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BV Continuity for the uncentered
Hardy–Littlewood maximal operator

Cristian González-Riquelme and Dariusz Kosz IMPA - Instituto de Matemática Pura e Aplicada
Rio de Janeiro - RJ, Brazil, 22460-320.
[email protected] Faculty of Pure and Applied Mathematics, Wrocław University of Science and Technology, Wyb. Wyspiańskiego 27, 50-370 Wrocław, Poland. [email protected]
Abstract.

We prove the continuity of the map fM~ff\mapsto\widetilde{M}f from BV()BV(\mathbb{R}) to itself, where M~\widetilde{M} is the uncentered Hardy–Littlewood maximal operator. This answers a question of Carneiro, Madrid and Pierce.

Key words and phrases:
Maximal operators; continuity; bounded variation;
2010 Mathematics Subject Classification:
26A45, 42B25, 39A12, 46E35, 46E39, 05C12.

1. Introduction

1.1. A brief historical perspective and background

The study of regularity properties for maximal operators has been an important topic of research over the last years. The most classical object in this context is the centered Hardy–Littlewood maximal operator, that for every fLloc1(d)f\in L^{1}_{\text{loc}}(\mathbb{R}^{d}) and xdx\in\mathbb{R}^{d} is defined as

Mf(x):=supr>0 B(x,r)|f|,Mf(x):=\sup_{r>0}\ \mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptB(x,r)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{B(x,r)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{B(x,r)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{B(x,r)}}|f|,

where  Bg:=Bgm(B)\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptB}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{B}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{B}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{B}}g:=\frac{\int_{B}g}{m(B)} and mm is the dd–dimensional Lebesgue measure. The uncentered Hardy–Littlewood maximal operator, denoted by M~\widetilde{M}, is defined analogously by taking the supremum over balls that contain the point xx but that are not necessarily centered at xx.

The program of studying the regularity of maximal functions was initiated by Kinnunen [15], who proved that the map

fMf\displaystyle f\mapsto Mf (1.1)

is bounded on W1,p(d)W^{1,p}(\mathbb{R}^{d}) if p>1.p>1. Since then, several authors have made many contributions in this direction, including connections with potential theory and partial differential equations (see, e.g., [4, 5, 7, 13, 16, 22, 23]).

An important direction has been the study of the continuity of the map (1.1) in several settings. Since it is not necessarily a sublinear map at the derivative level, its continuity does not follow from the boundedness. The first result answering a question of this kind was due to Luiro [18], who proved that the map (1.1) is in fact continuous from W1,p(d)W^{1,p}(\mathbb{R}^{d}) to itself for each p>1.p>1.

1.2. The endpoint case p=1p=1

This case is much more subtle. In fact, Kinnunen’s result certainly does not hold because for every non trivial fL1(d)f\in L^{1}(\mathbb{R}^{d}) we have M~fL1(d).\widetilde{M}f\notin L^{1}(\mathbb{R}^{d}). Nevertheless, since we are interested at the derivative level of the maximal functions, the natural version of the problem at p=1p=1 is to study the map

fMf,\displaystyle f\mapsto\nabla Mf, (1.2)

for fW1,1(d).f\in W^{1,1}(\mathbb{R}^{d}). In fact, the boundedness of the map (1.2) from W1,1(d)W^{1,1}(\mathbb{R}^{d}) to L1(d)L^{1}(\mathbb{R}^{d}) is an important open problem for d2.d\geq 2. In dimension d=1d=1 the answer is positive, as has been established for the uncentered version in [24] and for the centered version in [17]. In [19], Luiro proved that the map (1.2) is bounded when restricted to radial data in the uncentered setting. Other boundedness results in the radial setting have been achieved in [2, 3, 8, 14, 20].

An important work for our purposes is [1], where the boundedness

Var(M~f)Var(f),{\rm Var}\big{(}\widetilde{M}f\big{)}\leq{\rm Var}(f),

for every fBV(),f\in BV(\mathbb{R}), was proved. Moreover, the authors proved that M~f\widetilde{M}f is absolutely continuous for fBV(),f\in BV(\mathbb{R}), providing then an improvement over the original function.

1.3. Continuity at the endpoint case

In this article we are particularly interested in the continuity of the map (1.2) at the endpoint p=1.p=1. We notice that the methods outlined by Luiro [18] cannot be adapted to our case, since they depend on the fact that the map

fMff\mapsto Mf

is bounded on Lp(d)L^{p}(\mathbb{R}^{d}) for p>1p>1. In [12] the first continuity results at the endpoint were obtained. The authors proved, among other results, that the map

f(M~f)\displaystyle f\mapsto\big{(}\widetilde{M}f\big{)}^{\prime}

is continuous from W1,1()W^{1,1}(\mathbb{R}) to L1()L^{1}(\mathbb{R}) (cf. [12, Theorem 1]). This result was recently extended to the Wrad1,1(d)W^{1,1}_{\rm rad}(\mathbb{R}^{d}) setting (the subspace of W1,1(d)W^{1,1}(\mathbb{R}^{d}) consisting of radial functions) in [9], where a more general approach, that can be applied in the context of other maximal functions, is proposed.

In [12] the authors also consider the space BV()BV(\mathbb{R}) endowed with the norm fBV:=|f()|+Var(f).\|f\|_{BV}:=|f(-\infty)|+{\rm Var}(f). About this, they asked the following question:

Question 1.

(Question B in [12]) Is the map M~:BV()BV()\widetilde{M}:BV(\mathbb{R})\to BV(\mathbb{R}) continuous?

This question, in case of being answered affirmatively, would provide a generalization of [12, Theorem 1] (since W1,1()W^{1,1}(\mathbb{R}) embeds isometrically in BV()BV(\mathbb{R})). It is important to notice that, in general, the continuity in the BV()BV(\mathbb{R}) setting is more delicate that in the W1,1()W^{1,1}(\mathbb{R}) setting. An example of this is that in the fractional setting the analogue of [12, Theorem 1] holds (see [21]) but the answer to the analogue of the previous question is negative (see [12, Theorem 3]).

The main goal of the present manuscript is to answer this question. We prove the following.

Theorem 1.

The map M~:BV()BV()\widetilde{M}:BV(\mathbb{R})\to BV(\mathbb{R}) is continuous.

Several of the arguments in [12] (also the arguments in [9]) rely on the regularity of the original function, therefore they are not enough to conclude Theorem 1. In fact, the authors (also the authors of [9]) used in their work a reduction of the problem to the analogous question stated for “lateral” maximal operators, but in our case that reduction causes several problems. For instance, the one-sided maximal function of a function in BV()BV(\mathbb{R}) is not necessarily continuous, which brings additional difficulties. Therefore, a different idea is required in order to achieve the result. Our methods, although inspired by [12], provide a new approach to this kind of problems.

Having taken into account the main result obtained here, we summarize the situation of the endpoint continuity program (originally proposed in [12, Table 1]) in the table below. The word YES in a box means that the continuity of the corresponding map has been proved, whereas the word NO means that it has been shown that it fails. The remaining boxes are marked as OPEN problems. We notice that after this work the only open problems in this program are the ones related to the centered Hardy–Littlewood maximal operator in a continuous setting.

Table 1. Endpoint continuity program
———— W1,1W^{1,1}-continuity; continuous setting BVBV-continuity; continuous setting W1,1W^{1,1}-continuity; discrete setting BVBV-continuity; discrete setting
Centered classical maximal operator OPEN OPEN YES2 YES4
Uncentered classical maximal operator YES1 YES: Theorem 1 YES2 YES1
Centered fractional maximal operator YES5 NO1 YES3 NO1
Uncentered fractional maximal operator YES4 NO1 YES3 NO1

2. Proof of Theorem 1

2.1. Preliminaries

In this subsection we develop the main tools required in our work. We start by stating the following result which describes the behavior of the maximal function at infinity.

Lemma 2.

Given fBV()f\in BV(\mathbb{R}) let |f|():=limx|f|(x)|f|(\infty):=\underset{x\to\infty}{\lim}|f|(x) and |f|():=limx|f|(x).|f|(-\infty):=\underset{x\to-\infty}{\lim}|f|(x). Then

limxM~f(x)=limxM~f(x)=c,\lim_{x\to\infty}\widetilde{M}f(x)=\lim_{x\to-\infty}\widetilde{M}f(x)=c,

where c=max{|f|(),|f|()}c=\max\{|f|(\infty),|f|(-\infty)\}.

Proof.

Without loss of generality we assume that f0f\geq 0 and c=f()c=f(\infty). Observe that

M~f(x)limr (x1,x+r)f=c\widetilde{M}f(x)\geq\lim_{r\rightarrow\infty}\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt(x-1,x+r)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x-1,x+r)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x-1,x+r)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x-1,x+r)}}f=c

holds for any xx\in\mathbb{R}. Fix ε>0\varepsilon>0 and let N0>0N_{0}>0 be such that f(x)c+ε2f(x)\leq c+\frac{\varepsilon}{2} for |x|>N0|x|>N_{0}. We choose N1>N0N_{1}>N_{0} satisfying

2N0fN1N0ε2.\frac{2N_{0}\|f\|_{\infty}}{N_{1}-N_{0}}\leq\frac{\varepsilon}{2}.

Consider x0x_{0} satisfying |x0|>N1|x_{0}|>N_{1} and any interval Ix0I\ni x_{0}. If |I|<N1N0|I|<N_{1}-N_{0}, then clearly

 Ifc+ε2.\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}f\leq c+\frac{\varepsilon}{2}.

