C.Ü. Fen-Edebiyat Fakültesi
Fen Bilimleri Dergisi (2008)Cilt 29 Sayı 2
Some Estimates on Whitney Inequality for Differentiable Functions
Tuncay TUNÇ
Mersin University, Department of Mathematics, 33343, Mersin/TURKEY E-mail: ttunc77@hotmail.com
Received: 11.08.2008, Accepted: 06.11.2008
Abstract: In this study, we are interested in finding some estimates of the constants W k r( ), ( ,k r∈¥), in the well-known Whitney Inequality for differentiable functions on the closed interval [ ]−1,1 :
[ ]
(
)
( ) ( )[ ]
1 2 2 , 1,1 , , , 1,1 r r k r k E f W k r f k ω k + − − ≤ − .Key Words: Whitney Inequality, divided differences, interpolation.
Diferansiyellenebilir Fonksiyonlar için Whitney Eşitsizliği Üzerine Bazı Sonuçlar
Özet: Bu çalışmada,
[ ]
−1,1 kapalı aralığı üzerinde diferansiyellenebilir fonksiyonlar için Whitney Eşitsizliği olarak bilinen:[ ]
(
)
( ) ( )[ ]
1 2 2 , 1,1 , , , 1,1 r r k r k E f W k r f k ω k + − − ≤ − eşitsizliğindeki W k r( ) (, , k r, ∈¥ , sabitleri için üst sınırların bulunması üzerinde durulmuştur. )
1. Introduction and Main Results
Let ¥ denote the set of natural numbers, ¥0:= ∪¥
{ }
0 . We denote by 0, n n∈¥
P , the space of algebraic polynomials of total degree at most n, by C a b
[ ]
, the space of the real valued continuous functions on the closed interval[ ]
a b, equipped with the uniform norm:[ ], : max[ ],
( )
C a b x a bf f x
∈
=
and by Cr
[ ]
a b, , r∈¥0, the set all r−times continuously differentiable functions[ ]
,f ∈C a b ; 0
[ ]
[ ]
, : ,
C a b =C a b . The deviation of f ∈C a b
[ ]
, from P is defined by n[ ]
(
, ,)
: inf [ ], n n n n C a b P E f a b f P ∈ = − P .The purpose of the paper is to estimate the constants W k r
( )
, , k r, ∈¥, in the well known Whitney Inequality: If f ∈C a b[ ]
, , f ∈Cr[ ]
a b, , then[ ]
(
)
( )
( )[ ]
1 , , , , , , r r k r k b a b a E f a b W k r f a b k ω k + − ≤ − − where(
[ ]
)
[ , ]( )
, , , sup sup k k h h t x a b kh t g a b g x ω 0< ≤ ∈ − = ∆is the k−th modulus of smoothness of the function g, and
( )
( )
(
)
0 1 k k j k h j k g x g x jh j − = ∆ = − + ∑
is an k−th finite difference of g.Many mathematicians have studied to estimate the Whitney constants: see, say, [1-8] for the references. Burkill [1] obtained the only known precise result:
(2, 0) 1 2
W = . Whitney [2] proved that 1 2≤W k
( )
, 0 < ∞ for each k∈¥ and gavenumerical estimates for W k
( )
, 0 when k≤5. In 1982, Sendov [3] conjectured that( )
, 0 1W k ≤ for all k . However, this conjecture has been proved only for "small" k ’s:
Whitney [2] for k=3, Kryakin [4] for k=4 and Zhelnov [5-6] for k=5, 6, 7,8. In general case, the most recent result is due to Gilewicz, Kryakin and Shevchuk [7] who
( )
2 , 0 2 1W k ≤ + e .
It follows from Lemma 3 in [8] that
( ) ( )
,1 1 k ,W k ≤ eσ k∈¥
where σk = +1 1 2 ... 1+ + k. For r=2, 3, 4, the estimates of W k r
( )
, were obtained in [9]:( )
1 , , r k r r W k r k eσ + − ≤ ∈ ¥.Besides, in [9], the following estimates of W k r( , ) are obtained:
( )
( ) 2 1 2 1 1 1, , , !2 r cos r W r r r + π+ ≤ ∈¥( )
* 2 2 * 1 2, , !2 cosr r W r r r π ≤ ∈¥,where r* : 2= «©
(
r+1 2)
®¬+1, where§ ¨
a stands for the integral part of the number a. The main results of the paper are the following.Theorem 1. For any f ∈Cr
[ ]
−1,1 , there is a polynomial Pk r+ −1∈Pk r+ −1 such that( )
( )
(
2)
( )
(
( ))
1 3 1 2 , ,[ 1,1] . 2 ! ! 2 k r k r k k r k k f x P x x x k f k r ω + − + − ≤ − Π + − l[ ]
1,1 x∈ − , where l «:=©(
r+1 2)
¨ and( )
(
)
0 1 2 k j x = x j k Π =∏
+ − . Theorem 2. We have( )
(
)
1 1 1 , ! k k W k r r eσ + + + ≤ l l l if 1 2 1 k> l+ + .In Section 2, will be given some relevant facts on divided differences, and in Section 3, we shall prove the Theorems 1 and 2.
