Lp CONVERGENCE WITH RATES OF SMOOTH PICARD ... - Hindawi

L p CONVERGENCE WITH RATES OF SMOOTH PICARD SINGULAR OPERATORS GEORGE A. ANASTASSIOU

We continue with the study of smooth Picard singular integral operators on the line regarding their convergence to the unit operator with rates in the L p norm, p ≥ 1. The related established inequalities involve the higher-order L p modulus of smoothness of the engaged function or its higher-order derivative. Copyright © 2006 George A. Anastassiou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction The rate of convergence of singular integrals has been studied in [3–7, 9–11] and these articles motivate this work. Here we study the L p , p ≥ 1, convergence of smooth Picard singular integral operators over R to the unit operator with rates over smooth functions in L p (R). These operators were introduced and studied in [5] with respect to · ∞ . We establish related Jackson-type inequalities involving the higher L p modulus of smoothness of the engaged function or its higher-order derivative. The discussed operators are not in general positive. Other motivation derives from [1, 2]. 2. Results Next we deal with the smooth Picard singular integral operators Pr,ξ ( f ;x) defined as follows. For r ∈ N and n ∈ Z+ , we set ⎧ ⎪ ⎪ r −j r ⎪ (−1) j −n , ⎪ ⎪ ⎨ j αj = ⎪ r ⎪ r− j r ⎪ ⎪ 1 − ( − 1) j −n , ⎪ ⎩ j

j = 1,...,r, (2.1) j = 0,

j =1

that is

r

j =0 α j

= 1.

Hindawi Publishing Corporation Proceedings of the Conference on Differential & Difference Equations and Applications, pp. 31–45

32

L p convergence with rates of smooth Picard singular operators Let f ∈ L p (R) ∩ C n (R), 1 ≤ p < ∞, we define for x ∈ R, ξ > 0 the Lebesgue integral 1 Pr,ξ ( f ;x) = 2ξ

∞ r −∞

j =0

α j f (x + jt) e−|t|/ξ dt.

(2.2)

Pr,ξ operators are not positive operators, see [5]. ∞ We notice by (1/2ξ) −∞ e−|t|/ξ dt = 1, that Pr,ξ (c,x) = c, c constant, and

∞ r

1 αj f (x + jt) − f (x) e−|t|/ξ dt . Pr,ξ ( f ;x) − f (x) = 2ξ j =0 −∞

(2.3)

We use also that

∞ −∞

k −|t |/ξ

t e

⎧ ⎨0,

k odd,

dt = ⎩ 2k!ξ k+1 ,

k even.

(2.4)

We need the rth L p -modulus of smoothness

ωr f (n) ,h

p

:= sup Δrt f (n) (x) p,x , |t |≤h

h > 0,

(2.5)

where Δrt

f

(n)

(x) :=

r

(−1)

r− j

j =0

r (n) f (x + jt), j

(2.6)

see [8, page 44]. Here we have ωr ( f (n) ,h) p < ∞, h > 0. We need to introduce r

δk :=

j =1

α j jk,

k = 1,...,n ∈ N,

(2.7)

and denote by · the integral part. Call τ(w,x) :=

r j =0

α j j n f (n) (x + jw) − δn f (n) (x).

(2.8)

Notice also that −

r

(−1)

r− j

j =1

r r . = (−1)r j 0

(2.9)

According to [3, page 306] and [1], we get τ(w,x) = Δrw f (n) (x).

(2.10)

George A. Anastassiou 33 Thus τ(w,x)

p,x

≤ ωr f (n) , |w | p ,

w ∈ R.

(2.11)

Using Taylor’s formula, one has r j =0

α j f (x + jt) − f (x) =

n f (k) (x)

k!

k =1

δk t k + ᏾n (0,t,x),

(2.12)

where

t

᏾n (0,t,x) :=

0

(t − w)n−1 τ(w,x)dw, (n − 1)!

n ∈ N.

