# Distributional point values and delta sequences

Ricardo Estrada  Department of Mathematics
Louisiana State University
Baton Rouge
LA 70803
USA<EMAIL_ADDRESS>and  Kevin Kellinsky-Gonzalez  Department of
Mathematics
Louisiana State University
Baton Rouge
LA 70803
USA<EMAIL_ADDRESS>
###### Abstract.

Recently Sasane [18] defined a notion of evaluating a distribution at a point
using delta sequences. In this paper, we explore the relationship between
generalizations of his definition and the standard definition of
distributional point values. This allows us to obtain a description of
distributional point values via delta sequences and a characterization of when
a distribution is actually a regular distribution given by bounded function.
We also give a characterization of limits in a continuous variable by the
existence of the limits of certain sequences.

###### Key words and phrases:

Distributional point values, delta sequences

###### 2010 Mathematics Subject Classification:

46F10

##  1\. Introduction

Distributional point values were first defined in one variable by Łojasiewics
[11]. His definition is given as a distributional limit over a _continuous_
variable. In other words, if $f\in\mathcal{D}^{\prime}\left(\mathbb{R}\right)$
and $x_{0}\in\mathbb{R}$ then we say that $f$ has a distributional point
value, equal to $\gamma,$ at $x_{0}$ if

(1.1) $\lim_{\varepsilon\rightarrow 0}f\left(x_{0}+\varepsilon
x\right)=\gamma\,,$

in the distributional sense, that is, if for all test functions
$\phi\in\mathcal{D}\left(\mathbb{R}\right)$ we have that

(1.2) $\lim_{\varepsilon\rightarrow 0}\left\langle f\left(x_{0}+\varepsilon
x\right),\phi\left(x\right)\right\rangle=\gamma\int_{-\infty}^{\infty}\phi\left(x\right)\,\mathrm{d}x\,.$

Similarly [12] point values in several variables are defined as a
distributional limit over a continuous variable. Point values have been
studied extensively and are the first step in the study of distributional
asymptotic analysis and of the study of local properties of distributions [2,
5, 15, 16, 17, 23].

It is possible to find in the literature other definitions of distributional
point values, based on the use of _delta sequences._ For instance, in a recent
study, Sasane [18] uses the alternative definition
“$f\left(x_{0}\right)=\eta$” if for all positive and even test functions
$\phi$ with $\int_{-\infty}^{\infty}\phi\left(x\right)\,\mathrm{d}x=1$ one has

(1.3) $\lim_{n\rightarrow\infty}\left\langle
f\left(x_{0}+x\right),\phi_{n}\left(x\right)\right\rangle=\eta\,,$

where $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$ is the standard delta
sequence generated by $\phi,$ namely,
$\phi_{n}\left(x\right)=n\phi\left(nx\right).$ Naturally the question arises
if the two definitions are equivalent. More generally, if $\mathfrak{F}$ is a
family of test functions, we would like to consider the relationship between
the existence of the distributional point value and the existence of the limit
(1.3) whenever the sequence $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$ belongs
to $\mathfrak{F}.$ Interestingly, the two definitions are _not_ equivalent for
many classes of delta sequences, in particular for the family considered in
[18]. Nevertheless, we are able to show that for _some_ classes they are
actually equivalent.

In order to study this problem, we start by studying a very general question
about limits. Indeed, in a metric space $X,$ given a function
$f:X\setminus\left\\{x_{0}\right\\}\rightarrow\mathbb{R},$ then the limit

(1.4) $\lim_{x\rightarrow x_{0}}f\left(x\right)=L\,,$

exists if and only if

(1.5) $\lim_{n\rightarrow\infty}f\left(x_{n}\right)=L\,,$

for all sequences $\left\\{x_{n}\right\\}_{n=1}^{\infty}$ in
$X\setminus\left\\{x_{0}\right\\}$ that converge to $x_{0}.$ The question we
would like to consider is whether the existence of the limit
$\lim_{n\rightarrow\infty}f\left(x_{n}\right)$ for sequences
$\left\\{x_{n}\right\\}_{n=1}^{\infty}$ of a certain family implies that (1.4)
is satisfied. In Section 3, we consider the case where $X=(0,\infty],$
$x_{0}=\infty$ and $f$ continuous, showing that in such cases the existence of
the limit

(1.6) $\lim_{n\rightarrow\infty}f\left(na\right)=F\left(a\right)\,,$

for all $a>0$ implies that, in fact, $F$ is a constant function,
$F\left(a\right)=L,$ for all $a>0,$ and that
$\lim_{x\rightarrow\infty}f\left(x\right)=L.$ We give examples of other
families of sequences for which $\lim_{x\rightarrow\infty}f\left(x\right)=L$
might not hold true. We are also able to present, in Section 4, a
corresponding result when the function $f$ in (1.6) is not necessarily
continuous but just measurable and the limit holds almost everywhere.

The plan of the rest of the article is as follows. Basic results on
distributional point values are briefly discussed in Section 2, while delta
sequences are considered in Section 5. Section 6 gives several useful results
on the characterization of distributions and functions using normalized
positive test functions. The main results are given in Section 7, where we
give equivalent conditions to the existence of point values obtained from a
given family of delta sequences. In particular, we prove that the existence of
the distributional point value is equivalent to the existence of the point
value for the family of standard delta sequences generated by a positive
normalized test function. Then we show that for the family employed in [18]
the equivalence is the existence of the symmetric distributional point value.
We also consider radial delta sequences in several variables, and finish by
studying the family of all delta sequences of normalized positive test
functions.

