ANNALES
Universitatis Scientiarum
Budapestinensis
de Rolando Eotvos nominatae
SECTIO MATHEMATICA TOMUS XLIV.
REDIGIT A.´ CSASZ´ AR´ ADIUVANTIBUS L. BABAI, A. BENCZUR,´ K. BEZDEK., M. BOGNAR,K.B´ OR¨ OCZKY,¨ I. CSISZAR,´ J. DEMETROVICS, A. FRANK, J. FRITZ, E. FRIED, A. HAJNAL, G. HALASZ,´ A. IVANYI,´ I. KATAI,´ P. KOMJATH,´ M. LACZKOVICH, L. LOVASZ,´ GY. MICHALETZKY, J. MOLNAR,´ L. G. PAL,P.P.P´ ALFY,´ GY. PETRUSKA, A. PREKOPA,´ A. RECSKI, A. SARK´ OZY,¨ F. SCHIPP, Z. SEBESTYEN,´ L. SIMON, GY. SOOS,´ J. SURANYI,´ G. STOYAN, J. SZENTHE, G. SZEKELY,´ L. VARGA
2001
2019. majus´ 4. –23:07
ANNALES
Universitatis Scientiarum
Budapestinensis
de Rolando Eotvos nominatae
SECTIO BIOLOGICA incepit anno MCMLVII SECTIO CHIMICA incepit anno MCMLIX SECTIO CLASSICA incepit anno MCMXXIV SECTIO COMPUTATORICA incepit anno MCMLXXVIII SECTIO GEOGRAPHICA incepit anno MCMLXVI SECTIO GEOLOGICA incepit anno MCMLVII SECTIO GEOPHYSICA ET METEOROLOGICA incepit anno MCMLXXV SECTIO HISTORICA incepit anno MCMLVII SECTIO IURIDICA incepit anno MCMLIX SECTIO LINGUISTICA incepit anno MCMLXX SECTIO MATHEMATICA incepit anno MCMLVIII SECTIO PAEDAGOGICA ET PSYCHOLOGICA incepit anno MCMLXX SECTIO PHILOLOGICA incepit anno MCMLVII SECTIO PHILOLOGICA HUNGARICA incepit anno MCMLXX SECTIO PHILOLOGICA MODERNA incepit anno MCMLXX SECTIO PHILOSOPHICA ET SOCIOLOGICA incepit anno MCMLXII
2019. majus´ 4. –23:07 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 3–9
THE INFLUENCE OF S -QUASINORMALITY OF SOME SUBGROUPS OF PRIME POWER ORDER ON THE STRUCTURE OF FINITE GROUPS
By M. RAMADAN
Cairo University, Faculty of Science, Department of Mathematics, Giza-Egypt Received September
1. Introduction G
A subgroup of a group G which permutes with every subgroup of Kegel
is called a quasinormal subgroup of G . We say, following [7] that
S G a subgroup of G is -quasinormal in if it permutes with every Sylow
subgroup of G . Several authors have investigated the structure of a finite
group when some subgroups of prime power order of the group are well- G
situated in the group. Ito [6] proved that a finite group of odd order is G nilpotent provided that all minimal subgroups of G lie in the center of .
Buckley [3] proved that if all minimal subgroups of an odd order group
are normal, then the group is supersolvable. Shaalan [10] proved that if
S G G
every subgroup of G of prime order or 4 is -quasinormal in ,then is G
supersolvable. Recently, the authors [2, 8] proved the following: Put ( )=
fp p p g p p p P p
n n i
= 1 2 ,where 1 2 .Let i be a Sylow -subgroup
e
i
G P p i n
of and let the exponent of Ω( i )be ,where =1,2, , . Suppose that
i
e
i
fH j H P H H p i
all members of the family Ω( i ), =1,Exp = , =1,2, ,
i
g G G
n are normal (quasinormal) in .Then is supersolvable. The object of this
G fp p p g p p p n
paper is to get: Put ( )= 1 2 n ,where 1 2 .Let
e
i
P p G P p
i i
i be a Sylow -subgroup of and let the exponent of Ω( )be ,where
i
i n fH j H P
=1,2, , . Suppose that all members of the family Ω( i ),
e
i
H p i n g S G G
H =1,Exp = , =1,2, , are -quasinormal in .Then is i supersolvable. Throughout this paper the term group always means a group of finite order. Our notation is standard and taken mainly from [4].
2019. majus´ 4. –23:08 4 M. RAMADAN
2. Main results
We prove the following result:
Theorem Let p be the smallest prime dividing jG j and let P be
e
Suppose that all Sylow p subgroup of G of exponent p where e
a 1
e
H p g are S quasinormal in fH j P H
members of the family =1 Exp =
G Then G has a normal p complement
Proof G j H
We prove the Theorem by induction on j .Let be a cyclic
e
p H S
subgroup of P of order . Our hypothesis implies that is -quasinormal in
HQ G Q q
G . So it follows that is a subgroup of ,where is a Sylow -subgroup
q p HQ p
of G and .Then has a normal -complement by [5, Satz 2.8, p.420].
T
x
P H P G H P HQ G
We have that G = is normal in and so = is
x G
p
HQ H Q O G hQ j Q
normal in HQ. It follows that = . Thus ( )= is
q G q p i C H C H G G
a Sylow -subgroup of , G ( ). If ( )= for all cyclic
e
P P G G
subgroups of order p in , then it is easy to see that Z( )andso has
p C H G
a normal -complement by [4, Theorem 4.3, p.252]. Let G ( ) for some
e
H p C H p
cyclic subgroup of order .Then G ( ) has a normal -complement by
p p
jG j O G C G O G
induction on .Since ( ) H ( ), we have that ( ) has a normal G p -complement and so also does .
As an immediate consequence of Theorem 2.1, we have:
Corollary Put G fp p p g where p p p n
( )= 1 2 n 1 2
e
i
n Let P be a Sylow p subgroup of G of exponent p where i i
i =2 3
i
e
i
Suppose that all members of the family fH j H P H H p
i =1 Exp =
i
n g are S quasinormal in G Then G possesses an ordered Sylow
i =2 3
tower
Proof G p P n
By Theorem 2.1, has a normal n -complement. Let be a
p G K p P G
n n
Sylow n -subgroup of and be a normal -complement of in .By G induction, K possesses an ordered Sylow tower. Therefore, possesses an ordered Sylow tower too.
We need the following Lemma:
e
Lemma Let P be a normal Sylow p subgroup of G of exponent p
such that G P is supersolvable Suppose that all members of e
where 1
e
the family H H j H P H H p g are S quasinormal in G
= f =1 Exp =
Then G is supersolvable
2019. majus´ 4. –23:08
THE INFLUENCE OF S -QUASINORMALITY OF SOME SUBGROUPS 5
jG j H
Proof We prove the lemma by induction on . If every element in
G P Q G Q G is normal in , then by [2, Theorem 4(b), p.253], ( ), where ( )
is the largest supersolvably embedded subgroup of G (see[12]), and hence
hx j x P jx j i Q G G
, is a prime or 4 ( ). Therefore, is supersolvable
by Yokoyama [11]. Thus we may assume that there exists an element H
H G H
in H such that is not normal in . Our hypothesis implies that is
G HQ G q p
S -quasinormal in . So it follows that is a subgroup of and .
HQ H HQ
Clearly, H is a subnormal Hall subgroup of . Thus is normal in
p p
Q N H O G N H G L HO G N H
G G
and so G ( ). So ( ) ( ) .Let = ( ) ( ).
PL G P L L P L
Then G = .Since = ( ) is supersolvable, it follows that
p
G j O G
is supersolvable by induction on j .Then ( ) is a normal supersolvable
p
O G G
subgroup of G .Since ( ) is a normal supersolvable subgroup of ,it
p
G G
follows by [9, Exercise 7.2.23, p.159] that PO ( )= is supersolvable.
Theorem Put G fp p p g where p p p n
( )= 1 2 n 1 2
e
i
wherei Let P be a Sylow p subgroup of G of exponent p n i
i =1 2
i
e
i
Suppose that all members of the family fH j H P H H p
i =1 Exp =
i
i n g are S quasinormal in G Then G is supersolvable
=1 2 G
Proof By corollary 2.2, possesses an ordered Sylow tower. Then
G G p P is normal in . By Schur–Zassenhaus’ theorem, possesses a -Hall
1 i
P G K
subgroup K which is a complement to 1 in . Hence is supersolvable by
G j G
induction on j . Now it follows from lemma 2.3 that is supersolvable.
e
Theorem Let P be a normal p subgroup of G of exponent p where
e G P is supersolvable Suppose that all members of the family
1 such that
e
H p g are S quasinormal in G Then G is H j H P H
f =1 Exp =
supersolvable
Proof G j P
We prove the Theorem by induction on j .Let 1 be a Sylow G
p -subgroup of . We treat the following two cases:
P P G
Case = 1. Then by lemma 2.3, is supersolvable.
Case P P G fp p p g p p p n
1.Put ( )= 1 2 n ,where 1 2 . G P
Since G P is supersolvable, it follows by [1, Theorem 5, p.5] that
K P jG P H Pj
possesses supersolvable subgroups H P and such that : =
p jG P K P j p H P K P
= 1 and : = n .Since and are supersolvable, it
K jG j jG H j
follows that H and are supersolvable by induction on .Since : =
jG P H Pj p jG K j jG P K P j p = : = 1 and : = : = n , it follows by [1,
Theorem 5, p.5] that G is supersolvable.
2019. majus´ 4. –23:08
6 M. RAMADAN
Corollary Let K be a normal subgroup of G such that G K is
supersolvable Put K fp p p g where p p p and s
( )= 1 2 s 1 2
e
i
s where i let P be a Sylow p subgroup of K of exponent p i
i =1 2
i
e
i
Suppose that all members of the family fH j H P H H p
i =1 Exp =
i
i s g are S quasinormal in G ThenG is supersolvable
=1 2
jG j
Proof We prove the corollary by induction on . Theorem 2.4 implies
P K P
that K is supersolvable and so 1 is normal in ,where 1 is a Sylow
K p jK j P
p1-subgroup of and 1 is the largest prime dividing . Clearly, 1 is
G P K P G K
normal in G .Also,( 1) ( 1) = is supersolvable. Now we
jG j conclude that G P1 is supersolvable by induction on . Now it follows
from Theorem 2.5 that G is supersolvable. The corollary is proved.
For a finite group P , we write
p Ω1( P )if 2
Ω(P )= p Ω2( P )if =2
where, as usual,
i
i P hx P jjx jjp
Ω i ( )=
We are now in a position to prove the following results:
Theorem Let p be the smallest prime dividing jG j P be a Sylow
e
subgroup of G and let the exponent of P be p where e Suppose
p Ω( ) 1
e
P H H p g are all members of the family fH j H
that Ω( ) =1 Exp =
S quasinormal in G ThenG has a normal p complement
e
Proof P p
Let H be an abelian subgroup of Ω( ) of exponent .Our
S G HQ
hypothesis implies that H is -quasinormal in . So it follows that is a
Q q G q p H
subgroup of G ,where is a Sylow -subgroup of and . Clearly,
p
HQ Q N H O G N H G G
is normal in and so G ( ). Thus ( ) ( ) . Clearly,
p p p
HO G N H G HO G N H G HO G G
( ) G ( ) .If ( ) ( ) ,then ( )has
K K p
a normal p -complement, say, by induction. Thus is a normal -Hall
p p p
G K O G O G G
subgroup of O ( ). Since char ( )and ( ) is normal in ,it
p
G G O G p
follows that K is normal in .Since ( )isa -group, we have that
p G G p
K is a normal -Hall subgroup of and so has a normal -complement.
N H G H G
Thus we may assume that G ( )= . In particular, is normal in . p
If G has no normal -complement, then by Frobenius’ theorem, there exists
p L G N L C L p G
a nontrivial -subgroup of such that G ( ) ( ) is not a -group.
L P r jN L j
Clearly, we can assume that .Let be any prime dividing G ( )
r p R r N L R
with and let be a Sylow -subgroup of G ( ). Then normalizes
2019. majus´ 4. –23:08
THE INFLUENCE OF S -QUASINORMALITY OF SOME SUBGROUPS 7
L L R N L H G
andsoΩ( ) is a subgroup of G ( ). Since is normal in , hence
L R p
Theorem 2.1 implies that ( H Ω( )) has a normal -complement and so also
R R L r
does Ω( L) . By [5, Satz 5.12, p.437], centralizes . Thus for each prime
jN L j r p r R N L G
dividing G ( ) with , each Sylow -subgroup of ( ) centralizes
L N L C L p G G
and hence G ( ) ( )isa -group; a contradiction. Therefore has a
normal p -complement.
As an immediate consequence of Theorem 2.7 we have:
Corollary Put G fp p p g wherep p p n
( )= 1 2 n 1 2
e
i
Let P be a Sylow p subgroup of G and let the exponent of P be p
i i
i Ω( )
i
where i n Suppose that all members of the family fH j H
=2 3
e
i
i P H H p n g are S quasinormal in G
Ω( i ) =1 Exp = =2 3
i
Then G possesses an ordered Sylow tower
We need the following lemmas:
Lemma [M. Ezzat, Finite groups in which some subgroups of prime
power order are normal, M. SC. Thesis, Cairo University (1995)]. Suppose
P K is supersolvable P is a normal Sylow p subgroup of G and that
that Ω( )
where K is a p Hall subgroup of G ThenG is supersolvable
Lemma Let P be a normal p subgroup of G such that G P is
e
supersolvable and let the exponent of be p where e Suppose
Ω(P ) 1
e
that all members of the family fH j H H H p g are
Ω(P ) =1 Exp =
S quasinormal in G ThenG is supersolvable
jG j P
Proof We prove the lemma by induction on .Let 1 be a Sylow G
p -subgroup of . We treat the following two cases:
P P G
Case = 1. Then by Schur–Zassenhaus’ theorem, possesses a
K P G G P K
p -Hall subgroup which is a complement to in . Thus = is
P P G P
supersolvable. Since Ω( P )char and is normal in , it follows that Ω( )is
P K G P K G G P normal in G .ThenΩ( ) is a subgroup of .IfΩ( ) = ,then Ω( )
is supersolvable. Therefore G is supersolvable by Theorem 2.5. Thus we
K G P K P K
may assume that Ω(P ) .SinceΩ( ) Ω( ) = is supersolvable, it K follows by Theorem 2.5 that Ω( P ) is supersolvable. Applying lemma 2.9,
we conclude the supersolvability of G .
Case P P G fp p p g p p p n
1 put ( )= 1 2 n ,where 1 2 . G P
Since G P is supersolvable, it follows by [1, Theorem 5, p.5] that
K P jG P H Pj
possesses supersolvable subgroups H P and such that : =
p jG P K P j p H P K P = 1 and : = n .Since and are supersolvable, it
2019. majus´ 4. –23:08
8 M. RAMADAN
K jG j jG H j
follows that H and are supersolvable by induction on .Since : =
jG P H Pj p jG K j jG P K P j p = : = 1 and : = : = n , it follows by [1,
Theorem 5, p.5] that G is supersolvable.
As an immediate consequence of corollary 2.8 and lemma 2.10, we have:
Theorem Put G fp p p g where p p p n
( )= 1 2 n 1 2
e
i
Let P be a Sylow p subgroup of G and let the exponent of P be p
i i
i Ω( )
i
n Suppose that all members of the family fH j H i
where =1 2
e
i
P H H p n g are S quasinormal in G i
Ω( i ) =1 Exp = =1 2
i
Then G is supersolvable
jG j
Proof We prove the Theorem by induction on . By corollary 2.8,
P G
we have that G possesses an ordered Sylow tower. Then 1 is normal in .
p K
By Schur–Zassenhaus’ theorem, G possesses a 1-Hall subgroup which is
G K a complement to P1 in . Hence is supersolvable by induction. Now it
follows from lemma 2.10 that G is supersolvable.
We now obtain at once:
Corollary Put G fp p p g where p p
( )= 1 2 n 1 2
p Let P be a Sylow p subgroup of G and let the exponent of P
n
i i
Ω( i )
e
i
where i p n Suppose that all members of the family
be =1 2
i
e
i
P H H p i n g are S quasinormal in H j H
f Ω( ) =1 Exp = =2 3
i
G Then possesses an ordered Sylow tower
(i) G is supersolvable (ii) G P1
We can now prove:
Corollary Let K be a normal subgroup of G such that G K is
supersolvable Put K fp p p g where p p p and let s
( )= 1 2 s 1 2
e
i
P be p where P be a Sylow p subgroup of K and let the exponent of
i i
i Ω( )
i
i s Suppose that all members of the family fH j H P
=1 2 Ω( i )
e
i
H i H p s g are S quasinormal in G Then G is
=1 Exp = =1 2
i
supersolvable
jG j
Proof We prove the corollary by induction on . Theorem 2.11
P K P
implies that K is supersolvable and so 1 is normal in ,where 1 is a
K p K
Sylow p1-subgroup of and 1 is the largest prime dividing the order of .
K K G P Clearly, P1 char and since is normal in , it follows that 1 is normal
2019. majus´ 4. –23:08
THE INFLUENCE OF S -QUASINORMALITY OF SOME SUBGROUPS 9
G P K P G K G P
in G .Since( 1) ( 1) = is supersolvable, it follows that 1
G j G is supersolvable by induction on j . Therefore is supersolvable by lemma 2.10. The corollary is proved.
References
Annales Univ Sci Bu [1] M Asaad On the supersolvability of finite groups I,
dapest 18 (1975), 3–7. M Ezzat [2] M Asaad M Ramadan and Finite groups with normal subgroups
of prime power orders, Pure Math Appl 5 (1994), 251–256. Math Z [3] J Buckley Finite groups whose minimal subgroups are normal, 116
(1970), 15–17. Finite groups
[4] D Gorenstein Harper and Row, New York, 1968.
Endliche gruppen I [5] B Huppert Springer Verlag, Berlin–Heidelberg, New
York, 1967. Nagoya Math J [6] N Ito Uber¨ eine zur Frattini-gruppe duale Bildung, 9 (1955),
123–127. Math Z [7] O H Kegel Sylow-gruppen and Subnormalteiler endlicher gruppen, 78 (1962), 205–221.
[8] M Ramadan Finite groups in which some subgroups of prime power order
are quasinormal, J Egypt Math Soc 1 (1997), 1–7.
R Scott Group theory [9] W Prentice-Hall, Englewood Cliffs, New Jersey,
USA, 1964.
Shaalan [10] E The influence of -quasinormality of some subgroups on the
structure of a finite group, Acta Math Hung 56 (1990), 287–293. J [11] A Yokoyama Finite solvable groups whose -hypercenter contains all mini-
mal subgroups, Arch Math 26 (1975), 123–130.
Between nilpotent and solvable [12] M Weinstein (editor) Polygonal Publishing House, Passaic, 1982.
2019. majus´ 4. –23:08
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 11–26
TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR
By SZ. VATTAMANY´ and CS. VINCZE
Institute of Mathematics and Informatics, University of Debrecen, Debrecen Received March
Introduction Berwald In his outstanding work [2], Ludwig showed that a two-
dimensional Landsberg manifold reduces to a Berwald manifold if its Douglas
Landsberg spaces the extremals
tensor vanishes. In his own words: The
of which form a quasigeodesic system of curves are identical with the two
dimensional a nely connected Finsler spaces ([2], p. 110.) The notion of G a “quasigeodesic system of curves” was introduced by W Blaschke and
Bol in their book [3]. In modern language, the condition on the extremals states that the geodesics of a Landsberg manifold coincide with those of a linear connection on the underlying manifold. Since by the Douglas–Shen theorem ([7], 6.6) this property is equivalent to the vanishing of the Douglas tensor, our formulation is indeed a translation of Berwald’s theorem. The analogous result in the higher dimensional case was proved in [9] with the help of the projection of the Douglas tensor onto the indicatrix bundle. Unfortunately, this elegant method does not work in two dimensions, since the Douglas tensor then has components only along the Liouville vector field by formula (26a) of Remark 2.4 below. Since the Liouville vector field is
orthogonal to the unit sphere bundle, we infer immediately that the projected
Douglas tensor of a two dimensional Finsler manifold vanishes identically So we have to search for a completely different plan of attack in the two- dimensional case. Our approach is based on Berwald’s original ideas, and it may be considered as an intrinsic version of them.
We shall adopt throughout the notations, terminology and conventions of
positive de nite [9] with one restriction. In this paper (M E ) will denote a
two dimensional Finsler manifold (There is a generalization of the theorem
2019. majus´ 4. –23:07 12 SZ. VATTAMANY,´ CS. VINCZE for two-dimensional Finsler manifolds with nondegenerate Riemann–Finsler metric. It also needs a little modification of Berwald’s method as we can see in [1] using the machinery of classical tensor calculus.)
1. Two-dimensional Finsler manifolds C
The Berwald frame Starting from the Liouville vector field and M E the canonical spray S of ( ), let us first consider the unit vector fields
1 1
p p
C S S
C0 := and 0 := E 2 E 2
Next, using the Gram–Schmidt process we construct, at least locally, a g -
v 0
X X TM g T TM orthonormal basis (C0 0)of ( ), where 2( ) is the metric ten-
sor given by (20) in [9]. Applying the almost complex structure F associated g
with the Barthel endomorphism of (M E ), we obtain a local -orthonormal
h
S X TM
basis (FX0 0)of ( ). The quadruple
X FX S
( C0 0 0 0)
TM
constructed in this way is a (local) orthonormal basis of X( ); it is called M E the Berwald frame of the Finsler manifold ( ).
Note that
E (1) X0 =0
This can be seen by a straightforward calculation:
(20) [9] (15) [9]
g C X C F X d E FX
0= ( 0) = ( 0) ==J ( 0)=
(13) [6]
JF X dE vX dE X X E
= dE [ ( 0)] = [ 0)] = ( 0)= 0
Proposition The members of the Berwald frame have the follow
ing homogeneity properties
C
(2) [ C C0]= 0
(3) [ C S0]=0
X
(4) [ C X0]= 0 (5) [ C F X0]=0
2019. majus´ 4. –23:07 TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR 13
1 p Proof Taking into account the fact that the function is homoge-
2 E
neous of degree 1, i.e., that
1 1
p p
(6) C = E 2 E 2
we obtain
1 1 1
p p p
C C C C C C C
[ 0]= C = = = 0
E E
2 E 2 2
C S S
which shows (2). Similarly, since S is a spray and hence [ ]= ,
1 1 1 1 (6)
p p p p
C S C S S C S
[ C S0]= [ ]+ = + =0
E E E 2 E 2 2 2 so (3) is also true. In view of the definition (21) and the property (27) in [9] of the first
Cartan tensor C,
g C S F X X L J g FX FX
0=2 ( ( 0) 0)= C ( )( 0 0)=
1 1
X X g J C F X X g J C F X X = Cg( 0 0) 2 ( [ 0] 0] = 2 ( [ 0] 0)
In the same way we find that
C S F X C Cg X C g J C F X C g X J C S
0=2g ( ( 0) )= ( 0 ) ( [ 0] ) ( 0 [ ]) =
(1 1) (1 1)
g J C F X C g X C g J C F X C
= ( [ 0] ) ( 0 ) = ( [ 0] )
C F X C F X It follows from the last two relations that J [ 0]=0,andso[ 0]is
vertical. Thus,
(13) [6] (9) [6]
vX JFX J C FX X C J FX C C X X0 = 0 = 0 =[ ]( 0)=[ 0 ] [ 0 ]= [ 0]
which proves the relation (4). C Taking into account again the fact that the vector field [ FX0 ] is verti- cal, from the homogeneity of the Barthel endomorphism (see (18) in [9]) we
obtain
FX h FX C h FX C 0=[h C ]( 0)=[ ( 0) ] [ 0 ]=
(13) [6]
FX C F vX C FX C =[h ( 0) ] =[( 0) ]=[ 0 ] whence (5).
2019. majus´ 4. –23:07
14 SZ. VATTAMANY,´ CS. VINCZE
Proposition With the hypothesis above the following relations
hold
C FX v X FX v h
(7) ( FX0 0)= [ 0 0](:= 1 )
X S
(8) v [ 0 ]=0
Y X TM D X where D is the Cartan connection
(9) ( ): hY 0 =0
C Proof According to the definition of the second Cartan tensor (see
e.g. [9]. formula (23)),
g C FX FX X L g J FX J FX
h 2 ( ( ) )=( FX )( ( ) ( )) = 0 0 0 ( 0) 0 0
(13)/[6]
L g X X FX g X X g FX X X
=(FX )( )= ( ) 2 ([ ] )= 0 0 0 0 0 0 0 0 0
1.1
X FX X
=2g ([ 0 0] 0)=
v X FX X g h X FX X g v X FX X =2g ( [ 0 0] 0)+2 ( [ 0 0] 0)=2 ( [ 0 0] 0)
using the g -orthogonality of the vertical and horizontal subbundle in the last
step. From this we obtain the relation
C FX FX v X FX X (10) g ( ( 0 0) [ 0 0] 0)=0 Furthermore, by the properties (26), (27) in [9] of the second Cartan tensor,
we can write
C FX FX C
0=2g ( ( 0 0) )=
g X C g FX X C g X FX C = FX0 ( 0 ) ([ 0 0] ) ( 0 [ 0 ]) =
1.1, (5)
X FX C
= g ([ 0 0] )=
v X FX C g h X FX C g v X FX C = g ( [ 0 0] )+ ( [ 0 0] )= ( [ 0 0] )
This means that the equality
C FX FX v X FX C (11) g ( ( 0 0) [ 0 0] )=0
is valid automatically. The relations (10) and (11) show that the vertical vector
FX FX v X FX field C ( 0 0) [ 0 0] is orthogonal to two nowhere vanishing vertical vector fields. Hence it must be the zero vector field, which proves (7).
Similarly, on the one hand
C S F X X Sg X X g S X X
0=2g ( ( 0) 0)= ( 0 0) 2 ([ 0] 0)=
X S X
=2g ([ 0 ] 0)=
v X S X g h X S X g v X S X =2g ( [ 0 ] 0)+2 ( [ 0 ] 0)=2 ( [ 0 ] 0)
2019. majus´ 4. –23:07 TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR 15
on the other hand
C S F X C Sg X C g S X C g X S C
0=2g ( ( 0) )= ( 0 ) ([ 0] ) ( 0 [ ]) =
X S C g X S g X S C
= g ([ 0 ] )+ ( 0 )= ([ 0 ] )=
v X S C g h X S C g v X S C = g ( [ 0 ] )+ ( [ 0 ] )= ( [ 0 ] )
which imply the relation (8).
D X D X S
To prove the formula (9) it is sufficient to check that FX and 0 0 0
vanish. But this is immediate:
1 6 [9] (7)
D X v FX X C FX FX
FX = [ ]+ ( ) =0
0 0 0 0 0 0
1 6 [9] (8)
D X v S X
S 0 = [ 0] =0
Definition and Remark The function
g C FX FX X
(12) := ( ( 0 0) 0)
scalar M E
is said to be the main of ( ) with respect to the Berwald frame
X FX S X ( C0 0 0 0). – Actually, depends only on the choice of 0 and it is
uniquely determined up to sign.
Lemma With the help of the main scalar and the vector eld X
0
the rst Cartan tensors can be represented in the form
C i i X and C i i i
X b X X X
(13) = X0 0 0 = 0 0 0
where is the fundamental two form
C FX FX Proof The vertical vector field ( 0 0) can be uniquely repre-
sented as a linear combination
FX FX X C C TM C( 0 0)= 1 0 + 2 0 1 2 ( ) Since on the one hand
1 (27) [8]
p
g C FX FX C C FX FX S
( ( 0 0) 0)= b ( 0 0 ) =0
2 E
on the other hand
C FX FX C g X C C g ( ( 0 0) 0)= ( 1 0 + 2 0 0)= 2
it follows that 2 =0.So
g C FX FX X g X X
:= ( ( 0 0) 0)= ( 1 0 0)= 1
FX FX X hence C( 0 0)= 0,where is the main scalar. From this observation we infer immediately (13).
2019. majus´ 4. –23:07
16 SZ. VATTAMANY,´ CS. VINCZE
Corollary A positive de nite two dimensional Finsler manifold
is a Riemannian manifold if and only if its main scalar vanishes Proof This is an immediate consequence of (13), because (M E )isa
Riemannian manifold if and only if C =0.
Proposition A positive de nite two dimensional Finsler manifold
is a Berwald manifold if and only if the horizontal vector elds annihilate the
main scalar i e d
h =0
Proof We first recall that a Finsler manifold is a Berwald manifold if
and only if the h -covariant derivative, with respect to the Cartan connection, of the first Cartan tensor vanishes. (An intrinsic proof of this well-known fact
is available in [8].) Thus we can argue as follows:
M E is a Berwald manifold Y X TM D C
( ) ( ): hY =0
Y X TM D C FX FX
( ):0=( hY )( 0 0)=
(Fins2)/[6]
D C FX FX C D FX FX hY = hY [ )( 0 0)] 2 ( 0 0) =
(9) (13)
D C FX FX C FD X FX D C FX FX
hY hY = hY [ ( 0 0)] 2 ( 0 0) = [ ( 0 0)] =
(9)
D X hY X D X hY X d Y X
hY h
= hY ( 0)=[( ) ] 0 + 0 =[( ) ] 0 =[ ( )] 0
d
h =0
Corollary The second Cartan tensor of M E is completely
( )
determined by the formula
FX FX S X
(14) C ( 0 0)= ( ) 0
Proof D C C
Since S = (see [5], Prop. (A.12)), taking into account the
previous proof we obtain
C FX FX D C FX FX d S X h
( 0 0)= ( S )( 0 0)= [ ( )] 0 =
d hS X d S X S X
= [ ( )] 0 =[ ( ) ] 0 = ( ) 0
Proposition and Definition Let us consider the curvature tensor
1
R h h of M E Then
= 2[ ] ( )
FX S g R FX S X X
(15) R ( 0 0)= ( ( 0 0) 0) 0
and R is uniquely determined by this formula on the domain of the Berwald frame constructed in
1.1 The function
g R FX S X (16) := ( ( 0 0) 0)
2019. majus´ 4. –23:07
TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR 17
M E
is said to be the Gauss curvature of ( )
Proof
Since R is a semibasic tensor of type (1 2), it is uniquely de-
h
S X TM termined by its value on the local basis ( FX0 0)of ( ). So our only task is to verify the equality (15). Starting with the definition of the Nijenhuis torsion, we obtain
2
FX S hF X hS h FX S h FX hS h hF X S
R ( 0 0)= ([ 0 0]+ [ 0 0] [ 0 0) [ 0 0]) =
FX S h FX S h FX S v FX S = [ 0 0] [ 0 0]+2 [ 0 0]= [ 0 0]
Now, if
FX S f X f C f f C TM R ( 0 0)= 1 0 + 2 0 ( 1 2 ( ))
then on the one hand
R FX S C g f X f C C f g C C g ( ( 0 0) )= ( 1 0 + 2 0 )= 2 ( 0 )=
1 p
p
g C C Ef = f2 ( )= 2 2
2 E
d E
on the other hand, using repeatedly the fact that h =0,
R FX S C g v FX S C v FX S FC
g ( ( 0 0) )= ( [ 0 0] )= ( [ 0 0] )=
v FX S S = ( [ 0 0] )=
(16) [9]
i v FX S v FX S E
= S ( [ 0 0]) = [ 0 0]( )=
FX S E FX S E S FX E
= h [ 0 0]( ) [ 0 0]( )=[ 0 0]( )=
FX E FX S E
= S0[ 0( )] 0[ 0( )] = 0
R FX S f X
These imply that f2 =0, ( 0 0)= 1 0,and
g f X X g R FX S X f1 = ( 1 0 0)= ( ( 0 0) 0)
whence (15).
Theorem E Cartan For the Lie
( ’s “permutation formulas”) brackets of the members of the Berwald frame we have
1
p
S FX F X S X [ X0 0]= 0 0 ( ) 0
2 E
(17a–c) 1
p
X FX [ S0 0]= 0
2 E
S X
[ FX0 0]= 0
is the main scalar is the Gauss curvature of M E where ( )
2019. majus´ 4. –23:07 18 SZ. VATTAMANY,´ CS. VINCZE
Proof Since
(7) (14)
FX v X FX h X FX S X h X FX [ X0 0]= [ 0 0]+ [ 0 0] = ( ) 0 + [ 0 0] to prove the relation (17a) it remains to be shown that
1
p
X FX S F X (18) h [ 0 0]= 0 0
2 E First we observe that
(12)
g C FX FX X X g X X
2 =2( ( 0 0) 0)= 0 ( 0 0)
g J X FX X g J X FX X 2 ( [ 0 0] 0)= 2 ( [ 0 0] 0)
whence
J X FX X (19) g ( [ 0 0] 0)=
Similarly,
C FX FX C
0=2g ( ( 0 0) )=
g X C g J X FX C g X J X S
= X0 ( 0 ) ( [ 0 0] ) ( 0 [ 0 ]) =
g J X FX C g X J X S = ( [ 0 0] ) ( 0 [ 0 ]) =
Prop. I.7/[4]
g J X FX C g X X = ( [ 0 0] ) ( 0 0) Hence it follows that
1
p
J X FX C (20) g ( [ 0 0] 0)=
2 E
X FX
Now we use orthonormal expansion to express J [ 0 0]intermsofthe
v
C X TM
local basis ( X0 0)of ( ):
X FX g J X FX X X g J X FX C C J [ 0 0]= ( [ 0 0] 0) 0 + ( [ 0 0] 0) 0 =
(19) (20) 1
p
X C = 0 0
2 E
J h In view of the identity F = , from this we obtain (18) and hence (17a).
For the second formula (17b) we have
(8) 1
p
X h S X v S X h S X FJ X S [ S0 0]= [ 0 0]+ [ 0 0] = [ 0 0]= 0 =
2 E
(1) 1 Prop. I.7/[3] 1
p p
X S FX
= FJ[ 0 ] = 0 E 2 E 2
2019. majus´ 4. –23:07
TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR 19 S To prove (17c), first it will be shown that the vector field [FX0 0]is
vertical, i.e.,
FX S (21) h [ 0 0]=0
The vanishing of the h -horizontal torsion of the Cartan connection (see [9], 1.7 or [6], (M3)) yields
(9)
h FX S D S D FX D S FD X D S
S FX S FX
[ ]= FX = =
0 0 0 0 0 0 0 0 0 0 0 0
d E h
Using the fact that h = 0 and the vanishing of the -deflection of the Cartan
connection ([6], (M4)) we obtain
1 1 1
p p p
D S D S FX S D S
FX FX
FX = = + =
0 0 0 0 0
E E 2 E 2 2
1 1
p p
D FC FD C FX
= FX0 = 0 =0 E 2 E 2 thus (21) is true. By this observation and taking into account 1.9 it follows
that
S v FX S R FX S X [ FX0 0]= [ 0 0]= ( 0 0)= 0
completing the proof.
Proposition („Bianchi identity”)
X S S (22) + 0( )+ 0( )=0
Proof Starting with the Jacobi identity, taking into account that both
FX S d E
0 and 0 are horizontal vector fields and h = 0, finally using (17a–c)
we obtain
S FX FX X S S FX X
0=[X0 [ 0 0]] + [ 0 [ 0 0]] + [ 0 [ 0 0]] =
(17a–c) 1 1
p p
X FX FX S S F X S X
=[X0 0]+ 0 0 + 0 0 + 0 + ( ) 0 = E
2 E 2
X S FX S FX S S X S S X =(X0 ) 0 + [ 0 0]+ 0( ) 0 + ( )[ 0 0]+ 0( ) 0 =
(17b–c) 1
p
X X S FX S FX S S X =(X0 ) 0 + 0 + 0( ) 0 ( ) 0 + 0( ) 0 =
2 E
S S X =[X0( )+ + 0( )] 0 whence (22).
2019. majus´ 4. –23:07 20 SZ. VATTAMANY,´ CS. VINCZE
2. Two-dimensional Landsberg manifolds with vanishing Douglas tensor
Proposition A positive de nite two dimensional Finsler manifold
is a Landsberg manifold if and only if the main scalar is a rst integral of the
canonical spray i e
(23) S ( )=0
This is an immediate consequence of and in
Proof 2.1
Lemma Suppose that M E is a positive de nite two dimen
( )
sional Landsberg manifold Then the mixed curvature and the mixed Ricci
tensor of the Berwald connection are completely determined by the formulas
FX FX FX FX X (24) P ( 0 0) 0 = 0( ) 0
˜
FX FX FX
(25) P ( 0 0)= 0( )
is the main scalar of M E where ( )
Proof (25) is a trivial consequence of (24). To prove (24), we first recall
that the relation
X Y Z D C Y Z X Y Z X TM P
( ) = ( hX )( )( ( ))
holds in any Landsberg manifold (see e.g. [9], 2.1/(iv)). Thus, in our case
FX FX FX D C FX FX D C FX FX
P FX
( ) = ( FX )( )= ( )+ 0 0 0 0 0 0 0 0 0
(13)
C D FX FX D X FX
+2 ( FX ) = + 0 0 0 0 0
(9)
C FD X FX FX X
+2 ( FX ) = ( ) 0 0 0 0 0
whence (25).
Proposition The Douglas tensor of a positive de nite two dimensional Landsberg manifold is completely determined by
1
FX FX FX X FX F X C (26) D ( ) = [ ( ( )+2 ( )]
0 0 0 3 0 0 0 D
Proof As we know from 6.2 and 6.3 in [7], is semibasic, symmetric,
S i D D
and for any semispray , S = 0. Thus, in two dimensions is completely 0 0
2019. majus´ 4. –23:07
TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR 21
FX FX determined by its value on the triplet ( FX0 0 0). According to 6.2/(b)
in [7],
1 ˜
D D FX FX FX FX FX FX FX FX FX C
P P
( ) = ( ) J ( )
0 0 0 0 0 0 3 0 0 0
˜ (24) (25) 1 ˜
FX FX X D FX FX C
P P
( ) = X ( ) =
0 0 0 3 0 0 0
1 ˜ ˜ (25)
X FX FX D FX FX C
P P
= ( ) 2 X =
3 0 0 0 0 0 0
1 ˜
FX FX FX X D C P
= ( ( )) + 2 X 3 0 0 0 0 0 There remains only to calculate the second member of the right hand side.
Applying the rules of calculation of the Berwald connection ((27) and ( Brw1–
4) in [6]) we obtain
(17a)
D FX F D JFX FJ X FX h X FX X 0 0 = JFX0 0 = [ 0 0]= [ 0 0] =
1 1
p p
S S h F X S hX F X
= 0 ( ) 0 ( ) 0 = 0 0 E 2 E 2
Hence
1 ˜ ˜
p
D FX FX S FX FX FX
P P P
X = ( ) ( )= 0 0 0 0 0 0 0
2 E
(4.4a)/[7] ˜ (25)
FX FX F X = P( 0 0) = 0( ) and the result follows.
Remark Using the same technique, it may be proved that the effect
˜
D any P of the tensors P , and can be described analogously in (positive definite, two-dimensional) Finsler manifold. More precisely, the following
relations are fulfilled in the general case:
S ( )
p
FX FX FX FX X S X C (24a) P ( 0 0) 0 = [ 0( )+ 0( )] 0 +2 0
2 E
˜
FX FX FX X S (25a) P ( 0 0)= 0( ) 0( )
1
X FX FX FX FX X X S (26a) D ( ) = ( ( )) + ( ( ))+
0 0 0 3 0 0 0 0
S ( )
X S C
+2 F X0( )+2 0( )+3 E
2019. majus´ 4. –23:07
22 SZ. VATTAMANY,´ CS. VINCZE
Suppose that M E is a positive de nite two dimensional
Lemma ( )
Landsberg manifold with vanishing Douglas tensor The iterated Lie deriva
tives of the main scalar with respect to the vector elds X FX S up to
0 0 0
fth order can be expressed as follows
L L L
FX FX (27) 0 X0 = 0
1
p
L L L
X FX
(28) S = 0 0 0
2 E
2 2 1
L L L L
X FX
(29) FX = 0 X0 0 0 2 E
2 3
p
L L L FX
(30) S = 0 X0 0
2 E
3 3 2 2
L L L L L
FX X
(31) FX = +3 + X
0 X 0 0 0
0 E
3 1 2 1
p
L L L L
FX X
(32) S = 4 7 +
0 X 0 0 0 E
2 E 2
4 3 2 2 2
L L L L L L
X X
(33) FX = +4 +3( ) 6 +
X X
0 0 X0 0 0 0
4 11 2 4 1
L L FX
+ X + +
0 2 0
E E
2 (2E )
n
L L L n times X
:= X X0 0 0
Proof We shall verify only the first three formulas, the remaining ones can be handled in the same way. First we observe that the vanishing of the
Douglas tensor implies by (26) the relation
FX F X (34) X0( 0( )) = 2 0( ) From now on we calculate.
(34) (17a) (23)
L L FX X L L FX X F X
X X FX (a) FX0 0 =[ 0 0] + 0 0 =[ 0 0] 2 0( ) =
1 (23)
p
S FX FX FX L
= ( 0 )+ ( 0) 2 ( 0) = ( 0) = FX ,
2 E 0 thus (27) is proved.
(17b) (23) 1 1
p p
L L L S X X S FX
FX X
(b) S =[ 0 0] + 0( 0 ) = ( 0) = , E 0 0 2 E 2 0 so we have obtained (28).
2019. majus´ 4. –23:07 TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR 23
2 (17a) (23)
L L L L X FX X X X FX X
FX X
c) FX = ( )=[ ]( )+ [ ( )] = 0 X0 0 0 0 0 0 0 0 0 0
1 (28) (27) 1
p
L S X FX X X FX X
= ( )+ ( )( )+ [ ( )] = FX 0 0 0 0 0 0 0 2 E
2 E 0
2 (27) 1 2
L X FX X X FX L FX
+ [ ( )] = FX + ( ( ) )=
0 0 0 0 2 E 0 0 0
1 2 (34)
L X L L L
FX X FX
= FX ( ) =
2 E 0 0 0 0 0
1 2 2 2 1
L L L L L
FX FX FX X
= X +2 = , E 2 E 0 0 0 0 2 0
showing that (29) is also valid.
Theorem If a positive de nite two dimensional Landsberg mani fold has a vanishing Douglas tensor then it is a Berwald manifold
Proof (A) In the next, quite tedious calculations our aim is to show that
L (35) FX0 =0
Then on the one hand
FX F v X h FX d FX
0=( 0) =[( ) 0] =[ ( 0)] =( h )( 0) on the other hand
(23)
d S d hS d S S
( h )( )=( )( )=( ) = ( ) =0
d M E
so it follows that h = 0 and, in view of Proposition 1.7, ( )isaBerwald manifold.
Notice that our subsequent reasoning relies heavily on the fact that
X (36) = 0( ) This relation is an immediate consequence of the Bianchi identity (22) and the property (23). (B) We start with the “permutation formula” (17c) and apply both its
sides to the main scalar. Taking into account (23), we obtain
L L X FX
(37) S = ( )
0 0 0
X FX Now we evaluate the vector field [ S0 0] on the function 0( ).
