ANALOG OF AN INEQUALITY OF BOHR FOR INTEGRALS OF FUNCTIONS FROM Lp(Rn). I Текст научной статьи по специальности «Математика»

B. F.IVANQY

ANALOG OF AN INEQUALITY OF BOHR FOR INTEGRALS OF FUNCTIONS FROM Lp(Rn). I

Abstract. Let p € (2, +ro], n > 1 and A = (Ai,..., An), Afe > 0, 1 < k < n. It is proved that for functions y(t) € Lp(Rn) spectrum of which is separated from each of n the coordinate hyperplanes on the distance not less than Ak, 1 < k < n respectively, the inequality is valid:

y(t) dr

Et

< Cn(q)

n

1

A1/q k=1 Ak

IIy(t )y

LP (Rn)

where t = (ti,... ,tn) € Rn, Et = {t | t = (ti, ..., rn) € Rn,

Tj € [0,tj], if tj > 0, and Tj € [tj, 0], if tj < 0, 1 < j < n}, and

the constant C(q) > 0, —I— = 1 does not depend on y(t) and

pq

vector A.

Key words: Inequality of Bohr.

2010 Mathematical Subject Classification: 26D99.

Let us consider an arbitrary A > 0 and denote by P(A) the set of all finite trigonometric sums

N

p(t) =

m=l

the Fourier exponents of which satisfy the following condition

min \\m| > A.

1<m<N _

H. Bohr announced [1] and proved [2] that for such sums the next inequality, which later ([3-5] and etc.) was named after Bohr, is valid

n

\p(t) \ < 2A \\dP(t)/dt\\L^ (Ri) , P(t) e P (A)>

(g) Ivanov B. F., 2014

There exist several generalizations of this inequality - the details see in the author’s paper [6]. The author’s generalization was proposed in the article [6] mentioned above. Unlike other authors results in [6] the Bohr inequality was considered as the inequality which gives the estimation of the integral of the function via the norm of this function for some subclasses of the Lp(Rn) space for p E (1, 2]. The proof of the main results in [6] were significantly based on Hausdorff-Young inequality which did not permit to extend the obtained results to the case p > 2. In present paper the case p E (2, +ro] is considered.

Let p E (2, +ro], n > 1, A = (A1; A2,..., An) is any vector with positive coordinates and t = (t1 ,t2,... ,tn) E Rn.

Let us introduce the following designations:

n

1) Q(A) = U{y\y = (yi,V2,...,yn) E Rn, \yk\ < Ak}, i.e. Q(A) -

k=1

“cross” origin neighborhood in Rn;

2) r(Rn \ Q(A),p) - the set of all functions Y(t) E Lp(Rn), the Fourier transformations supporters of which are in Rn \ Q(A);

3) Et = {t \ t = (t1 ,t2,..., Tn) E Rn, Tj E [0, tj], if tj > 0, and Tj E [tj, 0], if tj < 0, 1 < j < n} - is the parallelepiped in Rn.

The main assertion of this paper is the theorem 3.2, section 3, which is the following.

Theorem 3.2. Let n > 1, p E (2, +ro] and A = (A1,..., An), Ak > 0, 1 < k < n. Then for any function j(t) E r(Rn \ Q(A),p) the next inequality is fulfilled:

Y(t)dT

Et

< Cn (q)

n

a 1/q Lk=1 Ak

IIY (t )\

LP(Rn)

(0.1)

where —I— = 1, and the constant C(q) > 0 does not depend on y(t) pq and vector A.

Let us note that if n =1 and p = then the inequality (0.1) may be obtained from [4] and [5].

The assertion of the theorem 3.2 for n = 1 was essentially employed by the author for the construction of the frequency criteria of the boundedness and smoothness in Frechet sense with respect to the parameters

1

of the ordinary differential equation systems solutions [7], and also for the boundedness of the nonlinear differential equations solutions [8].

The paper consists of the introduction and the three sections the first two of which are preliminary and the third contains theorem 3.2.

The proposed article contains §1, §2 and presents the first part of the paper; the main second part contains §3 and is prepared for publication.

The author expresses his deep gratitude to professor N. A. Shirokov for his attention to this research and valuable advices on it.

