CC BY
26DOI: 10.15393/j3.art.2014.2569 Issues of Analysis. Vol. 3(21), No. 2, 2014
B. F.IVANQY
ANALOG OF AN INEQUALITY OF BOHR FOR INTEGRALS OF FUNCTIONS FROM Lp(Rn). II
Abstract. Let p £ (2, n > 1 and A = (Ax,..., A„),
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)
L ^ (Rn)
n
1
k=1 ^k
i/q
l7(r )||
LP (Rn )
where t = (tx ,...,tn) £ Rn, Et = {t | t = (ri,...,r„) £ 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
p q
vector A.
Key words: inequality of Bohr.
2010 Mathematical Subject Classification: 26D99.
This work is the second and the final part of author's work [1]. It contains the proof of the statement announced in [1]. This statement provides conditions, sufficient for validity of a novel analog of Bohr inequality [2]. (In more detail about inequality of Bohr and it's generalizations - in author's paper [3]).
Let p £ (2, n > 1, A = (Ai, A2,..., An) is any vector with
positive coordinates and t = (t1 ,t2,... ,tn) £ Rn. Let us introduce the following designations:
n
1) Q(A) = U {y | y = (yi,y2 ,...,yn) £ Rn, \yk | < Afc}, i.e. Q(A) -k=1
"crosslike" neigborhood of zero in Rn;
(?) Ivanov B. F., 2014
2) r(Rn \ Q(A),p) - the set of all functions 7(t) G Lp(Rn), whose Fourier transformation support belongs to Rn \ Q(A);
3) Et = {t | t = (n ,T2 ,...,Tn) G Rn, Tj G [0, j ], if j > 0, and r0 G [tj, 0], if tj < 0, 1 < j < n} - is the parallelepiped in Rn.
The main result of this paper is the Theorem 3.2 from § 3, which is the following.
Theorem 3.2. Let n > 1, p G (2, and A = (Ai,..., An), Afc > 0, 1 < k < n. Then for any function 7(t) G r(Rn \ Q(A),p) the next inequality is fulfilled:
Y (t)dT
Et
< Cn (q)
L™ (Rn)
n
Ai/q k=i Ak
\\Y (t )ll
LP(Rn)
where —I— = 1, and the constant C(q) > 0 does not depend on 7(t) pq
and vector A.
Recall some notions and statements from the first part of the work. For simplicity, the author keeps numbering of lemmas and formulas used there.
Let n > 1. Designate the Fourier transformation of a function u(t) G G L1 (Rn) as U(y), where y G Rn, and choose U(y) in the following form
u(y) =
-i(y,t)
u(t)dt.
Rn
Let us denote the inverse Fourier transformation of a function v(y) G G L1 (Rn), by v(t), t G Rn. Then v(t) has a form:
®(t) = IT"
Ay,t)
v(y)dy.
Rn
Introduce following notations. Let S(Rn) be the space of infinitely differentiable functions which are rapidly decreasing at infinity, S'(Rn) -the space of slowly increasing distributions (also called the space of distributions of slow growth), i. e. the space of linear continuous functionals on S(Rn). Let us denote the space of finite infinitely differentiable functions on Rn, by D(Rn) and by D'(Rn) - the space of linear continuous functionals on D(Rn).
1
For every complex valued locally integrable function Y(t), t E Rn, we put into correspondence [4, p. 30, p. 32] the functional
(Y,p) = y Y(t)p(t)dt, p(t) E D(R).
Rn
The distributions from D'(Rn), generated by locally integrable functions, are called the regular functions. Since D(Rn) is densely embedded in S(Rn), then S'(Rn) C D'(Rn). As is known, for example, the func-tionals generated by functions y(t) E Lp(Rn), p > 1 are regular 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.
