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TO THE PROBLEM OF A STRONG DIFFERENTIABILITY OF INTEGRALS ALONG DIFFERENT DIRECTIONS

G. LEPSVERIDZE

Abstract. It is proved that for any given sequence (σn, n N) = Γ0Γ, where Γ is the set of all directions inR2(i.e., pairs of ortho- gonal straight lines) there exists a locally integrable functionfonR2 such that: (1) for almost all directions σ Γ\Γ0 the integral R

f is differentiable with respect to the family B of open rectangles with sides parallel to the straight lines fromσ; (2) for every direction σnΓ0the upper derivative ofR

fwith respect toBnequals +; (3) for every directionσΓ the upper derivative ofR

|f|with respect toB equals +.

§ 1. Statement of the problem. Formulation of the main result

Let B(x) be a differentation basis at the point x Rn (see [1]). The family{B(x) :x∈Rn}is called a differentiation basis inRn.

For f Lloc(Rn) and x Rn let us denote respectively by DB(f)(x) and DB(f)(x) the upper and the lower derivative of the integralR

f with respect toB at x[1]. When these two derivatives are equal their common value is denoted byDB(f)(x) and the basisB is said to differentiateR

f if the relationDB(f)(x) =f(x) holds almost everywhere.

LetB2 denote the differentiation basis in Rn consisting of all n-dimen- sional open intervals, and B2(x) be the family of sets fromB2 containing x.

Letσbe the union ofnmutually orthogonal straight lines inRn (n2) which intersect at the origin. The set of such unions will be denoted by Γ(Rn). Elements of this set will be called directions. Note that Γ(R2) corresponds in the one-to-one manner to the interval [0,π2) (see [2]).

For a fixed directionσwe denote byBthe differentiation basis in Γ(Rn) which is formed by alln-dimensional open rectangles with the sides parallel

1991Mathematics Subject Classification. 28A15.

Key words and phrases. Strong derivative of an integral, field of directions, negative result in the theory of differentiation of integrals.

157

1072-947X/98/0300-0157$15.00/0 c1998 Plenum Publishing Corporation

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to the straight lines fromσ. IfB differentiatesR

f atx, then the integral Rf is said to be strongly differentiable with respect to σatx.

The following problem was proposed by Zygmund (see [1], Ch. IV): Given a function f L(R2), is it possible to choose a direction σ such that R

f would be strongly differentiable with respect toσ?

Let W(Rn) (n 2) denote a class of locally integrable functions on Rn whose strong upper derivatives DB(f)(x) are equal to + almost everywhere along each fixed directionσ. When solving Zygmund’s problem, Marstrand [3] showed that the classW(R2) is not empty, and thus his answer to the above stated problem was negative. A stronger result was obtained by L´opez Melero [4] and Stokolos [5].

In connection with Zygmund’s problem we had the following question [2]: Given a pair of directions σ1 and σ2 differing from each other, does there exist an integrable function f such that the integral R

f is strongly differentiable a.e. with respect toσ1and strongly differentiable with respect toσ2on the null set only? Theorems 1 and 2 from [2] give a positive answer to this question.

It is known ([1], Ch. III) that if R

|f| is strongly differentiable almost everywhere, then the same holds for R

f. Papoulis [6] showed that the converse proposition does not hold in general. Namely, there exists an inte- grable functionf onR2 such that the integralR

f is strongly differentiable almost everywhere, whileR

|f|is strongly differentiable on the null set only.

Zerekidze [7] has obtained a stronger result from which it follows that for every function f from W(Rn) there exists a measurable function g such that|f|=|g| andR

g is strongly differentiable almost everywhere along all directions. In other words, changing the sign of the function on some set, we can improve the differentiation properties of the integral in all directions.

There arises a question whether the following alternative holds: Given function f from W(R2), can the differentiation properties of the integral Rf after changing the sign of the function be improved in all directions or they do not improve in none of them?

The following theorem gives a negative answer to this question and strengthens the results of Papoulis [6] and Marstrand [3].