##  2\. Preliminaries

We refer to the texts for the basic ideas about distributions [1, 8, 19, 22].
Ideas on the local behavior of distributions can be found in [2, 5, 16, 17,
23]. In this article, we will work mainly in the space
$\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$ of distributions on
$\mathbb{R}^{d},$ dual of the space $\mathcal{D}\left(\mathbb{R}^{d}\right)$
of standard test functions, that is, $C^{\infty}$ functions with compact
support, with its inductive limit topology [20].

If $f\in\mathcal{D}^{\prime}\left(\mathbb{R}\right)$ and $x_{0}\in\mathbb{R}$
then [11] we say that $f$ has a _distributional point value,_ equal to
$\gamma,$ at $x_{0}$ if $\lim_{\varepsilon\rightarrow
0}f\left(x_{0}+\varepsilon x\right)=\gamma,$ in the strong topology of
$\mathcal{D}^{\prime}\left(\mathbb{R}\right).$ Equivalently, since a sequence
of distributions converges strongly if and only if it converges weakly [20],
if for all test functions $\phi\in\mathcal{D}\left(\mathbb{R}\right)$ we have
that

(2.1) $\lim_{\varepsilon\rightarrow 0}\left\langle f\left(x_{0}+\varepsilon
x\right),\phi\left(x\right)\right\rangle=\gamma\int_{-\infty}^{\infty}\phi\left(x\right)\,\mathrm{d}x\,.$

Interestingly, the existence of the distributional limit
$\lim_{\varepsilon\rightarrow 0}f\left(x_{0}+\varepsilon x\right)$ implies
that this limit is a constant and that the point value exists. On the other
hand, if the limit

(2.2) $\lim_{\varepsilon\rightarrow 0^{+}}f\left(x_{0}+\varepsilon
x\right)=g\left(x\right)\,,$

exists, then $g$ does not have to be a constant, but it will have the jump
behavior [21], that is, $g$ is of the form

(2.3)
$g\left(x\right)=\gamma_{-}H\left(-x\right)+\gamma_{+}H\left(x\right)\,,$

where $H$ is the Heaviside function and $\gamma_{\pm}$ are some constants.
Distributions of the form (2.3) are the most general homogeneous distributions
of degree $0$ in one variable. Alternatively, (2.2) and (2.3) hold if and if
the lateral limits $f\left(x_{0}\pm 0\right)=\lim_{\varepsilon\rightarrow
0^{+}}f\left(x_{0}\pm\varepsilon x\right)=\gamma_{\pm},$ exist in
$\mathcal{D}^{\prime}\left(0,\infty\right)$ and $f$ does not have delta
functions at $x_{0}.$

In several variables point values are defined similarly [12], namely, if
$f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right),$ then the distributional
point value $f\left(\mathbf{x}_{0}\right)$ exists and equals $\gamma$ if
$\lim_{\varepsilon\rightarrow
0}f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right)=\gamma,$
distributionally. In several variables the limit $\lim_{\varepsilon\rightarrow
0}f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right)$ could exist without
being a constant. In fact, if

(2.4) $\lim_{\varepsilon\rightarrow
0^{+}}f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right)=g\left(\mathbf{x}\right)\,,$

then $g$ is homogeneous of degree $0.$ Homogeneous distributions of degree
zero are given by a formula of the type

(2.5) $\left\langle
g\left(\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle=\int_{0}^{\infty}\left\langle\alpha\left(\mathbf{w}\right),\phi\left(r\mathbf{w}\right)\right\rangle_{\mathcal{D}^{\prime}\left(\mathbb{S}\right)\times\mathcal{D}\left(\mathbb{S}\right)}r^{d-1}\,\mathrm{d}r\,,$

for a certain distribution
$\alpha\in\mathcal{D}^{\prime}\left(\mathbb{S}\right)$ [5, Thm. 2.6.2]. The
distribution $\alpha$ is the _thick distributional value_ [3] of $f$ at
$\mathbf{x}_{0},$ namely, $f$ has no delta functions at $\mathbf{x}_{0}$ and
$\alpha$ is the thick limit

(2.6) $\lim_{\varepsilon\rightarrow
0^{+}}f\left(\mathbf{x}_{0}+r\varepsilon\mathbf{w}\right)=\alpha\left(\mathbf{w}\right)\,,$

in the space
$\mathcal{D}^{\prime}\left(\left(0,\infty\right),\mathcal{D}^{\prime}\left(\mathbb{S}\right)\right),$
that is, for all $\rho\in\mathcal{D}\left(0,\infty\right),$

(2.7) $\left\langle\lim_{\varepsilon\rightarrow
0^{+}}f\left(\mathbf{x}_{0}+r\varepsilon\mathbf{w}\right),\rho\left(r\right)\right\rangle_{\mathcal{D}^{\prime}\left(0,\infty\right)\times\mathcal{D}\left(0,\infty\right)}=\left(\int_{0}^{\infty}\rho\left(r\right)\,\mathrm{d}r\right)\alpha\left(\mathbf{w}\right)\,,$

##  3\. The continuous case

We start with a general known result that will be useful in our analysis.