§1. Designations and Lemma on the Fourier Transformation Support of Several Functions Product

The section begins with the notation form choice of the Fourier transformation and the summary of several notations of the standard formulas following from such form. Thereafter the lemma 1.1 on the Fourier transformation support of several functions product in the case which is not suited for the direct employment of general theorems (for instance, [9, ch. 1, §5], [10, ch. 2, §7 ]) on the existence of resultant and on the support is proved. The assertion of lemma 1.1 will be used in §3.

Let n > 1. Following [11, p. 77], let us denote the Fourier transformation of the function by u(t) E L1 (Rn) as u(y), where y E Rn, but following [12, p. 425], let us choose u(y) as

also following [11, p. 77], let us designate as v(t), t E Rn, where v(t) according [12, p. 427] has the form:

Rn

The reciprocal Fourier transformation of the function v(y) E L1 (Rn),

Following [11, p. 73], [9, p. 31], let us denote the space of infinitely differentiable rapidly decreasing at infinity functions by S(Rn), and [11, p. 77] S'(Rn) — the space of slowly increasing distributions, i.e. [10,

p. 149], the space of slow growth distributions, i. e. the space of linear continuous functionals on S(Rn). According to [11, p. 73], [10, p. 15] let us denote the space of finite infinitely differentiable functions on Rn by D(Rn), and the space of linear continuous functionals on D(Rn) by D'(Rn).

Let us associate with each complex valued locally integrable function j(t), t £ Rn, the functional [9, p. 30, p. 32]

(Y,p) = J y(tMt)dt, p(t) £ D(Rn).

Rn

The distributions from D'(Rn), generated by locally integrable functions, are called [11, p. 75] the regular functions. Since D(Rn) is densely embedded in S(Rn) then S'(Rn) C D'(Rn). As is known [11, p. 77], for example, the functionals generated by functions y(t) £ Lp(Rn), p > 1, are the slowly increasing distributions.

The linear continuous functional on S(Rn), designated as y(v) and defined (with regard to the choice of the definition for (y, p) and the form of Fourier transformation) as

(Y, p) = (2n)n(y,P)

is called the slowly increasing function Fourier transformation.

According to the introduced designations we obtain the known formulas (see, for instance, [10, ch. II, §9] ) in the following form:

{yi (V) * Y2 (y)}~(t) = (2n)nYi(t) • Y2 (t)

{yi (V) • Y2 (V)}~(t) = Yi(t) * Y2 (t)

Let n > 2, 1 < k < n and Ak+1,..., An > 0. Let us designate

n

G(Ak+i,..., An) = U {V 1 V = (Vi ’...’Vn) £ Rn’ \Vp \ < AP }.

¡3=k+1

If n — k = 1 then G(An) is the direct product of Rn-i by the interval (—An, An). If n — k > 1 then G(Ak+1,..., An) is the direct product of Rk by “cross” neighborhood of zero in Rn-k = {y \ y = (yk+1,..., yn),

Vj £ R1, k + 1 < j < n}.

Lemma 1.1. Let n > 2, 1 < k < n, p £ [1, +to], the functions g\(6) £

£ Lq(R1), 1 + 1 = 1, 1 < A < k, y(t) £ Lp(Rn), A = (A1,..., An) - the P q

(1.1)

vector with positive coordinates and supp j(y) fl Q(A) = 0. Then ^supp j y(ti ,rk+1 ,...,rn) ]n 9x(rx ) jly^ P|

Pi G(Ak+i,..., An)=0.

Proof. Let us show that for any test function ((y) E S(Rn), such that

supp ((y) C G(Ak+i,An) (1.2)

the following equality is fulfilled

Y(r1 ,---,rk ,rk+1 ,...,rn) jn^A(rA ) j ly^^j =0,

which implies the lemma assertion.

Let us take the arbitrary test function ((y) E S(Rn) such that

supp ((y) C G(Ak+i,..., An),

and construct the sequence {(m(y)}^=1 of the test finite functions converging to ((y) in the sense of convergence in the space S(Rn) and satisfying for each m = 1, 2 ... the following conditions:

1) supp (m(y) c supp ifi(y), (1.3)

2) supp (m (y) C G(Ak+1,. . . , An) f{y \ y = (y1,. . .,yn), (1 4)

\yj \ < 3m, 1 < j < n}. ( . )

Following [9, p. 32], let us denote the infinitely differentiable function which is unit valued in the cube {y \ y = (y1,..., yn), \yj \ < 1, 1 < j < n}, and zero valued outside the cube {y \ y = (y1,... ,yn), \yj \ < 2, 1 < j <

