Acta Mathematica Scientia 2014,34B(4):1157–1165
http://actams.wipm.ac.cn
APPROXIMATION BY COMPLEX
SZÁSZ-DURRMEYER
OPERATORS IN COMPACT DISKS∗
Sorin G. GAL
Department of Mathematics and Computer Science, University of Oradea
Str. Universitatii No. 1, 410087 Oradea, Romania
E-mail : galso@uoradea.ro
Vijay GUPTA†
Department of Mathematics, Netaji Subhas Institute of Technology
Sector 3 Dwarka, New Delhi-110078, India
E-mail : vijaygupta2001@hotmail.com
Abstract In the present paper, we deal with the complex Szász-Durrmeyer operators and
study Voronovskaja type results with quantitative estimates for these operators attached
to analytic functions of exponential growth on compact disks. Also, the exact order of
approximation is found.
Key words
complex Szász-Durrmeyer operators; Voronovksaja type result; exact order of
approximation in compact disks; simultaneous approximation
2010 MR Subject Classification
1
41A25; 41A28; 30E10
Introduction
Concerning the convergence of the Bernstein polynomials in the complex plane, Bernstein
[10] proved that if f : G → C is analytic in the open set G ⊂ C, with D1 ⊂ G (with D1 = {z ∈
n
P
n k
n−k
C : |z| < 1}), then the complex Bernstein polynomials Bn (f )(z) =
f (k/n),
k z (1 − z)
k=0
uniformly converges to f in D1 .
Voronovskaja-type results with quantitative estimates for the complex Bernstein, complex
q-Bernstein, complex Baskakov, complex Szász (more precisely Favard-Szász-Mirakjan), complex Bernstein-Kantorovich, complex Balázs-Szabados and complex Stancu-Kantorovich operators attached to analytic functions on compact disks and the exact order of simultaneous
approximation by these complex operators were collected by the recent book Gal [1].
Recall that for f ∈ C[0, ∞), x ∈ [0, ∞) and n ∈ N, the classical Szász operator of real
variable is defined as
∞
ν X
Sn (f )(x) =
sn,ν (x)f
,
n
ν=0
∗ Received
March 25, 2013; revised December 9, 2013.
author: Vijay GUPTA.
† Corresponding
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ACTA MATHEMATICA SCIENTIA
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ν
where sn,ν (x) = e−nx (nx)
ν! .
Also, the Szász -Stancu operator of real variable x ∈ [0, ∞) is defined by
∞
X
ν +α
α,β
Sn (f )(x) =
sn,ν (x)f
,
n+β
ν=0
where α, β are two given parameters satisfying the conditions 0 ≤ α ≤ β.
For α = β = 0 we recapture the classical Szász operator.
In Gal [2], [5], Gupta-Verma [9], the overconvergence properties of the complex variants for
the above Szász type operators were studied.
Also, in Gal [3], [4], Gal-Gupta [6], [7], Gal-Gupta-Mahmudov [8], the overconvergence
properties of various complex variants for Durrmeyer-type operators were intensively studied.
In the present paper, we deal with the Szász-Durrmeyer operators
Z +∞
∞
X
SDn (f )(z) = n
sn,ν (z)
sn,ν (t)f (t)dt ,
0
ν=0
introduced and studied in the case of real variable in Mazhar-Totik [11]. Throughout the
article, we consider DR = {z ∈ C : |z| < R}. In the present paper, we study the rate of
approximation of analytic functions and the Voronovskaja type result for the Szász-Durrmeyer
operator SDn (f, z). Also, the exact order of approximation by this operator is obtained.
2
Auxiliary Results
First, we need the following auxiliary results.
S
S
Lemma 2.1 Suppose that f : [R, +∞) DR is continuous in [R, +∞) DR , analytic
in DR and there exists B, C > 0 such that |f (x)| ≤ CeBx , for all x ∈ [R, +∞). Denoting
∞
∞
P
P
f (z) =
ck z k , z ∈ DR , we have SDn (f )(z) =
ck SDn (ek )(z) for all z ∈ DR and n > B.
k=0
Proof
k=0
For any m ∈ N and 0 < r < R, let us define
fm (z) =
m
X
cj z j if |z| ≤ r and fm (x) = f (x) if x ∈ (r, +∞).
j=0
Since |fm (z)| ≤
∞
P
|cj | · rj := Cr , for all |z| ≤ r and m ∈ N, f is continuous on [r, R], from the
j=0
hypothesis on f it is clear that for any m ∈ N it follows |fm (x)| ≤ Cr,R eBx , for all x ∈ [0, +∞).
