$. Then $M_{2}f(y)\geq f_{Q^{\prime}}$ and we have

[TABLE]

Here we note that $|A|=|Q^{\prime}|-|Q|\approx l(Q_{0})l(Q)^{n-1}$, and $l(Q^{\prime})\approx l(Q)$. So from the above inequality and lemma 3.2 we get

[TABLE]

The reason for the last inequality is that $\frac{l(Q_{0})}{l(Q)}\leq c^{-1}$ and the function $-t\log t$ is increasing when $t<e^{-1}$. Finally, by taking the supremum over all such cubes $Q$, we obtain

[TABLE]

So for the nonlocal part we have

[TABLE]

By putting (8), (9) and (10) together we get

[TABLE]

and taking the supremum over all cubes $Q_{0}$ with $l(Q_{0})=\delta$ gives us

[TABLE]

To finish the proof, it’s enough to take the $\varlimsup\limits_{\delta\to 0}$ first and then let $c\to\infty$. ∎

Remark 3.4**.**

The above argument shows that for all functions in ${BMO}(\mathbb{R}^{n})$, if one chooses a sufficiently large localization of $M$, (10) holds, meaning that the mean oscillation of the nonlocal part is small. This also shows itself in the dyadic setting: if one considers the dyadic maximal operator $M^{d}$ and dyadic $BMO$, denoted by ${BMO_{d}}(\mathbb{R}^{n})$, then for a dyadic cube $Q_{0}$

[TABLE]

Hence $O(M_{2}f,Q_{0})=0$ and therefore no dilation is needed $(c=1)$.

4. Slices of BMO functions and unboundedness of directional and strong maximal operators

In this final section we discuss properties of slices of functions in ${BMO}(\mathbb{R}^{n})$, and for simplicity we restrict ourselves to ${BMO}(\mathbb{R}^{2})$. We begin by asking:

Q4. Suppose $\varphi,\psi$ are two functions of one variable, when does $f(x,y)=\varphi(x)\psi(y)$ belong to ${BMO}(\mathbb{R}^{2})$?

To answer this, we need the following lemma which is an application of Fubini’s theorem and its proof is found in [5].

Lemma 4.1**.**

Let $A,B\subset\mathbb{R}$ be two measurable sets with finite positive measure, and $f$ be a measurable function on $\mathbb{R}^{2}$. Then

[TABLE]

Now take two intervals $I,J$ with $l(I)=l(J)$. Then an application of the above lemma to $f(x,y)=\varphi(x)\psi(y)$ gives us

[TABLE]

Taking the supremum over all such $I,J$ we obtain

[TABLE]

When $f\in{BMO}(\mathbb{R}^{2})$ is not identically zero, the above condition implies that $\varphi,\psi\in{BMO}(\mathbb{R})$. To see this, note that if $\varphi$ is not identically zero, for some Lebesgue point of $\varphi$ like $x$, $\varphi(x)\neq 0$. Then from the Lebesgue differentiation theorem, for sufficiently small $\delta$ we must have $\sup_{l(I)=\delta}\,\ThisStyle{\ensurestackMath{\stackinset{c}{.2\LMpt}{c}{.5\LMpt}{\SavedStyle-}{\SavedStyle\phantom{\int}}}\kern-16.56937pt}\int_{I}{\varphi}\gtrsim|\varphi(x)|>0$. So $\psi$ has bounded mean oscillation on intervals with length less than $\delta$. For intervals $J$ with $l(J)\geq\delta$, $|\psi|$ has bounded averages because otherwise there is a sequence of intervals $J_{n}$ with $l(J_{n})\geq\delta$ and $\lim\,\ThisStyle{\ensurestackMath{\stackinset{c}{.2\LMpt}{c}{.5\LMpt}{\SavedStyle-}{\SavedStyle\phantom{\int}}}\kern-16.56937pt}\int_{J_{n}}{|\psi|}=\infty$. Then by dividing each of these intervals into sufficiently small pieces of length between $\frac{\delta}{2}$ and $\delta$, we conclude that $|\psi|$ has large averages over such intervals so $\sup_{l(J)=\delta}\,\ThisStyle{\ensurestackMath{\stackinset{c}{.2\LMpt}{c}{.5\LMpt}{\SavedStyle-}{\SavedStyle\phantom{\int}}}\kern-16.56937pt}\int_{J}{|\psi|}=\infty$. But then $\sup_{l(I)=\delta}O(\varphi,I)=0$, which means that $\varphi$ is constant. We summarize the above discussion in the following proposition.