Owing to the introduced notions we have the following version of well known formulas:
{yi(t) • Y2(t)}^(v) = (¿n) ^l(v) * ^2(v)
y(t) = 1, Y(v) = 2nS(v), t,v E R1
(1.1)
Let n > 2, 1 < k < n and Afc+i,..., An > 0. Denote
n
G(Ak+i,..., An)= |J {y | y = (yi ,...,yn) E Rn, \yp | < A^ }.
f3=k+1
If n — k = 1, then G(An) is the direct product of Rn-i and the interval (—An, An). If n — k > 1, G(Ak+1;..., An) - is the direct product Rk and the "crosslike" neighborhood of zero in Rn-k = {y \ y = (yk+1,... ,yn), y3 E R1 ,k + 1 < j < n}.
Lemma 1.1. Let n > 2, 1 < k < n, p E [1, +x>],the functions g\(0) E
E Lq(R1), 1 + 1 = 1, 1 < A < k, y(t) E Lp(Rn), A = (A1,..., An) -p q
the vector with positive coordinates and supp Y(y) H Q(A) = 0. Then ^sup^ Y (t1 , ... , Tk ,Tk+1,...,Tn ) Y\gx Q
p|G(Ak+i,..., An) = 0.
Let us take any /3 > 0 and denote by u(t, [3) such function, the Fourier transformation of which is
u (y,[) = /32 ^\-P/2,p/2](y) * i[-p/2,p/2] (y), (2.1)
where (y) is the characteristic function of the set M C R1.
Let a > /3. Denote by Q(t, [—a, a], [) such function, the Fourier transform of which is
ft(y, [—a,a],[) = C[-a,a](y) * u(y,[). (2.3)
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
n 1 r ]
K(y,a,b) = Y\_ -— 1 — ft(yk, [—ak,ak],bk) , (2.5)
k=
^ Wk
where y = (y1,..., yn). From (2.1) and (2.3) it follows that K(y, a, b) = 0 for y G Q(a — b).
Lemma 2.1. Let n > 1, t = (t1,..., tn) G Rn and a = (a1,..., an), b = (b1,...,bn) are any vectors which coordinates satisfy the following condition: 0 < bk < ak, 1 < k < n. Then
sin ( 2 ^ 0 ) sii
bk
n n si9ntksin (2^ 0) sin2 0
K (t,a,b) = ( n I -"—-d0. (2.6)
k=1 1tw
2 bk tk
Lemma 2.2. Let n = 1, a > 0, M G (1, and q G [1, then:
K^,a,Ma)
Lq (R1)
1/q
C1(M,q),
where
C1 (M,q) = < 4M
^ o
1 f sin 2M0 • sin2 0
n
03
d0
1/q
dx
1
q
00
0
§ 3. Main theorem.
We start with an auxiliary result.
Theorem 3.1. Let n > 1, p G (2, A = (A1,..., An), Ak > 0, 1 < k < n, M G (1, and t = (t1 ,...,tn) G Rn. Then for any function y(t) G r(Rn \ Q(A),p) equality is true:
(— l)n(t> / 7(t ) dT = / K(ru..., Tn, MI1 A, MI1 ■A) 7(t ) dT -
Et Rn
V [ K[t1,..., Ta - ta,..., Tn, M A,—1—A) Y (t ) dT+
Z^j , a ^ , n, M +1 'm +1 ) '
a=1 r.
n f m A 1 A )
+ > K t1 ,... ,Ta - ta,..., tr - tg,... ,Tn,—-A, —-A x
Z^ j I 1 ' a a' ' g g ' n m +1 'M +1
a,g=i Rn
a<g
xy (t ) dT + ... + (-1)n y K (ti — ti,..., Tn -
Rn
M A 1 . ) / \ ï , ^
-tn, ^-A,—-A Y(t) dT, (3.1)
n M + 1 'M + 1 ) /w ' v 7
where n(t) - number of negative coordinates of a point t.
Proof. Choose arbitrary point t G Rn and consider a subsidiary integral
J e-i(u'T)y(t) dT, u G Rn. (3.2)
Et
Since
{^Et (t)Y(v) = J e-i(y,T)E (t) dT =
Rn
tl t2 tn
= (-1)n(t) J j ...J e-i(yi Tl+y2 T2+-+yn Tn ) dTi dT2 ...dTn = 00 0
n 1 _ e-iy^tk
(-1)n(t) n
fe=1 y
where y = (y1,y2,..., yn) and t = (t1 ,t2,..., Tn), we obtain from (3.2) that:
(— 1)n(t) J e-i(u Et
T) dT = (— 1)n(t) j e-i(u,T)E (T)y(t) dT =
Rn
l ........{ El1—^ }
=e R
"(t) • Y (t) dT.