######  Proposition 3.1.

Let $f:\left(0,\infty\right)\rightarrow\mathbb{R}$ be continuous. Suppose that
for each $a>0$ the sequence $\left\\{f\left(an\right)\right\\}_{n=1}^{\infty}$
converges, to $F\left(a\right).$ Then $F\left(a\right)$ does not depend on
$a,$ that is,

(3.1) $F\left(a\right)=L\text{ for all }a>0\,,$

for some $L,$ and actually

(3.2) $\lim_{x\rightarrow\infty}f\left(x\right)=L\,.$

###### Proof.

Clearly the function $F$ is constant in each class of the quotient space
$\mathbb{R}/\mathbb{Q},$ $F\left(ra\right)=F\left(a\right)$ if
$r\in\mathbb{Q}.$ Also, $F$ is continuous or of the first Baire class [6, 13],
so that the set $D$ of points of continuity of $F$ is dense in
$\left(0,\infty\right).$ Let $\alpha\in D.$ Let $b>0.$ If $\varepsilon>0$ then
there exists $\delta>0$ such that $\left|a-\alpha\right|<\delta$ implies
$\left|F\left(a\right)-F\left(\alpha\right)\right|<\varepsilon$ and there
exist $r$ rational such that $\left|rb-\alpha\right|<\delta.$ Therefore

(3.3)
$\left|F\left(b\right)-F\left(\alpha\right)\right|=\left|F\left(rb\right)-F\left(\alpha\right)\right|<\varepsilon\,,$

and since $\varepsilon$ is arbitrary, $F\left(b\right)=F\left(\alpha\right).$

In order to prove (3.2), observe that if $\varepsilon>0,$ then for each
$a\in\left[1,2\right]$ there exists $n_{0}=n_{0}\left(a\right)$ such that
$\left|f\left(ka\right)-L\right|<\varepsilon$ for $k\geq n_{0}\left(a\right).$
This means that

(3.4) $\left[1,2\right]=\bigcup_{n=1}^{\infty}\bigcap_{k\geq
n}\left\\{a\in\left[1,2\right]:\left|f\left(ka\right)-L\right|<\varepsilon\right\\}\,.$

Therefore there exists $n_{0}$ such that $\bigcap_{k\geq
n_{0}}\left\\{a\in\left[1,2\right]:\left|f\left(ka\right)-L\right|<\varepsilon\right\\}$
contains an interval $I=\left[\alpha,\beta\right]$ with $\alpha\neq\beta.$
Observe now that $\bigcup_{k=n_{0}}^{\infty}kI$ contains a ray $(B,\infty),$
since in fact $\bigcup_{k=n_{1}}^{\infty}kI$ is a closed ray if
$n_{1}>1/\left(\beta-\alpha\right).$ Hence if $x>B$ then $x=ka$ for some
$k\geq n_{0}$ and some $a\in I$ and, consequently,
$\left|f\left(x\right)-L\right|=\left|f\left(ka\right)-L\right|<\varepsilon.$∎

It is interesting that there are sequences
$\left\\{\xi_{n}\right\\}_{n=1}^{\infty}$ with
$\lim_{n\rightarrow\infty}\xi_{n}=\infty$ such that for some continuous
functions $f:\left(0,\infty\right)\rightarrow\mathbb{R}$ the limit

(3.5) $\lim_{n\rightarrow\infty}f\left(a\xi_{n}\right)=G\left(a\right)$

exists for all $a>0$, but the function $G$ is not constant. Indeed, let
$f\left(x\right)=\sin\left(2\pi\ln x\right)$ and $\xi_{n}=e^{(n+1/n)}.$ The
limit $\lim_{n\rightarrow\infty}f\left(a\xi_{n}\right)=\sin\left(2\pi\ln
a\right),$ exists but it is not constant, of course.

##  4\. The measurable case

We shall now consider an extension of the results of Section 3 to measurable
functions.

######  Proposition 4.1.

Suppose $f:(0,\infty)\rightarrow\mathbb{R}$ is measurable. For $a>0$, suppose
that the function $F$ defined by

(4.1) $F(a)=\lim\limits_{n\rightarrow\infty}f(an)\,,$

is well-defined almost everywhere. Then $F$ is constant almost everywhere.