< n} by e1(y), y E Rn and set

(m(y) = ((y) • em(y), em(y) = eA — yU m = 2, 3,...

m

Then:

supp (m (y) c supp ((y) f supp eA — y) . (1.5)

m

From (1.5) we obtain:

1) the assertion (1.3),

2) supp (m (y) c supp e^ m y^ = {y \ y = (y1 ,...,yn), \yj \ < 2m,

1 < j < n}c{y \ y = (y1,...,yn), \yj \ < 3m, 1 < j < n}. (1.6)

Since according assumption supp ((y) C G(Ak+1,..., An), then (1.4) follows from (1.3) and (1.6). Besides, the definition of the function em(y) implies that (m (y) = ((y) in the cube {y \ y = (y1,...,yn), \yj \< m, 1 < < j < n}.

Let r = (r1 ,r2,..., rn) be any vector with nonnegative integer coordinates. Then as m ^ to :

QT! +r2+---+vn f / 1 \

sy? ay? ...dyn (y1y ,-,yn) = D' |((y) e^ mv)

= {DV<v)} J - v) + 0<1)

\m J

mm Thus we obtain:

1) the (m(y) derivatives of any order in any bounded domain uniformly converge to corresponding derivative of the function ((y) as m ^ to;

2) since we may point out U(0, b) — the ball of radius b > 0 such that supp ((y) C U(0,b), then (1.3) implies that supp (m(y) C U(0, b) for any m = 1,2,... and consequently for any k = (k1,k2 ,...,kn), r = (r1 ,r2,..., rn), where k1,k2,..., kn, r1 ,r2,... ,rn are the nonnegative integer numbers:

\yk DrVm<v)\ =

ki kn kn

Vi V“2 •••Vn-ETTl

dri +r2 + -+rn Vm <V)

<

dvl1 dv22 ■■■dfn

< bkl +k2+-+kn\DrVm<v)\< Ckr,

moreover the constants Ckr may be chosen independent of m.

Thus the sequence Vi<v), V2<v), ■ ■ ■, Vm<V), • • • converges to V<v) in the sense of convergence in S<Rn).

Further since the inclusion <1.4) is strict and the set supp Vm<v) is closed and bounded, then at each m = 1, 2,... there exists the set of n — k positive numbers Pk+i<m),pk+2<m),..., pn<m) such that Aj > 3pj<m) > > 0, k +1 < j < n and

supp vm<v) C G<Ak+i — 3pk+i<m),..., An — 3pn<m))n

f{y \ y = (y1,...,yn), \y0 \ < ^, 1 < j < n}. (1.7)

Thus for any test function ((y) E S(Rn), satisfying (1.2) we have:

Y(r1,...,rk,rk+1,...,rn) TTgx(rx) My),((yn =

>•00

,rn) Y[gx(rx^^(y),((y)j ) Ilgx(rx^^(y),(m(y^ ,

= lim <7 (t1 ,..., Tk ,Tk+1,... , Tn H gx (rx )}^(y),(m (y) h (1.8)

where (m(y), m =1, 2,... satisfies (1.7).

In the equality (1.8) the set supp (m(y) is contained, according (1.7), in the set of some complex form which is the intersection of the cube in Rn with the edge 6m and the set G(Ak+1 — 3pk+1 (m),..., An — 3pn(m)). For each m = 1, 2, . . . let us transform the expression, which is under the integral in the right-hand side of (1.8), changing the function (m(y) by the sum of functions the support of which is contained in the set more simple in some sense, namely in the parallelepiped which is the direct product of the cube in Rn_1 with the edge 6m by some small interval specified for each parallelepiped.

Let a,b > 0, m > 1 and k be from the lemma assumption. Let us introduce some designations:

1) q(0, a,b), 0 E R1 is the auxiliary infinitely differentiable function, satisfying the following conditions: n(0, a, b) = 1, if \0\ < a, and n(0, a, b) = 0, if \0\ > a + b;

2) (m(k + 1,y) = (m(y) • n(yk+1, Ak+1 — 3pk+1 (m), pk+1 (m));

3) (m (k + l,y) = [(m(y) — (m(k+1,y) — . . m (k+l — 1,y)] rjfyk+l, Ak+l — — 3pk+i(m),pk+i(m)), 2 < l < n — 1;

4) (m(n,y) = (m (y) — (m(k + 1,y) — ... — (m (n — 1,y).

