On degree of approximation of the Gauss-Weierstrass means for smooth $L^p({\mathbb{R}}^n)$ functions

Abstract

The notion of μ-smooth point of an $L^p({\mathbb{R}}^n)$-function f is introduced in terms of some ‘maximal function.’ Then the connection between the order of μ-smoothness of the function f and the rate of convergence of the Gauss-Weierstrass means to f, when ε tends to 0, is obtained.

MSC:41A25, 42B08, 26A33.

1 Introduction and formulations of main results

Let $\mathrm{\Phi }\in C_0({\mathbb{R}}^n)\cap L^1({\mathbb{R}}^n)$ and $\mathrm{\Phi }(0)=1$. The Φ-means of the integral ${\int }_{{\mathbb{R}}^n}f(x)dx$ are defined as [[1], p.6]

$$M_{\epsilon ,\mathrm{\Phi }}(f)={\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(\epsilon x)f(x)dx(\epsilon >0).$$

If ${lim}_{\epsilon \to 0^{+}}{M}_{\epsilon ,\mathrm{\Phi }}(f)=l$, then it is said that the (divergent) integral ${\int }_{{\mathbb{R}}^n}f(x)dx$ is summable to l. It is possible to obtain various summability methods by choosing a suitable function Φ. For example, by letting $\mathrm{\Phi }(x)=e^{-|x|}$, $\mathrm{\Phi }(x)=e^{-{|x|}^2}$ or for $\delta >0$, $\mathrm{\Phi }(x)=$ $\left\{\begin{array}{ll}{(1-{|x|}^2)}^{\delta };& |x|\le 1\\ 0;& |x|>1\end{array}\right\}$, the classical Abel, Gauss-Weierstrass and Bochner-Riesz means and corresponding summability methods are obtained. One of the important problems in classical harmonic analysis is to construct an (unknown) function f by means of its Fourier transform $\mathfrak{F}(f)$ defined as

$$\mathfrak{F}(f)(x)={\int }_{{\mathbb{R}}^n}f(t)e^{-2\pi ix\cdot t}dt.$$

However, $\mathfrak{F}(f)$ needs not be integrable for some $f\in L^p({\mathbb{R}}^n)$, and hence the formula

$$f(x)={\int }_{{\mathbb{R}}^n}\mathfrak{F}(f)(t)e^{2\pi ix\cdot t}dt$$

becomes incorrect. To overcome this difficulty, one may apply suitable summability methods (see, e.g., [1–3]).

Whenever a function Φ is radial, it is well known that [[1], p.8] for the Φ-means of the convergent or divergent integral ${\int }_{{\mathbb{R}}^n}\mathfrak{F}(f)(t)e^{2\pi ix\cdot t}dt$, the following equality holds:

$${\int }_{{\mathbb{R}}^n}\mathfrak{F}(f)(x)e^{2\pi ix\cdot t}\mathrm{\Phi }(\epsilon x)dx={\int }_{{\mathbb{R}}^n}f(x){\phi }_{\epsilon }(t-x)dx,$$

(1.1)

where ${\phi }_{\epsilon }(x)={(1/\epsilon )}^n{\phi }_{\epsilon }(x/\epsilon )$ and $\phi (x)=\mathfrak{F}(\mathrm{\Phi })$.

In particular, putting the function $e^{-{|x|}^2}$ instead of $\mathrm{\Phi }(x)$ in (1.1), the following formula for the Gauss-Weierstrass means of the integral ${\int }_{{\mathbb{R}}^n}\mathfrak{F}(f)(t)e^{2\pi ix\cdot t}dt$

$$S(x,\epsilon )={\int }_{{\mathbb{R}}^n}f(t){\phi }_{\epsilon }(x-t)dt(\epsilon >0)$$

(1.2)

is obtained. Here, the function ${\phi }_{\epsilon }$ is defined as

$${\phi }_{\epsilon }(x)\equiv W(x,\epsilon )={(4\pi \epsilon )}^{-(n/2)}{e}^{-{|x|}^2/4\epsilon },$$

(1.3)

and called the Gauss-Weierstrass kernel.