###### Proof.

Let us first suppose that $f\in L^{\infty}(0,\infty).$ Let
$\phi\in\mathcal{D}(0,\infty)$ be a test function. For $\lambda>0,$ let us
define

(4.2) $G(\lambda)=\int_{0}^{\infty}f(\lambda x)\phi(x)\,\mathrm{d}x\,.$

The function $G$ is continuous because $\phi\in\mathcal{D}(0,\infty)$. For a
fixed $\lambda,$ let us consider the sequence $\left\\{G(\lambda
n)\right\\}_{n=1}^{\infty}.$ Since $f$ is bounded, we can apply the dominated
convergence theorem to see that

$\lim\limits_{n\rightarrow\infty}G(\lambda
n)=\int_{0}^{\infty}\lim\limits_{n\rightarrow\infty}f(\lambda
nx)\phi(x)\,\mathrm{d}x=\int_{0}^{\infty}F(\lambda x)\phi(x)\,\mathrm{d}x\,,$

exists. The Proposition 3.1 then yields that $\int_{0}^{\infty}F(\lambda
x)\phi(x)\,\mathrm{d}x$ does not depend on $\lambda,$

(4.3) $\int_{0}^{\infty}F(\lambda
x)\phi(x)\,\mathrm{d}x=\int_{0}^{\infty}F(x)\phi(x)\,\mathrm{d}x\,.$

Therefore, the regular distribution $F$ is constant since $F(\lambda x)=F(x),$
$\lambda>0,$ and only the constants are homogeneous of degree $0$ in the
interval $\left(0,\infty\right)$ [5], that is, $F(x)=C,$ as _distributions._
Notice now that the locally integrable function that gives a regular
distribution is unique almost everywhere, so that $F(x)=C\ \ \
\left(\text{a.e.}\right).$

Let us now consider the case of a general measurable function $f$ for which
the limit $\lim\limits_{n\rightarrow\infty}f(an)=F(a)$ exists (a.e.). We can
then define the bounded function

(4.4) $h(x)=\arctan f{(x)}\,.$

Then $\lim\limits_{n\rightarrow\infty}h{(}an{)}=\arctan F(a)$ exists (a.e.).
Consequently, $\arctan F(a)$ is constant, and therefore so is $F(a).$∎

In Proposition 3.1, it is shown that in the continuous case not only is
$F\left(a\right)=L$ for all $a>0,$ but actually
$\lim_{x\rightarrow\infty}f\left(x\right)=L.$ This is no longer true in the
measurable case; for example, if $f=\chi_{B},$ the characteristic function of
a set $B$ of measure zero such that $B\cap\left(x,\infty\right)\neq\emptyset$
for all $x>0,$ then $\lim_{x\rightarrow\infty}f\left(x\right)$ does not exist.
Of course, this function $f$ is equal almost everywhere to a function
$\widetilde{f},$ the zero function, for which
$\lim_{x\rightarrow\infty}\widetilde{f}\left(x\right)$ exists. An example
where (4.1) exists for all $a>0$ but
$\lim_{x\rightarrow\infty}\widetilde{f}\left(x\right)$ does not exist for any
function $f$ such that $f\left(x\right)=\widetilde{f}\left(x\right)$ (a.e.)
can be constructed as follows. The strategy will be to construct an unbounded
set, $A$, with measure $0$ such that $f\left(x\right)\neq 0$ for infinitely
many $x\in A$

######  Example 4.2.

Let $\left\\{N_{k}\right\\}_{k=1}^{\infty}$ be a sequence of positive integers
such that

(4.5) $kN_{k}<N_{k+1}\,.$

For each $k$ let us choose a non empty open interval
$B_{k}\subset\left(N_{k}-1,N_{k}\right).$ Then if $j\in\mathbb{N},$
$j\left(\frac{1}{k},1\right)\cap B_{k}\neq\emptyset$ only if $N_{k}\leq
j<N_{k+1}.$ Therefore, if $x\in\left(\frac{1}{k},1\right),$ then

(4.6) $\chi_{B_{k}}\left(jx\right)=0\,\ \text{for all \ }j\in\mathbb{N}\,,\ \
x\notin A_{k}\,,$

where $A_{k}=\bigcup_{j=N_{k}}^{N_{k+1}-1}\frac{1}{j}B_{k}\,.$ Let now
$\left\\{\eta_{k}\right\\}_{k=1}^{\infty}$ be a sequence of strictly positive
numbers such that the series $\sum_{k=1}^{\infty}\eta_{k}$ converges and let
us further restrict the sets $B_{k}$ by requiring that
$\mu\left(A_{k}\right)<\eta_{k},$ for all $k.$ Let
$A=\limsup_{k\rightarrow\infty}A_{k}=\bigcap_{k=1}^{\infty}\bigcup_{q=k}^{\infty}A_{q}.$
Then $\mu\left(A\right)=0.$

Let us now define the function $f:(0,\infty)\rightarrow\mathbb{R}$ by

(4.7) $f\left(x\right)=\sum_{k=1}^{\infty}\chi_{B_{k}}\left(x\right)\,.$

If $\widetilde{f}\left(x\right)=f\left(x\right)$ almost everywhere, then the
limit $\lim_{x\rightarrow\infty}\widetilde{f}\left(x\right)$ does not exist.
On the other hand, $\lim_{n\rightarrow\infty}f\left(nx\right)$ exists almost
everywhere. Indeed, it is enough to show the existence almost everywhere in
$\left(0,1\right),$ and the limit of $f\left(nx\right)$ exists and equals $0$
if $x\in\left(0,1\right)\setminus A$ since if $x\notin A$ then there exists
$k_{0}$ such that $x\notin A_{k}$ for $k\geq k_{0}$ and consequently,
$f\left(nx\right)=0$ whenever $n\geq N_{k_{0}}.$

An example involving continuous functions can be obtained by a slight
modification.

