ON DIFFERENTIAL BASES FORMED OF INTERVALS

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1 GEORGAN MATHEMATCAL JOURNAL: Vol. 4, No., 997, 8-00 ON DFFERENTAL ASES FORMED OF NTERVALS G. ONAN AND T. ZEREKDZE n memory of young mathematician A. ereashvili Abstract. Translation invariant subbases of the differential basis (formed of all intervals, which differentiates the same class of all nonnegative functions as does, are described. A possibility for extending the results obtained to bases of more general type is discussed.. Definitions and Notation A mapping defined on R n is said to be a differential basis in R n if, for every x R n, (x is a family of open bounded sets containing the point x such that there exists a sequence {R } (x, diam R 0 (. For f L loc (R n the numbers ( D f, x = lim f diam R 0,R (x R and ( D f, x = lim diam R 0,R (x R are said to be respectively the upper and the lower derivative of the integral of f at the point x. f the upper and the lower derivative coincide, then their common value is called ( the derivative of the integral of f at the point x, and we denote it by D ( f, x. They say that the basis differentiates the integral of f if D f, x = f(x for almost all x. The set of those functions f L loc (R n, f 0, whose integrals are differentiable with respect to the basis will be denoted by F +. Under M we mean the maximal operator M (f(x = sup R (x R R f (f L loc (R n, x R n, 99 Mathematics Subject Classification. 28A5. Key words and phrases. Differential basis, translation invariant bases, interval, locally regularity X/97/ $2.50/0 c 997 Plenum Publishing Corporation R R f

2 82 G. ONAN AND T. ZEREKDZE corresponding to the basis. t will be assumed here that = (x. x R n is said to be a subbasis of (writing, if (x (x (x R n. The basis is said to be translation invariant or the T -basis, if (x = {x + R : R (O} (x R n (here O is the origin in R n. Let us have the bases and. We shall say that the family is locally regular with respect to the family (writing LR(, if there exist δ > 0 and c > 0 such that for any R, diam R < δ, there is R such that R R and R < c R. We shall agree that n = [0, ] n and f L( n, if f L(R n and supp f n. 2. T -ases Formed of ntervals Let be the basis in R 2 for which (x (x R 2 consists of all twodimensional intervals containing the point x. The theorem below characterizes, T -bases for which F + = F +. Theorem. Let be a T -basis. Then the following conditions are equivalent: (a F + = F + ; (b for every f L(R 2, f 0, a.e. on R 2 ( D f, x ( = D 2 f, x (c is locally regular with respect to. ( ( and D f, x = D 2 f, x ; The implication (b (a is evident. Therefore to prove Theorem it suffices to show that (a (c and (c (b. The implication (a (c follows from the following assertion. Theorem 2. Let be a T -basis. f is not locally regular with respect to, then there exists a function f L( 2, f 0, such that ( D 2 f, x = a.e. on 2, ( D f, x = f(x a.e. on 2. efore proving Theorem 2 we shall give several lemmas. For the interval we denote by α (α > 0 the interval H(, where H is the homothety with the coefficient α whose center is the center of. Lemma. Let and h >. Then {M 2 (hχ > } (2h + and {M 2 (hχ > } h(ln h.

3 ON DFFERENTAL ASES FORMED OF NTERVALS 83 The validity of this lemma can be shown by a direct checing. Projections of onto the ox - and ox 2 -axes will be denoted by pr and pr 2, respectively. For and h > 0 we have t is clear that R(, h 8h. R (, h = (2h + pr 3 pr 2, R 2 (, h = 3 pr (2h + pr 2, R(, h = R (, h R 2 (, h. Lemma 2. Let be a T -basis; h >. Suppose that for there is no J, J, such that J h. Then {M (hχ > } R(, h. Proof. Let J and (/ J J hχ we can easily find that either >. From the condition of the lemma pr J < pr or pr 2 J < pr 2, ( where is the Lebesgue measure on R. y Lemma, J (2h +, which by virtue of ( implies that either J R (, h or J R 2 (, h, so that we have J R(, h. t follows from the latter inclusion that {M (hχ > } R(, h. t can be easily seen that the following two lemmas are valid. Lemma 3. Let, J ; h >. f either pr J h pr J or pr 2 J h pr 2 J, then h J R(, h. Lemma 4. Let, J. f J and J\R(, h, then either pr J pr, or pr 2 J pr 2. The following lemma is also valid. Lemma 5. Let ; h >, and let,..., intervals. f be the equal {M (hχ m > } R( m, h (m =,, R( m, h R( m, h = (m = m, then {M ( m= hχ m } > R( m, h. (2 m=

