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Revista de la Unión Matemática Argentina

Print version ISSN 0041-6932On-line version ISSN 1669-9637

Rev. Unión Mat. Argent. vol.46 no.1 Bahía Blanca Jan./June 2005

 

Simultaneous approximation with linear combination of integral Baskakov type operators

Kareem J. Thamer, May A. Al-Shibeeb, A.I. Ibrahem

Abstract:
The aim of the present paper is to study some direct results in simultaneous approximation for the linear combination of integral Baskakov type operators.

1991 Mathematics Subject Classification. 41A28, 41A36.

Key words and phrases. Linear positive operators; Linear combinations; Simultaneous approximation.

1 INTRODUCTION

Agrawal and Thamer [1] introduced a new sequence of linear positive operators Mn  called integral Baskakov – type operators to approximate unbounded continuous functions on [0,∞) and it is defined as follow

Let α >  0  , f ∈ Cα [0,∞) =  {f ∈ C [0,∞)  : ∣f (t)∣ ≤ M (1 + t)αfor some  M  >  0} .

Then,

(1.1)                        ∑∞         ∞∫  Mn  (f (t);x) = (n - 1)     pn,υ (x)   pn,υ- 1(t) f (t) dt + (1 + x)-nf (0)                         υ=1        0  ,

where            (           )               n + υ - 1     υ        - (n+ υ)  pn,υ (x) =       υ        x  (1 + x)         , x ∈ [0,∞)  is the kernel of Lupas operators                ∑∞  (           )  Ln (f (t) ;x) =       n + υ - 1    xυ (1 + x)- (n+υ) f (υ/ )                 υ=0       υ                            n  . We may also write (1.1) as :

                 ∞∫    Mn  (f (t) ;x) =    Wn  (t,x) f (t) dt  ,                   0

where                      ∞∑                            -n  Wn  (t,x) =  (n -  1)    pn,υ (x) pn,υ-1(t) + (1 + x)  δ(t) ,  δ(t)                      υ=1  being the Dirac delta function.

The space C α[0,∞)  is normed by                              -α  ∥f ∥Cα =   sup  ∣f (t)∣ (1 + t)            0≤t≤ ∞  .

The operator (1.1) was used to study the degree of approximation in simultaneous approximation by Agrawal and Thamer [1]. It turned out that the order of approximation by the operator (1.1) is, at best, O (n -1)  , howsoever smooth the function may be. Thus, if we want to have a better order of approximation, we have to slacken the positivity condition. This is achieved by considering some carefully chosen linear combination introduced by May [6] and Rathore [7] of the operator (1.1). The linear combination is defined as follows:

Let d0, d1,..., dk  be (k + 1)  arbitrary but fixed distinct positive integers. Then, following Agrawal and Sinha [3] , the linear combination Mn  (f,k,x)  of Mdjn (f ;x)  , j = 0,1, 2,...,k  is given by

                                ∣                                 ∣                                  ∣∣ Md0n (f;x)  d-0 1 d -02  ... d-0 k ∣∣                               1  ∣∣ Md1n (f;x)  d-1 1 d -12  ... d-1 k ∣∣  (1.2)         Mn  (f,k,x) =  -- ∣ .           .    .         .    ∣,                               Δ  ∣∣ ..           ..    ..     ... ..    ∣∣                                  ∣ Mdkn (f;x)  d-k 1 d -k2  ... d-k k ∣

where Δ  is the Vandermonde determinant obtained by replacing the operator column of the above determinant by the entries 1. We have

                                 ∑k  (1.3)             Mn  (f,k,x) =      C (j,k) Mdjn (f ;x) ,                                   j=0

where

                           ∏k     dj  (1.4)         C (j,k) =         -------  ,  k ⁄=  0  and   C (0,0) = 1.                            i = 0 dj - di                              i ⁄= j

The object of the present paper is to investigate the degree of approximation of the operator M n(r) (f,k,x)  . First we establish a Voranovskaja type asymptotic formula and then obtain an error estimate in terms of the local modulus of continuity for the operator    (r)  M n   (f,k,x)  .

2 AUXILIARY RESULTS

Throughout our work, N  denotes the set of natural numbers,    0  N   integers, and 〈a,b〉 an open interval containing [a,b]  .