Proposition 4.2**.**

Let $f(x,y)=\varphi(x)\psi(y)$, $f\in{BMO}(\mathbb{R}^{2})$ if and only if (11) holds and if $f\neq 0$, then $\varphi,\psi\in{BMO}(\mathbb{R})$.

Remark 4.3**.**

When $\varphi$ and $\psi$ are not constants, the above argument shows that they belong to $bmo(\mathbb{R})$, the nonhomogeneous $BMO$, which is a proper subspace of ${BMO}(\mathbb{R})$. See [3] for more on $bmo(\mathbb{R})$.

Corollary 4.4**.**

Let $\log^{-}|x|=\max\{0,-\log|x|\}$ be the negative part of the logarithm and $p,q>0$ with $p+q\leq 1$. Then the function $f(x,y)=(\log^{-}|x|)^{p}(\log^{-}|y|)^{q}$ is in ${BMO}(\mathbb{R}^{2})$.

Proof.

A direct calculation shows that

[TABLE]

and the claim follows from proposition 4.2. ∎

Remark 4.5**.**

The above function $f$ doesn’t have bounded mean oscillation on rectangles, simply because the $BMO$-norm of the slices becomes larger and larger as we get close to the origin. See [8] (Example 2.32) for another example.

Now we answer the third question of this paper.

Theorem 4.6**.**

The operators $M_{e_{1}}$ and $M_{s}$ are unbounded on ${BMO}(\mathbb{R}^{n})$.

To prove this we need the following simple lemma.

Lemma 4.7**.**

Let $Q_{0}=[-1,1]^{n}$, $f\in{BMO}(\mathbb{R}^{n})$ with support in $Q_{0}$, and $x_{k}$ be a sequence in $\mathbb{R}^{n}$ with $|x_{k}-x_{m}|\geq 3\sqrt{n}$ for $k\neq m$. Then $g(x)=\sum f(x-x_{k})$ is in ${BMO}(\mathbb{R}^{n})$ and $\|g\|_{{BMO}(\mathbb{R}^{n})}\lesssim\|f\|_{{BMO}(\mathbb{R}^{n})}$.

Proof.

First, by comparing the average of $|f|$ on $Q_{0}$ with $Q_{0}+2e_{1}$ we have $\int_{Q_{0}}|f|\lesssim\|f\|_{{BMO}(\mathbb{R}^{n})}$. Next, take a cube $Q$ and suppose for some $k$, $Q\cap x_{k}+Q_{0}\neq\emptyset$. We note that the distance of the support of functions $f(\cdot-x_{k})$ from each other is at least $\sqrt{n}$ so if $l(Q)\leq 1$ then $O(g,Q)=O(f(\cdot-x_{k}),Q)\leq\|f\|_{{BMO}(\mathbb{R}^{n})}$. Otherwise we have

[TABLE]

Now to finish the proof, note that $\#\{k|Q\cap x_{k}+Q_{0}\neq\emptyset\}\lesssim|Q|$ which implies $O(g,Q)\lesssim\|f\|_{{BMO}(\mathbb{R}^{n})}$. ∎

Proof of Theorem 4.6.

We may assume $n=2$, since by a lifting argument similar to (6), we can conclude the theorem for higher dimensions. Let $f$ be as in Corollary 4.4, $N$ a positive integer, and consider the following function:

[TABLE]

$g_{N}$ has the following properties:

1. (i)

$\|g_{N}\|_{{BMO}(\mathbb{R}^{n})}\lesssim 1$. (Here our bounds only depend on $p,q$ but not $N$.)

This follows from Corollary 4.4 and Lemma 4.7 applied to $f$ with $x_{m,k}=(3\sqrt{2}m,\frac{k}{N})$. 2. (ii)

$M_{s}g_{N}\geq M_{e_{1}}g_{N}(x,y)\gtrsim(\log N)^{q}$ for $0\leq x,y\leq 1$.