One of the well-known and basic results for the Gauss-Weierstrass means is the following ([[4], p.5], [[5], p.223]).

Proposition 1.1 Let $f\in L^p({\mathbb{R}}^n)$ ($1\le p<\mathrm{\infty }$), and let the Gauss-Weierstrass means of f be defined as in (1.2). Then

• (a) ${lim}_{\epsilon \to 0}{\parallel S(x,\epsilon )-f\parallel }_{L^p}=0$;

• (b) ${lim}_{\epsilon \to 0}S(x,\epsilon )=f(x)$ at each x belonging to the Lebesgue set of f;

• (c) ${sup}_{\epsilon >0}|S(x,\epsilon )|\le c(Mf)(x)$, where $(Mf)(x)$ is the Hardy-Littlewood maximal function.

Various aspects of the Gauss-Weierstrass and Abel-Poisson type summability of the multiple Fourier series and integrals have been studied in Stein and Weiss [1], Golubov [6, 7] and Gorodetskii [8]; see also Weisz [2] and [9] and references therein.

The aim of the paper is to investigate the error of approximation of $f(x)$ by its Gauss-Weierstrass means $S(x,\epsilon )$ as $\epsilon \to 0$ at the so-called μ-smoothness point of f. Note that some problems of the Bochner-Riesz summability of Fourier transform of $f\in L_p({\mathbb{R}}^n)$ at the Dini-like points was studied in [10]. Also, the rate of convergence of the Gauss-Weierstrass means of relevant Fourier series at some kind of smoothness points was studied in [9].

Definition 1.2 Let $\mu (r)$ be a positive function on $(0,\mathrm{\infty })$, and assume that ${lim}_{r\to 0^{+}}\mu (r)=0$. If $\psi (t,x)$, defined on ${\mathbb{R}}^n\times {\mathbb{R}}^n$, is measurable, we define its μ-maximal function by

$$(M_{\mu }\psi )(x)=\underset{r>0}{sup}\frac{1}{\mu (r)r^n}{\int }_{|t|<r}|\psi (t,x)|dt.$$

(1.4)

Definition 1.3 Let, for a constant $\rho <1$, a function $\mu (r)$ be a continuous and positive function on the interval $(0,\rho )$, and assume that ${lim}_{r\to 0^{+}}\mu (r)=\mu (0)=0$. We say that a function $f\in L_{loc}^1({\mathbb{R}}^n)$ is μ-smooth of order $\mu (r)$ at $x\in {\mathbb{R}}^n$ if

$$D_{\mu }(x)=\underset{0<r<1}{sup}\frac{1}{r^n\mu (r)}{\int }_{|t|<r}|f(x-t)-f(x)|dt<\mathrm{\infty }.$$

(1.5)

The points $x\in {\mathbb{R}}^n$, for which (1.5) holds, are called μ-smoothness points of f.

Remark 1.4 Simple characterization of a μ-smoothness point is not known. However, most of the classes of ‘smooth’ functions in a classical sense have the μ-smoothness property. For example, if the modulus of continuity of f

$$w_f(r)=\underset{|x|\le r}{sup}{\parallel f(\cdot -x)-f(\cdot )\parallel }_{\mathrm{\infty }}$$

satisfies the inequality $w_f(r)\le c\mu (r)$ for $r\to 0$, then every point $x\in {\mathbb{R}}^n$ is a μ-smoothness point of f, as can easily be seen from (1.5). In particular, if f satisfies the local Lipschitz (Hölder) condition

$$|f(x-t)-f(x)|\le c{|t|}^{\alpha },0<\alpha \le 1,$$

then x is a μ-smoothness point of f, provided $\mu (r)=r^{\alpha }$.

From now on, we will assume that the function $\mu (r)$ is continued as a constant from $[0,\rho ]$ to $[\rho ,\mathrm{\infty })$, that is, $\mu (r)=\mu (\rho )$, $r\ge \rho$.