Then:

(m(y) = (m (k +1,y) + (m (k + 2,y) + . . . + (m(n — 1,y) + (m (n,y). (1.9)

Let us note some properties of the functions (m(k + l,y), 1 < l < n. The definition implies that the function n(yk+1, Ak+1—3pk+1(m), Pk+1 (m)) is infinitely differentiable with respect to yk+1 E R1, equals 1 for \yk+1\ <

< Ak+1 — 3pk+1(m) and equals zero for \yk+1 \ > Ak+1 — 2pk+1 (m). Hence the function (m (k + 1,y) is infinitely differentiable and (m (k + 1,y) = = (m (y), when

m

y E supp (m(y) f{y \ y = (У1, . . .,yn), \yk+1\ < Ak+1 — 3Pk+1(m)}.

Besides supp pm(k + 1, y) C supp pm (y) n {y \ y = (yi,...,yn), \yk+i\ <

< Ak+i - 2pk+i(m)}.

Thus (1.6) implies that supp pm (k + 1, y) C {y \ y = (yi,...,yn), \yi\ < 3m, 1 < i < n,i = k + 1, |yk+i| < Ak+i - 2pk+i(m)}.

Further for l = 2,... ,n : the functions pm(k + l,y) are, evidently, infinitely differentiable and

supp pm(k + l,y) C{y \ y = (yi,.. .,yn), \yi\ < 3m, 1 < i < n,

i = k + l, \yk+i\ < Ak+i - 2pk+i(m)}, 2 < l < n - 1; (1.10)

pm(n,y) is infinitely differentiable and

supp Pm(n, y) C{y \ y = (yi,. . .,yn), \yi\ < 3m, 1 < i < n, i = n,

if each of limits exists.

Let us show that for each k +1 < j < n the expression under the sign of limit in the right-hand side of (1.11) equals zero. We can take any natural numbers m > 1, j E {k + 1,... ,n}, the function (m(j,n) and, following the reasoning from [9, p. 134], let us at first construct the sequence {(m(l,j,y)}f=1 of the test functions from D(Rn), converging to (m(j,n) and such that each function (m(l,j,y) is the sum of the functions from D(Rn), which are the functions with the separated variables y1 ,У2, ...,yn.

The Weierstrass theorem (see, for [13, p. 348-349]) implies that for any natural number l we may point out such polynomial Pn(i)(y) of some degree N(l), that for each vector r = (r1,..., rn) with nonnegative integer coordinates satisfying the following condition 0 < \r\ < l, where \r\ =

\yn \ < An - 3pn(m)}. Hence (1.8) and (1.9) implies that

(1.11)

= r 1 + ... + rn, in the cube no = {y \ y = (y1,.. .,y„), \ys \ < 4m, 1 <

< s < n} the following inequality will be fulfilled

\Dr(m(j,y) - DrPn(i)(y)\ < 1. (1.12)

Let us designate:

N (l)

1) PN (l)(y) = Y avi V2-vn yV yV ...ynn , where v = (v1, vn) - the

vector with nonnegative integer coordinates, \v\ = v1 + v2 + ... + vn and

aVlV2...Vn - the constants;

2) (m(l,j,y) = pN(l) (y) \ n n(Уs, 3m,m) > X n(y3, Aj — 2Pj (m),Pj(m)).

s=1 . s=j

(1.13)

Then (m(l,j, y) is the sum of products of the functions depending only on one of the variables y1,y2,.............,yn :

N(l)

(m (l,j,y) = aviV2 ...vn < II yV n(ys , 3m,m) > X

|v|=0 s=1 .s=j

Xyjjny, Aj — 2Pj (m),Pj(m)) =

N(l) f n

^ avi v2...vn I n A1(vs ,ys)A2(vj , yj), (1.14)

|V|=0 lss=j

where

A1(vs,ys)= yvn(ys, 3m,m), s = j \

Vj / A ^/ \ / W * }

A2(vj /yj) = y<j n(yj, Aj — 2Pj (m), Pj (m)), s = j

(1.15)

Let us take any m > 1, j E {k + 1,... ,n}, the function (m(j, y) and show that the sequence

^m(l,j,y) = (m (l,j,y) — (m (j,У), l = 1, >2, . . .

converges to zero in the sense of convergence in D(Rn).