LEMMA 2.1 [4]. If for       0  m ∈ N   (the set of nonnegative integers), the   th  m   order moment of Lupas operators is defined by

           ∑∞         ( υ    )m  μn,m (x) =     pn,υ (x)  --- x   .             υ=0          n

Hence, μn,0(x) =  1,  μn,1(x) = 0  , and there holds the recurrence relation

n μn,m+1 (x) =  x(1 + x) [μ ′  (x) + m μn,m-1 (x)],    m ∈  N.                             n,m

Consequently

(i) μn,m (x)  is a polynomial in x  of degree at most m  .

(ii) For every                          (  -[(m+1)∕2])  x ∈ [0,∞)  , μn,m (x) =  O  n , where [β]  denotes the integral part of β  .

LEMMA 2.2 [1]. Let the function T    (x) ,  m ∈  N 0   n,m   be defined as

                   ∑∞         ∫∞  T    (x) = (n - 1)     p   (x)   p     (t) (t - x)m dt + (- x)m (1 + x)- n.    n,m                    n,υ        n,υ-1                     υ=1         0

Then,

                         -2x---  Tn,0(x) = 1,   Tn,1 (x) = n - 2

and

                                    ′  (n - m  - 2) Tn,m+1 (x) = x (1 + x)T n,m (x) + [(2x + 1) m + 2x] Tn,m (x)+                         + 2mx  (1 + x) T     (x) , m  ∈ N.                                         n,m -1

Hence,

(i) Tn,m (x)  is a polynomial in x  of degree m.

(ii) For every                           (  -[(m+1)∕2])  x ∈ [0,∞)  ,   Tn,m (x) = O   n .

(iii) The coefficients of n-(υ+1)   in T      (x)   n,2υ+2  and T      (x)   n,2υ+1  are given by                      υ+1  (2υ-+--2)! {x-(1 +-x)}--           (υ + 1)!  and (2υ-+-1)! {(υ + 1) (1 + 2x) - 1}  {x(1 + x)}υ      υ!   .

LEMMA 2.3 [5]. There exist polynomials qi,j,r (t)  independent of n  and υ  such that

           dr                ∑  tr (1 + t)r --r pn,υ (t) =           ni (υ - nt)j qi,j,r (t) pn,υ (t) .             dt                           2i + j ≤ r                           i,j ≥ 0

LEMMA 2.4 [6].If C  (j,k) , j = 0,1,2,...,k  are defined as in (1.4), then

∑k                 {      C (j,k) d-jm =    1,  m  = 0          .  j=0                  0,   m =  1,...,k

LEMMA 2.5 [8].Let f  be r  times differentiable on [0,∞)  such that f (r-1)(t) = O (tα)  for some α  as t -→ ∞ . Then for r = 1, 2,...  and   n > α + r  , we have

   (r)            (n-+-r---1)!-(n---r --1)!  M  n (f (t),x) =     (n - 1)! (n - 2)!    ×                                 ∞                   ∑∞           ∫                ×     pn+r,υ (x)   pn-r,υ+r- 1(t) f(r)(t) dt.                   υ=1          0

LEMMA 2.6 [2]. For r ∈ N  and n  sufficiently large, there holds

Mn  ((t - x)r , k,x) = n-(k+1) {Q (r,k,x) + o(1)},

where Q  (r,k, x)  is a certain polynomial in x  of degree r  .

3 MAIN RESULTS

In this section we shall state and prove the main results.

Theorem 3.1.Let f ∈ C α [0,∞)  and be bounded on every finite subinterval of [0,∞)  admitting a derivative of order 2k + r + 2  at a fixed point x ∈ (0,∞)  . Let f (t) = O (tα)  as t -→  ∞ for some α > 0  , then we have

                                                 2k+r+2                    k+1 [  (r)           (r)   ]    ∑      (i)  (3.1)     nL-i→m∞ n      M n  (f,k,x) - f   (x) =        f   (x) Q (i,k,r,x)                                                     i=r

and

                               [                          ](3.2)               Lim  nk+1   M (nr)(f,k + 1,x) - f(r)(x)  = 0,                     n-→  ∞

where Q (i,k,r,x)  are certain polynomials in x  .