We choose arbitrary v E (1, and designate fi(yk) =
= fi yk,
Mv
Ak
Mv
(M +1)v +1 ' (M +1)v +1 1 < k < n. Then
Ak
v
' (M + 1)v + 1
Ak
(— 1)n(t) I e-i(u,T
t ) dt =
Et
=e R
-i(u,T )
n {
k=1 k=1
1 _ e-iyk tk Wk
1 — fi (yk)
+
1 _ e-iyktk ^ +-:-fi (yk ) j ~(t ) • Y (t) dT =
iyk
=e R
-i(u,T )
in
k=1 k=1
1 _ e-iyk tk
iyk
}
1 — fi (yk) Ht) y(t) dT+
+ e-i(u,T) {L(y,t)} ~(t) • y(t) dT,
(3.3)
Rn
where L(y,t) - is the sum of 2n — 1 various items of the form:
n
Ji
1 _ e iya tc
1 — fi(y«)l • n
1 _ e-iyptv
ffe-h
Wp
fi (yp),
with J1 U J2 = {1, 2,...,n}, J1 H J2 = 0, J2 = 0.
We consider the first term in the right part (3.3). For the item standing in braces in (2.5) we obtain:
!n 1 _ e-iyk tk [ ] I
n^— [1— fi (yk )}[(T ) =
= (i)n / eeiy'r> n(1 — tk) n ^ dy =
v 7 Rn k=1 k=1 yk
-W J e-T'fi^ dy—
e
R
n
E(±) dy+
a=1 7 Rn k=1
+ E (-Y i e^e-^-"^ n 1 — °(yk) dy +
. 2: / / f-*- iyk
a,f=1 v 7 Rn k=1 yk
a<f
+ ... + (—W—1n f ei(y,T) e-iyi ti--iy-^TT 1 — fi (yk) dy =
2: k=1 iyk Rn k=1
( Mv v )
= K[T1 ,...,Tn, (M + 1)v +1 A (M + 1)v +1 AJ —
n Mv v
E K{T1,...,Ta — ta,...,Tn, (M + 1)v +1 A, (M + 1)v +1 AJ +
n Mv
+ ^ K{T1, ... , Ta — ta ,...,Tf — tf ,...,Tn, (m + 1)v +1 A,
a<f
v A^ + ... + (—1)n K (T1 — t1 ,...,Tn — tn,
(M + 1)v + 1
Mv
A,—-:- A
(M + 1)v +1 ' (M + 1)v +1
,
where A = (A1, . . . , An),
ts( Mv V
K T1,.. .,Tn,----A, ----A =
' n (M + 1)v + 1 ' (M + 1)v + 1 J
n rs( Mv V )
= 1=1 K , (M + 1)v +1 Ak, (M + 1)v +1 AV =
n n
k=1 k
1 — ft (yk U~(Tk).
k=1 k lyk 1
Hence,
!n 1 _ e iyk tk r ] I
n I1 — ft y)]}
1 — ft (yk) Wt) 7 (t ) dT =
Rn
v
= i e-^ )k(t1 ,..., Tn -A,-—^-a) • 7 (t ) dT—
J V (M +1)v +1 ' (M + 1)v +1 J n J
n
V / e-i(u,T)K
(T1.....
