Degree of Approximation of Continuous Fucntions by some Singular Integrals

Revue d’analyse numerique et de theorie de l’approximation

Tome XXVII, No.2, 1998, pp.251-261

1 Introduction

Let us denote

$C_{2\pi }=\{f:ℝ\to ℝ;f$is 2$\pi$-periodic and continuous on $ℝ\}$

and for $\alpha \in (0,1]$

$Lip\alpha =\{f\in C_{2\pi };\exists M>0$ with$|f\left(x\right)-f\left(y\right)|\le M|x-y|,\forall x,y\in ℝ\}.$

For $f\in C_{2\pi }$ and $\xi >0$ let us consider

$P(x,\xi )$ $={\left(2\xi \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}f\left(x+t\right)e^{-\left|t\right|/\xi }𝑑t,$

$Q(x,\xi )$ $=\left(\xi /\pi \right){\displaystyle {\int }_{-\pi }^{\pi }}\left[f\left(x+t\right)/\left(t^2+{\xi }^2\right)\right]𝑑t,$

$W(x,\xi )$ $={\left(\pi \xi \right)}^{-1/2}{\displaystyle {\int }_{-\pi }^{\pi }}f\left(x+t\right)e^{-t^2/\xi }𝑑t,$

called the Picard, Poisson-Cauchy and Gauss-Weierstrass singular integralsm respectively (see, e.g. [8]).

For $f\in C_{2\pi }$ and $p\in \mathbb{N}$, the $p-th$ modulus of smoothness of $f$ is defined by (see, e.g.,[5], p.47)

$${\omega }_p(f;t)=sup\{|{\mathrm{\Delta }}_h^{p}f\left(x\right)|;x,x+ph\in ℝ,0\le h\le t\},$$

where

$${\mathrm{\Delta }}_h^{p}f\left(x\right)=\sum _{k=0}^p{\left(-1\right)}^{p-k}\left(\genfrac{}{}{0pt}{}{p}{k}\right)f\left(x+kh\right)$$

The modulus ${\omega }_1(f;t)$ is denoted by $\omega (f;t)$

______________

AMS Subject Classification: 41A25, 41A35

___________

Regarding the approcimation by the previous singular itnegrals, the following estimates are obtained in ([8], [4]):

Theorem 1

If $f\in C_{2\pi }$ then as $\xi \to 0+$ we have

$\Vert f\left(x\right)-P(x,\xi )\Vert$ $=\mathrm{?}\left(\omega (f;\xi )\right),\Vert f\left(x\right)-Q(x,\xi )\Vert =\mathrm{?}\left(\omega (f;\xi )\left|\ln \left(1/\xi \right)\right|\right),$

$\Vert f\left(x\right)-W(x,\xi )\Vert$ $=\mathrm{?}\left(\omega (f;\xi ){\xi }^{-1/2}\right),$

where the uniform $\parallel \cdot \parallel$ is applied to $x$.

The main purpose of this paper is to obtain error bounds in terms of higher order moduli of smoothness, ${\omega }_n(f;\xi )$, for approximation by singular integrals of the previous type. Thus, if $f^{\left(p\right)}\in Lip\alpha$, then better approximation orders can be obtained. Also, in comparison with [8] and [4], the most estimates are obtained with explicit constants.

2 Approximation by singular integrals of Picard-Type

Firstly, we shall improve the estimate in Theorem 1

Theorem 2

If $f\in C_{2\pi }$ then we have

$$\Vert f\left(x\right)-P(x,\xi )\Vert \le \left(5/2\right){\epsilon }_2(f;\xi ),\forall \xi >0.$$

(ii)

If there exists $f^{\prime }\in Lip\alpha$ then

$$\Vert f\left(x\right)-P(x,\xi )\Vert \le \left(5/2\right){\xi }^{1+\alpha },\forall \xi >0.$$

Proof.