Further, the Limits (3.1) and (3.2) hold uniformly in [a,b]  , if   (2k+r+2)  f   exists and is continuous on (a - η,b + η) ⊂ (0,∞)  , η > 0  .

Proof. By the Taylor expansion, we have

        2k+r+2           ∑    f(i)(x)       i                2k+r+2  f (t) =           i!   (t - x) + ɛ(t,x) (t - x)      ,           i=0

where ɛ (t,x) -→ 0  as  t -→  x  .

Thus, using Lemma 2.5, we have for sufficiently large n

                                    [                                ] k+1[   (r)            (r)   ]     k+1  ∑k           (r)          (r)  n    M  n (f,k, x) - f  (x)  = n         C (j,k) M djn(f;x) - f   (x)                                        j=0

=  I1 + I2,

where

         [        k+1  2k+∑r+2 f(i)(x) ∑k         (djn - r - 2)! (djn + r - 1)!  I1 = n           --i-!--     C (j,k) ----(d-n---1)!-(d-n---2)!----              i=0          j=0              j          j                ∞             ∫∞                 r               ∑                                d--       i      (r)             ×     pdjn+r,υ (x)  pdjn-r,υ+r -1(t) dtr (t - x) dt - f  (x)               υ=1            0                                   ]                  2k+r+2 (n +  2k + r + 1) !      - n-2k-r-2             + (- 1)       -----------------(1 + x)          f (0) .                               (n - 1) !

         [k                      ∞        k+1  ∑                    ∑    (r)  I2 = n        C (j,k) (djn - 1)     pdjn,υ (x)             j=0                   υ=1

  ∫∞  ×   pdjn,υ- 1(t) ɛ (t,x) (t - x)2k+r+2 dt      0

             (n + 2k + r + 1) !                     ]+ (- 1)2k+r+2------------------(1 + x)-n-2k-r-2 f (0) .                    (n - 1) !

It's clear that

     2k+r+2 (n-+-2k-+-r-+-1)-!       -n- 2k-r- 2  (- 1)           (n -  1) !    (1 + x)          f (0) -→  0   as  n -→ ∞.

Let I1 = I3 + I4   , where

     [         k+1 2k∑+r+2f (i)(x) ∑ k         (djn - r - 2) ! (djn + r - 1) !  I3 =  n          ---i!--     C (j,k) ----(d-n----1)-! (d-n---2)-!---             i=r+1          j=0               j          j

                  ∞                             ⌋    ∑∞             ∫                 dr        i  ×     pdjn+r,υ (x)    pdjn-r,υ+r-1(t)--r (t - x) dt⌉ .     υ=1           0                dt

          [                                                           ]                  ∑ k         (djn - r - 2) ! (djn + r - 1) !  I4 = nk+1  f (r)(x)    C (j,k) ------------------------------ - f(r)(x) .                     j=0             (djn - 1) ! (djn - 2) !

Thus, by (1.4),

                 [k∑                                             ]I  = nk+1f (r)(x)      C (j,k) (djn---r --2)-! (djn-+-r---1)-!- 1 .   4                                (djn - 1) ! (djn - 2)!                    j=0

Now, in view of Lemma 2.4, we have

      (r)  I4 = f   (x) K (r,k) + o(1) ,  n -→  ∞ ,

where K  (r,k)  is a constant depending only on r  and k  .

Next, by Lemma 2.4 and Lemma 2.6,we get

     2k∑+r+2  I3 =       f (i)(x) Q (i,k, r,x) + o (1), n -→  ∞.        i=r+1

Thus

                        2k+∑r+2  I1 -→  f(r)(x) K (r,k) +       f(i)(x) Q (i,k,r,x)                          i=r+1

   2k+∑r+2  =        f(i)(x) Q (i,k,r,x)  as   n -→  ∞.      i=r

Now we must prove that I2 -→  0  as   n -→  ∞ . For this, it is sufficient to prove that

                (              2k+r+2   )  I ≡ xrnk+1M  (nr)  ɛ(t,x) (t - x)      ;x  -→  0   as  n -→  ∞.