Mv
^,^K(T1,...,Ta — t«,...,Tn, (m +1)v +1 A,
a=1Rn
V a\ • 7(t) dT + ^ i e-i(uT)
/ a R=1 J
(M + 1)v + 1 7 fl 1
a, P=1Rn
a</3
e-i(U,T) K (T1, ... , Ta — ta ,...,T0 —
Mv v
—tH,---,Tn, (M + 1)v + 1 A (M + 1)v + 1 A) • 7(T)dT+
+ . . . + (-1)n/ ^)K (T1 — .1, . . . , Tn — in, (M +M1Vv +1 A,
A^ • 7(t) dT. (3 . 4)
Rn
v
(M + 1)v + 1 As 7(t) G Lp(Rn) and owing to Lemma 2.2
K{T, (MtrVrr a, (m+t)"v+7A) G L(Rn),
M,v E (1, +ro), - + - = 1, then for every fixed t = (t1 ,...,tn) E Rn all pq
integrals given in the right part of (3.4), are continuous in point u = 0. So,
r ( n 1 — p-iyktk
1 — fi (yk) }~(t) • Y (t) dT =
IMK0 J
Rn
in 1 - e-iyktk r ^ ] 1
n—[1—fi y)]
lim / e~i(U'T) ■ " 1 — e-iyktk
= i K (t1 ,..., Tn, —-Mv-A, --v—-A ) y(t) dT—
1 V (M + 1) v +1 ' (M + 1) v +1 Jnj
R
n
Mv
K[ru...,^ — ta,...,Tn, M + 1)v +1 A,
a=1R„
A) y(t ) dr + ^ J K (T1,...,-
a, B=1r>n
v
Ta ta, ... , Tf
(M + 1)v +1 ,
a<f
Mv v
— tp ,...,Tn, (M + 1)v +1 A, (M + 1)v +1 AJ Y (T) dT+
+ ... + (— 1)nf K(T1 — t1 ,...,Tn — tn, (M ^ +1 A,
Rn
v
A ) y(t) dr. (3.5)
(M + 1) v +1 J We demonstrate that
lim i e-i(u,T) {L(y,t)} ~(t ) • y(t) dr = 0. (3.6)
|U|K0 J
Rn
If n = 1, then according to (3.3)
j e-iUT {L(y,t)} ~(t) • y(t) dr = Ri
= /e-iUT {fi(y)} ~(t) y(t) dr. (3.7)
Ri
If n > 2, then according to (3.3) integral standing after the sign of a limit in (3.6), equals to the sum of 2n — 2 integrals of the type
I(Ji ,J2 ,u,t) = J e-i(u'TU J]
Rn KaEJi
1 _ e iya ta
Wa
1 - ft (Va)
X
-„ 1 — e-iy?t? ^ X n -e-ft(VR)"(t) • y(t) dT,
PeJ2 %Vp
(3.8)
where J1 U J2 = {1, 2,..., n}, J1 H J2 = 0, J1 = 0, and of the integral
I(u,t) = J e-i(u,T) Rn
n 1 — e-iy?t? ^
n-e-ft (VR )"(t ) • Y (t ) dT. (3.9)
p=i Wp
Note that integral (3.7) results from (3.9) at n =1. We check now that each of integrals (3.8) and (3.9) tends to zero when \\u\\ ^ 0, from where the result (3.6) follows.
Consider any integral I(J1, J2,u,t). In that integral:
n
aE Ji
1 _ e-iya ta [ ^ ] 1 _ e — iy? t?
1 - ft (Va) • n -—- ft (VR )"(t) =
(¿)
i(y,T)
Rn
n
n
aEJi 1
1 _ e iyatc
oiy a Ta
PeJ2
1
Wp
ft (Va)] n
1 _ e-iypt? Wp
ft (Vp) dV =
1 _ e iyatc
r 2n J
aEJi ri
Wo
PeJ2
1 - ft (Va ) dVa X
-n- 1 f ■ 1 - e-iyPt? ~
X n 2T eiy? T?-:-ft (VP ) dVp.
PJ2n Ri V
(3.10)
Introduce the following notions for (3.10):
f (Ta,ta ) = 2n i eiyC TC
Ri
l(Tp,tp) = -J eiy?T? Ri
1 _ e iyatc
Wa
1 _ e-iy?t? Wp
1 - ft (Va ) dVa, a G Ji,
ft (Vp) dVp, p G J2.
e
Proposition 3.1.1. Let t = (t1,...,tn) G Rn, A = (Ai,...,An),
M G (1, +œ), v G (1, +œ) and r G [1, +œ). Then:
1) f (ra,ta) = K (ra-Mv-Aa, --V—-Aa ) -
(m +1)— +i a' (m +1)— +1 a)
f Mv — \
—K Ta — ta, ---r-Aa, —----Aa , 1 < a < n,
\a a (M +1)v +1 a' (M +1)v +1 J ' " " '
(M +1)v +1 '(M +1)v +1 f ( •, ta) G Lr (R1 ), 1 < a < n;
(M +1)v (M + 1)v
2) supp l (yp ,tp ) =
(M + 1)v + 1 ' (M + 1)v + 1
3) l( • ,tp) G Lr(R1), tp G R1, 1 < 3 < n.