This implies that for each fixed m, n ∈ N, n > B and z,
Z ∞
∞
j
X
(n|z|)j
−nz
−nt n j Bt
|SDn (fm )(z)| ≤ Cr,R |e
|
n
e
· t e dt
j!
j!
0
j=0
= Cr,R |e−nz |
∞
X
(n|z|)j
j=0
j!
·
nj+1
< ∞,
(n − B)j+1
since by the ratio criterium the last series is convergent. Therefore SDn (fm )(z) is well-defined.
Denoting fm,k (z) = ck ek (z) if |z| ≤ r and fm,k (x) =
f (x)
m+1
m
P
fm,k is of exponential growth on [0, ∞) and that fm (z) =
k=0
if x ∈ (r, ∞), it is clear that each
fm,k (z). Since from the linearity
No.4
S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS
of Sn we have
SDn (fm )(z) =
m
X
1159
ck SDn (ek )(z), for all |z| ≤ r,
k=0
it suffices to prove that lim SDn (fm )(z) = SDn (f )(z) for any fixed n ∈ N and |z| ≤ r. But
m→∞
this is immediate from lim kfm − f kr = 0, from kfm − f kB[0,+∞) ≤ kfm − f kr and from the
m→∞
inequality
|SDn (fm )(z) − SDn (f )(z)| ≤ |e−nz | · en|z| · kfm − f kB[0,∞) ≤ Mr,n kfm − f kr ,
valid for all |z| ≤ r. Here k · kB[0,+∞) denotes the uniform norm on C[0, +∞)-the space of all
complex-valued bounded functions on [0, +∞).
Lemma 2.2 Denoting ek (z) = z k and Tn,k (z) = SDn (ek )(z), we have the recurrence
formula
n
k+1
′
Tn,k (z) = Tn,k+1 (z) − n +
Tn,k (z).
z
z
We have
Proof
Tn,k (z) = n
=n
∞
X
ν=0
∞
X
ν=0
sn,ν (z)
Z
sn,ν (z)
+∞
e−nt
0
nν
·
ν!
Z
+∞
(nt)ν k
t dt
ν!
e−nt tν+k dt
0
This immediately implies
=
∞
1 X
(ν + k)!
sn,ν (z)
.
nk ν=0
ν!
′
∞
1 X (ν + k)! −nz (nz)ν
e
·
nk ν=0
ν!
ν!
∞
(nz)ν
(nz)ν−1
1 X (ν + k)!
−ne−nz ·
+ e−nz n ·
= k
n ν=0
ν!
ν!
(ν − 1)!
′
Tn,k
(z) =
∞
n
1 X (ν + k + 1)! −nz (nz)ν
ν
= −nTn,k (z) + · k+1
·e
·
·
z n
ν!
ν!
ν
+
k+1
ν=0
= −nTn,k (z) +
∞
n
n k + 1 X (ν + k + 1)! −nz (nz)ν
1
Tn,k+1 (z) − · k+1
·e
·
·
z
z n
ν!
ν!
ν +k+1
ν=0
n
k+1
Tn,k+1 (z) −
Tn,k (z),
z
z
which proves the recurrence in statement.
= −nTn,k (z) +
Remark 2.3 Tn,k (z) is a polynomial of degree k.
Indeed, by direct calculation we get Tn,0 (z) = SDn (e0 )(z) = 1 and since by Lemma 2.2 it
follows
z ′
nz + k + 1
(z) +
Tn,k (z),
Tn,k+1 (z) = Tn,k
n
n
taking above step by step k = 1, 2, · · · , by mathematical induction we easily get that Tn,k (z)
is a polynomial of degree k.
3
Main Results
Our first main result is the following theorem for upper bound.
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Theorem 3.1 Let 1 < R < +∞ and suppose that f : [R, +∞) ∪ DR → C is continuous
∞
P
in (R, +∞) ∪ DR , analytic in DR i.e., f (z) =
ck z k , for all z ∈ DR , and suppose that there
k=0
k
1
A
exist M > 0 and A ∈ ( R
, 1), with the property that |ck | ≤ M (2k)!