(2.13)

Using the above terminology, we obtain Δ(x) := Pr,ξ ( f ;x) − f (x) −

n/2

m=1

f (2m) (x)δ2m ξ 2m = ᏾n∗ (x),

(2.14)

where 1 ᏾n (x) := 2ξ ∗

∞ −∞

᏾n (0,t,x)e−|t|/ξ dt,

n ∈ N.

(2.15)

In Δ(x), see (2.14), the sum collapses when n = 1. We present our first result. Theorem 2.1. Let p, q > 1 such that 1/ p + 1/q = 1, n ∈ N, and the rest as above. Then Δ(x) ≤ p

(r p + 1)

1/ p

21/q τ 1/ p ξ n wr f (n) ,ξ p ,

1/q q2 (n − 1) + q (n − 1)!

(2.16)

where ⎡ ⎤ np

∞ 2 r p+1 np − 1 − (p/2)u u e du − Γ(np)⎦ < ∞. τ := ⎣ (1 + u)

p

(2.17)

p

n−1 |t | − w τ(w,x)dw e−|t|/ξ dt . (n − 1)!

(2.18)

0

Hence, as ξ → 0, Δ(x) p → 0. Proof. We observe that Δ(x) p =

∞

1 (2ξ) p

1 ≤ (2ξ) p

−∞

∞ −∞

p

᏾n (0,t,x)e−|t|/ξ dt

|t |

0

L p convergence with rates of smooth Picard singular operators

34

Hence we have I :=

∞ −∞

Δ(x) p dx ≤

1 (2ξ) p

∞

∞ −∞

−∞

γ(t,x)e−|t|/ξ dt

p

dx ,

(2.19)

where γ(t,x) :=

n −1

|t| |t | − w τ(w,x)dw ≥ 0. (n − 1)! 0

(2.20)

Therefore by using Hölder’s inequality, suitably we obtain 2 p −2 ξ −1 q p −1

R.H.S. (2.19) ≤

∞ ∞ −∞

−∞

γ p (t,x)e−| pt|/2ξ dt dx .

(2.21)

Again by Hölder’s inequality, we have γ p (t,x) ≤

|t|

τ(w,x) p dw |t |np−1 0

p

p/q . (n − 1)! q(n − 1) + 1

(2.22)

Consequently, we have R.H.S. (2.21) 2 p −2 ξ −1 ≤ q p −1

∞

−∞

|t|

τ(w,x) p dw |t |np−1 0 −| pt |/2ξ dt dx =: (∗),

p

p/q × e −∞ (n − 1)! q(n − 1) + 1 ∞

(2.23)

calling c1 :=

ξq

2 p −2

p

p/q , (n − 1)! q(n − 1) + 1

p −1

(2.24)

and

(∗) = c1

−∞

= c1

∞ −∞

≤ c1

∞

∞ −∞

∞

−∞

|t |

0

|t |

0

τ(w,x) p dw |t |np−1 e−| pt|/2ξ dx dt

|t |

0

∞ −∞

ωr f

r (n) p Δ f (x) dx dw |t |np−1 e−| pt|/2ξ dt w

(n)

p

np−1 −| pt |/2ξ

,w p dw |t |

e

dt .

(2.25)

George A. Anastassiou 35 So far we have proved

I ≤ c1

∞ −∞

|t |

0

ωr f

(n)

p

np−1 −| pt |/2ξ

,w p dw |t |

e

dt .

(2.26)

By [8, page 45], we have

R.H.S. (2.26) ≤ c1 ωr f (n) ,ξ

×

∞

p

p |t |

1+

−∞

0

w ξ

r p

(2.27)

dw · |t |np−1 e−| pt|/2ξ dt =: (∗∗).

But we see that

ξc1 (n) p ωr f ,ξ p ᏶, (∗∗) = rp+1

(2.28)

where ᏶=2

∞ 0

t 1+ ξ

r p+1

− 1 t np−1 e− pt/2ξ dt.