On the other hand, if |I|N1N0|I|\geq N_{1}-N_{0}, then

 If1|I|I[N0,N0]cf(x)𝑑x+1N1N0[N0,N0]f(x)𝑑xc+ε.\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}f\leq\frac{1}{|I|}\int_{I\cap[-N_{0},N_{0}]^{\rm c}}f(x)\,dx+\frac{1}{N_{1}-N_{0}}\int_{[-N_{0},N_{0}]}f(x)\,dx\leq c+\varepsilon.

Since ε>0\varepsilon>0 is arbitrary, the claim follows. ∎

The next goal is to use the BV()BV(\mathbb{R}) norm to control the difference between two BV()BV(\mathbb{R}) functions or between their maximal functions at a given point xx. The following estimates, although very basic, will be extremely useful later on.

Lemma 3.

Let f,gBV()f,g\in BV(\mathbb{R}). Then

|f(x)g(x)|2fgBVand|M~f(x)M~g(x)|2fgBV|f(x)-g(x)|\leq 2\|f-g\|_{BV}\quad{\rm and}\quad\big{|}\widetilde{M}f(x)-\widetilde{M}g(x)\big{|}\leq 2\|f-g\|_{BV}

hold for any xx\in\mathbb{R}.

Proof.

The first inequality follows since

|f(x)g(x)||(f(x)g(x))(f()g())|+|f()g()|.|f(x)-g(x)|\leq|(f(x)-g(x))-(f(-\infty)-g(-\infty))|+|f(-\infty)-g(-\infty)|.

Now, assume M~f(x)M~g(x)\widetilde{M}f(x)\geq\widetilde{M}g(x). By the first part of the lemma for any IxI\ni x we have

 I|g| I|f| I|gf| I|f|2fgBV.\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|g|\geq\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f|-\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|g-f|\geq\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f|-2\|f-g\|_{BV}.

Thus, M~g(x)M~f(x)2fgBV\widetilde{M}g(x)\geq\widetilde{M}f(x)-2\|f-g\|_{BV} and the second part follows as well. ∎

Contrasting with the W1,1()W^{1,1}(\mathbb{R}) setting (see [12, Lemma 14]), in our context to make the reduction to the case f0f\geq 0 is much more problematic. In order to deal with this issue we require several results describing the relations between ff and |f||f|.

In the following, for given gBV()g\in BV(\mathbb{R}) we define limyxg(y)=:g(x)\underset{y\uparrow x}{\lim}\ g(y)=:g(x^{-}) and limyxg(y)=:g(x+)\underset{y\downarrow x}{\lim}\ g(y)=:g(x^{+}). Also, for each a<b-\infty\leq a<b\leq\infty we introduce the quantity

Var(a,b)(g):=sup{i=1K|g(ai)g(ai1)|},{\rm Var}_{(a,b)}(g):=\sup\Big{\{}\sum_{i=1}^{K}|g(a_{i})-g(a_{i-1})|\Big{\}},

where the supremum is taken over all KK\in\mathbb{N} and all sequences a<a0<<aK<ba<a_{0}<\dots<a_{K}<b (notice that if gg is continuous at aa and bb, then the sequences satisfying a=a0<<aK=ba=a_{0}<\dots<a_{K}=b can be considered instead and the supremum will not change). For a given partition 𝒫={a0<a1<<aK}\mathcal{P}=\{a_{0}<a_{1}<\dots<a_{K}\} we denote Var(g,𝒫):=i=1K|g(ai)g(ai1)|{\rm Var}(g,\mathcal{P}):=\displaystyle\sum_{i=1}^{K}|g(a_{i})-g(a_{i-1})|. Finally, we write Dl(g):={x;g(x)g(x)}D_{l}(g):=\{x\in\mathbb{R};g(x)\neq g(x^{-})\} and Dr(g):={x;g(x)g(x+)}.D_{r}(g):=\{x\in\mathbb{R};g(x)\neq g(x^{+})\}.

Lemma 4.

Fix fBV().f\in BV(\mathbb{R}). Then for any a<b-\infty\leq a<b\leq\infty we have

Var(a,b)(f)Var(a,b)(|f|)\displaystyle{\rm Var}_{(a,b)}(f)-{\rm Var}_{(a,b)}(|f|) =xDl(f)(a,b)|f(x)f(x)|||f|(x)|f|(x)|\displaystyle=\sum_{x\in D_{l}(f)\cap(a,b)}|f(x)-f(x^{-})|-\big{|}|f|(x)-|f|(x^{-})\big{|}
+xDr(f)(a,b)|f(x)f(x+)|||f|(x)|f|(x+)|.\displaystyle\quad+\sum_{x\in D_{r}(f)\cap(a,b)}|f(x)-f(x^{+})|-\big{|}|f|(x)-|f|(x^{+})\big{|}.
Proof.

Fix a<b-\infty\leq a<b\leq\infty. We write Dl(f)(a,b)=:{xl,n;n}D_{l}(f)\cap(a,b)=:\{x_{l,n};n\in\mathbb{N}\} and Dr(f)(a,b)=:{xr,n;n}D_{r}(f)\cap(a,b)=:\{x_{r,n};n\in\mathbb{N}\}, assuming that both of these sets are infinite (the other cases can be treated very similarly). Given ε>0\varepsilon>0 we choose a partition 𝒫(a,b)\mathcal{P}\subset(a,b) such that

Var(f,𝒫)>Var(a,b)(f)ε{\rm Var}(f,\mathcal{P})>{\rm Var}_{(a,b)}(f)-\varepsilon

and

Var(|f|,𝒫)>Var(a,b)(|f|)ε.{\rm Var}(|f|,\mathcal{P})>{\rm Var}_{(a,b)}(|f|)-\varepsilon.

Then for fixed NN\in\mathbb{N} we construct 𝒫~=𝒫~(N)(a,b)\widetilde{\mathcal{P}}=\widetilde{\mathcal{P}}(N)\subset(a,b) by adding to 𝒫\mathcal{P} (if needed) some extra points. The procedure consists of the following three steps.

  1. (i)

    We set 𝒫1=𝒫{xl,n,xr,n;nN}.\mathcal{P}_{1}=\mathcal{P}\cup\{x_{l,n},x_{r,n};n\leq N\}.

  2. (ii)

    For each nNn\leq N we choose x~l,n<xl,n\widetilde{x}_{l,n}<x_{l,n} such that

    𝒫1(x~l,n,xl,n)=\mathcal{P}_{1}\cap(\widetilde{x}_{l,n},x_{l,n})=\emptyset

    and |f(xl,n)f(x~l,n)|<2nε.|f(x_{l,n}^{-})-f(\widetilde{x}_{l,n})|<2^{-n}\varepsilon. Similarly, we choose x~r,n>xr,n\widetilde{x}_{r,n}>x_{r,n} such that

    𝒫1(xr,n,x~r,n)=\mathcal{P}_{1}\cap(x_{r,n},\widetilde{x}_{r,n})=\emptyset

    and |f(xr,n+)f(x~r,n)|<2nε.|f(x_{r,n}^{+})-f(\widetilde{x}_{r,n})|<2^{-n}\varepsilon. Then we set 𝒫2=𝒫1{x~l,n,x~r,n;nN}.\mathcal{P}_{2}=\mathcal{P}_{1}\cup\{\widetilde{x}_{l,n},\widetilde{x}_{r,n};n\leq N\}.

  3. (iii)

    For K=K(𝒫2)K=K(\mathcal{P}_{2}), we let {{xk,yk};kK}\{\{x_{k},y_{k}\};k\leq K\} be the set of all pairs {x,y}𝒫2\{x,y\}\subset\mathcal{P}_{2} satisfying x<yx<y with (x,y)𝒫2=(x,y)\cap\mathcal{P}_{2}=\emptyset and f(x)f(y)<0,f(x)f(y)<0, which are not of the form {x~l,n,xl,n}\{\widetilde{x}_{l,n},x_{l,n}\} or {xr,n,x~r,n}\{x_{r,n},\widetilde{x}_{r,n}\}. Let kKk\leq K. If there exists zk(xk,yk)z_{k}^{\circ}\in(x_{k},y_{k}) such that |f(zk)|<2kε|f(z_{k}^{\circ})|<2^{-k}\varepsilon, then we just add zkz_{k}^{\circ} to 𝒫2.\mathcal{P}_{2}. If not, then at least one of the sets

    Ik,l:=(xk,yk]{z;sgn(f(z)f(yk))=sgn(f(z)f(xk))=1}I_{k,l}:=(x_{k},y_{k}]\cap\{z;\operatorname{\mathrm{sgn}}(f(z)f(y_{k}))=\operatorname{\mathrm{sgn}}(f(z^{-})f(x_{k}))=1\}

    and

    Ik,r:=[xk,yk){z;sgn(f(z+)f(yk))=sgn(f(z)f(xk))=1}I_{k,r}:=[x_{k},y_{k})\cap\{z;\operatorname{\mathrm{sgn}}(f(z^{+})f(y_{k}))=\operatorname{\mathrm{sgn}}(f(z)f(x_{k}))=1\}

    must be non-empty (here sgn(x)\operatorname{\mathrm{sgn}}(x) is the usual sign function taking the value of 1-1, 0, or 11, if x<0x<0, x=0x=0, or x>0x>0, respectively). Assume Ik,lI_{k,l}\neq\emptyset (the other case is similar) and choose zkIk,lz_{k}\in I_{k,l}. Then zk=xl,nz_{k}=x_{l,n} for some n>N.n>N. We find z~k(xk,zk)\widetilde{z}_{k}\in(x_{k},z_{k}) such that |f(z~k)f(zk)|<2kε|f(\widetilde{z}_{k})-f(z_{k}^{-})|<2^{-k}\varepsilon (in particular, we have sgn(f(xk))=sgn(f(z~k))\operatorname{\mathrm{sgn}}(f(x_{k}))=\operatorname{\mathrm{sgn}}(f(\widetilde{z}_{k}))), and add both zkz_{k} and z~k\widetilde{z}_{k} to 𝒫2.\mathcal{P}_{2}. The above process terminates after KK steps and we denote the final collection of points by 𝒫~.\widetilde{\mathcal{P}}.