Theorem. Let the sequence of directionsn)n=1 be given. There exists a locally integrable functionf onR2 such that:

(1)for almost all directions σ6=σn,n∈N), DB(f)(x) =f(x) a.e.;

(2)for every directionσn (n∈N),

DB2σn(f)(x) = + a.e.;

(3)

(3)for every directionσ,

DB(|f|)(x) = + a.e.

Remark. If the sequence (σn)n=1consists of a finite number of directions, then in item (1) instead of ”for almost all directions” it should be written

“for all directions”.

Corollary. There is a functionf ∈Lloc(R2)such that:

(a) the integral R

|f| is strongly differentiable a.e. in none of the direc- tions;

(b)for almost all irrational directions the integral R

f is strongly differ- entiable a.e., while for the rational directions it is strongly differentiable on the null set only.

§2. Auxiliary Assertions. Proof of the Main Result Before passing to the formulation of auxiliary assertions let us introduce some notation and definitions.

For the setG,G⊂R2, ∂G is assumed to be the boundary of the setG andGits closure. ByE we denote the unit square inR2.

Given a natural number n, let us construct two collections of straight lines: x=en1and y=en1,e= 0,1, . . . , n, which define the rectangular net En in the unit squareE and divide it into open square intervals Ekn, k= 1,2, . . . , n2, with sides of lengthn1.

For the rectifiable curvec denote byd(c) its length.

The set of measurable functions onRn taking only the values1 and 1 will be denoted byS(Rn).

For the measurable set G, G⊂R2, the number λ, 0< λ < 1, and the direction σ, denote byHσG, λ) the union of all those open rectangles R fromB for which

|R|1 Z

R

χG(y)dy≥λ,

where χG is the characteristic function of the set G. If, moreover, σ is a standard direction, then the setHσG, λ) will be denoted byHG, λ).

Furthermore, for the intervalI= (0, e1)×(0, e2) and the numbersλand c (0< λ <1, 1< c <∞) we define the intervalQ(I, λ, c) as follows:

Q(I, λ, c) =‚

−cλ1e1,(1 +1)e1

ƒ×‚

−e2,2e2

ƒ.

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Letσbe a fixed direction and letf ∈Lloc(Rn). In the present work we consider the following maximal Hardy–Littlewood functions:

MBf(x) = sup

RB(x)|R|1 Z

R

|f(y)|dy,

MBf(x) = sup

RB(x)|R|1

ŒŒ

ΠZ

R

f(y)dy

ŒŒ

Œ.

The validity of the following two assertions can be easily verified.

Lemma 1. Let 0 < ε <1 and nε= 9ε1. Let, moreover,c be a conti- nuous rectifiable curve in the unit squareEand letd(c)<1. Then for every natural number n,n≥nε, the following relation holds:

ŒŒ

Œ

kτn

EknŒŒŒ≤ε,

where τn is a collection of those natural numbers for which the square Ekn from the rectangular netEn intersects with the curvec.

Lemma 2. Letσ1 be a fixed direction fromΓ(R2). LetIσ1 be a rectangle fromB1 and B be a circle(on the plane). Then:

(1) ŒŒHσ1Iσ1, λ)ŒŒ> λ1ln(λ1)|Iσ1|, 0< λ <1;

(2)for every directionσ,

ŒŒHσB, λ)ŒŒ> λ1ln(λ1)|B|, 0< λ <21.

Lemma 3 (Zerekidze [7]). Let ε > 0. There exists a function s S(Rn)such that

ŒŒ

ΠZ

γ

s(x)d猌Œ< ε,

whereγis an arbitrary interval inRn anddηis the Lebesgue linear measure onγ.

Lemma 4 ([2]). Let I= (0, e1)×(0, e2)and letσ be an arbitrary non- standard direction fromΓ(R2). There exists a number c(σ),1< c(σ)<∞, such that for everyλ,0< λ <1,

HσI, λ)⊂Q(I, λ, c(σ)).

If, moreover,c(σ)λ1e1≤e2, then

ŒŒHσI, λ)ŒŒ9c(σ)λ1|I|.

Proof of the theorem. The proof of the theorem is divided into several parts.