, 1 < 3 < n; (3.11)
Proof of proposition 3.1.1. 1) Taken n =1, in (2.5) and (2.6) we
deduce from definition of function f (Ta,ta) that for any 1 < a < n :
f (Ta,ta) = K (ra, (M + 1—— +1 Aa' (M + 1)v +1 Aa) -( _ Mv A v A \
-K ^ - ta, (m + 1)v +1 Aa (m + 1)v + 1 Aa) '
and as owing to Lemma 2.2
K (Ta, (M +1 A", (M + 1)v + 1 ^ G ^ (Ri )
for any r E [1, then f (•, ta) E Lr(R1).
2) From definition of function l(rp,tp) we obtain, that
^ 1 — e-iypt<3
l (yp,tf) = -:-fi(yp).
iyp
Hence, by (2.4)
supp l (yp ,tp) = supp fi (yp) = [ Mv v Mv
+
(M + 1)v +1 (M + 1)v + 1' (M + 1)v +1 (M + 1)v +1_'
1 _ e-iyptt3
3) As-is bounded and owing to (2.4) fi (yp) E L1(R1), then
iyp
l (•, tp) E L1(R1). So with respect to the variable rp the function l(rp, tp)
v
is continuous and bounded. In addition, when t, = 0 as a result of double integration by parts, we have according to (2.4) and (3.11)
+ (M+ 1 )v + 1 A«
1 i' ■ (1 - e-iyP~
Krp ) = 2n J eiyp | y ft (y, )j dy, =
__(M+1)v A«
(M+1)v + 1 A«
(M +!)v A«
(M + 1 ) v+1 A«
1 / {H^M.--*=°(1)
2nT| J I y Jy2 TJ
(M+1)v A« ^
(M +1)v + 1 A«
-r( 7?1
Therefore, l( • ,tp) G Lr(R1 ) for any r G [1, +œ). □
Proposition 3.1.2. Let n > 2, t = (t1,...,tn), A = (A1,..., An ),
Afc > 0, 1 < k < n, u = (ui,...,un) and \uk\ < ,,, , ^—— Afc,
(M + 1)v +1
1 < k < n. Then I(J1, J2 ,u,t) = 0.
Proof of proposition 3.1.2. Changing, if necessary, in (3.8) numbering of variables y1,... ,yn, one can say without loss of generality that J1 = = {1,..., k}, J2 = {k + 1,..., n}, where 1 < k < n. Then according to (3.10) and proposition 3.1.1
f V , ÎA 1 - e-iya ta r ^ 1 I(J1 ,J2 ,u,t)= e-i(u'TH\[ -:- 1 - ft (ya) x
Rn U=1 y
n 1 - e-iy«t« ~ I
X n -:-ft y )> ) • 7 (T ) dT =
p=k+1 Wl3
= / e-i(u,T)
{ S K ^ (M + 1) v + 1 Aa (M + 1) V +1 Aa)
/ Mv v \ 1 '
-K \Ta - ta, (M +1)V +1 Aa, (M +1)v +1 Aa) J • JkJ+1 l(Tp^)
X7 (T1,..., Tk, Tk+1,... ,Tn ) dT 1... dTk dTk+1... dTn =
X
= e-i(u'T)F(r,t)$(r,t) dr,
Rn
where
F (r,t) = y(t i ,.. .,Tk ,rk+i, ...,rn)x
x
n
a=1
K Ta,
Mv v .
