COLLOQUIUM MATHEMATICUM

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1 COLLOQUIUM MATHEMATICUM VOL NO 2 ON WEAK TYPE INEQUALITIES FOR RARE MAXIMAL FUNCTIONS BY K HARE (WATERLOO,ON) AND A STOKOLOS (ODESSA ANDKANSASCITY,MO) Abstract The properties of rare maximal functions(ie Hardy Littlewood maximal functions associated with sparse families of intervals) are investigated A simple criterionallowsonetodecideifagivenraremaximalfunctionsatisfiesaconverseweaktype inequality The summability properties of rare maximal functions are also considered Introduction For a locally integrable function f : R d R the classic Hardy Littlewood maximal function M f is defined as Mf(x) = sup f(y) dy I x I where the supremum is taken over all bounded cubic intervals I R d containing x It is well known that the Hardy Littlewood maximal function does notmapfromltol, butonly fromltoweak L Inparticular, thefollowing weak type inequality holds for every f L and arbitrary positive : () c d {x:mf(x)>} f {x : Mf(x) > } C d I {x:mf(x)>} A well known theorem of Hardy and Littlewood [2] states that if f is supported on the unit cube I d and f Llog + L(I d ), then Mf L(I d ) Later, Stein [3] proved that the converse of this theorem also holds: if f L and Mf L(I d ), then f Llog + L(I d ) The proofs of these results are based on the weak type inequalities () stated above, and these, in turn, are proved by usingthe Vitali and Whitney covering lemmas Of course, the covering lemmas depend on properties of the family of sets to which the argument is applied, thus it is natural to consider araremaximal functionwherethesupremumis taken over a restricted set of intervals, and ask whether these inequalities and Stein s phenomenon remain true 2000 Mathematics Subject Classification: Primary 42B25 ResearchofKHaresupportedinpartbyNSERC [73] f

2 74 K HARE AND A STOKOLOS We shall see that this need not be the case Indeed, in this note we characterize the rare maximal functions which satisfy the weak type inequality We also show that Stein s phenomenon need not hold for individual functions f, but that rare maximal functions never map an entire Orlicz class larger than Llog + L into L 2 Weak type inequalities For simplicity we restrict ourselves to the one-dimensional case which is entirely typical Let l = {l k } where l k, l k 0, and let I = {intervals I R : I l} We define the rare maximal function M l f by the formula M l f(x) = sup I I, I x I I f(y) dy Define E {x : M l f(x) > } For every > 0, E is an open set and is the union of the intervals I such that I I f > Applying a Vitali covering argument to these intervals yields the weak type inequality E 2 f E The situation with the converse inequality is quite different Theorem Let l = {2 m k } with m k R + The rare maximal function M l f satisfies the weak type inequality {x:m l f(x)>} f C {x : M l f(x) > } for some constant C and every f L if and only if sup(m k+ m k ) < k Proof First, we will prove that if sup k (m k+ m k ) = then there exists a summable function f such that sup Ì {M l f>} f {x : M l f(x) > } = To do this we use the assumption that sup k (m k+ m k ) = to inductively define a subsequence {m kn } as follows: Choose k such that m k + m k > Given m k,,m kn, select k n so that m kn > +m kn + and

3 WEAK TYPE INEQUALITIES 75 m kn + m kn > n Let {α j } = {m k,m k +,m k2,m k2 +,} and set f(x) = a k χ [0;2 α k] (x) k= where a k > 0 will be specified later Observe that {2 α j } is a lacunary sequence Since f is a decreasing function for positive x, which vanishes for negative x, M l f( x ) M l f( x ) and M l f(x) decreases for positive x In particular, for x [2 α n+ ;2 α n ] we have M l f(x) M l f(2 α n ) Also, notice that [2 m k j + ;2 m k j + ] 2 m n for all n > k j, and that f is constant on the interval [2 m k j + ;2 m k j] It follows that if x [2 m k j + ;2 m k j] then m 2 kj M l f(x) = M l f(2 m k j) = 2 m kj These observations imply that and hence 0 f(y) dy [0;2 m k j + ] {x > 0 : M l f(x) > M l f(2 m k j)} [0;2 m kj + ], {x : M l f(x) > M l f(2 m k j)} 2 2 m kj + One can show by direct calculation (using the lacunarity of {2 α j }) that 2 α n 0 f a s 2 α n + s 2 s<n s na α s Thus if n is chosen such that m kj + = α n then {M l f(x)>m l f(2 m k j )} f 2 m k j + Furthermore, if we let = M l f(2 m k j), then Hence Ì {M l f>} f {x : M l f(x) > } n a s +2 m k j a s 2 α s s= 0 s=n C f C s na s 2 α s s n a s2 α s ( s<n a s +2 m k j s n a s2 α s )2 m kj + C s n a s2 α s (2 m kj+ m kj ) s<n 2 m k jas + s n a s2 α s