, for all k = 0, 1, · · · (which
A|z|
Bx
implies |f (z)| ≤ M e
for all z ∈ DR ) and |f (x)| ≤ Ce for all x ∈ [R, +∞).
1
. Then for all |z| ≤ r and n ∈ N, with n > B, we have
(i) Let 1 ≤ r < A
|SDn (f )(z) − f (z)| ≤
where Cr,A =
M
2r
∞
P
Cr,A
,
n
(rA)k < ∞;
k=1
(ii) If 1 ≤ r < r1 <
1
A
are arbitrary fixed, then for all |z| ≤ r and n, p ∈ N with n > B,
|SDn(p) (f )(z) − f (p) (z)| ≤
p!r1 Cr1 ,A
,
n(r1 − r)p+1
where Cr1 ,A is given as at the above point (i).
Proof
(i) By using the recurrence relation of Remark 2.3, we have
nz + k + 1
z ′
Tn,k (z) +
Tn,k (z)
n
n
for all z ∈ C, k ∈ {0, 1, 2, · · · }, n ∈ N . From this we immediately get the recurrence formula
Tn,k+1 (z) =
Tn,k (z) − z k =
nz + k
2k − 1 k−1
z
[Tn,k−1 (z) − z k−1 ]′ +
[Tn,k−1 (z) − z k−1 ] +
z
n
n
n
for all z ∈ C, k, n ∈ N . Now for 1 ≤ r < R, if we denote the norm-k.kr in C(Dr ), where
Dr = {z ∈ C : |z| ≤ r}, then by a linear transformation, the Bernstein’s inequality in the closed
unit disk becomes |Pk′ (z)| ≤ kr kPk kr , for all |z| ≤ r, where Pk (z) is a polynomial of degree ≤ k.
Thus from the above recurrence relation, we get
r
k − 1 nr + k
2k − 1 k−1
.kTn,k−1 − ek−1 kr
+
kTn,k−1 − ek−1 kr +
r
,
n
r
n
n
which implies the relation
2k − 1 k−1
2k − 1
kTn,k − ek kr ≤ r +
.kTn,k−1 − ek−1 kr +
r
.
n
n
kTn,k − ek kr ≤
In what follows we prove the result by mathematical induction with respect to k (with n ≥ 1
supposed to be fixed arbitrarily), that this recurrence implies
(2k)! k−1
r
for all k ≥ 1, n ≥ 1.
2n
Indeed for k = 1 and n ∈ N, the left hand side is 1/n and the right hand side is 1/n. Suppose
that it is true for k, the above recurrence relation implies that
2k + 1 (2k)! k−1 2k + 1 k
kTn,k+1 − ek+1 kr ≤ r +
.
r
+
r .
n
2n
n
kTn,k − ek kr ≤
It remains to prove that
2k + 1 (2k)! k−1 2k + 1 k
(2(k + 1) k
r+
.
r
+
r ≤
r ,
n
2n
n
2n
or after simplifications, equivalently to
2k + 1
r+
.(2k)! + 2r(2k + 1) ≤ (2(k + 1))!r.
n
No.4
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S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS
It is easy to see by mathematical induction that this last inequality holds true for all k ≥ 1 and
n ∈ N.
From the hypothesis on f (i.e. |f (x)| ≤ max{M, C}emax{A,B}x , for all x ∈ R+ ), it follows
that SDn (f )(z) is analytic in DR . Thus, by Lemma 2.1 we can write
SDn (f )(z) =
∞
X
ck SDn (ek )(z) =
k=0
∞
X
ck Tn,k (z), for all z ∈ DR , n > B,
k=0
which from the hypothesis on ck immediately implies for all |z| ≤ r
|SDn (f )(z) − f (z)| ≤
∞
X
|ck |.|Tn,k (z) − ek (z)| ≤
k=1
∞
X
k=1
∞
M X
Cr,A
=
(rA)k =
,
2nr
n
M
Ak (2k)! k−1
r
(2k)! 2n
k=1
where Cr,A =
M
2r
∞
P
(rA)k < ∞ for all 1 ≤ r <
k=1
1
A,
taking into account that the series
∞
P
uk is
k=1
uniformly convergent in any compact disk included in the open unit disk.
(ii) Denoting by γ the circle of radius r1 > r and center 0, since for any |z| ≤ r and v ∈ γ,
we have |v − z| ≥ r1 − r, by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N with
n > B, we have
Z
2πr1
p! SDn (f )(v) − f (v) Cr1 ,A p!