(M + 1)v + 1 a' (M + 1)v + 1 a
(
-K Ta - ta,
Mv
A
v
(M +1)v +1 a' (M +1)v +1
A,
l(TP ,tp ).
a=1 p=k+1
Consider functions F(u,t) and (u,t). Since, due to theorem conditions supp y(u) fl Q(A) = 0, then according to Lemma 1.1:
F(u,t) f G(Ak+iAn ) = 0.
Function 3(t, t) represents the direct product of functions 1,1 < a < k and functions l(Tp,tp), k + 1 < [ < n. It follows from definition l(Tp,tp), k +1 < [ < n that:
1 _ e-iuptp ^
{l(Ti ,tp )}>p ) =-:-ft (up ), up e R1, k +1 < [3 <
vup
Therefore according to [4, p. 240-241] and (1.1):
n.
3 (u,t) = (2n)k
Y[ S(ua)
,a=1
n 1 _ e-iuptp n -u-ft (up ),
p=k+1
:up
where u = (u1,..., un). In addition, considering (3.11), we obtain supp (u, t) = {u \ u = (u1;..., uk, uk+1,..., un), ua = 0, 1 < a < k,
\upI < (M +1)v Ap, k +1 < [ < n\. 1 pl - (M + 1)v + 1 p' " J
So the support of function $(u,t) is bounded. Hence (see, for example, [4, p. 135]), the convolution of functions FF(u,t), $(u,t) is defined and according to (1.1) we have:
I(Ji, J2,u,t) = F(u,t) *$ (u,t).
For A,B C Rn denote:
A + B = {x | x = a + b, a G A, b G B}.
Then (see [5, p. 69]) as the support of convolution of two functions is contained in closure of the sum of supports of these functions, so
supp I(J1,J2 ,u,t) C supp F(u, t) + supp $(u,t) C
C{x | x = (xi,... , Xk ,Xk+i, ...,Xn), Xa G R1, 1 < a < k, x, G r1 \ (-A,, A,), k +1 < p < n} + {y 1 y = (yi,..., yk, yk+i, ■■■,yn),
ya = 0, 1 < a < k, ly,I < + 1)V A,, k +1 < p < n} = i>a > - - > \yp\ - (M + 1)^ +1 P->
= {u I u = (ui,...,uk ,uk+i ,...,un), ua = xa + ya, xa G Ri, ya = 0, 1 < a < k u, = x, + y,, x, G Ri \ (-A,, A,), y,1 < (MM+1+IA,, k +1 < P < n} = {u | u = (ui,..., uk, uk+i,..., un), ua G Ri, 1 < a < k,
(M + 1)v +1 A ,k +1 < P < n}
Therefore, if u G {u | u = (ui,... ,un), lukI < (m + 1)v + 1 Ak, 1 <
< k < n}, then I(Ji,J2,u,t) = 0. □
Proposition 3.1.3. Let n > 1, t = (ti,...,tn), A = (Ai,..., An),
Ak > 0, 1 < k < n, u = (ui,...,un) and u| < (m + 1)v + 1 Ak, 1 <
< k < n. Then I(u, t) = 0.
Proof of proposition 3.1.3. Since the set
supp
n
p=1
1 _ e-iuptp
:up
ft (up )
= j u I u = (u1 ,...,un), \up \<
(M +1)v .
(.,, + ,)■ + 1A■ ' s ' << n
}
is bounded in Rn, then the convolution of functions
n 1 — e-iuptp ^
II -:-ft(up), 7(u1)
p 1 u
..,un )
is well-defined. Consequently
. n 1 — e-iup tp I (u,t) = < H-—-ft (up )* 7(u1,...,un ).
p=1
:up
Thus taking into account (3.11) we obtain:
supp I(u,t) Ç supp < H
1 _ e-iuptp
p=1
:up
ft (up)+ supp 7(u) =
= j u I u =(u1,...,un ), \up \<
, (M +1)v Ap, 1 < 3 < „> +
(M +1)v +1 p' '
+ {u \ u = (u1,..., un ), \up \> Ap, 1 < 3 < n} =
1
= j u \ u = (u1 ,...,un), \up \>
(M + 1)v + 1
Ap, 1 < [ < n
.
