General Mathematics Vol. 16, No. 4 (2008), 61–71
Voronovskaja’s theorem and the exact
degree of approximation for the derivatives
of complex Riesz-Zygmund means
Sorin G. Gal
Abstract
In this paper we obtain a quantitative Voronovskaja result and
the exact orders in approximation by the derivatives of complex
Riesz-Zygmund means in compact disks.
2000 Mathematical Subject Classification: Primary : 30E10 ;
Secondary : 41A25, 41A28.
1
Introduction
Of great importance in approximation theory, Fejér’s theorem states that
if f : R → C is a continuous function with period 2π, then the sequence
(Fn∗ (g)(x))n of arithmetic mean of the sequence of partial sums of the Fourier
series of g given by
Fn∗ (g)(x)
1
=
2π
π
−π
g(t)Kn (x − t)dt, x ∈ R,
61
Voronovskaja’s theorem and the exact degree of approximation...
where the kernel Kn (u) is given by Kn (u) =
1
n
sin(nu/2)
sin(u/2)
2
, converges uni-
formly to g on [−π, π] (see Fejér [2]).
Their complex form attached to an analytic function f (z) =
π
1
f (ze )Kn (u)du =
2πn
−π
iu
n−1
ck
k=0
∞
ck z k in
k=0
a disk DR = {z ∈ C, |z| < R} and given by
1
Fn (f )(z) =
2π
62
π
sin(nu/2)
f (ze )
sin(u/2)
−π
iu
2
du =
n−k k
z ,
n
(for the last formula see e.g. Gal [4, Theorem 3.1]) also have nice approximation properties, satisfying the estimate (see Gaier [3, Theorem 1])
Cr (f )
1
Mf r
:=
, for all n ∈ N,
≤
Fn (f ) − f r ≤ Mω1 f ;
n Dr
n
n
&
'
where M > 0, 0 < r < R, f r = sup{|f (z)|; |z| ≤ r} and ω1 f ; n1 Dr is the
classical modulus of continuity of f in Dr .
In fact, from the saturation result in Bruj-Schmieder [1, p. 161-162], it
follows that if f is not a constant then the approximation order by Fn (f ) is
exactly n1 .
The Fejér means, both trigonometric and complex cases were generalized
by Riesz and Zygmund, their complex form being defined for any s ∈ N by
Rn,s (f )(z) =
n−1
k=0
ck
s k
1−
z k , n ∈ N.
n
For s = 1 one recapture the Fejér means and for s = 2 one get the means
introduced by Riesz.
Voronovskaja’s theorem and the exact degree of approximation...
63
From the saturation result in Bruj-Schmieder [1, p. 161-162] it follows
that if f is not a constant then the approximation order by Rn,s (f ) is exactly
1
.
ns
In this note we prove that the approximation order by the derivatives
of Rn,s (f ) also is exactly
1
.
ns
Useful in the proof will be the Voronovskaja’s
formula for Rn,s (f )(z) with a quantitative estimate. It is worth noting that
our method is different from that in Bruj-Schmieder [1]. Also, in addition
we obtain a Voronovskaja result with a quantitative upper estimate and the
exact orders in simultaneous approximation by derivatives.
2
Main Results
First we prove an upper estimate in approximation by the derivatives of
Rn,s (f ).
Theorem 1 Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that
∞
|ck |k s r1k < ∞.
f : DR → C is analytic in DR . Also, denote Cr1 ,s (f ) =
If 1 ≤ r < r1 < R and s, p ∈ N then we have
(p)
Rn,s
(f ) − f (p) r ≤
k=0
r1 p!Cr1 ,s (f )
, n ∈ N.
(r1 − r)p+1ns
Proof. An estimate of the form
Rn,s (f ) − f r ≤
Cr,s (f )
, n ∈ N,
ns
essentially follows from Bruj-Schmieder [1]. Below we reprove this estimate
in a different and simple way, with an explicit constant Cr,s (f ). Thus,
Voronovskaja’s theorem and the exact degree of approximation...
k
denoting ek (z) = z and writing f (z) =
∞
64
ck ek (z) valid for all |z| < R, for
k=0
all |z| ≤ r we obtain
|Rn,s (f )(z) − f (z)| ≤
∞
|ck | · |Rn,s (ek )(z) − ek (z)| =
k=0
n−1
|ck | · |Rn,s (ek )(z) − ek (z)|+
k=0
∞
∞
n−1
1 s k
|ck | · |Rn,s (ek )(z) − ek (z)| ≤ s
|ck |k r +
|ck |r k .