(p)
(p)
|SDn (f )(z) − f (z)| =
dv ≤
p+1
2π
(v − z)
n 2π (r1 − r)p+1
γ
Cr ,A
p!r1
= 1
,
n (r1 − r)p+1
which proves (ii) and the theorem.
The following Voronovskaja type result holds.
Theorem 3.2 Let 2 < R < +∞ and suppose that f : [R, +∞) ∪ DR → C is continuous
∞
P
ck z k , for all z ∈ DR , and that there exist
in (R, +∞) ∪ DR , analytic in DR i.e., f (z) =
k=0
k
1
A
M > 0 and A ∈ ( R
, 1), with the property that |ck | ≤ M (2k)!
, for all k = 0, 1, · · · (which implies
A|z|
Bx
|f (z)| ≤ M e
for all z ∈ DR ) and |f (x)| ≤ Ce , for all x ∈ [R, +∞).
n
o
Ar
If 1 ≤ r < r + 1 < A1 then for all |z| ≤ r and n ∈ N with n > max 1−Ar
, B , we have
SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z) ≤ Cr,A,M (f ) ,
n
n
n2
where Cr,A,M (f ) =
2M
r(1−Ar)
+
4M
(r+1)2
∞
P
(k − 1)[A(r + 1)]k < ∞.
k=2
We denote ek (z) = z k , k = 0, 1, 2, · · · and Tn,k (z) = SDn (ek , z). By Lemma 2.1,
∞
P
we can write SDn (f )(z) =
ck Tn,k (z) for all n > B. Also
Proof
k=0
′′
′
zf (z) + f (z)
z
=
n
n
Thus
∞
X
k=2
ck k(k − 1)z k−2 +
∞
∞
k=1
k=1
1X
1X
ck kz k−1 =
ck [k(k − 1) + k]z k−1 .
n
n
∞
X
k 2 z k−1 SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z) ≤
|c
|
T
(z)
−
e
(z)
−
k n,k
k
n
n
n k=1
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ACTA MATHEMATICA SCIENTIA
Vol.34 Ser.B
for all z ∈ DR , n ∈ N, n > B.
By Remark 2.3, for all n ∈ N, z ∈ C and k = 0, 1, 2, · · · , we have
Tn,k+1 (z) =
z ′
nz + k + 1
Tn,k (z) +
Tn,k (z).
n
n
If we denote
k 2 z k−1
,
n
then it is obvious that Ek,n (z) is a polynomial of degree less than or equal to k and by simple
computation and the use of above recurrence relation, we are led to
Ek,n (z) = Tn,k (z) − ek (z) −
z ′
nz + k
2z k−2 (k − 1)3
Ek−1,n (z) +
Ek−1,n (z) +
n
n
n2
for all k ≥ 1, n ∈ N and |z| ≥ r.
Using the estimate in the proof of Theorem 3.1, we have
Ek,n (z) =
|Tn,k (z) − ek (z)| ≤
((2k)!)rk−1
2n
for all k, n ∈ N, |z| ≤ r, with 1 ≤ r.
For all k, n ∈ N, k ≥ 1 and |z| ≤ r, it follows
r ′
k
2|z|k−2 (k − 1)3
|Ek,n (z)| ≤ |Ek−1,n
(z)| + r +
|Ek−1,n (z)| +
.
n
n
n2
′
Now we shall find the estimation of |Ek−1,n
(z)| for k ≥ 1. Taking into account the fact
that Ek−1,n (z) is a polynomial of degree ≤ k − 1, we have
(k − 1)2 ek−2 k−1
k−1
′
|Ek−1,n (z)| ≤
kEk−1,n kr ≤
kTn,k−1 (z) − ek−1 (z)kr + r
r
n
r
k − 1 (2(k − 1))!rk−2
rk−2 (k − 1)2
2(k − 1)(2(k − 1))!rk−3
≤
+
≤
.
r
2n
n
n
Thus
and
r ′
2(k − 1)(2(k − 1))!rk−2
|Ek−1,n (z)| ≤
n
n2
2(k − 1)(2(k − 1))!rk−2
k
2rk−2 (k − 1)3
|Ek,n (z)| ≤
+
r
+
|E
(z)|
+
,
k−1,n
n2
n
n2
for all |z| ≤ r, k ≥ 1, n ∈ N, which implies
4rk−2 (2k)!
k
+ r+
|Ek−1,n (z)| for all |z| ≤ r.
|Ek,n (z)| ≤
n2
n
For 2 ≤ k ≤ n and |z| ≤ r, taking into account that r + k/n ≤ r + 1, we get
4rk−2 (2k)!