(i)

By the proof of Theorem 1 in [8] we have

$$P(x,\xi )-f\left(x\right)={\left(2\xi \right)}^{-1}{\int }_0^{+\mathrm{\infty }}{\varphi }_x\left(t\right)e^{-t/\xi }𝑑t,$$

where

$${\varphi }_x\left(t\right)=f\left(x+t\right)-2f\left(x\right)+f\left(x-t\right).$$

Hence

$\left|P(x,\xi )-f\left(x\right)\right|$ $\le {\left(2\xi \right)}^{-1}{\displaystyle {\int }_0^{+\mathrm{\infty }}}\left|{\varphi }_x\left(t\right)\right|e^{-t/\xi }𝑑t\le {\left(2\xi \right)}^{-1}{\displaystyle {\int }_0^{+\mathrm{\infty }}}{\omega }_2(f;t)e^{-t/\xi }𝑑t=$

$={\left(2\xi \right)}^{-1}{\displaystyle {\int }_0^{+\mathrm{\infty }}}{\omega }_2(f;\left(t/\xi \right)\xi )e^{-t/\xi }𝑑t\le {\left(2\xi \right)}^{-1}{\displaystyle {\int }_0^{+\mathrm{\infty }}}{\left[1+t/\xi \right]}^2{e}^{-t/\xi }𝑑t=$

$={\left(2\xi \right)}^{-1}{\omega }_2(f;\xi )\xi {\displaystyle {\int }_0^{+\mathrm{\infty }}}\left(1+2u+u^2\right)e^{-u}𝑑u=C{\omega }_2(f,\xi ),$

where by a simple calculus we have

$$C={\int }_0^{+\mathrm{\infty }}\left(1+2u+u^2\right)e^{-u}𝑑u/2=5/2.$$

Passing to supremum with $x\in ℝ$, we get the desired estimate.

(ii)

If $f^{\prime }\in Lip\alpha$ then we get

$$\Vert f\left(x\right)-P(x,\xi )\Vert \le \left(5/2\right){\omega }_2(f;\xi )\le \left(5/2\right)\xi \omega (f^{\prime };\xi )\le \left(5/2\right){\xi }^{1+\alpha }.$$

 

Remark 3

Obviously, the order of approximation in Theorem 2 $\left(ii\right)$ cannot be obtained by Theorem 1

Now, following the ideas in [5], p.57 relation (7), we shall generalize the Pcard’s singular integral in the following way.

For $p\in \mathbb{N}$ let us consider

$$P_p(x,\xi )=-{\left(2\xi \right)}^{-1}\sum _{k=1}^{p+1}{\left(-1\right)}^k\left(\genfrac{}{}{0pt}{}{p+1}{k}\right){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(x+kt\right)e^{-\left|t\right|/\xi }𝑑t,\xi >0.$$

We shall prove

Theorem 4

(i)

If $f\in C_{2\pi }$ then we have

$$\Vert f\left(x\right)-P_n(x,\xi )\Vert \le \left[\sum _{k=0}^{p+1}\left(\genfrac{}{}{0pt}{}{p+1}{k}\right)k!\right]{\omega }_{p+1}(f;\xi ),\forall \xi >0.$$

(ii)

If there exist $f^{\left(k\right)}\in C_{2\pi }$ then we get

$$\Vert \left(x\right)-P_p(x,\xi )\Vert \le \left[\sum _{k=0}^{p+1}\left(\genfrac{}{}{0pt}{}{p+1}{k}\right)k!\right]{\xi }^p\omega (f^{\left(p\right)};\xi ),\forall \xi >0.$$

Proof. We have

$f\left(x\right)-P_p(x,\xi )$ $=f\left(x\right){\left(2\xi \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}{e}^{-\left|t\right|/\xi }𝑑t+$

$+{\left(2\xi \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}\left[{\displaystyle \sum _{k=1}^{p+1}}{\left(-1\right)}^k\left(\genfrac{}{}{0pt}{}{p+1}{k}\right)\right]f\left(x+kt\right)e^{-\left|t\right|/\xi }𝑑t$ $=$

$={\left(2\xi \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}{\left(-1\right)}^{p+1}{\mathrm{\Delta }}_t^{p+1}f\left(x\right)e^{-\left|t\right|/\xi }𝑑t,$

where from

$\left|f\left(x\right)-P_p(x,\xi )\right|$ $\le {\left(2\xi \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}{\omega }_{p+1}\left(fl\left|t\right|\right)e^{-\left|t\right|/\xi }𝑑t=$

$={\xi }^{-1}{\displaystyle {\int }_0^{+\mathrm{\infty }}}{\omega }_{p+1}(f;t)e^{-\left|t\right|/\xi }𝑑t={\xi }^{-1}{\displaystyle {\int }_0^{+\mathrm{\infty }}}{\omega }_{p+1}(f;\left(t/\xi \right)\xi )e^{-t/\xi }𝑑t\le$