Having constructed 𝒫~\widetilde{\mathcal{P}} we see that

Var(a,b)(f)Var(a,b)(|f|)\displaystyle{\rm Var}_{(a,b)}(f)-{\rm Var}_{(a,b)}(|f|) Var(f,𝒫~)Var(|f|,𝒫~)ε\displaystyle\geq{\rm Var}(f,\widetilde{\mathcal{P}})-{\rm Var}(|f|,\widetilde{\mathcal{P}})-\varepsilon
n=1N|f(xl,n)f(x~l,n)|||f|(xl,n)|f|(x~l,n)|\displaystyle\geq\sum_{n=1}^{N}|f(x_{l,n})-f(\widetilde{x}_{l,n})|-\big{|}|f|(x_{l,n})-|f|(\widetilde{x}_{l,n})\big{|}
+n=1N|f(xr,n)f(x~r,n)|||f|(xr,n)|f|(x~r,n)|ε\displaystyle\quad+\sum_{n=1}^{N}|f(x_{r,n})-f(\widetilde{x}_{r,n})|-\big{|}|f|(x_{r,n})-|f|(\widetilde{x}_{r,n})\big{|}-\varepsilon
n=1N|f(xl,n)f(xl,n)|||f|(xl,n)|f|(xl,n)|\displaystyle\geq\sum_{n=1}^{N}|f(x_{l,n})-f(x_{l,n}^{-})|-\big{|}|f|(x_{l,n})-|f|({x}_{l,n}^{-})\big{|}
+n=1N|f(xr,n)f(xr,n+)|||f|(xr,n)|f|(xr,n+)|5ε.\displaystyle\quad+\sum_{n=1}^{N}|f(x_{r,n})-f(x_{r,n}^{+})|-\big{|}|f|(x_{r,n})-|f|(x_{r,n}^{+})\big{|}-5\varepsilon.

Also, we obtain

Var(a,b)(f)Var(a,b)(|f|)\displaystyle{\rm Var}_{(a,b)}(f)-{\rm Var}_{(a,b)}(|f|) Var(f,𝒫~)Var(|f|,𝒫~)+ε\displaystyle\leq{\rm Var}(f,\widetilde{\mathcal{P}})-{\rm Var}(|f|,\widetilde{\mathcal{P}})+\varepsilon
n=1|f(xl,n)f(xl,n)|||f|(xl,n)|f|(xl,n)|\displaystyle\leq\sum_{n=1}^{\infty}|f(x_{l,n})-f(x_{l,n}^{-})|-\big{|}|f|(x_{l,n})-|f|({x}_{l,n}^{-})\big{|}
+n=1|f(xr,n)f(xr,n+)|||f|(xr,n)|f|(xr,n+)|+6ε,\displaystyle\quad+\sum_{n=1}^{\infty}|f(x_{r,n})-f(x_{r,n}^{+})|-\big{|}|f|(x_{r,n})-|f|(x_{r,n}^{+})\big{|}+6\varepsilon,

since the only terms that contributes to Var(f,𝒫~)Var(|f|,𝒫~){\rm Var}(f,\widetilde{\mathcal{P}})-{\rm Var}(|f|,\widetilde{\mathcal{P}}) are those corresponding to the pairs {x~l,n,xl,n},\{\widetilde{x}_{l,n},x_{l,n}\}, {xr,n,x~r,n},\{x_{r,n},\widetilde{x}_{r,n}\}, {xk,zk},\{x_{k},z_{k}^{\circ}\}, {zk,yk}\{z_{k}^{\circ},y_{k}\} and {zk,zk~}.\{z_{k},\widetilde{z_{k}}\}. Letting NN\to\infty and ε0,\varepsilon\to 0, we obtain the claim. ∎

Now, we use Lemma 4 to show that the map fVar(a,b)(|f|)f\mapsto{\rm Var}_{(a,b)}(|f|) is continuous from BV()BV(\mathbb{R}) to [0,)[0,\infty).

Lemma 5.

Fix fBV()f\in BV(\mathbb{R}) and let {fj;j}BV()\{f_{j};j\in\mathbb{N}\}\subset BV(\mathbb{R}) be such that limjfjfBV=0.\underset{j\to\infty}{\lim}\|f_{j}-f\|_{BV}=0. Then for any a<b-\infty\leq a<b\leq\infty we have

limjVar(a,b)(|fj|)=Var(a,b)(|f|).\lim_{j\to\infty}{\rm Var}_{(a,b)}(|f_{j}|)={\rm Var}_{(a,b)}(|f|).
Proof.

It is possible to verify that fjff_{j}\to f implies Var(a,b)(fj)Var(a,b)(f).{\rm Var}_{(a,b)}(f_{j})\to{\rm Var}_{(a,b)}(f). Thus, it remains to show

limjVar(a,b)(fj)Var(a,b)(|fj|)=Var(a,b)(f)Var(a,b)(|f|).\lim_{j\to\infty}{\rm Var}_{(a,b)}(f_{j})-{\rm Var}_{(a,b)}(|f_{j}|)={\rm Var}_{(a,b)}(f)-{\rm Var}_{(a,b)}(|f|).

We define {xl,n;n}\{x_{l,n};n\in\mathbb{N}\} and {xr,n;n}\{x_{r,n};n\in\mathbb{N}\} as in the previous lemma. Given ε>0\varepsilon>0 we choose NN\in\mathbb{N} such that

n=N+1|f(xl,n)f(xl,n)|+|f(xr,n)f(xr,n+)|<ε.\sum_{n=N+1}^{\infty}|f(x_{l,n})-f(x_{l,n}^{-})|+|f(x_{r,n})-f(x_{r,n}^{+})|<\varepsilon.

We also denote Dlj,N:=Dl(fj){xl,1,,xl,N}D_{l}^{j,N}:=D_{l}(f_{j})\cap\{x_{l,1},\dots,x_{l,N}\} and Drj,N:=Dr(fj){xr,1,,xr,N}.D_{r}^{j,N}:=D_{r}(f_{j})\cap\{x_{r,1},\dots,x_{r,N}\}. By Lemma 3 we have that for jj big enough

|(Dlj,N|fj(x)fj(x)|fj|(x)|fj|(x)||f(x)f(x)|+f|(x)|f|(x)|)|<ε\Big{|}\Big{(}\sum_{D_{l}^{j,N}}|f_{j}(x)-f_{j}(x^{-})|-\big{|}|f_{j}|(x)-|f_{j}|(x^{-})\big{|}-|f(x)-f(x^{-})|+\big{|}|f|(x)-|f|(x^{-})\big{|}\Big{)}\Big{|}<\varepsilon

and

|(Drj,N|fj(x)fj(x+)|fj|(x)|fj|(x+)||f(x)f(x+)|+f|(x)|f|(x+)|)|<ε.\Big{|}\Big{(}\sum_{D_{r}^{j,N}}|f_{j}(x)-f_{j}(x^{+})|-\big{|}|f_{j}|(x)-|f_{j}|(x^{+})\big{|}-|f(x)-f(x^{+})|+\big{|}|f|(x)-|f|(x^{+})\big{|}\Big{)}\Big{|}<\varepsilon.

Moreover, we have

0\displaystyle 0 x(Dl(fj)(a,b))Dlj,N|fj(x)fj(x)|||fj|(x)|fj|(x)|\displaystyle\leq\underset{{}_{x\in\big{(}D_{l}(f_{j})\cap(a,b)\big{)}\setminus D_{l}^{j,N}}}{\sum}|f_{j}(x)-f_{j}(x^{-})|-\big{|}|f_{j}|(x)-|f_{j}|(x^{-})\big{|}
x(Dl(fj)(a,b))Dlj,N|fj(x)fj(x)|\displaystyle\leq\underset{{}_{x\in\big{(}D_{l}(f_{j})\cap(a,b)\big{)}\setminus D_{l}^{j,N}}}{\sum}|f_{j}(x)-f_{j}(x^{-})|
x(Dl(fj)(a,b))Dlj,N|f(x)f(x)|+4ffjBV<2ε\displaystyle\leq\underset{{}_{x\in\big{(}D_{l}(f_{j})\cap(a,b)\big{)}\setminus D_{l}^{j,N}}}{\sum}|f(x)-f(x^{-})|+4\|f-f_{j}\|_{BV}<2\varepsilon

and, similarly,

0x(Dr(fj)(a,b))Drj,N|fj(x)fj(x+)|||fj|(x)|fj|(x+)|<2ε.\displaystyle 0\leq\underset{{}_{x\in\big{(}D_{r}(f_{j})\cap(a,b)\big{)}\setminus D_{r}^{j,N}}}{\sum}|f_{j}(x)-f_{j}(x^{+})|-\big{|}|f_{j}|(x)-|f_{j}|(x^{+})\big{|}<2\varepsilon.

Finally, we observe that {xl,1,,xl,N}Dl(fj)\{x_{l,1},\dots,x_{l,N}\}\subset D_{l}(f_{j}) and {xr,1,,xr,N}Dr(fj)\{x_{r,1},\dots,x_{r,N}\}\subset D_{r}(f_{j}) for jj big enough, by the uniform convergence. Letting ε0\varepsilon\to 0 (and thus NN\to\infty) and applying Lemma 4, we obtain the claim. ∎

Let us now take a closer look at the properties of the maximal operator. Recall that the total variation of M~f\widetilde{M}f can be controlled by the total variation of ff. There is also a local version of this principle, where we focus on an interval (a,b)(a,b). However in this case some boundary terms must be included. Thus, to avoid the possibility that ff behaves badly at aa or bb, we use its adjusted version |f|¯\overline{|f|} defined by

|f|¯(x):=limsupIx;m(I)0 I|f|.\overline{|f|}(x):=\underset{I\ni x;m(I)\to 0}{\lim\sup}\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f|.