+ (r + 1)|Ek−1,n (z)|.
n2
But E0,n (z) = E1,n = 0, for any z ∈ C and therefore by writing last inequality for 2 ≤ k ≤ n,
we easily obtain step by step the following
|Ek,n (z)| ≤
|Ek,n (z)| ≤
k
4(r + 1)k−2 X
4(r + 1)k (k − 1)(2k)!
(2j)!
≤
.
n2
(r + 1)2 n2
j=2
No.4
S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS
1163
It follows that
SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z)
n
n
n
∞
X
X
≤
|ck | · |Ek,n (z)| +
|ck | · |Ek,n (z)|
k=2
≤
k=n+1
4M
(r + 1)2 n2
But
∞
X
|ck | · |Ek,n | ≤
k=n+1
≤
∞
X
k=n+1
∞
X
∞
X
(k − 1)(A(r + 1))k +
k=2
∞
X
|ck | · |Ek,n |.
k=n+1
|ck | · [|Tn,k (z) − ek (z)| + k 2 |z|k−1 /n]
|ck | ·
∞
∞
X
k 2 rk−1
(2k)!rk−1
2M X
+
|ck | ·
≤
(Ar)k
n
n
rn
k=n+1
k=n+1
k=n+1
2M
(Ar)n+1
2M
1
=
·
≤
· 2
r(1 − Ar)
n
r(1 − Ar) n
o
n
Ar
.
for all |z| ≤ r and n > max B, 1−Ar
Concluding, we obtain
SDn (f )(z) − f (z) − 1 f ′ (z) − z f ′′ (z)
n
n
≤
∞
X
4M
2M
1
(k − 1)(A(r + 1))k +
· 2,
2
2
(r + 1) n
r(1 − Ar) n
k=2
where for (r + 1)A < 1 we obviously have the series convergent.
The following exact order of approximation can be obtained.
Theorem 3.3 (i) In the hypothesis of Theorem 3.2, if f is not a polynomial of degree
≤ 0 then for all 1 ≤ r < r + 1 < R we have
1
Ar
kSDn (f ) − f kr ∼ , for all n > max
,B ,
n
1 − Ar
where the constants in the equivalence depend only on f and r.
(ii) In the hypothesis of Theorem 3.2, if r < r1 < r1 + 1 < 1/A and if f is not a polynomial
of degree ≤ p, (p ≥ 1) then
1
Ar
(p)
(p)
kSDn (f ) − f kr ∼ , for all n > max
,B ,
n
1 − Ar
where the constants in the equivalence depend only on f , r, r1 and p.
o
n
Ar
, B , we can write
Proof (i) For all |z| ≤ r and n ∈ N with n > max 1−Ar
1 ′
1
f ′ (z) + zf ′′ (z)
SDn (f )(z) − f (z) =
f (z) + zf ′′ (z) + · n2 SDn (f )(z) − f (z) −
n
n
n
Applying the inequality
kF + Gk ≥ | kF k − kGk | ≥ kF k − kGk,
we obtain
1
1 2
f ′ + e1 f ′′ ′
′′
.
kSDn (f ) − f kr ≥
kf + e1 f kr − · n SDn (f ) − f −
n
n
n
r
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Since f is not a polynomial of degree ≤ 0 in DR , we get kf ′ + e1 f ′′ kr > 0. Indeed, supposing
the contrary, it follows that
f ′ (z) + zf ′′ (z) = 0, for all |z| ≤ r.
The last equality is equivalent to [zf ′ (z)]′ = 0. Therefore zf ′ (z) = C, with C is a constant,
that is f ′ (z) = Cz , for all |z| ≤ r with z 6= 0.
