PUBLICATIONS DE L'INSTITUT MATHEMATIQUE
Nouvelle serie tome 53 (67), 1993, 73{80
ON THE RATE OF CONVERGENCE FOR MODIFIED
SZASZ-MIRAKYAN
OPERATORS ON FUNCTIONS
OF BOUNDED VARIATION
Ashok Sahai and Govind Prasad
Abstract. Recently Gupta and Agrawal [4] gave an estimate of the rate of convergence
for modied Szasz-Mirakyan operators for functions of bounded variation, using probabilistic
approach. Because of some mistakes in their paper, we were motivated to correct and improve
their results.
1. Introduction. Let f be a function dened on [0; 1). The SzaszMirakyan operator Sn applied to f is
Sn (f ; x) =
where
1
X
k=0
pk (nx)f (k=n);
pk (t) = e t tk =k!
Kasana et al. [5] proposed modied Szasz-Mirakyan operators to approximate
functions integrable on [0; 1):
Mn (f ; x) = n
1
X
k=0
pk (nx)
1
Z
0
pk (nt)f (t)dt:
We shall study Mn (f ; x) for functions of bounded variation on every nite
subinterval of [0; 1), giving quantitative estimates of the rate of convergence. Results of this type for Fourier series were obtained by Bojanic [2].
2. Auxiliary results. We shall need the following lemmas in the proof of
our theorem.
Lemma 1.
[1, pp. 104, 110] and [3, p. 304]: Berry{Essen Theorem. Let
X1 ; X2 ; . . . ; Xn be n indenpendently and identically distributed (i.i.d.) random
AMS Subject Classication (1985): Primary 41 A 30, 41 A 36
74
Ashok Sahai and Govind Prasad
variables with zero mean and a nite absolute third moment. If 2 E (X12 ) > 0,
then Supx2R jFn (x) (x)j (:41)l3;n , where Fn (x) is the distribution function of
(X1 + X2 + . . . + Xn )(n2 ) 1=2 ; (x) is the standard normal distribution function,
and l3;n is the Liapounov ratio l3;n = 3 2 3=2 n 1=2 ; 3 = E jX1 j3 .
We may note here that Gupta and Agrawal (1991) mistook 3 = E (X13 ) for
3 and since 3 > 3 for all nondegenerate distribution functions of i.i.d. Xi 's,
this led to an perror in the estimate, e.g., in their Lemma 1, it should have read
pk (nx) 5=2( nx).
p
For every x 2 (0; 1), we have pk (nx) (x)= n, where (x) =
2x 1=2 (4x2 + 3x + 1).
Proof. Let fk g be a sequence of i.i.d. random
variables all having the same
P
Poisson distribution with parameter x. Let n = nk=1 k ; then
Lemma 2.
p(n = k) = e nx(nx)k =k! = pk (nx);
wherefrom
2 = x; 3 = E j1
xj3 E (13 ) + 3xE (12 ) + 3x2 E (1 ) + x3
= 8x3 + 6x2 + x; and
p
l3;n = (8x3 + 6x2 + x)= n(x3=2 ):
So, we have
pk (nx) = P (k 1 < n k) = P
k 1 nx n nx
pnx < pnx
kpnxnx :
By using Lemma 1 we have
pk (nx)
p1
2
Also
nx)=pnx
Z (k
(k 1
p1
2
Therefore,
2
6x + 1 8x2 + 6x + 1
t2 =2 dt < 2(0:41) 8x +
p
pnx :
e
<
nx
nx)=pnx
Z (k
(k 1
nx)=pnx
e
nx)=pnx
pk (nx) t2 =2 dt 8x2 + 6x + 1
pnx
+
p1
2knx
p1nx :
p1nx < p(xn) :
x)2 ; x) (2x + 1)=n.
Proof. It is easily veried that: Mn (1; x) = 1; Mn (t; x) = (1 + nx)=n and
M2 (t2 ; x) = (x2 n2 + 4nx + 2)=n2. Hence, Mn((t x)2 ; x) = 2x=n + 2=n2, which
Lemma 3.
For n 2, we have 2x=n Mn ((t
leads to the assertion of the lemma.