(2.29)

Here we find ᏶ = 2ξ

⎡

∞

np ⎣

0

np

(1 + u)

r p+1 np−1 −(p/2)u

u

e

2 du − p

⎤

Γ(np)⎦ .

(2.30)

Thus by (2.17) and (2.30), we obtain ᏶ = 2ξ np τ.

(2.31)

Using (2.28) and (2.31), we get

(∗∗) =

ξc1 (n) p np ωr f ,ξ p 2ξ τ rp+1

p 2 p/q τξ np (n) =

p/q

p ωr f ,ξ p . (r p + 1) q2 (n − 1) + q (n − 1)!

(2.32)

This means that we have established that

I≤

2 p/q τξ np ωr f (n) ,ξ

(r p + 1)

p

p

p/q

p . 2 q (n − 1) + q (n − 1)!

That finishes the proof of the theorem.

(2.33)


36

The counterpart of Theorem 2.1 follows, case of p = 1. Theorem 2.2. Let f ∈ L1 (R) ∩ C n (R), n ∈ N. Then ⎛ ⎛ k

⎞⎞ r+1 (n − 1 + j)

= 1 j Δ(x) ≤ r! ⎝ ⎝ ⎠⎠ ξ n ωr f (n) ,ξ . 1 1 k!(r + 1 − k)! k =1

(2.34)

Hence, as ξ → 0, Δ(x)1 → 0. Proof. It follows that ∞ −|t |/ξ Δ(x) = 1 ᏾ (0,t,x)e dt n 2ξ −∞

1 ≤ 2ξ

∞ |t| −∞

0

n−1 |t | − w τ(w,x)dw e−|t|/ξ dt. (n − 1)!

(2.35)

Thus Δ(x) = 1

∞ −∞

1 ≤ 2ξ

Δ(x)dx

n −1

∞ ∞ |t| |t | − w τ(w,x)dw e−|t|/ξ dt dx =: (∗). (n − 1)! 0 −∞ −∞

(2.36)

But we see that

n −1

|t| n−1 |t | |t | − w τ(w,x)dw ≤ |t | τ(w,x)dw. (n − 1)! (n − 1)! 0 0

(2.37)

Therefore it holds that 1 (∗) ≤ 2ξ

∞ ∞ −∞

1 = 2ξ(n − 1)! 1 ≤ 2ξ(n − 1)!

−∞

|t |n−1 (n − 1)!

∞

−∞

∞

|t |

−∞

|t |

0

0

∞ −∞

0

τ(w,x)dw e−|t|/ξ dt dx

|t|

ωr f

τ(w,x)dx dw |t |n−1 e−|t|/ξ dt

(n)

(2.38)

n−1 −|t |/ξ

,w 1 dw |t |

e

dt .

That is we get Δ(x) ≤ 1

1 2ξ(n − 1)!

∞

−∞

|t |

0

ωr f

(n)

n−1 −|t |/ξ

,w 1 dw |t |

e

dt .

(2.39)

George A. Anastassiou 37 Consequently, we have Δ(x) ≤ 1

1 ωr f (n) ,ξ 1 2ξ(n − 1)!

ωr f (n) ,ξ 1 ξ n = (n − 1)!(r + 1)

∞ −∞

|t |

w 1+ ξ

0

∞

r

n−1 −|t |/ξ

dw |t |

e

dt

(2.40)

(1 + t)r+1 − 1 t n−1 e−t dt .

0

We have gotten so far (n) n Δ(x) ≤ ωr f ,ξ 1 ξ · λ , 1 (n − 1)!(r + 1)

(2.41)

where λ :=

∞

(1 + t)r+1 − 1 t n−1 e−t dt.

0

(2.42)

One easily finds that λ=

r+1

r +1 k

k =0

(n + k − 1)! − (n − 1)!.

(2.43)

But then one observes that λ (n − 1)!

