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(1)

SIMULTANEOUS APPROXIMATION

BY BETA-SZASZ OPERATORS

Manoj Kumar

Department of Applied Science,

Krishna Engineering College,

Mohan Nagar, Ghaziabad, Uttar Pradesh-201007,

India

ABSTRACT

The study of direct and inverse theorems remains an active area of research. This research now centers on approximation by members of various non-classical and nonlinear classes such as wavelets, shift-invariant subspaces, radial basis functions, ridge functions, neural nets, multivariate splines and the like. In the present paper, we study inverse approximation property of Beta-Szasz operators in simultaneous approximation.

Keywords-Simultaneous approximation, Summation-integral type operators, Linear positive

operators, Peetre’s K-functional.

MATHEMATICAL SUBJECT CLASSIFICATION:

41A25, 41A36.

I. INTRODUCTION

In approximation theory, the results, which determine structural characteristics of functions from their

degree of approximation, are known as inverse theorems. There are many-many generalizations of

direct and inverse results [1, 4, 6, 7, 9, 11]. The excellent textbooks of Timan [10] and of DeVore and

Lorentz [3] contain an abundance of information on direct and inverse theorems for approximation by

algebraic and trigonometric polynomials. In the present paper we study inverse approximation

estimate for these operators, which were defined as

)

,

0

[

)

(

)

(

)

(

1

)

,

(

0 , 1

,

 

x

dt

t

f

t

s

x

b

n

n

x

f

B

nv

v v n

n

(1.1)

where

f

C

[

0

,

)

{

f

C

[

0

,

)

:

f

(

t

)

Mt

for some

M

0

,

0

}

,

1

1 ,

)

1

)(

,

1

(

)

(

v n v v

n

x

v

n

B

x

x

b

,

!

)

(

)

(

,

v

nx

e

x

s

v nx v

n

and

B

(

n

1

,

v

)

n

!

(

v

1

)!

/(

n

v

)!.

It is easily checked that the operators

B

n

are linear positive operators and it is obvious

that

B

n

(

1

,

x

)

1

. Alternately the operators (1.1) may be written as

0

)

(

)

,

(

)

,

(

f

x

W

x

t

f

t

dt

(2)

where

(

)

(

)

1

)

,

(

,

1

,

x

s

t

b

n

n

t

x

W

nv

k v n

n

.

As far as the rate of approximation is concerned the operators (1.1) are just like exponential type operators [8]. But the operators

B

n are not exponential type operators, since they do not satisfy the following condition:

)

)(

,

(

)

(

)

,

(

W

x

t

t

x

x

P

n

t

x

W

x

n

n

,

P

(

x

)

is a function of x. (1.2)

The above equation (1.2) is the necessary condition for the operators to be of exponential type. The above condition (1.2) is frequently used in the analysis to prove the inverse theorem for exponential type operators. In the present paper we study an inverse result in simultaneous approximation for the operators (1.1).

By

C

0, we denote the class of continuous functions on the interval

(

0

,

)

having a compact support and

r

C

0 is r times continuously differentiable functions with

C

0

C

0

r

. Suppose

{

:

0 2

,

)

(r

r

C

g

g

G

supp

]

,

[

a

b

g

, where

[

a

,

b

]

(

a

,

b

)}

. For r times continuously differentiable functions f with supp f

]

,

[

a

b

, the Peetre’s K-functional are defined as

( 2) [ , ]

] , [ ) ( ]

, [ ) ( ) (

) (

inf

)

,

,

,

(

b a C r b

a C r b

a C r r G g

r

f

a

b

f

g

g

g

K

r  

 

 

,

0

1

.

II. PRELIMINARIES

This section consists of the following preliminary results, which will be helpful to prove the inverse approximation theorem in next section.

Lemma 2.1 [5]. For

m

N

{

0

}

, if the m-th order moment be defined as

m k

v n m

n

x

n

v

x

b

n

x

U





2

1

)

(

1

1

)

(

1 ,

, , then

U

n,0

(

x

)

1

,

U

n,1

(

x

)

0

and

)].

(

)

(

)[

1

(

)

(

)

2

(

n

U

n,m1

x

x

x

U

n(1,m)

x

mU

n,m1

x

Consequently,

U

n,m

(

x

)

O

n

[(m1)/2]

.