Using Lemma 2.3, we get

                                        ∞         k+1                   ∑        i∑                   j  ∣I∣ ≤ n    (n - 1) M (x)             n     pn,υ (x) ∣υ - nx∣                            2i + j ≤ r    υ=1                            i,j ≥ 0

   ∞     ∫                   ∣∣             ∣∣  ×    pn,υ-1(t) ∣ɛ (t,x)∣∣(t - x)2k+r+2 ∣ dt,     0

where M  (x) = sup ∣qi,j,r (x)∣ , and then applying the Schwarz inequality we get:

                                       {                      }                              ∑            ∞∑                      1∕2  ∣I∣ ≤ nk+1(n - 1) M (x)             ni      pn,υ (x) (υ - nx)2j                                           υ=1                           2i + j ≤ r                           i,j ≥ 0

  (  ∞          ( ∫∞                                   ) 2) 1∕2    { ∑                                 ∣∣       2k+r+2∣∣      }  × (     pn,υ (x) (   pn,υ-1 (t) ∣ɛ(t,x)∣∣(t - x)     ∣ dt)  )    .       υ=1          0

Since ɛ(t,x) -→  0  as   t -→ x  , for a given ɛ > 0 there exists a δ > 0  such that ∣ɛ (t,x) ∣ < ɛ  , whenever 0 < ∣t - x ∣ < δ  , and for ∣t - x∣ ≥ δ  there exists a constant C  such that ∣ɛ (t,x) ∣ ≤ C ∣t - x∣β , where β is an integer ≥ max (α,2k + r + 2)  .

Hence, as ∞∫    pn,υ- 1(t)dt = n1-1-  0   , we have

(                                      ) 2    ∫∞                  ∣             ∣  (    pn,υ-1 (t) ∣ɛ(t,x)∣∣∣(t - x)2k+r+2 ∣∣ dt)  ≤      0

   (               ) (                                      )       ∫∞                ∫∞  ≤  (   p     (t) dt) (    p     (t) (ɛ (t,x))2 (t - x)4k+2r+4dt) .           n,υ-1              n,υ-1       0                  0

         ⌊       1       ∫  ≤  ------|⌈        pn,υ-1(t) ɛ2(t - x)4k+2r+4 dt     n - 1            0< ∣t-x∣< δ

                                        ⌋      ∫              2       4k+2r+2β+4   |  +       pn,υ-1(t) C  (t - x)          dt⌉ .   ∣t-x∣≥δ

Now, by Lemma 2.2, we get

            (                                      )   ∞            ∫∞                  ∣            ∣     2  ∑   p   (x) (   p      (t) ∣ɛ (t,x)∣∣(t - x)2k+r+2∣ dt) ≤       n,υ          n,υ-1            ∣            ∣   υ=1           0

         ∑∞         ∫∞  ≤  --1---    pn,υ (x)   pn,υ-1 (t) ɛ2 (t - x)4k+2r+4 dt     n - 1 υ=1                      0

     2  ∑∞          ∞∫  + -C----    pn,υ (x)    pn,υ-1(t) (t - x)4k+2r+2β+4dt.    n - 1 υ=1                      0

           [                                     ]≤  ɛ2--1--- Tn,4k+2r+4 (x) - (- x)4k+2r+4 (1 + x)- n       n - 1

    C2  [                                           ]+ ------ Tn,4k+2r+2β+4(x) - (- x)4k+2r+2 β+4 (1 + x)-n  .    n - 1

       (         )     (            )  = ɛ2 O  n-(2k+r+2) + O  n -(2k+r+ β+2) .

By Lemma 2.1, we have

       k+1           ∑        i+j   ( -j∕2)   ( -(2k+r+2)∕2)  ∣I ∣ ≤ n   M  (x)              n   O  n     O   n                   2i + j ≤ r                   i,j ≥ 0

  {       (    )}1∕2  ×  ɛ2 + O  n- β     .

= O  (1) { ɛ2 + O (n-β)} 1∕2

≤ ɛ O (1).

Since ɛ > 0  is arbitrary, it follows that I -→  0 as n -→ ∞ . The assertion (3.2) follows along similar lines by using Lemma 2.4 for k + 1  in place of k  .

The last assertion follows, due to the uniform continuity of f(2k+r+2) on [a, b] ⊂ R+   (enabling δ  to became independent of x ∈ [a,b]  ) and the uniform of o(1)  term in the estimate of I3 and I4   (because, in fact, it is a polynomial in x)  .