(34) (37)
X FX S X FX X S FX [ S0 0]( 0( )) = 0[ 0( 0( ))] 0[ 0( 0( ))] =
(23)
S FX X X = 2 0[ ( 0)( )] 0( 0( )) =
(37) (36)
S FX X X X X
= 2 0[( 0) ] 0( ) 0( ) 0( 0 ) =
X X X X X X X = 2 0( )+ 0( ) 0( 0 )= ( 0( )+ 0( 0 ))
2019. majus´ 4. –23:07 24 SZ. VATTAMANY,´ CS. VINCZE
(17b) 1
p
FX X Since, on the other side, [ S0 0] = 0, it follows that
2 E
1 2
p
X X X (38) L = ( ( )+ ( )) FX0 0 0 0 2 E
Applying the vector field X0 to both sides of (38), owing to (1) we obtain
1 2
p
L L X X X X
X = ( )( ( )+ ( ))+ 0 FX0 0 0 0 0
2 E
h i
2 2 3 (36) 2 2 3
L L L L L L
X X
+ ( X ) + + = ( +( ) + )
X X 0 0 X0 0 0 0 i.e.
1 2 2 2 3
p
L L L L L
X X
(39) X = ( +( ) + ) FX 0 0 0 0 X0 2 E
The Lie derivatives of the two sides of (34) with respect to FX0 yield
1 1 2 2
p p
L L L L L
X FX FX
(40) FX = [ 2( ) 2 ]
0 0 0 0 FX0 E 2 E 2 Taking the difference of (39) and (40), and then substituting the term
1 2
p
L from the right hand side of (38) we obtain FX
2 E 0
1
p
FX FX [ X0 0]( 0( )) =
2 E
2 2 3 2 2 2
p
L L L L L
X FX
= ( X +( ) + +2 )+ ( ) X 0 0 X0 0 0 2 E The left hand side of this equality can also be written in the form
1 (17a) (23)
p
FX FX [ X0 0]( 0( )) =
2 E
1 1
p
S FX FX FX
= 0[ 0( )] ( 0)( 0( )) = E
2 2 E
(37) (38) 1 2
X X X X = 0( )+ 0( )+ 0( 0 )
2 E so it follows that
2 2 3 2 2 2 1
p
L L L L L L
FX X X
(41) ( ) + ( +3 +( ) +2 + X )=0 X
0 X 0 0 0 0 0 E 2 E 2
2019. majus´ 4. –23:07 TWO-DIMENSIONAL LANDSBERG MANIFOLDS WITH VANISHING DOUGLAS TENSOR 25
Now we apply the vector field X0 to (41). Taking into account that
2 2 (1) 2 2 4
p p p
X L L L L L L
X FX FX X FX
0 ( FX0 ) = 0( 0 ) = ( 0 ) 0 0 =
E E 2 E 2 2
(34) 8 2
p
L
= ( FX ) = 0
2 E
(41) 3 2 2 2 1
L L L L L
X X
=4 +3 +( ) +2 + X X X 0 0 0 0 0 2 E
we obtain the relation
3 2 2 2 1
L X L L L L
X X
( ) +3 +( X ) +2 + + X 0 X 0 0 0 0
0 2 E
4 2 3 2
L L L L L X
+ +5( X ) +7 +8 ( ) +
X X
0 0 0 X0 0
2 2 1 2 3 2
L L L L X
+14 + +8 + X =0 X
X0 0 0 0 E 2 E
Using (36), this takes the form
4 3 2 1 2
L L L L
(42) +6 + 5 X +11 + +
X X 0 X 0 0
0 2 E
2 3
L L X
+ 7 X +6 + =0 0 0 2 E
(C) To conclude the proof, we finally discuss the relation (42).
L (A)
(a) If = 0, then we see from (41) that FX = 0. This means by that 0
( M E ) is a Berwald manifold.
(b) In the case 0 the second factor has to vanish on the left hand side of (42). Then we take the Lie derivative of this factor with respect to
the vector field FX0. Applying the relations (27)–(33), after a somewhat
lengthy but quite straightforward calculation we obtain
3 2 2 2 1
L L L L L L
X X FX
(43) +3 +( X ) +2 + =0 X X 0 0 0 0 0
0 2 E
L
If FX = 0, then the process ends. Otherwise the first factor on the left 0
hand side of (43) is zero, but, owing to (41), this also yields the desired
L relation FX0 =0.
2019. majus´ 4. –23:07 26 SZ. VATTAMANY,´ CS. VINCZE
References
acso M Matsumoto [1] S B and Reduction theorems of certain Landsberg spaces
to Berwald spaces, Publ Math Debrecen 48 (1996), 337–366.
[2] L Berwald On Finsler and Cartan geometries, III, Two-dimensional Finsler
spaces with rectilinear extremals, Ann of Math , 42 (19–11), 84–112.
G Bol Geometrie der Gewebe [3] W Blaschke and Springer Verlag, Berlin,
1935.
Ann Inst Fourier [4] J Grifone Structure presque tangente et connexions I,
Grenoble 22, 1 (1972), 287–334.
Ann Inst Fourier [5] J Grifone Structure presque tangente et connexions II,
Grenoble 22, 3 (1972), 291–338.
Lect Mat [6] J Szilasi Notable Finsler connections on a Finsler manifold, 10
(1998), 7–34.
Sz Vattamany Di er [7] J Szilasi and On the projective geometry of sprays,
Geom Appl 12 (2000), 183–208.
Cs Vincze
[8] J Szilasi and A new look at Finsler connections and special
azi Finsler manifolds, Acta Math Acad Paedagog Nyh (N.S.) 16 (2000),
33–63.
any [9] Sz Vattam Projection onto the indicatrix bundle of a Finsler manifold,
Publ Math Debrecen accepted for publication.
2019. majus´ 4. –23:07 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 27–31
EXTREMAL FROBENIUS NUMBERS IN SOME SPECIAL CASES
By G. KISS
Kisujsz´ all´ as,´ Hungary Received August
1. Introduction
a a a a a n
Let 0 1 2 n be integers with gcd( 1 )=1.Itis
P
n
N x a i
well-known that the equation = i has a solution in non-negative
i =1
x N G a a a n integers i provided is sufficiently large. Denote ( 1 2 )the
greatest integer N for which the preceding equation has no such solution.
G a a a Frobenius number
( 1 2 n ) is called a The computation and estimate G
of G have given rise to many papers. The question of the estimation of t
naturally suggests the following extremal problem [2]. For integers n and , n t
define g ( )by
g n t G a a a
( )=max ( 1 2 n )
a a a t n
where the max is taken over all i satisfying 1 1 ,
a a g n t extremal Frobenius number
gcd( 1; ; n )=1. ( ) is called an
os Graham n Erd and proved in [2] that, for 2,
2 2
t t
t g n t
5 ( ) 2
n
n 1
n n k k n They found the exact value of g ( 2 + )forfixed ,if is sufficiently large
2 k
( n 20 ):
n
k k n k
2 n +4 1 for 1and 1(mod3)
n n k
(1) g ( 2 + )=
k k n k
2 n +4 +1 for 1and 1(mod3).
n t
Dixmier has proved [1, Thm.1] that, for 2
t
t 2 1
t n g n t t
( +1) 1 ( ) 1 1
n n 1 1
2019. majus´ 4. –23:07
28 G. KISS
os Graham thus proving a conjecture by Erd and [3, page 86] stating that
2
t
n t
g ( ) ( n 1)
In the same paper Dixmier improved the upper bound as follows
n t
[1, Thm.3]. For 2
n t v t r
(2) g ( ) ( 1)( 1) 1
v n r r n
where t 1= ( 1) and 0 1
g n t Dixmier gave the exact value of ( ) for some special cases.
In this paper we find the exact value of some further extremal Frobenius numbers and extend the validity of Erdos–Graham’s˝ formula (1) for any n
k + 2 using Dixmier’s upper bound.
2. Main result
Let d n k be integers such that d n k n d
Theorem 2 0
n k d or n k d then If 0(mod +1) 1(mod +1)
2
g k d d n dk d d ( n dn + )= ( 1) +2 + 1
Proof With the notations of (2), we obtain
k d n n k d d n n k d
dn + 1=( +1)( 1) + + =( +1)( 1) ( )
n d v d r n k d
if k .Wehave = +1 and = . We can apply formula (2):
n t v t r d dn k n k d
g ( ) ( 1)( 1) 1= [ + ( ) 1] 1=
dn k n k d = d ( + + + 1) 1=
2
d n k d d d n dk d d
= d [( 1) +2 + 1] 1= ( 1) +2 + 1
a a a t The proof will be complete if we find integers 0 1 2 n such that
2
G a a a d d n dk d d
( 1 2 n )= ( 1) +2 + 1
k d
We consider the cases n 0and1(mod + 1) separately.
k d n l d k
Case (i). Let n 0(mod + 1). Write = ( +1)+ ,then
k ld d dk k d ld k dn + = ( +1)+ + =( +1)( + )
2019. majus´ 4. –23:07
EXTREMAL FROBENIUS NUMBERS IN SOME SPECIAL CASES 29
A fa a a g d l
Let = 1; 2; ; n consist of all multiples of ( +1)andthe largest
d t
elements of the residue class ( 1) modulo ( +1)upto :
fd d d ld k d ld k d
A = +1;2( +1);3( +1) ;( + 1)( +1);( + )( +1);
k dn k d dn k d dn k l d g
dn + 1; + 1 ( +1); + 1 2( +1); ; + 1 ( 1)( +1)
Aj ld k l l d k n z dn k l d
It is clear that j =( + )+ = ( +1)+ = .Let = + 1 ( 1)( +1) d be the smallest element of A, which is not a multiple of ( + 1). Let us write
z in the form
dn k l d ld k d l d
z = + 1 ( 1)( +1)=( + )( +1) 1 ( 1)( +1)=
ld k l
=(d +1)( + +1) 1
z d z dz
We know that 0; z ;2 ; ;( 1) ; is a complete residue system mod A ( d + 1). Hence the largest integer, which has no representation by is (see
e.g.[4])
A dz d d d ld k l d d
G ( )= ( +1)= ( +1)( + +1) 1=
d d l k d = d ( + 1)[( 1) + +1] 2 1=
2
d d l d d k d d d
= d ( +1)( 1) + ( +1) + + 2 1
k l d Substituting n = ( + 1), we obtain the desired
2
A d d n dk d d
G ( )= ( 1) +2 + 1
k d n k l d
Case (ii). Suppose n 1(mod +1).Then = ( +1) 1, or
l d k dn k d dl dk d k d dl k d
n = ( +1)+ 1. We see that + =( +1) + + =( +1)( + ) ,
k d dl k d d dl k
hence dn + 1=( +1)( + ) 1=( +1)( + 1) is a multiple
d A fa a a g d
of ( +1).Let = 1; 2; ; n consist of all multiples of ( +1)andthe
d t
l largest elements of the residue class (1) modulo ( +1)upto :
fd d d ld k d
A = +1;2( +1);3( +1) ;( + 1)( +1);
k dn k d dn k d dn k l d g
dn + ; + ( +1); + 2( +1); ; + ( 1)( +1)
Aj ld k l l d k n
It is obvious that j =( + 1) + = ( +1)+ 1= .Let
dn k l d A
x = + ( 1)( + 1) be the smallest element of , which is in the x
residue class (1) mod ( d + 1). We write in the form
dn k l d ld k d l d
x = + ( 1)( +1)=( + 1)( +1)+1 ( 1)( +1)=
ld k l d d l k
=(d +1)( + 1 +1)+1=( + 1)[( 1) + ]+1
x d x dx The proof is carried out analogously to Case (i), since 0; x ;2 ; ;( 1) ;
is a complete system of residues mod ( d + 1). The largest integer, which has
no representation by A is
A d d d l d d k d d G ( )= ( +1)( 1) + ( +1) + 1=
2
d n k d d k d d n dk d d = d ( 1)( +1)+ ( +1) 1= ( 1) +2 + 1
2019. majus´ 4. –23:07 30 G. KISS
3. Some corollaries
First we apply our Theorem for d =2.
Let n k be integers such that n k n If
Corollary 2 0 2
n k n k then
0(mod3) or 1(mod3)
g n k n k
( n 2 + )=2 +4 +1
k This improves the case n 1(mod3)oftheErdos–Graham˝ result
(1) by omitting the premise “ n is sufficiently large”. For the greatest permis-
n
sible value k = 2, we have
n n n n n
g ( 3 2) = 2 +4( 2) + 1 = 6 7
n
The next exact value is obtained by Dixmier [1, Thm.4]. Since 1 divides
n
t 2=3 3, we have
n
(3n 1)(3 3)
n n n n n n
g ( 3 1) = (3 1) + 1 = 3(3 1) 3 +2=6 1 n 1
Now, take d = 3 in the Theorem. We obtain:
Let n k be integers such that n k n If
Corollary 2 0 3
k or n k then
n 0(mod4) 1(mod4)
g n k n k ( n 3 + )=6 +6 +5
We can continue this specification and get exact formulae for further
extremal Frobenius numbers
Corollary Let d n be integers such that d n Ifn
0(mod d +1)
n d then or 1(mod +1)
2
n dn d d n d d
g ( )= ( 1) + 1
n n n g n n n g n n n For example: g ( 2 )=2 +1; ( 3 )=6 +5; ( 4 )=12 + 11.
Acknowledgement
I would like to thank Professor Robert´ Freud for helpful discussions, suggestions.
2019. majus´ 4. –23:07 EXTREMAL FROBENIUS NUMBERS IN SOME SPECIAL CASES 31
References
[1] J Dixmier Proof of a conjecture by Erdos˝ and Graham concerning the problem
of Frobenius, J Number Theory, 34 (1990), 198–209.
os R L Graham [2] P Erd and On a linear diophantine problem of Frobenius,
Acta Arithmetica 21 (1972), 399–408.
os R L Graham
[3] P Erd and Old and New Problems and Results in Combi-
ematiques natorial Number Theory, Monographies de l Enseignement Math
28 UniversitedeGen´ eve,` (1980).
Journal fur [4] E S Selmer On the linear diophantine problem of Frobenius,
reine und angewandte Mathematik 293/294 (1) (1977), 1–17.
2019. majus´ 4. –23:07
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 33–38
ON THE “GOOD” NODES OF WEIGHTED LAGRANGE INTERPOLATION FOR EXPONENTIAL WEIGHTS
By L. SZILI
Department of Numerical Analysis L. Eotv¨ os¨ University, Budapest Received October
In [8] J Szabados established a relation between the “good” nodes of weighted Lagrange interpolation and the weighted Lebesgue constants. In this
note we give a generalization of his results for other exponential weights.
X fx x g I R n N
n n n
Let := 1 n := or ( 1 1) ( )be
I R
an interpolatory matrix and w : be a given weight function. It is
w X n
known that (see e.g. [7] and [10]) the weighted Lebesgue constants n ( )
N
( n ) play a fundamental role in the convergence-divergence behaviour
w X n of sequences of weighted Lagrange interpolatory polynomials. n ( )is
defined as the supremum norm on I of the weighted Lebesgue function
n n
X X
w x
( n )( )
w X x jq x j
n n
( ):= =: k n ( )
w x x x
n k n
(1) ( ) ( k n )( ) k
k =1 =1
I n N
( x )
n
Q
x c x x
n n
where ( ):= ( k n ).
k =1
Throughout this paper we shall assume that our weight has the form
Q
e Q I R n
w := ,where : is even, continuous and convex. The th
a a w n Mhaskar–Rahmanov–Saff number n := ( ) is the (unique) positive root
of the equation Z
1
a tQ a t n
2 n ( )
p
dt n N n = ( )
2 t 0 1
Research supported by the Hungarian National Scientific Research Foundation (OTKA) under Grant No. T032719, T32872 and by FKFP 0198/1999.
2019. majus´ 4. –23:07 34 L. SZILI
One of its property is that
kr w k j r w x j j r w x j r P n N
n n n n
n := max ( )( ) =max ( )( ) ( )
x I
jx j a
n
P n
where n denotes the polynomials of degree . Freud type weights
Our classes of weights on R are the so-called ,where
Erdos type weights
Q is even and polynomial growth at + ,andthe ,where
Q is even and of faster than polynomial growth at + . The exponential
Q
k
w e Q x k
weights on ( 1 1) that we discuss include k := ,where ( ):=
2
x k
=exp((1 ) ). Here 0 0andexp := exp(exp( exp()))
k k
x x
is the k 1-th iterated exponential and exp0( ):= . We shall denote these
R E R EXP classes of weights by F( ), ( )and [ 1 1], respectively. For a formal
definition of these classes, see [6, Definitions 1–3].
x xQ x Q x x
It is not difficult to see that the function T ( ):=1+( ( )) ( )(
I Q T I
(0 + ) ) guarantees the regular growth of .If is bounded on then I
Q is at most polynomial growth on . This is true for Freud-type weights.
x x I
In contrast, if T ( ) + as tends to the endpoints to (it is true for
E R EXP Q w ( )or [ 1 1]) then is faster than polynomial growth near the
endpoint of I . We shall derive the generalizations of results in [8]1from the following
statement.
Theorem Let w F R E R or EXP If r P satis es n
( ) ( ) [ 1 1] n
n
p
c
w
T
n
kr w k e T T a
n n
(2) n ( := ( ))
with a constant c and for a point y I we have r w y then n
w 0 ( )( )=1
2 3
kr w k
log n
jy j a d n
1+ w
nT
n
with some constant d depending only on w
w 0
p
Remarks n T n
1. For our weights we have n + if + .
F R A B A T x B x
Indeed, if w ( ) then there exist 1 such that ( ) (
F R w E R
(0 + )), see the definition of ( ) in [6, p. 153]. For ( ) we know
c n T cn
that for any 0 there exists 0 independent of such that n
1
a jx j ra r n Actually formula (2) in [8] holds only for n with some constant 1. That means the [8] Proposition 1 is true with some additional condition.
2019. majus´ 4. –23:07
ON THE “GOOD” NODES OF WEIGHTED LAGRANGE INTERPOLATION 35
N w EXP
( n ) (see [2, (2.7)]). Finally, if [ 1 1] then for some 0we
2
T O n n have n = ( )( + ), see [4, (3.8)].
2. Actually, the starting point of [8] was the problem of the behaviour
X n N
of the point systems n ( ) for which the weighted fundamental poly-
q jq x j A k n
nomials k n (see (1)) are uniformly bounded, i.e. ( ) uniformly in
k n
x , and . These point systems serve as a basis of constructing convergent weighted interpolation polynomials of degree at most n (1 + ) (see [12] and [9]).
From the Theorem and Remark 1 we immediately obtain
Let w F R E R or EXP and suppose that the
Corollary ( ) ( ) [ 1 1]
point system X n N is such that the corresponding weighted fundamen
n ( )
k n n N are tal polynomials of Lagrange interpolation q
k n ( =12 )
c such that
uniformly bounded Then there exists 0
1 2 3
jx j a c n N n
max k n 1+ ( )
nT
k n
1 n
w X n N n
3. Now let us consider the Lebesgue constants n ( )( ). Let
y R w X w X y
n n n n n n be such that ( )= ( ), and consider the weighted
polynomial
n
X
r w x q y q x
n n k n
( )( ):= sgn k n ( ) ( )
k =1
Evidently
j r w x j w X x w X r w y
n n n n n n
( n )( ) ( ) ( )=( )( )
kr w k w X j r w x j
n n n n
that is = ( ). Since ( )( k n ) = 1 thus from the Theorem
we obtain
Let w F R E R or EXP Suppose that the
Corollary ( ) ( ) [ 1 1]
constant c satis es and the point system X n N is such that w
0 (6) n ( )
n
p
c
w
T
n
w X e n N n
n ( ) ( )
c such that
Then there exists 0
2 3
w X n
log n ( )
c n N jx j a n
max k n 1+ ( )
nT
k n
1 n
2019. majus´ 4. –23:07 36 L. SZILI
It is known that for the above weights there are such point systems for
w X n n N n which n ( ) log ( ) (see [7], [2], [1]), and the order is the best possible (see [10], [11]). Thus the “best” weighted Lagrange interpolation
point systems satisfy
2 3
log log n
c n N jx j a n
max k n 1+ ( )
nT
k n
1 n w
with some constant c 0 depending only on .
Proof of Theorem w r P n
Fix a weight function . Then for every n
N
( n )wehave
jx j
nU
n a
n
a
n
j r w x j e x R kr w k n
(3) ( )( ) n ( ) U
where n R is the “majorizing function” (see [3, Lemma 7.1], [5, (4.11)], [4,
(5.11)]). D Now we show that there exist the constants c 0and 0 (they depend
only on w ) such that
(4)
D
3 2
T
n if 0
T n
2
U c
I R n a
(1 + ) n
+ if =
3 2 D
T
n if 1
T
n
I
1 if =( 1 1)
1+ a
n
D T
For the interval [0 n ) this statement is Lemma 7.1(d) in [3] if
F R w E R
w ( ); Theorem 4.3 of [5] for ( ) and Theorem 5.3 in [4] if
EXP
w [ 1 1].
D T a a I
n n Now let n . Then there is a ( (1 + )) such that
2
Q a
( )( n )
U
(5) n a (1 + ) n
2 n
R a
(see [5, p. 228], [4, p. 53–56], [3, Lemma 7.1] with = n and use the
x x
inequality log(1 + x ) if 0).
F R c c c Q Q
If w ( ) then there are 1 2 0suchthat 1 ( ) ( )
c Q a Q a n n N n 2 ( ) (see [6, Definition 1]). Since n ( ) ( ) (see [3, (5.5)])
thus we have
2
Q c a a Q a n ( ) n ( ) 1
n 3
c
1
n n 1+ 1+
2019. majus´ 4. –23:07
ON THE “GOOD” NODES OF WEIGHTED LAGRANGE INTERPOLATION 37
E R EXP
which proves (4) for Freud type weights. If w ( )or [ 1 1] then we
x x I obtain (4) in a similar way using that xQ ( ) is increasing for (0 + )
(see [5, Lemma 2.1(ii)], [4, Lemma 3.1(ii)]) and
2 3 2
a Q a nT n N n
( ) n ( ) n (see [5, Lemma 2.2(i)], [4, Lemma 3.2(i)]).
2
3 2 3 2
T n N D T T
n n
Since n ( )if = thus from (4) we obtain
1+
c
that there exists w 0 such that
n
p
c
jx j D
w
T
n
r P n N kr w k j r w x j e
n n n
(6) ( n )( ) ( 1 )
a T
n n
r P j r w x j jx j a D T
n n n n
If n satisfies (2) then ( )( ) 1for 1 ,i.e.
jy j a D T n
n 1 . From (3) and (4) we obtain that
3 2
jy j
nT c
1 n
a
n
j r w y j e kr w k n 1= ( n )( ) Hence, a simple rearrangement yields the statement of the Theorem.
Acknowledgment
The author expresses his thanks to Professor Peter´ Vertesi´ for pointing out this topic and for his helpful remarks.
References
[1] S Damelin The weighted Lebesgue constant of Lagrange interpolation for
Acta Math Hungar exponential weights on [ 1 1], , 81 (3) (1998), 223– 240.
[2] S Damelin The Lebesgue functions and Lebesgue constant of Lagrange inter-
polation for Erdos˝ weights, J Approx Theory, 94 (1998), 235–262. D S Lubinsky [3] A L Levin and Christoffel functions, orthogonal polynomials,
and Nevai’s conjecture for Freud weights, Constr Approx , 8 (1992), 463–
535.
D S Lubinsky Christo el Functions and Orthogonal Poly
[4] A L Levin and
nomials for Exponential Weights on [ 1 1], Mem. Amer. Math. Soc. 535, Vol. 111 (1994).
2019. majus´ 4. –23:07
38 L. SZILI T Z Mthembu
[5] A L Levin D S Lubinsky and Christoffel functions and or-
Rendiconti di Matem thogonal polynomials for Erdos˝ weights on ( ),
atica Serie VII, Vol. 14, Roma (1994), 199–289.
S Lubinsky D Mache C [6] D and ( 1) means of orthonormal expasion for
exponential weights, J Approx Theory, 103 (2000), 151–182.
[7] J Szabados Weighted Lagrange and Hermite–Fejer´ interpolation on the real
line, J of Inequal and Appl , 1 (1997), 99–112.
[8] J Szabados Where are the nodes of “good” interpolation polynomials on the
real line, J Approx Theory, 103 (2000), 357–359.
Szili P Vertesi
[9] L and An Erdos˝ type convergence process in weighted
Acta Math Hungar interpolation. II. (Exponential weights on [ 1 1]),
(submitted)
ertesi [10] P V On the Lebesgue function of weighted Lagrange interpolation. I.
(Freud-type weights) Constr Approx , 15 (1999), 355–367.
ertesi [11] P V On the Lebesgue function of weighted Lagrange interpolation II.,
J Austral Math Soc Series A , 65 (1998), 145–162.
ertesi [12] P V An Erdos-type˝ convergence process in weighted interpolation. I.
(Freud-type weights), Acta Math Hungar , 91 (3) (2001), 195–215.
2019. majus´ 4. –23:07 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 39–52
MODIFIED WEIGHTED (0; 0 2)-INTERPOLATION ON INFINITE
INTERVAL ( + )
By P. MATHUR and S. DATTA
Department of Mathematics and Astronomy, Lucknow University, Lucknow, India Received October
1. Introduction
al In 1975, L G P [8] introduced the following interpolation process.
Let
x x n n (1 1) 1 n +
be a system of distinct real points and put
n
Y
W x x x n
(1 2) ( )= ( i n )
i =1
y i n W x
The roots i n ( =1 1) of ( ) are interscaled between the roots
n
W x
of n ( ), i.e.,
x y x y x
n n
n n n n
(1 3) n 1 2 1 1 +
n
n 1
f g f g i n
Pal´ proved that forgiven arbitrary numbers i n and there i
=1 i =1
n
exists a unique polynomial of degree 2 1 satisfying the conditions:
R x i n R y i n
n
i n i n i n
(1 4) ( i n )= =1 ; ( )= =1 1
n
R a a
and an initial condition n ( )=0,where is a given point, different from
x i n n
the nodal points i n ( =1 , =1 2 ).
Szili W x H x n
[10], firstly applied this method in the case for n ( )= ( )
H n n
(here n denotes the th Hermite polynomial). Taking even, he proved the existence, uniqueness, explicit representation and the convergence theorem
2019. majus´ 4. –23:07
40 P. MATHUR, S. DATTA
R x for the polynomial n ( ) satisfying the conditions (1.4) together with an
additional condition
n 2
X H
n (0)
R n
(1 5) (0) = 2 i n
H x n
( i n )
i =1
I Joo If n is odd, the uniqueness fails to hold. Later [5] improved his results
by sharpening the estimates of the fundamental polynomials.
en In a recent paper, Z F Sebesty [9] modified the the above methods
by replacing the special condition (1.5) by an interpolatorial condition
R n n (1 6) (0) = 0 n for even
or
R n
(1 7) (0) = n n for odd n Let
(1 8)
x
Z
2
H x x x H H n
( ) n ( ) ( )
n
A x l x n l t dt
i n i n
i n ( )= ( )+2 ( ) 2 +
H x H x H x
n n n
i n i n
( i n ) ( ) ( )
0
H H x H H x
n n
2 n (0) ( ) (0) ( )
n
i n
+ + =1
2 2
H x x H x
n n
i n i n
( i n ) ( )
x
Z
H x
n ( )
B x L t dt i n i n
(1 9) i n ( ) ( ) =1 1
H y n
( i n ) 0
For n even, taking
H x
n ( )
A x
(1 10) 0 n ( )= H
n (0)
he showed that
n
n 1
X X
R x A x B x
n
i n i n i n
(1 11) ( )= i n ( )+ ( ) i
i =0 =1
n is the uniquely determined polynomial of degree 2 1 satisfying the
conditions (1.4) and (1.6). For n odd, taking
H x
n ( )
B x
(1 12) n n ( )=
H
n (0)
2019. majus´ 4. –23:07 MODIFIED WEIGHTED (0; 0 2)-INTERPOLATION 41
and
x
Z
2
H H H x x x n
( ) n ( ) ( )
n
A x l x n l t dt
i n i n
i n ( )= ( )+2 ( ) 2
H x H x H x
n n n
i n i n i n (1 13) ( ) ( ) ( )
0 n
i =1 x
for i n = 0, he showed that
n n
X X
R x A x B x
n
i n i n i n
(1 14) ( )= i n ( )+ ( ) i
i =1 =1
n
is the uniquely determined polynomial of degree 2 1 satisfying the conditions (1.4) and (1.7). He also proved that for a function f : R R which is continuously differentiable satisfying
2 2
r x x
2 2 2
f x e r x e f
(1 15) lim x ( ) =0 =0 1 2 lim ( ) =0
jx j jx j
together with, for n even
f x i n i n
i n = ( ) =0
(1 16)
f x i n
= ( i n ) =1 1 i n
and for n odd
f x i n i n
i n = ( ) =1
(1 17)
f x i n
= ( i n ) =1 i n
following estimate holds
2 1
x
p
e jf x R x j O f x
(1 18) ( ) n ( ) = (1) R
n
n x f
where O does not depend on and . ( ) is the Freud’s modulus of
continuity for f .
azs L Szili
J Bal [2] and [11] have earlier studied analogous modified
T F Xie L G P al problems for weighted (0 2)-interpolation and [12], [7]
have investigated such modifications in Pal´ type interpolation.
H x H x Sri
Considering the nodes as the interscaled zeros of n ( )and ( ),
n
vastava n
and Mathur [6], taking even, proved that there exists a unique
n
polynomial of degree 3 2 sates ing the conditions:
G x g i n
n i n
( i n )= =1
(1 19)
G y g wG y g i n
n n i n
( i n )= ( ) ( )= =1 1
i n i n
2019. majus´ 4. –23:07 42 P. MATHUR, S. DATTA
2
x
2
x e
where w ( )= and n
2
X H
(0) l (0)
n
i n
G g
(1 20) (0) = i n
n
H x n
( i n )
i =1
For n odd the uniqueness is not true. They also proved that for a function
f : R R satisfying the requirements (1.15), together with
g f x i n i n
i n = ( ) =1
(1 21) p 2
y 1
i n
p
g f y g O ne f
= ( i n )and =
i n i n n
the following estimate holds:
2 1 3
x
p
e jf x G f x j O n f
(1 22) ( ) n ( ) = (log )
n 2
In [6] results have been obtained under a special condition (1.20), which
looks to be artificial. Also it has been proved that for n odd, either the interpolatory polynomial does not exit or if it exists, they are infinitely many.
In this connection we raise the following:
n
Problem For each positive integer do there exist a weighted (0; 0 2)-
G n interpolatory polynomial n of degree 3 2 satisfying the conditions
(1.19) and
G g n (1 23) n (0) = 0 if is even
or
g n (1 24) G (0) = if is odd n 0
If it exists what will be its explicit form and does it converse? In this paper, we answer this problem in affirmative. In section 2, we give some preliminaries and state new results in section 3. The estimates of the fundamental polynomials and the convergence theorem have been proved in sections 4 and 5 respectively.
2019. majus´ 4. –23:07 MODIFIED WEIGHTED (0; 0 2)-INTERPOLATION 43
2. Preliminaries
th
H x n
Let n ( )bethe Hermite polynomial with usual normalization
+ Z
2 p
n t
H n t H t e dt n m
n n m
(2 1) ( ) m ( ) = 2 ! ( N)
which satisfies the differential equation;
H x xH x nH x
( ) 2 ( )+2 n ( )=0
n n
(2 2)
H x nH x
( )=2 n ( )
n 1
x H x n
It is well known that i n roots of ( ) satisfy the following relations:
n n
x x x n m x
n n
0 n + ( =2 ) n n 1
2 +1 2
x x x n m
n n n
+1 =0 1 n + ( =2 +1)
(2 3) 2 n
n
x x i
i n
i n = n =1 2
+1 2
l L i n
Let i n and denote the Lagrange fundamental polynomial corresponding
x y i n
to the nodal points i n and respectively, then
H x
n ( )
l x i n
i n ( )= =1 (2 4)
H x x x
n i n
( i n )( )
x
H ( )
n
L x i n
i n ( )= =1 1 (2 5)
H y x y
n i n
( i n )( )
H x
For the roots of n ( ), we have
2
i
x
(2 6) i n
n
p
p 2
x
1 4 n 3 2
H x O n jx j e x n
(2 7) n ( )= (1) 2 ! 1+ R
n
n 1
X X p
2 p 2
x y
i i
O n e O n
(2 8) e = ( )and = ( ) 0 i
=1 i =1
n
n 1 X
X 2 2 2
x y
x
i n
i n 2 2 2
l x O e e L x O e
(2 9) e ( )= ( )and ( )= ( )
i n i n i i =1 =1
n !
jH j n
(2 10) n (0) = for even n 2 !
2019. majus´ 4. –23:07
44 P. MATHUR, S. DATTA
n 2
2 n !
2 1 2
n (2 11)
( n +1)!
The above results have been taken from [6], we shall also require the follow-
en
ing estimates given by Sebesty [9]
n X
2 p 2
x
2 x
i n
e jA x j O n e
(2 12) i n ( ) = ( )
i =1
and n
X 2 2
x
2 x
i n
e jB x j O e
(2 13) i n ( ) = (1)
i =1
A x B x i n
where i n ( )and ( ) are given by (1.8) and (1.9) respectively. We shall
x c l l A A
i n i i n i i n
use the following notations in the sequel i = , = , = ,
B B i n
i = .
3. New Results
Considering as the roots of nodal points and the weight Theorem (1 3)
2
x
2
function w x e there exists a unique polynomial G of degree
( )= n
n satisfying the conditions and or according as n
3 2 (1 19) (1 23) (1 24)
is even or odd
Theorem Let
x
Z
H x l t x H t
H x H x H x
i i
n ( ) ( ) ( ) n
( ) ( ) ( ) n
n n
2 i
dt A x l x
i ( )= ( ) 2
i
3
t x
H x
H x
n
i i
( ) n (3 1) ( i )
0 i
=1 n
x
Z
L t y L t
H x H x H x
i i n
n ( ) ( ) ( ) ( ) ( ) n
2 i
B n L x dt
i ( )= ( )+
i 2
H y t y
nH y
n
i i
( ) n
2 ( i )
0
(3 2)
x Z
2
n y
2 +1 H (0)
n
i
i L t dt n
i ( ) 2 =1 1
H y
2 ny n
2 ( i ) 0 i
2019. majus´ 4. –23:07
MODIFIED WEIGHTED (0; 0 2)-INTERPOLATION 45
and
x
Z
H x H x
n ( ) ( )
n
C x L t dt i n i
(3 3) i ( )= 2 ( ) =1 1
nw y H y
n i
4 ( i ) ( )
0
For n even let
x
Z
2
H x H x H x H x H t nt H t
n n
n ( ) ( ) ( ) ( ) ( )+2 ( )
n n n
A x dt (3 4) ( )= +
0 2 2 3 2 3 3
n x H n H t n 4 n (0) 4 (0)
0
then
n n
n 1 1
X X X
G x g A x g B x g C x
n
i i i
(3 5) ( )= i ( )+ ( )+ ( )
i i
i i
i =0 =1 =1
the uniquely determined polynomial of degree n satisfying the
is 3 2
and
conditions (1 19) (1 23)
For n odd let
H x H x
n ( ) ( )
n
B x
(3 6) n ( )=
2 H
n (0)
then
n n
n 1
X X X
G x g A x g B x g C x
i i i
(3 7) ( )= i ( )+ ( )+ ( )
n
i i
i i
=1 =1 i =1
the uniquely determined polynomial of degree n satisfying the
is 3 2
conditions and
(1 19) (1 24)
If f is a continuously di erentiable function
Theorem : R R
and then
satisfying the requirements (1 15) (1 21)
2 1
x
3 2
p
e jf x G f x j O f for even n
(3 8) ( ) n ( ) =
n
and
2 1
x
3 2
p
e jf x G f x j O f for n odd
(3 9) ( ) ( ) =
n n
We will prove only our main Theorem 3 as the proof of other theorems is quite similar to that of theorems in [1].
2019. majus´ 4. –23:07 46 P. MATHUR, S. DATTA
4. Basic Estimates of Fundamental Polynomials ( n even)
Lemma For n even
2
3 x 2
x j O e A
(4 1) j 0( ) = (1)
n X
2 p 2
x
3 x 2
i
e x j O n e jA
(4 2) i ( ) = ( )
i =1
where A x and A x are given by and respectively
i ( ) 0( ) (3 1) (3 4)
Proof Integrating the last term of (3.4), by parts and using
H t nt H t H t nt H t nH t n
( )+2 n ( ) ( )+2 ( )+2 ( )
n n n
lim 2 = lim =
t
t t
0 t 0 2
H t ( n +1) ( )
= lim n =0
t t 0
together with (2.2) and (2.7), we get (4.1).
n A x i n
For even, i ( ), =1 given by (3.1), can be written in a
convenient form as:
x
Z
H x H x H x
( ) ( ) n ( )
n n
A x l x n l t dt
i i
i ( )= ( )+2 ( )
H x H x H x
n n n
i i
( i ) ( ) ( )
0
2
H x H H x H x
n n
n ( ) 2 (0) ( ) ( )
n
A x
2 + i ( )
2
H x H x
H x
n n i
( i ) ( ) n
( i )
A x
where i ( ) is given by (1.8). Thus by (2.2) and (2.12), we have
n n
X X
p
H x j
2 j 2 2 2
n ( )
x x x
x 1 2 2 3 2
i i i
e jA x j e e jA x j O n e i
(4 3) i ( ) = ( ) = ( )
H x j
i
n 1( ) i i =1 =1
which completes the proof of the lemma.
Lemma For n even
n X
2 p 2
y
3 x 2
i
e jB x j O n e
(4 4) i ( ) = ( )
i =1
where B x is given by
i ( ) (3 2)
2019. majus´ 4. –23:07 MODIFIED WEIGHTED (0; 0 2)-INTERPOLATION 47
Proof By (2.2) and (2.5), it follows that
L t y L t t L
H t i
( ) i ( ) ( ) ( )
n
i i
y L t n L t i
(4 5) = + i ( )+ +(1 ) ( )
i
t y
H y
n i
2 ( i )
B x y i
Hence i ( ), given by (3.2), can be written in a convenient form as, for 0
L x L
x H x H ( ) (0)
n ( ) ( )
n i
2 i
L B x x y L x
i i
i ( )= ( )+ + + ( )
i
H y nH y
n n i
( i ) 2 ( ) 2 2
x
Z
H ( x )
n 2
n y L t dt
+( +2 ) i ( ) =
i
nH y n
2 ( i )
0
(4 6)
x Z
2
x L
x H x H ( )
n ( ) ( ) n
i 2
n y L x dt
= +( +2 ) i ( ) +
i
H y nH y
n n i
( i ) 2 2 ( )
0
H x H x H n
n ( ) ( ) (0)
n
L x
+ i ( )+
2
H y
nH y
n i
2 ( ) n
4 ( i ) y
For i = 0, by (3.2) and (4.5), we have
2
L x
H x
n ( ) ( )
i
B x
i ( )= +
H y n
( i ) 2
(4 7) x
Z
H x H x
( ) n ( )
2 n
n y L t dt L x i
+( +2 ) i ( ) + ( )
i
nH y H y
n n i
2 ( i ) 2 ( ) 0
Thus
n
n 1
X X
x 2 H 2
n ( )
y y
2
i i
e jB x j L x e
i ( ) ( )+
i
H y n
2 ( i ) i
i =1 =1
x
2 Z
n
1 y
X
i
n jH x H x je
( +1) n ( ) ( )
n
L t dt
+ i ( ) +
2
n
H y
2 n
( i )
i =1 0
x
2 Z
n
1 y
X
i
x H x je jH
n ( ) ( )
n
2
j y L t dt j
+ 1 i ( ) +
i
2
nH y
n
i
2 ( ) i =1 0
2019. majus´ 4. –23:07 48 P. MATHUR, S. DATTA
2 2
n n y
1 y 1
X
X 2
i i
e H x jH x H x je
( ) n ( ) ( )
n n
jL t j jH j n
+ i ( ) + (0)
2 2
H y ny H y
n n
i i
2 ( i ) 4 ( ) i
i =1 =1
I I I I (4 8) 1 + I2 + 3 + 4 + 5
From (2.9), we have
n 1 X
2 2 p 2
y x
3 x 2 2 3 2
i
I O e e O n e (4 9) 1 = (1) = ( )
i =1
By (2.13), we have
n 1
X
p
jH x j
n 2 2
2 n ( )
x
1 y 3 2
i
e I O jB x j O n e
(4 10) 2 = (1) i ( ) = ( )
jH y j n
( i )
i =1
2
2 y 2
i
B x j y j O e
where i ( ) is given by (1.9). Since (1 ) = ( ), hence
i
n 1
X
p
x j H
2 j 2 2
n ( )
x y
y 1 3 2
i i
I O e e jB x j O n e
(4 11) 3 = (1) i ( ) = ( )
jH y j n
( i )
i =1
p 2
3 x 2
O n e (4 12) I4 = ( )
By (2.10), we have
n 1
X
jH x H x j
2 n 2
n
2 n ( ) ( ) !
x
y 1 3 2
i
I O e O e (4 13) = (1) = (1)
5 2 n
y H y
n i
i ( ) !
i =1 2
Thus by using (4.9)–(4.13) in (4.8), the lemma follows.
Lemma For n even
n 1
X 2 1 2
x
y 2 3 2
i
p
e jC x j O e
(4 14) i ( ) = n
i =1
where C x is given by
i ( ) (3 3)
Proof By (2.13), we have
2
n n
1 1 y X
X 2
i
jH x j
2 e 2 2
n ( ) 1
y y x
2 1 2 3 2
i i
p
e jC x j e jB x j O e i
i ( ) = ( ) =
jH y j
n n
2 ( i ) i i =1 =1
2019. majus´ 4. –23:07
MODIFIED WEIGHTED (0; 0 2)-INTERPOLATION 49
n even
Proof of Theorem By [3], Theorem 4 and [4] Theorem 1,
p x n
there exists a polynomial n ( )ofdegree , such that
2 1 1
x
2
p p
e jf x p x j O f
(4 15) ( ) n ( ) =
n n
2 1
x
2
p
e jf x p x j O f
(4 16) ( ) ( ) = (1)
n
n Further ([10], Lemma 4), we have for x R
2
x
2
e jp x j O
(4 17) n ( ) = (1)
2
x
2
jp x j O
(4 18) e ( ) = (1) n
and
p
2 p 1
x
2
p
e jp x j O n f jx j n
(4 19) ( ) = (1) for 2 +1.
n
n
G x
From the uniqueness of n ( ) in (3.5), it follows that
n n
n 1 1
X X X
p x p x A x p y B n wp y C x
n n n n
i i i i i
(4 20) ( )= ( i ) ( )+ ( ) ( )+ ( ) ( ) ( )
i i i =0 =1 =1 Using lemmas 1, 2, 3 and (4.15)–(4.20), we obtain
2 2
x x
3 2 3 2
e jf x G f x j e jf x p x j n
( ) n ( ) ( ) ( ) +
2 2 1 1
x x
3 2
p p
e jp x G f x j O e f n
+ n ( ) ( ) (1) +
n n n
2 X
x
3 2
e jp x f x jjA x j
n
i i
+ ( i ) ( ) ( ) +
i =0 n 1
2 X
x
3 2
e jp y f y jjB x j
n
i i
+ ( i ) ( ) ( ) +
i =1 n 1
2 X
x
3 2
e j wp y g x j jjC
n i
+ ( ) ( ) i ( ) i
i =1
n 1
1 2 X
x
3 2
p
f e jg C x j O
(4 21) (1) + i ( ) +
i n
i =1
2019. majus´ 4. –23:07
50 P. MATHUR, S. DATTA
n
n 1 1 X
2 X 2
x
3 x 2 3 2
e jw y p y jjC x j e jw y p y jjC x j
i i i i i
+ ( i ) ( ) ( ) + ( ) ( ) ( ) +
n n i
i =1 =1
n 1
2 X
3 x 2
e jw y p y jjC x j
n
i i
+ ( i ) ( ) ( )
i =1
By lemma 3 and (4.17)–(4.19), we have n 1
2 X 1
x
2
p
e jw y p y jjC x j O
n
i i
(4 22) ( i ) ( ) ( ) = (1) n
i =1 n 1
2 X 1
x
2
p
e jw y p y jjC x j O
i i
(4 23) ( i ) ( ) ( ) = (1)
n n
i =1
and
n 1
2 X 1
x
2
p
e jw y p y jjC x j O f
i i
(4 24) ( i ) ( ) ( ) = (1)
n n
i =1 Thus by using (4.22)–(4.24) in (4.21), the theorem follows.