n
k=n
k=0
k=n
Here we used that ek (z) =
∞
cj z j , with ck = 1 and cj = 0 for all j = k.
j=0
Taking into account that for k ≥ n we have 1 ≤
ks
, from the previous
ns
inequality we get
n−1
∞
∞
1 1 1 s k
s k
|Rn,s (f )(z) − f (z)| ≤ s
|ck |k r + s
|ck |k r = s
|ck |k s r k ,
n
n
n
k=0
k=n
for all |z| ≤ r, therefore we can take Cr,s (f ) =
∞
k=0
|ck |k s r k < ∞.
k=0
Now, 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, we have
p! Rn,s (f )(v) − f (v) (p)
(p)
dv ≤
|Rn,s (f )(z) − f (z)| =
2π γ
(v − z)p+1
2πr1
Cr1 ,s (f ) p!
·
·
,
s
n
2π (r1 − r)p+1
which proves the theorem.
Useful in the proof of the exact degree of approximation will be the
Voronovskaja’s formula for Rn,s (f )(z). For this purpose, first we need the
following simple lemma.
Voronovskaja’s theorem and the exact degree of approximation...
s
Lemma 1 Let k, s ∈ N. If we denote k =
s
65
αj,s k(k − 1)...(k − (j − 1)),
j=1
then the coefficients αj,s can be chosen independent of k and to satisfy α1,s =
αs,s = 1 for all s ≥ 1 and αj,s+1 = αj−1,s + jαj,s , j = 2, ..., s, s ≥ 2.
Proof. We have
k
s+1
=
s+1
αj,s+1k(k − 1)...(k − (j − 1)) = k
j=1
s
s
αj,s (k − j + j)k(k − 1)...(k − (j − 1)) =
j=1
s
s
αj,s k(k − 1)...(k − j)+
j=1
jαj,s k(k − 1)...(k − (j − 1)) =
j=1
s
αj,s k(k − 1)...(k − (j − 1)) =
j=1
s+1
αj−1,s k(k − 1)...(k − (j − 1))+
j=2
jαj,sk(k − 1)...(k − (j − 1)) = αs,s +
j=1
s
αj−1,s k(k − 1)...(k − (j − 1))+
j=2
s
jαj,s k(k − 1)...(k − (j − 1)) + α1,s ,
j=2
which implies
s+1
αj,s+1k(k − 1)...(k − (j − 1)) = α1,s+1 +
j=1
s
αj,s+1k(k − 1)...(k − (j − 1))+
j=2
αs+1,s+1 = αs,s +
s
αj−1,s k(k − 1)...(k − (j − 1))+
j=2
s
jαj,s k(k − 1)...(k − (j − 1)) + α1,s ,
j=2
and proves the lemma.
Now, the Voronovskaja’s theorem for Rn,s (f )(z) one states as follows.
Voronovskaja’s theorem and the exact degree of approximation...
66
Theorem 2 Let R > 1, DR = {z ∈ C; |z| < R} and let us suppose that
∞
f : DR → C is analytic in DR , that is we can write f (z) =
ck z k , for all
k=0
z ∈ DR . For any r ∈ [1, R) we have
,
,
s
,
,
1
Ar,s (f )
,
(j) ,
αj,s ej f , ≤
,n ∈ N
,Rn,s (f ) − f + s
,
,
n j=1
ns+1
r
∞
where Ar,s (f ) =
|ck |k s+1 r k < ∞, ej (z) = z j and αj,s are defined by
k=1
Lemma 1.
Proof. Denoting
Ek,n,s(z) = Rn,s (ek )(z) − ek (z) +
s
1 αj,s k(k − 1)...(k − (j − 1))z k ,
s
n j=1
we obtain
∞
s
1
αj,s z j f j (z) ≤
|ck | · |Ek,n,s (z)| =
Rn,s (f )(z) − f (z) + s
n j=1
k=0
n−1
k=1
|ck | · |Ek,n,s(z)| +
∞
|ck | · |Ek,n,s(z)| =
k=n
s s
k
1 |ck | · 1 −
αj,s k(k − 1)...(k − (j − 1)) · |z|k +
−1+ s
n
n j=1
k=1
s
∞
1
|ck | · −1 + s
αj,s k(k − 1)...(k − (j − 1)) · |z|k =
n j=1
k=n
∞
s
1
|ck | · −1 + s
αj,sk(k − 1)...(k − (j − 1)) · |z|k ,
0+
n j=1
k=n
n−1
where for the first sum we used Lemma 1.
