On the variation of maximal operators of convolution type
aa r X i v : . [ m a t h . A P ] S e p ON THE VARIATION OF MAXIMAL OPERATORS OFCONVOLUTION TYPE
EMANUEL CARNEIRO* AND BENAR F. SVAITER
Abstract.
In this paper we study the regularity properties of two maximal op-erators of convolution type: the heat flow maximal operator (associated to theGauss kernel) and the Poisson maximal operator (associated to the Poisson ker-nel). In dimension d = 1 we prove that these maximal operators do not increasethe L p -variation of a function for any p ≥
1, while in dimensions d > L -variation. Similar results are proved for thediscrete versions of these operators. Introduction
Background.
Let ϕ ∈ L ( R d ) be a nonnegative function such that Z R d ϕ ( x ) d x = 1 . We let ϕ t ( x ) = t − d ϕ ( t − x ) and consider the maximal operator M ϕ associated to thisapproximation of the identity M ϕ f ( x ) = sup t> (cid:0) | f | ∗ ϕ t (cid:1) ( x ) . (1.1)The Hardy-Littlewood maximal function, henceforth denoted by M , occurs when weconsider ϕ ( x ) = (1 /m ( B )) χ B ( x ), where B is the d -dimensional ball centered at theorigin with radius 1 and m ( B ) is its Lebesgue measure. In a certain sense, one couldsay that M controls other such maximal operators of convolution type. In fact, if our ϕ admits a radial non-increasing majorant in L ( R d ) with integral A , from [15, ChapterIII, Theorem 2] we know that M ϕ f ( x ) ≤ A M f ( x ) (1.2)for all x ∈ R d and thus we obtain the boundedness of M ϕ from L p ( R d ) to L p ( R d ) if p >
1, and from L ( R d ) to L , ∞ ( R d ) in the case p = 1.Over the last years there has been considerable effort in understanding the effectsof the Hardy-Littlewood maximal operator M , and some of its variants, in Sobolevfunctions. In [9] Kinnunen showed that M : W ,p ( R d ) → W ,p ( R d ) is bounded for p >
1. The paradigm that an L p -bound implies a W ,p -bound was later extended toa local version of M in [10], to a fractional version in [11] and to a multilinear versionin [5]. The continuity of M : W ,p ( R d ) → W ,p ( R d ) for p > p = 1 the issues become more subtle. The question on whether Date : June 26, 2018.2000
Mathematics Subject Classification.
Key words and phrases.
Maximal functions, heat flow, Poisson kernel, Sobolev spaces, regularity,bounded variation, discrete operators.*corresponding author email: [email protected]. the operator f
7→ ∇
M f is bounded from W , ( R d ) to L ( R d ), posed by Haj lasz andOnninen in [7], remains open in its general case (see also [6]). Partial progress wasachieved in the discrete setting in the work [3] for dimension d = 1 and in the work[4] for general dimension d >
1. In the continuous setting the only progress has beenin dimension d = 1. For the right (or left) Hardy-Littlewood maximal operator, whichwe call here M r (corresponding to ϕ ( x ) = χ [0 , ( x ) in (1.1)) Tanaka [17] was the firstto observe that, if f ∈ W , ( R ), then M r f has a weak derivative and k ( M r f ) ′ k L ( R ) ≤ k f ′ k L ( R ) , (1.3)which led to the bound for the non-centered Hardy-Littlewood maximal operator f M , k ( f M f ) ′ k L ( R ) ≤ k f ′ k L ( R ) . (1.4)This was later refined by Aldaz and Perez-L´azaro [1] who obtained, under the assump-tion that f is of bounded variation on R , that f M f is absolutely continuous and V ( f M f ) ≤ V ( f ) , (1.5)where V ( f ) denotes here the total variation of f . More recently, in the remarkablework [12], Kurka showed that if f is of bounded variation on R , then V ( M f ) ≤ C V ( f ) , (1.6)for a certain C > M : W ,p ( R d ) → W ,p ( R d ) is a boundedoperator for p >
1, can be applied to a general M ϕ of type (1.1) that admits a radialnon-increasing integrable majorant. In this case, if f ∈ W ,p ( R d ) with p >
1, we havethat ∇M ϕ f exists in the weak sense and k∇M ϕ f k p ≤ C k∇ f k p , (1.7)for a certain constant C >
1. Therefore, from (1.2) and (1.7), we already know that M ϕ : W ,p ( R d ) → W ,p ( R d ) is bounded for p > L p -variation) of a function. In other words, canwe prove inequalities like (1.6) (for d = 1) and (1.7) (for p ≥
1) with the constant C = 1? To our knowledge, the only known variation diminishing bounds for maximaloperators are the ones given in (1.3) and (1.5), and the ones given in the recent workof Aldaz, Colzani and P´erez-L´azaro [2] that show that the Lipschitz constant of afunction (or H¨older constant) actually decreases under the action of the non-centeredHardy-Littlewood maximal function.Here we give a pool of affirmative answers to this question for two classical maximaloperators of convolution type: the heat flow maximal operator (associated to the Gausskernel) and the Poisson maximal operator (associated to the Poisson kernel). We shallobtain positive results for these two maximal operators in dimension d = 1 for any p ≥
1, and in dimensions d > p = 2 or ∞ . We consider also the discrete versionsof these operators and prove similar results. We start by reviewing the definitions andmain properties of these convolution kernels, and as we move on the proofs, we shallsee that the key idea to achieve these results is to explore the nice interplay betweenthe maximal function analysis and the structure of the differential equations associated N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 3 to these kernels (heat equation for the Gaussian and Laplace’s equation for the Poissonkernel). 2.
Main results
The continuous setting.
We start by reviewing the definitions and stating theresults in the context of the Euclidean space R d . All the results presented here holdfor complex-valued functions, but for simplicity (since the maximal operators only seethe absolute value of a function) we will work with real-valued functions.2.1.1. Heat flow maximal operator.
Given u ∈ L p ( R d ), 1 ≤ p ≤ ∞ , we define u : R d × (0 , ∞ ) → R by u ( x, t ) = ( u ∗ K t )( x ) , where K t is the heat kernel given by K t ( x ) = 1(4 πt ) d/ e −| x | / t . (2.1)In this case we know that u ∈ C ∞ ( R d × (0 , ∞ )) and solves the heat equation ∂ t u − ∆ u = 0 in R d × (0 , ∞ ) , with lim t → + u ( x, t ) = u ( x ) a . e . x ∈ R d . We consider here the maximal function associated to this heat flow, henceforth denotedby ∗ to facilitate the notation, u ∗ ( x ) = sup t> ( | u | ∗ K t )( x ) . (2.2)We shall prove the following regularity results for this maximal operator. Theorem 1.
Let u ∗ be the heat flow maximal function defined in (2.2) . The followingpropositions hold. (i) Let < p ≤ ∞ and u ∈ W ,p ( R ) . Then u ∗ ∈ W ,p ( R ) and k ( u ∗ ) ′ k p ≤ k u ′ k p . (ii) Let u ∈ W , ( R ) . Then u ∗ ∈ L ∞ ( R ) and has a weak derivative ( u ∗ ) ′ thatsatisfies k ( u ∗ ) ′ k ≤ k u ′ k . (iii) Let u be of bounded variation on R . Then u ∗ is of bounded variation on R and V ( u ∗ ) ≤ V ( u ) . (iv) Let d > and u ∈ W ,p ( R d ) , for p = 2 or p = ∞ . Then u ∗ ∈ W ,p ( R d ) and k∇ u ∗ k p ≤ k∇ u k p . E. CARNEIRO AND B. F. SVAITER
Poisson maximal operator.
