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Testo completo

(1)

On certain modified

Szasz-Mirakyan operators

in polynomial weighted spaces

Zbigniew Walczak

Institute of Mathematics, Pozna´ n University of Technology, Piotrowo 3A, 60-965 Pozna´ n, Poland

[email protected]

Received: 21/2/2002; accepted: 1/5/2003.

Abstract. We consider certain modified Szasz-Mirakyan operators A

n

(f ; r) in polynomial weighted spaces of functions of one variable and we study approximation properties of these operators.

Keywords: Szasz-Mirakyan operator, degree of approximation, Voronovskaja type theorem.

MSC 2000 classification: 41A36.

Introduction

In the paper [1] M. Becker studied approximation problems for functions f ∈ C p and Szasz-Mirakyan operators

S n (f ; x) := e −nx X ∞ k=0

(nx) k k! f

 k n



, (1)

x ∈ R 0 = [0, + ∞), n ∈ N := {1, 2, . . . }, where C p with fixed p ∈ N 0 :=

{0, 1, 2, . . . } is polynomial weighted space generated by the weighted function w 0 (x) := 1, w p (x) := (1 + x p ) −1 , if p ≥ 1, (2) i.e. C p is the set of all real-valued functions f , continuous on R 0 and such that w p f is uniformly continuous and bounded on R 0 . The norm in C p is defined by the formula

kfk p ≡ kf (·) k p := sup

x∈R

0

w p (x) |f(x)|. (3)

In [1] theorems on the degree of approximation of f ∈ C p by the operators S n were proved. From these theorems it was deduced that

n→∞ lim S n (f ; x) = f (x), (4)

(2)

for every f ∈ C p , p ∈ N 0 and x ∈ R 0 . Moreover the convergence (4) is uniform on every interval [x 1 , x 2 ], x 2 > x 1 ≥ 0.

In this paper we shall modify the formula (1) and we shall study certain approximation properties of introduced operators.

Let C p be the space given above and let f ∈ C p 1 := {f ∈ C p : f 0 ∈ C p }, where f 0 is the first derivative of f .

For f ∈ C p we define the modulus of continuity ω 1 (f ; ·) as usual ([2]) by formula

ω 1 (f ; C p ; t) := sup

0≤h≤t k∆ h f ( ·)k p , t ∈ R 0 , (5) where ∆ h f (x) := f (x + h) − f(x), for x, h ∈ R 0 . From the above it follows that

t→0+ lim ω 1 (f ; C p ; t) = 0, (6) for every f ∈ C p . Moreover if f ∈ C p 1 then there exists M 1 = const. > 0 such that

ω 1 (f ; C p ; t) ≤ M 1 · t for t ∈ R 0 . (7) We introduce the following

1 Definition. Let R 2 := [2, + ∞) and let r ∈ R 2 and p ∈ N 0 be fixed numbers. For functions f ∈ C p we define the operators

A n (f ; r; x) := e −(nx+1)

r

X ∞ k=0

(nx + 1) rk

k! f

 k

n(nx + 1) r−1



, (8)

x ∈ R 0 , n ∈ N.

Similarly as S n , the operator A n is linear and positive. In § 2 we shall prove that A n is an operator from the space C p into itself for every fixed p ∈ N 0 .

From (8) we easily derive the following formulas

A n (1; r; x) = 1, (9)

A n (t; r; x) = x + 1

n , A n t 2 ; r; x 

=

 x + 1

n

 2 

1 + 1

(nx + 1) r



A n t 3 ; r; x 

=

 x + 1

n

 3 

1 + 3

(nx + 1) r + 1 (nx + 1) 2r



,

for every fixed r ∈ R 2 and for all n ∈ N and x ∈ R 0 .

(3)

1 Main results

From formulas (8), (9) and A n (t k ; r; x), 1 ≤ k ≤ 3, given above we obtain 2 Lemma. Let r ∈ R 2 be a fixed number. Then for all x ∈ R 0 and n ∈ N we have

A n (t − x; r; x) = 1 n , A n (t − x) 2 ; r; x 

= 1 n 2



1 + 1

(nx + 1) r−2

 ,

A n (t − x) 3 ; r; x 

= 1 n 3



1 + 3

(nx + 1) r−2 + 1 (nx + 1) 2r−3

 . Next we shall prove

3 Lemma. Let s ∈ N and r ∈ R 2 be fixed numbers. Then there exist positive numbers λ s,j , 1 ≤ j ≤ s, depending only on j and s, such that

A n (t s ; r; x) =

 x + 1

n

 s X s j=1

λ s,j

(nx + 1) (j−1)r (10) for all n ∈ N and x ∈ R 0 . Moreover λ s,1 = λ s,s = 1.