Voronovskaja’s theorem and the exact degree of approximation...
Taking into account that by Lemma 1 we have
s
67
αj,s n(n − 1)...(n −
j=1
(j − 1)) = ns , for all |z| ≤ r it follows
s
1 j j
αj,s z f (z) ≤
Rn,s (f )(z) − f (z) + s
n
j=1
-
∞
. |ck |r k
·
s
k=n
k(k − 1)...(k − (j − 1))
αj,s
ns
j=1
.
=
∞
∞
∞
1 1 1 s k
s+1 k
|c
|k
r
≤
|c
|k
r
≤
|ck |k s+1 r k ,
k
k
ns k=n
ns+1 k=n
ns+1 k=1
which proves the theorem.
By using Theorems 1 and 2 we are in position to obtain the exact degree
of approximation by the derivatives of Rn,s (f )(z).
Theorem 3 Let DR = {z ∈ C; |z| < R} be with R > 1 and let us suppose
∞
ck z k , for all z ∈ DR .
that f : DR → C is analytic in DR , i.e. f (z) =
k=0
Also, let 1 ≤ r < r1 < R and p, s ∈ N be fixed. If f is not a polynomial of
degree ≤ p − 1 then we have
(p)
(f ) − f (p) r ∼
Rn,s
1
,
ns
where the constants in the equivalence depend on f , r, r1 , s and p.
Proof. Denoting by Γ the circle of radius r1 > and center 0 (where r1 >
r ≥ 1), by the Cauchy’s formulas it follows that for all |z| ≤ r and n ∈ N
we have
(p)
(f )(z)
Rn,s
−f
(p)
p!
(z) =
2πi
Γ
Rn,s (f )(v) − f (v)
dv,
(v − z)p+1
Voronovskaja’s theorem and the exact degree of approximation...
68
where the inequality |v − z| ≥ r1 − r is valid for all |z| ≤ r and v ∈ Γ.
Taking into account Theorem 1, it remains to prove the lower estimate
(p)
for Rn,s (f ) − f (p) r . For this purpose, for all v ∈ Γ and n ∈ N we have
/
s
1
Rn,s (f )(v) − f (v) = s −
αj,s v j f (j) (v)+
n
j=1
0
.12
s
1 s+1
1 n
αj,s v j f (j) (v)
,
Rn,s (f )(v) − f (v) + s
n
n j=1
which replaced in the above Cauchy’s formula implies
⎧
s
⎪
⎪
αj,s v j f (j) (v)
⎪
⎪
−
⎨
1
p!
j=1
(p)
Rn,s
(f )(z) − f (p) (z) = s
dv+
n ⎪
2πi Γ
(v − z)p+1
⎪
⎪
⎪
⎩
1 p!
·
n 2πi
s+1
n
Rn,s (f )(v) − f (v) +
1 p!
·
n 2πi
⎫
⎪
⎪
⎪
⎪
⎪
⎬
.
αj,sv j f (j) (v)
j=1
(v −
Γ
1
ns
s
dv
z)p+1
⎪
⎪
⎪
⎪
⎪
⎭
⎧0
1(p)
s
1 ⎨
αj,s v j f (j) (v)
+
−
ns ⎩
j=1
.
s
s+1
Rn,s (f )(v) − f (v) + n1s
αj,s v j f (j) (v)
n
j=1
Γ
(v −
⎫
⎪
⎪
⎪
⎪
⎪
⎬
dv
z)p+1
Passing now to · r it follows
⎧,0
1(p) ,
, ,
s
⎨
,
1 ,
(p)
(p)
(j)
, −
Rn,s (f ) − f r ≥ s ,
α
e
f
j,s
j
,
n ⎩,
, j=1
,
r
=
⎪
⎪
⎪
⎪
⎪
⎭
Voronovskaja’s theorem and the exact degree of approximation...