######  Example 4.3.

Let $g$ be a continuous function in $\left(0,\infty\right)$ such that $0\leq
g\left(x\right)\leq f\left(x\right)$ and such that there exist points
$\xi_{k}\in B_{k},$ for all $k,$ such that $g\left(\xi_{k}\right)=1.$ Then
$\lim_{n\rightarrow\infty}g\left(nx\right)=0$ almost everywhere, _but not
everywhere_ , since $\lim_{x\rightarrow\infty}g\left(x\right)$ does not exist.

The examples show that it is possible for
$\lim\limits_{n\rightarrow\infty}f(an)$ to be equal to a constant $L$ almost
everywhere but without $\lim_{x\rightarrow\infty}f\left(x\right)$ existing. We
do have a convergence in measure type result.

######  Proposition 4.4.

Suppose $f:(0,\infty)\rightarrow\mathbb{R}$ is measurable. Suppose that

(4.8) $\lim\limits_{n\rightarrow\infty}f(an)=L\,\ \ \
\left(\text{a.e.}\right)\,.$

Then for all $\varepsilon>0$ and all $C>1,$

(4.9)
$\lim_{x\rightarrow\infty}\frac{\mu\left(\left\\{t\in\left[x,Cx\right]:\left|f\left(t\right)-L\right|>\varepsilon\right\\}\right)}{\mu\left(\left[x,Cx\right]\right)}=0\,,$

where $\mu$ denotes the Lebesgue measure of a set.

###### Proof.

Let us denote by $G\left(x\right)$ the quotient
$\mu\left(\left\\{t\in\left[x,cx\right]:\left|f\left(t\right)-L\right|>\varepsilon\right\\}\right)/\left(1-C\right)x.$
Notice that $G$ is a continuous function in $\left(0,\infty\right).$ Let $a>0$
be fixed and consider the sequence of functions
$f_{n}\left(x\right)=f\left(nx\right)$ in the interval $\left[a,Ca\right].$
Since $f_{n}$ converges to $L$ almost everywhere in this finite interval, it
converges to $L$ in measure. This means that for all $\varepsilon>0$ the
measure of the set
$\\{s\in\left[a,Ca\right]:\left|f_{n}\left(s\right)-L\right|>\varepsilon\\}$
tends to zero. But the transformation $t=ns$ gives

$\displaystyle\frac{\mu\left(\\{s\in\left[a,Ca\right]:\left|f_{n}\left(s\right)-L\right|>\varepsilon\\}\right)}{\mu\left(\left[a,Ca\right]\right)}$
$\displaystyle=\frac{\mu\left(\\{s\in\left[a,Ca\right]:\left|f\left(ns\right)-L\right|>\varepsilon\\}\right)}{\mu\left(\left[a,Ca\right]\right)}$
$\displaystyle=\frac{\mu\left(\left\\{t\in\left[na,Cna\right]:\left|f\left(t\right)-L\right|>\varepsilon\right\\}\right)}{\mu\left(\left[na,Cna\right]\right)}$
$\displaystyle=G\left(na\right)\,,$

so that $\lim_{n\rightarrow\infty}G\left(na\right)=0.$ Proposition 3.1 then
yields (4.9). ∎

##  5\. Delta sequences

A sequence $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ of distributions is called
a delta sequence if
$f_{n}\left(\mathbf{x}\right)\rightarrow\delta\left(\mathbf{x}\right)$ in
either the strong or the weak topology of
$\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right),$ since the two notions are
equivalent [20]. In other words, $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ is a
delta sequence if

(5.1) $\lim_{n\rightarrow\infty}\left\langle
f,\phi\right\rangle=\phi\left(\mathbf{0}\right)\,,$

for all $\phi\in\mathcal{D}\left(\mathbb{R}^{d}\right).$ In this article we
will be interested mostly in the case when the distributions $f_{n}$ are
actually smooth functions, but general delta sequences are also of interest,
of course. They have been employed in several problems [1], such as the
definitions of point values [18] or the definition of products of
distributions [9, 10, 14].

There are many ways to construct delta sequences. A simple one is the
following. Let $f$ be a fixed distribution of rapid decay at infinity, that
is, $f\in\mathcal{K}^{\prime}\left(\mathbb{R}^{d}\right).$ Then all the
moments $\mu_{\mathbf{k}}=\left\langle
f\left(\mathbf{x}\right),\mathbf{x}^{\mathbf{k}}\right\rangle,$ exist for
$\mathbf{k}\in\mathbb{N}^{d}$ since all polynomials belong to
$\mathcal{K}\left(\mathbb{R}^{d}\right),$ and the _moment asymptotic
expansion_

(5.2)
$f\left(\lambda\mathbf{x}\right)\sim\sum_{q=0}^{\infty}\sum_{\left|\mathbf{k}\right|=q}\frac{\mu_{\mathbf{k}}\mathbf{D}^{\mathbf{k}}\delta\left(\mathbf{x}\right)}{\mathbf{k}!}\frac{1}{\lambda^{q+d}}\,,\
\ \text{\ as }\lambda\rightarrow\infty\,,$

holds in $\mathcal{K}^{\prime}\left(\mathbb{R}^{d}\right)$ [5]. Therefore,
when $\mu_{\mathbf{0}}\neq 0$ if $\left\\{\xi_{n}\right\\}_{n=1}^{\infty}$ is
any sequence of positive numbers with
$\lim_{n\rightarrow\infty}\xi_{n}=\infty$ then

(5.3)
$g_{n}\left(\mathbf{x}\right)=\frac{\xi_{n}^{d}}{\mu_{\mathbf{0}}}f\left(\xi_{n}\mathbf{x}\right)\,,$

is a delta sequence, generated by $f$ and
$\left\\{\xi_{n}\right\\}_{n=1}^{\infty}.$ When $\xi_{n}=n$ for all $n,$ we
call this sequence the _standard_ delta sequence generated by $f.$

Another useful construction of delta sequences is provided by the ensuing well
known result.