(see [5], p.48)

$\le {\xi }^{-1}{\omega }_{p+1}(f;\xi ){\displaystyle {\int }_0^{+\mathrm{\infty }}}{\left(1+t/\xi \right)}^{p+1}{e}^{-t/\xi }𝑑t=$

$={\omega }_{p+1}(f;\xi ){\displaystyle {\int }_0^{+\mathrm{\infty }}}{\left(1+y\right)}^{p+1}{e}^{-u}𝑑u=C_{p+1}{\omega }_{p+1}(f;\xi ),$

where

$$C_{p+1}={\int }_0^{+\mathrm{\infty }}{\left(1+u\right)}^{p+1}{e}^{-u}𝑑u=\sum _{k=0}^{p+1}\left(\genfrac{}{}{0pt}{}{p+1}{k}\right)k!$$

Passing to supremum with $x\in ℝ$, we get the desired estimate.

By ${\omega }_{p+1}(f;\xi )\le {\xi }^p\omega (f^{\left(p\right)};\xi ),$ $\left(ii\right)$ is an immediate consequence of $\left(i\right)$.  

Remark 5

A natural question which arises refers to the construction of singular integrals of Picard type which approximate the continuous functions defined on compact intervals. Thus, for example, if $f$ is continuous on $[0,1]$ (we write $f\in C[0,1]$), then we can define

$$L(f,\xi )={\xi }^{-1}{\int }_0^{+\mathrm{\infty }}f\left(x{e}^{-t}\right)e^{-t/\xi }𝑑t,x\in [0,1],\xi >0.$$

In this case, the following pointwise estimate holds:

Theorem 6

If $f\in C[0,1]$ then

$$\left|L\left[f\right](x,\xi )-f\left(x\right)\right|\le 4\omega (f;\xi x),\forall x\in [0,1],\forall \xi >0.$$

where

$$\omega (f;t)=sup\{|f\left(x\right)-f\left(y\right)|;|x-y|\le 1,x,y\in [0,1]\}.$$

Proof. Denoting $e_i\left(t\right)=t^i,i=0,1,2,$ we get

$$L\left[e_0\right](x,\xi )=1,$$

$L\left[e_1\right](x,\xi )$ $=x{\xi }^{-1}{\displaystyle {\int }_0^{+\mathrm{\infty }}}{e}^{-t\left(1+1/\xi \right)}𝑑t=$

$=\left[x{\xi }^{-1}/\left(1+1/\xi \right)\right]\left[-{e^{-t\left(1+1/\xi \right)}|}_0^{+\mathrm{\infty }}\right]=x/\left(\xi +1\right),$

$$L\left[e_2\right](x,\xi )=x^2{\xi }^{-1}{\int }_0^{+\mathrm{\infty }}{e}^{-t\left(2+1/\xi \right)}𝑑t=x^2/\left(2\xi +1\right).$$

For fixed $x\in [0,1]$ we get

$L\left[{\left(e_1-x\right)}^2\right](x,\xi )$ $=x^2/\left(2\xi +1\right)-2x^2/\left(\xi +1\right)+x^2=$

$=2x^2{\xi }^2/\left[\left(\xi +1\right)\left(2\xi +1\right)\right]\le 2x^2{\xi }^2.$

Now, taking into account that $L$ isa positive linear operator, by [3], Theorem 6we immediately obtain

$$\left|L\left[f\right](x,\xi )-f\left(x\right)\right|\le 2\omega (f;\sqrt{2}\cdot \xi x)\le 4\omega (f;\xi x),$$

which proves the theorem.

At the end of this section we shall extendet the Picard’s singular integral to functions of two variables, in the following way.