It is known that |f|¯\overline{|f|} is upper semicontinuous and that |f|¯M~f\overline{|f|}\leq\widetilde{M}f (see [1, Lemma 3.3]).

Lemma 6.

Fix fBV().f\in BV(\mathbb{R}). Given a<b,-\infty\leq a<b\leq\infty, we have

Var(a,b)(M~f)Var(a,b)(|f|¯)+|M~f(a)|f|¯(a)|+|M~f(b)|f|¯(b)|.{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}\leq{\rm Var}_{(a,b)}\big{(}\overline{|f|}\big{)}+\big{|}\widetilde{M}f(a)-\overline{|f|}(a)\big{|}+\big{|}\widetilde{M}f(b)-\overline{|f|}(b)\big{|}.
Proof.

This follows by a slight modification of the proof of [1, Lemma 3.9]. ∎

The next result gives us the uniform control (with respect to jj) on the behavior of (M~fj)\big{(}\widetilde{M}f_{j}\big{)}^{\prime} near infinity, provided that {fj}j\{f_{j}\}_{j\in\mathbb{N}} is a converging sequence in BV()BV(\mathbb{R}). This, in turn, allows one to restrict the attention to a bounded interval, while dealing with the total variations of the maximal functions M~fj\widetilde{M}f_{j}. We point out that it is also possible to proceed without this reduction, but then for all considered functions the extended domain [,][-\infty,\infty] should be used instead of \mathbb{R}.

Lemma 7.

Fix fBV()f\in BV(\mathbb{R}) and let {fj;j}BV()\{f_{j};j\in\mathbb{N}\}\subset BV(\mathbb{R}) be such that limjfjfBV=0.\underset{j\to\infty}{\lim}\|f_{j}-f\|_{BV}=0. Then for any ε>0\varepsilon>0 there exist <a<b<-\infty<a<b<\infty such that

(a,b)|(M~f)|<ε\int_{\mathbb{R}\setminus(a,b)}\left|\big{(}\widetilde{M}f\big{)}^{\prime}\right|<\varepsilon

and

(a,b)|(M~fj)|<ε,\int_{\mathbb{R}\setminus(a,b)}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|<\varepsilon,

for every jj big enough.

Proof.

We prove that there exists b<b<\infty such that (b,)|(M~f)|<ε,\displaystyle\int_{(b,\infty)}\big{|}\big{(}\widetilde{M}f\big{)}^{\prime}\big{|}<\varepsilon, the symmetric case is treated analogously. First we deal with the case where M~f()>|f|()\widetilde{M}f(\infty)>|f|(\infty). Assume that M~f()|f|()>4ε\widetilde{M}f(\infty)-|f|(\infty)>4\varepsilon. Let us take bb big enough such that (we use here Lemma 3) we have |M~f(x)M~f()|<ε\big{|}\widetilde{M}f(x)-\widetilde{M}f(\infty)\big{|}<\varepsilon and ||f|(x)|f|()|<ε\big{|}|f|(x)-|f|(\infty)\big{|}<\varepsilon for every x(b,).x\in(b,\infty). Therefore, for jj big enough such that |fj||f|ε2\||f_{j}|-|f|\|_{\infty}\leq\frac{\varepsilon}{2} and M~fjM~fε2,\|\widetilde{M}f_{j}-\widetilde{M}f\|_{\infty}\leq\frac{\varepsilon}{2}, for each y(b,x)y\in(b,x) we have M~fj(y)M~fj(x).\widetilde{M}f_{j}(y)\geq\widetilde{M}f_{j}(x). This is the case because any interval IxI\ni x satisfying  I|fj|>M~fj(x)ε2\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f_{j}|>\widetilde{M}f_{j}(x)-\frac{\varepsilon}{2} contains y,y, since if I(y,)I\subset(y,\infty), in particular I(b,),I\subset(b,\infty), and then

 I|fj| I|f|+ε2|f|()+3ε2M~f()2εM~f(x)εM~fj(x)ε2.\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f_{j}|\leq\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f|+\frac{\varepsilon}{2}\leq|f|(\infty)+\frac{3\varepsilon}{2}\leq\widetilde{M}f(\infty)-2\varepsilon\leq\widetilde{M}f(x)-\varepsilon\leq\widetilde{M}f_{j}(x)-\frac{\varepsilon}{2}.

Therefore

(b,)|(M~fj)|=M~fj(b)M~fj()|M~f(b)M~fj(b)|+|M~fj()M~f()|+|M~f(b)M~f()|3ε\int_{(b,\infty)}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|=\widetilde{M}f_{j}(b)-\widetilde{M}f_{j}(\infty)\leq\big{|}\widetilde{M}f(b)-\widetilde{M}f_{j}(b)\big{|}+\big{|}\widetilde{M}f_{j}(\infty)-\widetilde{M}f(\infty)\big{|}+\big{|}\widetilde{M}f(b)-\widetilde{M}f(\infty)\big{|}\leq 3\varepsilon

for jj big enough, from where we conclude this case.

Now, we deal with the case where |f|()=M~f().|f|(\infty)=\widetilde{M}f(\infty). By Lemma 6 and [1, Lemma 3.3], assuming that bb is a continuity point for fjf_{j}, we obtain

(b,)|(M~fj)|\displaystyle\int_{(b,\infty)}\Big{|}\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\Big{|} Var(b,)(|fj|¯)+|M~fj(b)|fj|¯(b)|+|M~fj()|fj|¯()|\displaystyle\leq{\rm Var}_{(b,\infty)}\big{(}\overline{|f_{j}|}\big{)}+\big{|}\widetilde{M}f_{j}(b)-\overline{|f_{j}|}(b)\big{|}+\big{|}\widetilde{M}f_{j}(\infty)-\overline{|f_{j}|}(\infty)\big{|}
Var(b,)(|fj|)+|M~fj(b)|fj|(b)|+|M~fj()|fj|()|.\displaystyle\leq{\rm Var}_{(b,\infty)}(|f_{j}|)+\big{|}\widetilde{M}f_{j}(b)-|f_{j}|(b)\big{|}+\big{|}\widetilde{M}f_{j}(\infty)-|f_{j}|(\infty)\big{|}.

The analogous is obtained for ff instead of fjf_{j}. Let us assume that bb is a continuity point for ff and every fjf_{j}, such that Var(b,)(|f|)<ε.{\rm Var}_{(b,\infty)}\big{(}|f|\big{)}<\varepsilon. By Lemma 5 we have

Var(b,)(|fj|)<2ε,\displaystyle{\rm Var}_{(b,\infty)}(|f_{j}|)<2\varepsilon, (2.1)

for jj big enough. Also,

|M~fj(b)|fj|(b)||M~f(b)|f|(b)|+|M~fj(b)M~f(b)|+||fj|(b)|f|(b)|.\big{|}\widetilde{M}f_{j}(b)-|f_{j}|(b)\big{|}\leq\big{|}\widetilde{M}f(b)-|f|(b)\big{|}+|\widetilde{M}f_{j}(b)-\widetilde{M}f(b)|+\big{|}|f_{j}|(b)-|f|(b)\big{|}.

If bb is big enough to have |M~f(b)|f|(b)|<ε\big{|}\widetilde{M}f(b)-|f|(b)\big{|}<\varepsilon, then by Lemma 3 we get |M~fj(b)|fj|(b)|<2ε\big{|}\widetilde{M}f_{j}(b)-|f_{j}|(b)\big{|}<2\varepsilon and |M~fj()|fj|()|<ε\big{|}\widetilde{M}f_{j}(\infty)-|f_{j}|(\infty)\big{|}<\varepsilon for jj big enough. Combining this with (2.1) concludes the proof. ∎

Before we prove our key result regarding the variation of maximal functions, we need the following definition. A given partition 𝒫={a0<a1<<an},\mathcal{P}=\{a_{0}<a_{1}<\dots<a_{n}\}, with n2,n\geq 2, has property (V) with respect to ff if for each i{0,1,,n2},i\in\{0,1,\dots,n-2\}, we have sgn(f(ai+2)f(ai+1))sgn(f(ai+1)f(ai))<0\operatorname{\mathrm{sgn}}(f(a_{i+2})-f(a_{i+1}))\cdot\operatorname{\mathrm{sgn}}(f(a_{i+1})-f(a_{i}))<0.

Proposition 8.

Fix fBV()f\in BV(\mathbb{R}) and let {fj;j}BV()\{f_{j};j\in\mathbb{N}\}\subset BV(\mathbb{R}) be such that limjfjfBV=0.\underset{j\to\infty}{\lim}\|f_{j}-f\|_{BV}=0. Then

Var(,)(M~fj)Var(,)(M~f).{\rm Var}_{(-\infty,\infty)}\big{(}\widetilde{M}f_{j}\big{)}\to{\rm Var}_{(-\infty,\infty)}\big{(}\widetilde{M}f\big{)}.
Proof.

By Lemma 7 it is enough to prove that Var(a,b)(M~fj)Var(a,b)(M~f){\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}\to{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)} for every interval (a,b)(a,b)\subset\mathbb{R} with both aa and bb being points of continuity for ff and every fjf_{j}. In the following we fix <a<b<-\infty<a<b<\infty satisfying such assumption. Observe that Lemma 3 and Fatou’s lemma imply

liminfjVar(a,b)(M~fj)Var(a,b)(M~f).\underset{j\to\infty}{\lim\inf}\ {\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}\geq{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}.