To see this, let $0\leq x\leq 1$ and $\frac{l}{N}\leq y<\frac{l+1}{N}$ for some $l<N$. Then consider the average of $(g_{N})_{y}$ on $I=[0,3\cdot 2^{l+1}\sqrt{2}]$, which is bounded from below by

[TABLE]

Now note that for $2^{l}\leq m\leq 2^{l+1}-1$, $I$ contains the support of $f(\cdot-3\sqrt{2}m,y-\frac{l}{N})$ and since $0\leq y-\frac{l}{N}\leq\frac{1}{N}$ we have

[TABLE]

From this we get

[TABLE] 3. (iii)

$M_{e_{1}}g_{N}(x,y)=0$ for $y<-1$.

This holds simply because $g_{N}$ is supported on $[3\sqrt{2}-1,\infty)\times[-1,2]$. 4. (iv)

$M_{s}g_{N}(x,y)\lesssim 1$ for $0\leq x\leq 1,y\leq-2$.

To prove this final property of $g_{N}$, suppose $R=I\times J$ is a rectangle with $(x,y)\in R$. Then if $R\cap supp(g_{N})\neq\emptyset$ we have $l(I),l(J)\geq 1$, and we note that $\#\{(m,k)|R\cap supp(f(\cdot-3\sqrt{2}m,\cdot-\frac{k}{N}))\neq\emptyset\}\lesssim l(I)$ which implies

[TABLE]

Taking the supremum over all rectangles $R$ proves the last property of $g_{N}$.

To complete the proof, let us measure the mean oscillation of $M_{e_{1}}g_{N}$ on the square $[-3,3]^{2}$ by

[TABLE]

Then from the second and the third property of $g_{N}$ we obtain

[TABLE]

and the same is true for $M_{s}$. So by putting the first property of $g_{N}$ with (12) together, we conclude that neither $M_{s}$ nor $M_{e_{1}}$ is bounded on ${BMO}(\mathbb{R}^{2})$. ∎

By modifying the above function, one can construct a function in ${BMO}(\mathbb{R}^{2})$ such that none of its horizontal slices are in ${BMO}(\mathbb{R})$. To do this, let $r_{n}$ be an enumeration of rational numbers and consider the following function:

[TABLE]

Then we have

[TABLE]

So for all values of $y$, we get $\sup_{l(I)=2}O(h_{y},I)=\infty$, even though $h\in{BMO}(\mathbb{R}^{2})$.

This example shows that one can not control the maximum mean oscillation of the slices, when we look at intervals with a fixed length. However, in the following theorem, we show that there is a loose control when the length of intervals increases.

Theorem 4.8**.**

Let $f\in{BMO}(\mathbb{R}^{2})$ with $\|f\|_{{{BMO}(\mathbb{R}^{2})}}=1$. Then there exist constants $\lambda,c>0$, independent of $f$, such that for any sequence of intervals $I_{k}$ $(k\geq 1$) with $l(I_{k})=2^{k}$, and any interval $J$ with $l(J)=1$, we have

[TABLE]

Proof.

Let $E_{t}=\{y\in J|\sup\limits_{k\geq 1}\frac{O(f_{y},I_{k})}{k}>t\}$; then

[TABLE]

Now by taking the average over $E_{t,k}$ and applying Lemma 4.1 we get

[TABLE]

Next, let $J_{k}$ be the interval with the same center as J and with $l(J_{k})=2^{k}$, and note that $E_{t,k}\subset J\subset J_{k}$, so $I_{k}\times E_{t,k}\subset I_{k}\times J_{k}$. Then an application of Lemma 3.2 shows that

[TABLE]

So for an appropriate constants $a>0$, which is independent of $f$, we have $|E_{t,k}|\lesssim e^{-atk}$ for $t>0$. From this and (13) we get the estimate

[TABLE]

Now an application of Cavalieri formula gives us

[TABLE]

Hence (4.8) holds with $\lambda=\frac{a}{2}$, and this finishes the proof. ∎

Acknowledgements

I would like to thank G. Dafni and R. Gibara for suggesting these problems and valuable discussions and comments.