5. Basic Estimates of Fundamental Polynomials ( n odd)
Lemma For n odd
n X
2 p 2
x
x 3 2
i
e jA x j O n e
(5 1) i ( ) = ( )
i =1
where A x is given by
i ( )
x H x n x Proof
n n Since +1 = 0, is a zero of ( )( odd) thus for i 0, we 2 n
have
x
H ( )
n
A x A x i
(5 2) i ( )= ( )
H x n
( i )
Thus by (2.12), we have
n n
X X
p x
2 H 2 2
n ( )
x x
x 1 3 2
i i
e x j e A x j O n e jA j i
i ( ) = ( ) = ( )
H x
i
n 1( ) i i =1 =1
2019. majus´ 4. –23:07
MODIFIED WEIGHTED (0; 0 2)-INTERPOLATION 51 x
For i = 0, by (3.1), we have
2
Z
H x H x H x tH t H t n
( ) n ( ) ( ) ( ) ( )
n n
n 2
A x l x dt
(5 3) i ( )= ( ) 2
i
2 3
H x
H x t n
( i ) n
( i ) 0
Integrating the last term of (5.3), by parts, and using
tH t H t H t
( ) n ( ) ( ) n lim n = lim =0
2
t t 0 t 0 2
we have, by (1.13)
x
H ( )
n
A x A x i
(5 4) i ( )= ( )
H x n
( i )
Thus
n X
2 p 2
x
x 3 2
i
e x j O n e jA
i ( ) = ( )
i =1
hence the lemma is proved.
Lemma For n odd
n 1 X
2 p 2
x
y 3 2
i
e x j O n e jB
(5 5) i ( ) = ( )
i =1
1 2
3 x 2
p
B x j O e
(5 6) j 0( ) =
n
and
n 1
X 2 1 2
x
y 3 2
i
p
e x j O e jC
(5 7) i ( ) = n
i =1
The proof of this lemma is quite similar to that of lemmas 2 and 3, so we omit details.
The proof of Theorem 3 ( n odd) can be obtained following the same steps
as in the case of n even. We omit details.
2019. majus´ 4. –23:07 52 P. MATHUR, S. DATTA
References
azs J
[1] Bal Sulyozott´ (0 2)-interpolaci´ o´ ultraszferikus´ polinomok gyokein,¨
ozl
MTA III Oszt k , 11 (1961), 305–338.
azs J Approx Theory [2] Bal Modified weighted (0 2)-interpolation, Marcel
Dekker Inc., New York, 1998, 61–73. Acta Math Acad Sci Hungar [3] Freud G On two polynomial inequalities, 22
(1971), 109–116.
1 2 k
G x Acta [ ] Freud 4 On polynomial approximation with weight exp 2 ,
Math Acad Sci Hungar 24 (1973), 363–371.
o I Annales Univ Sci Budapest Sect Math [5] Jo On Pal´ interpolation, 37
(1994), 247–262.
Srivastava S [6] Mathur K K and Weighted (0; 0 2)-interpolation on infinite
interval, Acta Math Hung 70 1996, 57–73.
al L G Annales Univ [7] P A general lacunary (0; 0 1)-interpolation process,
Budapest Sect Comp 16 (1996), 291–301.
al L G Analysis Math [8] P A new modification to Hermite–Fejer´ interpolation,
1 (1975), 197–205.
en Z F [9] Sebesty Pal´ type interpolation on the roots of Hermite polynomials,
Pure Math Appl 9 (1998), 429–439.
[10] Szili L A convergence theorem for the Pal-method´ of interpolation on the
roots of Hermite polynomials, Analysis Math 11 (1985), 75–84.
L [11] Szili Weighted (0 2)-interpolation on the roots of classical orthogonal
polynomials, Bull of Allahabad Math Soc 8–9 (1993–94), 1–10. Chinese Quart J Math [12] Xie T F On Pal’s´ problem, 7 (1992), 48–52.
2019. majus´ 4. –23:07
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 53–64
r r ON MODIFIED WEIGHTED (0 1 2 )-INTERPOLATION ON AN ARBITRARY SYSTEM OF NODES
By P. MATHUR and S. DATTA
Department of Mathematics and Astronomy, Lucknow University, Lucknow, India Received October
1. Introduction
Balazs
Recently, J [2] showed that there exists a modified weighted
R x n
(0 2)-interpolation polynomial n ( )ofdegree 2 1 satisfying the con-
ditions:
R x y i n
n i n
( i n )= =1
R x y n n
( n n )=
n
i wR x y n n
( ) ( i n )= =1 1
n n
y y y x x
i n
n n where , , are arbitrary given real numbers, n are the n n
i n 1 1
W x
zeros of the polynomial n 1( ), i.e.,
n 1
Y
W x x x
i n n 1( )= ( )
i =1
(2)
x C a b where w ( ) ( ) is such a weight function which satisfies the condi-
tions:
wW x i n
i n ( n 1) ( )=0 =1 1
and
w x i n
( i n ) 0 =1 1
R x
Then n ( ) can be explicitly represented as:
n
n 1
X X
R x y A x y B x y A x
n n n
i n i n
( )= i n ( )+ ( )+ ( )
n n
i n i i =1 =1
2019. majus´ 4. –23:07
54 P. MATHUR, S. DATTA
A x B x A x
n n i n
where i n ( ), ( )and ( ) are basic interpolation polynomials each
n of degree 2 1.
This motivated us to consider the general problem of determining the
r r r modified weighted (0 1 2 )-interpolation ( 2) polynomial on an arbitrary system of nodes.
2. Definitions and New Results
r r r Modified weighted (0 1 2 )-interpolation ( 2) means the solution of the following problem:
Let the system of knots
b a x x n x x
n n N i i n (2 1) 1 n + ( := )
(r )
w x C a b
be given in the finite or infinite open interval (a b)andlet ( ) ( )
S x
be a weight function. Find a polynomial n ( )ofminimalpossibledegree
satisfying the conditions: m
(m ) ( )
S x y i n m r n
(2 2) ( i )= =1 =0 2
i
r
(r 1) ( 1)
S x y
n n (2 3) n ( )=
and
(r )
r
r 1 ( )
i w S x x y n n
n N
(2 4) ( ) ( i )= =1 1
i
r r
(m ) ( 1) ( )
y y y
where , n , are arbitrary given real numbers.
i i
W x n x
i n Let n 1( ) denote a polynomial of degree 1having ’s as the
zeros, i.e.,
n 1
Y
W x x x
i n (2 5) n 1( )= ( )
i =1
(r )
x C a b
If there exists a weight function w ( ) ( ) satisfying the conditions:
r
r
r 1 1 ( )
fw W x g x i n
(2 6) ( ) ( i )=0 =1 1 n 1
and
w x i n
(2 7) ( i ) 0 =1 1 then the following holds.
2019. majus´ 4. –23:07
r r
ON MODIFIED WEIGHTED (0 1 2 )-INTERPOLATION 55
r
If there exists a weight function w x C a b satisfying
Theorem ( ) ( )
the conditions and then there exists a unique modi ed weighted
(2 6) (2 7)
r r interpolation polynomial S x of degree nr corre
(0 1 2 ) n ( ) 1
sponding to the system of knots and satisfying the conditions
(2 1) (2 2) (2 3)
and (2 4)
W x
We note that, if the zeros of the polynomial n 1( ) are the zeros of the
r
x C a b classical orthogonal polynomials, then the weight functions w ( ) ( ) satisfying the conditions (2.6) and (2.7) do always exist. We will show this
in the sequel. To prove the above theorem we shall need the following
W x
n
(2 8) n 1( ) 0
q
q r
( ) 0 1
r 1
x W
r
(2 9) ( i )= 1
n
W x q r 1 r
( 1)! ( i ) = 1, n 1
and
q
q r
( ) 0 1
r 1
l x
r
(2 10) ( i )= 1
i
r l x q r
( 1)! ( i ) = 1, i
where
W x
n 1( )
l x i n
(2 11) i ( )= =1 1
x x W x
n i
( i ) ( )
3. Determination of fundamental polynomials
W x
Suppose that for the basic system of knots (2.1) the polynomial n 1( )
n w x
of degree 1 is given by (2.5) and there exists a weight function ( )
r
a b C
( ) satisfying the conditions (2.6) and (2.7).
A x k n t r
Let tk ( )( =1 , =0 ) denote polynomials of degree
nr 1 satisfying the conditions:
(3 1)
(m )
A x j k n m t r
mt
j jk
( )= =1 ; =0 2
tk
(r 1)
A x k n t r
( n )=0 =1 ; =0 2
rk
r
1 (r )
w A x k n j n t r j
( tk ) ( )=0 =1 ; =1 1; =0 2,
(m )
A x j n m r
( j )=0 =1 ; =0 2
n
(r 1)
(r 1)
x (3 2) A
( n )=1
r n
( 1)
r
r 1 ( )
w A x j n
n j ( (r 1) ) ( )=0 =1 1,
2019. majus´ 4. –23:07
56 P. MATHUR, S. DATTA
(m )
x j n k n m r A
j
( )=0 =1 1 =1 1 =0 2,
rk
(r 1)
x k n (3 3) A
( n )=0 =1 1,
rk
r
r 1 ( )
w A x k n t r
j jk
( rk ) ( )= =1 1; =0 2.
A x k n t r
We give.the explicit forms of tk ( ), =1 , =0 in the following:
If for basic system of knots there exists a weight function
Lemma (2 1)
r
w C a b satisfying the condition and then the polynomial
( x ) ( ) (2 6) (2 7)
x satisfying the conditions has the form A
tk ( ) (3 1)
r k n wehave t
For =0 2 =1 1
r
r 1
A x a x x x x l x
n
tk k
tk ( )= ( ) ( ) ( )+
k
x x
Z Z
r
1 r 1
l W x y x q y dy b y dy
n
k tk
+ ( ) ( ) tk ( ) + ( ) +
n 1
x
(3 4) x
n n
x
Z
r
r 2 2
X X
j
A
e y x W y dy d A x
n
jk jk jk
+ ( ) n 1( ) + ( )
j j t
x =0 = +1
n
t r where last summation is zero for = 2
1
a
(3 5) tk =
r 1
t x x n
!( k )
a
tk
l q x x k
tk ( )= ( )
k
t
r 1
W x x x k
( k )( )
n 1
r t
2
X
t
r 2
l c x x x l x
k k
(3 6) jk ( ) ( ) ( )
k
j =1
where c s are given by for s r t
ij =1 ( 2)
s
X
s
s s s u
( ) ( +1) ( )
l x l x l x c l x u
k k uk k
(3 7) ( k ) ( ) ( ) ! ( )=0
k
k k k u
u =1
2019. majus´ 4. –23:07
r r
ON MODIFIED WEIGHTED (0 1 2 )-INTERPOLATION 57
n o
t
(r )
a
tk
r r
r 1 1
b w x x l x n
tk = ( ) ( ) +
k
r
1 r 1
x
r t w W x k
( )! ( k )
n 1
r t
r
1 r 1 ( 1)
l r t x x x l x
n
k k
+ ( )( k ) ( ) ( )
(3 8) k
k
t
r 1
X
r t
t v r t
1 (r 1 ) ( )
c l x l x v
k k
! vk ( ) ( )
k k v
v =1
r are given by the equations s j
e =0 2 jk
(3 9)
i
X
s
(i )
i
s
r 1 ( 1)
b l x W n
tk ( )+
k
n 1
s
x n
s =1
s 1
X
s
m
1 (s 1 )
W x m e i r
x
mk
+ ( n ( )) ! =0 1 2;
1 n m
m =0
i r
and for = 1
t r
a r x x l x
n n k
(3 10) tk ( 1)!( ) ( )+
k
r 2
X
r
r m
r 2
r 1 ( 2) ( 2 )
W x x m W x b l e
n n
x
tk mk
+ ( ) ( )+ !( n ( )) =0
n
k
n 1 1 m
m =0
and d s j t r are given by equations for i t r
jk = +1 2 = +1 2
i
X
i
i t
r
r 1 ( )
t a f x x l x g d
n
x jk
(3 11) ik ! ( ) ( ) + =0
k
k
t t
j = +1
n we have when t k
For = =0
x
Z
r
1 r 2
X x
W ( )
n
r j
1 1
A
A x W x e y x W y dy
n
n (3 12) n ( )= + ( ) ( ) ( )
jk
0 n 1
r 1 1
W x
( n )
n
1 j
x =0 n
2019. majus´ 4. –23:07
58 P. MATHUR, S. DATTA
where e r are given by the relation j
s =0 2
jk
m
r 1 ( )
W x
( ( ))x
n
n 1 +
r 1
W x
( n )
n 1
(3 13)
s m
X X
m s
m s s i
r 1 ( ) ( )
i W x W x e
x x
+ ( ( )) ( n ( )) ! =0
n n
ik
n 1 1
s i i
s =1 =1
t r wehave
Also for =1 2
x
Z
r 2
X
r
1 j
A
A x W x y x y dy e W
m n
(3 14) ( )= ( ) ( ) n ( )
jk
n 1 1 t
j =
x
n
s j r t r are given by the relation
where = 2
jk
s m
X X
m s
m s s i
r
1 ( ) ( )
W x W x i e
mt
x x
(3 15) ( ( )) ( n ( )) ! =
n n
ik
n 1 1
s i
t s i
= =t
t r m
for = 2
Proof A x
r
First, we determine 2 k ( ), by (3.4), for which the last
A t r k n
summation vanishes. Then tk , =0 3, =1 1, can be
A x
k
determined, in the terms of r 2 ( ), from the recurrence relation (3.4).
q n A x tk
Since tk , given by (3.6), is a polynomial of degree 2, then ( ),
rn
given by (3.4), is a polynomial of degree 1. Obviously, by Leibnitz’s
x x j n
theorem, for = j , =1 1, by (2.9) and (2.10), we have for
n t m r
k =1 1, =0 2,
t m 0
(m )
t m
A x
jk =
( j )=
tk
m
0 t , due to (2.10).
x x
Now, for = n ,wehave
m
X
m
m s m s
t r
( ) r 1 ( ) ( )
A x a x x g x x l x g f f
n n
x x k
( )= tk ( ) ( ) ( ) +
n n
k
tk s
s =1
m
X
m s
m ( )
s
r 1 ( 1)
W b l x x n
+ ( ) kt ( )+
k
n 1
s
x n
s =1
s
X
s
i
1 (s 1 )
x e W i
ik
+ n ( ) !
1 x
n i
i =1
2019. majus´ 4. –23:07
r r ON MODIFIED WEIGHTED (0 1 2 )-INTERPOLATION 59
(m )
m r A x m r
For 1 2, by (3.9), we have ( n )=0.For = 1, we have
tk
r
r
( 1) 2 r
A x a r x x l x
n n n k
( )= tk ( 1)!( ) ( )+
k
tk
r 2
X
r
r m
r 2
r 1 ( 2) ( 2 )
W x x x e b l W i
n n
i tk
+ ( ) ( )+ n ( ) ! =0
k
n 1 x
1 n i
i =0
due to (3.10).
r
x C a b
If the weight function w ( ) ( ) satisfies the condition (2.6) and k
(2.7), then for j , (3.6), (2.9), we have
n o r
(r ) ( )
r r t r
1 1 r 1
w w A x a x x x x x l
n
tk k
tk ( ) = ( ) ( ) ( ) +
k
x x
j j
r r
r 1 1 1
w r W x x x q x
n
j tk
+ ! ( j ) ( ) ( )=
n 1
r t r
r 1 1
r a w x x x x l x
n
j j k j
= ! tk ( ) ( ) ( )
k
r 1
W x l x j
( j ) ( )
k n 1
=0
r t
r 1 1
W x x x
j k
( k ) ( ) n 1
due to
W x
( j )
n
1
l x j n
( j )= =1 1
k
x x W x
k k
( j ) ( )
n 1 k
In the case j = ,wehave
(r )
n
r 1 1
l x W x w W x k n
k j k
( )=1 n ( )=0and ( ) =0 =1 1
1 r 1
x k
Also,
r t 2
1 X
r t
2
l l x c x x x l x
k jk k
lim ( k ) ( ) ( ) ( ) =
k k
r t
x
x 1
x x k
( k )
j =1
t 1 r
( 1)
l x l x k
= ( k ) ( )
k
k
t
( r 1)!
t
r 1
X
r t
t v r t
1 (r 1 ) ( )
c l x l x v
k k
! vk ( ) ( )
k k v
v =1
2019. majus´ 4. –23:07 60 P. MATHUR, S. DATTA
Then by (2.9), we have
r
r
( ) a
r t
tk !
r r r
r 1 1 1 ( )
w A x fw x x l x g
n
x k
tk ( )= ( ) ( ) +
k
k
t
( r )!
r 1
a x x
n k
tk ( )
r
r 1 1
r w W x b tk
+ ! ( k ) +
n
1 r 1
r t W x
( 1)! ( k )
n 1
r t
( 1)
l x l x k
( k ) ( )
k
k
t
r 1
X
r t
t v r t
1 (r 1 ) ( )
v c l x l x
k k
! vk ( ) ( ) =0
k k v
v =1
b A x tk
This equality holds if we replace tk by the expression (3.8). Hence ( ),
r k n
t =0 2, =1 1, satisfies all the conditions given in (3.1).
k n t A x n
If = , then obviously, for =0, 0 n ( ) = 1. On differentiating (3.10),
m r x x
by Leibnitz’s theorem, =1 2 times, we have by (2.9), for = j ,
n j =1 1
(m )
A x j n m r
( j )=0 =1 1 =1 2
0 n
x x
For = n , by (3.11), we have
(m )
r 1 W
( x )
n 1
m x
( ) n
A x
( n )= +
n
0 r 1
W x
( n )
n 1
m s
X X
m s
( )
m
s i
r
1 ( )
W i x W x e
+ ( ) n ( ) ! =0
ik
n 1 x
1 n
s
x
n s =1 i =1
Also,
(r )
r 1
w A x j n
0 n ( ) =0 =1 1
x j
(m )
t r A x j n m r rn
Now, for =1 2, ( j )=0, =1 1, =0 2. For
x x m t r
= n and = 2, we have
(m )
A x
( n )=
tn
m s
X X
m s
s
m ( )
s i
r
1 ( )
i W W x x e
mt
= ( ) n ( ) ! =
ik
n 1 x
1 n
s i
x
n
s t t = i =
2019. majus´ 4. –23:07
r r
ON MODIFIED WEIGHTED (0 1 2 )-INTERPOLATION 61
(r )
r 1
w A x j n
by (3.13). Also, m ( ) =0, =1 1.
x
j
A x t r
Thus m ( ), =0 2, as given in (3.10) and (3.12) satisfies all
the conditions of (3.1).
Lemma A x k n has the form
rk ( ) =1 1
x
Z
r
1 r 2
A x W x t x W t dt
n
rk rk
( )= ( ) ( ) n ( ) +
n 1 1
x
n
(3 16)
x
Z
r 2
t x l t dt
n k
+ rk ( ) ( )
x
n where
1
(3 17) rk =
r r
r 1 2 1
r w x W x k
! ( k ) ( )
n 1
and
l x n
k ( )
rk
(3 18) rk =
W x
n n 1( )
(m )
A x j n m r Proof
Obviously, ( j )=0, =1 , =0 2. On
rk
r x x
differentiating (3.16), ( 1) times, by Leibnitz’s theorem, we have at = n
r
( 1) r 1
W x r W x x l x A
n n n
rk rk k
( )=( 2)! ( ) n ( )+ ( ) =0
rk n 1 1
due to (3.18).
r
x C a b If the weight function w ( ) ( ) satisfies, the conditions (2.6) and
(2.7), then by (2.9) and (3.17), we have
h i
(r )
r r r
1 r 1 1 2
w A x rw W x x x l x
n
rk j j k j jk
rk ( ) = ( ) ( ) ( ) =
n 1
x
j
A x k n
Hence rk ( ), =1 1, given by (3.16), satisfies all the conditions
given in (3.3).
Lemma A x is given by
n
(r 1) ( )
x
Z
r 1 x
W ( )
n
1 r 2
A x t x W t dt
n
n n (3 19) r ( )= ( ) ( )
( 1) r 1
r W x
( 1)! ( n )
n 1
x n
2019. majus´ 4. –23:07 62 P. MATHUR, S. DATTA
(m )
Proof A x j n m r
Obviously, ( j )=0, =1 , =0 2and
n
(r 1)
r
( 1) r
A x w x C a b
( n ) = 1. Since the weight function ( ) ( ) satisfies the
n
(r 1)
(r )
r 1
w A x A x
n r n
conditions (2.6) and (2.7), hence (r 1) ( ) = 0. Thus ( 1) ( )
x j given by (3.19), satisfies the conditions (3.2).
4. Proof of the Main Theorem
A x t r rn
As the polynomials tk ( ), =0 ,ofdegree 1arebasic
r r
interpolation polynomials, hence the modified weighted (0 1 2 )-
S x rn interpolation polynomial n ( )ofdegree 1 in Theorem 1, can be
explicitly represented in the form
r n
n 2 1
X X X
r r
(t ) ( ) ( 1)
A A S x y x y x y x A
n
n
tk rk
n ( )= ( ) + ( )+ r ( ) k
k ( 1)
t k
k =1 =0 =1
S x
Indeed, by Lemmas 1, 2 and 3, the polynomial n ( ) satisfies the conditions (2.2), (2.3) and (2.4), hence the Theorem is proved.
5. Remarks x
1. W
Now, we show that if the zeros of the polynomial n 1( )arethe
rn
zeros of the classical orthogonal polynomials of degree 1, then the
r
x C a b weight function w ( ) ( ) satisfying the conditions (2.6) and (2.7) always exist. It is known that the zeros of the classical orthogonal polynomials
are real and simple.
azs w
In [2] J Bal gave such weight functions for which
n
w x wW x k n
k k
(5 1) ( ) 0and n 1 ( )=0 =1 1
W where n 1 is the Jacobi, Laguerre of Hermite polinomial. So, we have only
to show that in the cases
(r )
r
r 1 1
w W x k n
(5 2) ( ) =0 =1 1
n 1
x k We prove it by induction.
2019. majus´ 4. –23:07
r r
ON MODIFIED WEIGHTED (0 1 2 )-INTERPOLATION 63 r
For r = 1, (5.2) is true, obviously. It is also true for = 2, by (5.1). Let i
(5.2) be true for r = , i.e.,
(i )
i
i 1 1
w W x k n
(5 3) ( ) =0 =1 1
n 1
x
k i
then we have to show that (5.2) holds for r = +1.
n o
i i
( +1) ( +1)
i
i i
i 1 1
w W x w W x wW x
( ) = ( ) n ( ) =
n
1 n 1 1
x x
k k
i
( +1)
i
i 1 1
w W x x wW
= ( ) n ( ) +
n
1 1 x
k
x
k
i
( )
i
i 1 1
i w W x x wW
+( +1) ( ) n ( ) +
n
1 1 x
k
x
k
i
( 1) i
+1
i
i 1 1
w W x wW x
+ ( ) n ( ) + =0
n
1 1 s
k
x 2 k due to (2.5), (5.3), (5.1) and (2.9). Hence if the nodes are the zeros of classical
orthogonal polynomials then by Theorem 1 there exists a modified weighted
r r S x rn
(0 1 2 )-interpolation polynomial n ( )ofdegree 1 satisfying
the conditions (2.2), (2.3) and (2.4).
azs r
2. (i) Bal ’s result [1] is a particular case of ours for =2.
2
x
x w x (ii) Taking the special weight functions w ( )=exp 2 and ( )=
2 th
x H x n
= exp( ) and the nodes as the zeros of n ( ), Hermite polynomial, the
Mathur r problem reduces to Datta and ’s problems [2], [4] for =2and
r = 3 respectively. 3. The convergence problem will be dealt in another paper.
References
azs J
[1] Bal Sulyozott´ (0 2)-interpolaci´ o´ ultraszferikus´ polinomok gyokein,¨
ozl
MTA III Oszt k , 11 (1961), 305–338.
azs J Approx Theory [2] Bal Modified weighted (0 2)-interpolation, Marcel
Dekker Inc., New York, 1998, 61–73.
P Mathur
[3] S Datta and On Weighted (0 2)-Interpolation on Infinite Interval
Annales Univ Sci Budapest Sect Math
(+ ), 42 (1999), 45–57.
P Mathur
[4] S Datta and On weighted (0 1 3)-interpolation on infinite interval
( + ), communicated for publication.
2019. majus´ 4. –23:07
64 P. MATHUR, S. DATTA
o I Annales Univ Sci Budapest Sect [5] Jo On weighted (0 2)-interpolation,
Math 38 (1995), 185–222. [6] Szili L Weighted (0 2)-interpolation on the roots of Hermite polynomials,
Annales Univ Sci Budapest Sect Math 27 (1984), 152–166.
2019. majus´ 4. –23:07
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 65–81
A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION ON INFINITE
INTERVAL ( + )
By S. DATTA and P. MATHUR
Department of Mathematics and Astronomy, Lucknow University, Lucknow, India Received September
1. Introduction
azs P Turan
J Bal on the suggestion of initiated the study of weighted
G x
(0 2)-interpolation, which means the determination of a polynomial n ( )of
n
degree 2 1suchthat
G a wG b i n
n n
i n i n i n
(1 1) ( i n )= ( ) ( )= ; =1 2
a b
i n i n
where i n , are arbitrary given numbers and are the zeros of the
th ( )
n P x w x
-ultraspherical polynomial n ( )( 1) with weight function ( )=
2 (1+ ) 2
x x
=(1 ) , [ 1 1]. He proved that generally there do not exist
n any polynomial of degree 2 1 satisfying the conditions (1.1). However,
taking n even, he proved the existence, uniqueness, explicit representation
G x n and convergence theorem for the polynomial n ( )ofdegree 2 satisfying
(1.1) together with n
X 2
G a l n
(1 2) (0) = i n (0) i n
i =1
L Szili If n is odd, the uniqueness is not true. [9] studied an analogous
th
n H x problem on the nodes as the zeros of the Hermite polynomial n ( ) with
2
x
2
x e I Joo weight function w ( )= . Later, [5] sharpened these results. In
an earlier paper [3] authors have improved I. Joo’s´ result by replacing the
G y y condition (1.2) with an interpolatory condition n (0) = 0,where 0 is an
arbitrary given number in the case of n even and obtained that the necessary
and sufficient condition for the existence of weighted (0 2)-interpolation in
n G y the case of odd is n (0) = 0.
2019. majus´ 4. –23:07
66 S. DATTA, P. MATHUR R B Saxena
K K Mathur and [7] extended the study of weighted
(0 2)-interpolation to the case of weighted (0 1 3)-interpolation, which means
T x n
to determine a polynomial n ( )ofdegree 3 1 such that
T x y T x y wT x y i n
n n
i n i n i n
(1 3) ( i n )= ( )= ( ) ( )= ; =1
n
i n i n
y y y w x
where i n , , are arbitrary given numbers and weight function ( )=
i n i n
2
x
e x x H x
R n
= ( )and i n ’s are the zeros of Hermite polynomial ( )given
by:
x x n
n n N
(1 4) 1 n ( )
n They proved that generally no polynomial of degree 3 1 satisfying the
conditions (1.3) exists, as such taking an additional condition:
n
h i X
2 3
R x l y x y
n i n
(1 5) (0) = 1+3 i n (0)
i n i n i n
i =1
where 0 is not a nodal point belonging to (1.4). They showed that there do
n n exists a unique polynomial of degree 3 ( even) and established its explicit
representation and convergence. If n is odd, uniqueness fails to hold.
The object of this paper is to get an analogous result by replacing the
R y
artificial looking, condition (1.5) by an interpolatory condition n (0) = 0, n where y0 is an arbitrary given number, in the case of even. Further, what
will be the necessary and sufficient condition for the existence and uniqueness
n of the (0 1 3)-interpolation in the case of odd? If it exists what will be its explicit form and does it converge? Here, we answer these questions in affirmative taking the nodes as the
th
n H x
zeros of Hermite polynomial n ( ). In section 2, we have given preliminaries. New results have been stated
in section 3. Sections 4 and 5 are devoted to the basic estimates of the n fundamental polynomials and the proof of our main theorem for n odd and even respectively.
2. Preliminaries
th
H n
Let n be the Hermite polynomial with usual normalisation
+ Z
2
t n
1 2
H t H t e dt n n m
n n m
(2 1) ( ) m ( ) = 2 ! N
2019. majus´ 4. –23:07 A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 67
which satisfies the differential equation:
H x xH x nH x
( ) 2 ( )+2 n ( )=0
n n
(2 2)
H x nH x
( )=2 n ( )
n 1
x H x n
It is well known that i n (the roots of ( )) satisfy the following relations:
n
x n m x x x
n n n 0 n ( =2 )
+1 n 1 2
2
x n m x x
n n n
(2 3) +1 =0 1 n ( =2 +1) n
2
n
x x i
i n
n i
= +1 n =1 2 2
x x H x H
i n i n
( i n )=2 ( )
n n
2
H x x n H x
i n i n
( )=2[2 ( 1)] ( )
n n
(2 4) i n
(4) 2
H x x n x H x
n
i n i n
( i n )=4 (3 2 +2 ) ( )
n
i n l
Let i n denote the fundamental polynomial of Lagrange interpolation corre- x
sponding to the nodal point i n .Then
H x
n ( )
l x i n
(2 5) i n ( )= ( =1 )
x x H x
n i n
( i n ) ( )
n j
0fori
l x j n
(2 6) i n ( )= j
1fori =
x H
n ( )
j n
j i
for
x x H x
x
l ( ) ( )
n
j n i n i n
(2 7) ( j n )=
i n
x j i
i n for =
h i
H x
2 n ( j n
1
x j i
j n
x x
x x H x
j n i n
( ) ( )
n
j n i n i n
l x
(2 8) ( j n )=
i n 2
x n
4 2( 1)
i n i 3 j =
and
x H
n ( )
j n
1
l x j i
j n
3 ( )
x x
i n
H x i n
j n ( )
n
i n
l x
(2 9) ( j n )= i n
2
x x n j i
i n (2 +3 2 ) = .
i n
H x
For the roots of n ( ), we have 2
2 i
x n i
(2 10) =1
i n n
2019. majus´ 4. –23:07
68 S. DATTA, P. MATHUR
p
p 2
x
1 4 n 2
H x O n n jx j e x
(2 11) n ( )= (1) 2 ! 1+3 R
2
n
x
X i n
e 1
(2 12) = O 0 1
2 n +1
H x n n
( i n ) 2 !
i =1 n
X 2 2 x
2 x
i n
l x O e
(2 13) e ( )= ( ) i n
i =1
n !
jH j n
(2 14) n (0) = for even n
2 !
n 2
2 n !
2 1 2
n (2 15)
( n +1)!
Also
x x
Z Z
x l t l t i n
i n ( ) ( )
i n 1
l x l xl x n l t dt dt i n
(2 16) = ( ( ) (0)) i n ( )+ ( )
i n i n
t x
i n 2 0 0
1 2
x n
If i n = ( 2( 1)), then
3 i n
x
Z
t x x l t l t
i n i n i n
( i n ( )+ ) ( ) ( )
i n
(2 17) 2 dt =
t x
( i n )
0
x
Z nx
i n 1
l t dt l x l
= i n ( ) ( ( ) (0))+ i n 3 6 i n
0 x
i n 1
2
l x l x n l x l i n
+ ( ( ) (0)) ( + +2)( i n ( ) (0))+
i n i n
2 i n 3 x
1 i n
H x H H x H n
+ ( ( ) (0)) ( n ( ) (0))
n n
H x H x
n n i n
3 ( i n ) 3 ( )
3. New Results
Let n be odd in Theorems 1, 2, 3.
th
Theorem Let the nodal points are the roots of n Hermite polynomial
2
x
H x and the weight function is w x e x
n ( ) ( )= ( R)
2019. majus´ 4. –23:07
A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 69
Then there exists a unique polynomial R x of degree n satisfying
n ( ) 3
the conditions and R y wherey is an arbitrary given number (1 2) (0) =
n 0 0 and
0 is a nodal point
For k n Theorem =1
3
x l x x B x A
k k
(3 1) k ( )= ( ) 3 ( )+
k
x Z
2 2
x t x t x l t l t
H x
k k k k k
( ) ( k + ( )+ ( ) ) ( ) ( )
n
k
dt
2 2
H x t x
n k
( k ) ( )
0 where
2 2
x n x x n
2( 1) k (2 3 )
k k
and k
(3 2) k = = ; 3 3
(3 3)
x Z
2
x S t x l t l t
H x
k k k
( ) ( k + ( )) ( ) ( ) n
3 k
dt B x x x l x k
k ( )=( ) ( )+ k
2
t x
H x
( k ) n
( k )
0 where
1 2
S n x
k = ( +2(1 ))
3 k x
2 Z
x 2
k
H x
e ( )
n
C x l t dt k
(3 4) ( )= k ( ) ;
2
H x n
6 ( k )
0 and
2 x
H ( )
n
D x (3 5) ( )=
0 2 H
2 n (0)
Then
n n n
X X X
R x y A x y B x y C x y D x
n
k k k
(3 6) ( )= k ( )+ ( )+ ( )+ ( )
k 0 0
k k
k =1 =1 =1
n satisfying the conditions
is a uniquely determined polynomial of degree 3 of Theorem 1
2019. majus´ 4. –23:07
70 S. DATTA, P. MATHUR
Let f be a twice continuously di erentiable
Theorem : R R function such that
2
x
2 r
x x e r
lim f ( ) =0; =0 1
jx j
(3 7)
2
r x
( )
f x r
lim e ( )=0; =1 2
jx j
Then the weighted interpolatory polynomials R x n
(0 1 3) n ( ) =35 7
together with
given by (3 6)
y f x y f x
k k
k = ( ) = ( )
k
2 1
x
2
k
p
O e n f
y =
k n
(3 8)
and
f
y0 = (0)
satisfy the estimate
2 1
x
p
e jf x R f x j O f
(3 9) ( ) n ( ) = (1)
n
O does not depend on n and x Here f denotes the Freud s
where ( )
modulus of continuity of f
In the case of n even, analogous to Theorem 1, there do exist a weighted
R x n
(0 1 3)-interpolatory polynomial ( )ofdegree 3 satisfying the condi-
n
y y tions (1.2) and R (0) = ,where is an arbitrary given number. Further, n 0 0 let
(3 10)
2
H x
n ( )
k x
2 for =0( 0 =0)
H (0)
n
2
H x
4 n ( )
l x x B x l k
( ) 3 k ( ) (0)+
A x k
( )= k 2 2
x H x
k ( )
n
k
k
x
2 R 2
x t x t x l t l t x
H ( + ( )+ ( ) ) ( ) ( )
k k k k
n k k
( ) k
k n
+ dt for =1 ,
2 2
x t x
H ( ) ( )
n k k 0
where
2
n
x 2( 1) 2
k 2
x n x n k
(3 11) = and = k =1
k k 3 3 k
2019. majus´ 4. –23:07 A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 71
2 x
H ( ) n
3
B x x x l x l k
( )=( k ) ( )+ (0)+
k k
2
x H x
n k
k ( )
x
Z
2
x S t x l t l t
H x
k k
( ) ( k + ( )) ( ) ( )
n
k k
k n (3 12) + dt ; =1
2
t x
H x
( k ) n
( k ) 0
where 1
2
n x
(3 13) S = +2(1 ) k k 3
and x
2 Z
x 2
k
H x
e ( )
n
C l t dt k n
(3 14) = k ( ) =1 k
2
H x n
( k ) 0
Then
n n n
X X X
A R x y x y x y x C
(3 15) ( )= ( )+ B ( )+ ( )
n
k k k k
k k
k =0 =1 =1 n
is a uniquely determined polynomial of degree 3 satisfying the conditions
R y
(1.2) and n (0) = 0 .
Let f be a twice continuously di erentiable
Theorem : R R
function satisfying the requirements and the numbers y y and y
(3 7)
k k k
are such that
y f x k n
= ( k ) =0
k
y f x k n
= ( k ) =1
k
(3 16) and
2
1
x
2
k
p
O e n f
y =
k
n
Then for interpolatory polynomial R n given by we
( x ) =2 4 6 (3 15)
n
have the estimate
2 1
x
p
e jf x R f x j O f x
(3 17) ( ) ( ) = (1) R
n
n
1
p
where O does not depend on n and x and f is the Freud s modulus
n of continuity
2019. majus´ 4. –23:07 72 S. DATTA, P. MATHUR
We shall prove only our main Theorem 3 and 4, as the proofs of the other theorems are quite similar to that of theorems in [1].
4. Basic Estimates with Respect to the Fundamental Polynomials
( n odd)
Lemma If i n are the Christo el
([6], Lemma 1.) i =1 numbers on Hermite nodes then
2 1 2 n
x x
i i
e e x i
n i
i Φ ( ) 1
1 6 1 3
i
n 2
x x x certainly x x x where
n
i i i
n i
Φ( )= i +1 1 2 = +1
Lemma Let n be odd then
2
x
D x j O e x
(4 1) j 0( ) = ( ) R n
2 X
3x 2
1
x 2
i
e jC x j e 2
(4 2) sup i ( )
n x R
i =1
Proof By using (2.2) and (2.11) in (3.5), (4,1) follows. Without loss of generality, we can assume that x 0. Using Lemma 1,
we get
q
1 4 n 1
jH x j n jH x j n n jh x j
i i i
n ( ) =2 n ( ) =2 2 ( 1)! ( ) = n 1 1
(4 3)
p p
p 2
x
1 2 1 2 n
1 4 n 2
i
x i n ce n n n
= 2 2 ! 2 !Φ ( i ); =1 i
We have by (3.4)
H x j 2 j
n ( )
x 2
i
jC x j e jB x j i
(4 4) i ( ) ( )
jH x j n
( k )
where x
2 Z x
2
i
e H x
n ( )
B x l t dt i
(4 5) i ( )= ( )
H x n
( i ) 0
and from [6]
n 2
x
X 2
2 e x
2
i
p
e jB x j O
(4 6) i ( ) = n
i =1
2019. majus´ 4. –23:07 A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 73
We remark that (4.6) also holds for n odd. Thus
2
n n
x X
X 2
i
jH x j 2 e 2
n ( )
x x
2 2
i i
e jC x j e jB x j i
i ( ) = ( ) =
jH x n
( i ) i
i =1 =1
X X
= + =
p p
n n jx j jx j
i i
2 2 2
x
x x X
X 1
i e
e 2 x
2 2
i
p
O jB x j O x e jB x j
e 2
n
i i
= (1) i ( ) + (1) Φ ( ) ( ) =
n 1 4
p p
n
jx j n jx j n
i i
2
3 x 2
e
= O (1)
n p
2 p
x
2 n 1 4
i
jH x j e n n jx j n i
where we used ( i ) 2 ! , , given in [5].
n
Lemma For n odd
n
X 2 2
x
x 3 2
i
jB x e x j O e R
(4 7) i ( ) =
i =1
Proof From (3.3), due to (2.16), we have
x
Z
H x n H x n
n ( ) 7 ( )
B x A x l t dt
i i
i ( )= ( ) ( )
H x H x
n n i ( i ) 3 ( )
0
(4 8)
x
Z
H x
5 2 n ( )
x l t dt
(1 ) i ( )
i
H x n
3 ( i ) 0
where
x x Z
2 Z
l x
H x nH x n ( ) n ( ) ( )
i 2
A x x l t dt l t dt
i i
i ( )= +(1 ) ( ) + ( ) +
i
H x H x
n n i
2 ( i ) ( )
(4 9) 0 0
H H H x x x n
( ) n ( ) ( )
n
l x x l x l x
i i
+ i ( ) ( ) (0) for 0
i
H x H x H x
n n n
i i
2 ( i ) ( ) 2 ( )
2019. majus´ 4. –23:07 74 S. DATTA, P. MATHUR
and
x x Z
2 Z
l x
H x nH x n ( ) n ( ) ( )
i 2
A x x l t dt l t dt
i i
i ( )= +(1 ) ( ) + ( ) +
i
H x H x
n n i
2 ( i ) ( )
(4 10) 0 0
H H x x
( ) n ( )
n
l x x l x x
i i
+ i ( ) ( )for =0.
H x H x
n n i
2 ( i ) ( )
By [[6], lemma 4], we have
n X
2 2 p
x
x 2
i
e jA x j O e n
(4 11) i ( ) = (1)
i =1
x
2 Z
n
x
X 2
i
p
H x je j 2
n ( )
x
n l t dt O e n
(4 12) i ( ) = (1)
jH x j n
( i )
i =1 0
and
x
2 Z
n x 2
X 2
i
p
x jjH x je j 2
(1 ) n ( )
x
i
l t dt O e n
(4 13) i ( ) = (1)
jH x j n
( i )
i =1 0
(4.11)–(4.13) also hold for n odd. Thus by (4.8), (4.7) follows.