Proof. We shall use the mathematical induction on s.

The formula (10) for s = 1, 2, 3 is given above.

Let (10) holds for f (x) = x j , 1 ≤ j ≤ s, with fixed s ∈ N. We shall prove (10) for f (x) = x s+1 . From (8) it follows that

A n (t s+1 ; r; x) = e −(nx+1)

r

X ∞ k=1

(nx + 1) rk (k − 1)!

k s

(n(nx + 1) r−1 ) s+1 =

= (nx + 1) r

(n(nx + 1) r−1 ) s+1 e −(nx+1)

r

X ∞ k=0

(nx + 1) rk

k! (k + 1) s =

= (nx + 1) r

(n(nx + 1) r−1 ) s+1 e −(nx+1)

r

X ∞ k=0

(nx + 1) rk k!

X s µ=0

 s µ

 k µ =

= (nx + 1) r (n(nx + 1) r−1 ) s+1

X s µ=0

 s µ



(n(nx + 1) r−1 ) µ A n (t µ ; r; x).

By our assumption we get

A n (t s+1 ; r; x) = (nx + 1) r

(n(nx + 1) r−1 ) s+1 ·

(4)

·

 

 1 + X s µ=1

 s µ



(nx + 1) X µ j=1

λ µ,j (nx + 1) (j−1)r

 

 =

=

 x + 1

n

 s+1 

 1 (nx + 1) rs +

X s j=1

X s µ=j

 s µ

 λ µ,j

(nx + 1) (s+j−µ−1)r

 

 =

=

 x + 1

n

 s+1 

 1

(nx + 1) rs + X s j=1

1 (nx + 1) (j−1)r

X s µ=s−j+1

 s µ



λ µ,µ+j−s

 

 =

=

 x + 1

n

 s+1 s+1 X

j=1

λ s+1,j

(nx + 1) (j−1)r

and λ s+1,1 = λ s+1,s+1 = 1, which proves (10) for f (x) = x s+1 . Hence the proof

of (10) is completed.

QED

4 Lemma. Let p ∈ N 0 and r ∈ R 2 be fixed numbers. Then there exists a positive constant M 2 ≡ M 2 (p, r), depending only on the parameters p and r such that

kA n (1/w p (t); r; ·)k p ≤ M 2 , n ∈ N. (11) Moreover for every f ∈ C p we have

kA n (f ; r; ·)k p ≤ M 2 kfk p , n ∈ N. (12) The formula (8) and inequality (12) show that A n , n ∈ N, is a positive linear operator from the space C p into C p , for every p ∈ N 0 .

Proof. The inequality (11) is obvious for p = 0 by (2), (3) and (9).

Let p ∈ N. Then by (2) and (8)-(10) we have

w p (x)A n (1/w p (t); r; x) = w p (x) {1 + A n (t p ; r; x) } =

= 1

1 + x p + + (x + 1/n) p 1 + x p

X p j=1

λ p,j

(nx + 1) (j−1)r

≤ 1 + X p µ=0

 p µ

 x µ 1 + x p

X p j=1

λ p,j

(nx + 1) (j−1)r ≤ M 2 (p, r), for x ∈ R 0 , n ∈ N and r ∈ R 2 , where M 2 (p, r) is a positive constant depending only p and r. From this follows (11).

The formula (8) and (3) imply

kA n (f (t); r; ·)k p ≤ kfk p kA n (1/w p (t); r; ·)k p , n ∈ N, r ∈ R 2 ,

for every f ∈ C p . By applying (11), we obtain (12).

QED

(5)

5 Lemma. Let p ∈ N 0 and r ∈ R 2 be fixed numbers. Then there exists a positive constant M 3 ≡ M 3 (p, r) such that

A n

 (t − ·) 2

w p (t) ; r; · 

p

≤ M 3

n 2 for all n ∈ N. (13)

Proof. The formulas given in 2 Lemma and (2), (3) imply (13) for p = 0.

By (2) and (9) we have A n (t − x) 2 /w p (t); r; x 

= A n (t − x) 2 ; r; x 

+ A n t p (t − x) 2 ; r; x  , for p, n ∈ N and r ∈ R 2 . If p = 1 then by the equality we get

A n (t − x) 2 /w 1 (t); r; x 

= A n (t − x) 2 ; r; x 

+ A n t(t − x) 2 ; r; x 

=

= A n (t − x) 3 ; r; x 

+ (1 + x)A n (t − x) 2 ; r; x  , which by (2) and (3) and 2 Lemma yields (13) for p = 1.