Let us consider

$$C_{2\pi ,2\pi }=\{f:ℝ\times 𝕍\to ℝ;f\text{ is continuous on }ℝ\times ℝ\text{ and }2\pi \text{-periodic in each variable}\}\text{,}$$

$\Vert f\Vert$ $=sup\{\left|f(x,y)\right|;x,y\in ℝ\},\forall f\in C_{2\pi ,2\pi },$

$\omega (f,\xi ,\eta )$ $=sup\{|f(x+h,y+k)-f(x,y)|;0\le h\le \xi ,0\le k\le \eta ,x,y\in ℝ\},$

$\xi ,\eta$ $>0,$

and for $f\in C_{2\pi ,2\pi }$

$$P(x,y,\xi ,\eta )={\left(4\xi \eta \right)}^{-1}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f(x+t,y+s)e^{0\left|t\right|/\xi }\cdot e^{-\left|s\right|/\eta }𝑑t𝑑s,\xi ,\eta >0.$$

We shall prove  

Theorem 7

$f\in C_{2\pi ,2\pi }$ then we have

where the uniform norm $\parallel \cdot \parallel$ is applied to $x$ and $y$.

Proof. We have

$\left|P(x,y,\xi ,\eta )-f(x,y)\right|$ $=$

$={\left(4\xi \eta \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}\left[f(x+t,y+s)-f(x,y)\right]e^{-\left|t\right|/\xi }\cdot e^{-\left|s\right|/\eta }𝑑t𝑑s\le$

$\le {\left(4\xi \eta \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}\omega (f;\left|t\right|,\left|s\right|)e^{-\left|t\right|/\xi }\cdot e^{-\left|s\right|/\eta }𝑑t𝑑s=$

$={\left(\xi \eta \right)}^{-1}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}\omega (f;\left(t/\xi \right)\xi ,\left(s/\eta \right)\eta )e^{-t/\xi }\cdot e^{-s/\eta }𝑑t𝑑s\le$

$\le {\left(\xi \eta \right)}^{-1}\omega (f;\xi ,\eta ){\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}\left[1+t/\xi +s/\eta \right]e^{-t/\xi }\cdot e^{-s/\eta }𝑑t𝑑s=3\omega (f;\xi ,\eta ),$

where from passing to supremum with $x,y\in ℝ$, we get our estimate.  

3 Poisson-Cauchy and Gauss-Weierstrass-Type Integrals

Some ideas in the previous section will b e considered in the case of the Poisson-Cauchy and Gauss-Weierstrass singular integrals, too.

Firstly, we shall prove

Theorem 8

(i)

If $f\in C_{2\pi }$ then we have

$\Vert f\left(x\right)-Q(x,\xi )\Vert$ $\le \left[1+\left(1/\pi \right)\ln \left({\pi }^2+1\right)\right]{\xi }^{-1}{\omega }_2(f;\xi )+\left(2/{\pi }^2\right)\xi \Vert f\Vert ,$ (1)

$\xi$ $\in (0,1]$

and

$\Vert f\left(x\right)-W(x,\xi )\Vert$ $\le \left(1/\sqrt{\pi }\right)\left(\sqrt{\pi }/2+1+\sqrt{\pi }/4\right){\xi }^{-1}{\omega }_2(f;\xi )+$ (2)

$+\left(\xi /{\pi }^{5/2}\right)\Vert f\Vert ,\xi$ $\in (0,1]$

If, moreover, $f\ne C$ (constant) then as $\xi \to 0+$ we get

$\Vert f\left(x\right)-Q(x,\xi )\Vert$ $=\mathrm{?}\mathrm{?}\left({\xi }^{-1}(f;\xi )\right)$

$\Vert f\left(x\right)-W(x,\xi )\Vert$ $=\mathrm{?}\mathrm{?}\left({\xi }^{-1}{\omega }_2(f;\xi )\right).$

(ii)

If tf$\ne C$ (constant) and $f^{\prime }\in Lip\alpha$ then as $\xi \to 0+$ we have

$$\Vert f\left(x\right)-Q(x,\xi )\Vert =\mathrm{?}\mathrm{?}\left({\xi }^{\alpha }\right)$$

and

$$\Vert f\left(x\right)-W(x,\xi )\Vert =\mathrm{?}\mathrm{?}\left({\xi }^{\alpha }\right).$$

Proof.