Lemma For n odd
n X
2 2 p
x
x 3 2
i
e jA x j O e n x R
(4 14) i ( ) = (1)
i =1
Proof A x
i ( ), given by (3.2), can be represented alternatively as follows.
x
For i 0
2 x H ( )
3 n
A x l x x B x I
i i
(4 15) i ( )= ( ) 3 ( )+ i
2
H x n
( i )
where
x
Z
x x i
2 i 1
2
I n x l t dt l x l l x l
= (2 ) i ( ) ( ) (0) + ( ( ) (0))
i i i i 3 6 i 2
(4 16) 0 x
1 1 i
2
x n l x H x H H x n
( + +2) i ( )+ ( ( ) (0)) ( )
n n
i
H x H x
n n i
3 3 ( i ) 3 ( )
2019. majus´ 4. –23:07
A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 75 x
and for i =0
2 x H ( )
3 n
A x l x x B x J
i i
(4 17) i ( )= ( ) 3 ( )+ i
2
H x n
( i )
where
x
Z
x x i
2 i 1
2
J n x l t dt l x l x
= (2 ) i ( ) ( )+ ( )
i i 3 6 i 2
(4 18) 0
x H x n
1 1 i ( )
2
x n l x H x H
( + +2) i ( )+ ( ( ) (0))
n n
i
H x H x
n n i
3 3 ( i ) 3 ( ) Hence
n 2
x
X
i
e jA x j I I I I I I I I I 2 (4 19) i ( ) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
i =1
where
n X
2 p 2
x
x 2 3 2
i
l I e x jl x j O ne
(4 20) = ( ) i ( ) = 1 i
i =1
n X
2 p 2
x
x 3 2
i
I jx j e jB x j O ne i (4 21) 2 =3 i + ( ) =
i =1
p
jx j O n
Using ( i = ( ) and Lemma 2
x
Z n
X 2
x
2 2 H ( )
n
x
2
i
I jx jj n x je l t dt i
= i (2 ) ( ) =
i 3
2 x
3 H
n
i
( ) i =1 0
(4 22)
n
X
p 2
x
3 2 3 2
O n jC x j O ne
= i ( ) =
i =1 n
X 2 x
1 2 H ( )
n
x
i
e jl x j
I = ( ) i
4 2 x
6 H n
( i )
i =1
2019. majus´ 4. –23:07 76 S. DATTA, P. MATHUR
From (2.6), we have
(4 23)
n
X 2
x x x x H x l x
2 H i
1 ( ) ( ( i ) 1) ( ) ( )
n n
x
i
I e nl x
= + i ( )
4 2 2 2
x x x H x x x
6 H
n n
i i i
( i ) ( ) ( ) ( )
i =1
n
X 2 x
2 H
1 ( ) n
x 2 2 3
i
e jx x x j l x n x x jl x j
i
( i ) 1 ( )+(1+ ( ) ) ( ) =
i i
2
x
6 H n
( i )
i =1
n n
X
X 2 2
H x H x j H x
2 j ( ) ( ) 2 ( )
n n n
x x
i i
jx je jl x j O n e jl x j i
= i ( ) + ( ) ( ) =
2 2
H x H x
n n i
( i ) ( ) i
i =1 =1
n n X
X 2
n jH x H x j
H x
n
2 n ( ) ( ) ( )
1 n
jx j x jl x j O n x jl x j
i i i
= Φ( i ) ( ) + ( ) Φ( ) ( )
n n n
2 n ! 2 ! i i =1 =1
2
x
p 2
jH x je
n cn
( )
p
x j n O e If j 2 ,then = (1) , therefore, we can assume that
nn !
p
n
x 2 . Hence
n X
2 p
x
I O e n x jl x j
n i 4 = (1) Φ ( i ) ( ) =
i =1
(4 24)
X X
p 2
x
O ne
= +
p p
jx j jx j n
2 log n 2 log
i i
Here
2
x X X 2
1 e
x jl x j c jl x j c
n
i i
(4 25) Φ ( i ) ( ) ( )
1 6 1 6
p p
n n
jx j jx j n
2 log n 2 log
i i
and X
X 1
p
x jl x j x j jl
n
i i
Φ ( ) ( ) i ( )
n
p p
jx j jx j n
2 log n 2 log
i i
(4 26)
n X
1 c 2
x 2
p p
jl x j n e
i ( ) log +
n n
i =1
2019. majus´ 4. –23:07 A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 77 where we used [5], Hilfssatz 4, Satz, 11. Thus by (4.24), (4.25) and (4.26),
we have
2 1
3 x 2
I O n
(4 27) 4 = (1) e 6
n
X 2
l H x x x 2 H
1 ( ) ( ) i ( )
n n
x
i
I e jx j
= i
5 2
x x
x x H x
H x
n
i i
2 ( i ) ( ) ) n
( i )
i =1
n n
X X
H x H x j jH x j
2 j 2 n
1 n ( ) ( ) 1 ( )
n
x x
2 i
(4 28) i
e jl x j jx je x jx j l
i i
i ( ) + ( )=
i
2
jH x j
H x n
2 2 ( i ) n
( i ) i
i =1 =1
2 p
n
3 x 2
e
= O (1) n
X 2 x
1 2 H ( ) n
x 2
i
I e n x jl x j
( +2+ ) i ( ) =
6 i
2 x
3 H n
( i )
i =1
(4 29)
n
X 2
p x
2 H ( ) 2
n
x
x 3 2
i
O n e jl x j O e n
= ( ) i ( ) =
2
H x n
( i )
i =1
(4 30)
n
X 2
H jH x j jH j
1 2 ( x ) 2 ( ) + (0)
n n n
x
i e I = =
7
2
jH x j
H x n
6 ( i ) n
( i )
i =1
2
n n
x X
X 2
i
H x jH x j
2 e
( ) n ( ) n
x 1 2
i
O n e O n H x jH x j
= ( ) = ( ) ( ) n ( ) =
n
2 1 3
H x jH x j jH x j
n n n
i i
( i ) ( ) ( ) i
i =1 =1
X
2 X
O n H x jH x
= ( ) ( ) n ( ) + =
n 1
p p
n jx j n jx j
i i
3 4
2 n
x n X
3 2 3 2 X 1 e
n 2
(2 !) x 2 3 2
n 2
i
e O e x
n
= + Φ ( i ) =
1 4 n 3 2
p p n
n (2 !)
n jx j n jx j
i i
2
3 x 2
e
p O
= n
2019. majus´ 4. –23:07
78 S. DATTA, P. MATHUR n
X 3 x
1 2 H ( )
n
x
i
I jx je
= i =
8 3 x
3 H n
( i )
i =1
(4 31)
n 2
x
X 3 2 p 2 1 e
3 x
i
p
O n jH x e O
= ( ) ( ) = n
3
n
jH x j n
( i )
i =1
(4 32) n
X 2 x
1 2 H ( )
n
x 2
i
e j x j jl j
I = 2 3 (0)
i i
9 2
x jH x
6 j
n i
i ( )
i =1
n n
X
X 2 2
H jH j H x jH j
1 2 ( x ) (0) 1 2 ( ) (0)
n n n n x
x 2
i i
e x e j j jx j
1 + i =
i
2 3 3
H x j
x x j H
3 j 6 n
( i ) n
( i )
i i i =1 =1
2
n
x
X 3 2 i
2 e
O n H x jH j
= ( ) ( ) n (0) + n
1 3
jH x j n
( i )
i =1 n
X 2 1 x
3 2 2
i
O n H x jH j e
+ ( ) ( ) n (0) = n
1 3
jH x j n
( i )
i =1
p
1 p 2 2
x
x 3 2
O ne C ne
= O 2 + n Using (4.19)–(4.32), we get the proof of (4.14).
1
If f C Lemma ([6], Lemma 5] (R)
2 r
f x w x r
lim x ( ) ( )=0 =0 1
x
and
f w x
lim ( x ) ( )=0
x
x of degree n such that for x then there exists a polynomial p
n ( ) R
1 1
p p
x jf x p x j O f w
( ) ( ) n ( ) = (1)
n n
1 1
p p
w jf x p x j O f
( x ) ( ) ( ) = (1)
n
n n
Further x
[9], Lemma 4 we have for R
w x jp x j O
( ) n ( ) = (1)
x jp x j O
w ( ) ( ) = (1) n
2019. majus´ 4. –23:07
A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 79
p
jx j n
and for 2 +1
p 1
p
w x jp x j O n f
( ) ( ) = (1)
n
n
p
we have for jx j n
Also [7], Lemma 4 2 +1
1
p
w x jp x j O n f
( ) ( ) = ( )
n
n Proof of Theorem
Let n be odd. From the uniqueness of polynomial
R x S x n n n ( ) in (3.6), it follows that every polynomial ( )ofdegree 3 satisfies
the relation:
n n
X X
S x S x A x S x B x
n n
i i i
( )= ( i ) ( )+ ( ) ( )+
n i
i =1 =1
n
X
wS x C x S D x
n i
+ ( ) ( i ) ( )+ (0) ( ) n 0
i =1
p x n
Let n ( ) be a polynomial of degree 3 satisfying Lemma 5, then we have
2 2 2
x x x
3 2 3 2 3 2
e jf x R x j e jf x p x j e jp x R x j
n n n
( ) n ( ) ( ) ( ) + ( ) ( ) =
n
2 2 2 X
x x x
2 3 2
O e jf x p x j e f x p x A x e
n n
i i
= (1) ( ) ( ) + ( ( i ) ( )) ( ) +
i =1
n
2 X
x
3 2
e f x p x B x
i i
+ ( ( i ) ( )) ( ) +
n
i =1
n
2 X 2
x x
3 2 3 2
e y wp x C x e jf p jjD x j
n i
+ ( ( ) ( i )) ( ) + (0) (0) ( ) n
i 0
i =1
Using lemmas 3, 4 and 5, we have
n
2 1 2 X
x x
3 2 3 2
p
C f e jf x R x j O e jy x j n
( ) ( ) = (1) + i ( ) +
i n
i =1
n
2 X
x
3 2
e j wp x C x j
n i
+ ( ) ( i ) ( )
i =1
2019. majus´ 4. –23:07 80 S. DATTA, P. MATHUR
By lemmas 2, 5 and (3.8), we have
2 1
x
3 2
p
e jf x R x j O f
( ) n ( ) = (1) +
n
n n X
1 2 2 X
x
3 x 2
i
p
O n f jC x je jC x j e i
+ ( ) i ( ) + ( ) =
n i
=1 i =1
1
p
f
= O (1) n Hence the theorem is proved.
5. Basic Estimates of Fundamental Polynomials
( n even)
A x B x
For n even, ( )and ( ), given by (3.10) and (3.12), can be written
i i in a convenient form as:
2 2
x H x
H ( ) ( )
n n
3
A x l x x B x I l i
(5 1) ( )= ( ) 3 i ( )+ (0)
i i
i
2 2 2
H x
x H x n
( i ) n
( i ) i
where I is (4.16) and
2 x
H ( )
n
B x B x l i
(5 2) ( )= i ( )+ (0) i
2
x H x
n i
i ( )
B x
where i ( ) is given by (4.8).
Lemma For n even
n X
2 2 p
x
3 x 2
i
jA x j O e n
(5 3) e ( ) = i
i =1
n
X 2 2
x
x 3 2
i
e jB x j O e
(5 4) ( ) = i
i =1
and
n 2
x
X 3 2
2 e
x
i
e x j O C
(5 5) j ( ) =
i n
i =1
Proof The proof of this lemma is similar to that of Lemmas 2, 3 and 4, so we omit details.
2019. majus´ 4. –23:07 A NOTE ON WEIGHTED (0 1 3)-INTERPOLATION 81
Proof of Theorem Following the same steps as in the proof of The- orem 3, the theorem follows. We omit details.
References
azs J
[1] Bal Sulyozott´ (0 2)-interpolaci´ o´ ultraszferikus´ polinomok gyokein,¨
ozl
MTA III Oszt k , 11 (1961), 305–338.
azs J Turan P Acta Math Acad [2] Bal and Notes on the interpolation VIII,
Sci Hung 12 (1961), 469–474.
Mathur P
[3] Datta S and On weighted (0 2)-interpolation on infinite interval
Annales Univ Sci Budapest Sect Math
( + ), 42 (1999), 45–57.
1 2 k
G x Acta [ ] Freud 4 On polynomial approximation with weight exp 2 ,
Math Acad Sci Hungar 24 (1973), 363–371.
[5] Freud G Lagrangesche interpolation uber¨ die Nullstellen der Hermiteschen
Orthogonalpolynome, Studia Sci Math Hung 4 (1969), 179–190.
o I Annales Univ Sci Budapest Sect [6] Jo On weighted (0 2)-interpolation,
Math 38 (1995), 185–222.
Saxena R B [7] Mathur K K and Weighted (0 1 3)-interpolation on the zeros
of Hermite polynomial, Acta Math Hung 62 (1993), 31–47.
o G Orthogonal Polynomials [8] Szeg Amer. Math. Soc. Coll. Publ. (New York,
1959). [9] Szili L Weighted (0 2)-interpolation on the roots of Hermite polynomials,
Annales Univ Sci Budapest Sect Math 27 (1984), 152–166.
2019. majus´ 4. –23:07
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 83–93
INTERVAL OSCILLATION CRITERIA FOR SECOND ORDER FORCED NONLINEAR DIFFERENTIAL EQUATIONS WITH DAMPING*
By W. LI and H-F. HUO
Department of Mathematics, Lanzhou University, Lanzhou, Gansu, Republic of China and Institute of Applied Mathematics, Gansu University of Technology Lanzhou, Gansu, People’s
Republic of China Received February
1. Introduction
In this paper we shall be concerned with the second order forced nonlin-
ear differential equation with damping
r t y t p t y t q t f y t g y t e t t t
(1 1) ( ( ) ( )) + ( ) ( )+ ( ) ( ( )) ( ( )) = ( ) 0
q f g e
where r , , , , are to be specified in the following text.
y t t t t
We recall that a function :[0 1) ( + ), i 0 is called a
t t t t
solution of Eq. (1.1) if y ( ) satisfies Eq. (1.1) for all [ 0 1). In the sequel
it will be always assumed that solutions of Eq. (1.1) exist for any t 0 0. A t solution x ( ) of Eq. (1.1) is called oscillatory if it has arbitrary large zeros,
otherwise it is called nonoscillatory.
t p t e t
When r ( ) 1, ( ) 0and ( ) 0, Eq. (1.1) reduces to the equation
y t q t f y t g y t
(1 2) ”( )+ ( ) ( ( )) ( ( )) = 0 Lalli which has been studied by Grace and [7]. They mentioned that though
stability, boundedness, and convergence to zero of all solutions of Eq. (1.2)
Grimmer Grace
have been investigated in the papers of Burton and [1],
Lalli Wong Burton and Spikes [5, 6], [11], and and [19], not much has
been known regarding the oscillatory behavior of Eq. (1.2) except for the Burton result by Wong and [19, Theorem 4] regarding oscillatory behavior
of Eq. (1.2) in connection with that of the corresponding linear equation
y t q t y t (1 3) ”( )+ ( ) ( )=0
*Supported by the NNSF of China and the Foundation for University Key Teacher by the Ministry of Education of China.
2019. majus´ 4. –23:07 84 W. LI, H-F. HUO
Recently, Rogovchenko [17] presented new sufficient conditions, which
ensure oscillatory character of Eq. (1.2). They are different from those of Lalli
Grace and [7] and are applicable to other classes of equations which Lalli are not covered by the results of Grace and [7]. However, all the
mentioned above oscillation results involve the interval of q and hence require
t the information of q on the entire half-line [ 0 + ). For related results refer
to [2, 10, 13–16].
t g y
When p ( ) 0and ( ) 1, Eq. (1.1) reduces to the equation
r t y t q t f y t e t (1 4) ( ( ) ( )) + ( ) ( ( )) = ( )
Numerous oscillation criteria have been obtained for Eq. (1.4); see Keener
Skidmore Bowers Skidmore Leighton [9], Rainkin [16], and [20], and
[21], and Teufel [22]. In these papers, the authors established oscillation
criteria for a more general nonlinear equation by employing a technique
e t
introduced by Kartsatos [8] where it is additionally assumed that ( )bethe t
second derivative of an oscillatory function h ( ) and their oscillation results
t require the information of q on the entire half-line [ 0 ).
However, from the Sturm Separation Theorem, we see that oscillation is
a b i
only an interval property, i.e., if there exists a sequence of subintervals [ i ]
t a i
of [ 0 ), as i , such that for each there exists a solution of equation
r t y t q t y t t t
(1 5) ( ( ) ( )) + ( ) ( )=0 0
a b i
that has at least two zeros in [ i ], then every solution of Eq. (1.5) is q
oscillatory, no matter how “bad” Eq. (1.5) is (or r and are) on the remaining
parts of [t0 ).
Ei Sayed [4] applied this idea to oscillation and established an interval
criterion for oscillation of a forced second order linear differential equation
r t y t q t y t e t t t
(1 6) ( ( ) ( )) + ( ) ( )= ( ) 0
Theorem A Suppose that there exist two positive increasing divergent
+ +
sequences fa g fa g and two sequences fc g fc g such that c c are
n n
n n
n n
positive numbers and
p
c
a +
n n
Z
q
2
c V r t c t a
= [1 ( )] cos n ( +
n n n
(1 7)
a
n
q
2
q t c c t a dt
+[ ( ) ]sin n ( ) =0
n n
2019. majus´ 4. –23:07
INTERVAL OSCILLATION CRITERIA 85
n n where n is a xed positive integer Suppose further that e t
for all 0 0 ( )
satis es
+ +
p
t a a
0 +
n
n +
c
n
e t
(1 8) ( )
p
t a a n
0 n +
c
n
n n Then Eq is oscillatory for all 0 (1 6)
We note that the result is not very sharp, because it was proved with the Coppel aid of a comparison theorem of Leighton [12] in the form given by
[3, Theorem 8, p11]. Recently, Wong [18] proved a more general oscillation
result for Eq. (1.6).
Suppose that for any T t there exist T s t
Theorem B 0 1 1
t such that s
2 2
t s t
0 [ 1 1]
e t
(1 9) ( )
t s t 0 [ 2 2]
1
Denote D s t fu C s t u t u s u t g i If
i i i i i
( i )= [ ]: ( ) 0 ( )= ( )=0 =12
there exists u D s t such that i
( i )
t
i Z
2 2
Q u qu ru
(1 10) i ( )= ( ) 0
s
i
then Eq is oscillatory i
for =1 2 (1 6) Wong Motivated by the ideas of Ei Sayed [4] and [18], in this paper we obtain, by using a generalized Riccati technique which is introduced by
Li [13] for the unforced equations and a new integral averaging technique, we
obtain several new interval criteria for oscillation, that is, criteria given by the
q f g e
behavior of Eq. (1.1) (or of r , , , and ) only on a sequence of subintervals
of [ t0 ). Finally, several examples that dwell upon the importance of our results are also included.
Hereinafter, we assume that
t p t
(H1) the functions r :[0 ), (0 )and :[0 ) R are continuous;
t q t
(H2) the function q :[0 ) [0 ) is continuous and ( ) 0onany
T t
ray [T )forsome 0;
yf y y (H3) the function f : R R is continuous and ( ) 0for 0;
2019. majus´ 4. –23:07
86 W. LI, H-F. HUO
g y K y
(H4) the function g : R R is continuous and ( ) 0for 0.
H t In the sequel we say that a function H := ( ) belongs to a function class
1
D s t fH C s t H t H s H t g i
i i i i i
( i )= [ ]: ( ) 0, ( )= ( )=0 , =1 2, denoted by
H D s t i i
( i ), =1 2.
2. Main Results
Let assumptions hold Suppose that
Theorem (H1) (H4)
f y for y
(2 1) ( ) 0 0
that for any T t there exist T s t s t such that and 0 1 1 2 2 (1 9)
1
holds If there exist H D s t and g C t such that i
( i ) ([ 0 ) (0 ))
t t
i
Z
1 Z 2 t 1 p ( )
2
H t t dt r t v t H t H t dt
(2 2) ( ) ( ) ( ) ( ) 2 ( )+ ( )
r t
4 ( ) s
1 s1
t
R
g for i where v t s ds and =1 2 ( )=exp( 2 ( ) )
2
v t Kq t r t g t p t g t r t g t
( t )= ( )[ ( )+ ( ) ( ) ( ) ( ) ( ( ) ( )) ]
is oscillatory
then every solution of Eq (1 1)
y t
Proof Suppose that ( ) is a nonoscillatory solution which is positive,
t t T T y t
say y ( ) 0when 0 for some 0 depending on the solution ( ). Denote
t
y ( )
u t v t r t g t t T
(2 3) ( ) ( ) ( ) + ( ) 0
y t
f ( ( )) t
It follows from (1.1) and (2.1) that u ( ) satisfies
t
u ( )=
2
t y t y t f y t
[ r ( ) ( )] [ ( )] ( ( ))
g t u t v t r t t r g t = 2 ( ) ( )+ ( ) ( ) +[ ( ) ( )]
2
f y t
y t
( ( )) f ( ( ))
y v t e t
( t ) ( ) ( )
v t r t v t g t t Kq t
2 g ( ) ( ) ( ) + ( ) ( ) ( )+
y t f y t
f ( ( )) ( ( ))
2
t y t
[ y ( )] ( )
t v t v t r t g t v t p t r
( ) ( ) 2 + ( )[ ( ) ( )] ( ) ( ) =
f y t
y t f ( ( )) ( ( ))
2019. majus´ 4. –23:07
INTERVAL OSCILLATION CRITERIA 87
t v t e t
y ( ) 2 ( ) ( )
g t v t r t r t v t g t v t Kq t
= 2 ( ) ( ) ( ) 2 ( ) ( ) ( ) ( ) ( )+
y t f y t
f ( ( )) ( ( ))
2
t y t
[ y ( )] ( )
v r t v t v t r t g t t p t
( ) ( ) 2 + ( )[ ( ) ( )] ( ) ( ) =
f y t
y t f ( ( )) ( ( ))
2
u t t v t e t
( ) p ( ) ( ) ( )
u t t
= ( ) ( )+
t r t r t f y t v ( ) ( ) ( ) ( ( )) That is,
2
t p t v t e t
u ( ) ( ) ( ) ( )
t u t u t
(2 4) ( ) ( ) ( )+
t r t r t f y t
v ( ) ( ) ( ) ( ( ))
t T e t
By assumption, we can choose s1, 1 0 so that ( ) 0ontheinterval
s t s t I u t I =[ 1 1] with 1 1. On the interval , ( ) satisfies by (2.4),
2
t p t
u ( ) ( )
t u t u t
(2 5) ( ) ( ) ( )
t r t r t v ( ) ( ) ( )
2
D s t H Let H ( 1 1) be given as in hypothesis. Multiplying (2.5) by and
integrating over I ,wehave
t t Z
Z 1 1 2
t p t u ( ) ( )
2 2
u dt H t t dt H t t
(2 6) ( ) ( ) ( ) ( )+ +
t v t r t
r ( ) ( ) ( ) s
s1 1
s H t
Integrating (2.6) by parts and using the fact that H ( 1)= ( 1) = 0, we obtain
t t Z
Z 1 1 2
t p t u ( ) ( )
2 2
u dt t t dt H t t
H ( ) ( ) ( ) ( )+ + =
t v t r t
r ( ) ( ) ( ) s
s1 1
s
t
2 r
Z 1
r t v t p t
1 ( ) ( ) ( )
H t u t H t H t dt
= ( ) ( )+ 2 ( )+ ( ) +
t v t r t r ( ) ( ) 2 ( )
s1
t
Z 1 2
t v t p t
r ( ) ( ) ( )
H t H t dt
+ 2 ( )+ ( )
r t 4 ( )
s1
t
Z 1 2
t v t p t
r ( ) ( ) ( )
H t H t dt
2 ( )+ ( )
r t 4 ( )
s1
2019. majus´ 4. –23:07
88 W. LI, H-F. HUO t which contradicts the condition (2.2). This contradiction proves that y ( )is
oscillatory.
t H D s t e t
When y ( ) is eventually negative, we see ( 2 2)and ( ) 0on t
[ s2 2] to reach a similar contradiction. The proof is complete.
t
When p ( ) 0, by Theorem 1, we have the following corollary.
Let assumptions and hold Suppose that
Corollary (H1) (H4) (2 1)
for any T t there exist T s t s t such that holds If 0 1 1 2 2 (1 9)
1
there exist H D s t and g C t such that i
( i ) ([ 0 ) (0 ))
t t
i i Z Z 1
2 2
H t t dt r t v t H t dt
(2 7) ( ) ( ) ( ) ( )[ ( )]
s s
i i
t
R
where v t g s ds and i for =1 2 ( )=exp 2 ( )
2
t v t Kq t r t g t r t v t
( )= ( )[ ( )+ ( ) ( ) ( ( ) ( )) ]
is oscillatory
then every solution of Eq (1 1)
t v t t q t
We remark that, if we take g ( )=0,then ( )=1, ( )= ( ). Hence
y y
Corollary 1 also reduces to Theorem B of Wong if f ( )= . y For the case when f ( ) is not monotonous or has no continuous deriva-
tive, the following result holds.
Suppose that hold Let assumption in The
Theorem (H1) (H4) (2 1)
orem be replaced by y
f ( )
c for y
(2 8) 0 0
y
c is a constant Suppose that q t and that for any T t there
where ( ) 0 0
exist T s t s t such that holds If there exist H D s t i 1 1 2 2 (1 9) ( i )
1
g C t such that
and ([ 0 ) (0 ))
t
t
u i
Z
Z 2 t 1 p ( )
2
r t t dt t v t H t H t dt H
(2 9) ( ) ( ) ( ) ( ) 2 ( )+ ( ) t
4 r ( )
s s
i i
t
R
for i where v t g s ds and =1 2 ( )=exp 2 ( )
2
v t Kcq t r t g t p t g t r t g t ( t )= ( )[ ( )+ ( ) ( ) ( ) ( ) ( ( ) ( )) ]
2019. majus´ 4. –23:07
INTERVAL OSCILLATION CRITERIA 89
is oscillatory
then every solution of Eq (1 1)
Proof t
Suppose that y ( ) is a nonoscillatory solution which is positive,
t t T T y t
say y ( ) 0when 0 for some 0 depending on the solution ( ). Denote
t
y ( )
u t v t r t g t t T
(2 10) ( )= ( ) ( ) + ( ) 0 t
y ( ) t
It follows from (1.1) and (2.8) that u ( ) satisfies
2
t y t y t
[ r ( ) ( )] [ ( )]
u t g t u t v t r t r t g t
( )= 2 ( ) ( )+ ( ) ( ) 2 +[ ( ) ( )]
y t t
( ) y ( )
t v t e t
y ( ) ( ) ( )
g t r t v t g t t Kcq t v
2 ( ) ( ) ( ) + ( ) ( ) ( )+
t y t
y ( ) ( )
2
t y t
[ y ( )] ( )
r t v t v t r t g t t p t v
( ) ( ) 2 + ( )[ ( ) ( )] ( ) ( ) =
y t t
y ( ) ( )
t v t e t
y ( ) 2 ( ) ( )
g t v t r t r t v t g t v t Kcq t
= 2 ( ) ( ) ( ) 2 ( ) ( ) ( ) ( ) ( )+
t y t
y ( ) ( )
2
t y t
[ y ( )] ( )
r v t v t r t g t v t p t
( t ) ( ) 2 + ( )[ ( ) ( )] ( ) ( ) =
y t t y ( ) ( )
2
u p t v t e t
( t ) ( ) ( ) ( )
u t t
= ( ) ( )+
t r t r t y t v ( ) ( ) ( ) ( ) That is,
2
u p t v t e t
( t ) ( ) ( ) ( )
t u t u t
(2 11) ( ) ( ) ( )+
t r t r t y t
v ( ) ( ) ( ) ( )
t T e t
By assumption, we can choose s1, 1 0 so that ( ) 0ontheinterval
s t s t I u t I =[ 1 1] with 1 1. On the interval , ( ) satisfies by (2.11),
2
t p t
u ( ) ( )
t u t u t
(2 12) ( ) ( ) ( )
t r t r t v ( ) ( ) ( ) Similar to the proof of Theorem 1, we can obtain a contradiction. The
proof is complete.
t
If p ( ) 0, then, by Theorem 2, we have the following corollary.
Let the assumptions and hold Suppose
Corollary (H1)–(H4) (2 8)
t and that for any T t there exist T s t s t such q that ( ) 0 0 1 1 2 2
2019. majus´ 4. –23:07 90 W. LI, H-F. HUO
1
that holds If there exist H D s t and g C t such i
(1 9) ( i ) ([ 0 ) (0 ))
that
t t
i i
Z Z
2 2
H t dt r t v t H t dt
( t ) ( ) ( ) ( )[ ( )]
s s
i i
t
R
i where v t g s ds and for =1 2 ( )=exp 2 ( )
2
t v t Kcq t r t g t r t g t
( )= ( )[ ( )+ ( ) ( ) [ ( ) ( )] ]
is oscillatory then every solution of Eq (1 1)
3. Examples
In this section we will show the applications of our oscillation criteria by two examples. We will see that the equations in the examples are os- cillatory based on the results in Section 2, though the oscillation cannot be
demonstrated by the results of Wong [18].
Example Consider the following nonlinear differential equation
p 5
4
ty t y t y t y t p
( ( )) 2 ( )+ p 4 ( )(1 + ( )) = t
4 t (1 + sin ) p
1 p
p
t t
(3 1) = (sin t cos ) 1
t
p p
1 2
p
t n Here the zeros of the forcing term (sin t cos )are + . Clearly,
t 4
4 4
y y y f y y
f ( )= (1 + )and ( )=1+5 1=
p
t t T n
Let H ( )=sin .Forany 1, choose sufficiently large so that
2 2 2
n n T s t n
+ 4 and set 1 = + 4 and 1 = ( +1) + 4 in (2.2).
t v t
Pick up g ( )=0,then ( )=1.Itiseasytoverifythat
2 2
n
(n +1) + ( +1) + Z Z 4 4
2 2 p 5
H t t dt t dt p
( ) ( ) = sin p 4 = t
4 t (1 + sin )
2 2 n n + 4 + 4
2019. majus´ 4. –23:07
INTERVAL OSCILLATION CRITERIA 91
n
(n +1) + ( +1) + Z Z 4 4
2 5 5 2 1
sds s ds
= sin s 4 2 = sin 4 =
s s
4 s (1 + sin ) 2 1+sin
n
n + 4 + 4
n
(n +1) + ( +1) + Z Z 4 4 2
5 2 1 5 sin s
ds ds = sin s =
2 2 2
s s
2 1+sin s (1 cos ) 2 1+sin
n
n + 4 + 4
n
(n +1) + ( +1) + Z
Z 4 2 4
5 sin s +1 1 5 5 1
ds
= 2 ds = 2 s
2 1+sin s 2 2 1+sin
n
n + 4 + 4
(n +1) +
Z 4
5 5 1 5 5 5
ds
=
s j
2 2 2 j sin 2 4 4 n + 4
2
t
where we have used the inequality 1 + sin t 2sin ,and
t
Z 1 2 t
1 p ( )
r t v t H t H t dt
( ) ( ) 2 ( )+ ( ) =
r t 4 ( )
s1
2 n
+ +
p
Z 4 2
p p p
1 cos t
p
t t dt = t 2 2 sin =
4 2 t
n + 4
n + +
Z 4 2
1 cos s 2
s s ds
= sin = s
4 s
n + 4
n + +
Z 4
1 2 2 5
s s ds = [cos s +2sin2 +4sin ] =
2 4
n + 4
which implies that (2.2) holds for i =1.
2019. majus´ 4. –23:07
92 W. LI, H-F. HUO
2 2
n t n Similarly, for s2 = ( +1) + 4 and 2 = ( +2) + 4 , we can
show that (2.2) holds. It follows from Theorem 1 that every solution of
p
t t Eq. (3.1) is oscillatory. Observe that y ( )=sin is such a solution. How-
ever, the results of Wong [18] fail for the oscillation of Eq. (3.1).
Example Consider the following nonlinear differential equation
p
2
p
t
10(1 + sin ) 1 1
ty t y t y t p
( ) 2 ( )+ p 2 ( ) + 2 =
y t t t (9 + sin ) 8 1+ ( )
(3 2) p
1 p
p
t t
= (sin t cos ) 1
t
h i
1 1
y y Let f ( )= + 2 ,then 8 1+y
2 2
( y 3)
y f ( )= 2 2
8(1 + y )
y y Clearly, the condition, f ( ) 0for 0 does not hold. Hence, Theorem
1 is not valid for Eq. (3.2). However, y
f ( ) 1 1 1
K
= + 2 = 0 y 8 1+y 8 Similar to the proof of Example 1, we see that the assumptions of Theorem
2 are satisfied. Hence, every solution of Eq. (3.2) is oscillatory. Observe that
p
t t Wong y ( )=sin is an oscillatory solution. Similarly, the results of [18] are not valid for the oscillation of Eq. (3.2).
References
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S W Wong T A Burton u
[19] J and Some properties of solution of +
t f u g u Monatsh Math
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[20] A Skidmore and Oscillatory behavior of solutions of + ( ) =
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2019. majus´ 4. –23:07
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 95–105
AREA-INRADIUS AND DIAMETER-INRADIUS RELATIONS FOR COVERING PLANE SETS
By MARIA´ A. HERNANDEZ´ CIFRE
Departamento de Matematicas,´ Universidad de Murcia, Murcia, Spain Received September
1. Introduction and first results
2 K Let K denote the family of compact convex sets in the Euclidean
2 2
K K A K r K D K K plane E .For ,let ( ), ( ), ( )and ( ) be the area, the inradius, the diameter and the minimal width (that is, the smallest distance
between two parallel support lines) of K , respectively.
K lattice point enumerator
For an arbitrary lattice L and a set ,the is
K L K L K lattice
denoted by G ( ) = card(int( ) ). A convex set is called a
point free convex set G K L K with respect to L if ( ) = 0. Further, is a
covering set if
2
L fK g g Lg E K + = + = A great number of results concerning covering sets with respect to the 2 integer lattice Z are known. However there are relatively few results on
covering sets with respect to an arbitrary lattice L ([6], [7], [8], [9], [10]). Scott The following elegant result was obtained by Awyong and in [2]: an inequality concerning the inradius and the area of a planar lattice-point-free
2 Z convex set, in the case where L is the integer lattice .
2
Let K be a compact planar convex set with G K Z
Theorem ( )=0
Then
p
A (1) (2r 1) 2( 2 1)
This work is supported in part by Direccion´ General de Investigacion´ (MCYT) BFM2001- 2871.
2019. majus´ 4. –23:10
96 M. A. H. CIFRE
with equality when and only when K is congruent to the diagonal square
shown in Figure
4
Figure
Awyong Henk Scott The d -dimensional analog was solved by , and in [1]. In this paper, we generalize Theorem 1 to rectangular lattices, using then this result to obtain, in the last section, a more general inequality for arbitrary lattices, and some other related results.
Before stating the main theorem, let us introduce some notation. We
u v
denote by Γuv the rectangular lattice generated by the vectors ( 0) and (0 ),
u v
with 0 . Q
Now, we define the rombhus r as follows:
2
P u v E B r P
Let =( 2 2) ,and P ( ) be the disc centered in and with
Figure r
radius r . Following the notation of , for each fixed we denote by
Q B r r
the rombhus with sides tangent to the disc P ( ), which pass through the
u v P v P P u
points P1 =( ), 2 =(0 ), 3 =(00) and 4 =( 0) respectively, and
u v
with angle arctan( ). P 2 P1
l1
r
P P P3 4
Figure Optimal rombhus
This set will have an important role in our results. It is easy to check that
2 2 2 2
v r
2( u + )
p p
A Q
( r )=
2 2 2 2 2 2
v u v r vr u u v r (2ur + 4 )(2 + + 4 )
2019. majus´ 4. –23:10
AREA-INRADIUS AND DIAMETER-INRADIUS RELATIONS 97
b r
Let us also denote by uv ( ) the function
2 2 2 2
v r v r
2( u + ) (2 )
p p
b r
(2) uv ( )=
2 2 2 2 2 2
v u v r vr u u v r (2ur + 4 )(2 + + 4 )
We prove the following result:
Theorem Let r denote the unique solution of the equation
p
2 2 2 2 2 2 2
r v u v r r v u v r r v u r (3) uv ( + ) + 4 + [ (3 + ) 8 ( + )+4 ]=0
2
For each K K with G K it holds
( Γuv )=0
1 2
v u then r K v A K uv
i) If 2 (2 ( ) ) ( ) 2
If v u then r K v A K b r
ii) 2 (2 ( ) ) ( ) uv ( ) These inequalities are tight in the following sense
1 2 1 2
cannot be replaced by uv because the equality would be
i) 2 uv 2
attained for the case r v when K is the in nite strip
( K )= 2
Now equality holds when K Q up to congruence
ii) = r
2. Proof of the Theorem
The proof of the theorem will be established by proving two previous lemmas, the second of which is a result from elementary calculus.
2
Lemma Let K K be a convex domain such that G K
( Γuv )=0
s 2
Then there exists another convex domain K K containing no points of such that
Γuv
s s
K A K and r K r K
i) A( )= ( ) ( ) ( )
s
is symmetric about the lines x u y v
ii) K = 2 = 2
K K
Proof Let be the convex domain which is obtained from by u Steiner symmetrization with respect to the line x = 2. It is well known
that Steiner symmetrization preserves the convexity and the area, and does
A K
not decrease the inradius [3]. Therefore, K is a convex domain with ( )=
A K r K r K G K
= ( ), and ( ) ( ). Now, we have to see that ( Γuv )=0.
K mu nv
Let us suppose that contains the lattice point of Γuv , + , with
Z K x u y nv
m n . Then, because of the symmetry of about = 2, the line =
u would intersect K in a line segment of length greater than . So, this line
2019. majus´ 4. –23:10 98 M. A. H. CIFRE
would also intersect K in a line segment with the same length, which implies
G K G K uv that ( Γuv ) 0, contradicting the hypothesis. Hence, ( Γ )=0.
We can use an analogous argument, but now symmetrizing about the line
s s s
v K A K A K r K r K
y = 2, to obtain a convex domain with ( )= ( ), ( ) ( )
s s
G K K
and ( Γuv ) = 0. By construction, is symmetric about the lines
u y v
x = 2and = 2, and the lemma is proved.
Lemma Let p
2 2 2 2 2 2 2
r uv r v u v r r v u v r r v u r h ( )=2 ( + ) + 4 +2 [ (3 + ) 8 ( + )+4 ]
p 2 2
with u v and r v u v Then
0 ( 2 + 2]
v u h r
i) If 2 ( ) 0 p
2 2
v u h r vanishes exactly in an only point r v u v ii) If 2 ( ) ( 2 + 2]
Proof Let us denote by p
2 2 2
r r v u v r h1( )=( + ) + 4
2 2 2 2
h r v u v r r v u r
2( r )= [ (3 + ) 8 ( + )+4 ]
r uv h r h r Then, h ( )=2 1( )+2 2( ).
p 2 2
v u v It is easy to compute that when r ( 2 + 2], it holds
3 2 2
u v r v
8 r ( + )(3 + )
r h1 ( )=4 0
2 2 2 3 2
v r
( u + 4 )
r h r so, h1( ) is a concave function, and moreover, 1( ) is strictly decreasing. Analogously, we can check that
2 2
r r vr u
h2 ( )= 96 48 +8 0
r h r
and then, h2( ) is a concave function and 1( ) is strictly decreasing. r
Hence, we can deduce that the original function h ( ) is concave and its
r h r h v
first derivative h ( ) is strictly decreasing. So, ( ) ( 2). Now,
p p p
2 2 2 2 2 2 2 2
u v v u u v u v v h + 2 = ( ) + + + 0
2 2 2
v v u v h ( 2) = 2 (4 )and
2 2
v v u v h ( 2) = 8 (2 3 )
Then we obtain that
2019. majus´ 4. –23:10
AREA-INRADIUS AND DIAMETER-INRADIUS RELATIONS 99
u h v h v h r
i) If v 2 , ( 2) 0andalso ( 2) 0. So, ( ) 0 and therefore r
h ( ) is strictly decreasing and negative. p
2 2
u h v r v u v
ii) If v 2 , ( 2) 0. Then, there exists an ( 2 + 2] such
r
that h ( ) = 0, and this point is unique because of the strict concavity r of h ( ).
Now we can prove our theorem.
K r K v A K
Let us define the functional f ( )=(2 ( ) ) ( ). Applying Lemma 1 s
s 2
K K K G K to the set , we may obtain a new convex set with ( Γuv )=0,
satisfying the conditions:
s s
K A K r K r K
i) A( )= ( )and ( ) ( ),
s
x u y v
ii) K is symmetric about the lines = 2, = 2.
s
K f K Then, it is clear that f ( ) ( ). So, it suffices to prove the theorem
2
K x u y v
for sets K which are symmetric about the lines = 2and = 2.
x u y v
To fully utilize the symmetry of K about the lines = 2and = 2,
u v we move the origin to the point P =( 2 2).
2
K Obviously, the area of a lattice-point-free convex set K with
respect to Γuv can be arbitrary large. However, the inradius of such a set
p 2 2
v r K v r K
is bounded above by u + 2. Besides, if ( ) 2, then (2 ( )
v A K ) ( ) 0 and the result is trivially true. Hence, we may assume that
p 2 2
r K u v v 2 ( ) + 2.
Since int( K ) does not contain the points
u u v v
P P = =
1 2 2 2 2 2
u u v v
P P = =
3 2 2 4 2 2
i K
it follows by the convexity of K that for each =1 4, is bounded by a
l P m l m l m l m l i
line i passing through , with slopes ( 2)= ( 4)= ( 1)= ( 3) 0
K Q
(by the symmetry of K ). So, lies within a rhombus determined by the
l i K Q A K A Q r K r Q
lines i , =1 4. Since , clearly ( ) ( )and ( ) ( )=
r f K f Q f K
= Q , and we have ( ) ( ). It is therefore sufficient to maximize ( )
2
Q K l i
over the set of all rhombi , determined by the lines i , =1 4.
l OX
But moreover; if the acute angle determined by the line 1 and the -axis
Figure u v Q Q
(see ) is not greater than arctan( ), then = r . Q
2019. majus´ 4. –23:10
100 M. A. H. CIFRE u v
But if arctan( ), it is not difficult to compute that the area of such
Q r
a rhombus in terms of its inradius Q takes the value
2 2 2 2
v r
2( u + )
Q
q q
A Q ( )=
2 2 2 2 2 2
ur v u v r vr u u v r Q
2 Q + + 4 2 + 4
Q Q
v
Since u , we can obtain easily that
q q
2 2 2 2 2 2
ur v u v r vr u u v r Q
2 Q + 4 2 + + 4
Q Q
q q
2 2 2 2 2 2
ur v u v r vr u u v r Q
2 Q + + 4 2 + 4
Q Q
Q Q
Then, the rhombus R (with the same inradius as ) has area strictly
Q
Q f Q f Q
greater than the area of . Therefore, again ( ) ( r ), and so, we have
Q
f K Q
just to maximize ( ) over the set of all rhombi r .
Q r
We had gotten the area of r , as a function of ,so
f Q r Q v A Q b r
r r uv ( r )=(2 ( ) ) ( )= ( )=
2 2 2 2
v r v r
2( u + ) (2 )
p = p
2 2 2 2 2 2
v u v r vr u u v r
2 ur + 4 2 + + 4
b r
But uv ( ) can be written in the following way
p
2 2 2 2
ur v u v r r 2 + + 4
2 2
b r u v p
uv ( )=2( + )
2 2 2
v vr u u v r (2r + ) 2 + + 4 and then, it is not difficult to see that
2 2 2
v r
2( u + )
b r h r
( )= ( )
uv p 2 p
2 2 2 2 2 2
v vr u u v r u v r
(2r + ) 2 + + 4 + 4 r
where h ( ) is the function defined in Lemma 2.
v u h r b r
i) If 2 , Lemma 2 assures us that ( ) 0; so uv ( ) is strictly monotonously decreasing, and then
1 2
b r b r uv uv
uv ( ) lim ( )=
v r 2 2
2019. majus´ 4. –23:10 AREA-INRADIUS AND DIAMETER-INRADIUS RELATIONS 101
1 2
Clearly, this bound can not be replaced by a 2 uv , because the
v K equality would be attained when r = 2, i.e., when is the infinite
strip.
u
ii) If v 2 , because of the proof of Lemma 2, we can assure that there
r b r
exists a unique solution of the corresponding equation (3); so uv ( )
r
attains its maximum value for r = . Hence, in this case,
b r b r uv
uv ( ) ( )
Q
and the equality holds when K = (up to congruence). r This completes the proof of the theorem.
3. Some results for arbitrary lattices
2 2
L E L Let us denote by L the set of lattices with det 0. Further, for
2 2
L L L L L B L
i i i
let i = ( ) be the successive minima of , i.e., ( )= ( )=
2
f j B L i g L i
=min 0 dim aff( ) ,andlet i = ( ) be the covering
2 2
L L B L f j B g g L i
minima of the lattice , i.e., i ( )= ( )=min 0 + ,
2
E F i g
meets every flat F of with dim( )=2 .
b b g L reduced We remember also that a basis f 1 2 of is (in the sense of
Minkowski) if
fv L nf g kv k g
i) b1 0 : is minimal
fv L nf g b v L kv k g
ii) b2 0 : 1 are a basis of , is minimal
b
iii) b1 2 0.