Given u ∈ L p ( R d ), 1 ≤ p ≤ ∞ , we now define u : R d × (0 , ∞ ) → R by u ( x, y ) = ( u ∗ P y )( x ) , where P y is the Poisson kernel for the upper half-space given by P y ( x ) = c d y ( | x | + y ) ( d +1) / , (2.3)with c d = Γ (cid:0) d +12 (cid:1) π ( d +1) / . In this case we know that u ∈ C ∞ ( R d × (0 , ∞ )) and that it solves Laplace’s equation∆ u = 0 in R d × (0 , ∞ ) , where here we take the Laplacian with respect to the ( d + 1) coordinates of ( x, y ) =( x , x , ..., x d , y ), as opposed to the notation used in the heat flow case, when we wrotethe Laplacian only on the x -variable. Note also thatlim y → + u ( x, y ) = u ( x ) a . e . x ∈ R d . In other words, u ( x, y ) is the harmonic extension of u to the upper half-space. We con-sider here the maximal function associated to the Poisson kernel (henceforth denotedby the star ⋆ , slightly different from the heat flow case) u ⋆ ( x ) = sup y> ( | u | ∗ P y )( x ) . (2.4)With respect to this maximal operator we shall prove the following regularity results,in analogy with Theorem 1. Theorem 2.
Let u ⋆ be the Poisson maximal function defined in (2.4) . The followingpropositions hold. (i) Let < p ≤ ∞ and u ∈ W ,p ( R ) . Then u ⋆ ∈ W ,p ( R ) and k ( u ⋆ ) ′ k p ≤ k u ′ k p . (ii) Let u ∈ W , ( R ) . Then u ⋆ ∈ L ∞ ( R ) and has a weak derivative ( u ⋆ ) ′ thatsatisfies k ( u ⋆ ) ′ k ≤ k u ′ k . (iii) Let u be of bounded variation on R . Then u ⋆ is of bounded variation on R and V ( u ⋆ ) ≤ V ( u ) . (iv) Let d > and u ∈ W ,p ( R d ) , for p = 2 or p = ∞ . Then u ⋆ ∈ W ,p ( R d ) and k∇ u ⋆ k p ≤ k∇ u k p . N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 5
The discrete setting.
Again, all the results presented in this section hold forcomplex-valued functions, but for simplicity (since the discrete maximal operators onlysee the absolute value of a function) we will keep working with real-valued functions.For a bounded discrete function f : Z d → R we define ∂ x i f ( n ) := f ( n + e i ) − f ( n ) , where n ∈ Z d and e i = (0 , ..., , ...,
0) is the canonical i -th base vector with 1 in theposition i . The discrete gradient is then the vector ∇ f ( n ) = ( ∂ x i f ( n ) , ∂f x ( n ) , ..., ∂ x d f ( n )) . We let k f k p = X n ∈ Z d | f ( n ) | p /p , if 1 ≤ p < ∞ and k f k ∞ = sup n ∈ Z d | f ( n ) | . We define the l p ( Z d )-norms of the discrete function n
7→ |∇ f ( n ) | in an analogous way(here | · | will always denote the usual Euclidean norm in R d ). For a discrete function f : Z d → R we define its discrete Laplacian ∆ f : Z d → R as∆ f ( n ) = 12 d X k m − n k =1 (cid:2) f ( m ) − f ( n ) (cid:3) , where k n k = | n | + | n | + ... + | n d | , if n = ( n , n , ..., n d ) ∈ Z d .2.2.1. Discrete heat flow maximal operator.
For a bounded function u : Z d → R , the heat flow in Z d with initial condition | u | is the unique (bounded and C in time)solution u : Z d × [0 , ∞ ) → R of the differential equation ∂ t u ( n, t ) = ∆ u ( n, t ); u ( n,
0) = | u ( n ) | . This solution can be given in terms of a convolution with the discrete heat kernel (seefor instance [8] and the references therein) u ( n, t ) = (cid:0) | u | ∗ K t (cid:1) ( n ) = X m ∈ Z d | u ( n − m ) | K t ( m ) , where K t ( m ) = e − t d Y i =1 I m i ( t/d ) , for m = ( m , m , ..., m d ) ∈ Z d , and where I k , for an integer k ≥ z , isthe I -Bessel function defined by I k ( z ) = ∞ X j =0 ( z/ j + k j ! ( j + k )! = 1 π Z π e z cos θ cos kθ d θ, and for a negative integer k we put I k := I − k . We note here that I satisfies thedifferential equation I k +1 ( z ) + I k − ( z ) = 2 I ′ k ( z ) , E. CARNEIRO AND B. F. SVAITER which plainly implies that ∂ t K t ( m ) = ∆ K t ( m ) . Moreover, for any θ ∈ R and z ∈ C , we also have [8, Lemma 7] ∞ X k = −∞ (cid:2) e − z I k ( z ) (cid:3) e − iθk = e z (cos θ − , which gives kK t k l ( Z d ) = X m ∈ Z d K t ( m ) = 1 . We then define the discrete heat flow maximal operator by u ∗ ( n ) = sup t> u ( n, t ) = sup t> ( | u | ∗ K t (cid:1) ( n ) . (2.5)Our next result shows that this maximal operator does not increase the l p -variationin two situations: (i) in dimension d = 1, for any 1 ≤ p ≤ ∞ ; (ii) in dimension d > p = 2. Theorem 3.
Let u ∗ be the discrete heat flow maximal function defined in (2.5) . Thefollowing propositions hold. (i) Let ≤ p ≤ ∞ and u : Z → R be a bounded discrete function such that k u ′ k p < ∞ . Then k ( u ∗ ) ′ k p ≤ k u ′ k p . (ii) Let u : Z d → R be a bounded discrete function such that k∇ u k < ∞ . Then k∇ u ∗ k ≤ k∇ u k . Discrete Poisson maximal operator.
Recall that the continuous Poisson kernel P y ( x ) for the upper half-space defined in (2.3) satisfies the semigroup property P y ∗ P y = P y + y (2.6)for any y , y >
0, and its Fourier transform verifies c P y ( ξ ) = Z R n P y ( x ) e − πixξ d x = e − πy | ξ | = (cid:16)c P ( ξ ) (cid:17) y . Given a bounded discrete function u : Z d → R , we aim to lift this function to aharmonic function on the discrete upper half-space, i.e. we want to construct u : Z d × Z + → R such that ∆ u ( n, y ) = 0 in Z d × N ; u ( n,
0) = u ( n ) . (2.7)Observe that here we use the ( d + 1)-dimensional discrete Laplacian, and that theparameter y is now also discrete. We aim to accomplish this by convolving the initialdatum u with a certain integrable discrete kernel P y (also denoted below by P ( · , y ))that satisfies the semigroup property (2.6). Let us proceed with a formal derivation ofthe kernel P y first. N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 7
The function P : Z d × Z + → R would have to satisfy ∆ P ( n, y ) = 0 in Z d × N ; P ( n,
0) = δ ( n ) , (2.8)where δ is the function that is 1, if n = 0, and zero otherwise. Writing the harmonicitycondition at the level y = 1 we have2( d + 1) P ( n ) = P ( n ) + P ( n ) + X k m − n k =1 P ( m ) . We now multiply the last expression by e − πinξ and sum over n ∈ Z d to get2( d + 1) c P ( ξ ) = 1 + c P ( ξ ) + c P ( ξ ) d X k =1 (cid:0) e πiξ k + e − πiξ k (cid:1) , where ξ = ( ξ , ξ , ..., ξ d ) belongs to the torus T d = [ − , ] d . From the semigroupproperty we have c P ( ξ ) = c P ( ξ ) and thus c P ( ξ ) − c P ( ξ ) d + 1) − d X k =1 cos(2 πξ k ) ! + 1 = 0 . (2.9)From (2.9) we conclude that c P ( ξ ) = ( d + 1) − d X k =1 cos(2 πξ k ) ! ± ( d + 1) − d X k =1 cos(2 πξ k ) ! − / , (2.10)where the choice of signs could (at least in principle) be taken in any measurable way.We shall pick the negative sign in (2.10) and define our discrete Posson kernel bythe following three expressions: c P ( ξ ) := ( d + 1) − d X k =1 cos(2 πξ k ) ! − ( d + 1) − d X k =1 cos(2 πξ k ) ! − / , (2.11) c P y ( ξ ) := (cid:16)c P ( ξ ) (cid:17) y , (2.12)for y ≥
0, and P y ( n ) = Z T d c P y ( ξ ) e πinξ d ξ. (2.13)Observe from (2.11) that 0 < c P ( ξ ) ≤ ξ ∈ T d , with c P ( ξ ) = 1 if and only if ξ = 0. Thus, for ξ ∈ T d \ { } the function y c P y ( ξ ) is decreasing (and goes to 0 as y → ∞ ). From (2.11), (2.12) and (2.13) it is clear that P : Z d × Z + → R satisfies (2.8),and therefore the maximum principle holds.From (2.13) we have | P y ( n ) | ≤ n, y ) ∈ Z d × Z + . We want to show nowthat we also have P y ( n ) ≥ n, y ) ∈ Z d × Z + . For this we use the maximumprinciple. Given ε > y > (cid:12)(cid:12) P y ( n ) (cid:12)(cid:12) ≤ Z T d d P y ( ξ ) d ξ < ε E. CARNEIRO AND B. F. SVAITER for any n ∈ Z d . Now, by the Riemann-Lebesgue lemma, there exists a radius r > (cid:12)(cid:12) P y ( n ) (cid:12)(cid:12) < ε for all 0 ≤ y ≤ y and | n | ≥ r . When we consider the cylindrical contour of radius r ,delimited by the hyperplanes y = 0 and y = y , by the maximum principle we have P y ( n ) > − ε (2.14)within this region. Since we could take y as large as we wanted (and then r large aswell) we conclude that (2.14) actually holds for all ( n, y ) ∈ Z d × Z + . Since ε > P y ( n ) ≥ n, y ) ∈ Z d × Z + as desired.Since c P y (0) = 1 for all y ≥
0, it follows from (2.15) (and a standard approximationargument using the smoothing F´ejer kernel) that P y is integrable and X n ∈ Z d P y ( n ) = 1for all y ≥
0. Therefore, given a bounded discrete function u : Z d → R the function u ( n, y ) = u ∗ P y ( n ) = X m ∈ Z d u ( n − m ) P y ( m )is well defined, and one now clearly sees that it satisfies (2.7).We now define our discrete Poisson maximal operator (keeping the notation ⋆ ) by u ⋆ ( n ) = sup y ∈ Z + ( | u | ∗ P y )( n ) . (2.16)By a classical result of Stein [14, Theorem 1] the maximal operator ⋆ verifies the l p -boundedness k u ⋆ k l p ( Z d ) ≤ C p k u k l p ( Z d ) (2.17)for 1 < p ≤ ∞ . With respect to this maximal function we prove the following regularityproperties, in analogy with Theorem 3. Theorem 4.