The next result provides an estimate of degree approximation in   (r)           (r)           0  M n (f ;x) -→ f   (x), r ∈ N   .

Theorem 3.2. Let 1 ≤ p ≤ 2k +  2  and f ∈  Cα [0,∞)  be bounded on every finite subinterval of [0,∞)  . Let             α  f (t) = O (t ) as -→  ∞  for some  α > 0.  If  (p+r)  f   exists and is continuous on (a - η,b + η) ⊂ (0,∞)  , η > 0  , then for n  sufficiently large

∥∥   (r)           (r)∥∥         (   - p∕2       ( - 1∕2)      -(k+1))    M n  (f, k,x) - f    ≤  max   C1n    ωf(p+r) n      , C2n       ,

where ω  (p+r) (δ)    f  denotes the modulus continuity of f(p+r)   on (a - η, b + η)  , C1 =  C1 (k,p,r) , C2 = C2 (k,p,r,f ) and ∥.∥ denotes the sup-norm on [a,b]  .

Proof: For x ∈ [a,b] and t ∈ [0, ∞)  ,  by the hypothesis we have

            p+r                   ( (p+r)       (p+r)   )             ∑   f-(i)(x)        i   -f-----(ξ) --f----(x)--       (p+r)  (3.3)f (t) =       i!   (t - x) +         (p + r)!         (t - x)     (1 - χ (t))              i=0

+h (t,x) χ (t) ,

where ξ  lies between t and x , and χ (t)  is the characteristic function of the set [0,∞)  \(a - η,b + η) , η > 0  .Operating on this equality by M (nr)(.,k, x)  and breaking the right hand side into three parts I1, I2 and I3   say, corresponding to the three terms on the right hand side of (3.3) as in the proof of Theorem 3.1, we have

    ∑p+r   (i)         (            )  I1 =    f---(x) M (nr)  (t - x)i ,k,x       i=0    i!        (r)        (  -(k+1))    = f   (x) + O  n        ,uniformly  for all x ∈ [a,b].

To estimate I2   , we have for every δ > 0

∣∣f (p+r)(ξ) - f(p+r)(x)∣∣ ≤ ω      (∣ξ - x∣)                            f(p+r)

≤  ωf(p+r) (∣t - x ∣)

                         (            )                                 ∣t - x∣  (3.4)                  ≤   1 + ---δ---  ωf(p+r) (δ) .

Since

     ∑ k                  ∑∞  I2 =     C (j,k)(djn - 1)     p(rd)n,υ (x)        j=0                  υ=0  j

  ∫∞            (f(p+r)(ξ) - f (p+r)(x))  ×    pdjn,υ-1(t) -----------------------(t - x)(p+r) (1 - χ (t)) dt.                         (p + r)!    0

Using (3.4) and Lemma 2.3, we have

               ∑k           ∑∞ ∣         ∣  ∣I2∣ ≤ ----1---     ∣C (j,k)∣    ∣∣p(r)  (x)∣∣        (p + r)! j=0          υ=0  djn,υ

  ∫∞            (           )  ×    pdjn,υ-1(t)  1 + ∣t --x∣  ∣t - x∣(p+r)ωf(p+r) (δ) dt                          δ    0

   ωf(p+r) (δ) ∑k              ∑            i ∣qi,s,r (x)∣  ≤  ----------    ∣C (j,k)∣             (djn) --r-------r     (p + r) ! j=0           2i + s ≤ r       x  (1 + x)                               i,s ≥ 0

  ∑∞                         ∫∞           (                          )  ×     p     (x) ∣(υ - d nx)∣s   p       (t)  ∣t - x∣p+r + 1-∣t - x ∣p+r+1 dt.         djn,υ           j         djn,υ- 1                 δ     υ=1                        0

Putting                        ∣qi,s,r(x)∣  K  = xs∈u[pa,b]    sup     xr(1+x)r              2i + s ≤ r              i,s ≥  0   , then applying Schwarz inequality for summation and for integral and Lemmas 2.1 and 2.2 as in the proof of theorem 3.1, we get

        [                           ]∣I ∣ ≤ K  O  (n-p∕2)+  1-O (n- (p+1)∕2)  ω  (p+r) (δ).   2                    δ                 f

Choosing      -1∕2  δ = n   , it follows that

            (     )   (     )  I2 = ωf(p+r) n-1∕2  O  n-p∕2 ,

where O - term holds uniformly in x ∈ [a,b]  .