(i)

By the proof of Theorem 1 in [8] (relations (4.3), (4.4), (4.8) and (4.0) we get

$Q(x,\xi )-f\left(x\right)$ $=\left(\xi /\pi \right){\displaystyle {\int }_0^{\mathrm{\wp }}}\left[{\varphi }_x\left(t\right)/\left(t^2+{\xi }^2\right)\right]𝑑t-f\left(x\right)E\left(\xi \right),$

$W(x,\xi )-f\left(x\right)$ $={\left(\pi \xi \right)}^{-1/2}{\displaystyle {\int }_0^{\pi }}{\varphi }_x\left(t\right)e^{-t^2/\xi }𝑑t-R(x,\xi ),$

where for all $\xi >0$ we have

$$\left|E\left(\xi \right)\right|=E\left(\xi \right)=1-\left(2\xi /\pi \right){\int }_0^{\pi }𝑑t/\left(t^2+{\xi }^2\right)=1-\left(2/\pi \right)arctg\left(\pi /\xi \right)\le \left(2/{\pi }^2\right)\xi$$

and

$$\left|R(x,\xi )\right|\le {\left(\sqrt{\pi }\right)}^{-1}\Vert f\Vert e^{-{\pi }^2/\xi }\le {\left(\sqrt{\pi }\right)}^{-1}\left(\xi /{\pi }^2\right)\Vert f\Vert .$$

Hence, for $x\in ℝ$ and $\xi >0$ we obtain

$$\left|Q(x,\xi )-f\left(x\right)\right|=\left(\xi /\pi \right){\int }_0^{\pi }\left[{\omega }_2(f;t)/\left(t^2+{\xi }^2\right)\right]𝑑t+\Vert f\Vert \cdot \left|E\left(\xi \right)\right|$$ (3)

and

$$\left|W(x,\xi )-f\left(x\right)\right|\le {\left(\pi \xi \right)}^{-1/2}{\int }_0^{\pi }{\omega }_2(f;t)e^{-t^2/\xi }𝑑t+\left|R(x,\xi )\right|.$$ (4)

But

$\left(\xi /\pi \right){\displaystyle {\int }_0^{\pi }}[{\omega }_2(f;t)/\left(t^2+{\xi }^2\right)]𝑑t$ $=\left(\xi /\pi \right){\displaystyle {\int }_0^{\pi }}\left[{\omega }_2(f;\left(t/\xi \right)\xi )/\left(t^2+{\xi }^2\right)\right]𝑑t\le$

$\le \left(\xi /\pi \right){\omega }_2(f;\xi ){\displaystyle {\int }_0^{\pi }}\left[{\left(1+t/\xi \right)}^2/\left(t^2+{\xi }^2\right)\right]𝑑t=$

$=\left(\xi /\pi \right){\omega }_2(f;\xi ){\displaystyle {\int }_0^{\pi }}\{1/{\xi }^2+2t/\left[\xi \left(t^2+{\xi }^2\right)\right]\}𝑑t=$

$=\left(\xi /\pi \right){\omega }_2(f;\xi )\left[\pi /{\xi }^2+\left(1/\xi \right)\ln \left(\left({\pi }^2+{\xi }^2\right)/{\xi }^2\right)\right]={\omega }_2(f;\xi )/\xi +$

$+\left(1/\pi \right){\omega }_2(f;\xi )\ln \left(\left({\pi }^2+{\xi }^2\right)/{\xi }^2\right)$ $\le \left[1+\left(1/\pi \right)\ln \left({\pi }^2+1\right)\right]{\xi }^{-1}{\omega }_2(f;\xi ),$

for all $\xi \in (0,1]$, since it is easy to prove that

$$\ln \left[\left({\pi }^2+{\xi }^2\right)/{\xi }^2\right]\le \left(1/\xi \right)\ln \left({\pi }^2+1\right),\forall \xi \in (0,1].$$

Then by (3) we immediately get (1).

Analogously, in the case of $W(x,\xi )$ we have

${\left(\pi \xi \right)}^{-1/2}{\displaystyle {\int }_0^{\pi }}\omega l2(f;t)e^{-t^2/\xi }𝑑t$ $\le {\left(\pi \xi \right)}^{-1/2}{\omega }_2(f;\xi ){\displaystyle {\int }_0^{\pi }}{\left(1+t/\xi \right)}^2{e}^{-t^2/\xi }𝑑t\ge \le$

$\le {\left(\pi \xi \right)}^{-1/2}{\omega }_2(f;\xi )\left\{{\left(\pi \xi \right)}^{1/2}/2+1+{\xi }^{-1/2}{\displaystyle {\int }_0^{+\mathrm{\infty }}}{u}^2{e}^{-u^2}𝑑u\right\}=$

(by [7],p.17, Problem 1.40,c))

$={\left(\pi \xi \right)}^{-1/2}{\omega }_2(f;\xi )\{{\left(\pi \xi \right)}^{1/2}/2+1+{\xi }^{-1/2})(\sqrt{\pi }/4)\}\le$

$\le \left(1/\sqrt{\pi }\right)\left(\sqrt{\pi }/2+1+\sqrt{\pi }/4\right){\xi }^{-1}{\omega }_2(f;\xi ),\text{ for all }\xi \in (0,1],$

which, together with (4), immediately proves (2).