Let u ⋆ be the discrete Poisson maximal function defined in (2.16) . Thefollowing propositions hold. (i) Let ≤ p ≤ ∞ and u : Z → R be a bounded discrete function such that k u ′ k p < ∞ . Then k ( u ⋆ ) ′ k p ≤ k u ′ k p . (ii) Let u : Z d → R be a bounded discrete function such that k∇ u k < ∞ . Then k∇ u ⋆ k ≤ k∇ u k . We now move on to the proofs of these results. We shall start with the discretecases, that are technically simpler but already give a good flavor of the main ideas thatshall be used in the continuous cases. The key insight here is that all of these maximalfunctions have the property of being subharmonic in the set where they disconnectfrom the original function. This shall be obtained by exploiting the structure of theunderlying partial differential equations.
N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 9 Proof of Theorem 3 - Discrete heat kernel
From now on we assume, without loss of generality, that u ≥
0, since k∇| u |k p ≤k∇ u k p for any 1 ≤ p ≤ ∞ .3.1. Preliminaries.
The essence of the following lemma is the fact that u ∗ is subhar-monic in the set where it disconnects from u . The statement in the following formatwill be more convenient later in the proof. Lemma 5.
Let u ∈ l ∞ ( Z d ) and u ∗ be its discrete heat flow maximal function. Let f ∈ l ∞ ( Z d ) be such that f ≥ u . Let I = { n ∈ Z d ; u ∗ ( n ) ≤ f ( n ) } and suppose that ∆ f ( n ) ≤ for all n ∈ I c . Then I = Z d .Proof. Suppose without loss of generality u is not identically zero. Let u : Z d × [0 , ∞ )be the heat flow in Z d with initial condition u . It suffices to prove thatsup t ≥ u ( n, t ) ≤ f ( n ) , for all n ∈ I c . It is easy to see that k u ( · , t ) k ∞ ≤ k u k ∞ for any t ≥
0. Therefore, | ∂ t u ( n, t ) | ≤ k ∆ u k ∞ ≤ k u k ∞ . (3.1)Take ε > T ε = sup { t ≥ u ( n, t ) ≤ f ( n ) + ε, ∀ n ∈ I c } . In view of (3.1) we have T ε ≥ ε k u k ∞ > . Suppose T ε < ∞ . For any n ∈ I c and 0 ≤ t ≤ T ε we have ∂ t u ( n, t ) = 12 d X k m − n k =1 (cid:2) u ( m, t ) − u ( n, t ) (cid:3) ≤ d X k m − n k =1 (cid:2) f ( m ) + ε − u ( n, t ) (cid:3) = f ( n ) + ε − u ( n, t ) + 12 d X k m − n k =1 (cid:2) f ( m ) − f ( n ) (cid:3) ≤ f ( n ) + ε − u ( n, t ) . Define y ( t ) = u ( n, t ) − ( f ( n ) + ε ). From the inequality above we have y ′ ( t ) ≤ − y ( t )and thus, for any t ∈ [0 , T ε ], y ( t ) ≤ e − t y (0) ≤ − εe − t ≤ − εe − T ε . Hence u ( n, T ε ) ≤ ( f ( n ) + ε ) − ε exp( T ε ) (3.2)for any n ∈ I c . Combining (3.2) with (3.1) we conclude that u ( n, t ) ≤ f ( n ) + ε for any t with T ε ≤ t ≤ T ε + ε k u k ∞ exp( T ε ) , which is in contradiction with the assumption that T ε < ∞ . Hence T ε = ∞ and u ( n, t ) ≤ f ( n ) + ε for all t ≥ n ∈ I c . To conclude the proof, take the limit ε → + . (cid:3) Proof of Theorem 3.
Step 1: Zorn’s lemma.
Recall that we are working here in the two cases: (i) d = 1 and1 ≤ p ≤ ∞ or (ii) d > p = 2. Consider the following family of functions: S = f : Z d → R ; u ( n ) ≤ f ( n ) ≤ u ∗ ( n ); ∀ n ∈ Z d ; k∇ f k p ≤ k∇ u k p . Note that S is non-empty since u ∈ S . We want to show ultimately that u ∗ ∈ S . Weput a partial order (cid:22) in S by considering the pointwise order (i.e. f (cid:22) g in S if andonly if f ( n ) ≤ g ( n ) for all n ∈ Z d ). Let us prove that ( S , (cid:22) ) is inductive, i.e. everytotally ordered subset has an upper bound in S . Let { f α } α ∈ Λ be a totally orderedsubset and define f ( n ) = sup α ∈ Λ f α ( n ) . We claim that f ∈ S . It is clear that u ( n ) ≤ f ( n ) ≤ u ∗ ( n ) for all n ∈ Z d . Let J ⊂ Z d be a finite set and define e J = { n ∈ Z d ; dist( n, J ) ≤ } (this distance is taken withrespect to the k · k -norm). There exists a sequence { f k } k ∈ N ⊂ { f α } α ∈ Λ such thatlim k →∞ f k ( n ) = f ( n )for all n ∈ e J . Thus, X n ∈ J (cid:12)(cid:12) ∇ f ( n ) (cid:12)(cid:12) p = lim k →∞ X n ∈ J |∇ f k ( n ) | p ≤ lim sup k →∞ k∇ f k k pp ≤ k∇ u k pp . Since this holds for any finite set J , we must have (cid:13)(cid:13) ∇ f (cid:13)(cid:13) p ≤ k∇ u k p , and thus f ∈ S . By Zorn’s lemma ( S , (cid:22) ) has (at least) one maximal element, whichwe call g . Step 2: Conclusion.