Thus, if u e {u \ u = (u1,..., un), \uk \ <
1
Ak, 1 < k <
(M + 1)v + 1
< n}, then I(u,t) = 0. Proposition 3.1.3 is proved. □
Continue the proof of Theorem 3.1. From (3.3), (3.5) and (3.6) we have:
(—1)n(t) y(t) dT = lim (—1)n(t) e-i(uT)j(t) dT = J IMK0 J
Et Et
= J K
Rn
(......
± ÎkL. .
i y \
Mv
. Tn.
A,
1
(M + 1)v +1 ' (M + 1)v +1
Mv 1
A 7(t)dT -
Ta ta. ... . Tn.
A,
(M + 1)v +1 ' (M + 1)v +1
A
x
7(t) dT + / K (T1..... Ta - t a, . . . ,TP
-tP. . . . ,Tn.
a, P=1Rn a<P
Mv
A,
1
(M + 1)v +1 ' (M + 1)v +1
a^ 7 (t ) dt+
T
+ ... + (-1)™ J K ( T1 - t1 ,...,Tn - tn,
Rn
1
Mv
A,
(M + 1)v + 1
(M + 1)v + 1 A^ 7(T) dT. (3.12)
□
Proposition 3.1.4. In the assumptions, formulated, above, we have:
lim
V^r +oo
K \T1, . . . ,Tn,
Mv
A,
1
(M + 1)v +1 ' (M + 1)v +1
A
M1
-K t1, ... ,Tn,-A,-A
1 ^ ' n M + 1 ' M + 1
)
= 0.
Lq (Rn )
Proof of proposition 3.1.4. We use the induction method by number of variables n.
Let at first n =1. According to (2.6)
sign t•œ
1 \ 1 /"sin 2M0 • sin2 0 ,
K \ T.a. — a \ = -J -— d0 T G R1
1
{T,a,~Ma)
then, taking K(t) =
d0, t G R1,
=K
T,
Mv
A,
1
(M + 1)v +1 ' (M + 1)v +1
A - K T,
M1
A, ——- A
M + 1 M + 1
)
2M
we obtain:
\\k (t )\l r) =
1/2 M+1
v/2
At
sin 2M0 • sin2 0 03
d0
(M +1) v + 1
At
<
4(M + 1) Anq
2M I 4(M + 1)
sin 2M0 • sin2 0 03
d0
(M + 1)v
(M + 1) v+1
q \ 1/q
dt
i/q dx\ <
n
A
sin 2M0 • sin2 0 2M0 • 02
d0
(M + 1)v
(M+1)v + 1
q \ 1/q dx
For 0> 0 and M > 1
sin 0 2 sin 2M0 2
< --<
0 1 + 0' 2M0 1 + 0'
then
\\K (T )\\Lq (R1) <
16M I 4(M + 1)
n
A
d0
(M+1)v
(1 + 0)3
(M +1) v + 1
<
16M I 4(M + 1)
n
<
A
x
1
(M + 1)v + 1 16M J 4(M + 1)
i + (M+1)y x
1 + (M + 1)v+1 x
q \ 1/q dx <
q \ 1/q dx <
l/q
(M + 1)v +1 n while v —> +oo.
A
x
(1 + x)3
dx 0
x
x
x
x
x
x
1
3
q
Let the statement of proposition 3.1.4 be true for n = k > 1. Then it is easy to prove that the statement of proposition 3.1.4 is true for n = k + 1:
K
^T1, . . . ,Tk ,Tk-K T1 , . . . , Tk ,
Mv
A,
1
' (M + 1)v +1 ' (M + 1)v +1
A
M1
Tk+1,-A,-A
k+1' M +1 'M +1
Mv 1
)
Lq (Rk+1)
k+1 /
nK [tj. (M , n a. (M, 1)v, n A i -
(M + 1)v +1 j (M + 1)v +1
k / H K yrj.