But since f is analytic in Dr and r ≥ 1, we necessarily have C = 0 (contrariwise, we would
get that f ′ (z) is not differentiable at z = 0, which is impossible because f ′ too is analytic in
Dr , with r ≥ 1), which implies f ′ (z) = 0 and f (z) = c for all z ∈ Dr , a contradiction with the
hypothesis. Now by Theorem 3.2, we have
f ′ + e1 f ′′ Ar
2
n SDn (f ) − f −
≤ Cr,A,M (f ), for all n > max B, 1 − Ar .
n
r
n
o
Ar
Thus, there exists n0 > max B, 1−Ar
(depending on f and r only) such that for all n ≥ n0 ,
we have
f ′ + e1 f ′′ 1 2
′
′′
≥ 1 kf ′ + e1 f ′′ k ,
kf + e1 f kr − · n SDn (f ) − f −
r
n
n
2
r
which implies that
kSDn (f ) − f kr ≥
1
kf ′ + e1 f ′′ kr
2n
for all n ≥ n0n
.
o
M (f )
Ar
For max B, 1−Ar
< n ≤ n0 − 1, we get kSDn (f ) − f kr ≥ r,n
with Mr,n (f ) =
n
n · kSDn (f ) − f kr > 0 (since kSDn (f ) − f kr = 0 for a certain n is valid only for f a constant
function, contradicting the hypothesis on f ).
Therefore, finally we have
Cr (f )
kSDn (f ) − f kr ≥
n
n
o
Ar
for all n > max B, 1−Ar , where
1
Cr (f ) =
min
Mr,n (f ), · · · , Mr,n0 −1 (f ), kf ′ + e1 f ′′ kr ,
n0 −1≥n>Ar/(1−Ar)
2
which combined with Theorem 3.1, (i), proves the desired conclusion.
(ii) The upper estimate is exactly Theorem 3.1, (ii), therefore it remains to prove the lower
estimate. Denote by Γ the circle of radius r1 and center 0. By the Cauchy’s formulas for all
|z| ≤ r and n ∈ N we get
Z
p!
SDn (f )(v) − f (v)
SDn(p) (f )(z) − f (p) (z) =
dv,
2πi Γ
(v − z)p+1
where |v − z| ≥ r1 − r for all |z| ≤ r and v n
∈ Γ.
o
Ar
For all v ∈ Γ and n ∈ N with n > max B, 1−Ar
, we get
1
1 2
f ′ (v) + vf ′′ (v)
SDn (f )(v) − f (v) =
f ′ (v) + vf ′′ (v) +
n Sn (f )(v) − f (v) −
,
n
n
n
which replaced in the Cauchy’s formula implies
SDn(p) (f )(z) − f (p) (z)
No.4
1165
S.G. Gal & V. Gupta: COMPLEX SZÁSZ-DURRMEYER OPERATORS

Z n2 SDn (f )(v) − f (v) − f ′ (v)+vf ′′ (v)

n
1
p!
f (v) + vf (v)
1 p!
=
dv
+
·
dv

n 2πi Γ (v − z)p+1
n 2πi Γ
(v − z)p+1


Z n2 SDn (f )(v) − f (v) − f ′ (v)+vf ′′ (v)

n
1 ′
1 p!
(p+1)
[zf (z)]
+ ·
dv
.
=

n
n 2πi Γ
(v − z)p+1
Z
′
′′
Passing to the norm k · kr , for all n ∈ N we obtain
kSDn(p) (f ) − f (p) kr

Z n2 SDn (f )(v) − f (v) −
1 
1
p!
(p+1) ≥
[e1 f ′ ]
− n
n
(v − z)p+1
r
2π Γ
f ′ (v)+vf ′′ (v)
n


dv ,
r
where by Theorem 3.2, it follows
Z n2 SDn (f )(v) − f (v) − f ′ (v)+vf ′′ (v)
p!
n
dv
2π
p+1
(v − z)
Γ
r
2
′
′′
2πr1 n
f
+
e
f
p!
1
SDn (f ) − f −
≤
·
2π (r1 − r)p+1 n
r1
!
∞
2M
4M X
p!r1
k
≤
+
(k − 1)[A(r + 1)]
·
.
2
r(1 − Ar) (r + 1)
(r1 − r)p+1
k=2
(p+1) > 0. Indeed, supposing the contrary it follows
Now, by hypothesis on f we have [e1 f ′ ]
r
′
that zf (z) is a polynomial of degree ≤ p, which by the analyticity of f obviously implies that
f is a polynomial of degree ≤ p, a contradiction with the hypothesis.
For the rest of the proof, reasoning exactly as in the proof of the above point (i), we
immediately get the required conclusion.
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