We now claim that g = u ∗ . Suppose this is not true and let I = { n ∈ Z d ; g ( n ) = u ∗ ( n ) } . By Lemma 5, we know that g cannot be superharmonicon I c , and thus there is a point n ∈ I c such that ∆ g ( n ) > d ≥ p = 2. Consider the function q n ( x ) = X k m − n k =1 ( g ( m ) − x ) . It is easy to see that x q n ( x ) is a strictly convex function with its unique minimizer x = x n given by x n = 12 d X k m − n k =1 g ( m ) = g ( n ) + ∆ g ( n ) . Therefore, if ∆ g ( n ) >
0, we can consider the function e g ( m ) = g ( m ) , if m = n ;min (cid:8) u ∗ ( n ) , g ( n ) + ∆ g ( n ) (cid:9) , if m = n. N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 11
Then g ≤ e g ≤ u ∗ pointwise and, since g ( n ) < e g ( n ) ≤ x n , the strict convexity of q n gives us q n ( g ( n )) > q n (cid:0)e g ( n ) (cid:1) ≥ q n ( x n ) . This plainly implies that k∇ g k = (cid:13)(cid:13) ∇ e g (cid:13)(cid:13) + q n ( g ( n )) − q n (cid:0)e g ( n ) (cid:1) > (cid:13)(cid:13) ∇ e g (cid:13)(cid:13) . Thus e g ∈ S and this contradicts the maximality of g .Now we deal with the case d = 1 and 1 ≤ p ≤ ∞ . The idea is the same as above. If1 ≤ p < ∞ we simply observe that the function (for fixed a, b ∈ R ) q ( x ) = | a − x | p + | b − x | p is convex (strictly convex if 1 < p ) with minimizer x = ( a + b ) /
2. If p = ∞ we notethat q ( x ) = max {| a − x | , | b − x |} is also convex with minimizer x = ( a + b ) /
2. This concludes the proof.4.
Proof of Theorem 4 - Discrete Poisson kernel
In this section we keep, without loss of generality, the assumption u ≥ k∇| u |k p ≤ k∇ u k p for any 1 ≤ p ≤ ∞ ).4.1. Preliminaries.
We start by proving the analogous statement to Lemma 5 for thediscrete Poisson maximal function.
Lemma 6.
Let u ∈ l ∞ ( Z d ) and u ⋆ be its discrete Poisson maximal function. Let f ∈ l ∞ ( Z d ) be such that f ≥ u . Let I = { n ∈ Z d ; u ⋆ ( n ) ≤ f ( n ) } and suppose that ∆ f ( n ) ≤ for all n ∈ I c . Then I = Z d .Proof. For y ∈ Z + we define β ( y ) = sup n ∈ Z d ( u ( n, y ) − f ( n )) + , where t + := max { t, } . Observe that β (0) = 0 and that, for any y , we have0 ≤ β ( y ) ≤ k u ⋆ k ∞ + k f k ∞ < ∞ and u ( · , y ) ≤ f + β ( y ) . (4.1)Suppose that I c = ∅ . In this case, β ( y ) = sup n ∈ I c ( u ( n, y ) − f ( n )) + . Take n ∈ I c and y >
0. Using the fact that u is harmonic in the discrete upperhalf-space, together with (4.1) and the hypothesis that ∆ f ( n ) ≤ n ∈ I c , we find2( d + 1) u ( n, y ) = u ( n, y −
1) + u ( n, y + 1) + X || m − n || =1 u ( m, y ) ≤ f ( n ) + β ( y −
1) + β ( y + 1) + X || m − n || =1 (cid:0) f ( m ) + β ( y ) (cid:1) ≤ d + 1) f ( n ) + β ( y −
1) + β ( y + 1) + 2 d β ( y ) . Therefore 2( d + 1) (cid:0) u ( n, y ) − f ( n ) (cid:1) ≤ β ( y −
1) + β ( y + 1) + 2 d β ( y ) . (4.2) Taking the supremum over n ∈ I c on the left hand side of (4.2) we find β ( y ) ≤ β ( y −
1) + β ( y + 1)2for any y >
0. Since the sequence y β ( y ) is bounded and β (0) = 0, we must have β ( y ) = 0 for all y >
0. Therefore I c would be empty, in contradiction with the originalassumption I c = ∅ . (cid:3) Proof of Theorem 4.
Once we have established Lemma 6, the proof of Theorem4 plainly follows by the argument based on Zorn’s lemma used in the proof of Theorem3 for the discrete heat flow maximal function. We will omit the details.5.
Proof of Theorem 1 - Continuous heat kernel
Preliminaries.
We begin this section with a selection of lemmas that will behelpful as we move on to the proof. Throughout this section we assume withoutloss of generality that u ≥
0, for if u ∈ W ,p ( R d ) we have | u | ∈ W ,p ( R d ) and |∇| u || = |∇ u | a.e. if u is real-valued (in the general case u complex-valued we have |∇| u || ≤ |∇ u | a.e), and if u is of bounded variation on R we have V ( | u | ) ≤ V ( u ).In what follows we write Lip( u ) = sup x,y ∈ R d x = y | u ( x ) − u ( y ) || x − y | for the Lipschitz constant of a function u : R d → R . Lemma 7. . (i) If u ∈ C ( R d ) ∩ L p ( R d ) , for some ≤ p < ∞ , then u ∗ ∈ C ( R d ) . (ii) If u is bounded and Lipschitz continuous then u ∗ is bounded and Lipschitzcontinuous with Lip( u ∗ ) ≤ Lip( u ) .Proof. (i) Recall that u ∗ ( x ) = sup t> ( u ∗ K t )( x )with the heat kernel K t defined in (2.1). Let us denote here τ h u ( x ) := u ( x − h ).Given ε >
0, there is a time t ε < ∞ such that | τ h u − u | ∗ K t ( x ) ≤ ( | τ h u − u | p ∗ K t ( x )) /p ≤ (cid:0) k τ h u − u k pp k K t k ∞ (cid:1) /p = k τ h u − u k p (4 πt ) d/ p ≤ k u k p (4 πt ) d/ p < ε whenever t > t ε , for all x, h ∈ R d . Note that we used Jensen’s inequality in the firstline above and Young’s inequality on the second line. On the other hand, given x ∈ R d and if 0 < t ≤ t ε , we can choose δ > | τ h u − u | ∗ K t ( x )= Z | y | < √ t ε | τ h u − u | ( x − y ) K t ( y ) d y + Z | y |≥√ t ε | τ h u − u | ( x − y ) K t ( y ) d y ≤ sup w ∈ B √ tε ( x ) | τ h u − u | ( w ) + k τ h u − u k p k χ {| y |≥√ t ε } K t k p ′ < ε N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 13 whenever | h | < δ , where we have used the fact that k χ {| y |≥√ t ε } K t k p ′ is bounded for0 < t ≤ t ε . Using the sublinearity, we then arrive at (cid:12)(cid:12) τ h u ∗ ( x ) − u ∗ ( x ) (cid:12)(cid:12) ≤ ( τ h u − u ) ∗ ( x ) ≤ ε for | h | < δ , which proves that u ∗ is continuous at x .(ii) It is easy to check that if u is bounded by M and has Lipschitz constant L , thenfor each time t > u ∗ K t is also bounded by M and admits the sameLipschitz constant L . In this case, the pointwise supremum of uniformly Lipschitzfunctions is still Lipschitz with (at most) the same constant. (cid:3) We will say here that a continuous function f is subharmonic in an open set A if,for every x ∈ A , and every ball B r ( x ) ⊂ A we have f ( x ) ≤ σ d − Z S d − f ( x + rξ ) d σ ( ξ ) , where σ d − denotes the surface area of the unit sphere S d − , and d σ is its surfacemeasure. Here B r ( x ) denotes the open ball of radius r and center x , and B r ( x ) denotesthe corresponding closed ball. Lemma 8 (Subharmonicity) . Let u ∈ C ( R d ) ∩ L p ( R d ) for some ≤ p < ∞ or u be bounded and Lipschitz continuous. Then u ∗ is subharmonic in the open set A = { x ∈ R d ; u ∗ ( x ) > u ( x ) } .Proof. Note that by Lemma 7 we have u ∗ continuous and thus the set A = { x ∈ R d ; u ∗ ( x ) > u ( x ) } is in fact open. Take x ∈ A and a radius r > B r ( x ) ⊂ A . Let h : B r ( x ) → R be the solution of the Dirichlet boundary valueproblem (cid:26) ∆ h = 0 in B r ( x ); h = u ∗ in ∂B r ( x ) . Since u ∗ is a continuous function, this problem does admit a unique solution h ∈ C ( B r ( x )) ∩ C (cid:0) B r ( x ) (cid:1) . Now let T > B r ( x ) × (0 , T ). Observe that v ( x, t ) := u ( x, t ) − h ( x ) ∈ C (Ω) ∩ C (cid:0) Ω (cid:1) and solves the heat equation in Ω. By themaximum principle for the heat equation, the maximum of v in Ω must be attainedeither in ∂B r ( x ) × [0 , T ] or in B r ( x ) × { t = 0 } . By construction, note thatmax ∂B r ( x ) × [0 ,T ] v ( x, t ) = max ∂B r ( x ) × [0 ,T ] u ( x, t ) − u ∗ ( x ) ≤ . Let y be such that max B r ( x ) v ( x,
0) = v ( y , . We claim that v ( y , ≤
0. In fact, let us suppose that v ( y , >
0. Then, by themaximum principle, v ( y , t ) ≤ v ( y ,
0) for any 0 ≤ t ≤ T , which in turn implies that u ( y , t ) ≤ u ( y ) for any 0 ≤ t ≤ T . Since T is arbitrary, we would have u ∗ ( y ) = u ( y )and thus y / ∈ A , contradiction. Thereforemax B r ( x ) v ( x, ≤ , which plainly gives u ( x , t ) ≤ h ( x )for any 0 ≤ t ≤ T . As T is arbitrary we conclude that u ∗ ( x ) ≤ h ( x ) , which is the desired result since h is harmonic and thus equal to its average over thesphere ∂B r ( x ), where h = u ∗ by construction. (cid:3) The next lemma will be important in the proof of the theorem for p = 2 and d ≥ Lemma 9.