=

r+1 k =1

r + 1 (n + k − 1)! . (n − 1)! k

(2.44)

We have proved (2.34). The case n = 0 is met next. Proposition 2.3. Let p, q > 1 such that 1/ p + 1/q = 1 and the rest as above. Then Pr,ξ ( f ) − f ≤ p

1/q

2 q

θ 1/ p ωr ( f ,ξ) p ,

(2.45)

where θ :=

∞ 0

(1 + x)r p e−(p/2)x dx < ∞.

Hence, as ξ → 0, Prξ → unit operator I in the L p norm, p > 1.

(2.46)

38


Proof. With some work, we notice that, see also [5], Pr,ξ ( f ;x) − f (x) =

1 2ξ

∞

−∞

Δrt f (x) e−|t|/ξ dt .

(2.47)

r Δ f (x)e−|t|/ξ dt.

(2.48)

And then Pr,ξ ( f ;x) − f (x) ≤ 1

∞

2ξ

−∞

t

We next estimate

∞ −∞

Pr,ξ ( f ;x) − f (x) p dx ≤

1 2pξ p

∞ −∞

1 4ξ = p p 2 ξ q

≤

≤

1 4ξ 2pξ p q 1

2 p −1 ξ p 2 q

−∞

p/q

r Δ f (x) p e−| pt|/2ξ dt

4ξ q

∞

∞

∞

−∞

p

r Δ f (x) p dx e−| pt|/2ξ dt t

p

ωr f , |t | p e−| pt|/2ξ dt

p ωr ( f ,ξ) p

ωr ( f ,ξ) p

−∞

∞ 0

e−|qt|/2ξ dt

−∞

p/q

∞

t

−∞

p/q

p/q =

∞

∞

0

t 1+ ξ

dx

r p

p/q

(2.49)

e

− pt/2ξ

dt

(1 + x)r p e−(p/2)x dx .

Clearly we have established (2.45). We also give the following. Proposition 2.4. It holds that Pr,ξ f − f ≤ er! ωr ( f ,ξ)1 . 1

(2.50)

Hence, as ξ → 0, Pr,ξ → I in the L1 norm. Proof. We do have again Pr,ξ ( f ;x) − f (x) ≤ 1

2ξ

∞ −∞

r Δ f (x)e−|t|/ξ dt. t

(2.51)

George A. Anastassiou 39 We estimate

∞ −∞

∞ ∞

Pr,ξ ( f ;x) − f (x)dx ≤ 1

2ξ

≤

1 2ξ

∞

−∞

−∞

ωr f , |t | 1 e−|t|/ξ dt ≤

= ωr ( f ,ξ)1

∞ 0

= ωr ( f ,ξ)1 r!

−∞

r Δ f (x)e−|t|/ξ dt dx t

ωr ( f ,ξ)1 ξ

∞

r −x

(1 + x) e dx = ωr ( f ,ξ)1

1+ 0

e−t/ξ dt

k

k!

(2.52)

= ωr ( f ,ξ)1 er! .

k!

k =0

r

r r k =0

r 1

t ξ

We have proved (2.50).

Next we consider f ∈ C n (R) ∩ L p (R), n = 0 or n ≥ 2 even, 1 ≤ p < ∞, and the similar smooth singular operator of symmetric convolution type 1 Pξ ( f ;x) = 2ξ

∞ −∞

f (x + y)e−| y|/ξ d y

∀x ∈ R, ξ > 0.

(2.53)

That is Pξ ( f ;x) =

1 2ξ

∞

0

f (x + y) + f (x − y) e− y/ξ d y

(2.53)∗

for all x ∈ R, ξ > 0. Notice that P1,ξ = Pξ . Let the central second-order difference

2y f (x) := f (x + y) + f (x − y) − 2 f (x). Δ

(2.54)

2− y f )(x) = (Δ 2y f )(x). When n ≥ 2 even using Taylor’s formula with Cauchy Notice that (Δ remainder we eventually find

n/2 (2ρ) f (x)

2y f (x) = 2 Δ

ρ =1

(2ρ)!

y 2ρ + ᏾1 (x),

(2.55)

(y − t)n−1 dt. (n − 1)!