1. We define some auxiliary sets.

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For any naturaln≥2 denote Γn

σ∈Γ(R2) : 0≤α(σ)<2(n+1)·n1‰

ˆ

σ∈Γ(R2) :π212(n+1)·n1< α(σ)< π21‰ ,1 cn= supˆ

c(σ) :σ∈Γ(R2)\Γn‰ , βn= max

š

exp(cnn222n); 2 nX1

n1=2

n1βn1+

n1

X

n1=2

λn1

‘›, (1)

λn= 2 n+

nX1 n1=2

(n1βn1+λn1

. (2)

LetIn = (0, en1)×(0, en2) be the interval for which en2 =cnn2nβnen1. Assume

Qn =Q(In,(n2nβn)1, cn).

Denote by Qn the interval with the same center of symmetry as Qn but with edges four times larger. Next, for e = 1,2, . . . , n denote by Ien, Qne, andQen those rectangles fromBe which are obtained from the intervals In, Qn, and Qn by rotation with respect to the center of symmetry (the centers of symmetry of the intervalsIn,Qn, andQn coincide).

Assume

Hen=HσeIn

e, βn1)∪Qen, e= 1,2, . . . , n, a2= 1/2, an=rn1(Mmnn11)1, n >2.

The sets Hen (e = 1,2, . . . , n) are compact. We use Lemma 1.3 from [1] and cover almost the whole unit square E by the sequence of non- intersecting sets H1jn, j = 1,2, . . ., homothetic to H1n such that all sets H1jn are contained in the unit square and have a diameter less thanan. By applying a similar treatment to the setsHen,e= 2, . . . , n, we obtain

ŒŒ

ŒE\

j=1HejnŒŒŒ= 0, diam(Hejn)≤an, e= 1,2, . . . , n, j= 1,2, . . . . (3) LetPejn (e= 1,2, . . . , n,j = 1,2, . . .) denote the homothety transforming the setHen toHejn. Assume

Iejn =Pejn(Ien), Qnej=Pejn(Qne) Qejn=Pejn(Qen).

1For the direction σ, the number 0 α < π2 is defined as the angle between the positive direction of the axisoxand the straight line fromσlying in the first quadrant of the plane.

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Denote bybn the circle with center at the point (21,21) and of radius rn, 0< rn<1. Choose a numberrnso small that the conditionsrn< rn1

andHbn, λn1)⊂Eare fulfilled.

For the direction σ denote the set Hσbn, λn1) by hn(σ) and for the standard direction use the notation hn =H(χbn, λn1). Let mn be a fixed natural number satisfying the condition

1≤mn|hn| ≤2. (4)

Let M1n, M2n, . . . , Mmnn be a collection of natural numbers such that M1n < M2n < · · · < Mmnn. Let us consider the rectangular nets EM1n, EM2n, . . . , EMmnn . Denote byqkin (k= 1,2, . . . , mn,i= 1,2, . . . ,(Mkn)2) the homothety transforming the unit square E to the square EM

n k

i from the rectangular netEMkn. Assume (σΓ(R2)),

Bkin =qkin(bn), hnki(σ) =qnki(hn(σ)), Gnk(σ) =(M

n k)2 i=1 hnki(σ), Ω2k(σ) = k1

k1=1G2k1(σ), k >1, Ωnk(σ) = n1

n1=2 mn1

k1=1Gnk11(σ) k1

k2=1Gnk2(σ), n >2, θk2=

kX1

k1=1

(Mk21)2, k >1,

θkn=

nX1 n1=2

mXn1

k1=1

(Mkn11)2+

k1

X

k2=1

(Mkn2)2, n >2, wkn= 2rn1Mkn1, k >1,

ωkn= 2θkn sup

σΓ(R2)

ˆ|E\nk(σ)|1‰ .

Choose numbers Mkn, k = 1,2, . . . , mn, increasing so rapidly that the relation

Mknmaxˆ

n1;d(∂hn); 9ωnk;Mmnn11;kn‰

(5) is fulfilled.

2. We shall construct the function sought for.

LetBkin be the circle with the same center as Bkin and a twofold larger radius. Assume

An1

1,2, . . . ,(M1n)2‰ ,

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Ank =

š

i:EiMk

E\ k1

k1=1 (Mkn1)2

i1=1 Bk1ni1‘‘

6

=

›

(k= 2, . . . , m2).