69
,
. ,
s
,
,
,
,
j (j)
1
s+1
n
(f
)(v)
−
f
(v)
+
α
v
f
(v)
R
, ,
n,s
j,s
ns
,
,
1 , p!
j=1
,
dv ,
,
p+1
,
n , 2π Γ
(v − z)
,
,
,
,
,
,
⎫
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎭
,
r
where by using Theorem 2 we obtain
,
. ,
s
,
,
,
,
αj,s v j f (j) (v)
, ns+1 Rn,s (f )(v) − f (v) + n1s
,
, p!
,
j=1
,
,
dv
,
, ≤
, 2π Γ
,
(v − z)p+1
,
,
,
,
,
,
r
,
,
s
,
p! 2πr1 ns+1 ,
1 Ar ,s (f )p!r1
,
(j) ,
·
(f
)
−
f
+
α
e
f
.
R
,
, ≤ 1
n,s
j,s j
p+1
s
,
,
2π (r1 − r)
n j=1
(r1 − r)p+1
r1
,
,0
1(p) ,
, ,
, s
(j)
, > 0.
α
e
f
But by hypothesis on f we have ,
j,s
j
,
,
,
, j=1
r
Indeed, supposing the contrary it follows that
s
αj,s z j f (j) (z) = Qp−1 (z),
j=1
for all z ∈ Dr , where Qp−1 (z) =
degree ≤ p − 1.
p−1
Aj z j necessarily is a polynomial of
j=1
Seeking now the analytic solution in the form f (z) =
∞
βk z k , replacing
k=0
in the differential equation and taking into account again Lemma 1, by
identification of the coefficients βk we easily obtain βk = 0, for all k ≥ p,
that is f (z) necessarily is a polynomial of degree ≤ p − 1, in contradiction
with the hypothesis.
Voronovskaja’s theorem and the exact degree of approximation...
70
Therefore, there exists an index n0 depending only on f , s, p, r and r1 ,
such that for all n ≥ n0 we have
,0
1(p) ,
, s
,
, ,
(j)
,
, −
αj,s ej f
,
,
, j=1
,
r
,
,
,
, ns+1 R (f )(v) − f (v) + 1 %s α v j f (j) (v)
n,s
s
j,s
,
,
j=1
n
1 , p!
, ≥
dv
,
p+1
n,
2π
(v
−
z)
Γ
,
,
r
,0
1(p) ,
,
, s
,
1,
(j)
, ,
,
α
e
f
j,s
j
,
2,
,
, j=1
r
which immediately implies
,0
1(p) ,
,
, s
,
1 1,
(p)
(p)
(j)
, ,
αj,s ej f
Rn,s (f ) − f r ≥ s · ,
,
,
n 2 , j=1
,
r
for all n ≥ n0 .
(p)
For n ∈ {1, ..., n0 − 1} we obviously have Rn,s (f ) − f (p) r ≥
(p)
with Mr,s,,p,n(f ) = n · Rn,s(f ) − f (p) r > 0, which finally implies
f (p) r ≥
Mr,s,p,n (f )
n
(p)
Rn,s (f ) −
Cr,s,p (f )
n
for all n, where
,0
⎫
⎧
1(p) ,
, ,⎬
s
⎨
,
1,
, .
αj,sej f (j)
Cr,s,p (f ) = min Mr,s,p,1(f ), ..., Mr,s,p,n0−1 (f ), ,
,
,⎭
⎩
2,
,
j=1
r
This completes the proof.
Acknowledgement. Supported by the Romanian Ministry of Education and Research, under CEEX grant, code 2-CEx 06-11-96.
Voronovskaja’s theorem and the exact degree of approximation...
71
References
[1] I. Bruj and G. Schmieder, Best approximation and saturation on domains bounded by curves of bounded rotation, J. Approx. Theory,
100(1999), 157-182.
[2] L. Fejér, Untersuchungen über Fouriersche Reihen, Mathematische
Ann., 58(1904), 51-69.
[3] D. Gaier, Approximation durch Fejér-Mittel in der Klasse A, Mitt.
Math. Sem. Giessen, 123(1977), 1-6.
[4] S.G. Gal, Convolution-type integral operators in complex approximation, Comput. Methods and Funct. Theory, 1(2001), No. 2, 417-432.
Sorin G. Gal
University of Oradea
Department of Mathematics and Computer Science
Str. Universitatii No. 1, 410087 Oradea, Romania
e-mail: galso@uoradea.ro