The condition $\ne C$ (constant) implies ${\omega }_2(f;\pi )\ne 0$. Indeed, if ${\omega }_2(f;\pi )=0,$ then by [5], p.52, Problem 4 we easily get that $f$ is linear on each interval, which combined with $f\in C_{2\pi }$, implies the contr5adiction $f\equiv C$ (cosntant) on $ℝ$.

Then by [2], p.488, Propery 7 we get

$$\xi ={\xi }^{-1}{\xi }^2={\xi }^{-1}\mathrm{?}\mathrm{?}\left({\omega }_2(f;\xi )\right)=\mathrm{?}\mathrm{?}\left({\xi }^{-1}{\omega }_2(f;\xi )\right),$$

which, together with (1) and (2), immediately gives

$\Vert f\left(x\right)-Q(x,\xi )\Vert$ $=\mathrm{?}\mathrm{?}\left({\xi }^{-1}{\omega }_2(f;\xi )\right),$

$\Vert f\left(x\right)-W(x,\xi )\Vert$ $=\mathrm{?}\mathrm{?}\left({\xi }^{-1}{\omega }_2(f;\xi )\right).$

(ii)

By $\left(i\right)$ we get

$$\Vert f\left(x\right)-Q(x,\xi )\Vert =\mathrm{?}\mathrm{?}\left({\xi }^{-1}{\omega }_2(f;\xi )\right)=\mathrm{?}\mathrm{?}\left({\xi }^{-1}\xi \epsilon (f^{\prime };\xi )\right)=\mathrm{?}\mathrm{?}\left({\xi }^{\alpha }\right).$$

The proof in the case of $W(x,\xi )$ is entirely analogous.

 

Remark 9

Obviosuly, the estimates in Theorem 8 cannot be obtained by Theorem 1 On the other hand, note that the same condition $f\equiv C$ (constant) is necessary for the validity of the estimate in Theorem 1, too, concerning the approximation by $Q(x,\xi )$ and $W(x,\xi )$.

Remark 10

The method in [5], p.57, relation (7) can be used in the Poisson-Cauchy and Gauss-Weierstrass integrals, too. As, for example, the Gauss-Weiterstrass singular integrals can be generalized by

$$W_p(x,\xi )=-\left[1/\left(2C\left(\xi \right)\right)\right]\sum _{k=1}^{p+1}{\left(-1\right)}^k\left(\genfrac{}{}{0pt}{}{p+1}{k}\right){\int }_{-\pi }^{\pi }f\left(x+kt\right)e^{-t^2/\xi }𝑑t,$$

where $p\in \mathbb{N}\cup \{0\},\xi >0,r>p/2+2$ and $C\left(\xi \right)={\int }_0^{\pi }{e}^{-t^2/{\xi }^2}𝑑t$, then an analogue with Theorem 4 (in Section 2) can be proved in this case, too.

Firstly, we need the following.

Lemma 11

We have

Proof. We can write (see, e.g., [7], p.17, Problem 1.40,c))

$${\int }_0^{\pi }{e}^{-t^2/\xi }𝑑t=\xi {\int }_0^{\pi /\xi }{e}^{-u^2}𝑑u\le \xi {\int }_0^{+\mathrm{\infty }}{e}^{-u^2}𝑑u=\xi \sqrt{\pi }/2,\forall \xi >0.$$

On the other hand, for $\xi \le 1$ we get

$$\xi {\int }_0^{\pi /\xi }{e}^{-u^2}𝑑u\ge \xi {\int }_0^{\pi }{e}^{-u^2}𝑑u,$$

since $e^{-u^2}>0$ and $\pi /\xi \ge \pi$, which proves the lemma.