Let p ≥ 2. By applying (10), we get w p (x)A n t p (t − x) 2 ; r; x 

= w p (x) 

A n t p+2 ; r; x 

− 2xA n t p+1 ; r; x  +

+x 2 A n (t p ; r; x)

= w p (x)

 

 x + 1

n

 p+2 p+2 X

j=1

λ p+2,j (nx + 1) (j−1)r +

−2x

 x + 1

n

 p+1 p+1 X

j=1

λ p+1,j

(nx + 1) (j−1)r +

+x 2

 x + 1

n

 p X p j=1

λ p,j (nx + 1) (j−1)r

 

 =

= w p (x)

 x + 1

n

 p 

 1 n 2 +

 x + 1

n

 2 p+2 X

j=2

λ p+2,j (nx + 1) (j−1)r +

−2x

 x + 1

n

 X p+1

j=2

λ p+1,j

(nx + 1) (j−1)r + x 2 X p j=2

λ p,j (nx + 1) (j−1)r

 

 which implies

w p (x)A n t p (t − x) 2 ; r; x 

≤ 1 n 2

(1 + x) p 1 + x p

 

 1 + 1

(nx + 1) r−2

p+2 X

j=2

λ p+2,j +

(6)

+ 2

p+1 X

j=2

λ p+1,j + X p j=2

λ p,j

 

 ≤ M 3 (p, r) n 2

for x ∈ R 0 , n ∈ N and r ∈ R 2 . Thus the proof is completed.

QED

Now we shall give approximation theorems for A n .

6 Theorem. Let p ∈ N 0 and r ∈ R 2 be fixed numbers. Then there exists a positive constant M 4 ≡ M 4 (p, r) such that for every f ∈ C p 1 we have

kA n (f ; r; ·) − f(·)k p ≤ M 4

n kf 0 k p , n ∈ N. (14) Proof. Let x ∈ R 0 be a fixed point. Then for f ∈ C p 1 we have

f (t) − f(x) = Z t

x

f 0 (u)du, t ∈ R 0 . From this and by (8) and (9) we get

A n (f (t); r; x) − f(x) = A n

Z t x

f 0 (u)du; r; x



, n ∈ N.

But by (2) and (3) we have

Z t

x

f 0 (u)du

≤ kf 0 k p

 1

w p (t) + 1 w p (x)



|t − x|, t ∈ R 0 , which implies

w p (x) |A n (f ; r; x) − f(x)| ≤ (15)

≤ kf 0 k p



A n ( |t − x|; r; x) + w p (x)A n

 |t − x|

w p (t) ; r; x



for n ∈ N. By the H¨older inequality and by (9) and 2, 4, 5 Lemmas it follows that

A n ( |t − x|; r; x) ≤ 

A n (t − x) 2 ; r; x 

A n (1; r; x) 1/2

√ 2 n , w p (x)A n

 |t − x|

w p (t) ; r; x



≤ w p (x)

 A n

 (t − x) 2 w p (t) ; r; x

 1/2  A n

 1 w p (t) ; r; x

 1/2

≤ M 4 n

for n ∈ N. From this and by (15) we immediately obtain (14).

QED

(7)

7 Theorem. Let p ∈ N 0 and r ∈ R 2 be fixed numbers. Then there exists M 6 ≡ M 6 (p, r) such that for every f ∈ C p and n ∈ N we have

kA n (f ; r; ·) − f(·)k p ≤ M 6 ω 1



f ; C p ; 1 n



. (16)

Proof. We use Steklov function f h of f ∈ C p

f h (x) := 1 h

Z h 0

f (x + t)dt, x ∈ R 0 , h > 0. (17) From (17) we get

f h (x) − f(x) = 1 h

Z h 0

t f (x)dt, f h 0 (x) = 1

h ∆ h f (x), x ∈ R 0 , h > 0, which imply

kf h − fk p ≤ ω 1 (f ; C p ; h) , (18) kf h 0 k p ≤ h −1 ω (f ; C p ; h) , (19) for h > 0. From this we deduce that f h ∈ C p 1 if f ∈ C p and h > 0.