######  Lemma 5.1.

Suppose $\left\\{\psi_{n}\right\\}_{n=1}^{\infty}$ is a sequence of normalized
positive test functions in $\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$
such that
$\operatorname*{supp}\psi_{n}\subset\\{\mathbf{x:}\left|\mathbf{x}\right|<r_{n}\\},$
where $\lim_{n\rightarrow\infty}r_{n}=0.$ Then
$\left\\{\psi_{n}\right\\}_{n=1}^{\infty}$ is a delta sequence.

###### Proof.

Let $\phi$ be any test function. Then by the first mean value theorem for
integrals,

(5.4)
$\left\langle\psi_{n},\phi\right\rangle=\int_{\operatorname*{supp}\psi_{n}}\psi_{n}\left(\mathbf{x}\right)\phi\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}=\phi\left(\mathbf{x}_{n}\right)\,,$

for some $\mathbf{x}_{n}\in\operatorname*{supp}\psi_{n}.$ Since
$\left|\mathbf{x}_{n}\right|\leq r_{n}\rightarrow 0,$ we obtain that
$\mathbf{x}_{n}\rightarrow 0,$ and consequently,
$\phi\left(\mathbf{x}_{n}\right)\rightarrow\phi\left(\mathbf{0}\right).$ Thus
$\psi_{n}\left(\mathbf{x}\right)\rightarrow\delta\left(\mathbf{x}\right).$∎

We now give a notion of point value of a distribution based on delta
sequences. Our definition applies to several spaces of distributions, but the
cases $\mathcal{A}=\mathcal{D},$ $\mathcal{E},$ or $\mathcal{S}$ seem the most
relevant.

######  Definition 5.2.

Let $\mathcal{A}\left(\mathbb{R}^{d}\right)$ be a space of test functions. Let
$\mathfrak{F}$ be a family of delta sequences whose elements belong to
$\mathcal{A}\left(\mathbb{R}^{d}\right).$ If
$f\in\mathcal{A}^{\prime}\left(\mathbb{R}^{d}\right)$ and
$\mathbf{x}_{0}\in\mathbb{R}^{d}$ we say that the value
$f\left(\mathbf{x}_{0}\right)$ exists and equals $\gamma$ with respect to
$\mathfrak{F}$ if

(5.5) $\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}\right\rangle=\gamma\,,$

for all $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}\in\mathfrak{F}.$ When this
holds we write

(5.6) $f\left(\mathbf{x}_{0}\right)=\gamma\ \ \ \left(\mathfrak{F}\right)\,.$

The definition of point value employed by Sasane [18] corresponds to the case
when $d=1,$ $f\in\mathcal{D}^{\prime}\left(\mathbb{R}\right),$ and
$\mathfrak{F}$ is the family of all delta sequences whose elements are the
standard sequences generated from a positive, normalized, and symmetric test
function of $\mathcal{D}\left(\mathbb{R}\right).$

##  6\. Several lemmas

In this section, we present several results on how positive test functions
allow us to study many properties of distributions. In particular, we see how
positive test functions tell us if a distribution is a regular distribution
given by a bounded measurable function and give us the essential supremum and
infimum of such a function. In this section, and only in this section, we will
make a notational difference between a regular distribution
$\mathsf{f}\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$ and the locally
integrable function $f\ $that generates it as

(6.1)
$\left\langle\mathsf{f}\left(\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle=\int_{\mathbb{R}^{d}}f\left(\mathbf{x}\right)\phi\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\,,\
\ \ \ \phi\in\mathcal{D}\left(\mathbb{R}^{d}\right)\,.$

In the rest of the article, we will use the same notation, $f,$ for the
distribution and the function.

Let us start with following simple result.

######  Lemma 6.1.

The set of functions of the form

(6.2) $\phi=c_{1}\psi_{1}-c_{2}\psi_{2}\,,$

where $c_{1}$ and $c_{2}$ are constants and where $\psi_{1}$ and $\psi_{2}$
are normalized positive test functions is the whole space
$\mathcal{D}\left(\mathbb{R}^{d}\right).$

When $d=1,$ the corresponding space with $\psi_{1}$ and $\psi_{2}$ normalized
positive symmetric test functions is the space of all even test functions.