According to the definition (see, for instance, [10, p. 85]), the sequence ^m(1,j,y), ^m(2,j,y), .. ., ^m(l,j,y), . . . converges to zero if:

1) there exists the ball U(0, R) of radius R > 0 with the center at origin such that: supp ^m(l, j,y) C U(0, R), l = 1, 2,...

2) for each r = (r1,..., rn), where ri, 1 < i < n are the integer nonnegative numbers, the sequence Dr^m(l,j,y) converges uniformly to zero as l ^ ^ +to.

Let us denote

n1j = {y \y = (y1,...^nX \ys\ < 4m, 1 < s < n s = j,

\yj \ < Aj — Pj(m)}

n2j = {y \ y = (y1,. . .^nX \ys \ < 3m, 1 < s < n s = j,

\yj \< Aj — 2Pj(m)}.

The first condition from the convergence definition is evidently true because according to (1.13):

supp (m(l,j,y) C n1j C noj C U(0, 4mv/n)

and according to (1.10):

supp (m(j,y) C n2j C U(0, 4m^/n).

Now we verify the second condition. Let y E n2j-, then according to the definition of the function (m(l,j,y), from (1.13) we obtain:

f n l

(m (l,j,y) = pN (l)(y) n(ys, 3m, m) f X Xn(yj, Aj — 2Pj (m),Pj(m)) = pN(l)(У), because n(ys, 3m, m) = 1 for \ys \ < 3m, 1 < s < n, s = j and n(yj, Aj — —2pj (m),pj (m)) = 1 for \yj \ < Aj — 2pj (m).

Thus according to (1.12) for y E n2j- the following inequality is true

\Dr ^m (l,j,y)\ = \Dr [(m (l,j,y) — (m (j,y)} \ =

= |Dr [PN (l)(y) — (m(j,y)] \ < J.

Let y E n1j \ n2j-, then (m(j,y) = 0 and hence for any l according to (1.13)

= 1, 2,

Dr ÿm(l,j,y) = Dr =

= Dr {PN(i)(y)

n(ys, 3m, m)

s=1

-s=j

n(yj, Aj - 2Pj(m),P1 (m)) > =

= Dr{Pn(!)(y) • F(m,j,y)},

where Y(m, j, y) =

Y\_v(ys, 3m, m)

s=1

s=j

n(yj, Aj - 2Pj (m),Pj(m))-

According to the formula of product differentiation the function

^rx+r2 + ...+r„

{PN(l) (y) • Y(m,j,y)}

dyl1 dy22 ••• dy is the sum 2lrl of the following addends

da1+a2 + ••• + an d bl +b2 + ... + bn

dy?1 dy? ••• dynn Pn(l)(y) • dybidy52 ••• dyt Y

where ai + bi = ri, a.2 + 62 = r2, ..., an + bn = rn.

Thus for y G n1j \ n2j we have

da 1+a2 +...+a

\Drÿm(l,j,y)\< 2|r| max

0<|a|<|r|

dya13ya22 ••• dya

PN(l) (y)

X

L~(nij\n2j )

x max

0<|b|<|r|

d b1+b2 +...+bn

drf1 dy2 ••• dyk

Y(m, j, y)

L^ (nij\n2j)

where \a\ = a1 + a2 + ... + an, \b\ = 61 + 62 + ... + bn, \r\ = \a\ + \b\. For l > \r\ according to (1.12) the inequality will be true

1

\Drÿm(l,j,y)\ < 2|r| - max

l 0<|b|<|r|

d ^

—r-----j— Y(m, j, y)

dyl1 ••• dyt

L~(nij\n2j )

n

n

d

dyl~dyn

and because the partial derivatives ^---------------Trm Y(m,j, y) are bounded

on nij \ n2j and independent of l, then as l to :

llD 0m (l,j, y) lli^(ni3-\n2j) ^ 0

and therefore the sequence 0m(1,j,y), 0m(2,j,y), ...

..., 0m(l,j, y), ... converges to zero in the sense of convergence in D(Rn).