For the sake of brevity we will represent by the maximum =
f g f g
=max 1 2 1 and by the minimum =min i 2 1 . Moreover, we
r u
denote by b ( ) the corresponding function defined by (2) when = and
v = . We will prove the following result.
Theorem Let r denote the unique solution of the equation
q
2 2 2 2 2 2 2
r r r r r r (4) ( + ) + 4 + [ (3 + ) 8 ( + )+4 ]=0
2 2
K K and L L withG K L it holds For each ( )=0
1 2
then r K A K
i) If 2 (2 ( ) ) ( ) 2
If then r K A K b r
ii) 2 (2 ( ) ) ( ) ( )
The inequalities are tight
2019. majus´ 4. –23:10
102 M. A. H. CIFRE p
3
Remark Because of 2 1 2 1 (see [5]), it will never hold 1
f g
2(2 1). So, when =max 1 2 1 = 1, we will have just the upper
r A b r
bound (2 ) ( ). Thus, for an arbitrary lattice, we will have the 2 1 1 following three possible cases:
2
r K A K
i) If 1 1 then (2 ( ) 2 1) ( ) 2 1 1 .
r K A K b r
ii) If 2 then (2 ( ) 2 ) ( ) ( ).
1 1 1 1 12 1
r K A K b r
iii) If 2 4 then (2 ( ) ) ( ) ( ).
1 1 1 1 2 1 1 r
with each solution of the corresponding equations we obtain from (4).
2
For instance, in the case of the integer lattice Z , 1 =2 1 =1,andwe
p
r r A
obtain = 2 2. So, the upper bound for the relation (2 1) takes the
p p
Awyong Scott value b11( 2 2) = 2( 2 1), which was proved by and in [2]. From this theorem, we can obtain as an obvious consequence the follow- ing corollary.
2 2
Corollary Let K K and L L be given such that
1 2
r K A K b r
(2 ( ) ) ( ) max ( )
2
Then K is a covering set We also prove an inequality relating the inradius and the diameter of a
2
K lattice-point-free convex set K .
2 2
Let K be a convex set of K and L L with G K L Proposition ( )=
=0 Then
K D K
(2r ( ) 2 1)( ( ) 1) 2 1 1 The limiting in nite strip shows that the stated bound is best possible
2 2
and L L be given such that Corollary Let K K
K D K
(2r ( ) 2 1)( ( ) 1) 2 1 1
Then K is a covering set
We observe that this inequality can be rewritten as
1 1
+ 1
K r K D ( ) ( ) So, the following corollary generalizes the previous one:
2019. majus´ 4. –23:10 AREA-INRADIUS AND DIAMETER-INRADIUS RELATIONS 103
2 2
Corollary Let K K and L L be given such that
1 1
k k Z
+
K r K D ( ) ( )
2 2
Then G k i e fK g j g Lg is at least a k fold covering ( K L) +
2
of E
3.1. Proof of Theorem 3
b b g L
Let f 1 2 be a reduced basis of in the sense of Minkowski, with
b k L b b
k 1 = 1( ), and let be the acute angle between 1 and 2 (so that
L kb k v b v 2 1( )= 2 sin ). Let 1 = 1,andlet 2 be a vector of length 2 1,which
is perpendicular to v1. Let now Γ denote the rectangular lattice determined
v v g by the basis f 1 2 .
We reduce the problem to rectangular lattices and symmetric convex
K
bodies. To this end, let K be the Steiner symmetral of with respect to
K A K A K
the line x = 1 2. Then, is a convex domain with ( )= ( ), and
K r K G K K
r ( ) ( ). We have to see that ( Γ) = 0. For, let us suppose that
nv m n Z
contains the Γ lattice point, mv1 + 2, with , . Then, the symmetry
x y n K
of K about = 1 2 assures that the line =2 1 intersects in a line K
segment of length greater than 1. Thus, this line also intersects in a line
K L
segment with the same length, which implies that G ( ) 0, contradicting
K
the hypothesis. Hence, G ( Γ) = 0. Now, we use an analogous argument
s
K
but symmetrizing about the line y = 1, and we obtain a convex set with
K the same area, greater or equal inradius, and such that G ( Γ) = 0.
Now, we may identify Γ with the rectangular lattice Γ generated by the
vectors ( 0) and (0 ) (note that either Γ =ΓorΓ = (Γ), where
2
s
K is the rotation by 2 at the origin). Then, applying Theorem 2 to 2
and Γ Γ, we obtain finally
1 2
if 2 s
s 2
K A K r K A K
(2r ( ) ) ( ) (2 ( ) ) ( )
b r
( )if 2 .
This concludes the proof of the theorem.
2019. majus´ 4. –23:10 104 M. A. H. CIFRE
3.2. The inradius-diameter results
of Proposition K r K r
Proof We will write ( )= , ( )= and
K D D ( )= .
In [9] the following result is proved:
D
(5) ( 2 1)( 1) 2 1 1 D
with equality when and only when K is a triangle of diameter and width
D D
=2 1 ( 1).
r
Since 2 ,wehave
D D (2r 2 1)( 1) ( 2 1)( 1) 2 1 1
Taking the infinite strip to be the limit of a sequence of triangles which r
give the equality in (5), when tends to 2 we have
D
lim (2r 2 1)( 1)=
r
2
1
r
= lim (2r 2 1) 1 = lim 2 1 =2 1 1
r r 2 2 1 2 2 1 So, the stated bound is best possible.
Proof of Corollary The proof of Corollary 3 follows an analogous
proof by Hammer [4], and we repeat it here for opportunity.
k
If k = 0 the result is trivial. So, let us suppose that 1, and consider
1 1
K D K D K K
the similarity transformation K = . Obviously, ( )= ( )
k k
1
r r K K fb b g L kb k i i
and ( )= ( ). Now let beabasisof with i = , =1,2,
k 1 2
R m b m b m k i i
and let = 1 1 + 2 2 be a lattice point with 0 i ( 1) , =1,2.
1
K K K R
Now, let us consider the translate K of given by = . k
We have:
k k
1 + 1 = 1 + 1 = 1 + 1 =
K r K D K r K D K r K
D ( ) ( ) ( ) ( ) ( ) ( )
1 1
= k + 1
K r K
D ( ) ( )
T K By Proposition 1, K contains a lattice point . Hence, contains
1
R K
the point T + , and so, the original domain contains the lattice point
k
1
R U k T kT R m k i
= + = + . But taking into account 0 i ( 1) ,we k
2019. majus´ 4. –23:10
AREA-INRADIUS AND DIAMETER-INRADIUS RELATIONS 105
m i k R
could have selected each i =1,2in different ways. So, might have
2 2
K k been chosen in k different ways. Therefore, contains at least, distinct
2
K L k lattice points in its interior, i.e. G ( ) .
Acknowledgements
The author wishes to thank Salvatore Vassallo for his valuable advices and helpful suggestions during the preparation of this paper. She also wishes to thank O. Perez´ Blaya for enlightening discussions.
References P Scott [1] W Awyong M Henk and Note on lattice-point-free-convex bodies,
Mh Math 126 (1998), 7–12. P Scott [2] W Awyong and New inequalities for planar convex sets with lattice
point constraints, Bull Austral Math Soc 54 (1996), 391–396.
W Fenchel Theorie der Konvexen Korper [3] T Bonnesen and Springer, Berlin
1934, 1974; Chelsea, New York 1948. Math Mag [4] J Hammer Lattice Points and Area-Diameter Relation, 52 (1979),
25–25.
L Lovasz [5] R Kannan and Covering minima and lattice-points-free convex
bodies, Ann Math 128 (1988), 577–602. J M Wills [6] U Schnell and Two Isoperimetric Inequalities with Lattice Con-
straints, Mh Math 112 (1991), 227–233. J M Wills [7] U Schnell and On Successive Minima and Intrinsic Volumes,
Mathematika 40 (1993), 144–147.
Geom Ded [8] S Vassallo A covering problem for plane lattices, 43 (1992), 321–335.
[9] S Vassallo Area-diameter and area-width relations for covering plane sets,
Bull Austral Math Soc 51 (1995), 163–169.
J M Wills Mh Math [10] S Vassallo and Covering sets for plane lattices, 121 (1996), 381–389.
2019. majus´ 4. –23:10
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 107–113
A CHARACTERIZATION OF ADJOINT 1-SUMMING OPERATORS
By QINGYING BU
Department of Mathematical Sciences, Kent State University, Kent, Ohio
Received January
X X B
For a Banach space ,let denote its dual and X denote its closed
p p p p
unit ball. For 1 ,let denote its conjugate, i.e., 1 +1 =1.For
p X p
1 ,let p ( ) denote the space of absolutely -summable sequences
on a Banach space X , i.e.,
1 p
X
N p
X x x X kx k kx k
p n n n
( )= =( ) : (p ) =
n =1
1 p
P
p
p kx k kx k X k k
n n p
where if = ,let =sup .Then( ( ) (p ))isa n n ]1=
Banach space (cf. [2, 7]). Let
n o
N
c X x x x X
n n
0( )= =( ) n : lim =0
n
c X X p X p
Then 0( ) is a closed subspace of ( ). For 1 ,let [ ] denote X
the space of weakly p -summable sequences on a Banach space , i.e.,
X
N p
X X x x jx x j x X
n n n p [ ]= =( ) : ( )
n =1
x X
and for p [ ], let
1 p
X
p
kx k jx x j x B n
p =sup ( ) :
[ ] X
n =1
2019. majus´ 4. –23:07
108 QINGYING BU
X k k p
Then ( [ ] [p ]) is a Banach space (cf. [2, 7, 8]). Here notice that if
1 p
P
p
p jx x j jx x j p hX i
n p
= ,let ( n ) =sup ( ) .For1 ,let n
n =1 X denote the space of strongly p -summable sequences on a Banach space ,
i.e.,
X
N
hX i x x X jx x j x X
n n n n
p = =( ) : ( ) ( ) [ ]
n n p
n =1
x hX i
and for p ,let
X
kx k x x x B
n
=sup ( ) :( ) n
X
n n
hp i
[ ]
p
n =1
hX i k k
Then ( p ) is a Banach space (cf. [2]). Let
hp i
n o
c hX i x x kx i n k hX i
n = =( ) n : lim ( ) =0
0 h i
n
x i n x x c hX i n
where ( )=(0 0 n +1 +2 ). Then 0 is a closed subspace
hX i
of p .
X Y
Definition Let , be Banach spaces. A Banach space operator
X Y c
u : is called strongly -summing if there exists a constant 0
x x x X y y y Y n
such that for any 1, 2, , n and 1 , 2 , , ,
n n
X X
kx k jhux y ij c jy y j k
(1) k sup sup ( )
k k
y B
k n
1
Y k
k =1 =1
D u c
Let ( ) denote the infimum taken over all possible as above. Then
D
( ) is a norm (see [2]).
X Y
Recall that a Banach space operator u : is called absolutely
x x
1-summing if there exists a constant c 0 such that for any 1, 2, ,
x X
n ,
n n
X X
kux k c jx x j x B
k X
(2) k sup ( ) : k
k =1 =1 Let 1( ) denote the absolutely 1-summing norm (see [4, p.31]). By Theorem 2.2.2 in [2], we have
2019. majus´ 4. –23:07
A CHARACTERIZATION OF ADJOINT 1-SUMMING OPERATORS 109
A Banach space operator u X Y is absolutely
Theorem (i) :
if and only if its adjoint operator u Y X is strongly
1 summing :
u D u summing In this case
1( )= ( )
u X Y is strongly summing if and
(ii) A Banach space operator :
Y X is absolutely summing In this u
only if its adjoint operator : 1
case u D u
1( )= ( )
Y u X
Let X , be Banach spaces. For a continuous linear operator : Y
,define
N N
u
y
: x
x ux
n n n
( n ) ( )
Then u is a linear operator. Define
N
x X x
:
0
n n
P P k
k ( )
s x s x
k k
i k
=1 k =1
i
X Y Z W
Then is well-defined linear map (see [1]). Let , , , be Banach
X Z v Y W spaces and let u : and : be Banach space operators.
Define
v X Y Z W
u :
n n
P P
x y ux vy
k k k
k ( ) ( ) k
k =1 =1
X c X
Let c0 denote the completion of 0 with respect to the injective
k k c Y c Y
tensor norm ,andlet 0 denote the completion of 0 with
k k
respect to the projective tensor norm (cf. [5, p. 223–227]).
Let u X Y be a Banach space operator Then the
Theorem :
following are equivalent
is strongly summing
(i) u
X hY i i e u sends each bounded sequence in X to
(ii) idc0( ( ))
strongly summable sequence in Y
u
X c hY i
(iii) ( c0( )) 0
X Y where u c c is the identity operator c
(iv) (idc0 )( 0 ) 0 id 0
c on 0
2019. majus´ 4. –23:07
110 QINGYING BU
u u
Furthermore in this case X hY i c X c hY i
: ( ) : 0( ) 0
and u c X c Y are continuous with idc0 : 0 0
(3)
u u
k k k k k u k D u
= = idc = ( )
X hy i c X c hY i
( ) 0( ) 0 0
c X c Y
0 0
x x X Proof
n
(i) (ii): It is easy to show that for =( ) n ( )and
y y Y
=( ) n [ ],
n 1
X
jhux y ij D u kx k ky k
n ( ) n ( ) [1]
n =1
u
y Y x ux hY i
n
Since is arbitrary in 1[ ], ( )=( n ) and
u
k D u k
(4) ( )
X hY i ( )
(ii) follows.
u (ii) (iii): By Closed Graph Theorem, we can show that is continu-
ous. So
u u
k x k k kx k x X k
(5) ( ) ( )
h i X hY i
( ) ( )
n N
Thus for ,
u u
k x i n k k kx i n k x X k
( )( ) ( ) ( )
h i X hY i
( ) ( )
u
x c X kx i n k k x i n k If ( ), then lim ( ) = 0. So lim ( )( ) = 0, i.e.,
0 ( ) h i
n n
u
c hX i
( x ) 0 . (iii) follows.
u (iii) (i): By Closed Graph Theorem, we can show that is continu-
ous. So
u u
k k k x k k x k x c X
(5) ( ) ( )
X c hY i c
h i 0( ) 0 ( ) 0
x x X y y x x x Y n Now for any , , n , , , ,let =( 0 0 )
1 1 n 1
y s y s y s hux y k n i
n k
and =( 0 0 )where k =sign for =1 . n
1 1 k
c X y Y
Then x 0( )and 1[ ]. So
n
X
u u
jhux y ij jh x y ij k x k ky k
k = ( ) ( ) h i k [1]
k =1
2019. majus´ 4. –23:07
A CHARACTERIZATION OF ADJOINT 1-SUMMING OPERATORS 111
u
k k kx k ky k
=
X c hY i
c0( ) 0 ( ) [1]
n
X
u
kx k k k jy y j
= sup k sup ( )
c X c hY i
0( ) 0 k
y B
k n
1 Y
k =1
Thus u is strongly -summing and
u
D y k k
(6) ( )
X c hY i c0( ) 0
(i) follows.
n P
(k )
z s x c X
(iii) (iv): For = k 0 ,
k =1
n n
X X k
(k ) ( )
u z s ux s ux
c k
((id 0 ) )= k = =
k k
=1 =1 i
(7)
n n
X X
k
( ) (k )
u u
s s x z u x k
= k = = ( )
i
k k
=1 i =1
X c c X By [7] (also see [3]), ( 0 )= 0( ) with the isometry .Soby(5)and
Theorem 9 in [1],
u
k u z k k u z k k z k
c
(idc ) = ((id ) ) = ( ) h i
0 0 h i
u u
k z k kz k k k k k
=
X c hY i X c hY i c
0( ) 0 ( ) c0( ) 0
u c X k k x Y k
So id c0 is a continuous operator from ( 0 )to(0
k y c X
). Thus idc0 can be norm-preserved extended to 0 . Therefore,
X Y u c c
(idc0 )( 0 ) 0 and
u
k k k u k
(8) idc
c c hY i
0 0(X ) 0
X c Y c0 0
(iv) follows.
x c X c X c X z c X
(iv) (iii): Let 0( ). Since ( 0 )= 0( ), 0
z x u Y c z such that ( ) . By (iv), (idc0 ) 0 . So by (7) and Theorem 9
in [1],
u u
x z u z c hY i ( )= ( )= ((idc0 ) ) 0
2019. majus´ 4. –23:07 112 QINGYING BU
Thus (iii) follows. Furthermore,
u
k x k k u z k c ( ) = (id )
h i 0
u k k u k x k k kz k k
c c
id 0 = id 0 ( )
c c X c Y X c Y 0 0 0 0
Therefore,
u
k u k k k
(9) id x
c c hY i
0(X ) 0 0
c X c Y
0 0
u
k
Now combining (4), (6), (8), (9) and noticing that k
X c hY i
c0( ) 0
u k
k , (3) holds. The proof is completed.
X hY i
( )
A Banach space operator u X Y is absolutely
Corollary (i) :
Y X In this case summing if and only if u c c
1 (idc0 )( 0 ) 0
u c Y c X is continuous and k u k u c
idc ) : 0 0 id 0 = 1( )
u X Y satis es that
(ii) A Banach space operator :
u c X c Y
(idc0 )( 0 ) 0
if and only if its adjoint operator u X is absolutely summing In
: Y 1
X c Y is continuous and k this case u c u k u c
idc0 : 0 0 id 0 = 1( ) Joe Acknowledgement I am grateful to my advisor, Professor
Diestel , for his help.
References
Bu J Diestel
[1] Q and Observations about the projective tensor product of
I p Quaestiones Math X
Banach spaces, p ,1 , 24 (2001),
519–533.
p p [2] J S Cohen Absolutely -summing, -nuclear operators, and their conjugates,
Math Ann 201 (1973), 177–200.
J Swart The Metric Theory of Tensor Products [3] J Diestel J Fourie and to
appear.
A Tonge Absolutely Summing Operators [4] J Diestel H Jarchow and Cam- bridge Univ. Press, Cambridge, 1995.
2019. majus´ 4. –23:07
A CHARACTERIZATION OF ADJOINT 1-SUMMING OPERATORS 113
J Uhl Amer Math [5] J Diestel and Vector Measures, Math. Surveys Vol. 15,
Soc Providence, 1977. Grothendieck [6] A Produits tensoriels topologiques et espaces nucleaires,´
Memoires Amer Math Soc 16 (1955). Grothendieck
[7] A Resum´ edelath´ eorie´ metrique´ des produits tensoriels
ao Paulo
topologiques, Bol Soc Mat S 8 (1956), 1–79.
Gupta Q Bu X J [ ] M
8 and On Banach-valued GAK-sequence spaces p [ ],
Analysis 2 (1994), 103–113.
2019. majus´ 4. –23:07
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 115–121
SHORT PROOFS FOR THE PSEUDOINVERSE IN LINEAR ALGEBRA
By CSABA J. HEGEDUS˝
Department of Numerical Analysis, Eotv¨ os¨ Lorand´ University, Budapest Received February
1. Introduction
We show how some theorems on projection matrices can be applied to give short proofs for the properties of the pseudoinverse and pseudosolution of under- or overdetermined linear system of equations. First, we recall some statements on projections and prove a maximum theorem for projections that map onto the same subspace. The symmetric projection is unique and the distance between a vector and a subspace can be given with the aid of a symmetric projection. As a by-product, one can give projections mapping onto the range space or null space of a matrix, and formulae for the distances from these subspaces. On the basis of these results it is then possible to give short derivations for the pseudoinverse and its properties. With the suggested simplifications it will be easy to teach a short but full pseudoinverse theory in undergraduate or graduate courses. The given remarks may be used as exercises.
Notations Matrices will be denoted by capital letters and vectors by
T a lower case letters, a denotes the transpose of . Real matrices and vectors are used throughout as it is straightforward to restate the results in complex variables.
2. Projections
We first recall some basic knowledge on projections and then prove our
n n n
P projection S theorems. Let S R be a subspace. R is a onto if
2019. majus´ 4. –23:07 116 CSABA J. HEGEDUS˝
2 T
S P P P P
range( P )= ,and = . Moreover, if = then the projection is called
orthogonal I symmetric or . The only invertible projection is the identity .To
2 1
P P P I P
see this, multiply P = with .Both and are projections. Px
Only the first application of P may produce a new vector , all subse-
k
x Px k
quent applications will leave it unchanged, P = , 1. This property
idempotent P explains the phrase: P is that is, any positive integer power of is equal to itself. Examples for projections: T
da T
P d a
1 = T =0
a d
P x
This is a rank one projection. The effect of I 1 on is the following: vector
T x
T a
d a x I P x x d
x is projected along line onto the plane =0:( 1) = T
a d
T
I P x
and a ( 1) = 0. Another example – generalizing the former – is
m n
T 1 T T
I A D A D D A D A
P2 = ( ) where R and is invertible.
x D P symmetric
Now P2 is orthogonal to the column vectors of . 2 is or
D A
orthogonal if = holds.
P P P P
Observe that P1 2 = 2 if 1 and 2 are mapping onto the same subspace.
Theorem The symmetric projection is unique among projections map
ping onto S
P P S Proof Indirectly assume 1 = 2 are two symmetric projections onto . Then
T T T
P P P P P P P P P P1 2 = 2 2 = 2 = 2 1 = 2 1 = 1
a contradiction. (Cf. [4], Sect. 2.6.1.)
Let H S be the set of projections that map onto the sub
Theorem ( )
m
space S of Moreover let P be the unique symmetric projection onto S
R s
x S and Px
Then for any vector =0
T
x Px
k P x k
max = s 2
kPx k
H S
P ( ) 2
The maximum is also reached by any projections P that satisfy Px
holds ˜ ˜ =
P x
= s 0
2019. majus´ 4. –23:07
SHORT PROOFS FOR THE PSEUDOINVERSE IN LINEAR ALGEBRA 117
x Px
Proof The theorem states that the angle between and =0is
P P P H S P P P P ˜ s smallest if = .If s , ( )then = , and applying Cauchy’s inequality leads to
T T
k P x k kPx k
x Px x P Px s
s 2 2
k P x k
= = s 2
Px k k Px k kPx k k 2 2 2
where equality still holds for the projections P˜ .
Theorem The Euclidean distance of vector x S from subspace S is
k I P x k where P is the symmetric projection onto S s
( s ) 2
P S x S
Proof If is a projection onto then the distance of from is given
I P x k P
by the minimum of k( ) 2 with respect to 0and .Wehave P x
seen in Theorem 2 that the smallest angle between x and is reached for
H S x
symmetric P ( ) hence the minimum distance takes place between and
P x
s . But we have
2 2 2 2
kx P x k k x k kP x k s s 2 = 2 (2 ) 2
from where it is seen by differentiation that the minimum with respect to is
I P x P x s reached for = 1so that the distance vector ( s ) and are mutually orthogonal to each other.
Now we recall two lemmas from standard linear algebra.
Let L be a matrix of full column rank If LB LC then
Lemma =
B follows
= C
L B C
Proof We have ( ) = 0. Any linear combinations of the columns
B C B of L may result in zero only if the columns of are zero, thus =
= C follows.
n
m T
Lemma Assume A R ThenA A is positive semide nite If A
T
is of full column rank then A A is positive de nite
T T T
Proof Ax x A Ax y y
Set y = ,then = 0. For definiteness observe
y A
that x = 0follows from =0if is of full column rank.
m n m k
m n k A A
Examples (i) Let , R and 1 R such A
that A1 has the maximal number of linearly independent columns of that
k
m T
k B B A is, rank( A)= holds. Then for R and invertible 1 define a projection by
T 1 T
A B A B A B P ( 1 )= 1( 1)
2019. majus´ 4. –23:07
118 CSABA J. HEGEDUS˝
A A Actually this is a mapping onto range(A1)=range( ). It is orthogonal if 1 =
T
A A
= B , and then the inverse of 1 exists by Lemma 2. The Euclidean
A k I P A A x k
distance of x from range( )isgivenby ( ( 1 1)) 2.
k n (ii) Similarly, let A2 R have the maximal number of linearly
T A
independent rows of A. Then the projection onto range( )isgivenby
n
T T T k
A C CA C P ( 2 )where 2 is assumed to be invertible and R .The
T T
I P C A y
projection onto nul(A)is ( 2 ) and the Euclidean distance of from
T T
P A A y nul( A)is ( 2 2 ) 2.
3. Short theory of the pseudoinverse
There is a lot of books on the theoretical and computational aspects of the pseudoinverse. We mention here [1], [2], [4], [6]. These works, of course, go into much deeper details on generalized inverse theory than what will be addressed here. The computational aspects are handled e.g. in [4]. Reference 64 of [2] contains a bibliography of 1775 items on the theory, so the interested reader may consult them for details. Among Hungarian student texts, we mention [5] that gives a comprehensive theory of the pseudoinverse in Ch. 2.
See also [3] and [7].
m n
A
Now assume that we have a rank factorization of matrix A R =
m r r n
L U A r
= LU ,where R and R so that rank( )= . We may think
L Q U R
of an LU factorization, but it is also possible to choose = and = A from the QR factorization of . + For the derivation of the pseudoinverse A we start from the four defining Penrose equations:
+ + + +
AA A A A AA A 1 = 2 =
+ + T + + T
AA AA A A A A 3 =( ) 4 =( )
+ +
Lemma AA and A A are symmetric projections
+ Proof Multiply eq. 1 by A from either side and observe equations 3
and 4. The same conclusions come from eq. 2 by multiplying with A
Theorem To every matrix A there exists uniquely the pseudoinverse
+ + + + T 1 T + T T 1
A L where L L L L U U UU and LU is an
= U =( ) = ( )
arbitrary rank factorization of A
Proof We begin with uniqueness. Assume indirectly that there are two
+ + A different pseudoinverses, A1 and 2. By applying Lemma 3 and Theorem
2019. majus´ 4. –23:07 SHORT PROOFS FOR THE PSEUDOINVERSE IN LINEAR ALGEBRA 119
+ + + +
A A A AA AA 1 on uniqueness, we have A1 = 2 and 1 = 2 and that leads to contradiction:
+ + + + + + + +
A A A A AA A AA A A1 =( 1 ) 1 = 2( 1)= 2 2 = 2 The unique pseudoinverse will be given constructively. Observe that
+ + T 1 T
L AA LL L L L L range( A) = range( ) hence = = ( ) is the unique
+ T 1 T
L L L symmetric projection to there and L =( ) follows by Lemma 1.
T T U Similarly, range( A ) = range( ) holds so that the symmetric projection to
+ T + T T + T + T T 1
A A A U U U U U UU U there is A = ( ) = ( ) = = ( ) from
+ T T 1
U UU where one gets U = ( ) .
+ + + +
L L I UU I L U r We have = r and = that is, is a left inverse and is a right inverse. With these
+ + + + + + + +
LL LU U L A A U U U L LU AA = = and = =
+ + +
U L
from where one concludes that A = .
Remarks A L A U I
If has full column rank, then = and = n is an
+ T 1 T
A A A QR appropriate choice, and A =( ) follows. With the factorization
+ 1 T
A QR A R Q A L I of = , one gets = .If has full row rank then = m
+ T T 1 T
A A A AA A QR and U = suffice, and then = ( ) . If now = then
+ T 1
Q R A
A = ( ) . Finally, if is rank deficient, then as a simple method, we
Q BQ Q Q B take A = 1 2 where 1, 2 are orthogonal matrices and is an upper
+ T 1 T
Q B Q bidiagonal matrix. In that case A = 2 ( ) 1 . Sometimes numerical rank
determination is a delicate process, for details, see [4].
Let P be a projection onto A Then the linear system
Theorem range( )
b is consistent i Pb b
Ax = =
Proof A Pb b Necessity. If the system is solvable then b range( )and =
T
UV P
should hold. For sufficiency assume P = is a rank factorization of .
U P A
Then range( A)=range( ) because is mapping onto range( ). Hence there
U AZ exist a matrix Z such that = .Then
T T
Pb UV b AZ V b b = = =
T
ZV b such that x = is a solution.
T
P UV Remark With the rank factorization of = , it is possible to give
T
A PA UV A U an explicit formula for Z .From = = it is seen that and
T T T
V A A B B = gives a rank factorization of ,i.e. is a full row rank matrix.
2019. majus´ 4. –23:07 120 CSABA J. HEGEDUS˝
T T T
B C C B Now choose a matrix C such that is invertible (for example, =
suffices). Then
T 1 T T 1
B C UB C B C U AC ( ) = ( ) =
T 1
C B C from where Z = ( ) . The pseudoinverse helps us to decide if a linear system is solvable. The
+ + +
AA b x A b
consistency condition b = yields the solution = immediately.
Assume the linear system Ax b is consistent Then the
Theorem = general solution is given by
+ n
x I A A t t x p g = +( ) R
+
where x is a particular solution and I A A t is the general solution of the
p ( )
Ax
homogeneous system =0
x x A x x Proof Assume 1, 2 are two solutions. Then ( 2 1) = 0 shows that the difference of two solutions is a solution of the homogeneous system
+ x and those solutions are in null( A). The pseudosolution may serve as a particular solution.
If the linear system is inconsistent, then we can make it consistent by A orthogonally projecting b onto range( ):
+ +
Ax Ax AA b AA = =
+ +
A b
The pseudosolution again is x = .
The pseudosolution has the following properties kb Ax k Theorem 2
+ +
x x kx k is minimal among the possible solutions is minimal for = 2
+ +
k k b AA b k k b Ax
Proof 2 = 2 is nothing else than the distance of A vector b from range( ) by Theorem 3. The general solution for both cases (consistent or inconsistent systems) is expressible by
+ + + + +
x x I A A t A AA b I A A t
g = +( ) = +( )
2 2
+
x k A b k so that it is the sum of two orthogonal vectors. Hence g 2 = 2 +
+ 2
I A A t k t + k ( ) 2 which is minimal if =0.
Remarks Demanding either the first or the second Penrose condition is
g g g
A A A
enough for AA or to be a projection, where denotes a generalized
g b inverse. If the first and third Penrose equation is fulfilled then A is a least squares solution. In case of a consistent system the fulfilment of the second and fourth condition is enough to get a minimum norm solution.
2019. majus´ 4. –23:07
SHORT PROOFS FOR THE PSEUDOINVERSE IN LINEAR ALGEBRA 121 Gisbert Stoyan Acknowledgements The author thanks Dr. for his
careful reading of the manuscript and his critical comments. Also, thanks are
el Galantai Lajos Laszlo due to Dr. Aur and Dr. for their remarks.
References
Israel A Greville T N E Generalized Inverses Theory and [1] Ben and
Applications John Wiley & Sons, New York, 1974.
Meyer Jr C D Generalized Inverses of Linear Trans [2] Campbell S L and
formations Pitman, London, 1979.
antai A Jeney A Numerical Methods Univ Text [3] Gal and (in Hungarian),
Miskolc University Press, Miskolc, 1998.
nd
G H Van Loan C Matrix Computations [4] Golub and 2 edition. John
Hopkins Univ. P. Baltimore and London, 1989.
oricz F Numerical Methods in Algebra and Analysis [5] M (in Hungarian), Poly-
gon, Szeged, 1997.
Mitra S K Generalized Inverse of Matrices and its Applica [6] Rao C R and
tions Wiley and Sons, Inc. New York, 1971.
nd
G Tako G Numerical Methods I Ed [7] Stoyan and (in Hungarian), Sec. 2.3 and Sec. 2.6.3., Typotex, Budapest, 2002.
2019. majus´ 4. –23:07
ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 123–128
INFINITELY MANY BI-IDEALS WHICH ARE NOT QUASI-IDEALS IN SOME TRANSFORMATION SEMIGROUPS
By YUPAPORN KEMPRASIT
Department of Mathematics, Faculty of Science, Chulalongkorn Univ., Bangkok, Thailand Received December
1. Introduction and Preliminaries
S S
A subsemigroup Q of a semigroup is called a quasi-ideal of if
QS Q bi ideal S B S
SQ , and by a of we mean a subsemigroup of
B such that BS B . Quasi-ideals are a generalization of left ideals and right ideals and bi-ideals are a generalization of quasi-ideals. The notion of
quasi-ideal was first introduced by O Steinfeld in [7]. In fact, the notion of
bi-ideal was given earlier. This can be seen in [3] and [2], page 84.
A S A A q
For a nonempty subset of a semigroup ,( ) and ( ) b denote the
S A A
quasi-ideal and the bi-ideal of generated by , respectively, that is, ( ) q
S A A
is the intersection of all quasi-ideals of containing and ( ) b is the A
intersection of all bi-ideals of S containing ([8], page 10 and 12).
For any nonempty subset A of S Proposition ([2], page 84–85)
1 1 1
A S A AS and A AS A A q
( ) = ( ) b =
Let BQ denote the class of all semigroups whose sets of bi-ideals and quasi-ideals coincide. It is known that the following semigroups belong
to BQ: regular semigroups ([6]), left[right] simple semigroups ([4]) and
left[right] 0-simple semigroups ([4]). Not only these semigroups are in BQ.
A nontrivial zero semigroup is an obvious example. In fact, J Calais [1]
S BQ
has characterized the semigroups in BQ as follows: A semigroup is in
x y x y x y S q
if and only if ( ) =( ) b for all , . It is not easy to see from
this characterization whether a given semigroup belongs to BQ. Since every
S x x q
quasi-ideal of a semi-group is a bi-ideal, it follows that ( ) b ( ) for
S every x . Therefore we have
2019. majus´ 4. –23:10
124 YUPAPORN KEMPRASIT
Proposition For an element x of a semigroup S if x is a quasi
( ) b
ideal of S then x x q
( ) b =( )
Let X be a nonempty set. It is well-known that the partial transformation X semigroup on X , the full transformation semigroup on and the one-to-one
partial transformation semigroup on X (the symmetric inverse semigroup on
X BQ T
) are all regular, so they all belong to ([6]). Let X denote the full
transformation semigroup on X . The author has shown in [5] that transfor-
fa T j X n X g X
mation semigroup X is infinite where is infinite, is not
regular and neither left simple nor right simple but always belongs to BQ.
G M E X
X X
Let X , and denote respectively the symmetric group on ,the
semigroup of all one-to-one transformations of X and the semigroup of all
X G BQ a T X
onto transformations of .Then X .For , is said to be
1
x X x fx g C x X
one to one at if ( ) = and let ( )bethesetofall
x T
such that is not one-to-one at . A transformation X is said to be
almost one to one C T
if ( ) is finite. Hence if X is almost one-to-one,
1
X x almost onto transformation
then for every x ,( ) is finite. By an
X T X n X AM AE
X X
of we mean X whose is finite. Let and denote
the set of all almost one-to-one transformations of X and the set of all almost
X G M AM
X X
onto transformations of , respectively. Note that X
G E AE X M E G
X X X X X
and X ,andif is finite, then = = and
AM AE T
X X
X = = . Also, we have
Proposition AM and AE are subsemigroups of T
X X X
1
Proof AM x X n C C x
Let X and ( ( ) ( ) ). Then
1 1
n C x X n C x fx g x
X ( )and ( ). Thus we have ( ) = and ( ) =
1 1 1
x g x x fx g x = f , and hence ( )( ) =( ) = . Therefore
1
X n C X n C C X n C
( ). This proves that ( ( ) ( ) ) ( ), so
1 1
C C C C X C ( ) ( ) ( ) = ( ) ( ( ) ) .Since and are
1
C C X AM
almost one-to-one, ( ) ( ( ) ) is finite. Thus X .
T
Since for X ,
n X X n X X n X X n X X n X
X =( ) ( ) ( ) ( )
AE T X
it follows that X is a subsemigroup of .
X M E AM AE
X X X
As was mentioned above, if is finite, then X , , and X
belong to BQ. The aim of this paper is to show that if is infinite, then
M E AM AE k
X X X
all X , , and contain at least bi-ideals which are not
jX j quasi-ideals where k = .
2019. majus´ 4. –23:10 INFINITELY MANY BI-IDEALS 125
2. Main Results
M AM X
We start by proving the following result for X and .
Theorem If X is an in nite set then the cardinality of the set of all
bi ideals in M which are not quasi ideals and the cardinality of the set of
X
all bi ideals in AM which are not quasi ideals are at least jX j
X
S M AM X Proof
X X
Assume that X is or .Since is infinite, then
X N j jX j N
j = where denotes the set of all positive integers. Therefore
N X
there is a bijection from X onto and for clarity in what follows, we
t n t X n N t X
write its images as s ( )for and . For each ,let
A s t N
t = ( )
fA j t X g X t X
and note that t forms a partition of . For each ,define
X X
t : by
s n x s t n n N
( t 2 ) if = ( )forsome , x
t =
xotherwise
M t
and note that X and
X X nfs t n j n g
t = ( ) is odd
X X t t X B
t t t
Clearly, for distinct , .Nowlet =( ) b , the bi-ideal of
t
S B S f g
t t t t t X
X generated by . Then by Proposition 1.1, = and note
X X B B B X X
t t t
that t for all . Thus if = ,then = and hence
t t
t t B B t t X jfB j t X gj jX j
t
= . Therefore t , for distinct , . Thus = .
t
B S t
We assert that no is a quasi-ideal of X . To show this by Proposition 1.2,
B t X X X
t t
that is, to show that ( ) q ,fix and define : by
s x s t n n N ( t n +1) if = ( )forsome ,
x =
x otherwise,
t n x s t n n N s ( +2) if = ( )forsome ,
x =
x otherwise.
M
Then X and
s t n s t n s t n y n N t
( ) t = ( 2 +2)= ( ) for all ,
x x x x X n A
t t
t = = for all .
S S B
q t t t t t t t t X
Therefore = X =( ) .If ,then
S
t t t t t
= for some X and hence = since is one-to-one.
2019. majus´ 4. –23:10
126 YUPAPORN KEMPRASIT
C S M C S AM
X X X
Thus ( )= if X = and ( ) is finite if = .Bythe
definitions of t and ,wehave
X nfs t g X X X nfs t n j n g
(2 1 1) ( 1) = = t =( ( ) is odd )
1 1
js t j S M s t S AM
X X X
Since ( 1) 1if X = and ( 1) is finite if = ,it
1
s t n j n gns t follows that f ( ) is odd ( 1) is an infinite set. From (2.1.1), we have
1
s t n j n gn s t X nfs t n j n g ( f ( ) is odd ( 1) ) ( ( ) is odd )
1
s t n j n gns t C
Consequently, f ( ) is odd ( 1) ( ) which is a contradic-
B fB j t X g S
t t t
tion. Hence .Thatis, is a family of bi-ideals in X as
required by the theorem.
M G AM T X
X X X
From Theorem 2.1 and the fact that X = and = if is
finite, the following corollary is obtained
Corollary Let S be M or AM Then S BQ if and only
X X X X
if X is nite
Theorem If X is an in nite set then the cardinality of the set of all
bi ideals in E which are not quasi ideals and the cardinality of the set of all
X
bi ideals in AE which are not quasi ideals are at least jX j
X
Proof S E AE t X n N A
t
X X
Assume that X be or .For and ,let
t n t X
and s ( ) be defined as in the proof of Theorem 2.1. Next, for ,define
X X
t : by
n
s t x s t n n N
2 if = ( ) for some even , x
t =
t x s t n n N
s ( 1) if = ( ) for some odd ,
x otherwise.
E t X t t X s
t t
Then X for every and for distinct , (since is
t
B S
t t t X
one-to-one). Now, let =( ) b , the bi-ideal of generated by .Then
B S f g B B
t t t t t
= X by Proposition 1.1. Observe that for distinct
t
t t X S x x
t t
. For, if not, then = for some X .Since = for
t t
x A
all t ,wehavethat
s t fs t n j n g
t t
( ( 1) ) t =( ( ) is odd ) =
fs t n j n g fs t n j n g
=( ( ) is odd ) = ( ) is odd t
2019. majus´ 4. –23:10
INFINITELY MANY BI-IDEALS 127
jfB j t X gj t
which is impossible because and t are functions. Hence =
jX j B S t
= . We assert that no is a quasi-ideal of X . To see this by Proposition
B t X X X
t t
1.2, that is, ( ) q ,fix and define , : by
s t n x s t n n N nf g
( 2) if = ( )and 1 2 ,
x
t x s t
= s ( 1) if = ( 2),
x otherwise,
t n x s t n n N nf g s ( 1) if = ( )and 1 ,
x =
x otherwise.
E
Then , X and
x x x x X n A
t t
t = = for all ,
s t n s t s t n n f g t
( ) t = ( 1) = ( ) for 1 2 3 ,
n 2
s t n s t s t n n N nf g t ( ) t = = ( ) for even 1 2 3 ,
2
s t n s t s t n n N nf g t
( ) t = ( 1) = ( ) for odd 1 2 3 ,
S S
t t t t t t t q X
Consequently, = and hence X =( ) .If
B S
t t t t t t
,then = for some X , and hence = since is
X n A
onto. It follows that must fix t pointwise. In addition, since
A nfs t s t g A nfs t s t g A nfs t g
t t t
( t ( 1) ( 2) ) =( ( 1) ( 2) ) = ( 1)
it follows from the definition of t that
A nfs t s t g fs t n j n n g
( t ( 1) ( 2) ) = ( ) is even and 2
Therefore we have
X nfs t s t g X n A fs t n j n n g
(2 3 1) ( ( 1) ( 2) ) =( t ) ( ) is even and 2
S X n X S E X n X
X X
Since X ,wehavethat = if = and is finite if
S AE X
X = . But we obtain from (2.3.1) that
n X fs n t j n g fs t g nfs t s t g X =( ( ) is odd (2 ) ) ( 1) ( 2)
which is an infinite set since is a function, so we have a contradiction.
B fB j t X g S
t t t
Hence .Thatis, is a family of bi-ideals in X as
required by the theorem.
E G AE T X
X X X
From Theorem 2.3 and the fact that X = and = if is
finite, we have
Corollary Let S be E or AE Then S BQ if and only if
X X X X
X is nite
2019. majus´ 4. –23:10 128 YUPAPORN KEMPRASIT
References
C R Acad Paris [1] J Calais Demi-groups quasi-inversif, 252 (1961), 1357–
2359.
G B Preston The Algebraic Theory of Semigroups [2] A H Clifford and Vol.
I., Amer. Math. Soc., (Providence, 1961).
D R Hughes Bull Amer [3] R A Good and Associated groups for a semigroup,
Math Soc 58 (1952), 624–625 (Abstract). Publ Math Debre [4] K M Kapp On bi-ideals and quasi-ideals in semigroups,
cen 16 (1969), 179–185.
[5] Y Kemprasit Some transformation semigroups whose sets of bi-ideals and
quasi-ideals coincide, Comm Algebra to appear. Acta Sci Math [6] S Lajos Generalized ideals in semigroups, 22 (1961), 217–
222. Publ Math Debrecen [7] O Steinfeld Uber¨ die quasiideale von halbgruppen, 4 1956), 262–275.
[8] O Steinfeld Quasi-ideals in Rings and Semigroups, Akademiai´ KiadoBu-´ dapest, (1978).