Lemma 2.2. Let the function

n,m

(

x

),

m

N

0, be defined as

 

0 , 1

,

,

(

)

(

)(

)

1

)

(

b

x

s

t

t

x

dt

n

n

x

nv m

v v n m

n

.

Then

n

x

x

x

n

n

)

1

(

2

)

(

,

1

)

(

,1 0

,

and

2

2 2

,

)

1

(

6

)

2

(

)

(

n

x

n

x

x

x

n

and there holds the recurrence relation

(3)

[( 1)/2]

,

(

)

 

m

m

n

x

O

n

.

Proof. The values of

n,0

(

x

),

n,1

(

x

)

easily follow from the definition. We prove the recurrence relation

1 0

, )

1 (

, )

1 (

,

(

1

)

(

)

(

)(

)

1

)

(

)

1

(

v

m v

n k n m

n

x

x

b

x

s

t

t

x

dt

n

n

x

x

x

dt

x

t

t

s

x

b

x

x

n

n

m

m

v n v

k n

  

0

1 ,

1

,

(

)

(

)(

)

)

1

(

1

Now using the identities

x

(

1

x

)

b

n(1,v)

(

x

)

((

v

1

)

(

n

2

)

x

)

b

n,v

(

x

)

and

)

(

]

[(

)

(

,

) 1 (

,

t

v

nt

s

t

s

t

nv

nv , we obtain

)]

(

)

(

)[

1

(

, 1

) 1 (

,

x

m

x

x

x

nm

nm

1 0

,

,

(

)

(

)(

)

)

)

2

(

1

(

1

v

m v

n v

n

x

s

t

t

x

dt

b

x

n

v

n

n

dt

x

t

t

s

x

x

t

n

nt

v

x

b

n

n

m

v n v

v

n

(

)

[(

)

(

)

(

1

2

)]

(

)

(

)

1

1 0 ,

,

 

b

x

t

s

t

t

x

dt

n

n

m

v n v

v

n

(

)

(

)

(

)

1

0

) 1 (

, 1

,

+

n

n,m1

(

x

)

(

1

2

x

)

n,m

(

x

)

b

x

s

t

t

x

dt

n

n

m

v n v

v n

1

0 ) 1 (

, 1

,

(

)

(

)

(

)

1

 

+

b

x

s

t

t

x

dt

n

nx

m

v n v

v

n

(

)

(

)

(

)

1

0

) 1 (

, 1

,

 

+

n

n,m1

(

x

)

(

1

2

x

)

n,m

(

x

)

.

(

m

1

)

n,m

(

x

)

n

n,m1

(

x

)

mx

n,m1

(

x

)

(

1

2

x

)

n,m

(

x

).

This completes the proof of recurrence relation. The values of

n,2

(

x

)

,

n,m

(

x

)

follow from the

recurrence relation.

Lemma 2.3 [5]. There exist the polynomials

Q

i,j,r

(

x

)

independent of n and k such that

{

(

1

)}

[

(

)]

(

2

)

(

1

(

2

)

)

, ,

(

)

,

(

)

0 , 2

,

x

n

v

n

x

Q

x

b

x

b

D

x

x

j i jr nv

j i

r j i

i v

n r

r

 

, where

dx

d

D

.

Lemma 2.4. Let 0<

<2 and

0

a

a

a



b



b

b

. If

f

C

0 with supp

f

[

a



,

b



]

and

/2

] , [ ) ( )

(

)

,

(

f

f

O

n



B

b a C r r

n , then

K

r

(

,

f

)

M

5

n

/2

n

K

r

(

n

1

,

f

)

(4)

Consequently

K

r

(

,

f

)

M

6

/2

,

M

2

0

.

Proof. It is sufficient to prove

(

,

)

)

,

(

f

M

5

n

/2

n

K

n

1

f

K

r r

, for sufficiently large n.

Because supp f

[

a



,

b



]

, therefore by [8, Th.3.2], there exists a function

h

(i)

G

(r)

,

i

r

,

r

2

such that 1 6 ] , [ ) ( ) (

)

,

(

f

h

M

n

B

b a C i i n Therefore ] , [ ) ( ) ( 1

7

(

,

)

3

)

,

(

b a C r r n

r

f

M

n

B

f

f

K

  

( 2) [ , ]

] , [ ) (

)

,

(

)

,

(

b a C r n b a C r

n

f

B

f

B

Next, it is sufficient to show that there exists a constant

M

6 such that for each

g

G

(r)

1 ( 2) [ , ]