4 76 K HARE AND A STOKOLOS Set now a s = 2 α s /s 2 Then a s 2 α s = s=n Since α n = m kj we have 2 m k j n s= n a s = 2 m k j Consequently, for all j N, and hence s= s=n s 2 n 2 α s s 2 2 m k j 2 αn n 2 = n 2 = o Ì {M l f>} f {x : M l f(x) > } C2m k j+ m kj C2 j, sup Ì {M l f>} f {x : M l f(x) > } = ( ) n Now assume sup(m k+ m k ) c 0 < Note that E is a disjoint union of intervals J, so E = J For every such J there is an index k and intervals J and J such that J J J, J J, J = 2 m k, J = 2 m k+ But J E, hence J f J Since m k+ m k c 0 for every k, E J 2 c 0 J 2 c 0 f J 2 c 0 f = 2 c 0 f = 2 c 0 f J J E and this is the desired inequality 3 Stein s phenomenon Standard arguments show that any rare maximal function satisfying the weak type inequality of Theorem has Stein s property (cf [, 6]) In contrast, our next result shows that suitable rare maximal functions do not Theorem 2 There exists a sequence l such that for some f L(I), f Llog + L(I) but M l f L(I)

5 WEAK TYPE INEQUALITIES 77 Proof We will demonstrate that if {m k } N is a strictly increasing sequence satisfying supm k /k = k N and l = {2 (mk+3) }, then Stein s phenomenon does not hold for the rare maximal function M l f To see this, set E k = [0;2 m k ] and G k = E k \E k+ = (2 m k+ ;2 m k ] Also, set G + k = (2 m k+ ;(2 m k+ +2 m k )/2], G k = ((2 m k+ +2 m k )/2;2 m k ] Notice G k 2 m k, while G + k = G k = 2 G k Choose an increasing subsequence m ni of m k such that m ni 2 i n i Set a k = 0 if k {n i } and a k = /m ni if k = n i Define The function f belongs to L(I) since Now f(x) = k 2 m k a k χ G + (x) k k 2 mk a k G + k k a k i 2 i < Next we show that f Llog + L(I) For this we first observe that a k loga k log(2 i n i ) 2 i < n i a k 0 i f log + f = 2 m k a k log(2 m k a k ) G + k a k 0 a k m k log2+ a k loga k 4 4 a k 0 a k 0 But clearly the first series diverges while the second is convergent So f Llog + L(I) Now we estimate M l f Let x G k and I be any interval of length I = 2 mn 3 containing x If I G + k+, then since the interval G k is separated from G + k+ by G k+, it follows that I G k+ 2 mk+ 2 As I = 2 mn 3 for some n and I > 2 mk+ 3 this means that I 2 mk 3 Thus I I f = I 2 m k+3 ( j>k,j {n i } I G + j f + j k,j {n i } I G + j j>k,j {n i }2 mj a j G + j + j k ) f 2 m j a j G + j I I

6 78 K HARE AND A STOKOLOS 2 m k+3 2 m k+3 j>k,j {n i } j>k,j {n i } a j + max j k,j {n i } 2m j a j a j +2 m β(k) a β(k) where β(k) = n i if k [n i ;n i+ ) (with the final inequality holding because {2 m n i a ni } is an increasing sequence) Otherwise I G + j = for all j > k Then we have f = f 2 m j a j max I I j k,j {n i } 2m j a j I j k,j {n i } I G + j In either case, if x G k then M l f(x) 2 m k+3 j k,j {n i } j>k,j {n i } a j +2 m β(k) a β(k) Thus M l f = ( M l f 2 m k+3 a j +2 m β(k) a β(k) ) G k k G k k j>k,j {n i } 8 j k j>k,j {n i }a + 2 m β(k) a β(k) 2 m k k By switching the order of summation one can see that a j n j a nj < Moreover, k j>k,j {n i } k 2 mβ(k) a β(k) 2 mk = i j {n i }ja j = j 2 m n iani k [n i ;n i+ ) 2 m k 2 i 2 m n iani 2 m n i < Hence M l f L(I) Let us note that the assumption sup k (m k+ m k ) = is weaker than sup k m k /k = It would be very interesting to investigate whether the condition sup k m k /k = is sharp in Theorem 2 Unfortunately, we are not able to answer this question If the answer were affirmative it would be a very unexpected fact Finally, we show that in the scale of Orlicz classes Φ(L) it is possible to prove a weak form of Stein s phenomenon Indeed, we have the following theorem