P1
Lemma 4. Let kn (x; t) = n
k=0 pk (nx)pk (nt). Then
On the rate of convergence for modied Szasz-Mirakyan operators . . .
(i) For 0 y < x, we have
y
Z
0
(ii) For x < z < 1, we have
1
Z
z
75
kn (x; t)dt 2x + 1
n(x y)2
(2.1)
kn (x; t)dt 2x + 1
n(z x)2
(2.2)
Proof. The results follow from straightforward application of Lemma 3 and
from the fact that (t x)2 (x y )2 in (i) and (t x)2 (z x)2 in (ii).
Lemma 5. For x 2 (0; 1), we have
n
Z
1
x
pk (nt)dt =
k
X
j =0
pj (nx):
Proof. The above equality can be obtained by repeated integration of its
left-hand side by parts. Also,
n
x
Z
0
pk (nt)dt = 1 n
Z
1
x
pk (nt)dt =
1
X
j =k+1
pj (nx):
3. Main Result. Our main result may be stated as follows.
Theorem. Let f be a function of bounded variation on every nite subinterval of [0; 1) and let f (t) = O(et ) for some > 0 as t ! 1. If x 2 (0; 1),
then for n suÆciently large we have
Mn (f ; x)
+
1
(x2 + 6x + 3)x
f
f (x+) + f (x )g 2
n
(4x j3x + 1j)x
2
pn
1=2
jf (x+) f (x )j + O(1)e
n
2 X
k=1
p
x=p k
Vxx+x=
k gx
x (2x + 1)x
n
2
(3.1)
;
Where Vab (gx ) is total variation of gx on [a; b] and
8
>
<
f (t) f (x+); x < t < 1;
gx gx(t) = 0;
t = x;
>
:
f (t) f (x ); 0 t < x:
Proof. First,
jMn (f ; x)
1
(f (x+) + f (x ))j
2
jMn (gx; x)j + 12 jf (x+)
f (x )jjMn (Sign(t x); x)j
(3.2)
76
Ashok Sahai and Govind Prasad
Thus to estimate the left-hand-side, we need estimates for Mn (gx ; x) and
Mn (Sign(t x); x). Now,
Mn (Sign(t x); x) =
1
Z
Z
0
=
1
Sign(t
x)kn (x; t)dt
kn (x; t)dt
x
= An (x)
x
Z
0
kn (x; t)dt
Bn (x); say:
Using Lemma 5, we have
An (x) =
Z
1
x
=n
kn (x; t)dt
1
X
k=0
pk (nx)
1
Z
x
pk (nt)dt =
1
X
k=0
pk (nx)
k
X
j =0
pj (nx)
= p20 + p1 (p0 + p1 ) + p2 (p0 + p1 + p2 ) + . . .
= p20 + p21 + p22 + . . . + p0 (p1 + p2 + . . . ) + p1 (p2 + p3 + . . . ) + . . .
Let 1 = (p0 + p1 + p2 +. . . )(p0 + p1 + p2 +. . . ). Then 1 An (x) = p0 (p1 + p2 +. . . )+
p1 (p2 +p3 +. . . )+. . . . Further, An (x) (1 An (x)) = 2An (x) 1 = p20 +p21 +p22 +. . . ,
and An (x) + Bn (x) = 1; so we get
1
1
X
X
jAn (x) Bn (x)j = j2An (x) 1j = p2k (nx) p(x) pk (nx) = p(x) :
n k=0
k=0
n
To estimate Mn (gx ; x), we rst decompose [0; 1) into three parts, as follows
1
1
pk (nx)
pk (nt)gx (t)dt =
gx (t)kn (x; t)dt
0
0
k=0
Z x x=pn
Z x+x=pn
gx (t)kn (x; t)dt
=
gx (t)kn (x; t)dt +
0
x x=pn
Z 1
+
gx (t)kn (x; t)dt
x+x=pn
Mn (gx; x) = n
Z
=
I1
1
Z
X
Z
+
I2
Z
+
I3
Z
gx (t)kn (x; t)dt
= E1 + E2 + E3 ; say:
First, we estimate E2 . For t 2 I2 , we have
and so
E2
p
jgx (t)j = jgx(t) gx (x)j Vxx+x=x=pnn (gx );
Z x+x=pn
x+x=ppn
=
gx(t)kn (x; t)dt Vx x= n (gx )
kn (x; t)dt
x x=pn
x x=pn
Z x+x=pn
Z t
p
x
+x=p n
= Vx x= n (gx )
d
(
x
;
t
);
where
;
(
x
;
t
)
=
kn (x; u)du:
t
n
n
x x=pn
0
Z
x+x=pn
On the rate of convergence for modied Szasz-Mirakyan operators . . .