Thus according to the definition of the convergence in D(Rn), the sequence pm(1,j,y), pm(2,j,n), ..., pm(l,j,n), ... converges to the function pm(j,n) and according to (l.ll):

Y (n, ...,Tk, Tfc+i, ...,rn) Y[ 9\ (tx ) J ~(y), p(y) j

V' lim lim ({y(Ti,...,Tk,Tk+i,...,Tn) x L' m^+^ +00

=k+i

x Y[g\ (tX ^A(y),Pm (l,j,y^j . (1.16)

For each l = 1, 2,... according to (1.14) we obtain: y(t1 i ... ,Tk j Tk+i , ... , Tn) H gx(Tx ^“(y) , pm

(l, j, y) =

N(l) k }

\ y(ti ,...,Tk ,Tk+i ,...,Tn)U gx(Tx)>^(yi,...,yn),

Irl =0 VI A=i J

\

n Ai (rs ,ys ) > A2 (rj y )

s=i I /

s=j J /

(1.17)

Assertion 1.1.1. Let n > 2, k +1 < j < n and l > 1. Then according to the designations and assumptions introduced above:

Y (ti ,... ,Tk ,Tk+i,

,Tn) n gx(Tx^(y)’

A=i )

ari ...rn

\

f[Ai(rs ,ys) > A2 (rj ,yj )

s=i

= 0. (1.18)

s=i

s=j J /

The proof of assertion 1.1.1. Changing, if it is necessary, the numeration of the variables yi,y2,... ,yn, we may without the loss of generality suppose that in the equality (1.18): j = n. According to the assumption

0 = (supp 7(y)) n Q(A),

then

0 = (supp Y(y)) n{y \ y = (yi,.. .,yn), \yn\ < A4 =

= supp Y(y) n G(An).

Let ip1 (yi,..., yn-i) be the arbitrary test function from S(Rn-i) and 0(y) = '4>i(yi,... ,yn-i)A2(rn,yn), where, as it is evidently follows from (1.15), A2(rn,yn) is the test function. Then ^(y) £ S(Rn), supp ^(y) C C G(An) and

^(t) = (2^) J el(y’T) ^(y)dy = 0i(Ti,... ,Tn-i)^42 (rn ,Tn),

Rn

where T = (ti, . . . ,Tn), y = (yi,. . .,yn).

Hence

0= (Y(y),^(y)) = (¿) / Y (ti ,...,Tn) 0i (Ti,...,Tn-i)x

Rn

XA2 (rn, Tn )dT i ...dTn-idTn = (2^ J 0l(Tl ,...,Tn-i)x

Rn-1

X ^ J y(t i,..., Tn-i ,Tn) A2 (rn ,Tn )dTndTi ...dTn-i. (1.19)

Ri

Since -01 (ti,... ,rn-i) G S(Rn-1) is the arbitrary test function then from the equality to zero of the integral in the right-hand side of (1.19), according to Du Bois-Reymond lemma [10, p. 95] we obtain that for almost all the points (t1 ,..., rn-1) G Rn-1 the following equality is true

1

0 = — Y (t1,...,t„-1 ,rn) A2(rn,rn)drn

(1.20)

R1

Therefore according to (1.20):

y(t1 ,.. .,rk,rk+1,... ,rn) g\(r\)

xA2 (rn,Vn)) =

A=1

}™{ n A1 (rv ’Vv ^ X

= (^tt) I Y(t1 ’...’rk’rk+1 ’...’rn)

Rn

n gA(ta)

,A=1

j'il-Ajr.)j X

(¿)

X A2 (rn ’ Tn )dr 1... drk drk+1... drn =

n—1

Rn

k

ng^(r^)

,A=1

jn A1 (r,’T,)} X

(^r) / y(t1 ’...’Tk ’Tk+1 ’ .. . ’Tn)A2(rn’Tn)drn R1

The assertion 1.1.1 is proved. □

The proved assertion and (1.17) implies that

Y(t1 ’ ... ’ rk ’ rk+1’ . . . ’ rn) IlgAr) jlv) (l’j’V)Sj

dr1 . . . drn 1 = 0.

and (1.16) implies the assertion of lemma 1.1. □

n

§2. The Kernel K(t,a,b)

This section begins with the construction of the auxiliary function Q(r’ [-a.’a\’fi)’ r G R1, with the help of which the kernel K(t,a’b) t G G Rn used in §3 is defined Then the estimation of the kernel norm is obtained in lemma 2.2.

Let us take any 3 > 0 and denote by u(t, 3) such function, the Fourier transformation of which is

s(V’3) = 32 £[-P/2, p/2] (V) * £[-p/2, p/2] (V)’ (2.1)

where £m (v) is the characteristic function of the set M C R1. Then according to (2.1):

S(V’3) = ( W (3 -|V|)’ lvl< 3 (2.2)

[ 0’ |vl > 3

and

J s(V’3) dV = 1.