7 WEAK TYPE INEQUALITIES 79 Theorem 3 Let l k 0 and Φ : [0; ) [0; ) be some increasing function such that Φ(L) L(I) If M l f L(I) for all functions f Φ(L)(I), then Φ(L) Llog + L(I) Proof Assume that Φ(L) L(log + L) This is equivalent to the assumption that for ψ(t) Φ(t) tlogt there exist b k as k such that ψ(b k ) +0 as k We will show how to construct a function f Φ(L) with M l f L Without loss of generality we may assume that l k =2 m k with m k N and m k Let r j (t), j = 0,,, denote the standard Rademacher functions and define E k = {t [0;] : r mj (t) = ; j = 0,,,k} Notice that E k is a union of dyadic-rational intervals of length 2 m k, E k = 2 k and E k E k+ Let G k = E k \E k+ By construction the sets G k are pairwise disjoint Clearly each G k is a union of dyadic-rational intervals of length 2 m k+ and has measure 2 k Finally, we define the function f(x) = a k χ Gk (x) k= where a k are positive numbers which will be specified later It is obvious that Φ(f) = Φ(a k ) G k = k )a k log(a k )2 k k ψ(a k We will now estimate from below the rare maximal function Let x G k and let I be the unique dyadic-rational interval which contains x and has length I = 2 m k Notice I is contained in E k The crucial fact is that due to the dyadic structure of G j the value of the fraction G j I I = 2 j +k does not depend on the concrete choice of m k for j k Thus f(y)dy = j χ Gj (y)dy I I I j= Ia G j I a j = j 2 I j k j ka j +k

8 80 K HARE AND A STOKOLOS This means M l f(x) j ka j 2 j +k χ Gk (x), and hence M l f j 2 k j ka j +k G k k a j 2 j +k 2 k j k Changing the order of summation we have M l f j 2 j k ja j 2 = a j 2 j 2 j j These calculations show that if we can find a k such that a k 2 k k = and k ψ(a k )a k log(a k )2 k <, k then f Φ(L) but M l f L(I) Without loss of generality we may assume that ψ(b k ) 2 k and b k+ 2b k The second assumption ensures that there exists a strictly increasing sequence {n j } of positive integers such that 2 n j n j b j < 2nj+ n j + Set a k = b j if n j k < n j+ Then it is easy to check that a k 2 k k diverges Furthermore, k )a k log(a k )2 k ψ(a k ψ(b j )b j log(b j ) 2 k j n j k But and ψ(b j ) 2 j, thus ( b j logb j C 2n j 2 n j ) log C2 n j n j n j k ψ(a k )a k log(a k )2 k C j ψ(b j ) < Corollary Let l k 0 and α be a positive number If M l f L for all functions f L(log + L) α, then α

9 WEAK TYPE INEQUALITIES 8 The theorem above shows that there are no conditions in terms of the growth of theindividualfunctionf, exceptthetrivial condition f Llog + L, which guarantee the summability of the rare maximal operator However, it is easy to see that such a condition may be found in terms of the integral smoothness of f Namely, assume that the function f is defined on the unit torus and introduce the modulus of continuity of f in the standard way: Then M l f(x) sup k ω(f;h) = sup f( +t) f( ) t h l k l k l k f(x+t) f(x) dt+f(x) f(x+t) f(x) dt+f(x) l k k l k Thus M l f l k f( +t) f( ) dt l k + f k l k 2 k l k l k 0 l k f( +t) f( ) dt+ f 2 k ω(f;l k )+ f Recall that thecase l k = 2 k correspondsto the Hardy Littlewood maximal function Mf So the condition (2) ω(f;2 k ) < k is sufficient for the summability of Mf for the individual function f Thus (2) implies that f Llog + L and this condition is sharpin thesense that for an arbitrary modulus of continuity ω(δ) with ω(2 k ) =, there exists a function f such that ω(f;δ) ω(δ), while f Llog + L (for details see [4]) Thus there is no improvement of the class of summability for the Hardy Littlewood maximal operator of smooth functions However, if l k is a more rare sequence, such that (2) is not true, but (3) ω(f;l k ) <, k then (3) is a sufficient condition for the summability of the rare maximal function for the individual function, which is weaker than the inclusion of f in Llog + L

10 82 K HARE AND A STOKOLOS REFERENCES [] MdeGuzmán,DifferentiationofIntegralsinR n,lecturenotesinmath48, Springer, 975 [2] GHHardyandJELittlewood,Amaximaltheoremwithfunction-theoretic applications, Acta Math 54(930), 8 6 [3] EMStein,NoteontheclassLlogL,StudiaMath32(969), [4] PLUl yanov,embeddingofsomefunctionclassesh ω p,izvakadnauksssr Ser Mat 32(968), (in Russian) Department of Pure Mathematics University of Waterloo Waterloo, Ontario N2L 3G Canada kehare@mathuwaterlooca Department of Mathematics Odessa State University Petra Velikogo, Odessa, Ukraine Current address: Department of Mathematics UMKC Kansas City, MO , USA stokolos@excitecom Received 5 May 999 (382)