2019. majus´ 4. –23:10 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 129–150
L-COMMUTATIVITY OF THE OPERATORS IN SPLITTING METHODS FOR AIR POLLUTION MODELS
By IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
Central Laboratory of Parallel processing, Bulgarian Academy of Sciences, Sofia, Bulgaria Eotv¨ os¨ Lorand´ University, Budapest Eotv¨ os¨ Lorand´ University, Budapest National Environmental Research Institute, Department of Atmospheric Environment,
Roskilde, Denmark Received March
1. Introduction
Splitting procedures can lead to a substantial reduction of the compu- tational work when large-scale problems are to be treated. Therefore, such procedures are often used in the numerical solution of many boundary value problems for differential equations describing real-life processes, see [3, 9, 5, 11, 16, 21, 13, 18]. A detailed theoretical study and analysis of the splitting procedure can be found in [8, 10]. An important example of real-life modelling is the problem of large-scale air pollution transport. Mathematical models of this kind are usually presented as a system of three-dimensional time-dependent partial differential equations which describe the processes of advection, diffusion, deposition, pollutant emission sources and chemical reactions. The environmental problems are becoming more and more important for the modern society, and their impor- tance will certainly increase in the near future. High pollution levels (high concentrations and/or depositions of certain chemical species) may cause damages to plants, animals and humans. Moreover, some ecosystems can also be damaged (or even destroyed) when the pollution levels are very high. This is why the pollution levels must be carefully studied and controlled in the efforts to make it possible (i) to predict the appearance of high pollution levels and/or (ii) to decide what can be done to prevent the exceedance of prescribed critical levels. The control of the pollution levels in different highly developed and densely populated regions of Europe and North America is an important task for the modern society. Its importance has been steadily increasing during the last two decades. The necessity to establish reliable control strategies for air pollution levels will become even more important in
2019. majus´ 4. –23:07 130 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV the next two-three decades. Large scale air pollution models can be used to design reliable control strategies. The numerical treatment of such mathematical models includes operator or time splitting. This procedure has several advantages: 1. the obtained sub-systems are easier to treat numerically than the original system; 2. we can exploit the special properties of the different sub-systems and apply the most suitable numerical method for each; 3. if each numerical method preserves the main qualitative properties then so does the global model. It is known that splitting procedures work well in the computer treatment of many air pollution models, [22, 12, 23]. At the same time, little attention has been devoted to the analysis of splitting procedures used in practice and to the question why splitting usually leads to good results. As it was mentioned above, splitting procedures are used in order to facilitate the choice of efficient numerical methods in the treatment of the different operators involved in the model under consideration. Assume that the selected methods are not only efficient, but also sufficiently accurate. Then the success of splitting is determined by the splitting error. The recent paper [5] presents an analysis of operator splitting in air pollution models. By using the Lie operator formalism, a general expression is derived for a three-term splitting procedure in the pure initial value case. The procedure is called Strang splitting procedure in [5], however it has been introduced independently in 1968 by [6] and [16]. Therefore, it is reasonable to call it Marchuk–Strang splitting procedure. The splitting error for the advection– diffusion–reaction problem is analysed in the above mentioned reference [5]. Different conditions for reducing the errors, which are caused by the splitting procedures, are discussed there. Sufficient conditions, for which the splitting errors vanish, are also derived. These conditions are too strong and, thus, rather unrealistic when large models are used to treat practical problems. For example, it is obtained that if the velocity field u and diffusion matrix k are
independent of the space coordinates x, then there is no splitting error when di usion advection and are splitted. The recent work [20] presents numerical
methods, which are proposed for several splitted problems. errors The splitting in the numerical treatment of the splitted problem
are closely related to the requirement for L-commutativity of the operators involved in the splitting procedure. More precisely, the errors due to the ap-
plication of splitting procedures disappear when the corresponding operators L L-commute. This is why the -commutativity of operators will be a major tool in the derivation of the results. The goals of this work are:
2019. majus´ 4. –23:07
L-COMMUTATIVITY OF THE OPERATORS 131 L to analyse the -commutativity of operators used in the mathematical model for studying large-scale air pollution transport,
to formulate conditions under which the splitting errors vanish,
to investigate the splitting errors of two widely used splitting procedures. The paper is organized as follows. In Section 2 the definitions of the
commutator operator and the L-commutativity are given. In Section 3 we
define the operators associated with processes of advection di usion depo
chemistry emission sition and The necessary and sufficient conditions for
the L-commutativity of the operators are studied in Section 4. In Section 5 we introduce two different splitting procedures: the splitting procedure based on the separation of the physical processes involved in the Danish Eulerian Model (it will be called the DEM splitting in this paper, but this is done only in order to facilitate the references to it and, at the same time, to keep in mind that this procedure is used in a particular air pollution model) and the Marchuk–Strang splitting. It is also shown how the Lie operator formalism can be used to analyse the structure of the splitting error. Some concluding remarks are given in Section 6.
2. Background definitions
Throughout this paper we use the following notations. Let S denote
4 m
R some normed space of functions of type R with the variables x =
3 +
x x R t R t
=(x1 2 3) and 0 . Clearly, any element f(x ) S can be
t l m f x T
identified with the set of functions l ( ) , =12 ,wherethe
4
R notation T stands for the set of mappings of type R . The notation Slin will be used for the linear functions in S.
Assume that A : S S is a given operator. Such an operator can be identified with m operators of type S T, called components of the operator
A. We always assume that the operators A : S S are differentiable in
Frechet sense [14] and the derivative operator is denoted by A . In the sequel
t t t k t k t k t t g S, 1(x ), 2(x ), , m (x ) T, 1(x ), 2(x ), 3(x ) T, k(x )
4 3 3
R
is a mapping of type R and has the form of a diagonal matrix
k t k t k t
k(x t)=diag( 1(x ) 2(x ) 3(x ))
t u t u t
The functions u1(x ), 2(x ), 3(x ) T define a vector field
u t u t u t u(x t)=( 1(x ) 2(x ) 3(x ))
4 3
R of type R .
2019. majus´ 4. –23:07
132 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
m
R R l m p x p
Let . The functions l ( ), =1 2 are mappings of type
m +3
R
R , therefore the mapping
R R R
R(x p)=( 1(x p) 2(x p) m (x p)) m
m +3
R
is a mapping of type R .
t f f x T
As usual, for the scalar valued function ( ) the notation i means r the partial derivative w.r.t. the i -th component. The differential operator
will be applied also in the usual way. It acts always with respect the space
x x f t
variables x1, 2 and 3. That is, for a scalar valued function (x ) T the f
symbol r means the gradient operator w.r.t. x in the sense
f t f t f t f t
r (x )=( 1 (x ) 2 (x ) 3 (x ))
f t f t f t
For a three-dimensional vector field f(x t)=(1(x ) 2(x ) 3(x )) the sym- f
bol r yields the divergence operator that is
t f t f t f t
r f(x )= 1 1(x )+ 2 2(x )+ 3 3(x ) r
We remark that for an elements f S the operator acts componentwise,
r r f l m
rf S f l
that is and ( ) l = ( ), =12 . The same notation is used
2 r
for the Laplace operator Δ = r . The use of the operator to the function
f p p p of type (x 1(x) 2(x) m (x)) may lead to some misunderstanding. To
avoid this, we introduce the operator rx as
r x x x p p p f f f
(1) x( 1 2 3 1(x) 2(x) m (x)) = ( 1 2 3 )
rf that is it acts w.r.t. the first three variables of the function f , while means
the gradient vector of the composite function
h f p p p (x)= (x 1(x) 2(x) (x) m (x))
that is
f f f
rf x x x p p p
(2) ( 1 2 3 1(x) 2(x) m (x)) =
x x x1 2 3 We would like to emphasize that the right-hand sides of expressions (1) and
(2) are usually different because
m
X
f f p
k
f i
= i + =1 2 3
x p x
i k i
k =1 3
The multiplication of two elements of the space R means the standard scalar
rf rh f h f h f h product. E.g. for f h T the notation yields 1 1 + 2 2 + 3 3 which will be applied in the sequel.
2019. majus´ 4. –23:07 L-COMMUTATIVITY OF THE OPERATORS 133
The following properties of the r operator can be easily verified:
f
For a scalar function T and a vector field g the relation
f rf f r (3) r ( g)=( )g + g holds.
Due to (3) we have
f rf f r (4) r ( (Mg)) = ( )(Mg)+ (Mg)
where M is any matrix.
f p p p For a scalar function (x 1(x) 2(x) m (x)) and a vector function
g the following relation holds:
m
X
f f r r f f rp
r g g g g j (5) ( )= + x + ( j +3 )( )
j =1
R R R
For the function R(x p)=( 1(x p) 2(x p) m (x p)) we introduce
x x
two Jacobi matrices: the first is defined w.r.t. the variables x 1, 2, 3 and
p p p
denoted by Rx(x p), the second one w.r.t. the variables 1 2 m ,and
m 3 R
denoted by Rx(x p). Consequently, they are matrices of type and
m m
R , respectively, and are defined by the formulas
R i m j
x p x p j i (Rx( ))i j = ( ) =1 2 and =1 2 3
and
R i j m
R x p x p i j i (6) ( p( )) = 3+j ( ) =1 2 Here and further we assume the required smoothness of the functions in the definitions.
3. Operators used in air pollution models and their L-commutativity
i A S S
First we define the operators i : ,( =12 3 4 5), appearing in
air pollution models as follows:
r c l m
A c u c S l ( 1( ))l := ( ), =1 2 , , which is associated with the
process advection
rc l m
A c r k c S l ( 2( ))l := ( ), =1 2 , , which is associated with the
process di usion
c l m
A c c S l l ( 3( ))l := , =12 , , which is associated with the
process deposition
2019. majus´ 4. –23:07
134 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
g l m
A c c S l
( 4( ))l := , =1 2 , , which is associated with the process
emission
R l m
A c x c c S l ( 5( ))j := ( ), =12 , , which is associated with the
process chemistry commutativity In the following the L of two differentiable operators plays
a central role.
Assume that A B : S S are differentiable on S. We define the operator E : S S as follows:
A B
(7) E (s):=(B (s) A)(s) (A (s) B)(s) s S
A B Definition The operator E is called the commutator of the op-
A B erators A and B. The element E (s) S is called the commutator error of A B
the operators A and B for the element s S.
E
Obviously, E = .Let denote the subspace of those
A B B A A B
elements in S for which the commutator error turns into zero, that is =
A B
E g = fs S : (s)=0 .
A B
L
Definition We say that the operators A and B -commute on 0 if
L
= .If = S, then we say that the operators A and B -commute, 0 A B A B
that is the operators A and B L-commute if the relation
(8) E (s)=0 s S A B
holds.
Remark If A and B are linear operators then A (s)=A and B (s)=B
for any s S. In this case (8) turns into the formula B A = A B, hence the
L-commutativity is equivalent to the usual commutativity.
Our goal is to analyse the L-commutativity of any pairs of the operators
A i
i , =1 2 5. To this aim we compute their derivatives.
A i
The operators i , =1 2 3 are linear. Therefore, the following relations
hold for their derivatives:
i A c A c S
(9) ( )= i =1 2 3 i
Furthermore, the following relation follows from the fact that the operator A4
is constant:
(10) A4(c)=0 c S The derivative of A5 at the point c S is the Jacobi matrix Rc(x c)andit
acts as follows:
(11) A5(c)(c)=Rc(x c)c c S
2019. majus´ 4. –23:07 L-COMMUTATIVITY OF THE OPERATORS 135
It is necessary to emphasize here the following fact: due to the special struc-
tures of the first four operators (the l -th component of these operators depend
c L
only on l ), it is sufficient to study componentwise the -commutativity
A A i j j
properties of the pairs ( i ), where =12 3 4. This observation is
used to facilitate the proofs of some of the following theorems. L Some su cient conditions for the -commutativity of some of the oper- ators defined above can be found in the literature. For instance,
1. if u and k are independent of x, then the operators A1 and A2;
2. if r u =0andR is independent of x, then the operators A1 and A5; 3. if R is independent of x and linear in c, then the operators A2 and A5
L-commute [5].
However, the necessity of these strong and unrealistic conditions is not clear. For example, the condition 1 (especially, the independence of u of x) is very unrealistic because the velocity field u can strongly depend on both
x and t . Therefore, it is worthwhile to examine the possibility to relax these assumptions by replacing them by some weaker, more realistic conditions.
E.g. the condition r u = 0 is much more realistic because it describes the continuity principle in the lower layers of the atmosphere. We shall analyse the commutativity in the next section under this natural condition, too. We
shall also give some exact (necessary and sufficient) conditions for the L-
A A i j j
commutativity of the operators i and ( =1 5).
4. Condition of L-commutativity of the operators in air pollution models
In this section, we shall derive conditions for the L-commutativity of
A A i j j
different pairs of the operators i , , =12 3 4 5. For the sake of
E E i j
brevity, we shall use the notations i j := and := .
A Aj A A
i i j
4.1. L-commutativity of the advection and diffusion operators
As was stated in the previous section, it is sufficient to treat these op- erators componentwise. This means that if an arbitrary component of the
advection operator L-commutes with the corresponding component of the
diffusion operator, then the operators will be L-commutative. By use of this
fact, we obtain that the commutator operator reads
r r r c r r rc
E c k u u k l l l (12) ( 1 2( )) := [ ( ( ))] + [ ( ( ))]
2019. majus´ 4. –23:07
136 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
m for all l =1 2 . In the following we examine the commutator under the condition
(13) k is independent of x and r u =0.
Then, by using the properties of the r operator, after straightforward, but tedious calculations we obtain that (12) yields the relation
3 3 3 3
X X X
X 2
E k c u k c u
s r s r s r r
c s l l l (14) ( ( )) = 2 ( )( ) ( )( )
1 2 s
r s r
s =1 =1 =1 =1
l m
E c l
The equations ( 1 2( )) =0, =12 define a system of second order S PDE’s and the set of its solution is 1 2 . As one can easily check in case u is independent of x, the velocity field
S is divergencefree (continuity assumption) and the relation 1 2 = (that is
the L-commutativity of the operators A1 and A2) holds. On the other hand, if one of the following conditions are satisfied:
u is linear,
k k k u u u 1 = 2 = 3 = const. and the functions 1, 2 and 3 are harmonic
functions w.r.t. x [4], i.e., Δu =0,
L S A A then lin 1 2 that is the operators 1 and 2 -commute on the linear elements. (The latter choices can be interpreted as an approximation to the general case.)
4.2. L-commutativity of the advection and deposition operators
Due to property (3), in a similar way as in Subsection 4.1, we obtain
r c r c c r
E c u u u l l l l l l l (15) ( 1 3( )) = [ ( )] + [ ( )] = ( )
By using the relation (15) we get: the operators A1 and A3 are L-commuting if and only if the gradient of each deposition function is orthogonal to the
velocity field, that is the condition r u
(16) ( l ) =0
m holds for l =1 2 .
2019. majus´ 4. –23:07 L-COMMUTATIVITY OF THE OPERATORS 137
4.3. L-commutativity of the advection and emission operators
By using the formula (3), we obtain
r g rg g r
E c u u u l l l l (17) ( 1 4( )) = ( )=( ) +
which implies the following: The operators A1 and A4 are L-commuting if
and only if the condition g r u
(18) ( l )=0
m r holds for l =12 . If the continuity condition u = 0 is assumed, then the commutativity holds if and only if the gradients of each emission
functions are orthogonal to the velocity field, that is the condition g r u
(19) ( l ) =0
m holds for l =1 2 .
4.4. L-commutativity of the advection and chemistry operators
For the commutator of the advection and chemistry operators we can
write:
m
X
E c r R
c R x c r u x c u j l l l j (20) ( 1 5( )) = ( c( )) ( )+ ( ( ) )
j =1
Using (3) and the notation (6), we obtain
m m
X X R x c
l ( )
c rc c r
R x c r u u u j j j
(21) ( c( ))l j ( )= (( ) + )
c
j j j =1 =1
Further, applying the formula (5) we get
m
X R x c
l ( )
R R r rc r R
r x c u x c u u u x c l j l
(22) ( l ( ) )= ( ) + ( ) + x ( )
c j
j =1
Combining (21) and (22) with (20), for the l -th component of the commutator
we obtain
m
X R x c
l ( )
E R r r R r
c c u x c u u x c l l l j
(23) ( 1 5( )) = + ( ) + x ( )
c j
j =1
2019. majus´ 4. –23:07 138 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
Consequently, the operators A1 and A5 L-commute if the relations
r R r u u x c
(24) =0 and x l ( )=0
m
hold for all l =12 and c S. Obviously, under the continuity
r u c S A assumption = 0, a fixed element belongs to 1 5 (that is 1 and
A5 are L-commuting on this element) if and only if the conditions
3
X
t R t l m
u x x c x i l
(25) i ( ) ( ( )) = 0 =1 2
i =1 R
are satisfied. Therefore, in case of explicit independence of the functions l
R
x R x c c l
of the variable , that is in the case l ( )= ( ), the conditions in (25) are fulfilled for all c S, so, under these assumptions the operators A1 and
A5 are L-commuting.
4.5. L-commutativity of the diffusion and deposition operators
By use of the relation (4) we obtain
r rc r r c
E c k k l l l l l
( 2 3( )) = [ ( )] [ ( )] =
r rc c r r rc r
k k k l l l l l
(26) = ( l )( ) [ ( )] ( )
This means that the operators A2 and A3 L-commute if the condition
r
(27) l =0
l m
is satisfied for all =12 . Therefore in case l = const the operators
A2 and A3 L-commute on any element of S.
4.6. L-commutativity of the diffusion and emission operators
For this case we get the relation
r rg
E c k l l (28) ( 2 4( )) = [ ]
which means the following: The operators A2 and A4 are L-commuting if
and only if the condition
g g g
l l l
k k
(29) k1 + 2 + 3 =0
x x x x x
x1 1 2 2 3 3
l m rg l m
is satisfied for all =12 . Clearly, if l =0forall =12 ,
then the operators A2 and A4 are L-commuting.
2019. majus´ 4. –23:07 L-COMMUTATIVITY OF THE OPERATORS 139
On the base of (29) we can formulate an important for the practice corol-
lary, which gives the necessary and sufficient condition of L-commutativity
t k t k t
of the diffusion and emission operators: if k1(x )= 2(x )= 3(x ) = const r then the operators A2 and A4 are L-commuting if and only if g = 0 that is
the components of the given g S are harmonic functions.
4.7. L-commutativity of the diffusion and chemistry operators
For the commutator operator we have
m
X
R r rc r r R
E c x c k k x c l l j l j (30) ( 2 5( )) = +3 ( ) ( ) ( x ( ))
j =1
A cumbersome calculation gives the following result:
m
X
E r r R r R rc
c k x c x x c k l l l j j ( 2 5( )) = ( x ( )) ( +3 ( ))( )
j =1
m m
X X
R rc rc
x c r k l j j
(31) (( r +3 +3 ( )) )( ) j
=1 r =1
R
By use of (31) clearly we have: under the conditions 1 1(x c)) =
R R R r j l
x c x c x c l l l j
= 2 ( )) = 3 ( )) = 0 and r +3 +3 ( )=0forall =
m L
=1 2 the operators A2 and A5 -commute. Consequently, in case
L R(x c)=R(c)andR(c) Slin the operators -commute.
4.8. L-commutativity of the deposition and emission operators
m
Obviously, the following relationships are valid for all l =1 2 :
g
E c l l l (32) ( 3 4( )) =
Consequently, the operators A3 and A4 are not L-commuting in any realistic case.
2019. majus´ 4. –23:07 140 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
4.9. L-commutativity of the deposition and chemistry operators
Clearly, by definition
m
X
R c R
E c x c x c l j j l l l j (33) ( 3 5( )) = [ +3 ( )] ( )
j =1
Assume that 1 = 2 = = m = .Thenwehave
m
X
R c R
E c x c x c l l j l j (34) ( 3 5( )) = ( +3 ( )) ( )
j =1
Obviously, the splitting error turns into zero for all c S if and only if the
relation
m
X R x c
l ( )
R
c x c l
(35) j = ( )
c j
j =1
m R R
l x c c l
is satisfied for all =12 . Let us examine the case l ( )= ( ).
Then for all fixed l , (35) yields a partial differential equation of first order.
This equation has the general solution
c c c
1 2 m 1
R c c c c
m m l
(36) l ( 1 2 )=
c c c
m m m
m 1
R R
where l : is any continuously differentiable function for all
m
l =1 . Therefore, we obtained: under the conditions 1 = 2 = =
L
= m = and R(x c)=R(c) the operators A3 and A5 are -commuting if and
R c c l m m only the functions l ( 1 ) have the form (36) for all =1 .
4.10. L-commutativity of the emission and chemistry operators
By definition
m
X
R g
E c x c l l j j (37) ( 4 5( )) = [ +3 ( )]
j =1
Consequently, the operators A4 and A5 L-commute if and only if g lies in the
null space of the Jacobian Rc(x c), that is g ker Rc(x c).
2019. majus´ 4. –23:07 L-COMMUTATIVITY OF THE OPERATORS 141
4.11. Summarizing the L-commutativity of the operators
Here we give a short summary of the results obtained in the previous
sections.
E
The commutator error 3 4 does not vanish in any realistic case.
Under the assumptions
r r r u g R x c R c
(38) =0 l =0 =0 ( )= ( )
E E E E E
the commutator errors 1 3, 1 4, 1 5, 2 3 and 2 4 vanish.
Under the further conditions
(39) k(x) = const u R c are linear, 1 = = m = const.
E E E
the commutator errors 1 2, 2 5 and 3 5 are also zero.
L If in addition g ker JR, then even the operators A4 and A5 -commute. Clearly, these results cover those of [5], however, for certain pairs of
operators they give more general conditions for the L-commutativity than the requirements formulated there. For instance, if u is linear (not necessarily constant), then for concentration functions of a special from (linear functions) the operators A1 and A2 L-commute.
Finally we remark that the assumptions of the linearity can be interpreted
A i
in the following way, too. The operators i , =1 2 5, are defined on the
linear finite elements and the derivation is understood in generalized sense.
A
We define the operators i ,12 5, as the mappings which are obtained after the semidiscretization of the weak form of the original fully continuous PDE’s, in the linear finite element spaces. Then the functions g and u in the definition of the operators can be considered as the projections of the original functions into the linear finite element subspace.
5. Application of the splitting error analysis in air pollution models
In this section we present two examples of air pollution models in which the above results can be applied: the Danish Eulerian Model (DEM) [22] and the advection – diffusion – reaction model as defined in [5]. For the first model a splitting procedure based on a separation of the physical processes is used. This way of implementing a splitting procedure in air pollution modelling was first proposed in [8]. We shall use the abbre-
viation DEM splitting in this paper in order (i) to facilitate the references to
2019. majus´ 4. –23:07 142 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV this splitting and (ii) to emphasize the fact that it has been used in the Danish Eulerian Model. For the second model we shall use a symmetrical splitting proposed simultaneously in [6, 7] and [16]. We refer to this splitting procedure as the Marchuk–Strang splitting procedure in this paper. A very good description of this way of splitting, which is particularly oriented to air pollution models, is given in [12].
5.1. Air pollution models and mathematical formulation of the long-range transport of air pollutants
The air pollution models must satisfy several important requirements [22, 12, 23]: 1. The mathematical models must be defined on large space domains, be- cause the long range transport of air pollution is an important environ- mental phenomenon, and high pollution levels are not limited to the areas where the high emission sources are located. 2. All relevant physical and chemical processes must be adequately de- scribed in the models used. 3. Enormous files of input data (both meteorological data and emission data) are needed. 4. The output files are also very big, and fast visualization tools must be used in order to represent the trends and tendencies, hidden behind many megabytes (or even many gigabytes) of digital information, so that even non-specialists can easily understand them. All important physical and chemical processes must be taken into account when an air pollution model is to be developed. Systems of partial differential equations (PDE’s) are often used to describe mathematically an air pollution
model. Consider a three-dimensional space domain Ω and assume that x
x x x ( 1 2 3) Ω. Then the PDE systems are of the following type:
(40) t c x
l ( )
c t f t t T c c l m
A x t x x x x l l
= ( ) ( )+ ( ) [0 ] ( 0)= l 0( ) =1 t
where
c r c r rc c l m
A x t u k l l l
(41) ( ) l ( )+ ( ) ( 1 + 2) =1
u c u c u c
l l
( 1 l ) ( 2 ) ( 3 )
c l m r u
(42) ( l )= + + =1
x x x1 2 3
2019. majus´ 4. –23:07
L-COMMUTATIVITY OF THE OPERATORS 143
c c c
l l l
r k rc k k k l m
(43) ( l )= 1 + 2 + 3 =1
x x x x x
x1 1 2 2 3 3 t
The vector-function f (x )isdefinedasasum
t g R (44) f (x )= +
where
T
g g g
( 1 m )
T
R R R ( 1 m )
and
R R c c c l m
x m l (45) l = ( 1 2 ) =1 2 The different quantities that are involved in the mathematical model have the
following meaning:
c l
l denotes the concentration of the -th species;
u u u
1, 2 and 3 are velocities;
k k k
1, 2 and 3 are diffusion coefficients;
g
the functions l describe the emission sources in the space domain;
l
1 l and 2 are the dry and wet deposition coefficients, respectively;
R c c c m
the nonlinear functions l ( 1 2 ) describe the chemical reactions.
R l
The functions l , representing the chemical reactions in which the -th
pollutant is involved are of the form
m m m
X X X
R c c c c c c l m
m
lj j lj k j k
l ( 1 2 )= + =12
j k j =1 =1 =1 This is a special kind of nonlinearity (it is seen that the chemical terms are described by quadratic functions), but it is not clear if this property can efficiently be exploited. To the authors’ knowledge, it is not exploited in the existing large scale air pollution models. The models defined by (40)–(45) are traditionally used to calculate some concentration fields by using both meteorological and emission data as input [23]. This gives an answer to the question: what are the concentration lev- els and/or the deposition levels caused by the existing emissions under the particular meteorological conditions that take place in the time-period under consideration? However, it is much more important to study the question: how can the concentrations be kept under certain critical levels?
2019. majus´ 4. –23:07 144 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
5.2. Splitting and its role
It is very difficult to treat directly the system (40). Therefore, some kind Splitting of splitting is to be used. , or problem decomposition, is commonly used during the first step of the numerical treatment of large air pollution models (as in many other large-scale scientific and engineering problems). The big problem, the model described by the system of equations (40), is divided into several smaller problems through some splitting procedure. These smaller problems might have special properties that can be exploited in the numerical solutions. For example, the systems of linear algebraic equations that arise, after splitting, from the diffusion part of the model normally have banded, symmetric, and positive definite matrices. On the other hand, it is not easy to evaluate the error that arises from the splitting techniques used. Splitting according to the major physical processes is very popular; see, for example, [9], [12] and [22]. Such splitting procedures lead often to a num- ber of sub-models which are to be treated cyclicly at every time-step [22]. In the DEM [22] these sub-models are describing the horizontal advection (46), the horizontal diffusion (47), the chemical reactions including the emissions (48), the deposition (49) and the vertical exchange (50), so there are five splitted systems of the form
(1) (1) (1)
u c u c
c ( ) ( )
l l
l 1 2
(46) =
x x
t 1 2
(2) (2) (2)
c c c
l l l k
(47) = k1 + 2
x x x x t 1 1 2 2
(3) c
l (3) (3) (3)
g R c c c
m l
(48) = l + ( 1 2 ) t
(4) c
l (4)
c
l (49) = ( l + )
1 2 l
t
(5) (5) (5)
c u c c
( )
l l
l 3
k
(50) = + 3
x x x t 3 3 3
(j )
j
The values c , =1 5 are connected through the initial conditions, that
l j
(j ) ( +1)
c j
is c is used as an initial condition for , =1 4, and for the next
l l time-step the process continues cyclicly. We shall call the splitting procedure
(46)–(50) DEM splitting procedure
2019. majus´ 4. –23:07 L-COMMUTATIVITY OF THE OPERATORS 145
An alternative of the above splitting procedure can be the already men-
tioned symmetrical Marchuk Strang splitting scheme [6,7,16].Usually,this splitting scheme is applied to a model of air pollution transport, where the deposition and emission parts are included into the reaction part of the prob- lem operator. So, instead of (40) we consider an advection-diffusion-reaction
problem t
c(x )
t t T
(51) = A(c(x )) (0 ] c(x 0) = c0(x) t
where
t A t A t A t
A(c(x )) = 1(c(x )) + 2(c(x )) + 5(c(x )) t anditisassumedthatintoA5( c (x )) the deposition and emissions are in-
cluded. In the Marchuk–Strang splitting the problem (51) is split by ordering
A A
the operators A1, 2 and 5 symmetrically in the following way: i
(1)
t
c (x ) (1) (1)
t t (52) = A1(c (x )) 0 c (x 0) = c0(x)
t 2
i
(2)
t
c (x ) (2) (2) (1)
t t (53) = A2(c (x )) 0 c (x 0) = c x
t 2 2
(3)
t
c (x ) (3) (3) (2)
A t t (54) = 5(c (x )) (0 ] c (x 0) = c x
t 2
i
(4)
t
c (x ) (4) (4) (3)
t t (55) = A2(c (x )) 0 c x = c x
t 2 2 2
i
(5)
t
c (x ) (5) (5) (4)
A t t (56) = 1(c (x )) c x = c x
t 2 2 2 Let us now suppose that one can solve both the original problem and the splitted subproblems exactly. In this case it is possible to express the splitting error with the help of the so-called Lie operator formalism, as will be shown in the next chapter.
5.3. Lie operator formalism and splitting error
In this chapter, following the technique of [15] and [5], we shall derive the local splitting error of the Marchuk-Strang splitting procedure and give the results of a similar error analysis for the DEM splitting. We will see that in terms of the local error the order of the Marchuk–Strang splitting scheme is higher than that of the DEM splitting.
2019. majus´ 4. –23:07 146 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
First we need to introduce the concept of Lie operator, since it plays an
important role in the derivation of the error formula. Let A be a generally non-linear operator of type S S. With this
given operator we associate a new operator, which we will denote by A
and call it the Lie operator associated to A. This operator acts on the space F
of differentiable operators S S and maps each operator into the new
F c
operator A , such that for any element S,
F c F c A c (57) ( A )( )=( ( ) )( ) It is easy to see that the Lie operator is linear.
Let us consider the initial value problem
c
A c t T c c
(58) (x t)= ( (x )) on (0 ] (x 0) = 0(x)
t A
and denote by A the Lie operator associated to the particular operator
of problem (58). Let F be any differentiable operator S S. Applying
F c t the operator A to the solution (x ) of (51) and using the chain-rule of
differentiating, one obtains the relation
F c t F c t
(59) ( A )( (x )) = ( (x )) t
from which by induction follows also that
i
c t A F c t i
(60) F ( (x )) = ( )( (x )) =2 3
i t
Assume that the solution c(x t) of (51) is an analytic function. Then, using
its Taylor series expansion, one can easily show that
A
e I c
(61) c (x )=( )( 0(x)) where I is the identity operator S S. Applying now to each of the subproblems the corresponding exponen- tiated Lie operator and composing them in the order defining the Marchuk–
Strang splitting procedure (52)–(56), for the solutionc ˆ of the splitted problem
at time one can get
1 1 1 1 1
A A A A A
1 2 5 2 1
c e e e e I c ˆ(x )= e 2 2 2 2 2 ( 0(x))
In order to compute the product of exponentials on the right-hand side, we can use the well-known Baker–Campbell–Hausdorff (BCH) theorem [19]. This
2019. majus´ 4. –23:07
L-COMMUTATIVITY OF THE OPERATORS 147
X Y
Y e e
claims that for any pair of linear operators X , the product can Z locally be written as the exponential e of the Lie operator
1 1 1
X Y X Y X X Y Y Y X X Y Y X (62) Z = + + [ ]+ ([ ]+[ ])+ [ ]+
2 12 24
X Y XY YX X X Y where [ X Y ] is the commutator [ ]= and [ ]is
1
X X Y X A recursively defined by [X X Y ]=[ [ ]] etc. Substituting = 2 1 etc. and applying (62) four times, we obtain that the Marchuk–Strang solution
cˆ can be expressed as
ˆ
A
I e c cˆ(x )= ( 0(x)) ˆ where the new Lie operator A has the form
ˆ 2 4
A A A A E O
(63) = 1 + 2 + 5 + A + ( ) with
1 1
E A A A A A A
A = [ ] [ ]+ 24 1 1 2 24 1 1 5
1 1 1
A A A A A A A A (64) + [ A ] [ ]+ [ ]+ 12 2 2 1 24 2 2 5 12 5 5 1
1 1 1
A A A A A A A A + [ A ]+ [ ]+ [ ]
12 5 5 2 12 2 5 1 12 5 2 1
A A A A
Remark If 1 and 2 are Lie operators, then [ 1 2] = 0 is equiv-
E A A A
alent to A =0,where and are the operators belonging to the Lie
1 2 1 2 A operators A1 and 2, respectively.
In order to characterize the error at time that arises if we apply operator
splitting on the interval [0 ], we can use the notion of the local splitting error, defined as the difference between the exact solution of the splitted problem and the exact solution of the original problem [17, 20]. According to the above considerations, for the Marchuk–Strang splitting scheme this local error
can be given as
ˆ
A A
e I e I c x
(65) ErrS ( ):= ( ( )) p 0
Applying now (63) and the definition of the exponential, we obtain that the behaviour of the error function as 0isasfollows:
3 4
E I c o
x A
Err S ( )= ( )( ( )) + ( ) p 0
2019. majus´ 4. –23:07 148 IVAN DIMOV, ISTVAN´ FARAGO,´ AGNES´ HAVASI, ZAHARI ZLATEV
3 i.e. the local splitting error of the Marchuk–Strang scheme is o ( ). Therefore, we say that the Marchuk–Strang splitting is a second order splitting scheme [17]. A similar analysis shows that the local error of the DEM splitting is only
2 o ( ), which means that in the above sense the Marchuk–Strang splitting scheme provides a one order higher approximation to the original problem
than the DEM splitting.
k
We remark that if we apply operator splitting on the interval [ k ( +1) ],
k =1 2 , then (65) becomes
ˆ
A A
I e I e c k x
(66) ErrS ( ):= (ˆ( ))
p k where clearlyc ˆ(x ) contains some error due to applying splitting in the first
k steps.
6. Concluding remarks
Analyzing the splitting error both for the DEM and the Marchuk–Strang
splitting procedure, one can conclude:
A A i j i j j
If all the pairs ( i ), where =12 3 4 5and ,intheDEM
splitting procedure L-commute, then no splitting error occurs.
A A i j i j j
If all the pairs ( i ), where =12 5and , in the Marchuk– Strang splitting procedure applied to the advection – diffusion – reaction problem commute, then no splitting error occurs.
The splitting error in the DEM splitting procedure is of first order if at
A A i j i j j
least one pair ( i ), where =1 2 5and , does not commute.
The splitting error in the Marchuk–Strang splitting procedure is of second
A A i j i j j
order if at least one pair ( i ), where =12 5and , does not commute.
As we proved in Section 4, for the realistic situations the splitting errors for the operators A3 and A4 do not vanish. On the other hand, for the
other cases under the assumptions
r r r u g R x c R c
=0 l =0 =0 ( )= ( )
the splitting errors are equal to zero for each pair of operators in the
air pollution modeling with the exception of (A1 A2), (A2 A5), (A3 A5)
and (A4 A5). If additionally we assume the linearity of u, R and the c k x
solution , l = = const. and ( ) = const. then the splitting errors
2019. majus´ 4. –23:07 L-COMMUTATIVITY OF THE OPERATORS 149
exist only for the operators A4 and A5. Since for the linear elements
m m R JR := Rc(c) is a constant matrix therefore under the condition
g ker JR even the last commutator is equal to zero.
The diurnal cycle strongly influences the commutators leading to a rel- atively small local splitting error over nightly periods. Specific circum- stances determine actual values of the splitting error.
References
Testing the sensitivity of Air Pollution Levels to Vari
[1] I Dimov Z Zlatev
ations of some Chemical Rate Constants Computations
in: Large Scale
in Engineering and Environmental Sciences Notes on Numerical Fluid
Vol Mechanics (M. Griebel, O. Iliev, S. Margenov and P. S. Vas- silevski, Eds), Ind Workshop on “Large-Scale Scientific Computations”,
Varna, Bulgaria, June, 1997, Proceedings, 1997, pp. 167–175. Economical Estimation
[2] I Dimov Tz Ostromsky I Tzvetanov Z Zlatev Large Scale Compu
of the Loses of Crops Due to High Ozone Levels in:
tations in Engineering and Environmental Sciences Notes on Numerical Mechanics Fluid (M. Griebel, S. Margenov and P. Yalamov, Eds), IInd Workshop on “Large-Scale Scientific Computations”, Sozopol, Bulgaria, June 2–6, 1999, Proceedings, Vieweg, Braunschweig, Germany, pp. 273– 280.
[3] E G Dyakonov Difference schemes with splitting operators for multidimen-
of Computational Math and Math sional stationary problems, Journal
Phys 2 (1962), No 4 (in Russian). Analytic function theory [4] E Hille Vol 1., Blaisdell Publishing Company, Lon-
don, 1959. J G Verwer
[5] D Lanser and Analysis of operators splitting for advection-
Comput Appl diffusion-reaction problems in air pollution modelling, J
Math 111 (1999), 201–216.
[6] G I Marchuk Some application of splitting-up methods to the solution of
mathematical physics problems, Applik Mat 13 (1968), No. 2.
On the theory of the splitting up method Numerical so
[7] G I Marchuk In:
of partial di erential equations lution II. SYSSPADE, 1970, Academic
Press, New York–London, 1971.
I Marchuk [8] G Methods of computational mathematics, Moscow, Nauka, 1980. (in Russian)
[9] G I Marchuk Mathematical modelling for the problem of the environment,
Studies in Mathematics and Applications No North-Holland, Amster- dam, 1985.
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[10] G I Marchuk Methods of Splitting, Moscow, Nauka, 1988 (in Russian).
I Marchuk Applications of Adjoint Equations to Problems of Global
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Environmental Protection Change and HERMIS’94, Vol. 1, (E.A. Lipi-
takis, Edr), 1994, Hellenic Mathematical Society, pp. 1–20. J H Seinfeld
[12] G J McRae W R Goodin Numerical solution of the at- of mospheric diffusion equations for chemically reacting flows, Journal
Computational Physics 45 (1984), 1–42.
[13] V V Penenko N N Obraztsov A variational initialization model for the
Meteorol Hydrol fields of meteorological elements, Soviet , 11 (1976),
1–11. Functional Analysis [14] W Rudin Tata McGraw-Hill Publishing Company Ltd.,
New Delhi, 1973. Numerical Hamiltonian Problems [15] J M Sanz Serna M P Calvo Chapman
and Hall, 1994. SIAM [16] G Strang On the construction and comparison of difference schemes,
Journal Numer Anal 5 (1968), 505–517.
[17] B Sportisse An Analysis of Operator Splitting Techniques in the Stiff Case, CERMICS-INRIA, 2004, Route des Lucioles – BP 93 – Sophia-Antipolis
CEDEX.
A A Samarski Equations of the Mathematical Physics [18] A N Tikhonov
Moscow, Nauka, 1977. (in Russian)
S Varadarajan Lie Groups Lie Algebras and Their Representations [19] V
Prentice-Hall Inc., EnglewoodCliffs, New Jersey, 1974. J G Blom [20] J G Verwer W H Hundsdorfer and Numerical time integration for air pollution models, CWI Report MAS-R9825, 1998.
[21] N N Yanenko On convergence of the splitting method for heat equation with
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2019. majus´ 4. –23:07 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 151–156
THIN COMPLETE SUBSEQUENCE
By NORBERT HEGYVARI´
Department of Mathematics, TFK, Eotv¨ os¨ Lorand´ University, Budapest Received December
1. Introduction
The well known theorem of Lagrange states that every non-negative
S f integer n is the sum of four squares. In other words the sequence = 1 4
2
g Wirsing A
n is bases of order four. defined the notion of thin bases;
h
1
h A x c x c A x is thin bases of order if ( ) ;( 0), where ( ) is the counting
1 h
A h A x x
function of A. Let us note if is a bases of order then ( ) .Bya
S
non-constructive method Wirsing proved [1] that S has a subbases which
1 4 1 4
x O x x J Spencer is almost thin, proving S ( )= ( (log ) ). Later [2] gave a short proof for it, using the Janson’s inequality, which is an important tool
of probabilistic method.
S
Let us mention it is not even known an explicit subsequence S of for
p
x o x
which S ( )= ( ).
A fa a
A related question would be the following: let = 1 n
g N
. A is said to be complete if there exists ΔA such that for every
n
ΔA we have
X
n A fS B S B S g
Σ( )= ( ): ( )= b ; B is a finite subset of A ( )=0
b B
k
Aj k j A j A
Clearly if j = then Σ( ) 2 . This implies if is complete then x
A( )
x x x A x x A 2 ΔA i.e. for 0 ( ) log2( Δ ).
1 Research partially supported by Hungarian National Foundation for Scientific Research, Grant No. T025617 and No. TO29759 and by DIMACS (Center for Discrete Mathematics and Theoretical Computer Science) NSF-STC-91-19999.
2019. majus´ 4. –23:07 152 NORBERT HEGYVARI´
The related notion of thin bases is the following
A thin complete subsequence A A Definition is said to be of if is
complete and
x o x A ( )=(1+ (1)) log2 We shall show that a wide class of complete sequences have a thin
complete subsequence and not merely the sequence of squares S . We prove
Let A fa a g be a complete sequence of inte
Theorem = 1 2
gers Assume that a a Then A contains a thin complete
n n
lim n +1 =1 subsequence
The proof will be completely constructive.
fx x g
Let X = 1 2 . Let us denote
X fx x g i
( )=sup i +1
i
X X (So that if G ( ) then it indicates the size of the biggest gap in )
Proof of the Theorem We need some lemmas.
Lemma Let X fx x g N Assume that for every
= 1 n
i =1 2
x x x i
(1) i +1 1 + +
X x G Then (Σ( )) 1
The proof of Lemma 1 is easy or see [3].
Lemma Let A be a complete sequence of integers Let A f
1 = 2ΔA
a a g be an in nite subsequence of A for which a a i
2 = i
1 2 i +1
Let A A a and assume there are elements b b
=1 2 2 = [1 1) 1 2
A such that b Furthermore assume the sets b a A
ΔA 2 1 1 Δ
S S S
A A fb g fb g are pairwise disjoint Then B A g fb g fb
1 2 1 2 = 1 A2 1 2
is complete
Proof of Lemma A f a g a
Let 1 = 2ΔA 1 2 .
First we prove that for every i =1 2
a a a
(2) a + + + i i +1 1 2
2019. majus´ 4. –23:07
THIN COMPLETE SUBSEQUENCE 153
i a a
by induction on i .For = 1 (2) is trivial. Furthermore by 2 we i i +2 +1
get
a a a a a
2 i ( + + )+
i i i +2 +1 1 +1
which provides the inductive steps. So A1 fulfills (1), hence
A a
(3) G (Σ( 1)) 1
n a b b n B B
We claim that for every ΔA + 1 + 1 + 2 := Δ , Σ( ) Assume now
n n B t
contrary to the assertion there exists an ΔB and Σ( ) Let be the
subsript defined by
a b b a n
(4) ( + ) t
1 2 t +1
b n b b b
(clearly the equality cannot hold). Now (n 1) ( 2)= 2 1, thus
n b n b a A
(5) ΔA ( 1) ( 2) 1 Δ
A a A A A a
Furthermore Σ(A2)=Σ( ) [1 1)andsoΣ( 1)+Σ( 2) Σ( 1)+[1 1). A
Thus by (3) we conclude that Σ( A1)+Σ( 2) contains a set which is the union
a
of blocks of consecutive integers with length at lest 1 ΔA and gaps at most
i n b A A A
Δ . So (5) implies that there is an = 0 or 1 such that i Σ( 1)+Σ( 2)
n A A b B
and thus Σ( 1)+Σ( 2)+ i Σ( ) a contradiction.