Let f, g ∈ C ( R d ) ∩ W , ( R d ) with g Lipschitz. Suppose that g ≥ and that f is subharmonic in the open set J = { x ∈ R d ; g ( x ) > } . Then Z R d ∇ f . ∇ g d x ≤ . Proof.
Formally, the identity Z R d ∇ f . ∇ g d x = Z R d ( − ∆ f ) g d x would imply the result, since f subharmonic in the set { g > } would mean − ∆ f ≤ g has compactsupport. To see this let Ψ ∈ C ∞ c ( R d ) be a non-negative function such that Ψ( x ) ≡ B and supp(Ψ) ⊂ B . For N ∈ N , let Ψ N ( x ) := Ψ( x/N ) and consider g N ( x ) := g ( x ) Ψ N ( x ). If we could prove the result for each g N (note that f is subharmonic in theset { g N > } and each g N is Lipschitz since g is bounded), since g N → g in W , ( R d )we would obtain Z R d ∇ f . ∇ g d x = lim N →∞ Z R d ∇ f . ∇ g N d x ≤ . From now on we assume that supp( g ) ⊂ B R for some R > ϕ ∈ C ∞ c ( R d ) with support on the unit ball B andintegral 1. For ε > ϕ ε ( x ) = ε − d ϕ ( x/ε ) and write f ε = f ∗ ϕ ε . We see that f ε ∈ C ∞ ( R d ) and it is not hard to check that ∂ x i f ε = ( ∂ x i f ) ∗ ϕ ε = f ∗ ( ∂ x i ϕ ε ) , and ∂ x i x i f ε = ( ∂ x i f ) ∗ ( ∂ x i ϕ ε ) = f ∗ ( ∂ x i x i ϕ ε ) . (5.1)Let us define the set J ε = { x ∈ J ; dist( x, ∂J ) > ε } . A simple computation shows that f ε is subharmonic on J ε . In fact, if x ∈ J ε and B r ( x ) ⊂ J ε , we have f ε ( x ) = Z B ε f ( x − y ) ϕ ε ( y ) d y ≤ Z B ε σ d − Z S d − f ( x − y + rξ ) d σ ( ξ ) ϕ ε ( y ) d y = 1 σ d − Z S d − f ε ( x + rξ ) d σ ( ξ ) . Since f ε ∈ C ∞ ( R d ), this implies that ( − ∆ f ε )( x ) ≤ x ∈ J ε .For ε > ψ ∈ C ∞ c ( R d ) we can apply integration by parts to get Z R d ∇ f ε . ∇ ψ d x = Z R d ( − ∆ f ε ) ψ d x. N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 15
Now since |∇ f ε | ∈ L ( R d ) and ∆ f ε ∈ L ( R d ), we might approximate our function g ∈ W , ( R d ) by such ψ ∈ C ∞ c ( R d ) (in the W , -norm) to obtain Z R d ∇ f ε . ∇ g d x = Z R d ( − ∆ f ε ) g d x = Z J \ J ε ( − ∆ f ε ) g d x + Z J ε ( − ∆ f ε ) g d x ≤ Z J \ J ε ( − ∆ f ε ) g d x. (5.2)Let C be the Lipschitz constant of g . Then, for any x ∈ J \ J ε we have | g ( x ) | ≤ Cε. (5.3)From (5.1) we observe that ∂ x i x i f ε = ( ∂ x i f ) ∗ (cid:0) ε − ( ∂ x i ϕ ) ε (cid:1) , (5.4)where ( ∂ x i ϕ ) ε ( x ) = ε − d ( ∂ x i ϕ )( x/ε ). From (5.3) and (5.4), using H¨older’s inequalityand Young’s inequality we get Z J \ J ε (cid:12)(cid:12) ( − ∆ f ε ) g (cid:12)(cid:12) d x ≤ C Z J \ J ε d X i =1 (cid:12)(cid:12) ( ∂ x i f ) ∗ ( ∂ x i ϕ ) ε (cid:12)(cid:12) d x ≤ C d Z J \ J ε |∇ f | ∗ ( |∇ ϕ | ) ε d x ≤ C d Z J \ J ε (cid:12)(cid:12) |∇ f | ∗ ( |∇ ϕ | ) ε (cid:12)(cid:12) d x ! / m ( J \ J ε ) / ≤ C d k∇ f k k∇ ϕ k m ( J \ J ε ) / . (5.5)Since supp( g ) ⊂ B R , we know that m ( J \ J ε ) → ǫ →
0. Finally, since f ε → f in W , ( R d ) we use (5.2) and (5.5) to get Z R d ∇ f . ∇ g d x = lim ε → Z R d ∇ f ε . ∇ g d x ≤ lim ε → Z J \ J ε ( − ∆ f ε ) g d x = 0 . (cid:3) Lemma 10 (Reduction to the Lipschitz case) . In order to establish Theorem 1 - parts(i) and (iv) - it suffices to consider the initial datum u Lipschitz.Proof. If p = ∞ , recall that a function u ∈ W , ∞ ( R d ) can be modified on a set ofmeasure zero to become bounded and Lipschitz continuous.If 1 < p < ∞ , we take ε > u ε ( x ) = u ∗ K ε ( x ) . It is clear that u ε is Lipschitz continuous. Suppose that the result is true for u ε , i.e.that u ∗ ε ∈ W ,p ( R d ) and k∇ u ∗ ε k p ≤ k∇ u ε k p , (5.6)where u ∗ ε ( x ) = sup τ> u ε ∗ K τ ( x ) = sup t>ε u ∗ K t ( x ) . From Young’s inequality we have k u ε k p ≤ k u k p (5.7)and, together with Minkowski’s inequality, we also have k∇ u ε k p ≤ k∇ u k p (5.8)for any ε >
0. From (5.6), (5.7) and (5.8) we find that u ∗ ε is uniformly boundedin W ,p ( R d ). Note that u ∗ ε converges pointwise to u ∗ as ǫ →
0. From the weakcompactness of W ,p ( R d ) we then conclude that u ∗ ∈ W ,p ( R d ) and u ∗ ε ⇀ u ∗ as ε → k∇ u ∗ k p ≤ lim inf ε → k∇ u ∗ ε k p ≤ lim inf ε → k∇ u ε k p ≤ k∇ u k p , which completes the proof. (cid:3) Proof of part (iv).