] , [ ) ( ) ( 8 ] , [ ) 2 (

)

,

(

b a C r b a C r r b a C r

n

f

M

n

f

g

n

g

B

        

(2.1)

Also using linearity property, we have

] , [ ) 2 ( ] , [ ) 2 ( ] , [ ) 2 (

)

,

(

)

,

(

)

,

(

b a C r n b a C r n b a C r

n

f

B

f

g

B

g

B

(2.2)

Applying Lemma 2.3, we get

2 0 1 , , 2 2 , , 0 , 2 1 0 2 2

)

1

)(

1

)...(

1

)(

(

)

(

)

(

)}

1

(

{

)

(

)

,

(

             

 

r n k n k n r r j i j j i r j i k i n r r

x

r

n

n

n

dt

t

b

x

p

x

x

x

Q

nx

k

n

dt

t

x

W

x

Therefore by Schwarz inequality and Lemma 2.1, we obtain ] , [ ) ( ) ( 9 ] , [ ) 2 (

)

,

(

b a C r r b a C r

n

f

g

M

n

f

g

B

(2.3)

where the constant

M

9 is independent of f and g. Next by Taylor’s expansion, we have

,

)

(

)!

2

(

)

(

)

(

!

)

(

)

(

2 ) 2 ( 1 0 ) (    

r

r i r i i

x

t

r

g

x

t

i

x

g

t

g

where

lies between t and x. Using above expansion we get

] , [ 2 2 2 ] , [ ) 2 ( ] , [ ) 2 (

)

)(

,

(

)!

2

(

1

)

,

(

b a C r n r r b a C r b a C r

n

W

x

t

t

x

dt

x

g

r

g

B

          

(2.4)

(5)

] , [ ) 2 ( 8 ] , [ )

2 (

)

,

(

b a C r b

a C r

n

g

M

g

B

(2.5)

Combining the estimates of (2.2)-(2.5), we get (2.1). The other consequence follows from [2].

This completes the proof of the lemma.

Lemma 2.5. Let

a

a

a



b



b

b

and

f

(r)

C

0 with supp

f

[

a



,

b



]

then if

),

,

,

1

,

(

0

a

b

C

f

r

we have

f

(r)

Liz

(

,

1

,

a

,

b

).

Proof. Let

h

and

g

G

(r), then for

f

C

0

(

,

1

,

a

,

b

)

r

)

(

)

(

)

(

2 ( ) ( ) 2 () )

( 2

x

g

g

f

x

f

r r r r

2 10

9 ]

, [ ) 2 ( 2 ] , [ ) ( ) ( 2

)

,

(

4

2

f

g

g

M

K

r

f

M

b a C r b

a C r

r

   

It follows that

f

Liz

(

,

1

,

a

,

b

)

i.e.

f

Lip

(

,

a

,

b

).

III.MAIN RESULT

In this section, we shall prove our main result, namely, inverse approximation theorem, which is stated as Theorem: Let

0

2

,

0

a

1

a

2

b

2

b

1

and suppose

f

C

[

0

,

)

. Then in the following statements

(

i

)

(

ii

)

(i)

/2

] , [ ) ( )

(

1 1

,.)

(

f

f

O

n



B

b a C r r

n

(ii)

f

(r)

Lip

(

,

a

2

,

b

2

).

Proof. Let us choose

a

,

a



,

b

,

b



in such a way that

a

1

a

a



a

2

b

2

b



b

b

1. Also suppose

g

C

0 with supp

g

[

a



,

b



]

and

g

(

x

)

1

on

[

a

2

,

b

2

]

. For

x

[

a

,

b

]

with

dx

d

D

, we have

))

),

)(

(

)

)(

((

(

)

(

)

(

)

,

(

()

) (

x

x

fg

t

fg

B

D

x

fg

x

fg

B

n

r r

r

n

))

)),

(

)

(

)(

(

(

(

))

)),

(

)

(

)(

(

(

(

B

f

t

g

t

g

x

x

D

B

g

x

f

t

f

x

x

D

n

r n

r

2 1

J

J

( say)

By Leibnitz theorem, we have

0

1

W

(

x

,

t

)

f

(

t

)(

g

(

t

)

g

(

x

))

dt

x

J

r n

r

 





0 ) ( 0

))]

(

)

(

)(

(

[

)

,

(

f

t

g

t

g

x

dt

x

t

x

W

i

r

i r

i r i

n r

(6)

 





0 ) ( )

( ) ( 1

0

))

(

)

(

)(

(

)

,

(

)

,

(

)

(

x

B

f

x

W

x

t

f

t

g

t

g

x

dt

g

i

r

r

n i

n i r r

i

J

3

J

4, say. Following [8, Th. 3.3], we obtain

)

(

)

(

)

(

() /2 )

( 1

0 3

  





g

x

f

x

o

n

i

r

J

r i i

r i

, uniformly in

x

[

a

,

b

]

Next following [8, Th. 3.2], Schwarz inequality, Taylor’s expansion of f and g and Lemma 2.2, we get

)

(

!