Lemma A fa a a g
Let = 1 2 n be an infinite sequence
a a
n n
of integers. Assume that lim n +1 =1 Then for every 1 real
K N A fK a a k
number and there exits a subsequence = k
1 2
g a A
k of such that
n
a a K k
(6) k n n +1
and
a k
n +1
(7) lim =1
n
a
k n
Proof of Lemma Let
A fa a a g
K mK
(8) = 2 K
a K a K m a
mK K K It is obvious that . Furthermore for every , (m +1)
and
a a
a
m K m K
( +1) mK +1 ( +1)
lim = lim =1
n n
a a a
mK mK
K (m +1) 1
2019. majus´ 4. –23:07
154 NORBERT HEGYVARI´
A a a K
Hence every subsequence of satisfies (6). Let now k = and assume
1
a a a
k k
that the elements k have been defined. Then let
1 2 n 1
a fa ja a a Ag a
t t t
k k
k =min . By the definition of we have
n n
n 1
a a a
k k
(9) k n
n +1
n 1
a
k
n
Hence .Furthermore
a
k
n 1
a k
n +1
a a
k k
n
n 1
a
k n
and so
a a
k k
n n
a a
k k
n +1
n 1
a
k
n
Since limn = 1 we prove the lemma.
a k
n +1
Proof of the Theorem K
Let =5ΔA and = 2. By Lemma 3 we can
A A A f a a a
select a subsequence of for which = 2ΔA
1 1 1 2 n
g a
and a 2 . Furthermore by (6) and (8) we can choose elements n
n +1
b b A a b b a b a
1 2 of for which 1 1 2 1 ΔA (say let 1 = 5Δ+1 and
b a b b A A a A
2 = 10Δ1)Then 2 1 Δ . Finally let 2 = [1 1). Clearly the
b A A
elements b1 2 and the sequences 1 2 satisfies the conditions of Lemma
S S S
A A fb g fb g 2 and hence B = 1 2 1 2 is complete.
In the rest of the proof we shall show tha B is thin.
Since A2 is a finite sequence, thus
x jA j A X
B ( ) 2 +2+ 1( ) B
which means that if A1 is thin so is .
a A a
By Lemma 3 for the elements of we have limn =2, n 1 n +1 which implieses the theorem.
Concluding remarks
We can now apply the theorem for some “classical” sequences which have thin complete subsequences.
We shall investigate the following three type of sequences: let
P f p g
= 2 3 n
2019. majus´ 4. –23:07 THIN COMPLETE SUBSEQUENCE 155
be the sequence of prime numbers, let
k m
p q fp q p q p q p q m k N g B ( )= :( )=1; 1
be the sequence the Birch-sequence.
g x Z x g x m
Let m ( ) [ ]. Assume ( ) has positive leading coeffitient and
fg g n g m g c d m (1) ( ) =1
and finally consider the sequence
G fg g n g m = m (1) ( )
The sequence of P .
Richert proved in [8] that ΔP = 7 and it is well
p p n n
known that n +1 1as .
p q B p q
The sequence B ( ). Erdos˝ conjectured and Birch proved that ( ) q is complete sequence (see [5]). By the irrationality of (log p log )weinfer
that the quotient of consecutive terms of this sequence tends to 1. G The sequence G . Finally the completeness of the sequences were
investigated by many authors. In 1948 Sprague proved for the sequence of k
squares that ΔS = 129 [6]. Further he proved in [7] that for every the
k
n n N g sequence f : is complete. A far-reaching generalization of Birch’s and Sprague’s results was published by J. W. Cassels (see [4] and (5)). This
result gives in the general case that the sequence G is complete. Furthermore
g n g n
n m since lim ( +1) m ( ) = 1 we conclude that for these sequences
fulfills the conditions of the Theorem. Hence we obtain the following:
The sequences P B p q and G have thin complete sub
Corollary ( ) sequence
Certainly there exists complete sequence which has no thin complete
F g F F subsequence. For instance if Φ = f 1 where 1 =1, 2 =2is
the sequence of Fibonacci then it is well known that Φ is complete and
x c x c F ( )= log2 ; 1. But if we omit at least two elements from Φ then the remaining sequence cannot be complete (see [5]).
2019. majus´ 4. –23:07 156 NORBERT HEGYVARI´
References Analysis [1] E Wirsing Thin Subbases, 6 (1986), 285–308.
[2] J Spencer Four Squares with Few Squares, Number Theory, (New York,
1991–1995) 295–297, Springer Verlag, New York
Duke Math J [3] R L Graham Complete sequences of polynomial values, 31 (1964), 275–286.
[4] J W Cassels On the representation of integers as the sums of distinct sum-
mands taken from a fixed set, Acta Sci Math 21 (1960), 111–124. On sums of integers taken from a xed sequence [5] R L Graham Proc. Wash.
Univ. Conf. on Number Theory, (1971) 22–40.
Sprague Math Z [6] R Uber¨ Zerlegungen in ungleiche Quadratzahlen, 51
(1948), 289–290.
Sprague n [7] R Uber¨ Zerlegungen in -te Potenzen mit lauter verschidenen
Grundzahlen, Math Z 51 (1948), 466–468.
Math Z [8] H E Richert Uber¨ Zerfallungen¨ in ungleiche Primzahlen, (1949), 342–343.
2019. majus´ 4. –23:07 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 157–170
WARPED PRODUCT OF FINSLER MANIFOLDS
By LASZL´ O´ KOZMA, RADU PETER and CSABA VARGA
Institute of Mathematics and Informatics, University of Debrecen, Debrecen, Hungary Technical University of Cluj-Napoca, Cluj-Napoca, Romania
Babes¸ Bolyai University, Faculty of Mathematics and Informatics, Cluj-Napoca, Romania Received March
1. Introduction
Recently the notion of warped product is playing an important role in Riemannian geometry (see [3, 8, 9, 11, 14, 16]), moreover in geodesic metric spaces [9]. This construction can be extended for Finslerian metrics with some minor restriction. This is motivated by Asanov’s papers ([4, 5]) where some models of relativity theory are described through the warped product
of Finsler metrics. For example, Asanov [4] studied the property of the
M generalized Schwarzschild metric on R .
2. Preliminaries
n TM M
Let M be a real manifold of dimension and ( ) the tangent M
bundle of M . The vertical bundle of the manifold is the vector bundle
i
M V d T TM x
( V )givenby =Ker ( ). ( ) will denote local coordinates
i
i 1
M x y U
on an open subset U of ,and( ) the induced coordinates on ( )
a
TM x y y
. The radial vector field is locally given by ( )= a .
x
F TM R A Finsler metric on M is a function : + satisfying the
following properties: f
2 f
F TM n M
1. is smooth on M where = (0)
f
u u M
2. F ( ) 0forall
u j jF u u TM R
3. F ( )= ( )forall ,
p M I fu T M j F u g p 4. For any the indicatrix p = ( ) 1 is strongly convex.
2019. majus´ 4. –23:07
158 LASZL´ O´ KOZMA, RADU PETER, CSABA VARGA F AmanifoldM endowed with a Finsler metric is called a Finsler
manifold (M F ).
2
F x y
1 ( )
g x y
Condition 4. implies that the quantities ij ( )= forms a
i j y
2 y
i
positive definite matrix so a Riemannian metric h can be introduced in the
e
T M vertical bundle ( V ). On a Finsler manifold there does not exist, in general, a linear metrical connection. The analogue of the Levi-Civita connection lives just in the vertical bundle, however, there are several ones.
In this paper we use the Cartan connection which is a good vertical R
connection on V,i.e.an -linear map
v
f
M X X V X V r : ( ) ( ) ( )
having the usual properties of a covariant derivations, metrical with respect
f
i T M V
to h , and ’good’ in the sense that the bundle map : defined by
v v
Z r r V
( )= is a bundle isomorphism when is restricted to . The latter
Z
f
H u M
property induces the horizontal subspaces u =Ker for all which
V d u
are direct summands of the vertical subspaces =Ker( ) u :
f
M H V
T =
V
M X
For a tangent vector field X on we have its vertical lift and its
H
f X
horizontal lift to M .
H
Θ:V denotes the horizontal map associated to the horizontal
bundle H. Using Θ, first we get the radial horizontal vector field =Θ .
H v
r
In our case = ( ˙). Secondly we can extend the covariant derivation
f r of the vertical bundle to the whole tangent bundle of M . Denoting it with ,
for horizontal vector fields we have
v
1 f
r H r H Z X M
Z =Θ( (Θ ( ))) ( )
Z
f
Y X An arbitrary vector field ( M ) is decomposed into vertical and
horizontal parts:
V H
r Y r Y r Y
Z Z
Z = +
f f f f
X T M X T M X T M M Thus r : ( ) ( ) ( ) is a linear connection on
induced by a good vertical connection. Its torsion and curvature Ω are
defined as usual:
r Y r X X Y X Y Y
X =[ ]+ ( )
R X Y r r Z r r Z r Z
X Y Y X X Y Z ( )= [ ]
2019. majus´ 4. –23:07
WARPED PRODUCT OF FINSLER MANIFOLDS 159 X Y and the torsion has the property that for horizontal vectors ( )isa vertical vector [2]. The metrical property of the Cartan connection is also
important [2]:
X hY Z i hr Y Z i hY r Z i X
= X + The Cartan connection does not verify the Koszul formula for all vectors,
but this formula is true for the horizontal ones, as is shown in the next Lemma:
Let M F be a Finsler manifold with its Cartan connection
Lemma ( )
r For X Y Z H the following relation holds
hr Y Z i
2 X =
hY Z i Y hZ X i Z hX Y i hX Y Z i hY Z X i hZ X Y i = X + [ ] + [ ] + [ ]
Proof For the first three terms we use the metrical property of the Cartan connection, and for the last three terms we use the relation satisfied by the
torsion as follows:
X hY Z i hr Y Z i hY r Z i X
= X + ;
Y hZ X i hr Z X i hZ r X i Y
= Y + ;
Z hX Y i hr X Y i hX r Y i Z
= Z + ;
Y Z r Z r Y Y Z Z
[ ]= Y ( );
Z X r X r Z Z X X
[ ]= Z ( );
X Y r Y r X X Y Y
[ ]= X ( )
X Y Z i
Summing up and using the fact that for horizontal vectors h ( ) is
Y Z Y Z zero because ( ) is vertical for horizontal vectors we obtain the Koszul formula. We are interested in some properties of the curvature of Cartan connec-
tion listed below.
Let M F be a Finsler manifold The curvature of the
Lemma ( )
Cartan connection satis es the following properties for horizontal vectors
X Y Z V W
X Y R Y X
1. R ( )= ( )
hR X Y W i hR X Y V i W
2. V ( ) = ( )
R X Y R Y Z R Z X
X Y
3. Z ( )+ ( )+ ( )=0
hR X Y W i hR V W X Y i X
4. V ( ) = ( ) The proof of the previous Lemma can be found in [2, p. 31], and [12, p. 72].
2019. majus´ 4. –23:07
160 LASZL´ O´ KOZMA, RADU PETER, CSABA VARGA
M p n
Let P be a submanifold of of dimension and let us consider
F F j P
= TP ; it is a Finsler metric and thus becomes a Finsler space. Let
e
e
P P h i T TP T TM
x and let be the orthogonal complement of in .
ex ex ex
ex
e e e
e
P P P x P
Let P be the disjoint union of all , and let :
ex
e
P
the natural projection. Then ( P ) admits a natural structure of real
n p differentiable vector bundle, rank P = . It is the normal bundle of the
submanifold P .
e e
Y P
Let X be respectively a tangent vector field on and a cross section
e e f
P X Y T M r Y
in T and prolongations to . Then the restriction of
e
X
e P
to T does not depend upon the choice of prolongations and is denoted by
Y
r . The bundle direct sum decomposition
e
X
f e
M T P P T =
leads to the Gauss–Weingarten formulae:
e
r Y r Y I Y
= + (X )
e
e
X
X
e e
A X r
r = +
e
e
X
X
e
P P
Here Sec( ) and a similar argument (independence of extensions of
e e
X T P r r
to ) leads to the notation .Then is the induced connection,
e
X
e
A r
I the second fundamental form, the operators of Weingarten and is the normal connection ([7, 1, 10]). Next we define the umbilical point of a
Finsler submanifold and the umbilical submanifold.
q P
Definition A point is an umbilical point if there exists a vector
H P I X Y hX Y iZ P Z ( ) such that ( )= . The submanifold is said to be
totally umbilical if every point of P is an umbilical point.
3. Construction of the warped product N F Let ( M F1)and( 2) be Finsler manifolds with their Cartan con-
1 2
r f M R
nections r and ,andlet : + be a smooth function. Let
M N M p M N N
p1 : ,and 2 : . We consider the product manifold
f e
M N R N F
M endowed with the metric : , q
2 2 2
v v F v f v F v F ( 1 2)= 1 ( 1)+ ( 1( 1)) 2 ( 2)
2019. majus´ 4. –23:07 WARPED PRODUCT OF FINSLER MANIFOLDS 161
We show that the metric defined above is a Finsler metric. First it is clear that
f e
M N F F F
F is smooth on , because 1 and 2 are. is not necessarily smooth
v TM TN F
at the vectors of the form ( v1 0) and (0 2) .Thismeansthat
N
is not a really Finsler metric on the product manifold M , therefore the
f e
N F
study should be restricted to the domain M . Secondly is homogeneous F with respect to the vector variables because F1 and 2 are. Third, the Hessian
of F with respect to the vector variables is of the form:
A 0
2 B
0 f
B F F
where A and are the Hessians of the Finsler metrics 1 and 2.Sothe
F F Hessian of F is positive because the Hessians of 1 and 2 are. It means that
the indicatrix of F is strongly convex. The difference between this metric and
a classical Finsler metric is that it is not smooth at the vectors of the form
v
( v1 0) and (0 2).
N F v F v v
The product manifold M with the metric ( )= ( 1 2), for
f e
v v M N
v =(1 2) defined above will be called the warped product of
N f
the manifolds M , ,and will be called the warping function. We denote
M N M N F f
this warped product by f . We just showed that ( )isaFinsler manifold in the restricted sense above.
Our goal is to express the geometry of warped product by the geometries f of M N and the warping function . The study follows the line adopted in Riemannian and semi-Riemannian cases [13], with the specific situation due
to the Finslerian context. N The manifold M will be called base and the manifold will be called fiber as in [13].
4. The gradient of a function in Finsler geometry
M R
In this section we define the gradient of the smooth function f : +
df Shen rf x with x 0. We follow the line of [15, p. 43]. Define by
1
rf L df x
x := ( )
x
L T M T M x
where x : is the Legendre transformation. Shen proves that
x
H
b
rf f
r =
b
f f where r is the gradient of with respect to Riemannian metric induced by
the Finsler metric, and
q
b b
F rf hrf rf i
f ( )= r
2019. majus´ 4. –23:07
162 LASZL´ O´ KOZMA, RADU PETER, CSABA VARGA
H
f rf
We work with r , the horizontal lifting of which has the property that H
2 H
F rf hrf rf i
( )= H .
rf
Next we define the Hessian of a function.
f F M
Definition The Hessian of a function ( ) is its second covari-
f
r rf
ant differential H = ( ).
f
Lemma The Hessian H satis es the following relation
f H
H X Y XY f r Y f hr rf Y i X
( )= ( X ) = ( )
for X Y H
Proof
H H f
X Y r df X Y hr rf Y i H
( )= ( )( )= X
H
hrf Y i
since Yf = and it follows that
H H H
XY f X hrf Y i hr rf Y i hrf r Y i X
= = X +
H
hr rf Y i r Y f X
= X ( ) +( )
which implies the assertion.
M f M R f M N
If f is smooth on (i.e. : is smooth), the lift of to
b
b
f p M N R a T M q N a f
is the map := 1 : .If p and then the lift
b
a p q T M q dp a a
of to ( ) is the unique vector in (p q )( ) such that 1( )= .If
b
X M X M N X
X ( ) the lift of to is the vector field whose value at each
p q X p q
( ) is the lift of p to ( ). Because of the product coordinate systems it
b
X X M
is clear that X is smooth. It follows that the lift of ( ) is the unique
M N p X p
element of X( )thatis 1-related to and 2-related to the zero vector N
field on N . The same method could be used to lift objects defined on to
N M .
Now we prove a Lemma needed in what follows:
Lemma If h is a smooth function on M then the gradient of the lift
h p of h to M N is the lift to M N of the gradient of h on M f
1 f
H H
v TN hr h p v i v h p
Proof Let .Now ( 1) = ( 1) = 0. Next for
TM
x we have that
H
e e
d p r h p dp x i
h 1(( ( 1)) ) 1( ) =
H H H H H
x i dp i r h p x h p h rh x = h( ( 1)) =( ( 1)) = ( ) 1( )
2019. majus´ 4. –23:07 WARPED PRODUCT OF FINSLER MANIFOLDS 163
From these two properties we obtain the assertion in the lemma.
rh
Due to this lemma there will be no confusion if we denote h and
p r h p instead of h 1 and ( 1), resp.
5. Properties of warped metrics N F
Let ( M F1)and( 2) be two Finsler manifolds, with Finsler metrics
F M N
F1, 2 resp. We consider the product manifold and the warped metric
M N M p M
defined above. We consider the projections p1 : and 2 :
TM M TN N N N
and the canonical projections 1 : and 2 : .
p dp TM TN TM
The projections p1 2 resp. generate the projections 1 :
dp TM TN TN v v v TM TN dp v v v i
and 2 : ,for =( 1 2) i ( 1 2)= , i =1 2.
1
N p p p M It is obvious that the fibers p = 1 ( ) and the leaves
1
M q p q q N M N = 2 ( ) are Finsler submanifolds of F and the warped
metric has the properties:
q N p j M
q
1. for each the map 1 (M ) is an isometry onto .
p M p j N
N 2. for each the map 2 (p ) is a positive homothety onto with
scale factor 1 .
f
M N M q p N 3. for each ( p q ) the leaf and the fiber are orthogonal with respect to the Riemannian metrics induced by the Finsler metrics.
The canonical projection 1 gives rise to the vertical bundle
f
TM
( V1 1 )
f
d d TTM TM where V1 =Ker( 1)and 1 = 1 : . The same is true for
the manifold N . Now we have that
d d TTM TTN T TM TN TM TN
d 1 2 = ( 1 2): = ( )
d d and Ker d ( 1 2)=Ker 1 Ker 2. It follows that the vertical space
1
N V V V h i of the manifold M , = 1 2, so the Riemannian metrics and
2
i V V h ,definedon 1 and 2 as in the introduction give rise to a Riemannian
1 2 2
h i V h i h i f v h i H
metric on as follows: v = + ( ( )) .Nowlet v v1 1 1 2 1
1 r and H2 be the horizontal spaces with respect to the Cartan connections
2
M F N F and r on the Finsler manifolds ( 1)and( 2), resp.
2019. majus´ 4. –23:07 164 LASZL´ O´ KOZMA, RADU PETER, CSABA VARGA
We have the direct sum decomposition
M N TTM TTN V H V H
TT( )= = 1 1 2 2
F M N Next the Finsler metrics F1 2 on the manifolds and resp. generate
1 2
i h i V V the Riemannian metrics h and on the vertical spaces 1 and 2,
resp. By the horizontal maps these Riemannian metrics are mapped onto
H
horizontal spaces H1 2 resp. Finally these Riemannian metrics generates
TM TN
a Riemannian metric on T ( ). In what it follows we work mostly
H H H
on the direct sum H1 2 the direct sum of the liftings of 1 and 2 to the
TTN
TTM . N
The following theorem relates the Cartan connections of M and to the
M N
Cartan connection of f .
Theorem On B M N if X Y X H and V W X H the
= f ( 1) ( 2)
following relations are true
r Y on H H is the lift of r Y on H X
1. X 1 2 1
r V r X Xf f V V
2. X = =( )
H
r W I V W hV W i f rf
3. nor V = ( )= ( )
X V V X
4. ( )= ( )=0
r W X N is the lift of r W on N V
5. tan V ( )
Proof hr Y V i
We apply the Koszul formula (see Lemma 1) for 2 X
V hX Y i hV X Y i X V
and we obtain that it is equal to + [ ] because [ ]=
X Y M hX Y i
=[Y V ] = 0. Because are lifts from , is constant on
e
V T N V hX Y i
fibers (liftings on N ), and because follows that =0.
hV X Y i hr Y V i V X N
Analogously [ ] = 0. Thus X =0forall ( )andit follows formula (1).
First we prove the first equality from (2). The second one will be
X hV Y i hr V Y i hV r Y i X
proved after (3). We have that = X + =0,so
hr V Y i hV r Y i hr V W i
X X
X = . We apply the Koszul formula for 2 ,
hV W i and we observe that all the terms vanish except X .
It follows from the expression of the Riemannian metric induced by
2
hV W i v w f v hV W i w the warped metric that ( )= 1( ) w .Thisterm
2
X hV W i X f v hV W i w
is constant on leaves. Thus = ( ( 1( )) w )=
Xf
fX f v hV W i hV W i w =2 ( ( ( ))) w =2 . From these relations we have
1 f
2019. majus´ 4. –23:07
WARPED PRODUCT OF FINSLER MANIFOLDS 165
Xf
r V V r V r X X V X V
X V
that X = .Now =[ ]+ ( ). We can f
assume that [X V ]=0.
hW X i
It is obvious that V = 0. But this means that
hr W X i hW r X i hW Xf f V X V i Xf f hV W i V
V = = ( ) + ( ) = ( )
H
X V hrf Xi Xf
because ( ) is vertical. Now = . Thus
H
hr W X i h hV W i f rf Xi
V = ( )
This yields (3).
H
hr X W i hX r W i hX hV W i f rf i V
V = =
1
H H
hhX rf i f V W i X rf ihV W i
= h = f
The above gives the second part of (2) and it follows that
Xf
r X r V V X
V = =
f
X V V X and the mixed part of the torsion vanishes ( )= ( )=0.Thelast assertion (5) is trivial.
It is a remarkable fact that the torsion vanishes on the mixed part. This will let us to compute the curvature of warped product.
Now the next Corollary easily follows:
Corollary The leaves M q of a warped product are totally
geodesic the bers p M are totally umbilical
Proof By the claim (1) in Theorem 7 it follows that for a geodesic
M M N
in its lifting on f is also a geodesic. The second assertion comes from (3) of Theorem 7.
2019. majus´ 4. –23:07 166 LASZL´ O´ KOZMA, RADU PETER, CSABA VARGA
6. Geodesics of warped product manifolds
s
In a warped product manifold a curve can be written as ( )=
s s M =( ( ) ( )) where the curves and are the projections of into
and N , resp. Now we give conditions for a curve in the warped product to
be geodesic with respect to the warped metric.
Theorem A curve in M N is a geodesic if and only if
=( ) f H
2
H
jj jj
H
rf r
1. H =
f
H
H H
d f
2 ( ( ))
r
2. H =
f ds
s
Proof We work in an interval around =0.
Case T M T N (0) is neither in nor in .Then (0) 0and
(0) (0)
X M
(0) 0. So we can suppose that is an integral curve for in and is an
V N X V M N
integral curve for in .Alsowedenoteby and the lifts on f .
H H
r X V H
It follows that is a geodesic curve if and only if H ( + )=0. X +V
Butthismeansthat
H H H H
r r r r X V X X
H H H
H + + + =0
X X V V Now we use Theorem 7 from the previous section and we have that
H 2
jjV jj
H H
X r rf
H =0
X f
and
H
X f
H
r V V
2 + H =0
V
f
Case T M M
Suppose that (0) (0) .If is a geodesic, because
M (0) is totally geodesic, it follows that remains in (0). Thus is
constant and the assertions of the theorem are trivial. Conversely if condition
(2) from Theorem 7 holds, since (0) = 0 it follows that is constant. Then
condition (1) in Theorem 7 implies that is a geodesic, and so is .
T N rf Case Suppose that (0) and nonzero. Suppose that
(0)
N
is not zero, because otherwise (0) is totally geodesic and the conclu-
sion follows as in Case .Nowif is a geodesic, it follows that on no
p N interval around 0 remains in the totally umbilical fiber . It follows
2019. majus´ 4. –23:07
WARPED PRODUCT OF FINSLER MANIFOLDS 167
fs g i s i
that there is a sequence i 0 such that for all , ( ) is neither in
T M T N
s or in . The assertions in the theorem follows by continuity s ( )
i ( ) i
from the first case. Conversely, if (1) in the theorem is true it follows that
H
r fs g H
(0) 0 hence there exists a sequence i as above, and using
(0)
again the first case it follows that is a geodesic.
7. Curvature of warped product manifolds
Now we express the curvature of the warped product. The curvature
tensor is defined by the relation
R X Y r r Z r r Z r Z
Z X Y Y X ( )= [X Y ]
Because the projection p1 is an isometry it follows that the lift of the curvature
on M is equal to the curvature of the warped product when is computed for
vectors from on H1.
Theorem Let M N be a warped product of Finsler manifolds with
f
M
curvature tensor R and let X Y Z H and U V W H Let R and
1 2 Z
N
M F and N F resp denote the curvature tensors of the manifolds R ( ) ( )
U 1 2
The following relations are true
M
R X Y X H is the lift of R X Y on M
1. Z ( ) ( ) ( )
1 Z
f
H X Y
( ) f
R V X V where H is the Hessian of f
2. Y ( )=
f
R V W Xf f V W
3. X ( )=( ) ( )
hV W i
R X V r rf X
4. W ( )= ( )
f
hrf rf i
N
R V W R V W fhV U iW hW U iV g
5. U ( )= ( )
U 2
f p Proof (1) This is true because the projection 1 is an isometry and the
leaves are totally geodesic.
V X r r Y r r Y
X X V
(2) Because [ ] = 0 it follows that V =
Y f
(r )
X
R V X r r Y V
V X
= Y ( ). By Theorem 7 we have that = because f
2019. majus´ 4. –23:07
168 LASZL´ O´ KOZMA, RADU PETER, CSABA VARGA
r Y X H
X ( 1). The second term
Yf
r r Y r X Yf f V Yf f r V V
V X X
X = = ( ) +( )
f
f f Yf X f V Yf f Xf f V =[(XY ) + (1 )] +( )( )
2
f Xf f XY f f V Because X (1 )= the last expression reduces to ( ) .
Thus
f
R V X XY f r Y f f V H X Y f V X
Y ( )= [( ( ) ) ] = ( ( ) )
(3) We can assume that [V W ] = 0. It follows that
R V W r r X r r X
V W W V
X ( )=
But
r r X r Xf f W V Xf f W Xf f r W
W V V
V = (( ) )= ( ) +( )
Xf f Xf f
Now V ( ) = 0 because is constant on the fibers. This implies that
R V W Xf f r W r V Xf f V W
V W
X ( )=( )[ ]=( ) ( )
R V W V
We note that X ( ) 2 by the properties of the Cartan connection.
hR X Y W i hR V W Y i X
By the symmetry of curvature V ( ) = ( ) =0
R V W
because X ( ) is vertical. Now we use (2), the curvature symmetries,
and then we obtain that relation (3) is true.
hR X V Ui hR W U Wi X
(4) We have that W ( ) = ( ) = 0 because of the
R V W
point above. We use here the properties from Lemma 2. Now X ( )is
vertical and it follows that
f
hR V X Y i hR V X W i H X Y hV W i Y
W ( ) = ( ) = ( )
hV W i f hr rf Y i
=( ) X ( ) which gives assertion (4).
2019. majus´ 4. –23:07 WARPED PRODUCT OF FINSLER MANIFOLDS 169
(5) Again we can assume that [U V ] is zero. V W
R ( )=
H N
r r U r r U r f hW U i f rf r U g
W W V V
= V = ( ) +
V
H N
hV U i f rf r U g hr W U i r f V
W ( ) + = (
V
H H
hW r U i rf f hW U i f r rf V
+ V )( ) ( ) ( )
N H
r r hr V U i hV r U i rf f U
W W
+ V +( + )( )
W
H N
hV U i f r rf r r U hr V r W U i
W W V
+( ) W ( ) =(
V
N N N H N
U r U r r hW r U i hV r U i rf f r W
V )( )+
V W W V
N H N H
hV r rf hW r rf i f U i
( U ) +( ( ) V
W )
H H H H
V U i f hrf rf i f hW U i f hrf rf i f V
+(h )( ) ( )( )
H h
hrf rf i
N
V W U hV U iW hW U iV
= R ( ) + 2 ( )
f
N
hV r U i hV r U i
We use that W = , and the properties from Theorem 7. W
Thus we have
H h
hrf rf i
N
R V W R V W hV U iW hW U iV
U ( )= ( )+ ( )
U 2 f
References Shun ichi Hojo [1] L Maria Abatangelo Sorin Dragomir and On submani-
folds of Finsler Spaces, Tensor N S , 47 (1988), 272–285.
G Patrizio Finsler Metrics A Global Approach [2] M Abate and , Lecture Notes in Mathematics, 1591, Springer Verlag, Berlin, Heidelberg, 1994.
[3] Stephanie B Alexander Richard L Bishop Warped products of Hadamard
spaces, Manuscripta Math 96 (1998), 487–505.
R M [4] G S Asanov Finslerian metric functions over the product and their
potential applications, Rep Math Phys 41 (1998), 117–132. Fortschr Phys [5] G S Asanov Finslerian extensions of Schwarzschild metric,
40 (1992), 667–693. An Introduction to Riemann Finsler Geometry [6] D Bao S S Chern Z Shen , Springer Verlag, 2000.
2019. majus´ 4. –23:07
170 LASZL´ O´ KOZMA, RADU PETER, CSABA VARGA Meetings of Minds [7] Aurel Bejancu On the theory of Finsler submanifolds, , eds. P. L. Antonelli, Kluwer Academic Press, 1999, 111–129.
[8] Bang Yen Chen Geometry of warped product CR-submanifolds in Kaehler
manifolds, Monatsh Math 133, (2001), 177–195.
[9] Chien Hsiung Chen Warped products of metric spaces of curvature bounded
from above, Trans Amer Math Soc , 351 (1999), 4727–4740.
Conf Sem Mat Univ Bari [10] S Dragomir Submanifolds of Finsler Spaces. ,
271 (1986), 1–15. Proc Japan
[11] Byung Hak Kim Warped producs with critical Riemannian metric, Ser A Math Sci
Acad 71 (1995), 117–118. Foundations of Finsler geometry and special Finsler spaces [12] M Matsumoto ,
Kasheisha Press, Japan, 1986. Semi Riemannian geometry with applications to relativity [13] Barrett O Neill ,
Pure and Applied Mathematics. Academic Press, New-York, 1983.
olker Di Geom Appl [14] S N Isometric immersions of warped products, 6
(1996), 1–30. Lectures on Finsler Geometry
[15] Z Shen , World Scientific, 2001.
Gen Relativity [16] M Ulanovskii Lorentzian warped products and singularity,
Gravitation, 31 (1999), 1813–1820.
2019. majus´ 4. –23:07 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 171–182
K-STRUCTURES AND FOLIATIONS
By LUIGIA DI TERLIZZI, JERZY KONDERAK, ANNA MARIA PASTORE and ROBERT WOLAK
Dipartimento di Matematica, Universita` di Bari, Italy
Instytut Matematyki, Uniwersytet Jagiellonski,´ Poland Received March
1. Preliminaries
n s f Let M be 2 + dimensional manifold on which is defined an -structure
of rank 2n with complemented frames. This means that there exist vector
M s
fields 1 s on such that if 1 are dual 1-forms then
f
(1) ( i )=0
f
(2) i =0 s
for any i =1 and s
2 X
f I i
(3) = + i
i =1 TM
Let Γ( TM ) be the module of differentiable sections of .Itiswell M
known that in such conditions we can define a Riemannian metric g on
TM
such that for any X Y Γ( ) the following equality holds:
s
X
g X Y g fX fY X Y i
(4) ( )= ( )+ i ( ) ( )
i =1
P
s
f K f f
We suppose also that the -structure is a -structure, i.e. [ ]+ i
i =1
d f f f
i =0,where[ ] is the Nijenhuis torsion of (cf. [2]) and the fundamental
F X Y g X f Y dF
2-form, F defined as ( )= ( ) is closed, i.e. =0.
n
d d F d
s s
If 1 = = = and 1 ( i ) 0 we say that the
K S M S d
-structure is an -structure and is an -manifold. Finally, if i =0for
s K C M all i =1 then the -structure is called a -structure and is said a
C-manifold.
2019. majus´ 4. –23:07 172 LUIGIA DI TERLIZZI, JERZY KONDERAK, ANNA MARIA PASTORE, ROBERT WOLAK
We recall some facts that will be used in the sequel (cf. [2]).
K M
1. On a -manifold the vector fields 1 s are Killing.
S Y TM i s 2. If f is an -structure then for any Γ( ) and for any =1 we have that
1
r fY i
(5) Y = ;
2 C
if f is a -structure then
r i
(6) Y =0
where we denote by r the Levi–Civita connection of the Riemannian
metric g .
C rF rf 3. A K-structure is a -structure if and only if =0or =0.
4. On S-manifolds we have:
s
X
1
r F Y Z X g X Z Z g X Y
i i
( X )( )= ( ) ( ) ( ) ( ) 2
i =1
s
X
1
X Y Z Z Y
i j i j
j ( ) ( ) ( ) ( ) ( ) 2
i j =1
2. Transversally holomorphic foliations
Let M be a n s manifold with an f structure of rank
Theorem 2 +
Then f is a K if and only if the foliation f is a transversely Kahler
2 n ker
s
foliation given by an isometric action of R
f K
Proof Let be a structure. We know that the vector fields 1 2
i j s
s j
are Killing vector fields and [ i ]=0, =12 . Therefore
the subbundle spanned by 1 2 s is integrable and defines the foliation K
F . In fact, the condition for the -structure, i.e.
s
X
f f f d g i
0= [ ]+ i =0
i =1
i j s j
easily implies that [ i ]=0forany =1 . Moreover, the fact that
r i j j
the vector fields i are Killing ensures that =0forany =1
i
s F
. As the vector fields i are Killing the foliation is Riemannian and
2019. majus´ 4. –23:07
K-STRUCTURES AND FOLIATIONS 173 F
the Riemannian metric g is bundle-like. The normal bundle of can be
F
identified with Imf. First, notice that Imf . In fact,
X
2 s
g f X g f f X f X
i j i j
( i ( )) = ( ( ) ( )) + ( ) ( ( )) = 0
j =1
T F
Moreover, rank(Imf) = codim F so indeed Imf is the normal bundle of .
j
Now we should prove that f Imf is foliated, which is equivalent to the property
f j i s L ( )=0 =1
Imf
i
f Y Y
i.e. L ( ) = 0 for any section of Imf. We may assume that is an
i
P
s
F f f f d g i
infinitesimal automorphism of . First compute [ ]+ i =0
i =1
X F
on r and an infinitesimal automorphism of , a section of Imf. Thus
X T F
[ r ] is a section of . Therefore
s s
X X
f f f d g X f f X d X
r r
i i i
[ ]+ i ( )= ([ ( )]) + ( ) i
i =1 =1
f f X d X f X T F X
r r r
Hence ([ ]) = 0 and i ( )=0.So[ ( )] for any an
Y
infinitesimal automorphism of F .Let be an infinitesimal automorphism
of F . Then:
f L f Y Y f f Y i
( ( )) = [ i ] ([ ( )] = 0
i
In fact, we may assume that Y is an infinitesimal automorphism commuting
f j i
with i as the subbundle Imf is -invariant. So Imf is constant along leaves
F d d
i j
of (“foliated”). Moreover, the forms i are base-like as ( )=0.
So is the form F as
F Y g f Y i
( i )=0 ( ( )) = 0
F g
and dF = 0. So our foliation is transversely Kahler¨ as the metric ,
F
tensor field f and the 2-form project along leaves to the Riemannian metric
J g J g , tensor field and the 2-form Ω, respectively. The structure ( Ω) is
Kahlerian.¨
s M
Now, assume that we have an R isometric action on which is trans-
versely Kahlerian.¨ Then we have Killing vector fields 1 s pair-wise
F commuting which define a transversely Kahler¨ foliation .As s are
1
T F
Killing vector fields they leave invariant the subbundle Q orthogonal to .
We define the tensor field forms i as follows:
j
j f
j i Q i
i ( )= =0 ( )=0 i
2019. majus´ 4. –23:07
174 LUIGIA DI TERLIZZI, JERZY KONDERAK, ANNA MARIA PASTORE, ROBERT WOLAK
Q f f U N i
On the tensor field is defined as follows: let i : be a local submer-
F d f j Q T N
x x
i Q
x sion defining the foliation ;then : f is an isomorphism
( )
i
x U X Q d f X X X T N
x x
i i f
for any .If then ( )by ;if (x ) we denote i
1
j X dx f ˆ Q
( i ) by . With this notation in mind we put
d
X J X
f ( )= ( )
X Q X Q f X Q x for any .Soforany x ( ) as well. If necessary we can
modify the Riemannian metric by assuming on Q the following values:
X Y g X Y
g ( )= ( )
X Y Q g f s for all . With 1 s , 1 , and defined as above it is
straightforward to verify that they define a K-structure.
Example We are going to use the well-known construction of a sus-
F g J pension to produce examples of K-manifolds, cf. [11]. Let ( ¯ ¯)bea
compact Kahler¨ manifold. Let
s s
T Z Iso F g J h : 1( )= ( ¯ ¯)
be a homomorphism of groups, which is equivalent to choosing s commuting
s s
g J Z R F
holomorphic isometries of ( F ¯ ¯). The group acts on the product
s s
Z v R w F p v w v p h p w
as follows. Let p , , , ( )( )=( + ( )( )). The
s R
action is locally free and commutes with the standard action of .Ifwe
s
F g g g g
endow R with the product metric ˜ = 0 ¯ ( 0-the Euclidean metric
s s
R T F
of ), then the action is isometric. The quotient manifold h is a
s F
compact fibre bundle over T with standard fibre . The Riemannian metric
s s
g g T F R
˜ defines a Riemannian metric on h for which the induce -action
is isometric. Moreover, the foliation defined by this action is transversely
s
T F K
Kahler,¨ so any such manifold h is equipped with a -structure. We finish the section with a very useful proposition whose proof is
straightforward.
Proposition Let W be a foliated submanifold of M i e if x W
then the leaf L W Let fU f g g be a cocycle de ning foliation
x
i i ij i I
F and N the transverse manifold H the holonomy pseudogroup associated
an H invariant submanifold of N such W to this cocycle Then there exists 0
1
that W j f W for any index i I
U = ( )
i 0 i
This proposition can be used to find properties of geodesics orthogonal to
W W F . In fact, the submanifold is totally geodesic iff is totally geodesic 0
in the orthogonal direction to F .
i
2019. majus´ 4. –23:07 K-STRUCTURES AND FOLIATIONS 175
3. General properties
Having proved the fact that our structure is a transversely Kahler¨ foliation let us draw some general conclusions. The set of points of leaves without holonomy is open and dense in the
manifold M , cf. [14]. Unless all leaves are compact, there is no compact leaf among them. If the foliation has a compact leaf without holonomy, then this leaf covers any other leaf, thus all leaves are compact, cf. [13, 10, 17].
Leaves without holonomy are stable, which means that for any leaf L without
holonomy there exists an 0 such that any leaf at the distance smaller than
L L from is diffeomorphic to . The holonomy pseudogroup of our foliations consists of local isomor- phisms of the Kahler¨ structure of the transverse manifolds, i.e. hermitian isometries which preserve the complex structure. The Molino structure the- orem for Riemannian foliations, cf. [10], has its hermitian version, cf. [16,
19].
Let F be a transversely hermitian codimension q foliation
Theorem 2
on a compact manifold M Then the bundle of transverse orthonormal frames
B q F admits an U q reduction B M U q F which is its foli
( M O(2 ); ) ( ) ( ( ); )
ated subbundle The lifted foliation F is transversely parallelisable The
U
closures of its leaves are bres of locally trivial bration called the basic
foliation onto a compact manifold The foliation of any closure by leaves of
F is a Lie foliation The projections on M of the bres of the basic bration
U
are the closure of the foliation F
The structure theorem permits us to define the commuting sheaf, cf. [20, 19]. Local foliated transformations which preserve the complex structure and the hermitian metric of the normal bundle, or equivalently which define local
isomorphisms of the induced Kahler¨ (hermitian) structure on any traverse
M U q F manifold are lifted to the bundle B ( ( ); ) and preserve the transverse parallelism. So local foliated infinitesimal automorphisms of the complex
structure and the hermitian metric of the normal bundle are the foliated vector
M U q F fields which when lifted to bundle B ( ( ); ) commute with the trans- verse parallelism. Therefore we can formulate the following proposition, cf.
[19, 20].
Let F be a transversely hermitian codimension q folia
Proposition 2
tion on a compact manifold M Then the closures of leaves are submanifolds
2019. majus´ 4. –23:07
176 LUIGIA DI TERLIZZI, JERZY KONDERAK, ANNA MARIA PASTORE, ROBERT WOLAK
are orbits of the commuting sheaf of the foliation and form a singular foliation
Now we apply the above results to our K-structures. The vector fields
s
R
i are defined by an isometric action of , which therefore defined a repre-
s M g sentation of R into the group Isom( ) of isometries of the Riemannian
manifold (M g). Therefore the closures of leaves are the orbits of some
r s
R M g
toral action – the action of T -the closure of in Isom( )-the smallest
s
R V
compact abelian subgroup of Isom( M g) containing the image of .Let
r r s s
Li e T R Li e R
be the orthogonal complement in R = ( )of = ( ). As the flows
i s
of i , =1 preserve the hermitian metric and the complex structure on the normal bundle, so the flows of the characteristic vector fields of the toral
action. Thus the vector fields corresponding to vectors of V define the global
trivialisation of the commuting sheaf
As the vector fields i commute it is most natural to recall the notion of the rank of a manifold, cf. [4, 15]. This fact will help us determine the
dimension of the closures of leaves.
Proposition Let F be a codimension q foliation determined by an
s
isometric action of the group R Then the closures of leaves have at most
dimension s q where rk q is the rank of the q sphere + rk( 1) ( )
Let us choose a leaf L whose closure is of maximal dimension. The same property have neighbouring leaves. The fact that the foliation is Riemannian
ensures that these neighbouring leaves live on sphere bundles of a tubular
q
neighbourhood of L, So do their closures. Therefore on the ( 1)-spheres
s r s
there are (r ) commuting vector fields. Thus must be smaller or equal
q
to rk( 1). F
The closures of leaves define a singular Riemannian foliation b .The
set of points M0 where the closures are of maximal dimension is open and
dense in M , and on this set the closures form a regular Riemannian foliation. M
Let us look closer The tangent bundle TM on 0 admits the orthogonal
T F Q Q T F T F Q
N b b b
decomposition b where = . Denote by the
Q f b
orthogonal projection of TM onto N . Then define the (1 1)-tensor field
f f M by b . It is a kind of -structure on the open set 0. We will study its
properties in relation to the initial K-structure in a subsequent paper.