Let Sn S(R2). The functions ψn, gn, and fn will be defined for n= 2,3, . . ., as follows:

ψn(x) =βn

Xn e=1

Nn

X

j=1

χIn

ej(x), gn(x) =λnSn(x)

mn

X

k=1

X

iAnk

χBn

ki(x), fn(x) =gn(x) +ψn(x).

The index function is defined as the seriesf(x) =P

n=2fn(x).

Let us prove thatf ∈L(R2). We havekfk1P

n=2

€nk1+kgnk1 . First we estimatenk1. Using Lemma 2(a) and formula (1), we obtain

nk1ln1n) Xn e=1

X j=1

βnln(βn)|Iejn| ≤

ln1n) Xn e=1

ŒŒ

Œ

j=1HejnŒŒŒ=nln1n)<2n.

Similarly, applying Lemma 2(b) and relations (1), (2), (4) we have

kgnk1ln1n)Xmn

k=1 (Mkn)2

X

i=1

λnln(λn)|Bkni|‘

ln1n)mn|hn|<2n. The two latter relations yield the desired inclusion.

3. Here we shall prove that for almost all directionsσ6=σn, n∈N) the integralR

f is strongly differentiable a.e.

(a) Let us first estimate the maximal function MBgn. Introduce the sets

Bn = supp(gn) = mn

k=1

iAnkBkin, Bn= mn

k=1

iAnkBkin and letρn =Pmn

k=1(Mkn)2.

By Lemma 3 we can assume that sup

γ

ŒŒ

ΠZ

γ

sn(x)d猌Œ≤rn(2n+1λnρnMmnn)1, (6)

whereγis an arbitrary interval of an arbitrary straight line inR2andis the Lebesgue measure onγ.

(8)

Let us show that for every direction σ and for all x from R2\Bn the inequality

MBgn(x)2n, n= 2,3, . . . . (7) is fulfilled. This inequality will be proved only for the case where σ is a standard direction, since the general case has a similar proved.

Let us fix a natural number n (n 2), a point x from R2\Bn, and a intervalR fromB2(x). We assume thatR∩Bn6=and (k1, i) (1≤k1 mn, 1≤i≤(Mkn1)2) is a pair of natural numbers for which

R∩Bkn1i1 6=∅. (8)

From the inclusionx∈R2\Bn we have

dist(x, Bnk1i1)dist(∂Bkn1i1, Bkn1i1)≥rn(Mkn1)1≥rn(Mmnn)1. Taking also the inclusionx∈Rand (8) into consideration, we get

diam(R)≥rn(Mmnn)1.

LetR=R1×R2. It follows from the last relation that at least for one p(p= 1,2) the length of the intervalRp is underestimated as follows:

|Rp|121rn(Mmnn)1. (9) Without loss of generality we assume thatp= 1. We have (see (9), (6))

|R|1ŒŒŒ Z

R

gn(y)dyŒŒŒ≤λn|R|1

mn

X

k=1

X

iAnk

ŒŒ

ΠZ

R

sn(y)χBn

ki(y)dyŒŒŒ

≤λn|R|1

mn

X

k=1

X

iAnk

Z

R2

ŒŒ

ΠZ

R1

sn(y1, y2R1(y1, y2Bn

ki(y1, y2)dy1

ŒŒ

Œdy2

≤λn|R1|1ρnsup

γ

ŒŒ

ΠZ

γ

sn(y)d猌Œ2n.

To complete the proof of relation (7) it remains to note that R∈B2(x) and is arbitrary.

Let us now show that |B|= 0, whereB= limn→∞supBn. Indeed, using relations (4) and Lemma 2 (b), we obtain

X n=2

|Bn| ≤ X n=2

βn1ln1n)

mn

X

k=1 (Mkn)2

X

i=1

λnln(λn)|Bkin| ≤

4 X n=2

βn1ln1n)

mn

X

k=1

|hn| ≤8 X n=2

βn1ln1n)<∞.