Hence we can write

w p (x) |A n (f ; x) − f(x)| ≤ w p (x) {|A n (f − f h ; x) | +

+ |A n (f h ; x) − f h (x) | + |f h (x) − f(x)|} := L 1 (x) + L 2 (x) + L 3 (x), for n ∈ N, h > 0 and x ∈ R 0 . From (12) and (18) we get

kL 1 k p ≤ M 2 kf h − fk p ≤ M 2 ω 1 (f ; C p ; h) , kL 3 k p ≤ ω 1 (f ; C p ; h) .

By 6 Theorem and (19) it follows that kL 2 k p ≤ M 4

n kf h 0 k p ≤ M 4

nh ω 1 (f ; C p ; h) . Consequently

kA n (f ; r; ·) − f(·)k p ≤



1 + M 2 + M 4 nh



ω 1 (f ; C p ; h).

Now, for fixed n ∈ N, setting h = n 1 , we obtain kA n (f ; r; ·) − f(·)k p ≤ M 6 (p, r)ω 1



f ; C p ; 1 n



and we complete the proof.

QED

(8)

From 6 Theorem and 7 Theorem we derive the following two corollaries:

8 Corollary. For every fixed r ∈ R 2 and f ∈ C p , p ∈ N 0 , we have

n→∞ lim kA n (f ; r; ·) − f(·)k p = 0.

9 Corollary. If f ∈ C p 1 , p ∈ N 0 and r ∈ R 2 , then kA n (f ; r; ·) − f(·)k p = O(1/n).

Finally, we shall give the Voronovskaya type theorem for A n . 10 Theorem. Let f ∈ C p 1 and let r ∈ R 2 be fixed number. Then,

n→∞ lim n {A n (f ; r; x) − f(x)} = f 0 (x) (20) for every x ∈ R 0 .

Proof. Let x ∈ R 0 be a fixed point. Then by the Taylor formula we have f (t) = f (x) + f 0 (x)(t − x) + ε(t; x)(t − x)

for t ∈ R 0 , where ε(t) ≡ ε(t; x) is a function belonging to C p and ε(x) = 0.

Hence by (8) and (9) we get

A n (f ; r; x) = f (x) + f 0 (x)A n (t −x; r; x)+A n (ε(t)(t −x); r; x), n ∈ N, (21) and by H¨older inequality

|A n (ε(t)(t − x); r; x)| ≤ 

A n ε 2 (t); x  1/2 

A n (t − x) 2 ; x  1/2

. By 8 Corollary we deduce that

n→∞ lim A n ε 2 (t); r; x 

= ε 2 (x) = 0.

From this and by 2 Lemma we get

n→∞ lim nA n (ε(t)(t − x); r; x) = 0. (22) Using (22) and 2 Lemma to (21), we obtain the desired assertion (20).

QED

11 Remark. It is easily verified that the operators A n (f ; r; x) := e −(nx+1)

r

X ∞ k=0

(nx + 1) rk

k! n(nx+1) r−1

Z (k+1+r)/(n(nx+1)

r−1

) (k+r)/(n(nx+1)

r−1

)

f (t)dt,

p ∈ N 0 , x ∈ R 0 , n ∈ N and r ∈ R 2 , have analogous approximation properties

in the space C p .

(9)

12 Remark. In [1] it was proved that if f ∈ C p , p ∈ N 0 , then for the Szasz- Mirakyan operators S n (defined by (1)) is satisfied the following inequality

w p (x) |S n (f ; x) − f(x)| ≤ M 9 ω 2

 f ; C p ;

r x n



, x ∈ R 0 , n ∈ N 0 , where M 9 = const. > 0 and ω 2 (f ; ·) is the modulus of smoothness defined by the formula

ω 2 (f ; C p ; t) := sup

0≤h≤t k∆ 2 h f ( ·)k p , t ∈ R 0 ,

where ∆ 2 h f (x) := f (x) − 2f(x + h) + f(x + 2h). In particular, if f ∈ C p 1 , p ∈ N 0 , then

w p (x) |S n (f ; x) − f(x)| ≤ M 10

r x n , for x ∈ R 0 and n ∈ N (M 10 = const. > 0).

7 Theorem and 10 Theorem and 9 Corollary in our paper show that operators A n , n ∈ N, give better degree of approximation of functions f ∈ C p and f ∈ C p 1

than S n .

References

[1] M. Becker: Global approximation theorems for Szasz-Mirakjan and Baskakov operators in polynomial weight spaces, Indiana Univ. Math. J., 27(1)(1978), 127–142.

[2] R. A. De Vore, G. G. Lorentz: Constructive Approximation, Springer-Verlag, Berlin,

1993.

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