4 84 G. ONAN AND T. ZEREKDZE Proof. nequality (2 is equivalent to the following inequality: if x m= Let x R( m, h and (x, then m= m= hχ m. (3 R( m, h and (x. Then we may have three cases. Let us consider each of them separately. (i intersects none of the intervals m (m =, ; in this case the validity of (3 is evident. (ii intersects only one interval m (m =,. Let n be the interval for which n. x R( m, h, and therefore by the condition of the lemma, hχ n M (hχ n (x, from which, taing into account the equality m = (m n, we obtain (3. (iii intersects more than one interval m (m =,. t follows from the equality R( m, h R( m, h = (m m that (2h + m m = (4 Denote P = {m [, ] : m = }. y (4 we have which gives either m, \(2h + m = (m P, pr h pr m or pr 2 h pr 2 m (m P. Now according to Lemma 3 we can conclude that h m R( m, h (m P. From the obtained inequality, taing into consideration the pairwise nonintersection of R( m, h (m =,, we write m= hχ m = m P hχ m = m P h m m P Thus inequality (3 and Lemma 5 are proved. R( m, h.

5 ON DFFERENTAL ASES FORMED OF NTERVALS 85 Lemma 6. Let, h >, and for every m [, ] let { m,q } q m be a family of equal intervals. f {M (hχ m,q > } R m,q (R m,q = R( m,q, h; m =, q =, q m, R m,q R m,q = (m, q (m, q, (6 pr m, h pr m+,, pr 2 m, h pr 2 m+,, (5 (m =, (7 then {M ( q m hχ m,q m= } > 2 q m m= R m,q. Proof. The inclusion we have to prove is equivalent to the following inequality: if x q m R m,q and (x, then m= Assume x m= q m hχ m,q m= 2. (8 q m R m,q and (x. t is clear that (8 is fulfilled for m,q = (m =,, q =, q m. Denote otherwise n = min{m [, ] : m,q (q =, q m, m,q } and consider first the case with < n <. y virtue of Lemma 5 (see (5 and (6, we write which implies q n M ( q n hχ n,q = hχ n,q q n (x, h n,q. (9 For some q [, q n ] n,q, it is clear that \R n,q =. Therefore, according to Lemma 4, we have either pr pr n,q or pr 2 pr 2 n,q. Taing into account that the intervals { m,q } qm and using (7, we can write that either pr h pr m,q (n < m, q =, q m (m =, are equal

6 86 G. ONAN AND T. ZEREKDZE or pr 2 h pr 2 m,q (n < m, q =, q m, whence by Lemma 3 we get h m,q R m,q (n < m ; q =, q m. (0 Clearly, m,q = (n m < n; q =, q m, so that by (6, (9, (0 and m < n we have q m m= hχ m,q = + q n q m m= h n,q + = A + A 2 + A h m,q = q m m=n+ q m m=n+ n q m m= h m,q = h m,q + R m,q + = 2. nequality (8 can be proved in a simpler way for n = or n =, since in these cases we do not have the terms A and A 3. For the basis let us define the operator M(f(x = f (f L loc (R n, x R n. R sup R (x R Lemma 7. Let the basis differentiate the integrals of the functions f L(R n ( N, f <. f {M(f > λ } e <, where λ > 0 ( N, λ < and e is an outer measure, then differentiates the integral of the function f. Note that since M (f M (f (f L(R n, the conclusion of Lemma 7 will be the more so valid when the inequality {M (f > λ } e < is fulfilled. Proof. Let us note that: (i f < ; therefore the set A 0 = {x R n : f (x < } is of measure 0; (ii differentiates f ( N, i.e., the sets A ={x R n :D ( f, x= f(x} ( N have a complete measure on R n. (iii {M (f > λ } e < ; therefore the set lim {M (f > λ } is of measure 0.