Similar with Theorem 4  

Theorem 12

We have

(5)

Proof. We get

$$f\left(x\right)-W_p(x,\xi )=\left[1/\left(2C\left(\xi \right)\right)\right]{\int }_{-\pi }^{\pi }{\left(-1\right)}^{p+1}{\mathrm{\Delta }}_t^{p+1}f\left(x\right)e^{-t^2/{\xi }^2}𝑑t,$$

which implies

$\left|f\left(x\right)-W_p(x,\xi )\right|$ $\le \left[1/C\left(\xi \right)\right]{\displaystyle {\int }_0^{\pi }}{\omega }_{p+1}(f;t)e^{-t^2/{\xi }^2}𝑑t\le$

$\le \left[1/C\left(\xi \right)\right]{\omega }_{p+1}(f;\xi ){\displaystyle {\int }_0^{\pi }}{\left[1+t/\xi \right]}^{p+1}{e}^{-t^2/{\xi }^2}𝑑t=$

$=\left[\xi /C\left(\xi \right)\right]{\omega }_{p+1}(f;\xi ){\displaystyle {\int }_0^{\pi /\xi }}{\left[1+u\right]}^{p+1}{e}^{-u^2}𝑑y\le$

Lemma 11

$\le \left[\xi /\left(\xi {\displaystyle {\int }_0^{\pi }}{e}^{-u^2}𝑑u\right)\right]{\epsilon }_{p+1}(f;\xi ){\displaystyle {\int }_0^{+\mathrm{\infty }}}{\left[1+u\right]}^{p+1}{e}^{-u^2}𝑑u=$

$=\left[1/{\displaystyle {\int }_0^{\pi }}{e}^{-u^2}𝑑u\right]\left[{\displaystyle {\int }_0^{+\mathrm{\infty }}}{\left[1+u\right]}^{p+1}{e}^{-u^2}𝑑u\right]{\omega }_{p+1}(f;\xi )=$

which proves the theorem.  

4 Final Remarck

Remark 13

Related with $Q(x,\xi )$, the Poisson-Cauchy singular integral in Introduction, it is the well-known Poisson integral defined by

$$I(x,\xi )=\left(\xi /\pi \right){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\left[f\left(x+t\right)/\left(t^2+{\xi }^2\right)\right]𝑑t,\xi >0.$$

As concerns this integral, Th. Angheluţă proved in [1] the estimate

$$\Vert \left(x\right)-I(x,\xi )\Vert =\mathrm{?}\mathrm{?}\left(\omega (f;\xi )\left|\ln \left(1/\xi \right)\right|\right),\text{ as }\xi \to 0+,f\in C_{2\pi }.$$

Comparing with Theorem 1, we note that although $Q(x,\xi )$ and $I(x,\xi )$ differ in their limits of integration, they give the same order of approximation.

Remark 14

It is not difficult to verify that, for example, $Q(x,\xi )$ and $W(x,\xi )$ are positive linear operators on $C_{2\pi }$, satisfying the conditions in the classical Korovkin’s result.

Remark 15

However, it is easy to verify that the estimates which can be derived by, e.g., [3] are weaker than those given by our previous results.

Remark 16

As regards the Poisson singular integral $I(x,\xi )$ in Remark 13, a saturation theorem is proved in [6]. Then it would be of interest to obtain saturation theorems for $P(x,\xi ),P_p(x,\xi ),Q(x,\xi ),W(x,\xi )$ and $W_p(x,\xi ),$ too.

Remark 17

With respect to the Poisson singular integral $I(x,\xi ),$ it is known the following Hardy-Littlewood;s result (see, e.g., [9], p.101):

A question which arises is to give an analogous characterization for $\partial P(x,\xi )/\partial x,$ $\partial P_p(x,\xi )/\partial x,\partial Q(x,\xi )/\partial x,\partial W(x,\xi )/\partial x$ and $\partial W_p(x,\xi )/\partial x,$ too.

Remark 18

Direct and converse approximation results in uniform approximation by linear combinations of Gauss-Weierstrass-type operators obtained in $W(x,\xi )$ by replacing $\pi$ with +$\mathrm{\infty }$ and $-\pi$ with $-\mathrm{\infty }$, were given in [10].

Remark 19

Also, the results in [10] are given in terms of the $L^p$-norm in [11].

Remark 20

Then it would be of interest of obtain the estimates in the present paper by replacing the uniform norm with the $L^p$-norm, $p>0$.

Received 10 II 1993        Department of Mathematics

Timişoara University

B-dul V. Pârvan

R-1900 Timişoara, România