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) = Γ0⊂Γ, 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 B2σ of open rectangles with sides parallel to the straight lines fromσ; (2) for every direction σn∈Γ0the upper derivative ofR
fwith respect toB2σnequals +∞; (3) for every directionσ∈Γ the upper derivative ofR
|f|with respect toB2σ 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 (n≥2) 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 byB2σthe 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 c1998 Plenum Publishing Corporation
to the straight lines fromσ. IfB2σ 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 DB2σ(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 directions(σn)∞n=1 be given. There exists a locally integrable functionf onR2 such that:
(1)for almost all directions σ (σ6=σn,n∈N), DB2σ(f)(x) =f(x) a.e.;
(2)for every directionσn (n∈N),
DB2σn(f)(x) = +∞ a.e.;
(3)for every directionσ,
DB2σ(|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=en−1and y=en−1,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 lengthn−1.
For the rectifiable curvec denote byd(c) its length.
The set of measurable functions onRn taking only the values−1 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 byHσ(χG, λ) the union of all those open rectangles R fromB2σ 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 setHσ(χG, λ) will be denoted byH(χG, 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) =
−cλ−1e1,(1 +cλ−1)e1
×
−e2,2e2
.
Letσbe a fixed direction and letf ∈Lloc(Rn). In the present work we consider the following maximal Hardy–Littlewood functions:
MB2σf(x) = sup
R∈B2σ(x)|R|−1 Z
R
|f(y)|dy,
MB∗2σf(x) = sup
R∈B2σ(x)|R|−1
Z
R
f(y)dy
.
The validity of the following two assertions can be easily verified.
Lemma 1. Let 0 < ε <1 and nε= 9ε−1. 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:
∪
k∈τn
Ekn≤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 fromB2σ1 and B be a circle(on the plane). Then:
(1) Hσ1(χIσ1, λ)> λ−1ln(λ−1)|Iσ1|, 0< λ <1;
(2)for every directionσ,
Hσ(χB, λ)> λ−1ln(λ−1)|B|, 0< λ <2−1.
Lemma 3 (Zerekidze [7]). Let ε > 0. There exists a function s ∈ S(Rn)such that
Z
γ
s(x)dη< 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,
Hσ(χI, λ)⊂Q(I, λ, c(σ)).
If, moreover,c(σ)λ−1e1≤e2, then
Hσ(χI, λ)≤9c(σ)λ−1|I|.
Proof of the theorem. The proof of the theorem is divided into several parts.
1◦. We define some auxiliary sets.
For any naturaln≥2 denote Γn=
σ∈Γ(R2) : 0≤α(σ)<2−(n+1)·n−1
∪
∪
σ∈Γ(R2) :π2−1−2−(n+1)·n−1< α(σ)< π2−1 ,1 cn= sup
c(σ) :σ∈Γ(R2)\Γn , βn= max
exp(cnn222n); 2 nX−1
n1=2
n1βn1+
n−1
X
n1=2
λn1
, (1)
λn= 2 nβn+
nX−1 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 Q∗n 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, andQ∗en those rectangles fromB2σe which are obtained from the intervals In, Qn, and Q∗n by rotation with respect to the center of symmetry (the centers of symmetry of the intervalsIn,Qn, andQ∗n coincide).
Assume
Hen=Hσe(χIn
e, βn−1)∪Q∗en, e= 1,2, . . . , n, a2= 1/2, an=rn−1(Mmn−n−11)−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
E\ ∞∪
j=1Hejn= 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) Q∗ejn=Pejn(Q∗en).
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.
Denote bybn the circle with center at the point (2−1,2−1) and of radius rn, 0< rn<1. Choose a numberrnso small that the conditionsrn< rn−1
andH(χbn, λ−n1)⊂Eare fulfilled.
For the direction σ denote the set Hσ(χbn, λ−n1) by hn(σ) and for the standard direction use the notation hn =H(χbn, λ−n1). 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 (σ∈Γ(R2)),
Bkin =qkin(bn), hnki(σ) =qnki(hn(σ)), Gnk(σ) =(M
n k)2 i=1∪ hnki(σ), Ω2k(σ) = k∪−1
k1=1G2k1(σ), k >1, Ωnk(σ) = n∪−1
n1=2 mn1
k1∪=1Gnk11(σ)∪ k−∪1
k2=1Gnk2(σ), n >2, θk2=
kX−1
k1=1
(Mk21)2, k >1,
θkn=
nX−1 n1=2
mXn1
k1=1
(Mkn11)2+
k−1
X
k2=1
(Mkn2)2, n >2, wkn= 2r−n1Mkn−1, k >1,
ωkn= 2θkn sup
σ∈Γ(R2)
|E\Ωnk(σ)|−1 .