2. Some Relevant Facts
In this section we shall give some auxiliary facts and notations which we will need in the proofs of the theorems.
Let k∈¥ and { }yj kj=0be a collection of distinct points yj∈[ , ]a b . Recall, the divided difference of a function g:[ , ]a b →¡ at the points {yj}kj=0 is defined by
[
]
(
)
0 1 0 0, ( ) , , , ; . k j k k j j i i i j g y y y y g y y = = ≠ = −∑∏
LBy the definition, it can be easily seen that the equality
[
0, 1, , ; (· , )]
[ 0, 1, , ; (·, ) ]d d
k k
c y y y h y dy = y y y c h y dy
∫
L L∫
(2.1)holds for any continuous function h defined on the rectangle R: [ , ] [ , ]= a b × c d .
Denote by L x g y y
(
; ; 0, 1,L,yk)
the Lagrange interpolation polynomial of degreek
≤ that interpolates the function g at the points yj, j=0,k. Then, as well known
(
0 1)
[
0 1]
(
)
0 ( ) ; ; , , , , , , , ; . k k k j j g x L x g y y y x y y y g x y = − L = L∏
−Now, let n∈¥ and { } 0 n i i
x = be a collection of points xi∈[ , ]a b that may coincide. Let 0
{ }yj kj= be a collection of distinct points yj∈[ , ]a b such that each of n+1 points x i coincides with one of the points yj. Let a point yj coincides exactly with sj pointsx , i
then the number pj = −sj 1 is called multiplicity of the point yj. Clearly,
0 1 k j j= s = +n
∑
, that is 0 k j j= p = −n k∑
. Let a function g∈C a b[ , ] have pj first derivatives at a neighborhood of each point yj. The generalized divided difference of order n of the function g at the pointsx , i i=0,1,...,n, is defined by
[
]
[
]
0 1 0 1 0 1 0 0 1 1 , , , ; : , ,..., ; . ! k n k k n p p p k j j k x x x g y y y g p y y y − = ∂ = ∂ ∂ ∂ ∏
L LFor n=0, set
[
x g0;]
:=g x( )
0 . The generalized divided differences possess the same properties as the ordinary divided differences. Say, if x0 ≠xn, then
(
)
1 2 0 1 1 0 1 0 [ , , , ; ] [ , , , ; ] [ , , , ; ] n n , n n x x x g x x x g x x x g x x − − = − L L L (2.2)and let L x g x x
(
; ; 0, ,1 L,xn)
be the Hermite-Lagrange interpolation polynomial of degree ≤n, that interpolates the function g at the points y y, ,L,y and interpolatesL( )s
(
x g x x; ; 0, ,1 L,xn)
=g( )s (yj), j=0, ,k s=0,pj where (0) ( ) : ( ) g x =g x , then(
0 1)
[
0 1]
(
)
0 ( ) ; ; , ,..., , , , , ; . k k k j j g x L x g x x x x x x x g x x = − = L∏
− (2.3)The following lemma which proved in [9] enables to generalize the Lemma 3 of Zhuk and Natanson in [8].
Lemma 1. Let r0∈¥ , n0 ∈¥, r0 ≤n and
{ }
xi ni=0 be an arbitrary collection of points[ , ] i
x ∈ a b . If a function f ∈C a b[ , ] has the r0−1-st absolutely continuous derivative on
[ , ]a b , then
[ , ,x x0 1 L,xn; ] [ ,f = x xr r+1,L,xn;fr], (2.4)
holds for each r=0,1,L , where ,r0 f x0( ) := f x( ),
1 1( ) : 0 ( (1 ) 0) f x =
∫
f xt′ + −t x dt and, forr>1, 1 1 1 ( ) 1 1 1 0 1 0 0 0 ( ) : ( ( ) (1 ) ) . r t t r r r r r r r f x f xt t t x t x dt dt − − − =∫ ∫ ∫
L + − + + −L L 3. Proofs of TheoremsThroughout this section, [ , ] : [ 1,1]a b = − , l «:=©
(
r 1 / 2+)
¬® where§ ¨
a stands for the integral part of a, and( 1) , , 1, 2, , 2 2 1 , 0,1, , . s j j k s s x j j k k − = + = = − + = L l L
To shorten notation, we write ‖ ‖ and g ωk( )g instead of ‖ ‖g C[ 1,1]− and (2 / , ,[ 1,1])
k k g
ω − , respectively.