###### Proof.

It is enough to show that the real valued elements of
$\mathcal{D}\left(\mathbb{R}^{d}\right)$ have the form (6.2) for some positive
constants $c_{1}$ and $c_{2}.$ Let
$\zeta_{1}\in\mathcal{D}\left(\mathbb{R}^{d}\right)$ be such that
$\zeta_{1}\left(\mathbf{x}\right)\geq\max\left\\{\phi\left(\mathbf{x}\right),0\right\\}$
and let $\zeta_{2}=\zeta_{1}-\phi.$ Then we write $\zeta_{j}=c_{j}\psi_{j}$
where the $\psi_{j}$ are normalized positive test functions and
$c_{j}=\int_{\mathbb{R}^{d}}\zeta_{j}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}.$
In the symmetric case we just also ask $\zeta_{1}$ to be even.∎

Our first characterization using positive normalized test functions is the
following.

######  Lemma 6.2.

Let $\mathsf{f}\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right).$ Then
$\mathsf{f}$ is a regular distribution in an open set
$U\subset\mathbb{R}^{d},$ given by a bounded function $f\in
L^{\infty}\left(U\right)$ if and only if there exists a constant $M>0$ such
that for all positive, normalized test functions
$\phi\in\mathcal{D}\left(U\right)$ we have

(6.3)
$\left|\left\langle\mathsf{f}\left(\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle\right|\leq
M\,.$

###### Proof.

If $f\in L^{\infty}\left(U\right).$ Then when
$\phi\in\mathcal{D}\left(U\right),$

$\left|\left\langle\mathsf{f}\left(\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle\right|=\left|\int_{U}f\left(\mathbf{x}\right)\phi\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\right|\leq\left\|f\right\|_{L^{\infty}\left(U\right)}\left\|\phi\right\|_{L^{1}\left(U\right)}\,,$

so that if $\phi$ is normalized,
$\left|\left\langle\mathsf{f}\left(x\right),\phi\left(x\right)\right\rangle\right|\leq\left\|f\right\|_{L^{\infty}\left(U\right)}.$
Therefore (6.3) holds with $M=\left\|f\right\|_{L^{\infty}\left(U\right)}.$

Conversely, if (6.3) is satisfied for some $M>0$ for all positive, normalized
test functions of $U$ then
$\left|\left\langle\mathsf{f}\left(x\right),\psi\left(x\right)\right\rangle\right|\leq
2M\left\|\psi\right\|_{L^{1}\left(U\right)}$ for all real test functions
$\psi\in\mathcal{D}\left(U\right),$ because of Lemma 6.1 (or $4M$ if complex).
This means that $\mathsf{f}$ is continuous in $\mathcal{D}\left(U\right),$ a
dense subspace of $L^{1}\left(U\right)$ with the topology induced by
$L^{1}\left(U\right)$ in its subspace. Hence, $\mathsf{f}$ admits an extension
$f\in\left(L^{1}\left(U\right)\right)^{\prime}\simeq
L^{\infty}\left(U\right),$ and this means that

(6.4)
$\left\langle\mathsf{f}\left(\mathbf{x}\right),\psi\left(\mathbf{x}\right)\right\rangle=\int_{U}f\left(\mathbf{x}\right)\psi\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\,,$

for all $\psi\in\mathcal{D}\left(U\right).$ Therefore, $\mathsf{f}$ is a
regular distribution given by the bounded function $f$ in the open set $U.$∎

In the proof we can see that $\inf\left\\{M:\text{ (\ref{SLe 1})
holds}\right\\}\leq\left\|f\right\|_{L^{\infty}\left(U\right)}.$ In fact, we
have more.

######  Lemma 6.3.

If $\mathsf{f}\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$ is a regular
distribution in $U,$ given by a bounded function $f\in
L^{\infty}\left(U\right)$ then

(6.5) $\left\|f\right\|_{L^{\infty}\left(U\right)}=\inf\left\\{M:\text{
(\ref{SLe 1}) holds for all positive, normalized test functions}\right\\}\,,$

and

(6.6)
$\left\|f\right\|_{L^{\infty}\left(U\right)}=\sup\left\\{\left|\left\langle\mathsf{f},\phi\right\rangle\right|:\text{
}\phi\in\mathcal{D}\left(U\right)\text{ positive, normalized test
function}\right\\}\,.$

###### Proof.

Clearly $\inf\left\\{M:\text{ (\ref{SLe 1}) holds for all positive, normalized
test functions}\right\\}$ is equal to
$\sup\left\\{\left|\left\langle\mathsf{f},\phi\right\rangle\right|:\text{
}\phi\in\mathcal{D}\left(U\right)\text{ positive, normalized test
function}\right\\};$ let us call this $K.$ We know that
$K\leq\left\|f\right\|_{L^{\infty}\left(U\right)}.$ To prove the converse
inequality, let $s<\left\|f\right\|_{L^{\infty}\left(U\right)}.$ Then there
exists $\mathbf{x}_{0}\in U$ such that the distributional point value
$\mathsf{f}\left(\mathbf{x}_{0}\right)$ exists and
$s<\left|\mathsf{f}\left(\mathbf{x}_{0}\right)\right|.$ If $\phi$ is a
positive normalized test function, then so are the test functions
$\varphi_{\lambda}\left(\mathbf{x}\right)=\lambda^{d}\phi\left(\mathbf{x}_{0}+\lambda\mathbf{x}\right)$
for all $\lambda>0,$ and if $\lambda$ is big enough,
$\varphi_{\lambda}\in\mathcal{D}\left(U\right).$ Since
$\lim_{\lambda\rightarrow\infty}\left\langle\mathsf{f},\varphi_{\lambda}\right\rangle=\mathsf{f}\left(\mathbf{x}_{0}\right),$
we can find $\lambda$ such that
$\left|\left\langle\mathsf{f},\varphi_{\lambda}\right\rangle\right|>s.$
Consequently, $K>s,$ and because
$s<\left\|f\right\|_{L^{\infty}\left(U\right)}$ is arbitrary,
$K\geq\left\|f\right\|_{L^{\infty}\left(U\right)}.$∎

In fact, the same argument in the proof of Lemma 6.3 allows us to obtain the
ensuing.