In the case p = ∞ , we know that u ∈ W , ∞ ( R d ) canbe modified on a set of measure zero to become Lipschitz continuous with Lip( u ) ≤k∇ u k ∞ . From Lemma 7, u ∗ will also be bounded and Lipschitz continuous, withLip( u ∗ ) ≤ Lip( u ), and the result follows, since in this case u ∗ ∈ W , ∞ ( R d ) with k∇ u ∗ k ∞ ≤ Lip( u ∗ ).If p = 2, we are essentially done as well. In fact, from Lemma 10 it suffices toconsider the case u Lipschitz continuous in W , ( R d ). In this case, from Lemma 7we know that u ∗ is also Lipschitz continuous, and from Lemma 8 we have that u ∗ issubharmonic in the open set A = { x ∈ R d ; u ∗ ( x ) > u ( x ) } . Recall from our discussionin the introduction of the paper that we already have u ∗ ∈ W , ( R d ), and thus thehypotheses of Lemma 9 apply to f = u ∗ and g = ( u ∗ − u ). Therefore, k∇ u k = Z R d |∇ u | d x = Z R d |∇ u ∗ − ∇ ( u ∗ − u ) | d x = Z R d |∇ ( u ∗ − u ) | d x − Z R d ∇ u ∗ . ∇ ( u ∗ − u ) d x + Z R d |∇ u ∗ | d x ≥ Z R d |∇ u ∗ | d x = k∇ u ∗ k , which concludes the proof.5.3. Proof of part (i) - case < p < ∞ . Step 1: Set up.
The initial considerations are the same as before. From Lemma 10 itsuffices to consider the case u Lipschitz continuous in W ,p ( R ). From Lemma 7 weknow that u ∗ is also Lipschitz, and from Lemma 8 we have that u ∗ is subharmonic inthe open set A = { x ∈ R ; u ∗ ( x ) > u ( x ) } . Let us explore the structure of R to writethe open set A as a countable union of disjoint open intervals A = [ j I j = [ j ( α j , β j ) . Note also that subharmonicity is equivalent to convexity in each ( α j , β j ), when we dealwith continuous functions. N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 17
Step 2: Zorn’s lemma.
As we did in the discrete case, we use here an argument basedon Zorn’s lemma. Define the family of Lipschitz continuous functions S = f : R → R ; u ( x ) ≤ f ( x ) ≤ u ∗ ( x ); ∀ x ∈ R ;Lip( f ) ≤ Lip( u ); k f ′ k p ≤ k u ′ k p ;The family S is non-empty since u ∈ S . We put a partial order (cid:22) in S by consideringthe pointwise order (i.e. f (cid:22) g in S if and only if f ( x ) ≤ g ( x ) for all x ∈ R ). Let usprove that ( S , (cid:22) ) is inductive. Let { f α } α ∈ Λ be a totally ordered subset and define f ( x ) = sup α ∈ Λ f α ( x ) . We claim that f ∈ S . Being a pointwise supremum of uniformly Lipschitz functions,we have Lip( f ) ≤ Lip( u ). Let us write L = Lip( u ). For each N ∈ N consider the2 N + 1 points { j/N } with − N ≤ j ≤ N , j ∈ Z . For each of these j ′ s choose afunction f j,N ∈ Λ such that f ( j/N ) − f j,N ( j/N ) < N . Then choose f N = sup − N ≤ j ≤ N f j,N . It is clear that for each x ∈ [ − N, N ], since | x − j/N | ≤ / N for some j , we have f ( x ) − f N ( x ) ≤ N + LN .
Therefore f N → f pointwise as N → ∞ . From the conditions on the family S we knowthat k f N k W ,p is uniformly bounded, and from the weak compactness of W ,p ( R ) wemust have f ∈ W ,p ( R ) and f N ⇀ f . We then arrive at the bound (cid:13)(cid:13)(cid:0) f (cid:1) ′ (cid:13)(cid:13) p ≤ lim inf N →∞ k f ′ N k p ≤ k u ′ k p , which shows that f ∈ S . From Zorn’s lemma we guarantee the existence of (at least)one maximal element in ( S , (cid:22) ), which we call g . Step 3: Finding an appropriate segment to cut.
We want to show that g = u ∗ . Supposethis is not the case, i.e. that the open set B = { x ∈ R ; u ∗ ( x ) > g ( x ) } ⊂ A is non-empty.Let us write B as a countable union B = [ l Q l = [ l ( γ l , δ l ) . We claim that g cannot be superharmonic on B . In fact, if one of the intervals ( γ l , δ l )is finite, since u ∗ ( γ l ) = g ( γ l ) and u ∗ ( δ l ) = g ( δ l ), the maximum principle would give us u ∗ ≡ g in [ γ l , δ l ], a contradiction. If the interval is of type ( γ l , ∞ ) (resp. ( −∞ , δ l )),we would have ( u ∗ − g ) strictly positive and convex in ( γ l , ∞ ), with ( u ∗ − g )( γ l ) = 0.This is a contradiction since ( u ∗ − g ) is Lipschitz and belongs to L p ( R ), and thus musttend to zero at infinity. To conclude the proof of the claim, note that we cannot have B = ( −∞ , ∞ ), since u ( x ) = g ( x ) = u ∗ ( x ) at the global maximum x of u . Therefore, there exists an interval [ a, b ] ⊂ B such that g (cid:18) a + b (cid:19) < g ( a ) + g ( b )2 . (5.9)Let ℓ ( x ) be the equation of the line connecting the points ( a, g ( a )) and ( b, g ( b )), i.e. l ( x ) = g ( b ) − g ( a ) b − a ( x − a ) + g ( a ) . Let us consider the functions f u ∗ ( x ) := u ∗ ( x ) − ℓ ( x ) and e g ( x ) := g ( x ) − ℓ ( x ). Let y be the point of minimum of e g when restricted to [ a, b ]. From (5.9) we have e g ( y ) ≤ e g (( a + b ) / <
0. We claim that there exists a line e ℓ parallel to the x -axis such thatthe graph of f u ∗ is above e ℓ and the graph of e g is below e ℓ in a neighborhood of y . Tosee this start by noting that f u ∗ ( y ) − e g ( y ) = C >
0. For each − e g ( y ) > ε > a ε = max { a ≤ x ≤ y ; e g ( x ) ≥ e g ( y ) + ε } and b ε = min { y ≤ x ≤ b ; e g ( x ) ≥ e g ( y ) + ε } . From this we have e g ( x ) ≤ e g ( y )+ ε , for each x ∈ [ a ε , b ε ], with equality on the endpoints.Suppose that for each ε > z ε ∈ [ a ε , b ε ] such that f u ∗ ( z ε ) < e g ( y )+ ε .There will be a subsequence of { z ε } ε> that accumulates around a certain z ∈ [ a, b ]thus giving f u ∗ ( z ) ≤ e g ( y ) ≤ e g ( z ) < f u ∗ ( z ) , a contradiction. Therefore, we can find an ε > f u ∗ ( x ) ≥ e g ( y ) + ε for each x ∈ [ a ε , b ε ].If we undo the ∼ operation and return to the original picture, we have found afinite interval [ a ε , b ε ] such that g is below the line connecting ( a ε , g ( a ε )) to ( b ε , g ( b ε ))in [ a ε , b ε ] (being strictly below in ( a ε , b ε )) and u ∗ is above this line. Step 4: Conclusion.