Now, we state the main results of the paper.

Theorem 1.5 Let $f\in L^p({\mathbb{R}}^n)$, $1<p<\mathrm{\infty }$, be μ-smooth at $x\in {\mathbb{R}}^n$. Then the following estimate holds:

$$|S(x,\epsilon )-f(x)|\le c_1{\int }_0^{\mathrm{\infty }}{r}^{n+1}{e}^{-r^2/4}\mu (\epsilon r)dr+c_2{\epsilon }^{-n/2}{e}^{-1/4\epsilon }(\epsilon \to 0^{+}),$$

(1.6)

where $c_1$ and $c_2$ are constants independent of ε.

Corollary 1.6 Let $f\in L^p({\mathbb{R}}^n)$, $1\le p<\mathrm{\infty }$, have the μ-smoothness property at x, and let $\mu (r)$ be a modulus of continuity (see [[11], p.40]) on $[0,\rho ]$ and continued as a constant to $[\rho ,\mathrm{\infty })$, i.e., $\mu (r)=\mu (\rho )$, $r\ge \rho$ ($0<\rho <1$). Then, under the conditions of Theorem  1.5, we have

$$|S(x,\epsilon )-f(x)|\le c\mu (\epsilon )(\epsilon \to 0).$$

(1.7)

Corollary 1.7 Let $\alpha >0$ and $\mu (r)={(\frac{1}{ln\frac{1}{r}})}^{\alpha }$, then

$$|S(x,\epsilon )-f(x)|\le c{\left(\frac{1}{ln\frac{1}{\epsilon }}\right)}^{\alpha }(\epsilon \to 0).$$

(1.8)

Corollary 1.8 Let $\alpha >0$ and $-\mathrm{\infty }<\beta <\mathrm{\infty }$ be fixed parameters. If we take $\mu (r)=r^{\alpha }{|lnr|}^{\beta }$ for $0<r\le \rho <1$ and $\mu (r)=\mu (\rho )$ for $\rho <r<\mathrm{\infty }$, then under the conditions of Theorem  1.5,

$$|S(x,\epsilon )-f(x)|\le c{\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }(\epsilon \to 0).$$

(1.9)

In particular, for $\beta =0$ in (1.9), we obtain $|S(x,\epsilon )-f(x)|\le c{\epsilon }^{\alpha }$ as $\epsilon \to 0$.

The following lemma plays a crucial role in the proof of the main results.

Lemma A (cf. [[9], Lemma A])

Suppose that φ is differentiable on $(0,\mathrm{\infty })$, and that the following limits exist:

$$\underset{r\to \mathrm{\infty }}{lim}{r}^n\mu (r)\phi (r)=0\mathit{\text{and}}\underset{r\to 0^{+}}{lim}{r}^n\mu (r)\phi (r)=0.$$

(1.10)

Let $\psi (t,x)$ be measurable on ${\mathbb{R}}^n\times {\mathbb{R}}^n$ and $(M_{\mu }\psi )(x)<\mathrm{\infty }$, then

$${\int }_{{\mathbb{R}}^n}|\psi (t,x)\phi (|t|)|dt\le (M_{\mu }\psi )(x){\int }_0^{\mathrm{\infty }}{r}^n\mu (r)|{\phi }^{\prime }(r)|dr,$$

(1.11)

where $(M_{\mu }\psi )(x)$ is a μ-maximal function defined by (1.4) and ${\phi }^{\prime }$ is a derivative of φ.

We need also the following lemmas on the well-known properties of the Gauss-Weierstrass kernel and the upper incomplete gamma function.