Now, we prove the remaining inequality, that is,

limsupjVar(a,b)(M~fj)Var(a,b)(M~f).\underset{j\to\infty}{\lim\sup}\ {\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}\leq{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}.

Given ε>0\varepsilon>0 we show that

Var(a,b)(M~fj)<Var(a,b)(M~f)+4ε{\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}<{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}+4\varepsilon

holds if jj is big enough. Let 𝒫={a=a0<a1<<aK=b},\mathcal{P}=\{a=a_{0}<a_{1}<\dots<a_{K}=b\}\subset\mathbb{R}, K,K\in\mathbb{N}, be a partition satisfying

Var(|f|,𝒫)>Var(a,b)(|f|)ε\displaystyle{\rm Var}\big{(}|f|,\mathcal{P}\big{)}>{\rm Var}_{(a,b)}\big{(}|f|\big{)}-\varepsilon

and

Var(M~f,𝒫)>Var(a,b)(M~f)ε.\displaystyle{\rm Var}\big{(}\widetilde{M}f,\mathcal{P}\big{)}>{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}-\varepsilon.

Also, by the uniform convergence and Lemma 5 we conclude that

Var(|fj|,𝒫)>Var(a,b)(|fj|)2ε\displaystyle{\rm Var}\big{(}|f_{j}|,\mathcal{P}\big{)}>{\rm Var}_{(a,b)}\big{(}|f_{j}|\big{)}-2\varepsilon (2.2)

and

Var(M~fj,𝒫)>Var(a,b)(M~f)2ε\displaystyle{\rm Var}\big{(}\widetilde{M}f_{j},\mathcal{P}\big{)}>{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}-2\varepsilon (2.3)

hold for jj big enough. Now, we take 𝒫~=𝒫~(j)\widetilde{\mathcal{P}}=\widetilde{\mathcal{P}}(j) such that 𝒫𝒫~[a,b]\mathcal{P}\subset\widetilde{\mathcal{P}}\subset[a,b] and

Var(M~fj,𝒫~)>Var(a,b)(M~fj)ε.{\rm Var}\big{(}\widetilde{M}f_{j},\widetilde{\mathcal{P}}\big{)}>{\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}-\varepsilon.

Without loss of generality we can assume that 𝒫~\widetilde{\mathcal{P}} is such that for each i{1,,K}i\in\{1,\dots,K\} the set 𝒫~[ai1,ai]={ai1=ai,0<<ai,ni=ai}\widetilde{\mathcal{P}}\cap[a_{i-1},a_{i}]=\{a_{i-1}=a_{i,0}<\dots<a_{i,n_{i}}=a_{i}\} satisfies property (V) with respect to M~fj\widetilde{M}f_{j} unless it consists of two elements. For each such ii we denote 𝒫~i={ai,1,,ai,ni1}\widetilde{\mathcal{P}}_{i}=\{a_{i,1},\dots,a_{i,n_{i}-1}\} and claim that it is possible to find a partition 𝒫~i={ai,1,,ai,ni1}(ai1,ai)\widetilde{\mathcal{P}}_{i}^{*}=\{a_{i,1}^{*},\dots,a_{i,n_{i}-1}^{*}\}\subset(a_{i-1},a_{i}) such that

Var(|fj|,𝒫~i)Var(|fj|,{ai,1,ai,ni1})>Var(M~fj,𝒫~i)Var(M~fj,{ai,1,ai,ni1})εK.\displaystyle{\rm Var}\Big{(}|f_{j}|,\widetilde{\mathcal{P}}_{i}^{*}\Big{)}-{\rm Var}\Big{(}|f_{j}|,\{a_{i,1}^{*},a_{i,n_{i}-1}^{*}\}\Big{)}>{\rm Var}\left(\widetilde{M}f_{j},\widetilde{\mathcal{P}}_{i}\right)-{\rm Var}\left(\widetilde{M}f_{j},\{a_{i,1},a_{i,n_{i}-1}\}\right)-\frac{\varepsilon}{K}.

Indeed, for ni2n_{i}\leq 2 we use the convention that all the variation terms above are equal to 0,0, so the inequality holds (we set 𝒫~i=\widetilde{\mathcal{P}}_{i}^{*}=\emptyset or 𝒫~i={ai,1}\widetilde{\mathcal{P}}_{i}^{*}=\{a_{i,1}\} if n=1n=1 or n=2n=2, respectively). It remains to consider the case ni3n_{i}\geq 3 in which property (V)(V) is guaranteed. We assume that M~fj(ai,0)<M~fj(ai,1)\widetilde{M}f_{j}(a_{i,0})<\widetilde{M}f_{j}(a_{i,1}) (the opposite case can be treated analogously). Then 𝒫~i\widetilde{\mathcal{P}}_{i}^{*} shall be chosen in such a way that given k{1,,ni1}k\in\{1,\dots,n_{i}-1\} we have

|fj|(ai,k)>max{M~fj(ai,k1),M~fj(ai,k)ε2niK,M~fj(ai,k+1)},|f_{j}|(a_{i,k}^{*})>\max\Big{\{}\widetilde{M}f_{j}(a_{i,k-1}),\widetilde{M}f_{j}(a_{i,k})-\frac{\varepsilon}{2n_{i}K},\widetilde{M}f_{j}(a_{i,k+1})\Big{\}},

for kk odd, and

|fj|(ai,k)M~fj(ai,k)|f_{j}|(a_{i,k}^{*})\leq\widetilde{M}f_{j}(a_{i,k})

for kk even. We describe in detail the procedure for selecting the points ai,k.a_{i,k}^{*}. If kk is odd, then we find an interval Iai,kI\ni a_{i,k} such that

 I|fj|>max{M~fj(ai,k1),M~fj(ai,k)ε2niK,M~fj(ai,k+1)}.\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f_{j}|>\max\Big{\{}\widetilde{M}f_{j}(a_{i,k-1}),\widetilde{M}f_{j}(a_{i,k})-\frac{\varepsilon}{2n_{i}K},\widetilde{M}f_{j}(a_{i,k+1})\Big{\}}.

Clearly, I(ai,k1,ai,k+1)I\subset(a_{i,k-1},a_{i,k+1}) and we can find ai,kIa_{i,k}^{*}\in I satisfying |fj|(ai,k) I|fj|.|f_{j}|(a_{i,k}^{*})\geq\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f_{j}|. For kk even we take I=(ai,k1,ai,k+1)I=(a_{i,k-1}^{*},a_{i,k+1}^{*}) if kni1k\neq n_{i}-1 or I=(ai,ni2,ai)I=(a_{i,n_{i}-2}^{*},a_{i}) otherwise. Since  I|fj|M~fj(ai,k),\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f_{j}|\leq\widetilde{M}f_{j}(a_{i,k}), there exists ai,kIa_{i,k}^{*}\in I satisfying |fj|(ai,k)M~fj(ai,k).|f_{j}|(a_{i,k}^{*})\leq\widetilde{M}f_{j}(a_{i,k}). We note that the appropriate configuration of the sets II guarantees that the inequalities ai1<ai,1<<ai,ni1<aia_{i-1}<a_{i,1}^{*}<\dots<a_{i,n_{i}-1}^{*}<a_{i} hold.

Observe, also, that the partition {ai,1<<ai,ni1}\{a_{i,1}^{*}<\dots<a_{i,n_{i}-1}^{*}\} either consists of 22 elements or satisfies property (V) with respect to |fj||f_{j}|. Thus, regardless of which case occurs, we obtain

Var(|fj|,𝒫~i)Var(|fj|,{ai,1,ai,ni1})=k=1ni1αk|fj|(ai,k),{\rm Var}\Big{(}|f_{j}|,\widetilde{\mathcal{P}}_{i}^{*}\Big{)}-{\rm Var}\Big{(}|f_{j}|,\{a_{i,1}^{*},a_{i,n_{i}-1}^{*}\}\Big{)}=\sum_{k=1}^{n_{i}-1}\alpha_{k}|f_{j}|(a_{i,k}^{*}),

where αk=2(1)k+1\alpha_{k}=2(-1)^{k+1} for k{2,,ni2}k\in\{2,\dots,n_{i}-2\} and αk{0,2(1)k+1}\alpha_{k}\in\left\{0,2(-1)^{k+1}\right\} for k{1,ni1}k\in\{1,n_{i}-1\} (the boundary values depend on sgn(|fj|(ai,1)|fj|(ai,ni))\operatorname{\mathrm{sgn}}\big{(}|f_{j}|(a_{i,1}^{*})-|f_{j}|(a_{i,n_{i}}^{*})\big{)} and the parity of nin_{i}). Similarly,

Var(M~fj,𝒫~i)Var(M~fj,{ai,1,ai,ni1})k=1ni1αkM~fj(ai,k){\rm Var}\left(\widetilde{M}f_{j},\widetilde{\mathcal{P}}_{i}\right)-{\rm Var}\left(\widetilde{M}f_{j},\{a_{i,1},a_{i,n_{i}-1}\}\right)\leq\sum_{k=1}^{n_{i}-1}\alpha_{k}\widetilde{M}f_{j}(a_{i,k})

(we eventually change the sign of the second term on the left-hand side in order to get the boundary coefficients equal to α1\alpha_{1} and αni1\alpha_{n_{i}-1}). Consequently, the claim follows since for each kk we have

αk(|fj|(ai,k)M~fj(ai,k))εniK.\alpha_{k}\left(|f_{j}|(a_{i,k}^{*})-\widetilde{M}f_{j}(a_{i,k})\right)\geq\frac{-\varepsilon}{n_{i}K}.