77
R
Since ab dt n (x; t) 1 for all [a; b] [0; 1), we have
n
X
p
jE2 j = Vxx+x=x=pnn (gx ) n1
k=0
p
x=p k
Vxx+x=
k (gx ):
(3.3)
Let uspestimate E1 . By using Lebesgue-Stieltjes integration by parts, with y =
x x= n, we get
E1 = gx(y+)n (x; y)
Since, jgx (y +)j = jgx (y +)
y
Z
0
n (x; t)dt gx (t):
gx(x)j Vyx+ (gx ), it follows that
jE1 j Vyx+ (gx )n (x; y) +
y
Z
n (x; t)d( Vtx gx ):
0
By (2.1) of Lemma 4, we have
jE1 j Vyx+ (gx) n(2xx +y1)2 + 2xn+ 1
y
Z
0
(x
1
d ( V x(g )):
t)2 t t x
Integration by parts leads to the following
Z
0
y
Z y
x
x
V^tx(gx )
x (g )) = Vy+ (gx ) + V0 (gx ) + 2
d
(
V
dt;
t
x
t
2
2
2
(x t)
(x y )
x
t)3
0 (x
1
where V^tx (gx ) is the normalized form of Vtx (gx ) and V^tx (gx ) = Vtx (gx ). Consequently, we have
Z y
i
x
2x + 1
2x + 1 h Vyx+ (gx ) V0x (gx )
x
jE1 j Vy+ (gx ) n(x y)2 + n (x y)2 + x2 + 2 (Vxt (gtx))3 dt
0
Z x x=pn
h x
i
2x + 1 V0 (gx )
1
=
+2
Vtx (gx)
2
3 dt :
n
Substituting x
x
(x
0
p
t)
x= t for the variable t in the integral above, we get
Z n
2x + 1 x
jE1 j nx2 (V0 (gx ) + Vxx x=pt (gx )dt)
1
n
X
2x + 1 x
nx2 (V0 (gx ) + Vxx x=pk (gx ))
k=1
x + 1)
2(2nx
2
n
X
k=1
Vxx x=pk (gx):
p
Finally, we evaluate E3 . Setting, Z = x + x= n, we obtain
E3 =
Z
z
1
gx (t)kn (x; t)dt =
Z
z
1
gx(t)dt n (x; t):
(3.4)
78
Ashok Sahai and Govind Prasad
We dene Qn (x; t) on [0; 2x] as
Qn(x; t) = 1
=0
So
E3 =
Z 2x
z
gx (t)dt Qn (x; t) gx(2x)
x (x; t ); 0 t < 2x
; t = 2x;
Z
1
2x
kn (x; u)du +
= E31 + E32 + E33 ; say:
Now, using partial integration for the rst term, we get
E31 = gx(z )Qn(x; z ) +
Z 2x
z
Z
1
2x
gx(t)dt n (x; t)
(3.5)
Q^ n (x; t)dt gx (t);
^ n (x; t) is the normalized form of Qn (x; t). Since Qn (x; z ) = Qn (x; z ) and
where Q
jgx(z )j Vxz (gx), we have
jE31 j Vxz (gx)Qn (x; z ) +
Z 2x
z
Q^ n (x; t)dt Vxt (gx ):
^ n (x; t) Qn (x; t) on [0; 2x), we
Further, by using Lemma 4(ii) and the fact that Q
have
Z
(2x + 1)
(2x + 1) 2x
1
z
jE31 j Vx (gx) n(x z )2 + n
d V t (g )
(
t
x )2 t x x
z
Z 1
1 2x
+ V2x (gx )
kn (x; u)du
2
2x
Z
(2x + 1) 2x
1
(2
x
+
1)
z
d V t (g )
Vx (gx) n(x z )2 + n
(
t
x )2 t x x
z
1
2x + 1 + V22xx (gx )
2
nx2
Z
(2x + 1)
(2x + 1) 2x
1
t
z
Vx (gx) n(z x)2 + n
2 dt Vx (gx ):
z (x t)
(2x + 1)
(2x + 1) n Vx2x (gx ) Vxz (gx )
=Vxz (gx )
+
n(z x)2
n
x2