Prof of Theorem 1. Let Lk+2l be the Hermite-Lagrange interpolation polynomial of degree ≤ +k 2l, which interpolates the function f at the pointsx x0, ,1L,xk+2l. By Newton's Formula, the coefficients of k 2
x +l and k 2 1
x + −l in the polynomial Lk+2l are Ak+2l=[ , ,x x0 1 L,xk+2l; ],f
2 1 0 1 2 1 2 0 1 2 1 0 1 2 2 2 [ , , , ; ] [ , , , ; ] [ , , , , ; ] , 2 k k k k k k A x x x f A x x x f x x x x f + − + − + + − + − + = + + = l l l l l l L L L respectively.
Consider the polynomial
2 2 1 1( ) 2 ( ) 2 1 2 ( ) (2 ) 2 2 2 1( ), 2 2 k k k r k k k k k A A P+ − x =L+ l x − + −+ll T+ l x − l−r + −+ −ll T+ −l x
of degree ≤ + −k r 1, where T xn( )=cos
(
narccosx)
is the n-th Chebyshev polynomial. The polynomial Pk r+ −1 is the desired one in Theorem 1. Let r be odd, i.e.r=2l−1. Since Tn =1, we conclude that( ) 1( ) ( ) 2 ( ) 22 1 22 21 :1 2 3. 2 2 k k k r k k k A A f x −P+ − x ≤ f x −L+ l x + + −+ll + + −+ −ll = + +i i i
First we estimate i2+i3. By using (2.2), (2.4) and (2.1), we obtain
(
)
1 1 2 0 1 2 2 2 0 1 2 1 1 0 1 0 1 1 0 0 0 1 [ , , , , ; ] [ , , , ; ] 2 1 ([ , , , ; ] , 2 r k k k k t t k r A x x x x f x x x f x x x g g dt dt − + = + − + − + − =∫ ∫ ∫
− l L l l L l L L Land similar arguments provide 1 1 1 2 1 0 1 0 1 1 0 0 0 1 [ , , , ; ] 2 r t t k k r A x x x g g dt dt − + −l =
∫ ∫ ∫
L L + L where 1 ( ) 1 1 2 ( ) : (1 ) 2 ( 1) (1 )( 1) r r j i i r j j g x f x t t t t − = = + − − + + − − ∑
, i=0,1. Since, for bothi=0,1,[
]
( )
( )
( )
0 1 2/ 0 ( ) ( ) , , , ; 2 ! 2 ! 2 , ,[ 1,1] 2 ! 2 ! k k k k i k k i k k i k k r r k r k k k k k x x x g g x g k k k k t f f k k k ω ω ω = ∆ ≤ = − ≤ L Then( )
( )
1 1 2 2 1 2 3 2 1 2 2 1 2 2 3 r 3 k k k k t t k k A A i i k −ω k ω + + − + − + − + = + ≤∫ ∫ ∫
= l l l l L LLet us now estimate i . By using 1 (2.2), (2.3), (2.4) and (2.1), we obtain
(
)
[
]
(
)
(
[
]
[
] [
]
)
2 2 0 1 2 2 0 1 2 4 2 2 2 0 1 2 2 0 1 2 3 2 1 ( ) ( ) 1 ( ) , , , , ; 1 ( ) , ,..., , , , ; 4 2 , ,..., , ; , ,..., , , ; k k k k k k k k f x L x x x x x x x f x x x x x x x x f x x x x f x x x x x f + + + − + − + + − + − + − − = − Π − Π = − + l l l l l l l l l l L( )
2 1 1 1[
]
0 1 2 3 4 1 0 0 0 1 ( ) , ,..., ; 2 . 4 r t t k r x x x x x g g g dt dt − − Π =∫∫ ∫
− + l L L where g ui( )= f( )r(
utr−2(tr−1−tr−2+ +L t4)+ −(1 x t)3+ai)
2, 3, 4i= , where a2 = − −1 (1 x t)2, a3 = −1 2t1+ +(1 x t)2 and a4 = +(1 x t)2−1. Therefore, as in the estimation ofi2+i3, we obtain
(
)
( )
( )
2 ( ) 1 2 1 ( ) ( ) , 2 ! ! k r k k k k x x i f x L x f k r ω + − Π = − ≤ l lwhich completes the proof for the case r is odd, but the same conclusion can be drawn for the case r is even, in this manner, Theorem 1 is proved.
