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Ul’yanov Type Inequalities For Moduli Of Smoothness

Sadulla Jafarov

y

Received 8 May 2012

Abstract

LetT denote the interval[ ; ]. In this work we investigate the inequality of Ul’yanov type for moduli of smoothness of an integer order in theLp(T); p 1 spaces. In particular, we study(p; q) inequalities for moduli of smoothness of a derivative of a function via the modulus of smoothness of the function itself.

1 Introduction

Letf be2 -periodic and letf 2Lp[0;2 ] =Lp forp 1. Throughout this work,k kp

will denote theLp-norm and will be de…ned by

kfkp= 8<

: 1 2

Z2 0

jf(x)jpdx 9=

;

1=p

; f2Lp; 1 p <1:

The modulus of smoothness!k(f; )pof a functionf 2Lp;1 p 1, of fractional orderk >0are de…ned by

!k(f; )p= supjhj khf(x) p (1) where

k hf(x) =

X1

=0

( 1) k

f(x+ (k )h); k >0:

Note that, the following(p; q)inequalities between moduli of smoothness, nowadays called Ul’yanov-type inequalities, are known:

!k f(r);

q C

0

@Z

0

u !k+r(f; t)p q1 du u

1 A

1=q1

; (2)

Mathematics Sub ject Classi…cations: 26D15, 41A25, 41A63, 42B15.

yDepartment of Mathematics, Faculty of Art and Sciences, Pamukkale University, 20017, Deni- zli, Turkey; Mathematics and Mechanics Institute, Azerbaijan National Academy of Sciences, 9, B.Vahabzade St., Az-1141, Baku, Azerbaijan

221

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where

r2N[ f0g; 0< p < q 1; = 1 p

1 q; q1= q ifq <1;

1 ifq=1:

In the case r= 0; p 1 the inequality (2) was proved by Ul’yanov [15]. In other cases, (p; q) estimates (the modulus of smoothness!k(f; )p of an integer order, the r-the derivative, r 2 N and the fractional derivative of order r >0 of the function) were obtained in references [3], [4], [14].

Note that the inequality between moduli of smoothness of various orders in di¤erent metrics was investigated by [6].

We denote byEn(f)p the best approximation off 2Lp(T)by trigonometric poly- nomials of degree not exceedingn, i.e.,

En(f)p:= infTn2 nkf Tnkp; n= 0;1;2; ::::;

where n denotes the class of trigonometric polynomials of degree at mostn.

LetWpr[0;2 ] =Wpr;(r= 1;2; :::)be the linear space of functions for whichf(r 1) is absolutely continuous andf(r)2Lp(T); p >1. It becomes a Banach space with the norm

kfkWpr :=kfkp+ f(r)

p: Letf 2Lp. For >0, theK-functional is de…ned by

K ; f;Lp; Wpr := inf kf kp+ (r)

p: 2Wpr : Let1< p <1. We de…ne an operator on Lp(T)by

( hg)(x) := 1 2h

Zh h

g(x+t)dt; 0< h < ; x2T:

Thek-modulus of smoothness k(; g)p;(k= 1;2; :::), ofg2Lp(T)is de…ned by

k( ; g)p := sup0<hi<

1 i k

Yk i=1

(I hi)g

Lp(T)

; >0; (3)

where Iis the identity operator [1], [5], [7].

In the case ofk= 0we set k( ; g)p:=kgkLp(T)and ifk= 1we write ( ; g)p:=

1( ; g)p.

It can be shown easily that the modulus of smoothness k(; g)pis a nondecreasing, nonnegative, continuous function satisfying the conditions

lim !0 k( ; g)p = 0;

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k( ; f+g)p k( ; f)p+ k( ; g)p forf; g2Lp(T),

f (f) := a0(f)

2 +

X1 k=1

(ak(f) coskx+bk(f) sinkx) (4)

is the Fourier series of the functionf 2L1(T):

Then-thpartial sums and de La Vallée-Poussin sum of the series (4) are de…ned, respectively, as

Sn(x; f) :=a0(f)

2 +

Xn k=1

(ak(f) coskx+bk(f) sinkx)

and

Vn(f) :=Vn(x; f) := 1 n

2nX1

=n

S (x; f):

The following Lemma holds.