Let us define a new function h ( x ) = g ( x ) , if x / ∈ [ a ε , b ε ] g ( b ε ) − g ( a ε ) b ε − a ε ( x − a ε ) + g ( a ε ) , if x ∈ [ a ε , b ε ] . From the previous step we know that u ≤ h ≤ u ∗ and it is also clear that Lip( h ) ≤ Lip( g ) ≤ Lip( u ). Finally, by Jensen’s inequality we obtain k g ′ k pp = Z [ a ε ,b ε ] c | g ′ ( x ) | p d x + Z [ a ε ,b ε ] | g ′ ( x ) | p d x ≥ Z [ a ε ,b ε ] c | g ′ ( x ) | p d x + ( b ε − a ε ) Z b ε a ε | g ′ ( x ) | d x ( b ε − a ε ) ! p ≥ Z [ a ε ,b ε ] c | g ′ ( x ) | p d x + ( b ε − a ε ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z b ε a ε g ′ ( x ) d x ( b ε − a ε ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p = Z [ a ε ,b ε ] c | g ′ ( x ) | p d x + ( b ε − a ε ) (cid:12)(cid:12)(cid:12)(cid:12) g ( b ε ) − g ( a ε )( b ε − a ε ) (cid:12)(cid:12)(cid:12)(cid:12) p N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 19 = Z R | h ′ ( x ) | p d x = k h ′ k pp . Therefore we get that h ∈ S but this is a contradiction since h is strictly bigger thanthe maximal element g in ( a ε , b ε ). This shows that g = u ∗ and the proof is concluded.5.4. Proof of part (ii).
The argument we shall use for this part is inspired in Tanaka’s[17]. Recall that when u ∈ W , ( R ), after adjusting on a set of measure zero, u maybe taken to be absolutely continuous. From Lemma 7 we see that u ∗ is also continuousand the set A = { x ∈ R ; u ∗ ( x ) > u ( x ) } is open. Let us again write A as a countableunion of disjoint open intervals A = [ j I j = [ j ( α j , β j ) . From Lemma 8 we know that u ∗ is subharmonic (thus convex) in each subinterval I j = ( α j , β j ). Therefore, u ∗ must be locally Lipschitz on each I j and, in particular,it is absolutely continuous on each compact subinterval of I j . From this we concludethat u ∗ is differentiable a.e. on each I j , with derivative that we will denote by v .We observe now that in each subinterval I j the variation of u ∗ is smaller than thevariation of u . In fact, since u ∗ is convex on I j , let γ j ∈ [ α j , β j ] be a minimum of u ∗ on [ α j , β j ] (note that we might have γ j = α j or γ j = β j ). The crucial observation isthat u ∗ is monotone in [ α j , γ j ] and in [ γ j , β j ], thus leading to (using the continuity of u ∗ and approaching by compacts from inside I j ) Z I j | v ( x ) | d x = (cid:2) u ∗ ( α j ) − u ∗ ( γ j ) (cid:3) + (cid:2) u ∗ ( β j ) − u ∗ ( γ j ) (cid:3) ≤ (cid:2) u ( α j ) − u ( γ j ) (cid:3) + (cid:2) u ( β j ) − u ( γ j ) (cid:3) ≤ Z γ j α j | u ′ ( x ) | d x + Z β j γ j | u ′ ( x ) | d x = Z I j | u ′ ( x ) | d x. (5.10)Note that in case α j = −∞ (resp. β j = ∞ ) we have u ( α j ) = 0 and u ∗ ( α j ) = 0 (resp. u ( β j ) = 0 and u ∗ ( β j ) = 0) due to the fact that u ∈ W , ( R ) and u ∗ ∈ L weak ( R ) andis convex on I j . In particular, since u ∈ W , ( R ), we have that v ∈ L ( A ).We shall prove now that u ∗ is weakly differentiable with( u ∗ ) ′ = χ A c u ′ + χ A v, (5.11)where χ A and χ A c denote the indicator functions of the sets A and A c . In fact, let ϕ ∈ C ∞ c ( R ). Observe first that Z I j u ∗ ( x ) ϕ ′ ( x ) d x = (cid:2) u ( β j ) ϕ ( β j ) − u ( α j ) ϕ ( α j ) (cid:3) − Z I j v ( x ) ϕ ( x ) d x, (5.12) obtained again by the continuity of u ∗ and a limiting argument approaching by com-pacts from inside I j . From (5.12) we have Z R u ∗ ( x ) ϕ ′ ( x ) d x = Z A c u ∗ ( x ) ϕ ′ ( x ) d x + Z A u ∗ ( x ) ϕ ′ ( x ) d x = Z A c u ( x ) ϕ ′ ( x ) d x + X j (cid:2) u ( β j ) ϕ ( β j ) − u ( α j ) ϕ ( α j ) (cid:3) − Z A v ( x ) ϕ ( x ) d x = Z A c u ( x ) ϕ ′ ( x ) d x + (cid:20)Z A u ( x ) ϕ ′ ( x ) d x + Z A u ′ ( x ) ϕ ( x ) d x (cid:21) − Z A v ( x ) ϕ ( x ) d x = − Z R u ′ ( x ) ϕ ( x ) d x + Z A u ′ ( x ) ϕ ( x ) d x − Z A v ( x ) ϕ ( x ) d x = − Z R ( χ A c u ′ + χ A v )( x ) ϕ ( x ) d x, as we wanted to show.We are now in position to conclude. From (5.10) and (5.11) we have k ( u ∗ ) ′ k = Z R | ( u ∗ ) ′ ( x ) | d x = Z A c | ( u ) ′ ( x ) | d x + Z A | v ( x ) | d x ≤ Z A c | ( u ) ′ ( x ) | d x + Z A | ( u ) ′ ( x ) | d x = k ( u ) ′ k . Proof of part (iii). If u has bounded variation, then u is bounded. Thedistributional derivative Du is a Radon measure with | Du | ≤ V ( u ), where | Du | denotes the total variation of Du . For ε > u ε ( x ) = u ∗ K ε ( x ) . Note that u ε ∈ C ∞ ( R ) is bounded and Lipschitz. We let u ∗ ε ( x ) = sup τ> u ε ∗ K τ ( x ) = sup t>ε u ∗ K t ( x ) . From Lemma 8 we know that u ∗ ε is subharmonic in the open set A = { x ∈ R ; u ∗ ε ( x ) >u ε ( x ) } . Let us write again A = [ j I j = [ j ( α j , β j ) , and thus we will have u ∗ ε convex in each I j = ( α j , β j ). Now consider a partition P = { x , x , ..., x N } . Refine this partition by including the endpoints α j and β j for which x k ∈ [ α j , β j ] for k = 1 , , ..., N . Thus we obtain a new partition P ′ = { y , ..., y M } ⊃ P , where M ≥ N . We now estimate the variation V P ( u ∗ ε ) of the function u ∗ ε with respectto the partition P by observing that V P ( u ∗ ε ) = N − X i =1 (cid:12)(cid:12) u ∗ ε ( x i +1 ) − u ∗ ε ( x i ) (cid:12)(cid:12) ≤ V P ′ ( u ∗ ε ) = M − X j =1 (cid:12)(cid:12) u ∗ ε ( y j +1 ) − u ∗ ε ( y j ) (cid:12)(cid:12) ≤ V ( u ε ) , (5.13) N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 21 where in the last inequality we used the convexity of u ∗ ε in each I j = ( α j , β j ) and thefact that u ∗ ε and u ε agree at the endpoints α j and β j (minor modifications are neededfor the cases α j = −∞ or β j = ∞ ). Since Du ε ( x ) = Du ∗ K ε ( x ) , from Young’s inequality we have V ( u ε ) = | Du ε | ≤ | Du | k K ε k = | Du | ≤ V ( u ) . (5.14)We now observe that as ε →
0, we have u ∗ ε → u ∗ pointwise. From (5.13) and (5.14)we find that V P ( u ∗ ) = N − X i =1 (cid:12)(cid:12) u ∗ ( x i +1 ) − u ∗ ( x i ) (cid:12)(cid:12) = lim ε → N − X i =1 (cid:12)(cid:12) u ∗ ε ( x i +1 ) − u ∗ ε ( x i ) (cid:12)(cid:12) ≤ V ( u ) . Since this holds for any partition P we obtain V ( u ∗ ) ≤ V ( u ) , and the proof is concluded.6. Proof of Theorem 2 - Continuous Poisson kernel
Throughout this proof we will again assume without loss of generality that u ≥ Preliminaries.
The first result of this section is analogous to Lemma 7.