Lemma 1.9 [[1], p.9]

The Gauss-Weierstrass kernel, $W(t,\epsilon )={(4\pi \epsilon )}^{-(n/2)}{e}^{-{|t|}^2/4\epsilon }$, has the following property:

$${\int }_{{\mathbb{R}}^n}W(t,\epsilon )dt=1(\mathit{\text{for all }}\epsilon >0).$$

(1.12)

Lemma 1.10 [[12], p.948]

The upper incomplete gamma function, defined as

$$\mathrm{\Gamma }(s,\tau )={\int }_{\tau }^{\mathrm{\infty }}{u}^{s-1}{e}^{-u}du(s>0,\tau >0),$$

has the following asymptotic property:

$$\mathrm{\Gamma }(s,\tau )=O(1){\tau }^{s-1}{e}^{-\tau }\mathit{\text{as }}\tau \to \mathrm{\infty }.$$

(1.13)

2 Proof of the main results

Proof of Lemma A Changing variables to polar coordinates $t\to (r,\theta )$, $0<r<\mathrm{\infty }$, $\theta \in S^{n-1}$ ($S^{n-1}$ is the unite sphere of ${\mathbb{R}}^n$), the left side of (1.11) becomes

$$\begin{array}{rcl}I(x)& =& {\int }_{{\mathbb{R}}^n}|\psi (t,x)\phi (|t|)|dt\\ =& {\int }_0^{\mathrm{\infty }}{r}^{n-1}[{\int }_{S^{n-1}}|\psi (r\theta ,x)\phi (r)|d\sigma (\theta )]dr\\ =& {\int }_0^{\mathrm{\infty }}{r}^{n-1}|\phi (r)|[{\int }_{S^{n-1}}|\psi (r\theta ,x)|d\sigma (\theta )]dr.\end{array}$$

Now, denoting

$$\lambda (t)={\int }_{S^{n-1}}|\psi (t\theta ,x)|d\sigma (\theta ),0\le t<\mathrm{\infty },$$

(2.1)

$$\mathrm{\Lambda }(r)={\int }_0^r\lambda (t)t^{n-1}dt,0\le r\le \mathrm{\infty },$$

(2.2)

we get

$$\begin{array}{rcl}I(x)& =& {\int }_0^{\mathrm{\infty }}{r}^{n-1}|\phi (r)|\lambda (r)dr={\int }_0^{\mathrm{\infty }}|\phi (r)|d\mathrm{\Lambda }(r)\\ =& |\phi (r)|\mathrm{\Lambda }(r){|}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\mathrm{\Lambda }(r)sgn\phi (r){\phi }^{\prime }(r)dr.\end{array}$$

Using (1.10) and considering the inequality

$$\mathrm{\Lambda }(r)={\int }_0^r\lambda (t)t^{n-1}={\int }_{|t|\le r}|\psi (t,x)|dx\le r^n\mu (r)(M_{\mu }\psi )(x),$$

(2.3)

we have

$$|\phi (r)|\mathrm{\Lambda }(r){|}_0^{\mathrm{\infty }}=0.$$

Thus,

$$\begin{array}{rcl}I(x)& =& -{\int }_0^{\mathrm{\infty }}\mathrm{\Lambda }(r)sgn\phi (r){\phi }^{\prime }(r)dr\\ \le & {\int }_0^{\mathrm{\infty }}\mathrm{\Lambda }(r)|{\phi }^{\prime }(r)|dr\stackrel{(\text{2.3})}{\le }(M_{\mu }\psi )(x){\int }_0^{\mathrm{\infty }}{r}^n\mu (r)|{\phi }^{\prime }(r)|dr.\end{array}$$

That completes the proof. □

Proof of Theorem 1.5 Let us fix x, a μ-smoothness point of f, and consider the difference

$$\begin{array}{rl}|S(x,\epsilon )-f(x)|& =|{\int }_{{\mathbb{R}}^n}[f(x-t)-f(x)]W(t,\epsilon )dt|\\ \le {\int }_{|t|\le 1}|f(x-t)-f(x)|W(t,\epsilon )dt+{\int }_{|t|>1}|f(x-t)-f(x)|W(t,\epsilon )dt\\ =A_1(\epsilon )+A_2(\epsilon ).\end{array}$$

(2.4)