j = 1 V
M
Aj,
A
j M + 1 j M + 1 j
)
x
(
xK Tk+1,
Mv
A
1
(M + 1)v +1 k+1. (M + 1)v +1
k
+
Ak+1 +
)
j=1
M
A,,
1
A,
j M + 1 j M + 1 j
)
x
(
xK Tk+1,
Mv
A
(M + 1)v +1 k+1. (M + 1)v +1
A
k+1
)
k+1 / KT
,=1
-Il Kl-, , —A, A,
j M + 1 M + 1 j
)
<
Lq (Rk+1 )
<
Mv
nK (-,. (M ,,Aj. (M, 1)v,,Aj i -,=1
(M + 1)v +1 (M + 1)v +1
k
K
,=1
X
T
-| IKIT, , A, A,
Il \ j M + 1 M + 1 j
1
)
x
Mv
K -k+1, , -— Ak+1
(M +1)v +1 k+^ (M +1)v +1
Lq (Rk )
Ak+1
)
+
Lq (R1 )
1
1
+
x
3=1 K ( rk+1,
S K(
K\r:i A
1
A:
: M + 1 : M + 1 3 Mv 1
)
A
(M +1)v +1 k+1' (M +1)v +1
x
Lq (Rk)
Ak+i
)
M1
-K lvTk+1' m+1 Ak+1 > MTT Ak+1)
->• 0
Lq (R1)
while v ^ □
Continue the proof of Theorem 3.1. Proceeding in (3.12) to the limit for v ^ we obtain the statement (3.1). □
Theorem 3.2. Let n > 1, p E (2, +x>] and A = (A1?..., An), Ak > 0, 1 < k < n. Then for any function y(t) E T(Rn \ Q(A),p) the next inequality is fulfilled:
Y (r)dr
Et
< Cn (q)
L™ (Rn)
n
A1/q k=1 Ak
\\Y (r)\\
LP(Rn)
where —I— = 1, and the constant C(q) > 0 does not depend on y(t) pq
and the vector A.
Proof. From (3.1) we obtain:
Y(r)dr
Et
< 2n
K
M A 1 A"
A, —-A
M + 1 M + 1
L^ (Rn)
Lq (Rn)
Wy (r )ll
LP (Rn)
Owing to (2.5), (2.6) and Lemma 2.2
K
r,
M1
A, ——- A
/
n k(
k=i v
M
K Tk, „ , „ Afc
M + 1 ' M + 1 1
M + 1 ' M + 1
A
Lq (Rn)
< C2(M,q)
where C2(M,q) = i 4(M + 1) J 1 J
Lq (R1)
1 f sin 2M9 • sin2 e
n
1
k=i Ak
CC CC
e3
de
1 < q <
i/q k
q \ 1/q dx
1
Hence, Y (r)dr
Et
< 2nCn(M,q)
n
a 1/q
Lk=1 Ak
IIy(t )II
LP (Rn ) •
(3.13)
The inequality (3.13) is correct for any M > 1, therefore taking
C(q) = 2 inf C2 (M,q),
M>1
we obtain: Y(r)dr
Et
< 2n = [C (q)f
inf C2(M,q) M>1
(n A*)
\k=1 Ak /
№ A*)
U=1 Ak /
II Y(r )||
LP (Rn )
IIY (r )II
LP(Rn)
Theorem 3.2 is proved. □
1
n
References
[1] Ivanov B. F. Analog of an inequality of Bohr for integrals of functions from Lp(Rn). The Issues of Analysis, 2014, vol. 3(21), no. 2, pp. 16-34.
[2] Bohr H. Un théoréme général sur l'intégration d'un polynome trigonomé-trigue. Comptes Rendus De L'Academie des sciences, 1935, vol. 200, no. 15, pp.1276-1277.
[3] Ivanov B.F. On a generalization of an inequality of Bohr. The Issues of Analysis, 2013, vol. 2 (20), no. 2, pp. 21-57.
[4] Gelfand I.M., Shilov G.E. The generalized functions and the operations over them, iss. 1. M., PhM, 1959, 470 p.
[5] Vladimirov V. S. Generalized functions in mathematical physics. M., Nauka, 1979, 318 p.
Received July 14, 2014.
Saint Petersburg State Technological University of Plant Polymers, Str. Ivan Chernykh, 4, 198095 Saint Petersburg, Russia. E-mail: ivanov-bf@yandex.ru