(z x)2
Z 2x
o
1
t (g )dt :
+2
V
x x
z (t x)3
Thus
Z 2x
o
1
(2x + 1) n Vx2x (gx )
t (g )dt :
+
2
V
jE31 j n
x
x
x2
z (t p x)3
Replacing the variable in the last integral by x + x= u, we have
Z n
Z 2x
1
dt
x+x=pu (g )du
t
V
(
g
)
=
V
x
x
x
x
(t x)3 2x2 1
x+x=pn
21x2
n
X
k=1
p
Vxx+x= k (gx ):
On the rate of convergence for modied Szasz-Mirakyan operators . . .
Therefore
jE31 j (2xnx+2 1) Vx2x (gx ) +
n
2(2x + 1)
nx2
n
X
k=1
n
X
k=1
p
x+x= k
Vx
79
o
p
Vxx+x= k (gx )
(3.6)
(gx ):
Further, for evaluating E32 , using Lemma 4(ii) we obtain
jE32 j gx (2x) (2xnx+2 1) (2xnx+2 1)
n
X
k=1
p
Vxx+x= k (gx):
(3.7)
Finally, using Lemma 4(ii) and the assumption that f (t) = O(et ), ( > 0) as
t ! 1, we nd that for n suÆciently large,
jE33 j Mn
1
X
k=0
1
pk (nx)
Z
1
2x
1
et
e nt (nt)k
dt
k!
(nt)k
e (n )t
dt
k!
2x
k=0
Z 1
1
X
n
= mM
e x pk (mx)
pk (mt) ( )2k+1 dt; where m = n : (3.8)
m
2
x
k=0
Z 1
1
X
2k+1
x
M e m (1 + m ) pk (mx) pk (mt)dt
2x
k=0
x
M 0 (2x +nx1)2e :
Using (3.5) to (3.8), we have, for n suÆciently large,
n
2
x
2
p
X
jE3 j 3x (2nx + 1) Vxx x= k (gx ) + O(1) x (2x +n 1)e : (3.9)
k=1
= nM
X
pk (nx)
Z
Using (3.2) to (3.4) and (3.9) we are lead to the proposition (3.1) of the theorem
which can easily be checked to be asymptotically the best.
REFERENCES
[1] R.N. Bhattacharya and R.R. Rao, Normal Approximation and Asymptotic Expansions ,
Wiley, New York, 1976.
[2] R. Bojanic, An estimate of the rate of convergence for Fourier series of functions of bounded
variation, Publ. Inst. Math. (Belagrade) 26(40) (1979), 57-60.
[3] Y.S. Chow and H. Teicher, Probability Theory , Springer-Verlag, New York, 1988.
[4] V. Gupta and P.N. Agrawal, An estimate of the rate of convergence for Modied SzaszMirakyan operators of functions of bounded variation , Publ. Inst. Math. (Belagrade)
49(63) (1991), 101-107.
[5] H.S. Kasana, G. Prasad, P.N. Agrawal and A. Sahai, On modied Szasz operators , in:
Proceedings of the International Conference on Mathematical Analysis and its Applications
80
Ashok Sahai and Govind Prasad
(Kuwait) Eds. S.M. Mazhar, A. Hamoui and N.S. Faour; Pergamon Press, Oxford, 1985;
pp. 29-41.
Department of Mathematics
University of Roorkee
Roorkee 247 667 India
(Received 05 02 1992)
(Revised 03 03 1993)