Now, we apply the claim in order to get the following estimate

Var(a,b)(|fj|)Var(|fj|,𝒫)\displaystyle{\rm Var}_{(a,b)}(|f_{j}|)-{\rm Var}(|f_{j}|,\mathcal{P}) Var(|fj|,𝒫i=1K𝒫~i)Var(|fj|,𝒫i=1K{ai,1,ai,ni1})\displaystyle\geq{\rm Var}\Big{(}|f_{j}|,\mathcal{P}\cup\bigcup_{i=1}^{K}\widetilde{\mathcal{P}}_{i}^{*}\Big{)}-{\rm Var}\Big{(}|f_{j}|,\mathcal{P}\cup\bigcup_{i=1}^{K}\{a_{i,1}^{*},a_{i,n_{i}-1}^{*}\}\Big{)}
=i=1KVar(|fj|,𝒫~i)Var(|fj|,{ai,1,ai,ni1})\displaystyle=\sum_{i=1}^{K}{\rm Var}\Big{(}|f_{j}|,\widetilde{\mathcal{P}}_{i}^{*}\Big{)}-{\rm Var}\Big{(}|f_{j}|,\{a_{i,1}^{*},a_{i,n_{i}-1}^{*}\}\Big{)}
i=1KVar(M~fj,𝒫~i)Var(M~fj,{ai,1,ai,ni1})εK\displaystyle\geq\sum_{i=1}^{K}{\rm Var}\left(\widetilde{M}f_{j},\widetilde{\mathcal{P}}_{i}\right)-{\rm Var}\left(\widetilde{M}f_{j},\{a_{i,1},a_{i,n_{i}-1}\}\right)-\frac{\varepsilon}{K}
Var(M~fj,𝒫~)Var(M~fj,𝒫~~)ε,\displaystyle\geq{\rm Var}\Big{(}\widetilde{M}f_{j},\widetilde{\mathcal{P}}\Big{)}-{\rm Var}\Big{(}\widetilde{M}f_{j},\widetilde{\widetilde{\mathcal{P}}}\Big{)}-\varepsilon,

where 𝒫~~:={ai,k;i{1,,K},k{0,1,ni1,ni}}.\widetilde{\widetilde{\mathcal{P}}}:=\{a_{i,k};i\in\{1,\dots,K\},k\in\{0,1,n_{i}-1,n_{i}\}\}. In particular, we note that 𝒫~~\widetilde{\widetilde{\mathcal{P}}} consists of at most 3K+13K+1 elements and thus

Var(M~fj,𝒫~~)<Var(M~f,𝒫~~)+12KfjfBV<Var(a,b)(M~f)+ε{\rm Var}\Big{(}\widetilde{M}f_{j},\widetilde{\widetilde{\mathcal{P}}}\Big{)}<{\rm Var}\Big{(}\widetilde{M}f,\widetilde{\widetilde{\mathcal{P}}}\Big{)}+12K\|f_{j}-f\|_{BV}<{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}+\varepsilon

follows by Lemma 3 for jj big enough. Combining the above inequalities with (2.3), we arrive at

Var(a,b)(|fj|)Var(|fj|,𝒫)>Var(a,b)(M~fj)Var(a,b)(M~f)4ε,{\rm Var}_{(a,b)}(|f_{j}|)-{\rm Var}(|f_{j}|,\mathcal{P})>{\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}-{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}-4\varepsilon,

which, in view of (2.2), gives

Var(a,b)(M~fj)<Var(a,b)(M~f)+6ε,{\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}<{\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}+6\varepsilon,

provided that jj is big enough. Consequently, limjVar(a,b)(M~fj)=Var(a,b)(M~f).\underset{j\to\infty}{\lim}{\rm Var}_{(a,b)}\big{(}\widetilde{M}f_{j}\big{)}={\rm Var}_{(a,b)}\big{(}\widetilde{M}f\big{)}.

Having obtained Proposition 8 we continue with the remaining tools required. Our general purpose in the next few lemmas is to get more information about the derivative of maximal function. In particular, we are interested in studying the behavior of the sequence {(M~fj)(x)}j\big{\{}\big{(}\widetilde{M}f_{j}\big{)}^{\prime}(x)\big{\}}_{j\in\mathbb{N}} for a given point xx.

Lemma 9.

Fix fBV()f\in BV(\mathbb{R}) and let {fj;j}BV()\{f_{j};j\in\mathbb{N}\}\subset BV(\mathbb{R}) be such that limjfjfBV=0.\underset{j\to\infty}{\lim}\|f_{j}-f\|_{BV}=0. If  Ix,j|fj|=M~fj(x)\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI_{x,j}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j}}}|f_{j}|=\widetilde{M}f_{j}(x) with Ix,jxI_{x,j}\ni x, and χIx,jχI\chi_{I_{x,j}}\to\chi_{I} a.e. with 0<m(I)<0<m(I)<\infty, then we have  I|f|=M~f(x).\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f|=\widetilde{M}f(x).

Proof.

Follows a slight modification in [12, Lemma 12]. ∎

Let us now define E:={x;M~f(x)>|f|¯(x)}.E:=\big{\{}x\in\mathbb{R};\widetilde{M}f(x)>\overline{|f|}(x)\big{\}}. Since M~f\widetilde{M}f is absolutely continuous and |f|¯\overline{|f|} is upper semicontinuous, we have that EE is open. We notice that if xE{x;M~f(x)=M~f()}x\in E\setminus\big{\{}x;\widetilde{M}f(x)=\widetilde{M}f(\infty)\big{\}}, then there exists a finite interval IxI\ni x such that  I|f|(x)=M~f(x).\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I}}|f|(x)=\widetilde{M}f(x). Indeed, there exists a sequence {(ak,bk)}k\{(a_{k},b_{k})\}_{k\in\mathbb{N}} such that x(ak,bk)x\in(a_{k},b_{k}) and  (ak,bk)|f|M~f(x).\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt(a_{k},b_{k})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(a_{k},b_{k})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(a_{k},b_{k})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(a_{k},b_{k})}}|f|\to\widetilde{M}f(x). Since M~f(x)>{|f|¯(x),M~f()}\widetilde{M}f(x)>\big{\{}\overline{|f|}(x),\widetilde{M}f(\infty)\big{\}}, we have {bkak;k}(ϵ,ϵ1)\{b_{k}-a_{k};k\in\mathbb{N}\}\subset(\epsilon,\epsilon^{-1}) for some ϵ>0\epsilon>0. Thus, by taking a subsequence (if required), we get akaa_{k}\to a and bkbb_{k}\to b, with ba(0,).b-a\in(0,\infty). By the boundedness of ff we conclude that  [a,b]|f|=M~f(x).\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt[a,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,b]}}|f|=\widetilde{M}f(x). Also, let us observe that max{ [a,x]|f|, [x,b]|f|} [a,b]|f|,\max\big{\{}\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt[a,x]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,x]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,x]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,x]}}|f|,\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt[x,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[x,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[x,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[x,b]}}|f|\big{\}}\geq\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt[a,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,b]}}|f|, therefore M~f(x)= [a,x]|f|\widetilde{M}f(x)=\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt[a,x]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,x]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,x]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[a,x]}}|f| or M~f(x)= [x,b]|f|.\widetilde{M}f(x)=\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt[x,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[x,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[x,b]}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{[x,b]}}|f|.

The next result states that for a.e. xEx\in E the derivative of the maximal function M~f\widetilde{M}f can be described by an explicit formula.

Lemma 10.

Let fBV()f\in BV(\mathbb{R}). Assume that M~f\widetilde{M}f is differentiable and |f||f| is continuous at xx (that happens a.e. because M~f\widetilde{M}f and |f||f| have bounded variation). Let us suppose that xEx\in E is such that there exists an interval IxxI_{x}\ni x with m(Ix)<m(I_{x})<\infty such that  Ix|f|=M~f(x)\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI_{x}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x}}}|f|=\widetilde{M}f(x) and Ix[x,)I_{x}\subset[x,\infty) or Ix(,x]I_{x}\subset(-\infty,x]. Then

(M~f)(x)={Ix|f|(m(Ix))2|f|(x)m(Ix)=1m(Ix)(M~f(x)|f|(x))ifIx[x,),|f|(x)m(Ix)Ix|f|(m(Ix))2=1m(Ix)(|f|(x)M~f(x))otherwise.\big{(}\widetilde{M}f\big{)}^{\prime}(x)=\begin{cases}\frac{\int_{I_{x}}|f|}{(m(I_{x}))^{2}}-\frac{|f|(x)}{m(I_{x})}=\frac{1}{m(I_{x})}\left(\widetilde{M}f(x)-|f|(x)\right)\quad{\rm if}\ I_{x}\subset[x,\infty),\\ \frac{|f|(x)}{m(I_{x})}-\frac{\int_{I_{x}}|f|}{(m(I_{x}))^{2}}=\frac{1}{m(I_{x})}\left(|f|(x)-\widetilde{M}f(x)\right)\quad{\rm otherwise.}\end{cases}

Also, if M~f(x)=M~f()\widetilde{M}f(x)=\widetilde{M}f(\infty), then we have (M~f)(x)=0\big{(}\widetilde{M}f\big{)}^{\prime}(x)=0.

Proof.