)!

(

!

)

(

)

(

1/2

1

) ( ) ( 4

 

r

o

n

i

r

i

x

f

x

g

J

r i

i r i

)

(

)

(

)

(

( ) /2

) ( 1

  





g

x

f

x

o

n

i

r

i r i

r i

, uniformly in

x

[

a

,

b

]

Finally applying Leibnitz theorem, we obtain

dt

x

f

t

f

t

g

x

t

x

W

i

r

J

r i

i r r

i

i

n

(

,

)

[

(

)(

(

)

(

))]

0 0 ) (

2





 

 





r

i

r i

n i r

x

fg

x

f

B

x

g

i

r

0

) ( )

( ) (

)

(

)

(

)

,

(

)

(

)

(

)

(

)

(

)

(

)

(

( ) /2

0

) ( )

( 





g

x

f

x

fg

x

o

n

i

r

r

r i

i i r

)

(

/2

O

n

, uniformly in

x

[

a

,

b

]

.

Combining the estimates of

J

1,

J

2 ,

J

3and

J

4 , we get

/2

] , [ ) ( )

(

)

(

,.)

(



 

fg

O

n

fg

B

b a C r r

n .

Thus by Lemma 2.4 and Lemma 2.5, we have

(

fg

)

(r)

Lip

(

,

a

,

b

)

, since

g

(

x

)

1

on

[

a

2

,

b

2

]

, it follows that

f

(r)

Lip

(

,

a

2

,

b

2

)

. This proves implication

(

i

)

(

ii

)

for the case

0

1

.

Now to prove the implication for

1

2

, for any interval

[

a

1

,

b

1

]

(

a

1

,

b

1

)

and let

a

2

,

b

2 be such that

(

2

,

2

)

(

2

,

2

)

 

a

b

b

a

and

(

2

,

2

)

(

1

,

1

)

  

b

a

b

a

. Let

0

we shall prove the assertion for

2

. From the previous case it implies that

f

(r) exists and belongs to

Lip

(

1

,

a

1

,

b

1

)

.

Let

g

C

0 be such that

g

(

x

)

1

on

[

a

2

,

b

2

]

and supp

g

(

a

2

,

b

2

)

. Then for characteristic function

2

(

t

)

of the interval

[

1

,

1

]

 

b

a

, we have

] , [ ]

, [ ) ( )

(

2 2 2

2

(.),.)

)

(

(.)(

(

[

)

(

,.)

(

 

 

b a C n

r b

a C r r

n

fg

fg

D

B

g

f

t

f

B

] , [ 2 2

(.)),.)

)

(

)(

(

(

[

 

b a C n

r

(7)

2 1

I

I

, say.

Following [5, Th. 3.3], we have

] , [ ) ( 1 2 2

)

(

),.)]

(

)

(

(

[

 

b a C r n m

fg

t

f

x

g

B

D

I

] , [ ) ( ) ( ) ( 0 2 2

)

(

,.)

(

 





  

b a C r i n i r i

fg

f

B

g

i

r

)

(

)

(

/2

] , [ ) ( ) ( ) ( 0 2 2    





 

g

f

fg

O

n

i

r

b a C r i i r r i

)

(

/2

O

n

.

Next following [8, Th. 3.2] and Leibnitz theorem, we have

] , [ 2 ) ( ) ( ) ( 1 0 2 2 2

),.)

(

(.))

)

(

)(

(

(

,.)

(

 





  

b a C r n i n i r r i

t

g

t

g

t

f

B

f

B

g

i

r

I

3 4 [ , ]

(

1

)

2 2

I

I

Cab

O

n

(say).

Again following [8, Th. 3.3], we get

)

(

)

(

)

(

() /2 ) ( 1 0 3     





g

x

f

x

O

n

i

r

I

r i i

r i

, uniformly in

[

2

,

2

]

 

a

b

x

.