R1

Let a > 3. Denote by Q(r’ [—a’ a]’ 3) such function, the Fourier transformation of which is

Q(V’ [-a’a]’3) = i—aM(v) * s(V’3). (2.3)

From (2.1), (2.2) and (2.3) we obtain that

0 < Q(V’ [—a’a]’3) < 1 Q(V’ [-a’ a]’ 3) = 0’ y G (—a — 3a + 3)

Q(V’ [—a’ a]’ 3) = 1’ V G [—a + 3a — 3] Q"2(V’ [—a’a]’3) G L“(R1)

. (2.4)

For arbitrary vectors b = (b1 ’... ’ bn) and a = (a1 ’... ’ an) such that

0 < bk < ak’ 1 < k < n let us denote by K(t’ a’b)’ t G Rn the function, the Fourier transformation of which is

n1

k=1

K(V’a’b) = rT-— 1 — Q(Vk’ [—ak’ak]’bk) ’ (2.5)

iVk L -1

where v = (v1’ .. . ’Vn). From (2.4) it follows that K(y^b) = 0 for y G G Q(a — b).

Lemma 2.1. Let n > 1’ t = (t1 ’...’tn) G Rn a = (a1’...’an)’ b = (b1 ’... ’bn) are any vectors which coordinates satisfy the following condition: 0 < bk < ak’ 1 < k < n. Then

sign tk•tt sin ( 2 0 ) sin2 0

k=1 1h t

2 hk tk

03

d0. (2.6)

Proof. From (2.5) we obtain:

k (t’a’b) = n

JVktk .

k=1

2n

R1

iVk

1 — Q(yk’ [—ak’ak]’bk) dyk

Let us choose k G {1’... ’ n}. Since

1 f giVktk

2n J ivk

R1

dVk =

2 sign tk

and for any p > 0:

[£[-p,p] (Vk)] ~(tk) = f eiyktk £[-p,p] (Vk) dVk =

R1

then from (1.1), (2.1) and (2.3) we have:

1

2n

e

iyktk

R1

1

iVk

1 — Q(yk’ [—ak’ ak]’ bk) dyk =

= 1 sign tk — 2 sign tk * 2n {[£[-akak](Vk)] ~(tk)x

X ( b") ' ( [£[-h‘/2'hk/2] (Vk)] ~(tk ))H =

12 = 2sign tk — bin

S1gn(tk — r)sin akT ^ sil?2bT/2) dr

r3

R1

1

1

1 2 f sin akr- sin2 (bkr/2) ,

= - sign tk — sign tw -----------------3--------------dr+

2 b2k n J r3

k R1

sign tk •tt 42

sin akr • sin (bkr/2)

dr.

r

tk

After changing the variable in the right-hand side integrals bk r/2 = 0’ we obtain:

1

K(t’a’b) = II i 2 sign tk — sign tk •

1 f sin(2ak0/bk) • sin2 0

k=1

03

R1

sign tk •tt

1 f sin(2ak0/bk) • sin 0

+n J 03

2 hk tk

d0.

d0+

(2.7)

Since according to [14, p. 193, the formula 858.731]: ntt

f sin(2ak0/bk) • sin 0 n

03

d0 =- ’ 2

(2.8)

then substituting (2.8) in (2.7) we obtain the assertion of this lemma. □ Lemma 2.2. Let n = 1’ a > 0’ M G (1’ +to) and q G [1’ then:

1

K t’a’ — a

V M )

Lq (R1)

-/ c1(m’ q)’

where

C1 (M’ q) = 1 4M

1 f sin 2M0 • sin2 0

03

d0

q \1/q dx

OO

0

Proof. Let us put n = 1 and b = — in (2.6). Then

M

sign t-X

1 \ 1 f sin 2M0 ■ sin2 0 in

K I t,a, — a I = — ------------0^-------------d0,

O at

(t'a'Ma)

whence it follows that

Lq (R1)

1

k t,a, — a

V M )

sign t-X

1 r sin 2M0 • sin2 0

at

□

a1/q

• < 4M

1 f sin 2M0- sin2 0

03

d0

03

q \ 1/q dx

d0

q \ 1/q dt

a1/q

Ci (M, q).

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