The last claim follows because xx is a local minimum of M~f.\widetilde{M}f. Assume without loss of generality that Ix=(x,ax)I_{x}=(x,a_{x}), ax>xa_{x}>x (the other case is similar). We have, for h>0h>0, that

M~f(x)M~f(xh)h (x,ax)|f| (xh,ax)|f|h=xax|f|axxxax|f|+xhx|f|axx+hhIx|f|(axx)2|f|(x)axx\frac{\widetilde{M}f(x)-\widetilde{M}f(x-h)}{h}\leq\frac{\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt(x,a_{x})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a_{x})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a_{x})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a_{x})}}|f|-\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt(x-h,a_{x})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x-h,a_{x})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x-h,a_{x})}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x-h,a_{x})}}|f|}{h}=\frac{\frac{\int_{x}^{a_{x}}|f|}{a_{x}-x}-\frac{\int_{x}^{a_{x}}|f|+\int_{x-h}^{x}|f|}{a_{x}-x+h}}{h}\to\frac{\int_{I_{x}}|f|}{(a_{x}-x)^{2}}-\frac{|f|(x)}{a_{x}-x}

as h0.h\to 0. Therefore (M~f)(x)Ix|f|(axx)2|f(x)|axx.\big{(}\widetilde{M}f\big{)}^{\prime}(x)\leq\frac{\int_{I_{x}}|f|}{(a_{x}-x)^{2}}-\frac{|f(x)|}{a_{x}-x}. Also, for h>0h>0 we have

M~f(x+h)M~f(x)hx+hax|f|axxhxax|f|axxhIx|f|(axx)2|f|(x)axx\frac{\widetilde{M}f(x+h)-\widetilde{M}f(x)}{h}\geq\frac{\frac{\int_{x+h}^{a_{x}}|f|}{a_{x}-x-h}-\frac{\int_{x}^{a_{x}}|f|}{a_{x}-x}}{h}\to\frac{\int_{I_{x}}|f|}{(a_{x}-x)^{2}}-\frac{|f|(x)}{a_{x}-x}

as h0.h\to 0. This concludes the proof. ∎

Now, we can use the obtained formula to prove the following result regarding pointwise convergence.

Lemma 11.

Fix fBV()f\in BV(\mathbb{R}) and let {fj;j}BV()\{f_{j};j\in\mathbb{N}\}\subset BV(\mathbb{R}) be such that limjfjfBV=0.\underset{j\to\infty}{\lim}\|f_{j}-f\|_{BV}=0. Then

(M~fj)(M~f)\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\to\big{(}\widetilde{M}f\big{)}^{\prime}

a.e. in E.E.

Proof.

The claim is trivial if EE has measure zero, so assume this is not the case. We define EjE_{j} as the analogue of EE for fj.f_{j}. Let us take xEx\in E such that M~fj\widetilde{M}f_{j} and M~f\widetilde{M}f are differentiable at xx for every jj and ff and fjf_{j} is continuous at xx. By the uniform convergence we have that xEjx\in E_{j} for jj big enough. We also make the following observation. If there are intervals Ix,jxI_{x,j}\ni x such that  Ix,j|fj|=M~fj(x)\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI_{x,j}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j}}}|f_{j}|=\widetilde{M}f_{j}(x), then the quantities m(Ix,j)m(I_{x,j}) are bounded below uniformly. Indeed, if for a sequence {jk}k\{j_{k}\}_{k\in\mathbb{N}} we have m(Ix,jk)0m(I_{x,j_{k}})\to 0, then we would have  Ix,jk|fjk||f|(x)<M~f(x)\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0ptI_{x,j_{k}}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j_{k}}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j_{k}}}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{I_{x,j_{k}}}}|f_{j_{k}}|\to|f|(x)<\widetilde{M}f(x) by the uniform convergence and continuity of ff at xx, contradicting the pointwise convergence of the maximal functions.

Assume first that xE{y;M~f(y)=M~f()}x\in E\setminus\big{\{}y;\widetilde{M}f(y)=\widetilde{M}f(\infty)\big{\}} and take ε>0\varepsilon>0 such that M~f(x)>M~f()+2ε.\widetilde{M}f(x)>\widetilde{M}f(\infty)+2\varepsilon. Then for jj big enough we have M~fj(x)>M~fj()+ε.\widetilde{M}f_{j}(x)>\widetilde{M}f_{j}(\infty)+\varepsilon. Also, there exists N>|x|N>|x| such that for jj big enough and each y[N,N]y\in\mathbb{R}\setminus[-N,N] we have |fj|(y)<M~f()+ε<M~fj(x)|f_{j}|(y)<\widetilde{M}f(\infty)+\varepsilon<\widetilde{M}f_{j}(x). We can observe then that Ix,j[N,N]I_{x,j}\subset[-N,N] for jj big enough. Let us assume that we have δ>0\delta>0 and a sequence {jk}k\{j_{k}\}_{k\in\mathbb{N}} such that

|(M~fjk)(x)(M~f)(x)|>δ.\displaystyle\left|\big{(}\widetilde{M}f_{j_{k}}\big{)}^{\prime}(x)-\big{(}\widetilde{M}f\big{)}^{\prime}(x)\right|>\delta. (2.4)

Without loss of generality assume that Ijk=(x,ajk)I_{j_{k}}=(x,a_{j_{k}}) (the other case is treated analogously). Since x<ajk<Nx<a_{j_{k}}<N, there exists a subsequence (that we also denote by jkj_{k}) such that ajka[x,N].a_{j_{k}}\to a\in[x,N]. Moreover, in view of the previous observation, we have axa\neq x. Thus, Lemma 9 gives  (x,a)|f|=M~f(x)\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt(x,a)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a)}}|f|=\widetilde{M}f(x) and consequently, in view of Lemma 10, we obtain (M~f)(x)=(x,a)|f|(ax)2|f|(x)ax\big{(}\widetilde{M}f\big{)}^{\prime}(x)=\frac{\int_{(x,a)}|f|}{(a-x)^{2}}-\frac{|f|(x)}{a-x}. Also, (M~fjk)(x)=Ix,j|fjk|(ajkx)2|fjk|(x)ajkx\big{(}\widetilde{M}f_{j_{k}}\big{)}^{\prime}(x)=\frac{\int_{I_{x,j}}|f_{j_{k}}|}{(a_{j_{k}}-x)^{2}}-\frac{|f_{j_{k}}|(x)}{a_{j_{k}}-x} holds. However, by the uniform convergence we have

Ix,j|fjk|(ajkx)2|fjk|(x)ajkx(x,a)|f|(ax)2|f|(x)ax,\frac{\int_{I_{x,j}}|f_{j_{k}}|}{(a_{j_{k}}-x)^{2}}-\frac{|f_{j_{k}}|(x)}{a_{j_{k}}-x}\to\frac{\int_{(x,a)}|f|}{(a-x)^{2}}-\frac{|f|(x)}{a-x},

reaching a contradiction with (2.4). Thus, we conclude this case.

Now, if M~f(x)=M~f(),\widetilde{M}f(x)=\widetilde{M}f(\infty), then by Lemma 10 we have (M~f)(x)=0.\big{(}\widetilde{M}f\big{)}^{\prime}(x)=0. Also, if for a subsequence jkj_{k} we have M~fjk(x)=M~fjk()\widetilde{M}f_{j_{k}}(x)=\widetilde{M}f_{j_{k}}(\infty), then (M~fjk)(x)=0.\big{(}\widetilde{M}f_{j_{k}}\big{)}^{\prime}(x)=0. Therefore, this subcase follows and we can assume that xEj{M~fj(x)=M~fj()}.x\in E_{j}\setminus\big{\{}\widetilde{M}f_{j}(x)=\widetilde{M}f_{j}(\infty)\big{\}}. It is now enough to prove that Ix,j|fj|(ajx)2|fj|(x)ajx0.\frac{\int_{I_{x,j}}|f_{j}|}{(a_{j}-x)^{2}}-\frac{|f_{j}|(x)}{a_{j}-x}\to 0. Let us suppose that for some δ>0\delta>0 and a subsequence jkj_{k} we have |(M~fjk)|>δ.\big{|}\big{(}\widetilde{M}f_{j_{k}}\big{)}^{\prime}\big{|}>\delta. As before, we assume the case Ijk=(x,ajk)I_{j_{k}}=(x,a_{j_{k}}). We claim that there exists C(δ,f)>0C(\delta,f)>0 such that for jkj_{k} big enough we have m(Ix,jk)<C(δ,f)<.m(I_{x,j_{k}})<C(\delta,f)<\infty. Indeed, in view of

2fjkm(Ix,jk)>|Ix,jk|fjk|(ajkx)2|fjk|(x)ajkx|>δ,\frac{2\|f_{j_{k}}\|_{\infty}}{m(I_{x,j_{k}})}>\left|\frac{\int_{I_{x,j_{k}}}|f_{j_{k}}|}{(a_{j_{k}}-x)^{2}}-\frac{|f_{j_{k}}|(x)}{a_{j_{k}}-x}\right|>\delta,

we have 2fjkδ>m(Ix,jk)\frac{2\|f_{j_{k}}\|_{\infty}}{\delta}>m(I_{x,{j_{k}}}) and thus fjkf\|f_{j_{k}}\|_{\infty}\to\|f\|_{\infty} gives our claim. Now, since m(Ix,jk)<C(δ,f),m(I_{x,j_{k}})<C(\delta,f), we have that ajk(x,x+C(δ,f))a_{j_{k}}\in(x,x+C(\delta,f)) for jj big enough. Consequently, there exists a subsequence (that we also denote by jkj_{k}) such that ajkaa_{j_{k}}\to a for some a(x,x+C(δ,f)].a\in(x,x+C(\delta,f)]. Then by Lemma 9 we have that  (x,a)|f|=M~f(x).\mathchoice{\mathop{\vrule width=6.0pt,height=3.0pt,depth=-2.5pt\kern-8.0pt\intop}\nolimits_{\kern-6.0pt(x,a)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a)}}{\mathop{\vrule width=5.0pt,height=3.0pt,depth=-2.6pt\kern-6.0pt\intop}\nolimits_{(x,a)}}|f|=\widetilde{M}f(x). Therefore, Lemma 10 gives

(x,a)|f|(ax)2|f|(x)ax=(M~f)(x)\frac{\int_{(x,a)}|f|}{(a-x)^{2}}-\frac{|f|(x)}{a-x}=\big{(}\widetilde{M}f\big{)}^{\prime}(x)

and the left-hand side must be equal to 0. Since we have

(M~fjk)(x)=Ix,jk|fjk|(ajkx)2|fjk|(x)ajkx(x,a)|f|(ax)2|f|(x)ax=0\big{(}\widetilde{M}f_{j_{k}}\big{)}^{\prime}(x)=\frac{\int_{I_{x,j_{k}}}|f_{j_{k}}|}{(a_{j_{k}}-x)^{2}}-\frac{|f_{j_{k}}|(x)}{a_{j_{k}}-x}\to\frac{\int_{(x,a)}|f|}{(a-x)^{2}}-\frac{|f|(x)}{a-x}=0

by the uniform convergence, we reach a contradiction. This concludes the proof. ∎

It remains to take a look at the set C:=EC:=\mathbb{R}\setminus E.