In order to estimate $A_1(\epsilon )$, we let

$$\psi (t,x)=\{\begin{array}{ll}f(x-t)-f(x),& |t|\le 1;\\ 0,& |t|>1,\end{array}$$

and then

$$A_1(\epsilon )={\int }_{{\mathbb{R}}^n}W(t,\epsilon )|\psi (t,x)|dt,W(t,\epsilon )={(4\pi \epsilon )}^{-(n/2)}{e}^{-{|t|}^2/4\epsilon }.$$

(2.5)

Now, by Lemma A, taking $\phi (|t|)={(4\pi \epsilon )}^{-n/2}{e}^{-{|t|}^2/4\epsilon }$, we have

$$A_1(\epsilon )\le (M_{\mu }\psi )(x){\int }_0^{\mathrm{\infty }}{r}^n\mu (r)|{\phi }^{\prime }(r)|dr,\text{where }\phi (r)={(4\pi \epsilon )}^{-(n/2)}{e}^{-r^2/4\epsilon }.$$

(2.6)

Since f is μ-smooth at the point $x\in {\mathbb{R}}^n$, we have $(M_{\mu }\psi )(x)\equiv D_{\mu }(x)<\mathrm{\infty }$ (see (1.5)). So we get

$$A_1(\epsilon )\le c_1{\int }_0^{\mathrm{\infty }}{r}^n\mu (r)|{\phi }^{\prime }(r)|dr\le c{\int }_0^{\mathrm{\infty }}{r}^{n+1}{e}^{-r^2/4}\mu (\epsilon r)dr.$$

(2.7)

To estimate $A_2(\epsilon )$, we first apply Hölder’s inequality for $p>1$ and observe that

$$\begin{array}{rcl}{A}_2(\epsilon )& \le & |f(x)|{\int }_{|t|>1}W(t,\epsilon )dt\\ +{\left({\int }_{|t|>1}{|f(x-t)|}^{p}dt\right)}^{1/p}{\left({\int }_{|t|>1}{|W(t,\epsilon )|}^{q}dt\right)}^{1/q}(\frac{1}{p}+\frac{1}{q}=1).\end{array}$$

(2.8)

Let us estimate the first term on the right of (2.8). Changing variables to polar coordinates yields

$$\begin{array}{rcl}{\int }_{|t|>1}W(t,\epsilon )dt& =& k_1{\int }_1^{\mathrm{\infty }}{r}^{n-1}[{\int }_{S^{n-1}}{\epsilon }^{-n/2}{e}^{-r^2/4\epsilon }d\sigma (\theta )]dr\\ =& k_2{\int }_1^{\mathrm{\infty }}{r}^{n-1}{\epsilon }^{-n/2}{e}^{-r^2/4\epsilon }\\ =& k_3{}_{\int }^{{}_{(1/2\sqrt{\epsilon })}{}^{}\mathrm{\infty }}{r}^{n-1}{e}^{-r^2}dr\text{(we set }u=r^2\text{, }du=2rdr\text{)}\\ =& k_3{}_{\int }^{{}_{(1/4\epsilon )}{}^{}\mathrm{\infty }}{u}^{n/2-1}{e}^{-u}du\\ =& k_3\mathrm{\Gamma }(\frac{n}{2},\frac{1}{4\epsilon }),\end{array}$$

where $\mathrm{\Gamma }(s,\tau )$ is the upper incomplete gamma function. Now, using asymptotic formula (1.13), we get

$${\int }_{|t|>1}W(t,\epsilon )dt=O\left({\epsilon }^{1-\frac{n}{2}}{e}^{-1/4\epsilon }\right)\text{as }\epsilon \to 0^{+}.$$

(2.9)