Applying Taylor’s expansion of f, we have

I

W

nr

(

x

,

t

)[

f

(

t

)(

g

(

t

)

g

(

x

))

2

(

t

)]

dt

0 ) (

4

W

x

t

t

x

g

t

g

x

t

dt

i

x

f

r i

n r p i i

)

(

))

(

)

(

(

)

)(

,

(

!

)

(

2 0 ) ( 1 0 ) (

 

 

dt

t

x

g

t

g

x

t

r

x

f

f

t

x

W

r r r r

n

(

)

(

(

)

(

))

(

)

!

)

(

)

(

(

)

,

(

2 ) ( ) ( 0 ) (





(

lying between t and x )

I

5

I

6, say.

Next following [5, Th. 3.2], we get

)

(

))

(

)

(

(

)

)(

,

(

!

)

(

1 0 ) ( 0 ) ( 5   

W

x

t

t

x

g

t

g

x

dt

O

n

i

x

f

I

nr i

r i

i

( uniformly in

[

2

,

2

]

 

a

b

x

)

I

7

O

(

n

1

)

(say) Since

g

C

0, therefore we can write

dt

x

t

t

x

W

m

x

g

i

x

f

I

nr i m

(8)

dt

x

t

x

t

t

x

W

i

x

f

r i r

n r

i i

2

0 ) ( 0

) (

)

)(

,

(

)

,

(

!

)

(

  

( where

(

t

,

x

)

0

as

t

x

)

I

8

I

9 ( say)

Following [5, Th. 3.2], we get

)

(

!

)!

(

)

(

!

)

(

( ) 1

1 ) ( 8

 

r

O

n

m

r

x

f

m

x

g

I

m r r

m m

(

)

(

)

(

1

)

0

) ( )

( 





g

x

f

x

O

n

m

r

r m

m r m

. Also

I

9

O

(

n

/2

)

uniformly in

x

[

a

2

,

b

2

]

.

Finally using mean value theorem and Lemma 2.3, we obtain

 

 

 

0

1 ,

,

0 , 2 ] , [

6

(

,

)

)}

1

(

{

)

(

2 2

r n

r r s m s

m r s m

s m b

a

C

W

x

t

t

x

x

x

x

Q

n

I

] , [ 2

) ( )

(

2 2

)

(

)

(

!

)

(

)

(

 

b a C r

r

dt

t

g

r

x

f

f

O

(

n

/2

)

where

is chosen in such a way that

0

2

. Finally combining the above estimates we get

/2

] , [ ) ( )

(

2 2

)

(

,.)

(

fg

fg

 

O

n



B

b a C r r

n .

Since supp

fg

(

a

2

,

b

2

)

, it follows from Lemma 2.4 and Lemma 2.5 that

)

,

,

(

)

(

2 2

)

(

  

b

a

Lip

fg

r

. Furthermore, since

g

(

x

)

1

on

[

a

2

,

b

2

]

, we have

(

,

2

,

2

)

)

(

b

a

Lip

f

r

.

This completes the proof of the theorem.

REFERENCES

[1] P. N. Agrawal and V. Gupta, “ Inverse theorem for linear combinations of modified Bernstein polynomials, ” Pure and Applied Math Sci. 40, 29-38, 1995.

[2] H. Berens and G. G. Lorentz, “ Inverse theorems for Bernstein polynomials, ” Indiana University Mathematics Journal, 21, 693-708, 1972.

[3] R. A. De-Vore and G. G. Lorentz, “ Constructive Approximation, ” Spriger-Verlag, Berlin, 1993.

[4] Z. Finta and V. Gupta, “ Direct and inverse estimates for Phillips type operators, ” J. Math Anal Appl. 303(2), 627-642, 2005.

(9)

[6] V. Gupta, P. Maheshwari and V. K. Jain, “ Lp-inverse theorem for modified beta operators, ” Int. J. Math.

Sci., 33(20), 1395-1403, 2003.

[7] H. S. Kasana, P. N. Agrawal and V. Gupta, “ Inverse and saturation theorems for linear combination of modified Baskakov operators, ” Approx Theory and its Appl. 7 (2), 65-81, 1991.

[8] C. P. May, “ Saturation and inverse theorems for combinations of a class of exponential type operators, ” Canad. J. Math. 28, 1224-1250, 1976.

References

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