7 ON DFFERENTAL ASES FORMED OF NTERVALS 87 t follows from (i (iii that the set A = A \ ( lim {M (f > λ } A 0 is of complete measure. Let us show that for every x A we ( have D f, x = f (x, which will prove the lemma. Let x A and ε > 0. From the inclusion x A we can conclude that: there is 0 N ( 0 = (x, ε such that x {M (f > λ }, = 0 = 0 λ < ε/3 and = 0 f (x < ε/3; 2 there is δ > 0 such that for every R (x, diam R < δ R 0 R 0 f f (x < ε/3. According to and 2 we can write that for every R (x, diam R < δ, 0 0 f f (x f R R f (x R R + + f + f (x < ε/3 + M R (f (x + = R 0 = 0 = 0 +ε/3 2ε/3 + λ < ε. = 0 Since ( ε > 0 is arbitrary, we conclude that the equality D f, x = f (x is valid. Proof of Theorem 2. Let 3. Choose h such that ln h Let α = ln h /2 2(+2 h. Obviously, 0 < α <. For and h > 0 denote Q 0 (, h = (2h +. Let 2 m < pr Q 0 (, h 2 m+ and 2 m < pr 2 Q 0 (, h 2 m +, where m, m N. Denote by Q(, h the interval concentric with Q 0 (, h, and pr Q(, h = 2 m+, pr 2 Q(, h = 2 m +. For the basis denote by M (r (r > 0 the operator M (r (f(x = sup f (f L(R n, x R n. R (x,diam R<r R LR(, since there is such that: there is no J, J for which J < 2 h ; 2 diam Q(, 2 h < /. Divide 2 into intervals equal to Q(, 2 h and denote them by Q,q ( q q. Tae,q ( q q equal to T q (, where T q is a shift translating Q(, 2 h to Q,q. R

8 88 G. ONAN AND T. ZEREKDZE Let the families {,q } q... { m,q} q m consisting of equal intervals be already constructed. Consider the sets A m = A 2 m = m q j {M (/ j= m q j j= (h, χ j,q > }, R j,q (R j,q = R( j,q, 2 h. f A m >, then we stop the construction. f A m, then we shall construct the family { m+,q } q m+ as follows: Consider the set A m = 2 = 2 \(Am A 2 m which, obviously, can be represented as G G 2, where G is open and G 2 consists of a finite number of smooth closed lines. t is clear that there is δ (0, / such that if we divide 2 into equal intervals {J j } with diameters less than δ, then A m ( α A m. 4 J j A m J j LR(, since there is such that: there is no J, J for which J 2 h ; 2 diam Q(, 2 h < δ; 3 pr m, 2 h pr and pr 2 m, 2 h pr 2. Divide 2 into intervals equal to Q(, 2 h. Denote the intervals included in A m by Q m+,q ( q q m+. Tae m+,q ( q q m+ equal to T q (, where T q is a shift translating Q(, 2 h to Q m+,q. y our construction we obtain q m+ ( A m Q m+,q y Lemma,- for q =, q m+ α 4 A m. ( ( {M 2 (h χ m+,q > } 2 h m+,q Q m+,q, (2 {M2 (h χ m+,q > } h > ln h m+,q. (3 Since diam Q m+,q < / (q =, q m+, because of (2 we have {M 2 (h χ m+,q > } = {M (/ (h χ m+,q > } (q =, q m+. (4

9 ON DFFERENTAL ASES FORMED OF NTERVALS 89 t can be easily seen that (see (3, (4 (/ {M (h χ m+,q > } α Q m+,q (q =, q m+, which by ( readily implies q m+ q m+ {M (/ (h χ m+,q > } α Q m+,q ( α α A m > α 4 2 A m. (5 According to the construction and Lemma 2 we have for q =, q m+ {M (h χ m+,q > /2 } R m+,q (R m+,q = R( m+,q, 2 h. On account of our choice of h, because of (3 and (4 we write {M (/ (h χ m+,q > } > 2 R m+,q (q =, q m+, whence q m+ q m+ (h χ m+,q > } > 2. R m+,q (6 {M (/ Let us show that for sufficiently large m the construction ceases, i.e., we shall have the inequality m m= q m {M (/ (h χ m+,q > } >. (7 Assume the contrary, i.e., A m (m N. ntroduce the notation A 3 m = A 4 m = q m {M (/ q m R m,q. (h χ m,q > }, (m N Clearly, (5 and (6 will be valid for all m N, and therefore we write A 3 m+ > α 2 A m = α ( 2 α ( 2 m j= ( + 2 Aj 3 α 2 m j= Aj 3 A 4 j ( 2 m j= Aj 3,