Choose numbers Mkn, k = 1,2, . . . , mn, increasing so rapidly that the relation
Mkn≥max
2ηn−1;d(∂hn); 9ωnk;Mmn−n−11;kn
(5) is fulfilled.
2◦. We shall construct the function sought for.
LetB∗kin be the circle with the same center as Bkin and a twofold larger radius. Assume
An1 =
1,2, . . . ,(M1n)2 ,
Ank =
i:EiMk∗∩
E\ k−1
k1∪=1 (Mkn1)2
i1∪=1 Bk∗1ni1
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
i∈Ank
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 havekfk1≤P∞
n=2
kψnk1+kgnk1 . First we estimatekψnk1. Using Lemma 2(a) and formula (1), we obtain
kψnk1≤ln−1(βn) Xn e=1
X∞ j=1
βnln(βn)|Iejn| ≤
≤ln−1(βn) Xn e=1
∞∪
j=1Hejn=nln−1(βn)<2−n.
Similarly, applying Lemma 2(b) and relations (1), (2), (4) we have
kgnk1≤ln−1(λn)Xmn
k=1 (Mkn)2
X
i=1
λnln(λn)|Bkni|
≤
≤ln−1(λn)mn|hn|<2−n. The two latter relations yield the desired inclusion.
3◦. Here we shall prove that for almost all directionsσ (σ6=σn, n∈N) the integralR
f is strongly differentiable a.e.
(a) Let us first estimate the maximal function MB∗2σgn. Introduce the sets
Bn = supp(gn) = m∪n
k=1 ∪
i∈AnkBkin, B∗n= m∪n
k=1 ∪
i∈AnkBki∗n and letρn =Pmn
k=1(Mkn)2.
By Lemma 3 we can assume that sup
γ
Z
γ
sn(x)dη≤rn(2n+1λnρnMmnn)−1, (6)
whereγis an arbitrary interval of an arbitrary straight line inR2anddηis the Lebesgue measure onγ.
Let us show that for every direction σ and for all x from R2\B∗n the inequality
MB∗2σgn(x)≤2−n, 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\B∗n, 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\B∗n we have
dist(x, Bnk1i1)≥dist(∂Bk∗n1i1, 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|1≥2−1rn(Mmnn)−1. (9) Without loss of generality we assume thatp= 1. We have (see (9), (6))
|R|−1 Z
R
gn(y)dy≤λn|R|−1
mn
X
k=1
X
i∈Ank
Z
R
sn(y)χBn
ki(y)dy≤
≤λn|R|−1
mn
X
k=1
X
i∈Ank
Z
R2
Z
R1
sn(y1, y2)χR1(y1, y2)χBn
ki(y1, y2)dy1
dy2≤
≤λn|R1|−1ρnsup
γ
Z
γ
sn(y)dη≤2−n.
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→∞supB∗n. Indeed, using relations (4) and Lemma 2 (b), we obtain
X∞ n=2
|B∗n| ≤ X∞ n=2
β−n1ln−1(βn)
mn
X
k=1 (Mkn)2
X
i=1
λnln(λn)|B∗kin| ≤
≤4 X∞ n=2
βn−1ln−1(βn)
mn
X
k=1
|hn| ≤8 X∞ n=2
βn−1ln−1(βn)<∞.
Thus|B∗|= 0, and hence for everyx∈R2\B∗ there exists a numberp1(x) such that
x∈R2\B∗n for n≥p1(x). (10) This and inequality (7) imply that for every directionσ and for allxfrom R2\B∗ the following relation is fulfilled:
MB∗2σgn(x)≤2−n for n≥p1(x). (11) (b) We will now proceed to the estimation of MB2σψn. Taking into account Lemma 4, we find that each one of the following inclusions are fulfilled:
Hσ(χIn,(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)n−1 .
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:
Hσ(χIn
ki,(n2nβn)−1)⊂Qnki for σ∈r(R2)\e=1∪n Γne, (12) k= 1,2, . . . , n, j = 1,2, . . . .