(2.56)

where ᏾1 (x) :=

y 0

2t f (n) (x) Δ

Notice that Pξ ( f ;x) − f (x) =

1 2ξ

∞ 0

2y f (x) e− y/ξ d y. Δ

(2.57)


40

Furthermore by (2.4), (2.55), and (2.57) we easily see that K(x) :=Pξ ( f ;x) − f (x) − 1 = 2ξ

∞ y 0

0

2t f Δ

n/2 ρ =1

f (2ρ) (x)ξ 2ρ

(y − t)n−1 (x) dt e− y/ξ d y. (n − 1)!

(n)

(2.58)

Therefore we have

∞ y

K(x) ≤ 1

2ξ

0

0

2 (n) (y − t)n−1 Δ t f (x) dt e− y/ξ d y. (n − 1)!

(2.59)

Here we estimate in L p norm, p ≥ 1, the error function K(x). Notice that we have ω2 ( f (n) , h) p < ∞, h > 0, n = 0, or n ≥ 2 even. Operators Pξ are positive operators. The related main L p result here comes next. Theorem 2.5. Let p, q > 1 such that 1/ p + 1/q = 1, n ≥ 2 even, and the rest as above. Then K(x) ≤ p

τ1/ p ξ n ω2 f (n) ,ξ p ,

1/q (4p + 2)1/ p q2 (n − 1) + q (n − 1)!

(2.60)

where ⎛

τ := ⎝

∞ 0

⎞

np

(1 + x)

2p+1 np−1 −(p/2)x

x

e

2 dx − p

Γ(np)⎠ < ∞.

(2.61)

Hence, as ξ → 0, K(x) p → 0. Proof. We observe that K(x) p ≤

1 2pξ p

∞

0

p (y − t)n−1 − y/ξ Δ 2t f (n) (x) dy . dt e (n − 1)!

y

0

(2.62)

Call γ(y,x) :=

y 0

2 (n) (y − t)n−1 Δ t f (x) dt ≥ 0, (n − 1)!

(2.63)

then we have K(x) p ≤

1 2pξ p

∞ 0

γ(y,x)e− y/ξ d y

p

.

(2.64)

George A. Anastassiou 41 And hence Λ :=

∞ −∞

1 ≤ p p 2 ξ =

K(x) p dx ≤

∞

∞

−∞

1 2ξq p/q

∞ ∞ −∞

p

γ(y,x) e

0

∞ ∞ −∞

1 2pξ p

0

− py/2ξ

γ(y,x)e− y/ξ d y

dy

∞ 0

p

e

p

−qy/2ξ

dx

dy

(by Hölder’s inequality)

p/q

dx

γ(y,x) e− py/2ξ d y dx =: (∗).

0

(2.65) By applying again Hölder’s inequality, we see that y 2 (n) p 1/ p Δ f (x) dt

γ(y,x) ≤

t

0

(n − 1)!

y (n−1+1/q) q(n − 1) + 1

1/q .

(2.66)

Therefore it holds that (∗) ≤

∞ ∞ y

1

q(n − 1) +1

p/q

p

(n − 1)! 2ξq p/q

−∞

0

·y

0

2 (n) p Δ f (x) dt

t

pn−1 − py/2ξ

e

(2.67)

dx d y =: (∗∗).

We call c2 :=

2ξq p/q

1

p

(n − 1)!

q(n − 1) + 1

p/q .

(2.68)

And hence (∗∗) = c2 = c2 = c2 ≤ c2

∞ ∞ y −∞

0

0

∞ y ∞

−∞

0

0

∞ y ∞ ∞ y 0

0

≤ c2 ω2 f

−∞

0

0

(n)

,ξ

2 (n) p Δ t f (x) dt dx y pn−1 e− py/2ξ d y 2 (n) p Δ t f (x) dx dt y pn−1 e− py/2ξ d y 2 (n) Δ f (x − t) p dx dt y pn−1 e− py/2ξ d y t

p

ω2 f (n) ,t p dt y pn−1 e− py/2ξ d y

p p

∞

0

y 0

t 1+ ξ

2p

dt y

pn−1 − py/2ξ

e

dy .