10 90 G. ONAN AND T. ZEREKDZE which by the equality A m = m j= A3 m implies that A m > 2 > for sufficiently large m. From this contradiction we conclude that (7 is valid. Consider the function f = m and f 0. From (7 we get m= qm h χ m,q. Clearly, f L( 2 {M (/ (f > } >. (8 From the construction we can easily see that 2 h and the families {,q } q,..., { m,q} qm satisfy all the conditions of Lemma 6. Therefore by Lemma 6 we write {M (f > /2 } = {M (2 f > 2} = ( m q m } m q m = {M h χ m,q > 2 R m,q. (9 m= m= Obviously, (6 is fulfilled for all m [, m ], so that taing the construction into account, we have m q m m q m 2 R m,q < {M (/ 2 (h χ > } m,q. (20 m= m= Due to (9 and (20 we write {M (f > /2 } < 2. (2 t can be easily seen that h ln h m q m m m,q < m= m= q m {M (/ Hence, due to our choice of h, we obtain m f = h q m m= (h χ > }, m,q / m,q < ln h < 2. (22 Consider the function f = =3 f. Clearly, f L( 2 (see (22 and f 0. ( (/ t is obvious that if x lim {M 2 (f > }, then D 2 =3 f, x = (/. ecause of (8 lim {M 2 (f > } has a complete measure on 2. Therefore the latter equality holds almost everywhere on 2.

11 ON DFFERENTAL ASES FORMED OF NTERVALS 9 t is clear that differentiates f ( N. Moreover, owing to (2 we get {M =3 (f > /2 } < =3 <, which by Lemma 7 2 implies that differentiates f. After maing some technical changes in the proof of Theorem 2 we can obtain the following generalization. Theorem 3. Let be a T -basis. f is not locally regular with respect to then for any function f L\L ln + L( 2, f 0, there is a Lebesgue measure-preserving and invertible mapping ω : R 2 R 2, {x : ω(x x} 2 such that ( D 2 f ω, x = a.e. on 2, ( D f ω, x = (f ω(x a.e. on 2. To prove the implication (c (b let us show the validity of Lemma 8. Let be a T -basis. f is locally regular with respect to, then for every f L(R 2 {M (r (f > λ} 25 { M (cr (f > λ 4c} (λ > 0, 0 < r < δ, where δ and c are the constants from the definition of local regularity of with respect to. Proof. Note first that if and ( are the one-dimensional intervals and is either the left or the right half of the interval, then for the point y the set E = {z R : ( + z y } possesses the following properties: E is an interval; 2 E = ; 3 E = 2 2. t follows from these properties that 4 5E is liewise valid. Let f L(R 2. Denote by P the set of all two-dimensional intervals { E such that E M (cr (f > λ } and show the inclusion 4c {M (r (f > λ} 5E (λ > 0, 0 < r < δ. (23 E P

12 92 G. ONAN AND T. ZEREKDZE Let x {M (r (f > λ} (λ > 0, 0 < r < δ. Then there is (x such that diam < r and f > λ. (24 Since LR(, there is an interval from such that and c. Then it is obvious that diam c diam < cr. Let y be the point for which (y. Divide the interval into four equal intervals. ecause of (24 the integral mean will be greater than λ at least on one of these intervals. Denote this interval by. Thus f > λ. (25 Consider the set E = {z R 2 : ( + z y }. y virtue of the above discussion we can easily note that E ; 2 E ; 3 E 4 ; 4 5E. Since x, then to prove (23 it suffices to show that E P. Let z E. Then ( + z y. Since is the T -basis and (y, we have ( + z y (z. Moreover, owing to (25, we have + z y +z y f f c f = c4 f > λ 4c, from which by the inequality diam( + z y < cr, we conclude that { z M (cr (f > λ }. Hence E P. Thus inclusion (23 is proved. 4c Now, using (23 and Lemma from [2], we will have (r {M (f > λ} 5E 25 E 25 {M (cr 4c} (f > λ for λ > 0, 0 < r < δ. E P E P The basis is said to be regular if there is a constant c < such that for every R there exists a cubic interval Q with the following properties: Q R, Q c R. According to the well-nown Lebesgue theorem on differentiability of integrals, the regular basis differentiates L, i.e., D ( f, x = f(x a.e. on R n for every f L(R n. We shall say that the basis possesses property (E if for every f L(R n, D ( f, x f(x D ( f, x a.e. on R n.