Lemma 11. . (i) If u ∈ C ( R ) ∩ L p ( R d ) , for some ≤ p < ∞ , then u ⋆ ∈ C ( R d ) . (ii) If u is bounded and Lipschitz continuous then u ⋆ is bounded and Lipschitzcontinuous with Lip( u ⋆ ) ≤ Lip( u ) .Proof. Just follow the proof of Lemma 7. (cid:3)
We now investigate the set where u ⋆ disconnects from u . As in the heat kernelcase, we will show that u ⋆ is subharmonic in this set. The main tool for this will bethe structure of the underlying Laplace equation (namely, the mean value property). Lemma 12 (Subharmonicity II) . Let u ∈ C ( R ) ∩ L p ( R d ) for some ≤ p < ∞ or u be bounded and Lipschitz continuous. Then u ⋆ is subharmonic in the open set A = { x ∈ R d ; u ⋆ ( x ) > u ( x ) } .Proof. From Lemma 11 we see that u ⋆ is continuous and A is in fact an open set. Inwhat follows we keep denoting by B r ( x ) the open d -dimensional ball centered in x withradius r , and now we introduce B r ( x, y ) to denote the open ( d + 1)-dimensional ballcentered in ( x, y ) with radius r .Let x ∈ A . Since u ( x ) < u ⋆ ( x ) andlim y → + u ( x , y ) = u ( x ) , there exists δ = δ ( x ) > y < δ then u ( x , y ) < u ⋆ ( x ) − ( u ⋆ ( x ) − u ( x )) . (6.1) Now let y ≥ δ ( x ). Choose a radius 0 < r < δ such that B r ( x ) ⊂ A . For any r < r we have B r ( x , y ) ⊂ A × (0 , ∞ ). Recall that the function u ( x, y ) is harmonicin R d × (0 , ∞ ) and therefore, by the mean value property, we have u ( x , y ) = 1 r d +1 ω d +1 Z B r ( x ,y ) u ( x, y ) d x d y , (6.2)where ω d +1 denotes the volume of the ( d + 1)-dimensional unit ball. From (6.2) wearrive at u ( x , y ) ≤ r d +1 ω d +1 Z B r ( x ,y ) u ⋆ ( x ) d x d y = 1 r d +1 ω d +1 Z B r ( x ) p r − | x − x | u ⋆ ( x ) d x. (6.3)By (6.1) we know that u ⋆ ( x ) = sup y ≥ δ u ( x , y ) and since (6.3) holds for any y ≥ δ ,we have u ⋆ ( x ) ≤ r d +1 ω d +1 Z B r ( x ) p r − | x − x | u ⋆ ( x ) d x (6.4)for any r < r = r ( x ).Condition (6.4) can be viewed as a “weighted” subharmonicity. We shall provethat it actually implies subharmonicity in the usual sense. First observe that (6.4) issufficient to establish the maximum principle for u ⋆ in the domain A (in each connectedcomponent to be precise), i.e. if Ω is an open connected set such that Ω ⊂ A , and u ⋆ has an interior maximum in Ω then u ⋆ must be constant in Ω. With this in hand,consider any ball B s ( x ) ⊂ A and let h : B s ( x ) → R be the solution of the Dirichletboundary value problem (cid:26) ∆ h = 0 in B s ( x ); h = u ⋆ in ∂B s ( x ) . Consider the function g = u ⋆ − h . Let us prove that g satisfies the same local “weighted”subharmonicity (6.4) in B s ( x ). In fact, for a given x ∈ B s ( x ), we know that (6.4)holds in a neighborhood of x . Therefore, we can find a radius r such that B r ( x ) ⊂ B s ( x ) and (6.4) holds for r < r . Using the fact that h is harmonic and that ourweight is a radial function we arrive at g ( x ) = u ⋆ ( x ) − h ( x ) ≤ r d +1 ω d +1 Z B r ( x ) p r − | x − x | u ⋆ ( x ) d x ! − h ( x )= r d +1 ω d +1 Z B r ( x ) p r − | x − x | (cid:8) u ⋆ ( x ) − h ( x ) (cid:9) d x ! = r d +1 ω d +1 Z B r ( x ) p r − | x − x | g ( x ) d x ! . By the maximum principle, since g is continuous in B s ( x ), the maximum of g in B s ( x ) must be attained on the boundary. However, g = 0 in ∂B s ( x ), and therefore u ⋆ ( x ) ≤ h ( x ) , N THE VARIATION OF MAXIMAL OPERATORS OF CONVOLUTION TYPE 23 which shows that u ⋆ is subharmonic (in the usual sense, defined after Lemma 7) since h is harmonic and thus equal to its average over the sphere ∂B s ( x ), where h = u ⋆ byconstruction. (cid:3) Proof of Theorem 2.
Having established the subharmonicity of u ⋆ in the dis-connecting set (Lemma 12), the proof of Theorem 2 follows essentially in the sameway as the proof of Theorem 1, by observing that the Poisson kernel also satisfies thesemigroup property P y ∗ P y = P y + y to reduce to the Lipschitz case as in Lemma10. We shall omit the details. 7. Acknowledgements
We would like to thank the following colleagues for helpful remarks during thepreparation of this manuscript: William Beckner, Luis Caffarelli, Kevin Hughes, Mil-ton Jara, Diego Moreira, Roberto Oliveira, Lillian Pierce, Luis Silvestre and RalphTeixeira. E. C. acknowledges support from CNPq-Brazil grants 473152 / − / −
2, and FAPERJ grant E − / . / / −
5, 474944 / −
7, FAPERJ grant E − / . / References [1] J. M. Aldaz and J. P´erez-L´azaro, Functions of bounded variation, the derivative of the onedimensional maximal function, and applications to inequalities, Trans. Amer. Math. Soc. 359(2007), no. 5, 2443–2461.[2] J. M. Aldaz. L Colzani and J. P´erez-L´azaro, Optimal bounds on the modulus of continuity of theuncentered Hardy-Littlewood maximal function, J. Geom. Anal. 22 (2012), 132–167.[3] J. Bober, E. Carneiro, K. Hughes and L. B. Pierce, On a discrete version of Tanaka’s theorem formaximal functions, Proc. Amer. Math. Soc. 140 (2012), 1669–1680.[4] E. Carneiro and K. Hughes, On the endpoint regularity of discrete maximal operators, to appearin Math. Res. Lett.[5] E. Carneiro and D. Moreira, On the regularity of maximal operators, Proc. Amer. Math. Soc.136 (2008), no. 12, 4395–4404.[6] P. Haj lasz and J. Maly, On approximate differentiability of the maximal function, Proc. Amer.Math. Soc. 138 (2010), 165–174.[7] P. Haj lasz and J. Onninen, On boundedness of maximal functions in Sobolev spaces, Ann. Acad.Sci. Fenn. Math. 29 (2004), no. 1, 167–176.[8] A. Karlsson and M. Neuhauser, Heat kernels, theta identities, and zeta functions on cyclic groups,Topological and asymptotic aspects of group theory, Contemp. Math. 394 (2006), 177–189.[9] J. Kinnunen, The Hardy-Littlewood maximal function of a Sobolev function, Israel J. Math. 100(1997), 117–124.[10] J. Kinnunen and P. Lindqvist, The derivative of the maximal function, J. Reine Angew. Math.503 (1998), 161–167.[11] J. Kinnunen and E. Saksman, Regularity of the fractional maximal function, Bull. London Math.Soc. 35 (2003), no. 4, 529–535.[12] O. Kurka, On the variation of the Hardy-Littlewood maximal function, preprint athttp://arxiv.org/abs/1210.0496.[13] H. Luiro, Continuity of the maximal operator in Sobolev spaces, Proc. Amer. Math. Soc. 135(2007), no. 1, 243–251.[14] E. M. Stein, On the maximal ergodic theorem, Proc. Nat. Acad. Sci. U.S.A. 47 (1961), 1894–1897.[15] E. M. Stein,
Singular Integrals and Differentiability Properties of Functions , Princeton UniversityPress, 1970.[16] F. Temur, On regularity of the discrete Hardy-Littlewood maximal function, preprint athttp://arxiv.org/abs/1303.3993.[17] H. Tanaka, A remark on the derivative of the one-dimensional Hardy-Littlewood maximal func-tion, Bull. Austral. Math. Soc. 65 (2002), no. 2, 253–258.
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