Lemma 12.

Let fBV()f\in BV(\mathbb{R}). Then for a.e. xCx\in C we have (M~f)(x)=0.\big{(}\widetilde{M}f\big{)}^{\prime}(x)=0.

Proof.

Assume that |f|¯\overline{|f|} and M~f\widetilde{M}f are differentiable at xx (this happens a.e. because |f|¯\overline{|f|} and M~f\widetilde{M}f have bounded variation). Then, since M~f(x)=|f|¯(x)\widetilde{M}f(x)=\overline{|f|}(x) and M~f|f|¯\widetilde{M}f\geq\overline{|f|}, we have (M~f)(x)=|f|¯(x).\big{(}\widetilde{M}f\big{)}^{\prime}(x)=\overline{|f|}^{\prime}(x). Now, assume, in order to get a contradiction, that |f|¯(x)>0\overline{|f|}^{\prime}(x)>0 (the other case is analogous). Then there exist h0,L>0h_{0},L>0 such that |f|¯(x+h)|f|¯(x)+Lh\overline{|f|}(x+h)\geq\overline{|f|}(x)+Lh for every 0<h<h0.0<h<h_{0}. Thus, for a.e. 0<h<h00<h<h_{0} we have |f|(x+h)|f|¯(x)+Lh|f|(x+h)\geq\overline{|f|}(x)+Lh, which implies M~f(x)0h0|f|¯(x)+Lhh0=|f|¯(x)+Lh02>|f|¯(x),\widetilde{M}f(x)\geq\frac{\int_{0}^{h_{0}}\overline{|f|}(x)+Lh}{h_{0}}=\overline{|f|}(x)+\frac{Lh_{0}}{2}>\overline{|f|}(x), a contradiction. ∎

Combining the previous results we obtain the following.

Corollary 13.

Fix fBV()f\in BV(\mathbb{R}) and let {fj;j}BV()\{f_{j};j\in\mathbb{N}\}\subset BV(\mathbb{R}) be such that limjfjfBV=0.\underset{j\to\infty}{\lim}\|f_{j}-f\|_{BV}=0. Then

((M~fj)(M~f))χE10.\left\|\left(\big{(}\widetilde{M}f_{j}\big{)}^{\prime}-\big{(}\widetilde{M}f\big{)}^{\prime}\right)\chi_{E}\right\|_{1}\to 0.
Proof.

By the classic Brezis–Lieb lemma [6], the boundedness of the map fM~ff\mapsto\widetilde{M}f from BV()BV(\mathbb{R}) to itself and Lemma 11, we just need to prove the following,

(M~fj)χE1(M~f)χE1.\displaystyle\left\|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\chi_{E}\right\|_{1}\to\left\|\big{(}\widetilde{M}f\big{)}^{\prime}\chi_{E}\right\|_{1}. (2.5)

By Fatou’s Lemma, Proposition 8 and Lemma 12, we have

E|(M~f)|liminfjE|(M~fj)|limsupjE|(M~fj)|limj|(M~fj)|=|(M~f)|=E|(M~f)|,\int_{E}\left|\big{(}\widetilde{M}f\big{)}^{\prime}\right|\leq\underset{j\to\infty}{\lim\inf}\int_{E}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|\leq\underset{j\to\infty}{\lim\sup}\int_{E}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|\leq\underset{j\to\infty}{\lim}\int_{\mathbb{R}}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|=\int_{\mathbb{R}}\left|\big{(}\widetilde{M}f\big{)}^{\prime}\right|=\int_{E}\left|\big{(}\widetilde{M}f\big{)}^{\prime}\right|,

from where (2.5) follows. ∎

2.2. Proof of Theorem 1

Finally, we are ready to prove the main result. In what follows CjC_{j} denotes the set analogous to CC defined for fjf_{j} instead of f.f.

Proof.

Since by Lemmas 2 and 3 we have M~fj()M~f()\widetilde{M}f_{j}(-\infty)\to\widetilde{M}f(-\infty), it remains to prove that

(M~fj)(M~f)\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\to\big{(}\widetilde{M}f\big{)}^{\prime}

in L1().L^{1}(\mathbb{R}). We make the following claim

CEj|(M~fj)|0.\displaystyle\int_{C\cap E_{j}}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|\to 0. (2.6)

Indeed, by Proposition 8, Lemma 12 and Corollary 13 we have

limjE|(M~fj)|=E|(M~f)|=|(M~f)|\displaystyle\underset{j\to\infty}{\lim}\int_{E}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|=\int_{E}\left|\big{(}\widetilde{M}f\big{)}^{\prime}\right|=\int_{\mathbb{R}}\left|\big{(}\widetilde{M}f\big{)}^{\prime}\right| limsupj(EjC|(M~fj)|+E|(M~fj)|)\displaystyle\geq\underset{j\to\infty}{\lim\sup}\left(\int_{E_{j}\cap C}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|+\int_{E}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|\right)
=limsupjEjC|(M~fj)|+limjE|(M~fj)|\displaystyle=\underset{j\to\infty}{\lim\sup}\int_{E_{j}\cap C}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|+\underset{j\to\infty}{\lim}\int_{E}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|

and the claim follows. Consequently, by (2.6) and Corollary 13 we get

|(M~fj)(M~f)|\displaystyle\int_{\mathbb{R}}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}-\big{(}\widetilde{M}f\big{)}^{\prime}\right| =CEj|(M~fj)(M~f)|+CCj|(M~fj)(M~f)|+E|(M~fj)(M~f)|\displaystyle=\int_{C\cap E_{j}}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}-\big{(}\widetilde{M}f\big{)}^{\prime}\right|+\int_{C\cap C_{j}}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}-\big{(}\widetilde{M}f\big{)}^{\prime}\right|+\int_{E}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}-\big{(}\widetilde{M}f\big{)}^{\prime}\right|
=CEj|(M~fj)|+E|(M~fj)(M~f)|0\displaystyle=\int_{C\cap E_{j}}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}\right|+\int_{E}\left|\big{(}\widetilde{M}f_{j}\big{)}^{\prime}-\big{(}\widetilde{M}f\big{)}^{\prime}\right|\to 0

as j,j\to\infty, from where we conclude our result. ∎

2.3. Concluding remarks

We end our discussion by showing that the assumptions f,fjBV()f,f_{j}\in BV(\mathbb{R}) are important, not only ffjBV().f-f_{j}\in BV(\mathbb{R}).

Example.

Let A=k=1(4k2,4k)A=\cup_{k=1}^{\infty}(4k-2,4k) and take

f=χ(,0)A,andfn=f+1nχ(0,4n+2).f=\chi_{(-\infty,0)\cup A},\quad{\rm and}\quad f_{n}=f+\frac{1}{n}\chi_{(0,4n+2)}.

Then we have fnfBV0\|f_{n}-f\|_{BV}\rightarrow 0, while M~fnM~fBV↛0\|\widetilde{M}f_{n}-\widetilde{M}f\|_{BV}\not\rightarrow 0.

Indeed, the first claim is obvious and for the second one we argue as follows. We observe that M~f1\widetilde{M}f\equiv 1 and M~fn(x)=1+1n\widetilde{M}f_{n}(x)=1+\frac{1}{n} for x{3,7,,4n1}x\in\{3,7,\dots,4n-1\}. Moreover, if n3n\geq 3, then for any x{1,5,,4n+1}x\in\{1,5,\dots,4n+1\} we have

M~fn(x)max{1,23+1n}=1,\widetilde{M}f_{n}(x)\leq\max\Big{\{}1,\frac{2}{3}+\frac{1}{n}\Big{\}}=1,

which is due to the fact that for any interval IxI\ni x we have |IA(0,4n+2)|23|I(0,4n+2)|.|I\cap A\cap(0,4n+2)|\leq\frac{2}{3}|I\cap(0,4n+2)|. Thus, for 𝒫n={1,3,,4n+1}\mathcal{P}_{n}=\{1,3,\dots,4n+1\} we have Var(M~fnM~f,𝒫n)2n1n↛0{\rm Var}\big{(}\widetilde{M}f_{n}-\widetilde{M}f,\mathcal{P}_{n}\big{)}\geq 2n\cdot\frac{1}{n}\not\rightarrow 0.

Acknowledgements.

Cristian González-Riquelme was supported by CAPES-Brazil. Dariusz Kosz was supported by the National Science Centre of Poland, project no. 2016/21/N/ST1/01496. The authors are thankful to Emanuel Carneiro for helpful discussions during the preparation of this manuscript.

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