(2.69)


42

That is so far we proved that

Λ ≤ c2 ω2 f

(n)

,ξ

p

p

∞

y

0

0

t 1+ ξ

2p

dt y

pn−1 − py/2ξ

e

dy .

(2.70)

But R.H.S. (2.70) =

p c2 ξ ω2 f (n) ,ξ p (2p + 1)

∞

1+ 0

y ξ

2p+1

− 1 y pn−1 e− py/2ξ d y . (2.71)

Call M :=

∞

y 1+ ξ

0

2p+1

− 1 y pn−1 e− py/2ξ d y.

(2.72)

Thus M=ξ

⎛

∞

pn ⎝

0

⎞

np

(1 + x)

2p+1 pn−1 −(p/2)x

x

e

2 dx − p

Γ(np)⎠ .

(2.73)

That is we get M = ξ pn τ.

(2.74)

Therefore it holds that

Λ≤

τξ pn ω2 f (n) ,ξ

p

2(2p + 1) (n − 1)!

p

p

p/q . 2 q (n − 1) + q

(2.75)

We have established (2.60). The counterpart of Theorem 2.5 follows, p = 1 case. Theorem 2.6. Let f ∈ L1 (R) ∩ C n (R), n ≥ 2 even. Then

K(x) ≤ n (n + 1)(n + 2) + (n + 1) + 1 ξ n ω2 f (n) ,ξ . 1 1

6

2

2

(2.76)

Hence, as ξ → 0, K(x)1 → 0. Proof. Notice that 2t f (n) (x) = Δ2t f (n) (x − t), Δ

(2.77)

all x,t ∈ R. Also it holds that

∞ −∞

2 (n) Δ f (x − t)dx = t

∞ −∞

2 (n)

Δ f (w)dw ≤ ω2 f (n) ,t , t 1

all t ∈ R+ .

(2.78)

George A. Anastassiou 43 Here we obtain K(x) = 1 ≤

∞

1 |K(x)|dx ≤ 2ξ −∞

1 2ξ

(2.59)

∞ ∞ −∞

0

∞ ∞ y −∞

y n −1 (n − 1)!

y 0

0

0

2 (n) (y − t)n−1 − y/ξ Δ t f (x) d y dx dt e (n − 1)!

2 (n) Δ t f (x)dt e− y/ξ d y dx

∞ ∞ y

n −1 2 (n) y Δ t f (x)dt dx e− y/ξ d y (n − 1)! −∞ 0 0 ∞ y ∞ n −1 2 (n) y (2.77) 1 − y/ξ Δ f (x − t)dx dt = dy e t 2ξ 0 (n − 1)! 0 −∞ ∞ y n −1 (2.78) 1

y ≤ ω2 f (n) ,t 1 dt e− y/ξ d y 2ξ 0 (n − 1)! 0

2

=

1 2ξ

≤

ω2 f (n) ,ξ 2ξ

ξ n ω2 f (n) ,ξ = 6(n − 1)!

y

∞

1

1+ 0

0

∞ 1

0

t ξ

dt

(1 + x) − 1 x 3

y n−1 − y/ξ dy e (n − 1)!

n −1 −x

e dx

(n + 1)(n + 2) (n + 1) 1 n (n) ξ ω2 f ,ξ 1 . + + =n 6 2 2 (2.79)

We have proved (2.76). The related case here of n = 0 comes next. Proposition 2.7. Let p, q > 1 such that 1/ p + 1/q = 1 and the rest as above. Then P ξ ( f ) − f ≤ p

ρ1/ p ω2 ( f ,ξ) p , 21/ p q1/q

(2.80)

where ρ :=

∞ 0

(1 + x)2p e−(p/2)x dx < ∞.