13 ON DFFERENTAL ASES FORMED OF NTERVALS 93 Obviously, the basis, containing a regular subbasis, possesses the property (E. t directly follows from the relation LR( that contains a regular subbasis, and hence possesses property (E. Now, to prove the implication (b (c, it suffices to consider the following assertion. Lemma 9. Let be a subbasis of with property (E. Suppose that there exist positive constants c, c 2, c 3 and δ such that for every f L(R 2, {M (r (f > λ} { c M (c 2r (f > λ } (λ > 0, 0 < r < δ. c 3 Then for every f L(R 2, f 0, ( ( D f, x = D 2 f, x ( and D f, x ( = D 2 f, x, a.e. on R 2. The proof of Lemma 9 is based on the well-nown esicovitch theorem on possible values of upper and lower derivative numbers (see [], Ch.V, 3 and is carried out analogously to the proof of Theorem from [2]. 3. On a Possibility of Extending Theorem to More General ases t should be noted that beyond the scope of T -bases Theorem becomes invalid. Moreover, even for bases which are similar enough to T -bases the local regularity of with respect to is insufficient for the equality F + = F + to be fulfilled. A basis will be called a T -basis if for every x R n, R (x and y R n there is a translation T such that T (R (y. The following theorem is valid. Theorem 4. There is a T -basis with the properties: is locally regular with respect to, and (O = (O; 2 there is a function f L( 2, f 0 such that ( D 2 f, x = a.e. on 2, ( D f, x = f(x a.e. on 2. Proof. Consider the sequences α, α > 0 and h, h > 0 with the properties α h <, ln h h =. (26

14 94 G. ONAN AND T. ZEREKDZE Let q = 2 m, where m N ( N and m. For every, we divide 2 into q 2 equal square interval and denote them by 2,q. The length of square intervals sides will be denoted by. For every,q 2 let us consider the square interval,q concentric with,q 2 having the side length δ = /(2h +. We shall assume q so that δ > + ( N. Let f = sup{α h χ,q : q =, q 2} ( N and f = f. Clearly, f 0. t can be easily checed that 2 q f = α h,q = α h which according to (26 implies 2 q (2h + 2 2,q < α, h α f f < <, h i.e., f L( 2. y Lemma for N, q =, q 2 we have {M 2 (α h χ,q > α } (2h +,q = 2,q, (27 and {M2 (α h χ,q > α } h ln h,q. Thus we write q 2 q 2 {M 2 (α h χ,q > α } = {M2 (α h χ,q > α } q 2 q 2,q 2 h ln h,q = h ln h (2h + 2 > ln h (28 9h for N. For every N let us consider those intervals 2 +,q (q =, q2 + which are contained in q 2 {M 2 (α h χ,q > α }. Denote their union by O. t can be easily seen that if {q } tends rapidly enough to, then O 2 q 2 {M 2 (α h χ,q > α } ( N. (29

15 ON DFFERENTAL ASES FORMED OF NTERVALS 95 Using the property of dyadic intervals, we can see that {O } is the sequence of independent sets, i.e., for every n 2 and for arbitrary pairwise different natural numbers, 2..., n, n m= O m = n m= O m. Owing to (26, (28, and (29, O > ln h 2 9h now orel Kantelly s lemma, we get =. Using lim O, has a complete measure in 2. (30 From (27 we have for N, q =, q 2 {M 2 (α h χ,q > α } = {M ( (α h χ,q > α }, which for N implies = q 2 O {M ( q 2 {M 2 (α h χ,q > α } = (α h χ,q > α } {M ( (f > α }. ecause of (30, from the latter inclusion we obtain lim {M ( 2 (f > α } has a complete measure in 2. (3 Since α, 0 (, we can easily see that if x lim {M ( ( 2 (f > α }, then D 2 f, x =. y virtue of (3 we conclude that the latter equality holds a.e. on 2. Assume to be fixed. For every,q (q =, q 2 let us consider the intervals J,q (pr J,q = 3 pr,q, pr 2 J,q = (0, and J,q 2 (pr J,q 2 = q 2 (0,, pr 2 J,q 2 = 3 pr 2,q. Denote G i = J,q i (i =, 2, G = G G2. t can be easily verified (see (26 that G < which implies that lim G has zero measure. ( Determine now an unnown basis. f x lim G Q 2 (R 2 \ 2 (Q is a set of rational numbers, then we put (x = (x. For x ( 2 \ lim G Q 2 choose (x such that if d is the length of the lesser