The same is true for the second term of (2.8):

$$\begin{array}{rcl}{\left({\int }_{|t|>1}{|W(t,\epsilon )|}^{q}dt\right)}^{1/q}& =& k_4{\left({\int }_1^{\mathrm{\infty }}{r}^{n-1}[{\int }_{S^{n-1}}{\left({\epsilon }^{-n/2}{e}^{-r^2/4\epsilon }\right)}^{q}d\sigma (\theta )]dr\right)}^{1/q}\\ =& k_5{\epsilon }^{\frac{n}{2q}-\frac{n}{2}}{\left({}_{\int }^{{}_{(\sqrt{q}/2\sqrt{\epsilon })}{}^{}\mathrm{\infty }}{r}^{n-1}{e}^{-r^2}dr\right)}^{1/q}\\ =& k_5{\epsilon }^{\frac{n}{2q}-\frac{n}{2}}{\left({}_{\int }^{{}_{(q{/}_{4\epsilon })}{}^{}\mathrm{\infty }}{u}^{n/2-1}{e}^{-u}du\right)}^{1/q}.\end{array}$$

Now, using formula (1.13), we get

$${\left({\int }_{|t|>1}{|W(t,\epsilon )|}^{q}dt\right)}^{1/q}=O\left({\epsilon }^{\frac{1}{q}-\frac{n}{2}}{e}^{-1{/}_{4\epsilon }}\right)\text{as }\epsilon \to 0^{+}.$$

(2.10)

Collecting estimates (2.9) and (2.10), and taking into account that

$${\left({\int }_{|t|>1}{|f(x-t)|}^p\right)}^{1/p}\le {\parallel f\parallel }_p<\mathrm{\infty },$$

we have

$$A_2(\epsilon )=O\left({\epsilon }^{-n/2}{e}^{-1/4\epsilon }\right)\text{as }\epsilon \to 0^{+}.$$

(2.11)

By (2.7) and (2.11) we have shown that inequality (1.6) holds, as desired.

To complete the proof, we have to show that the conditions of Lemma A are satisfied; that is, for $\phi (r)={(4\pi \epsilon )}^{-n/2}{e}^{-r^2/4\epsilon }$,

$$\underset{r\to \mathrm{\infty }}{lim}{r}^n\mu (r)\phi (r)=0\text{and}\underset{r\to 0^{+}}{lim}{r}^n\mu (r)\phi (r)=0.$$

But this is obvious. □

Proof of Corollary 1.6 Let $\mu (r)$, $r\in [0,\mathrm{\infty })$ be a modulus of continuity, i.e., (a) $\mu (r)\to 0$ as $r\to 0^{+}$; (b) $\mu (r)$ is non-negative and non-decreasing on $(0,\mathrm{\infty })$; (c) $\mu (r)$ is continuous and subadditive $(0,\mathrm{\infty })$.

It follows from the subadditivity of $\mu (r)$ that

$$\mu (\epsilon r)\le (1+r)\mu (\epsilon )\text{for all }\epsilon ,r>0.$$

By employing this in (1.6), we get

$$\begin{array}{rcl}|S(x,\epsilon )-f(x)|& \le & c_1\mu (\epsilon ){\int }_0^{\mathrm{\infty }}(1+r)r^{n+1}{e}^{-r^2}dr+c_2{\epsilon }^{-n/2}{e}^{-1/4\epsilon }\\ \le & c_3\mu (\epsilon )+c_2{\epsilon }^{-n/2}{e}^{-1/4\epsilon }.\end{array}$$

(2.12)

Now, since the function $\mu (r)$ is a modulus of continuity, it cannot tend to zero too rapidly as $\epsilon \to 0$, that is, for instance, if $\frac{\mu (\epsilon )}{\epsilon }\to 0$ as $\epsilon \to 0$, then $\mu (\epsilon )\equiv 0$. Therefore

$${\epsilon }^{-n/2}{e}^{-1/4\epsilon }\le c_4\mu (\epsilon ),\epsilon \to 0$$

for some constant $c_4$. Taking into account this in (2.12), we obtain

$$|S(x,\epsilon )-f(x)|\le c\mu (\epsilon ),\epsilon \to 0,$$

where the constant c does not depend on $\epsilon >0$. □

Proof of Corollary 1.7 Let us show that for some $0<\rho <1$ the function

$$\mu (r)=\left\{\begin{array}{ll}0,& r=0\\ {(1/ln1/r)}^{\alpha },& 0<r<\rho <1\\ {(1/ln1/\rho )}^{\alpha },& \rho \le r<\mathrm{\infty }\end{array}\right\}(0<\alpha <\mathrm{\infty })$$