(9)

Thus|B|= 0, and hence for everyx∈R2\B there exists a numberp1(x) such that

x∈R2\Bn for n≥p1(x). (10) This and inequality (7) imply that for every directionσ and for allxfrom R2\B the following relation is fulfilled:

MBgn(x)2n for n≥p1(x). (11) (b) We will now proceed to the estimation of MBψn. Taking into account Lemma 4, we find that each one of the following inclusions are fulfilled:

HσIn,(n2nβn)1)⊂Q(In,(n2nβn)1, c(σ))⊂Qn for σ∈Γ(R2)\Γn.

Without loss of generality, we assume that every directionσn,n∈N, is not standard. Suppose (k= 1,2, . . . , n),

Γnk

σ∈Γ(R2);|α(σ)−α(σk)|<2(n+1)n1‰ .

Since the rotation is a measure-preserving transformation and the centers of symmetry of the intervalsInandQncoincide, from the previous inclusion it follows (by virtue of the homothety properties) that the following inclusions hold:

HσIn

ki,(n2nβn)1)⊂Qnki for σ∈r(R2)\e=1n Γne, (12) k= 1,2, . . . , n, j = 1,2, . . . .

The definition of the rectangles Qnej andQejn immediately implies that for every directionσthere exists a rectangleEejn(σ)∈Bpossessing the prop- erty

Qnej⊂Eejn(σ)⊂Qejn. (13) We have |Qejn| = 16|Qnej| ≤ 144cnβn2nn|Iejn|. On the other hand, by Lemma 2 (a) we have

ŒŒHσeIn

ej, βn1)ŒŒ≥βnln(βn)|Iejn| ≥cnn222nβn|Iejn|. The last two relations imply

X n=2

ŒŒ

Πn

e=1

j=1QejnŒŒŒ144 X n=2

n12n Xn e=1

X j=1

|Hejn|=

= 144 X n=2

n12n Xn e=1

ŒŒ

Œ

j=1HejnŒŒŒ144 X n=2

2n <∞.

(10)

Hence|Q|= 0, whereQ= limn→∞sup n

e=1

j=1Qejn.

This in turn implies that for all pointsxfromE\Qthere exists a number P2(x) such that

x∈E\ n

e=1

j=1Qejn for n≥P2(x). (14) Further we have

ŒŒ

Œe=1n Γne

ŒŒ

ΠXn e=1

|Γne|= Xn e=1

n12n= 2n.

Consequently|Γ|= 0, where Γ = limn→∞supe=1n Γne.

This implies that for every directionσfrom Γ(R2)\Γ there exists a num- bern(σ) such that

σ∈Γ(R2)\ n

e=1Γne for n≥n(σ). (15) Now let us show that ifσ∈Γ(R2)\Γ and x∈E\Q, then

MBψn(x)2n for n≥P2(x, σ), (16) whereP2(x, σ) = maxˆ

P2(x);n(σ)‰ .

Indeed, let us fix a direction σ from Γ(R2)\Γ, a point x from E\Q and a rectangle Rσ from B(x). Let n be a fixed natural number and n≥P2(x, σ). Sinceσ∈Γ(R2)\Γ andn≥P2(x, σ)≥n(σ), it follows from (12), (13), and (15) that for all e,j (e= 1,2, . . . , n, j= 1,2, . . .) the chain of inclusions

HσIn

ej,(n2nβn)1)⊂Qnej⊂Eejn(σ)⊂Qejn (17) is fulfilled. Sincex∈E\Q andn≥P2(x, σ)≥P2(x), from (14) and (13) we have

x∈E\e=1n

j=1Qejn⊂E\e=1n

j=1Enej(σ).

Let{j1, . . . , jse} (e= 1,2, . . . , n) be a set of natural numbers for which

|Rσ ∩Iejni| > 0, i = 1,2, . . . , se. If we observe that the set Rσ ∩Eejni (i= 1,2, . . . , se, e= 1,2, . . . , n) is a rectangle fromB containing at least one point fromE\HσIn

ej,(n2nβn)1) (see (17)), then we obtain

ŒŒRσ∩Eejni(σ)ŒŒ1 Z

RσEn

eji(σ)

χIn

eji(y)dy=

=ŒŒRσ∩Eejni(σ)ŒŒ1ŒŒRσ∩IejniŒŒ(n2nβn)1,

(11)

and consequently

ŒŒRσ∩IejniŒŒ(n2nβn)1ŒŒRσ∩Eejni(σ)ŒŒ, e= 1,2, . . . , n, i= 1,2, . . . , se. Next, since the rectanglesQejn,j= 1,2, . . ., do not intersect for every fixed e(and hence the rectanglesEejn(σ),j= 1,2, . . .), we have

ŒŒ

Πse

i=1

€Rσ∩Eejni(σ)ŒŒŒ≤ |Rσ|, e= 1,2, . . . , n.