LEMMA 1. Forf 2Lp;1 p 1;andk= 1;2; :::we have

c1(p; k) k

1 n; f

p

n 2k Vn(2k)(f; x)

p+kf(x) Vn(f; x)kp

c2(p; k) k 1 n; f

p

:

PROOF. Considering reference [7], the inequality

k

1 n; Tn

p

c3(p; k)n 2k Tn(2k)

p (5)

holds, where Tn is a trigonometric polynomial of order n. Using the properties of smoothness k(; f)p [5 ], [7] and (5), we have

k

1 n; f

p

c4(p; k) k 1 n; Tn

p

+kf Tnkp

!

c5(p; k) n 2k Tn(2k)

p+kf Tnkp : By reference [7] the Jackson inequality

En(f)p c6 k

1 n+ 1; f

p

; k= 1;2; :::; (6)

holds, with a constantc6>0 independent ofn.

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Note that, to estimate k

1 n; f

p

from below we shall use the following inequality in [7]

n 2k Tn(2k)

p c7(p; k) k 1

n; Tn p: (7)

LetVn(f)be de La Vallée-Poussin sum of the series (4).

We denote byTn(x; f)the best approximating polynomial of degree at most nto f in Lp(T). In this case, from the boundedness ofVn inLp(T);we obtain

kf Vn(f)kp kf(x) Tn(x; f)kp+kTn(x; f) Vn(x; f)kp

c7(p)En(f)p+kVn(x; Tn(x; f) f(x))kp

c8(p; k)En(f)p: (8)

Using (7) and (8) we reach

n 2k Vn(2k)(x; f)

p+kf(x) Vn(x; f)kp

c9(p; k) k

1 n; Vn

p

+En(f)p

!

c10(p; k) k 1 n; f

p

+ k 1

n; f Vn

p

!

c11(p; k) k

1 n; f

p

: Thus the proof of Lemma 1 is completed.

In this work we study(p; q)-inequalities of Ul’yanov type for the modulus of smooth- ness k f(r);

p; k= 1;2; :::,r= 1;2; :::de…ned in the form (3). To prove we use the method of the proof given in the study [14].

Main result in the present work is the following theorem.

THEOREM 1. Letf 2Lp;1 < p < q <1; = 1 p

1

q. Then for anyk= 1;2; :::, r= 1;2; :::the following estimate holds:

k ; f(r)

q C

0

@Z

0

u ( +r) r+k(u; f)p q du u

1 A

1=q

: (9)

2 Proof of the Main Result

According to reference [7] for1< q <1the following equivalence holds:

k

1 2n; f(r)

q

K 1

2n; f(r); Lq(T); Wq2k(T)

= inf f(r)

q+ 2 2nk (2k)

q : 2Wq2k(T) : (10)

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IfVn is the de La Vallée-Poussin sum of the functionf using Lemma 1 we get

K 1

2n; f(r); Lq(T); Wq2k f(r) V2n f(r)

q+ 2 2nk V2(2k)n

q :=I1+I2: (11) Taking account of (8) we have

kf Vn(f)kp c12En(f)p: (12)

Considering [16] and [4], the following(p; q)-inequality holds:

(V2l)(r) (V2n)(r)

q c13 l 1

X

m=n

2m q (V2m+1)(r) (V2m)(r)

q p

!q

: (13)

Using the Bernstein-type inequality [7], [9], [14] we obtain

V2(r)m+1 V2(r)m

p c142mrkV2m+1 V2mkp: (14) Taking into account the relations (13), (14) and Jackson inequality [6] we have

I1 = f(r) V2n f(r)

q

c15 X1 m=n

2m q2mqrE2qm(f)

c16 X1 m=n

2m q2mqr k+r 1 2m; f

q

p

!1=q

c17 0 B@

2 n

Z

0

u ( +r) k+r(u; f)p qdu u

1 CA

1=q

: (15)

On the order hand, for 1 2 the following equivalence holds:

k( 1; f)p k( 2; f)p: (16)

It is known that for trigonometric polynomials of degreenthe following Nikol’skii inequality holds [4], [8], [10] :

kTnkq c18n1p 1qkTnkp; 0< p q 1: (17)

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Use of inequality (17) gives us I2 = 2 nk V2(k)n

(r)

q

c192 nk2n V2(k+r)n

p

c202n 2nr k+r 1 2n; f

p

c21

0 B@

2 n

Z

0

u u r k+r(u; f)p

q du u

1 CA

1=q

= c21 0 B@

2 n

Z

0

u ( +r) k+r(f; u)p q du u

1 CA

1=q

: (18)

Using (10), (11), (15) and (18), we have (9).

Acknowledgment. The author wishes to express deep gratitude to the referee for valuable suggestions.

References

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