The definition of the rectangles Qnej andQ∗ejn immediately implies that for every directionσthere exists a rectangleEejn(σ)∈B2σpossessing the prop- erty
Qnej⊂Eejn(σ)⊂Q∗ejn. (13) We have |Q∗ejn| = 16|Qnej| ≤ 144cnβn2nn|Iejn|. On the other hand, by Lemma 2 (a) we have
Hσe(χIn
ej, βn−1)≥βnln(βn)|Iejn| ≥cnn222nβn|Iejn|. The last two relations imply
X∞ n=2
∪n
e=1
∞∪
j=1Q∗ejn≤144 X∞ n=2
n−12−n Xn e=1
X∞ j=1
|Hejn|=
= 144 X∞ n=2
n−12−n Xn e=1
∞∪
j=1Hejn≤144 X∞ n=2
2−n <∞.
Hence|Q∗|= 0, whereQ∗= limn→∞sup ∪n
e=1
∞∪
j=1Q∗ejn.
This in turn implies that for all pointsxfromE\Q∗there exists a number P2(x) such that
x∈E\ ∪n
e=1
∞∪
j=1Q∗ejn for n≥P2(x). (14) Further we have
e=1∪n Γne
≤ Xn e=1
|Γne|= Xn e=1
n−12−n= 2−n.
Consequently|Γ|= 0, where Γ = limn→∞supe=1∪n 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
MB2σψn(x)≤2−n for n≥P2(x, σ), (16) whereP2(x, σ) = max
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 B2σ(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
Hσ(χIn
ej,(n2nβn)−1)⊂Qnej⊂Eejn(σ)⊂Q∗ejn (17) is fulfilled. Sincex∈E\Q∗ andn≥P2(x, σ)≥P2(x), from (14) and (13) we have
x∈E\e=1∪n ∞∪
j=1Q∗ejn⊂E\e=1∪n ∞∪
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 fromB2σ containing at least one point fromE\Hσ(χIn
ej,(n2nβn)−1) (see (17)), then we obtain
Rσ∩Eejni(σ)−1 Z
Rσ∩En
eji(σ)
χIn
eji(y)dy=
=Rσ∩Eejni(σ)−1Rσ∩Iejni≤(n2nβn)−1,
and consequently
Rσ∩Iejni≤(n2nβn)−1Rσ∩Eejni(σ), e= 1,2, . . . , n, i= 1,2, . . . , se. Next, since the rectanglesQ∗ejn,j= 1,2, . . ., do not intersect for every fixed e(and hence the rectanglesEejn(σ),j= 1,2, . . .), we have
s∪e
i=1
Rσ∩Eejni(σ)≤ |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
Rσ∩Iejni≤
≤βn|Rσ|−1 Xn e=1
se
X
i=1
(n2nβn)−1Rσ∩Eejni(σ)=
=n−12−n|Rσ|−1 Xn e=1
s∪e
i=1
Rσ∩Eejni(σ)≤2−n.
To complete the proof of (16), it remains to note thatRσ∈B2σ(x) and is arbitrary.
(c) Let us show that for almost all directions σ the maximal function MB∗2σf 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 B2σ(x).
We have
|Rσ|−1 Z
Rσ
f(y)dy≤
PX(x,σ)
n=2
|Rσ|−1 Z
Rσ
|fn(y)|dy+
+|Rσ|−1 Z
Rσ
X
n=p(x;σ)+1
fn(y)dy=a1(x, Rσ) +a2(x, Rσ)
and
a1(x, Rσ)≤
P(x,σ)
X
n=2
MB2σψn(x) +
P(x,σ)
X
n=2
MB2σgn(x)≤
≤
P(x,σ)
X
n=2
kψnkL∞+
P(x,σ)
X
n=2
kgnkL∞ ≤
P(x,σ)
X
n=2
(nβn+mnλn).
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|−1 Z
Rσ
fn(y)dy≤
≤ X∞ n=p(x,σ)+1
(MB2σψn(x) +MB∗2σgn(x))≤2.
Hence
|Rσ|−1 Z
Rσ
f(y)dy≤
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 fromB2σ(x), we get
MB∗2σf(x)<∞, σ∈Γ(R2)\Γ, x∈E\(Q∗∪B∗).
Consequently forσ∈Γ(R2)\Γ and forx∈E\(Q∗∪B∗) we have
−∞< DB2σ(f)(x)≤DB2σ(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 directionσ(σ6=σn,n∈N) the relationDB2σ(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→∞supN∪n
j=1Hσs(χIn
sj, β−n1). Indeed, 1 = lim
n→∞supN∪n
j=1Hsjn≤ lim
n→∞supN∪n
j=1Hσs(χIn
sj, β−n1+ + lim
n→∞supN∪n
j=1Q∗sjn=|Js|+|Q∗|=|Js|. LetDs=n=2∞∩ Dsn, whereDns =
y∈E:DB2,σs(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)