The two last relations yield

|Rσ|1 Z

Rσ

ψn(y)dy=βn|Rσ|1 Xn e=1

X j=1

ŒŒRσ∩IejniŒŒ

≤βn|Rσ|1 Xn e=1

se

X

i=1

(n2nβn)1ŒŒRσ∩Eejni(σ)ŒŒ=

=n12n|Rσ|1 Xn e=1

ŒŒ

Πse

i=1

€Rσ∩Eejni(σ)ŒŒŒ2n.

To complete the proof of (16), it remains to note thatRσ∈B(x) and is arbitrary.

(c) Let us show that for almost all directions σ the maximal function MBf is finite a.e. onR2. Suppose

P(x, σ) = max{P1(x);P2(x, σ)}.

Fix a direction σ from Γ(R2)\Γ, a point x from E\(Q∪B), and a rectangleRσ from B(x).

We have

|Rσ|1ŒŒŒ Z

Rσ

f(y)dyŒŒŒ

PX(x,σ)

n=2

|Rσ|1 Z

Rσ

|fn(y)|dy+

+|Rσ|1ŒŒŒ Z

Rσ

X

n=p(x;σ)+1

fn(y)dyŒŒŒ=a1(x, Rσ) +a2(x, Rσ)

and

a1(x, Rσ)

P(x,σ)

X

n=2

MBψn(x) +

P(x,σ)

X

n=2

MBgn(x)

P(x,σ)

X

n=2

nkL+

P(x,σ)

X

n=2

kgnkL

P(x,σ)

X

n=2

(nβn+mnλn).

(12)

Estimate nowa2(x, Rσ). Using the theorem on the passage to the limit under the integral sign as well as relations (7) and (16), we obtain

a2(x, Rσ) X

n=p(x,σ)+1

|Rσ|1ŒŒŒ Z

Rσ

fn(y)dyŒŒŒ

X n=p(x,σ)+1

(MBψn(x) +MBgn(x))2.

Hence

|Rσ|1ŒŒŒ Z

Rσ

f(y)dyŒŒŒ

p(x,σ)X

n=2

(nβn+mnλn) + 2<∞.

Since the right-hand side of this inequality does not depend on a choice of rectangles fromB(x), we get

MBf(x)<∞, σ∈Γ(R2)\Γ, x∈E\(Q∪B).

Consequently forσ∈Γ(R2)\Γ and forx∈E\(Q∪B) we have

−∞< DB(f)(x)≤DB(f)(x)<+∞.

Using now the Besicovitch theorem on possible values of upper and lower derivatives (see [1], Ch. V), we obtain (|Γ|=|Q∪B|= 0) and for almost every directionσ6=σn,n∈N) the relationDB(f)(x) =f(x) is fulfilled a.e.

4. It will now be shown that for every direction σs(s∈N) the strong upper derivative of the integral R

f is equal to +a.e. onE.

To this end we fix a natural number sand notice that |Js| = 1, where Js= limn→∞supNn

j=1HσsIn

sj, βn1). Indeed, 1 =ŒŒŒ lim

n→∞supNn

j=1HsjnŒŒŒŒŒŒ lim

n→∞supNn

j=1HσsIn

sj, βn1ŒŒŒ+ +ŒŒŒ lim

n→∞supNn

j=1QsjnŒŒŒ=|Js|+|Q|=|Js|. LetDs=n=2 Dsn, whereDns

y∈E:DB2s(fn)(y) =fn(y)‰ .

Since the basis B2 differentiates the integrals of the bounded functions (see [1], Ch. III), it is evident that|Dns|= 1 for everyn= 2,3, . . ..

Let us fix a pointxfromJs∩Ds\(Q∪B) and prove that

DB2σs(f)(x) = +∞. (18)

参照

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