2019. majus´ 4. –23:07 K-STRUCTURES AND FOLIATIONS 177
4. Submanifolds in K-manifolds
The theory of submanifolds tangent to the characteristic foliation devel-
oped for various types of f -structures can be also treated in a “foliated” way, so very often these results are straightforward generalisations of properties of submanifolds of Kahler¨ manifolds. We assume that the ambient manifold
M is compact, although in most cases this condition can be weakened to
“complete”.
m s M
Let W be an + dimensional submanifold of tangent to the char-
F x W T F T W x
acteristic foliation ,i.e.forany , x or equivalently
x T W i s x
i ( ) for =1 . The following lemma is a simple generalization
of the Frobenius theorem.
Lemma Let x be a point of a submanifold W of dimension m s
to the foliation F Then there exists an adapted chart V
tangent :
s
2 n +
R atx such that the set U fy V j y
s m s
=( 1 2 n + ) = + +1( )=
y g is a connected component of V W containing x and s
= 2 n + ( )=0
jU is an adapted chart for the induced foliation of W m ( 1 +s
As a corollary we obtain the following:
Proposition Let W be a submanifold tangent to the characteristic
foliation of a K manifold M Then for any point x of W there exist neigh
bourhoods U and V of x in W and M respectively having the following
properties
is a connected component of V W containing x
i) U
is a foliated subset of V for the characteristic foliation
ii) U
f V N
iii) there exists a Riemannian submersion with connected bres : 0
ahler manifold N de ning the characteristic foliation onto a K 0
1
W of N such that U f W iv) there exists a submanifold ¯ 0 = ( ¯ )
Now we turn our attention to CR-submanifolds, cf. [21, 9, 12, 3, 7, 5, 6].
K M
Let W be a connected submanifold of a -manifold . Assume that W
W is tangent to the characteristic foliation. Then is called a contact
D W
CR-submanifold of M if there exists a differentiable distribution on of
D x D T W x
constant dimension, : x , satisfying the following conditions:
D f x W f D D x
i) is invariant with respect to ,i.e.forany ( x ) ;
D x D T W
ii) the complementary orthogonal distribution : x is
x
f x W f D T W
anti-invariant with respect to ,i.e.forany ; ( ) x . x
2019. majus´ 4. –23:07
178 LUIGIA DI TERLIZZI, JERZY KONDERAK, ANNA MARIA PASTORE, ROBERT WOLAK
D h
A contact CR-submanifold W is non-trivial if dim = 0and
D q O f D T W f T W D
x x
dim = ; cf. [21] p. 48. Let ( x )= ( )= 0 and
f D T W f T W D x ( )= x ( ). Then the distribution has constant dimension
x 0
D D D T F D D T F D
and D = 0 or = 0 ,and = 0 or 0 , respectively, where
D T F
D0 is the orthogonal complement of 0 . This means that the tangent
W T F D D bundle TW of admits the following decomposition: and
0 0
j imf j D D TW
ker TW = = 0 0 . For the rest of the paper we assume that
D T F D = 0 .
Our previous considerations lead to the following:
Proposition Let W be a submanifold tangent to the characteristic
foliation of a K manifold Then W is a contact CR submanifold i the
corresponding submanifolds in any transverse manifold are the characteristic
foliation CR submanifolds
D
Our distributions D and have the following properties.
Theorem Let W be a contact CR submanifold of a K manifold M
Then the distribution D T F is completely integrable and its integral
submanifolds are anti invariant submanifolds tangent to the characteristic foliation
For the proof see Theorem III.3.1 of [21] or [9], Theorem 3.1. Similarly
we have the following version of Theorem III.3.2 of [21], or [9], Th.3.5 ,
W M
where B is the second fundamental form of the submanifold in :
Theorem Let W be a contact CR submanifold of a K manifold M
X f Y B Y f X for any the distribution D is integrable i B
Then ( )= ( )
X Y D Its integral submanifolds are invariant submanifolds of M
Remark As the properties described by the above theorems are local, they can be derived from the corresponding theorems for CR-submanifolds of Kahler¨ manifolds, compare Theorems IV.4.1 and IV.4.2 of [21]. Having proved these basic properties let us turn our attention to
geodesics:
Proposition Let W be a contact CR submanifold tangent to the char
foliation F of a K manifold M If g B X Y f Z for any
acteristic ( ( ) )=0
D Z D then any geodesic of W tangent to D at one point
X Y 0 0 0
at any point of its domain D remains tangent to 0
2019. majus´ 4. –23:07
K-STRUCTURES AND FOLIATIONS 179
F jW
Proof The foliation is a Riemannian foliation and the distribution
D
D0 0 is the orthogonal complement of the bundle tangent to the foliation.
D D Therefore a geodesic orthogonal to F , i.e. tangent to at one point
0 0
D
is tangent to D0 0 at any point of its domain. Moreover, such orthogonal
D
geodesics are D0 0 -horizontal lift of the corresponding geodesics in the
a b W
transverse manifold, cf. [10]. Let us consider a geodesic :( )
A ft a b t D g A
tangent to D0 at 0 and the set = ( ): ˙ ( ) 0 .Theset A is closed and 0 . We shall show that it is also open. As the problem is local we can reduce our considerations to a foliated submanifold of a
K-anifold with the characteristic foliation given by a global submersion with
1
M N W h W
connected fibres, i.e. the characteristic fibration f : and = ( ¯ )
N T W
where W¯ is a CR-submanifold of the Kahler¨ manifold . Therefore ¯
D
admits a decomposition into orthogonal distributions D¯ and ¯ such that
1 1 1
h D D h D D h D B D ¯ ¯ ¯
= ( )and 0 =ker ( ) 0 =ker ( .Let be the
M B
second fundamental form of the submanifold W in and ¯ be the second
N B X Y B X Y
fundamental form of the submanifold W¯ in .Then ( )= ¯ ( ) ,
W X cf. [21], p. 101, where for any vector X tangent to ¯ is its ker
¯ ¯ ¯
D M g B X Y fZ X Y D
( D0 0 )-lift to , and hence ¯( ( ) )=0forany and
D D Z ¯ . Then Proposition IV.4.2 of [21] ensures that ¯ is a totally geodesic
¯ ¯ ¯
W D
foliation of W .Let ¯ be the geodesic in corre then ¯ is tangent to at
this point. Since the foliation D¯ is totally geodesic ¯ must be contained in
D D D¯
some leaf of . Hence being the 0 0 -horizontal lift of ¯ ,itmustbe
A A a b tangent to D0. Therefore the set is open, and thus =( ). Taking as a model Kahler¨ manifolds we can introduce the following
notions: W
Definition We say that a contact CR-submanifold is:
B X Y X Y D
i) D0-totally geodesic iff ( )=0forany 0;
X Y X D Y D
ii) contact mixed foliate if B ( )=0forany and ,and
PX Y B X PY X Y D B ( )= ( )forany 0.
It is not difficult to verify the following:
Lemma
is a D totally geodesic i W is D totally geodesic
i) W 0 ¯ ¯
is contact mixed foliate i W is mixed foliate ii) W ¯ Then we can prove:
2019. majus´ 4. –23:07
180 LUIGIA DI TERLIZZI, JERZY KONDERAK, ANNA MARIA PASTORE, ROBERT WOLAK
Proposition Let W be a contact CR submanifold tangent to the char
acteristic foliation of a K manifold M IfW is D D is
0 totally geodesic then
W tangent to D at one point remains tangent
a foliation and any geodesic of 0
D at any point of its domain to 0
Proof It is a consequence of Lemma 2, Corollary IV.4.3 of [21] and of the considerations similar to those of the second part of the proof of Proposition 4. Another property of Kahler¨ manifolds gives us the following theorem, cf.
Theorem IV.6.1 of [21] or [1].
Theorem Let W be a contact totally umbilical non trivial contact CR
submanifold of a K manifold M If then a geodesic orthogonal
dim D0 1
to F and tangent to W at one point has this property on an open subset of
its domain W Proof The corresponding submanifold ¯ in the transverse manifold is totally umbilical. Since the characteristic foliation is Riemannian we have
to show that the geodesic is tangent to W on an open subset of its domain. This property is a local one and therefore we can reduce our considerations to
the canonical fibration. The geodesic is the ker -horizontal lift of a geodesic W in N . Therefore it is sufficient to know that the submanifold ¯ is totally geodesic. This is precisely the fact which Bejancu’s theorem ensures. Finally we have the following theorem about totally geodesic CR-
submanifolds, cf. Theorem 3.4 of [7] for S-structures.
Theorem Let W be a totally geodesic contact CR submanifold of a
K manifold M ThenD and D T F are Riemannian foliations and locally
s
is di eomorphic to R N N
i) W 0 1
s
D is given by the projection R N N N N
ii) the foliation 0 1 1
s
D T F is given by the projection R N N N N iii) the foliation 0 1 0
¯
W N is a Riemannian product of N N of a totally
iv) the submanifold 0 1
and a totally geodesic anti invariant invariant submanifold N
geodesic 0
N of N submanifold 1
Proof The problem is local and we can reduce our considerations to the
1
f W
case of canonical fibration. Therefore we can assume that W = ( ¯ )for N
some CR-submanifold W¯ of the Kahler¨ manifold and that the submersion
M N T F f : is a Riemannian submersion. The orthogonal complement of
2019. majus´ 4. –23:07
K-STRUCTURES AND FOLIATIONS 181
1
D D D df jW D W ¯
on is equal to ker = 0 0 . Therefore 0 =( ) ( ) ker and
1
df jW D D D D ¯ ¯ ¯
0 =( ) ( ) ker where and are invariant and anti-invariant N distributions, respectively, of the CR-submanifold W¯ of .
Since W¯ is totally geodesic, cf. [21], Prop. V.2.5, Theorem IV.6.2 N
of [21] assures that the submanifold W¯ of is a Riemannian product of
N N
N0 1 of a totally geodesic invariant submanifold 0 and a totally geodesic N
anti-invariant submanifold N1 of . Therefore it remains to prove that the
D T F W
foliations D and are Riemannian foliations of the submanifold .
D
The subbundle D0 is the orthogonal complement of , therefore the foliation
W D
D is Riemannian iff any a geodesic of which is tangent to 0 at one point
remains tangent to D0 at any point of its domain, cf. [22, 10]. Likewise
D T F
the subbundle D0 is the orthogonal complement of , therefore the
T F W
foliation D is Riemannian iff any a geodesic of which is tangent D
to D0 at one point remains tangent to 0 at any point of its domain.
W D x
Let us take a geodesic of which is tangent to 0 at one point .
Since f is a Riemannian submersion is a horizontal geodesic, i.e. tangent
f W D
to ker . Its image is a geodesic in ¯ , cf. [8], which is tangent to ¯ at
D
one point. As both distributions, D¯ and ¯ are totally geodesic, the geodesic
D
f remains tangent to ¯ throughtout its domain. The ker -orthogonal lift
f M W
passing through the point x of is a geodesic in and which is tangent
to D0 . Both geodesics, and , have the same tangent vector at the point
x , therefore they must be equal. Similar considerations are valid for the other distribution.
References Rend Mat [1] A Bejancu Umbilical CR-submanifolds of a Kahler¨ manifold, 15
(1980), 29–45.
U n O s J Di [2] D E Blair Geometry of manifolds with structural group ( ) ( ),
Geom 4 (1970), 155–167. S [3] J L Cabrerizo Some results on normal CR-submanifolds of -manifolds,
Bull Math Soc Sci Roum 36(84) (1992), 103–107.
Geometric Theory of Foliations [4] C Camacho A Neto Birkhauser,¨ Boston
1985.
C S [5] L Di Terlizzi On invariant submanifolds of -manifolds and -manifolds,
Acta Math Hungar 85 (1999), 229–239.
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S Ianus A M Pastore f [6] K Duggal Maps interchanging -structures and
their harmonicity, preprint 2000
S Port Math [7] L M Fermandez CR-products of -products, 47 (1990), 167– 181.
[8] R Hermann A sufficient condition that a map of Riemannian manifolds be a
fiber bundle, Proc AMS 11 (1960), 236–242.
f Stud [9] I Mihai CR-subvarietati ale unei -varietati cu repere complementare,
Cerc Mat 35 (1983), 127–136.
Riemannian Foliations Progress in Math [10] P Molino vol. 73, Birkhauser¨
1988. Global Analysis on Foliated Spaces [11] C C Moore Cl Schochet MSRI LN
no 9, Springer 1988.
S Stud Cerc [12] L Ornea Subvarietati Cauchy–Riemann generice in -varietati,
Mat 36 (1984), 435–443.
Ann of Math [13] B Reinhart Foliated manifolds with bundle-like metrics, 69
(1959), 119–132. Di erential Geometry of Foliations
[14] B Reinhart Springer Verlag 1983.
Foliations on Riemannian Manifolds and Submanifolds [15] V Rovenskii Birk-
hauser¨ 1998.
r G Suppl Rend Cir Mat Palermo [16] R A Wolak On -foliations, 6 (1984), 329–341.
[17] R A Wolak Foliations admitting transverse systems of differential equations,
Comp Math 67 (1988), 89–101.
[18] R A Wolak Foliated and associated geometric structures on foliated mani-
folds, Ann Fac Sc Toulouse 10 (1989), 337–360. Geometric Structures on Foliated Manifolds [19] R A Wolak Santiago de Com-
postela 1989.
G Manus Math [20] R A Wolak Foliated -structures and Riemannian foliations,
66 (1989), 45–59.
CR submanifolds of Kaehlerian and Sasakian Manifolds [21] K Yano M Kon
Progress in Math vol 30, Birkhauser¨ 1983. Yorozu [22] S Behaviour of geodesics in foliated manifolds with bundle-like
metrics, J Math Soc Japan 35 (1983), 251–272.
2019. majus´ 4. –23:07 ANNALES UNIV. SCI. BUDAPEST., 44 (2001), 183–198
CONVERGENCE OF MOVING AVERAGE PROCESSES DEDUCED BY NEGATIVELY ASSOCIATED RANDOM VARIABLES
By H.-Y. LIANG and JONG-IL BAEK
Department of Applied Mathematics, Tongji University, Shanghai P. R. China
School of Mathematics and Informational Statistics, Wonkwang University, South Korea
Received October Revised April
1. Introduction
fY i g
Let i be a sequence of identically distributed random
P
fa i g ja j i
variables and a sequence of real numbers with i
i =
.Put
X
a X Y k
i k i
= +k 1 i
=
fY i g
When i is a sequence of independent random variables,
there have been some authors who studied limit properties for the moving av-
fX k g Ibragimov
erage process k 1 . In particular, (1962) had established
fX k g Burton Dehling
the Central Limit Theorem for k 1 , and (1990)
fX k g E tY
had obtained large deviation priciple for k 1 assuming exp( 1)
Li for all t ,and et al. (1992) had obtained the following result on
complete convergence.
Theorem A Suppose fY i g is a sequence of independent
i
g be and identically distributed i i d random variables Let fX k
k 1
2 t
t Then EY and E jY j imply
de ned as above and 1 2 1 =0 1
n
X X
1 t
n P X
(1 1) k 0
k n =1 =1
2019. majus´ 4. –23:10 184 H.-Y. LIANG, JONG-IL BAEK
Recently, Zhang (1996) gave a general version of Theorem A under
identically distributed -mixing assumptions. Clearly, Theorem A implies
n
X X
1 t
P Y n
(1 2) i 0
i n =1 =1
While, by Kolmogorov’s law of iterated logarithm, we know
n
P
Y
i
i =1 lim sup = a.s.
1 2
n n t Therefore, (1.2) is not true for t = 2, further (1.1) does not hold for =2. The main aim of this note is to extend and generalize Theorem A to NA
random variables; discuss the result for t = 2 in NA setting, which had not
been settled by Li et al. (1992) in i.i.d. setting.
fX i n g
A finite family of random variables i 1 is said to be B
negatively associated (NA) if for every pair of disjoint subsets A and of
ng
f1 2 ,
f X i A f X j B j
Cov( 1( i ) 2( )) 0 f whenever f1 and 2 are coordinatewise increasing and such that the covariance
exists. An infinite family of random variables is NA if every finite subfamily Saxena
is NA. This definition is introduced by Alam and (1981) and care-
Proschan Block Savits
fully studied by Joag Dev and (1983) and , and
Joag Dev Proschan Shaked (1982). As pointed out and proved by and (1983), a number of well known multivariate distributions possess the NA property, such as (a) multinomial, (b) convolution of unlike multinomials, (c) multivariate hypergeometric, (d) Dirichlet, (e) Dirichlet compound multi- nomial, (f) negatively correlated normal distribution, (g) permutation distri- bution, (h) random sampling without replacement, and (i) joint distribution of ranks. Because of its wide applications in multivariate statistical analysis and
reliability, the notion of NA have received considerable attention recently. Proschan
We refer to Joag Dev and (1983) for fundamental properties, Matula Newman (1984) for the central limit theorem, (1992) for the three
series theorem, Su et al. (1997) for a moment inequality, a weak invariance principle and an example to show that there exists infinite family of non-
degenerate non-independent strictly stationary NA random variables, Shao
Liang Su
and Su (1999) for the law of the iterated logarithm, and (1999a) for
Su Liang convergence rates of law of the logarithm, Liang and (1999b) and
2019. majus´ 4. –23:10 CONVERGENCE OF MOVING AVERAGE PROCESSES 185
(2000) for complete convergence of weighted sums, Roussas (1994) for the central limit theorem of random fields, some examples and applications.
2. Main Result
fY i g
Here, let i be a sequence of identically distributed
EY fX k g
NA random variables (r.v.’s) with 1 =0and k 1 be defined as in
x x
section 1. Denote by L( )=max(1 log ).
Let h x be a slowly varying function as x and
Theorem ( ) 0
rt t
t h x is non decreasing when r IfE jY j h jY j
r 1 1 2 ( ) =1 1 ( 1 )
then
0
n
X X
t
r 2 1
n h n P X n
( ) k
k n =1 =1
2 r
If E Y L jY j Theorem Let r then there exsits
1 [ 1 ( 1 )]
such that
some 0 0 0
n
X X
r 2 1 2
n P X nL n
k ( ( ))
n
=1 k =1
h i
2 1
Theorem For E Y L jY j then
0 if 1 ( ( 1 )) 0
n
X X 1
1 2
P X nL n
k ( ( ))
n n =1 k =1
Remark Since i.i.d. r.v.’s are a special case of NA r.v.’s, Theorem
2.1 generalizes and extends Theorem A. Theorems 2.2–2.3 complement the Li results for t = 2 in NA setting, which had not been discussed by et al.
(1992) in i.i.d. setting.
Remark Gut fY g
(1980) conjectured that under i is a sequence of
h i
2 1
E Y L jY j
i.i.d. symmetric random variables, for 0, if 1 ( ( 1 )) ,
then 0,
n
X X 1
1 2
P Y n L n
i ( ( ( ))
n i n =1 =1
2019. majus´ 4. –23:10
186 H.-Y. LIANG, JONG-IL BAEK
a a j
Clearly, Theorem 2.3 extends and generalizes (taking 0 =1, j =0, 0)
Gut’s (1980) above conjecture.
3. Proof of Main Result
a b a O b C C q
In this section, means = ( ). and q ( 1) will represent
positive constants, their value may change from one place to another.
P
Lemma Burton Dehling Let a be an absolutely
( and (1990)). i i
=
P P
convergent series of real numbers a a b ja j Suppose i
= i =
i
i = =
b b is a function satisfying the following conditions
Φ:[ ] R
a
(i) Φ is bounded and continuous at
and C such that for all jx j j x j C jx j
(ii) There exist 0 0 Φ( )
Then
i n
+ X
1 X
A
a a
lim Φ j =Φ( )
n
n
i i
= j = +1
q
x jx j q
Remark Taking Φ( )= , 1, from Lemma 1 we have
q
i n
+
X X
1
q
a ja j
(3 1) lim j =
n
n
i j
= =i +1
Su Shao Su Let p and let
Lemma ( et al. (1997), and (1999)). 2
p
fX i g be a sequence of with EX and E jX j Then
i i
i 1 NA r.v.’s =0
there exist constant A and B such that p
0 p 0
p
p 2
n n n
X X X
2 p
E X A EX E jX j
p i
i +
i
i i
i =1 =1 =1
p
p 2
n k n
X X X
2 p
B X EX E jX j E
p i
max i +
i
k n
1
i i i =1 =1 =1
2019. majus´ 4. –23:10
CONVERGENCE OF MOVING AVERAGE PROCESSES 187
Lemma Shad Su Let fX j n g be mean zero
( and (1999)). j 1 NA n
P 2
nite variance Denote by B EX Then for any x
r.v.’s n = 0 0 j
j =1
and 0 1
P jS j x jX j P k
max k 2 max +
k n k n
1 1
2
x x
2 2
+ exp 1+ ln 1+
x B B n
1 2( + n ) 3
fZ i g Remark
If i ; is a sequence of identically
jZ j
distributed mean zero NA r.v.’s with E 1 , finite variance and
P
ja j fa i g i
i ; a sequence of real numbers with .Put i
=
P 2
B E ja Z j i
= i . From Lemma 3, we have
i =
m
X X X
A A A
x x a Z P P a Z x a Z P
i i i i i
i 2 +
i i m
i j m
= = j
2 X
E jZ j
2 x 1
P ja j ja Z j
i i
2 sup i + exp +
x B x
1 2( + )
i
ji j m
P P
jZ j
E 1
ja j m ja j i
Since i , 0, choose such that . Thus,
x
i
i j m = j
we get
X
A
P x a Z i
(3 2) i 2
i
=
2
2 x
P ja Z j i
2 sup i + exp
x B
1 2( + ) i
Proof of Theorem Note that
n n
X X X X
X a Y a Y
k i k i i
(3 3) = + = ni
i i k =1 = k =1 =
2019. majus´ 4. –23:10 188 H.-Y. LIANG, JONG-IL BAEK
It suffices to show that for every 0,
X X
r t
2 + 1
A
n h n P a n Y
(3 4) ( ) i
ni
i
n =1 =
X X
r t
2 1
A
n h n P a Y n
(3 5) ( ) i
ni
i n =1 =
+
a a a a ni
where = ni 0, =( ) 0. We prove only (3.4), the proof of (3.5)
ni ni
is analogous. Let Y
ni =
t t t t
1 t + 1 + + 1 1 + 1
I I n a Y n a Y I ja Y j n n a Y n
i i i
= ( i )+ ( )+ ( )
ni ni NI ni
Then
X X
r t
2 + 1
A
n h n P Y n a
( ) i
ni
i
n =1 =
X X
r t
2 1
A
n h n P n Y
( ) ni +
2
i
n =1 =
X X
+ 1 t
P h n ja Y j n I I
+ ( ) ( i =: +
ni 1 2
i
n =1 =
P +
n
From (3.1) we can assume, without loss of generality, that a ,
ni
i =
t t
+ 1 + 1
j a I fi Z j a g
1 and denote by nj = :( +1) . It is easy to
ni ni
verify from Lemma 1 that
k X
1 t
I Cn k
(3 6) # nj ( +1)
j =1
For I2,using(3.6)wehave
X X X
t
r 2
I n h n I k jY j k P
2 ( ) (# nj ) ( 1 +1)
k nj j
n =1 =1 =
k n
[ ]
X X X
t
r 2
n h n I P k jY j k
( ) (# nj ) ( 1 +1)
n k j n =1 = =1
2019. majus´ 4. –23:10
CONVERGENCE OF MOVING AVERAGE PROCESSES 189
X X
t t t
r 1 1 1
n h n n P k jY j k
( ) k ( 1 +1)
k n
n =1 =
rt t
E jY j h jY j [ 1 ( 1 )]
By EY1 =0,weget
P
EY
ni
i = t
t 1
jY j I jY j n n 2 E 1 ( 1 ) 0 as
1 t
n
Thus, to prove I1 , we need only to show that
X X
r t
2 1
A
I n h n P Y EY n ni
=: ( ) ( ni ) 0
1
i n =1 =
In fact, we use the Markov’s inequality for a suitably large M , which will be
n fY i g
determined later, Lemma 2 and note that for each 1, ni
is still a sequence of NA r.v.’s from the definition, we have
M 2
X X X
M r M t
2 2
A
E jY j E jY j I n h n n ni
1 ( ) ni + =
i i
n =1 = =
I
=: I3 + 4
P
2 q
r E jY j q ja j Cn
If 2, note that 1 and ni for 1, taking
i =
t r t
M 2 ( 1) (2 ), we get
h
X X
M t t t
r 2 2 + 1
Y j n n I n h n n P ja
( ) ( i )+
3 ni
i
n =1 =
M 2 i
+ 2 + 1 t
E ja I ja Y j Y j n i
+ i (
ni ni
X
t M
r 2 (1 1 2)
n h n ( )
n =1
2019. majus´ 4. –23:10
190 H.-Y. LIANG, JONG-IL BAEK
r rt s r s
If 1 2and1 2, then there exists some such that 1,
r s
taking M 2( 1) ( 1), we have
h
X X
t t M t
r 2 2 + 1
P n I n h n n ja Y j n
( ) ( i )+
3 ni
i
n =1 =
M i 2
+ 2 + 1 t
E ja Y j I ja Y j n i
+ i ( =
ni ni
M 2
2 t
Z
X X
M t
r 2 + 2
n h n P ja Y j x dx
= ( ) ( i )
ni
i
n =1 = 0
X
s M
r 2 ( 1) 2
n h n ( )
n =1
M I
If r = 1. Choose = 2. Similarly to the below proof of 4 ,weget
I3 .
As to I4,wehave
X X
t
r 2 + 1
Y j n P I n h n ja
( ) ( i )+
4 ni
i
n =1 =
X X
M t M t
r 2 + + 1
E ja n h n Y j I ja Y j n I I i
+ ( ) i ( )=: + ni
ni 5 6
i
n =1 =
I
From the proof of I2 we know 5 .
n
2
X X X
M t M t M t
r 2
I n h n I j E jY j I k jY j k
6 ( ) (# nj ) 1 ( 1 1 )+
j k
n =1 =1 =1
j n
( +1)
X X X
M t M t M t
r 2
h n n I j E jY j I k jY j k
+ ( ) (# nj ) 1 ( 1 1 )=
k n j
n =1 =1 =2 +1
I
=: I7 + 8
k
Note that for M 1and 1, we have
X
M t M t
( 1)
I j Cnk
(# nj ) k j =
2019. majus´ 4. –23:10
CONVERGENCE OF MOVING AVERAGE PROCESSES 191
rt
Hence, taking M ,weget
X X
M
r M t
1
t
I n n E jY j I k jY j k
7 h ( ) 1 ( 1 1 )
k k
=1 n =[ 2]
X
M
r M t
t
k k E jY j I k jY j k h ( ) 1 ( 1 1 )
k =1
rt t
E jY j h jY j
[ 1 ( 1 )]
M
( 1)
X X
t
M
k
M t
r 2
t
h n n n E jY j I k jY j k
I8 ( ) 1 ( 1 1 )
n
k n
n =1 =2 +1 k
[ 2]
X X
M
M t
r 1
t
h n n E jY j I k jY j k
( ) 1 ( 1 1 ) n
k =2 =1
rt t
E jY j h jY j
[ 1 ( 1 )]
Proof of Theorem We need only to prove that for 0 2,
X X
r
2 + 1 2
A
Y a n P nL n
(3 7) i ( ( ))
ni
i
n =1 =
X X
r
2 1 2
A
n P nL n a Y
(3 8) i ( ( ))
ni
i n =1 =
We give proof of (3.7), the proof of (3.8) is analogous. Let
p p
10
n L n nL n n
n = ( ) = ( ) N 2 4
(1) + + + +
Y I a a ja I a Y Y I Y j Y
n n n n n
i i i
= ( i )+ ( )+ ( )
ni ni ni ni ni
(2) + +
Y a I a Y Y
n n n i
=( i ) ( )
ni ni ni
(3) + +
Y Y Y a I a
n n n i
=( i + ) ( )
ni ni ni
(4) + + + +
Y Y Y Y Y a I a a I a
n n n n
i i i
=( i + ) ( )+( ) ( )
ni ni ni ni ni
where N is some large positive integer, which will be specified later on. Then
X X
r
2 + 1 2
A
Y n P nL n a
i ( ( ))
ni
i n =1 =
2019. majus´ 4. –23:10
192 H.-Y. LIANG, JONG-IL BAEK
X X
r (1)
2 1 2
A
Y n P nL n
( ( )) +
ni 4
i
n =1 =
X X
r (2)
2 1 2
A
P Y nL n
+ n ( ( )) +
ni 4
i
n =1 =
X X
r (3)
2 1 2
A
P Y nL n
+ n ( ( )) +
ni 4
i
n =1 =
X X
r (4)
2 1 2
A
Y P nL n J J J J
+ n ( ( )) =: + + +
ni 4 1 2 3 4
i
n =1 =
+ 1 2
a L
From (3.1), we can assume (2 (2)) , denote by
ni
1 2 + 1 2
I fi j L j a j L j g Z
nj = :(( +2) ( +2)) (( +1) ( +1))
ni
k P
1 2
I Cn k L k I
Note that # nj (( +2) ( +2)) . Similarly to the proof of 2 ,
j =1
we get
X X
r
2 + 1 2
J n P ja Y j nL n
i ( ( )) 4 ni
4 N
i
n =1 =
X X X
r 2 2
P n I jY j L jY j Cnj
(# nj ) ( 1 ( 1 ) )
nj k n
=1 j =1 =
k X
X 2
jY j
r
1 2 3 2 1 2 1
k n L k n P k k
( (3 )) +1
jY j
CL( 1 ) n k =1 =1
2
r L x Cx x k
Choose 0suchthat 1 2 , ( ) when 2 0 for some
k0 0. Hence,
k
Z
k
X
r
r 3 2 1 2 3 2 1 2
L k n x L k x dx n ( (2 )) ( (2 ))
n =1 1
2019. majus´ 4. –23:10
CONVERGENCE OF MOVING AVERAGE PROCESSES 193
k k
k Z
0 Z
r r r
3 2 3 2 1 2 1 2
x k x x L k dx k ( ) dx + ( (2 0))
1 k k0
Therefore,
X 2
jY j r
1 2 r
J k P k k jY j L jY j E
4 +1 [ 1 ( 1 )]
jY j CL( 1 )
k =1
2 r
r r E jY j Choose r0 such that 0 1, hence 1 0 . From the definition
(2) (2)
Y N r r
of Y , we know that 0, taking ( 1) ( 1), by the property ni ni 0
of NA, we have
X X
r 2 (2) 1 2
A
J n nL n = Y ( ( ))
2 ni 4
i
n =1 =
X
r 2 (2)
P i Y
n (there are at least N ’s such that 0) ni
n =1
N
X X
r 2 +
Y P n a n
( i )
ni
i
n =1 =
X
r N Nr
r 2 ( 0 1) 0
n L n ( ( ))
n =1
(3) 2 r
J EY E jY j Similarly, Y 0and .By =0and 0 ,
ni 3 1 1
P
+ 2 r
Cn
( a ) 0 ,wehave
ni
i =
X
(1)
1 2
EY n
( nL( ))
ni
i =
X
+ + + 1 2
Y j I jI Y j P ja E ja ja nL n
n n n
i i
[ ( i )+ ( )] ( ( ))
ni ni ni
i =
r r
( 0 1) ( 0 1)
n L n n 1 ( ( )) 0 az
2019. majus´ 4. –23:10
194 H.-Y. LIANG, JONG-IL BAEK
Therefore, to prove J1 , it suffices to show that
X X
r
(1) (1)
2 1 2
A
EY n P Y nL n
J =: ( ) ( ( ))
ni
1 ni 5
n =1 i =
(1)
fY i g
Note that for each n 1, is still a sequence of NA
ni
p p
10 1
nL n x nL n r.v.’s. Taking = ( ), = ( ), = and it is easy to verify 10 2
that
(1) (1) X (1) (1) 2 2
jY EY j B Y EY C nE Y E
sup 2 n = = ( )
ni ni ni
ni 0 1
i
i =
P 2
nC a where C satisfies 2. Hence, by using (3.2) we get
0 ni 0
i =
2
1 r
X
X 2
nL n
( ) 2
C EY
r
2 2 100 400(1+ ) A
0
n n J 1
1 4 exp 2 =4
C nE Y
2( n + 0 ) n
n =1 1 =1 q
2
r C EY here 0 =40 ( 1)(1 + 0 1 ).
Proof of Theorem Similarly to the proof of Theorem 2.2, we prove
only that
X X
1
+ 1 2
A
P nL n a Y
i ( ( )) 0
ni
n
i
n =1 =
We mas assume 1 and choose 0 such that . Denote by
1 2 (1 ) 2
n L n
n = ( ( )) ,
(1) + + + +
Y Y I Y j Y Y I a a ja I a
n n n n n
i i i
= ( i )+ ( )+ ( )
ni ni ni ni
ni
(2) + + 1 2
Y Y Y a I a nL n
n n i
=( i ) ( ( ))
ni ni ni
4 N
(3) + + 1 2
Y Y Y a I a nL n
n n i
=( i + ) ( ( ))
ni ni ni
4 N
(4) + + 1 2
Y Y Y a I a nL n
n i
=( i + ) ( ( )) +
ni ni ni
4 N
+ + 1 2
a I a nL n Y Y
n i
+( i ) ( ( ))
ni ni
4 N
2019. majus´ 4. –23:10 CONVERGENCE OF MOVING AVERAGE PROCESSES 195
where N is some large positive integer, which will be specified later on. Then
X X
1
+ 1 2
A
P a Y n L n
i ( ( ( )) )
ni
n
n i
=1 =
X X
1
(1)
1 2
A
Y nL n
P ( ( )) +
ni
n 4
i
n =1 =
X X
1
(2)
1 2
A
Y nL n
+ P ( ( )) +
ni
n 4
i
n =1 =
X X
1
(3)
1 2
A
P nL n
+ Y ( ( )) +
ni
n 4
i
n =1 =
X X
1
(4)
1 2
A
nL n Q Q Q Q Y P
+ ( ( )) =: + + +
ni 1 2 3 4
n 4
i
n =1 =
Q From (3.1), similarly to the proof of J4 we can get 4 .Fromthe
(2) (2)
Y N
definition of Y we know that 0, hence taking 1 ( )and
ni ni
P
I j n I nj
noticing that (# nj ) for 0 (the definition of is as in the
j =1
proof of Theorem 2.2),
X X
1
(2) 1 2
A
Y P nL n Q = ( ( ))
2 ni
n 4
i
n =1 =
N
X
X 1 +
a P Y n
( i )
ni
n
i
n =1 =
X
X 2
1 Y
1
I P
(# nj )
1
n
jY j
( L( 1 )) j
n =1 =1
N
1
CnL j L j
( n )( +1) ( +1)
1 1
nL n L j L j ( L( ( )) + (( +1) ( + 1)))
2019. majus´ 4. –23:10
196 H.-Y. LIANG, JONG-IL BAEK
h X
X 1
1 ( )
I nj L j L n
(# nj ) ( ( +1)) ( ( )) +
n j
n =1 =1
i
1 (1 )
j L n
+(nj L ( +1)) ( ( ))
i h
X 1
N N
( ( ) (1 )
L n L n
( ( )) +( ( )) n
n =1
(3) 2 1
Y Q EY E Y L jY j Similarly, 0and .By =0and [ ( ( )) ]
ni 3 1 1 1
,weget
X
(1)
1 2
EY nL n
( ( ))
ni
i =
1 X
1 2 (1 ) 2 + 1 2 (1 ) 2
n L n P ja Y j n L n
[ ( ) ( i ( ))+ ni
1 2 n
( nL( ))
i =
+ + 1 2 (1 ) 2
E ja ja L n Y jI Y j n i
+ i ( ( ))]
ni ni
1 X
1 2 (1 ) 2 2 1
I n L n P Y nL n j L j (# nj )[ ( ) ( 1 ( )( +1) ( + 1))+
1 2 n ( nL( ))
j =1
1 2 2 1
L j E jY jI Y nL n j L j
+((j +1) ( +1)) 1 ( 1 ( )( +1) ( +1))
( 2) (1 2)
L n L n n
( ( )) +( ( )) 0as
Thus, to prove Q1 , it suffices to show that
X X 1
(1) (1)
1 2
A
Q P Y nL N EY
=: ( ) ( ( )) 0
ni
1 ni
n
i
n =1 =
n fY i g
Since, for each 1, ni remains a sequence of NA r.v.’s,
f g
using Lemma 2, choose p max 2 2(1 ) +2 we have
p 2
X X
1 X
p p
2 (1) 2 (1)
A
E jY j E jY j nL n
Q ( ( )) + =: ni
1 ni
n
i i
n =1 = =
Q =: Q5 + 6
2019. majus´ 4. –23:10 CONVERGENCE OF MOVING AVERAGE PROCESSES 197
While
Q5
X
1 X
p
2 1 + 1 2 (1 ) 2
nL n nL n P ja Y j n L n
( ( )) [ ( ) ( i ( ))+
ni
n
i
n =1 =
p 2
+ 2 + 1 2 (1 ) 2
E ja I ja L n Y j Y j n i
+ i ( ( ))]
ni ni
X
1 X
p
2 1 2 1
nL n I nL n P Y CnL n jL j
( ( )) (# nj )[ ( ) ( 1 ( ) ( ))+
n n
=1 j =1
p 2
1 2 1 1 2 1
I j E Y L jY j L jY j Y CnL n jL j
+( jL( )) ( 1 ( ( 1 )) ) ( ( 1 )) ( 1 ( ) ( ))]
X 1
p p
2 2
L n L n
[( ( )) +( ( )) ] n
n =1
X
X 1
p
2 + p + 1 2 (1 ) 2
Q nL n E ja ja L n Y j I Y j n i
( ( )) [ i ( ( ))+ ni
6 ni
n
i
n =1 =
p
p 2 (1 ) 2 + 1 2 (1 ) 2
n L n P ja Y j n L n
+ ( ( ) ( i ( ))]
ni
X
1 X
p p p
2 2
nL n I jL j E jY j
( ( )) (# nj )[( ( )) 1
n n
=1 j =1
p p
2 1 2 (1 ) 2 2 1
I CnL n jL j n L n P Y CnL n jL j
( Y1 ( ) ( )) + ( ) ( 1 ( ) ( ))]
X X 2
1 Y
p p
2 2
L n I jL j E
( ( )) (# nj ) ( ( ))
1
n
jY j
( L( 1 )) j
n =1 =1
p p
p 2 1 2 1 2 (1 ) 2
j L jY j I Y CnL n jL j n L n Y
j 1 ( ( 1 )) ( 1 ( ) ( )) + ( )
2 1
Y
n jL j
1 CnL ( ) ( )
P
1 1
jY j
( L( )) 1
nL n L jL j
1 L ( ) + ( ( ))
n o
X
p p
1 +( ) +(1 )
L n L n
( ( )) 2 +( ( )) 2 n
n =1
2019. majus´ 4. –23:10 198 H.-Y. LIANG, JONG-IL BAEK
Acknowledgement This research was partially supported by the Na- tional Natural Science Foundation of China (No. 10171079) and No. 2000- 2-10400-1001-3 from the Basic Research Program of the KOSEF as well as Wongwang University Grant in 2002.
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2019. majus´ 4. –23:10
INDEX
Jong Il Liang H Y Baek : Convergence of moving average processes
deduced by negatively associated random variables 183
Bu Qingying: A characterization of adjoint 1-summing operators 107
ia A Hernandez Cifre Mar : Area-inradius and diameter-inradius relations
for covering plane sets 95
Datta S Mathur P : Modified weighted (0; 0 2)-interpolation on infinite
interval ( + ) 39
r r Datta S Mathur P : On modified weighted (0 1 2 )-interpolation
on an arbitrary system of nodes 53
Datta S Mathur P : A note on weighted (0 1 3)-interpolation on infinite
interval ( + ) 65
o Istvan Havasi Agnes Zlatev Zahari Dimov Ivan Farag :
L-commutativity of the operators in splitting methods for air pollution
models 129
o Istvan Dimov Ivan Havasi Agnes Zlatev Zahari Farag :
L-commutativity of the operators in splitting methods for air pollution
models 129
Agnes Dimov Ivan Farago Istvan Zlatev Zahari Havasi :
L-commutativity of the operators in splitting methods for air pollution
models 129
us Csaba J
Heged : Short proofs for the pseudoinverse in linear algebra 115
ari Norbert Hegyv : Thin complete subsequence 151
Huo H F Li W : Interval oscillation criteria for second order forced nonlin-
ear differential equations with damping 83
Kemprasit Yupaporn: Infinitely many bi-ideals which are not quasi-ideals
in some transformation semigroups 123 Kiss G : Extremal Frobenius numbers in some special cases 27
2019. majus´ 4. –23:07
aszlo Peter Radu Varga Csaba Kozma L : Warped product of Finsler
manifolds 157
Li W Huo H F : Interval oscillation criteria for second order forced nonlin-
ear differential equations with damping 83
H Y Baek Jong Il Liang : Convergence of moving average processes
deduced by negatively associated random variables 183
Mathur P Datta S : Modified weighted (0; 0 2)-interpolation on infinite
interval ( + ) 39
r r Mathur P Datta S : On modified weighted (0 1 2 )-interpolation
on an arbitrary system of nodes 53
Mathur P Datta S : A note on weighted (0 1 3)-interpolation on infinite
interval ( + ) 65
aszlo Varga Csaba Peter Radu Kozma L : Warped product of Finsler
manifolds 157 S Ramadan M : The influence of -quasinormality of some subgroups of prime
power order on the structure of finite groups 3
Szili L : On the “good” nodes of weighted Lagrange interpolation for expo-
nential weights 33
Terlizzi Luigia Di Konderak Jerzy Pastore Anna Maria Wolak
Robert
: K-structures and foliations 171
aszlo Peter Radu Varga Csaba Kozma L : Warped product of Finsler
manifolds 157
any Sz Vincze Cs Vattam : Two-dimensional Landsberg manifolds with
vanishing Douglas tensor 11
any Sz Vincze Cs Vattam : Two-dimensional Landsberg manifolds with
vanishing Douglas tensor 11
o Istvan Havasi Agnes Zlatev Zahari Dimov Ivan Farag :
L-commutativity of the operators in splitting methods for air pollution
models 129
ISSN 0524-9007 Address: MATHEMATICAL INSTITUTE, EOTV¨ OS¨ LORAND´ UNIVERSITY ELTE, TTK, KARI KONYVT¨ AR,´ MATEMATIKAI TUDOMANY´ AGI´ SZAKGYUJTEM˝ ENY´ PAZM´ ANY´ PETER´ SET´ ANY´ 1/c, 1117 BUDAPEST, HUNGARY Muszaki˝ szerkeszto:˝ Dr. SCHARNITZKY VIKTOR Akiadas´ ert´ felelos:˝ az Eotv¨ os¨ Lorand´ Tudomanyegyetem´ rektora Akezirat´ a nyomdaba´ erkezett:´ 2002. majus.´ Megjelent: 2002. junius´ Terjedelem: 18,3 A/4 ´ıv. Peld´ anysz´ am:´ 500 Kesz´ ult¨ az EMTEX szedoprogram˝ felhasznal´ as´ aval´ az MSZ 5601–59 es´ 5602–55 szabvanyok´ szerint Az elektronikus tipografal´ as´ Juhasz´ Lehel es´ Fried Katalin munkaja´ Nyomda: Haxel Bt. Felelos˝ vezeto:˝ Korandi´ Jozsef´
2019. majus´ 4. –23:07