The following lemma will be needed in the proof Theorem 2. Set h=2 /k, and recall, the logarithm with base a> ≠0 ( 1), is defined by loga x: log / log= x a,x>0. Lemma 2. Let k ≥2. For l+ <1 log (2 k−1), the equality
(
2)
(
2)
[ 1,1] [ 1, 1 2 ] max 1 ( ) max 1 ( ) x∈ − x − Π x =x∈ − − + h x − Π x l l holds.Proof. For− + ≤ ≤ −1 h y h/ 2, consider the function
(
)
(
)
(
)
2 2 2 ( ) 1 ( ) (1 ) (2 ) ( ) : 1 . (1 ) 1 1 ( ) y h y h y h h y h H y y y y y + − Π + + + + = = − − − − Π l l lSince H(−h/ 2)=1, H′ −( h/ 2)>0 and H′ has only one zero in [ 1− + −h, h/ 2] forl≤ −(k 2)(k+1) / (2 )k , it is sufficient to show that H( 1− + ≤h) 1. Indeed,
( 1 ) 2 2( 2) 1, 1 1 k H h k k − − + = ≤ − − l
for l+ <1 log (2 k−1). The proof is complete, since log (2 k− − < −1) 1 (k 2)(k+1) / (2 )k , for allk≥2.
Proof of Theorem 2. In order to prove Theorem 2 it is enough to check the inequality
(
)
1 2 3 1 ( 1) 1 ( ) , 2 ! ! 2 ! k r k r k r k k k x x k r r eσ + + + + + + − Π + ≤ l l l l (3.1)[ ]
1,1x∈ − . First, we prove the estimate
( )
(
)
(
)
2 1 1 2 1 ( ) 1 4( 1) 1 4( 1) max , , 2 ! 2 / 2 / k k k k x x k k ek σ k e ek σ k + + − − Π + + ≤ + − + l l l l l l l (3.2)for all x∈ −[ 1,1] and l+ <1 log (2 k−1). Let Ck,l denote the right hand of (3.2). If − < < − +1 x 1 h and u=k x( +1) / 2 then 0< <u 1 and
(
)
(
)
( )(
)
2 1 2 1 1 1 2 1 1 1 1 / 2 1 1 1 1 ( ) 2 1 1 1 1 2 ! 1 2 1 / 2 1 2 1 4( 1) . 2 / k k k k k k u k x x u u u u u k k k k k u k u k k u e k ek k σ σ σ + + + − + + + + + + − + + + + − Π = − − − − − + ≤ − + + ≤ ≤ + l l l l l l l l l l l l l l L l l l lOn the other hand, applying similar arguments to the case − + < < − +1 h x 1 2h, and using Lemma 2, we obtain (3.2).
Now, taking into account (3.2) and the inequality !k ≥k ek −k 2πk which follows from Stirling's formula, we get
, 3 ! ( , ) . 2 4 2 r k r k k r k e k r W k r C k π + ≤ + l
It is easy to check that 1 , 3 ( 1) , 2 4 2 r k r k k r k k e k k C e k σ π + + + + + ≤ l l l l forl+ <1 log (2 k−1). Thus, Theorem 2 is proved.
References
[1] H. Burkill, H., Proc. Lond. Math. Soc., 1952, 3, 150-174. [2] H. Whitney, J. Math. Pures Appl., 1957, 36, 67-95. [3] Bl. Sendov, C. R. Acad. Bulgare. Sci., 1982, 35, 1-11.
[4] Yu. V. Kryakin, Izv. Ross. Akad. Nauk Ser. Mat. 1997, 61(2), 95-100. [5] O.D. Zhelnov, East J. Approx., 2002, 8(1), 1-14.
[6] O.D. Zhelnov, Whitney inequality and its generalization, (Dissertation), Inst. of Math.Nat. Ac. of Sci. of Ukraine, Kiev, 2004, p. 129.
[7] Z.J. Gilewicz, Yu.V. Kryakin, and I. A. Shevchuk, J. Approx. Theory, 2002, 119, 271-290.
[8] V.V. Zhuk, G.I. Natanson, Vestnik Leningrad Univ., 1984, 1, 5-11.
[9] T. Tunc, Methods of Functional Analysis and Topology, 2007, 13(1), 95-100, [10] R.A. DeVore, G.G. Lorentz, Constructive Approximation, Springer-Verlag, Berlin Heidelberg, 1993, p.452.