16 96 G. ONAN AND T. ZEREKDZE side of the interval x, and d (x [δ, /2] (for x lim G, (x is the number with the property x G for (x, then (x, while if d [δ, /2] for some (x, then (x iff either G = or G 2 =. t can be easily verified that for every and x R 2 there is a translation T, T ( (x; this implies that is the T -basis. The local regularity of with respect to is also easily checed. y proving the inclusion f F + the proof of the theorem will be completed. Denote A = { ( x R n : D f, x = f (x } ( N. Clearly, A ( N has a complete measure on R 2. Denote A = ( 2 ( A \ lim G Q 2. Since f(x = 0 for x 2, differentiates f on R 2 \ 2 ; A is the set of a complete measure on 2. Therefore to prove the inclusion f F +, it suffices to show the differentiability of f with respect to on the set A. Let x A and ε > 0. Find δ > 0 for which (/ f f(x < ε when (x, diam < δ. Consider (x (x for which 6 f < ε/2. (32 (x t is clear that f (x = (x f (x, and differentiates (x f at the point x. This implies that there is δ (0, (x/2 such that (x (x f f (x < ε/2, for (x, diam < δ. (33 Consider an arbitrary interval (x, diam < δ. For N let K +/2 d /2. Clearly, (x. y virtue of the inequality + < δ ( N and the condition x G ( (x, we have supp f = for (x, which gives f = 0 for (x. (34 Analogously, f = 0 if d < δ, while if δ d /2, then = 0 because supp f = by construction of the basis. Thus f = 0. (35 f Let is show that supp f 6 supp f for +. (36

17 ON DFFERENTAL ASES FORMED OF NTERVALS 97 To this end it is sufficient to consider the following easily verifiable facts: d /2 for + ; 2 let P be a set of vertices of the squares,q 2 (q =, q2. f the vertices of the interval J 2 belong to the set P, then J supp f = supp f ; 3 for every interval J 2, d J /2 there is an interval J J, J 6 J with vertices at the points of the set P. y (36 we can write that for +, f = α h supp f 6α h supp f = 6 f, which, owing to (32 and the inequality (x, implies f = f 6 f < ε/2. ( y virtue of relations (33 (35 and (37 (it should also be noted that f (x = 0 for (x, we have + f f (x = (x f + + f + (x (x (x f f (x + f (x < ε ε = ε. Due to the arbitrariness of x A and ε > 0, we conclude that D ( f, x = f(x for every x A. 4. Remars ( Let us consider one application of the obtained results. Let Z be the basis in R 2 for which Z (x (x R 2 consists of all intervals x, D 2 d D, where D and d are the lengths, respectively, of the greater and of the lesser side of. This basis was introduced by Zygmund and it was he who initiated (see [], Ch. V, 4 the study of the differential properties of Z. Morion showed (see [], Appendix V that for the integral classes Z behaves lie, i.e., Z does not differentiate a wider integral class than L ln + L. A question arises: does there exist in general a function whose integral does not differentiate and differentiates Z? From the above proven theorems the answer is positive. Z is the T -basis. The fact that LR( z is easily verified. y Theorem this implies that the strict inclusion F + F + Z holds. Moreover,