For x ∈ [a,b]  and t ∈ [0,∞) \ (a - η,b + η)  , we can choose a δ > 0  in such a way that ∣t - x∣ ≥ δ  . Hence

      ∑k                         ∑              ∣q   (x)∣  ∣I3∣ ≤     ∣C  (j,k)∣ (djn - 1)             (djn)i--ri,s,r----r        j=0                                      x  (1 + x)                                2i + s ≤ r                                i,s ≥ 0

   ∞                          ∫  × ∑   p    (x) ∣(υ - d nx)∣s      p       (t) ∣h(t,x)∣ dt.         djn,υ           j            djn,υ-1    υ=1                       ∣t-x∣≥s

Now, for ∣t - x∣ ≥ δ  we can find a positive constant M  such that                      γ  ∣h (t,x) ∣ ≤ M ∣t - x∣  , where γ  is any integer≥ max  (α,2k + r + 2)  .

Hence, by Schwarz inequality, Lemmas 2.1 and 2.2 we have

         ∑k                         ∑              ∣q    (x)∣  ∣I3∣ ≤ M      ∣C (j,k)∣ (djn - 1)            (djn)i -ri,s,r---r-           j=0                                      x  (1 + x)                                  2i + s ≤ r                                  i,s ≥ 0

   ∞                          ∫    ∑                        s                        γ  ×     pdjn,υ (x) ∣(υ - djnx) ∣     pdjn,υ-1(t) ∣t - x∣ dt.     υ=1                      ∣t-x∣≥s

    (       )        (       )  = O  n(r-γ)∕2   =  O  n- (k+1)  uniformly in x ∈ [a,b].

The required result follows on combining the estimates of I1, I2   and I3   . e o

Acknowledgement. The authors are thankful to the referee for making substantial improvements in the paper.

References:

[1] P.N. Agrawal and Kareem J. Thamer, Approximation of unbounded functions by a new sequence of linear positive operators, J. Math. Anal. App. 225(1998), 660-672.         [ Links ]

[2] P.N. Agrawal and Kareem J. Thamer, Degree of approximation by a new sequencf linear operators, Kyungpook Math. J., 41(1) (2001), 65-73.         [ Links ]

[3] P.N. Agrawal and T.A.K. Sinha, A saturation theorem for a combination of modified Lupas operators in Lp-spaces, Bull. Inst. Math. Academia Sinica 24 (1996), 159-165.         [ Links ]

[4] H.S. Kasana, P.N. Agrawal and V. Gupta, Inverse and Saturation theorems for linear combination of modified Baskakov operators, Approx. Theory Appl. 7(2)(1991), 65-82.         [ Links ]

[5] H.S. Kasana, G. Prasad, P.N. Agrawal and A. Sahai, On modified Szasz operators, Proc. Int. Conf. Math. Anal. And its Appl. Kuwait (1985), 29-41, Pergamon Press, Oxford (1988).         [ Links ]

[6] C.P. May, Saturation and Inverse theorem for combinations of a class of exponential type operators, Canad. J. Math. 28(1976), 1224-1250.         [ Links ]

[7] R.K.S. Rathore, Linear Combinations of Linear Positive Operators and Generating Relations in Special Functions, Ph.D. Thesis I.I.T. Delhi (India) (1973).         [ Links ]

[8] A. Sahai and G. Prasad, On simultaneous approximation by modified Lupas operators, J. Approx. Theory 45 (1985), 122-128.         [ Links ]

Kareem J. Thamer
Department of Mathematics,
College of Education-Amran,
Sana'a University,
Maeen Post Office, Box (13475), Sana'a – Republic of Yemen.
k_alabdullah2005@yahoo.com

May A. Al-Shibeeb
Rayed, P.O.Box (46379),
Post Code 11532, Saudia Arabia King Dom.
maey9999@hotmail.com

A.I. Ibrahem
Department of Mathematics,
College of Science,
Basrah University,
Basrah – Iraq.

Recibido: 26 de diciembre de 2002
Aceptado: 25 de agosto de 2005

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