is a modulus of continuity, i.e., it is continuous, non-decreasing, subadditive on $[0,\mathrm{\infty })$ and tends to zero as $r\to 0^{+}$. The continuity and ${lim}_{r\to 0}\mu (r)=0$ are obvious. To prove the other properties, it suffices to show that (see [11], p.41)

$${\mu }^{\prime }(r)\ge 0\text{and}{(\mu (r)/r)}^{\prime }\le 0(0<r<\rho ).$$

Simple calculations show that the above inequalities are fulfilled if one takes $\rho =e^{-\alpha }$. □

Proof of Corollary 1.8 Let us substitute the function

$$\mu (r)=\left\{\begin{array}{ll}0,& r=0\\ r^{\alpha }{|lnr|}^{\beta },& 0<r\le \rho \\ {\rho }^{\alpha }{|ln\rho |}^{\beta },& r>\rho \end{array}\right\}$$

in (1.6), where $\alpha >0$ and $\beta \in (-\mathrm{\infty },\mathrm{\infty })$ are given numbers and $0<\rho <1$.

If $\beta \ge 0$, we have, for sufficiently small $\epsilon >0$,

$$\mu (\epsilon r)\le {\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{r}^{\alpha }{(1+\frac{|lnr|}{|ln\epsilon |})}^{\beta }\le {\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{r}^{\alpha }{(1+|lnr|)}^{\beta }.$$

By making use of this estimate in (1.6), we have for $\epsilon \ll 1$ that

$$\begin{array}{rcl}|S(x,\epsilon )-f(x)|& \le & c{\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{\int }_0^{\mathrm{\infty }}{r}^{n+1}{e}^{-r^2/4}{r}^{\alpha }{(1+lnr)}^{\beta }dr+O\left({\epsilon }^{-n/2}{e}^{-1/4\epsilon }\right)\\ \le & c_1{\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }+c_2{\epsilon }^{-n/2}{e}^{-1/4\epsilon }\le c_3{\epsilon }^{\alpha }{|ln\epsilon |}^{\alpha },\epsilon \to 0.\end{array}$$

Let now $\beta <0$. By setting $\delta =-\beta >0$, we have for $\epsilon \ll 1$

$$\begin{array}{rcl}\mu (\epsilon r)& =& {\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{r}^{\alpha }|\frac{ln\epsilon r}{ln\epsilon }{|}^{\beta }={\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{r}^{\alpha }|\frac{ln\epsilon }{ln\epsilon r}{|}^{\delta }\\ =& {\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{r}^{\alpha }|1-\frac{lnr}{ln\epsilon r}{|}^{\delta }\le {\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{r}^{\alpha }{(1+\frac{|lnr|}{|ln\epsilon r|})}^{\delta }.\end{array}$$

Since $\epsilon r<\rho <1$, it follows that

$$\mu (\epsilon r)\le {\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{r}^{\alpha }{(1+\frac{|lnr|}{|ln\rho |})}^{\delta }.$$

Using this in (1.6), we get

$$\begin{array}{rcl}|S(x,\epsilon )-f(x)|& \le & c{\epsilon }^{\alpha }{|ln\epsilon |}^{\beta }{\int }_0^{\mathrm{\infty }}{r}^{n+1}{e}^{-r^2/4}{r}^{\alpha }{(1+\frac{|lnr|}{|ln\rho |})}^{\delta }{e}^{-r}dr+c_2{\epsilon }^{-n/2}{e}^{-1/4\epsilon }\\ \le & c{\epsilon }^{\alpha }{|ln\epsilon |}^{\beta },\end{array}$$

which is the desired result. □