18 98 G. ONAN AND T. ZEREKDZE Theorem 3 implies that by perturbing the values ω of any function f L\L ln + L( 2, f 0, one can get the function f ω with the following properties: D 2 ( f ω, x = a.e. on 2, and Z differentiates f ω. Thus, for the integral classes the bases and Z behave similarly, while for individual nonnegative functions the basis Z behaves better than the basis. (2 The bases and are said to be positive equivalent ( + if for every f L(R n, f 0, D ( f, x = D ( f, x and D ( f, x = D ( f, x a.e. on R n (i.e., condition (b in Theorem means that the bases and are positive equivalent. is said to be a usemann Feller type basis (F -basis if for any R and x R we have R (x. We shall say that exactly differentiates ϕ(l (writing D(ϕ(L if differentiates ϕ(l and does not differentiate a wider integral class than ϕ(l. For integral classes the behavior of, F, T -bases was studied by Stooloc in [3], where he introduced the property (S and proved that if possesses the property (S, then D(L ln + L, and if does not possess this property, then D(L. One can easily see that ignoring the F property, this result remains valid. Thus, for the integral classes, the T -basis behaves lie or ( is the basis formed of square intervals. The Z -basis illustrates that an analogous fact does not hold for nonnegative individual functions and, generally speaing, if we combine Stooloc s assertion and Theorem, then we shall have: if LR(, then + ; 2 if LR( and possesses the property (S, then D(L ln + L and F + F + (strictly; 3 if does not possess the property (S, then D(L. (3 Let be a T -basis. Consider the intervals of the type = (0, x (0, x 2. Denote the set of points (x, x 2 by A. The set A indicates how rich the family is. One can easily prove the following criterion of local regularity of with respect to : ( LR( ( m, 0 N : A ([/m, /m [/m, /m for, > 0. (4 The basis is said to be invariant (H-basis with respect to homotheties if for every x R n and every homothety H centered at x we have (x = {H(R : R (x}. f is locally regular with respect to and, moreover, if δ is equal to, then is called regular with respect to. For the basis define the sets: R, = {r > :, pr = r pr 2 }, R 2, = {r > :, pr 2 = r pr }.

19 ON DFFERENTAL ASES FORMED OF NTERVALS 99 For the basis which is simultaneously the T - and H basis one can easily verify that ( LR( ( is regular with respect to ( m N : R i, [m, m m+, N, i =, 2. (5 Let ( = (R n be the basis in R n for which (x (x R n consists of all n-dimensional intervals containing the point x. Theorems 4 are also valid for (R n (n 3, T -bases. Proofs for the n-dimensional case are similar to those for the two-dimensional case. (6 Let δ 0,..., δn 0 (. Denote i,m = [(m δi, mδi (i =, n;, m N. Let be the basis in R n for which (x = { n i= i i,m : i n i= i i,m x; i, m i N. Such bases are sometimes called nets. We i call them N-bases. Let be the N-basis in R n (n 2. Denote by T the least T -basis containing. For the following analogue of Theorem is valid: the following conditions are equivalent: (i F + = F + ; (ii + ; (iii is locally regular with respect to T. Obviously, (ii (i, the implication (i (iii follows directly from Theorem. Thus it remains only to show that (iii (i. To this end, note that using Lemma of [2] one can easily obtain the inequality {M (r T (f > λ} 3 n { M (r (f > λ 2 n } (f L(R n ; λ, r > 0. From this and Lemma 5 we get the upper bound of the distribution function of M (r (f by means of the distribution function of M (cr (f. t easily follows from the relation LR( that contains a regular subbasis. Hence possesses the property (E. Next, we obtain the implication (iii (ii from Lemma 6. Note that for δ i = /2 (i =, n; N the relation + was proved earlier in [2]. (7 The basis constructed in Theorem 4 is not a F -basis, which is not a casual fact. n particular, the following assertion is valid: let ( = (R n be a F -basis and let LR(. Then +. To prove the above assertion we have to consider the following facts which easily follow from the usemann-feller property of the basis and from the relation LR(: (i contains a regular subbasis; (ii for every f L(R n, {M (r (f > λ} {M (cr (f > λ } (λ > 0, 0 < r < δ, c where c and δ are the constants from the definition of local regularity of with respect to.

20 00 G. ONAN AND T. ZEREKDZE From (i we have that possesses the property (E. Now, taing into account (ii and using Lemma 6, we can conclude that the relation + is valid. Acnowledgement The authors express their gratitude to the referee for his useful remars. References. M. de Guzmán, Differentiation of integrals in R n. Lecture Notes in Math. 48, Springer, T. Sh. Zereidze, Convergence of multiple Fourier Haar series and strong differentiability of integrals. (Russian Trudy Tbilis. Mat. nst. A. Razmadze 76(985, A. M. Stooloc, On the differentiation of integrals of functions from Lϕ(L. Studia Math. 88(988, (Received Authors address: Faculty of Mechanics and Mathematics. Javahishvili Tbilisi State University 2, University St., Tbilisi Republic of Georgia