(2.81)

Hence, as ξ → 0, Pξ → I in the L p norm, p > 1. Proof. From (2.57), we get Pξ ( f ;x) − f (x) p ≤

1 2pξ p

∞ 0

2 Δ f (x)e− y/ξ d y y

p

.

(2.82)

44


We then estimate

∞ −∞

Pξ ( f ;x) − f (x) p dx

1 ≤ p p 2 ξ

1 = 2ξq p/q 1 = 2ξq p/q 1 ≤ 2ξq p/q ≤

∞

∞ −∞

0

−∞

∞ ∞ −∞

0

∞

0

0

0

e

−qy/2ξ

dy

p/q

dx

2 Δ f (x − y) p dx e− py/2ξ d y y 2 Δ f (x) p dx e− py/2ξ d y y

p ω2 ( f , y) p e− py/2ξ d y

p ∞

ω2 ( f ,ξ) p 2q p/q

∞

y

∞ ∞

0

2 Δ f (x) p e− py/2ξ d y

(2.83)

(1 + x)2p e−(p/2)x dx .

The proof of (2.80) is now evident. Also we give the following. Proposition 2.8. It holds that Pξ f − f ≤ 5 ω2 ( f ,ξ)1 . 1

2

(2.84)

Hence, as ξ → 0, Pξ → I in the L1 norm. Proof. From (2.57), we have Pξ ( f ;x) − f (x) ≤ 1

∞

2ξ

2 Δ f (x)e− y/ξ d y. y

0

(2.85)

Hence we get

∞ −∞

Pξ ( f ;x) − f (x)dx ≤ 1

2ξ

=

1 2ξ

1 ≤ 2ξ ≤

We have established (2.84).

∞ ∞ −∞

0

∞ ∞

−∞

0

∞ 0

2 Δ f (x − y)dx e− y/ξ d y y 2 Δ f (x)dx e− y/ξ d y y

(2.86)

ω2 ( f , y)1 e

ω2 ( f ,ξ)1 2

∞ 0

− y/ξ

dy

5 (1 + x)2 e−x dx = ω2 ( f ,ξ)1 . 2

George A. Anastassiou 45 References [1] G. A. Anastassiou, Rate of convergence of nonpositive generalized convolution type operators, Journal of Mathematical Analysis and Applications 142 (1989), no. 2, 441–451. , Sharp inequalities for convolution-type operators, Journal of Approximation Theory 58 [2] (1989), no. 3, 259–266. , Moments in Probability and Approximation Theory, Pitman Research Notes in Mathe[3] matics Series, vol. 287, Longman Scientific & Technical, Harlow, 1993. , Quantitative Approximations, Chapman & Hall/CRC, Florida, 2001. [4] , Basic convergence with rates of smooth Picard singular integral operators, Journal of Com[5] putational Analysis and Applications 8 (2006), no. 4, 313–334. [6] G. A. Anastassiou and S. G. Gal, Convergence of generalized singular integrals to the unit, multivariate case, Applied Mathematics Reviews, Vol. 1, World Scientific, New Jersey, 2000, pp. 1–8. , Convergence of generalized singular integrals to the unit, univariate case, Mathematical [7] Inequalities & Applications 3 (2000), no. 4, 511–518. [8] R. A. DeVore and G. G. Lorentz, Constructive Approximation, Fundamental Principles of Mathematical Sciences, vol. 303, Springer, Berlin, 1993. [9] S. G. Gal, Remark on the degree of approximation of continuous functions by singular integrals, Mathematische Nachrichten 164 (1993), 197–199. , Degree of approximation of continuous functions by some singular integrals, Revue [10] d’Analyse Numérique et de Théorie de l’Approximation 27 (1998), no. 2, 251–261. [11] R. N. Mohapatra and R. S. Rodriguez, On the rate of convergence of singular integrals for Hölder continuous functions, Mathematische Nachrichten 149 (1990), 117–124. George A. Anastassiou: Department of Mathematical Sciences, The University of Memphis, Memphis, TN 38152, USA E-mail address: [email protected]