######  Lemma 6.4.

If $\mathsf{f}\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$ is a real
regular distribution in $U,$ given by a function $f\in L^{1}\left(U\right)$
then the essential supremum and infimum of $f$ are also given as

(6.7) $\operatorname*{esssup}_{\mathbf{x}\in
U}f\left(x\right)=\sup_{\phi\in\mathcal{D}\left(U\right),\phi\geq
0,\int\phi=1}\left\langle\mathsf{f}\left(\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle\,,$

and

(6.8) $\operatorname*{essinf}_{\mathbf{x}\in
U}f\left(\mathbf{x}\right)=\inf_{\phi\in\mathcal{D}\left(U\right),\phi\geq
0,\int\phi=1}\left\langle\mathsf{f}\left(x\right),\phi\left(x\right)\right\rangle\,.$

We notice that when $f\in L^{1}\left(U\right)$ then (6.7) could be $+\infty$
and (6.8) could be $-\infty.$

##  7\. Comparison of definitions

We will now study whether the existence of the distributional point value
$f\left(\mathbf{x}_{0}\right)$ is equivalent to the existence of
$f\left(\mathbf{x}_{0}\right)$ $\left(\mathfrak{F}\right)$ for several
families of delta sequences $\mathfrak{F}.$

###  7.1. Standard delta sequences generated by a positive normalized test
function

In this section we consider the family $\mathfrak{F}$ of standard delta
sequences generated by a positive normalized test function of
$\mathcal{D}\left(\mathbb{R}^{d}\right).$

######  Proposition 7.1.

Let $f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right).$ Then $f$ has a
thick distributional point value at $\mathbf{x}_{0}$ if and only if for all
standard delta sequences generated by a positive normalized test function of
$\mathcal{D}\left(\mathbb{R}^{d}\right),\
\left\\{\phi_{n}\right\\}_{n=1}^{\infty},$ the limit

(7.1) $\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\right\rangle=\gamma_{\left\\{\phi_{n}\right\\}}\,,$

exists.

###### Proof.

A standard delta sequences generated by a normalized positive test function
$\phi$ is of the form
$\phi_{n}\left(\mathbf{x}\right)=n^{d}\phi\left(n\mathbf{x}\right).$ If the
distributional thick point value
$f_{\mathbf{x}_{0}}\left(\mathbf{w}\right)=\gamma\left(\mathbf{w}\right)$
exists, $\gamma\in\mathcal{D}^{\prime}\left(\mathbb{S}\right),$ then

$\displaystyle\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\right\rangle$
$\displaystyle=\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),n^{d}\phi\left(n\mathbf{x}\right)\right\rangle$
$\displaystyle=\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\left(1/n\right)\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle$
$\displaystyle=\int_{0}^{\infty}\left\langle\gamma\left(\mathbf{w}\right),\phi\left(r\mathbf{w}\right)\right\rangle_{\mathcal{D}^{\prime}\left(\mathbb{S}\right)\times\mathcal{D}\left(\mathbb{S}\right)}r^{d-1}\,\mathrm{d}r\,,$

exists. Conversely, let $\phi$ be a normalized positive test function. If the
limit (7.1) exists for all standard delta sequences generated by a positive
normalized test function, it will exist for
$\phi_{n}^{\left\\{a\right\\}}\left(\mathbf{x}\right)=n^{d}a^{d}\phi\left(na\mathbf{x}\right)$
for all $a>0.$ Consequently, if the function $\Phi$ is defined as

(7.2) $\Phi\left(a\right)=\left\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),a^{d}\phi\left(a\mathbf{x}\right)\right\rangle\,,\
\ \ a>0\,,$

then

(7.3)
$\lim_{n\rightarrow\infty}\Phi\left(na\right)=\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}^{\left\\{a\right\\}}\left(\mathbf{x}\right)\right\rangle=\gamma_{\left\\{\phi_{n}^{\left\\{a\right\\}}\right\\}}\,,$

exists for all $a.$ Since $\Phi$ is continuous, Proposition 3.1 yields that
$\gamma_{\left\\{\phi_{n}^{\left\\{a\right\\}}\right\\}}=\gamma_{0}\left(\phi\right)$
is independent of $a$ and actually
$\lim_{\lambda\rightarrow\infty}\Phi\left(\lambda\right)=\gamma_{0}\left(\phi\right).$
Hence,

(7.4) $\lim_{\varepsilon\rightarrow 0^{+}}\left\langle
f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle=\gamma_{0}\left(\phi\right)\,,$

for all normalized positive test functions. Therefore, Lemma 6.1 yields that
the limit $\lim_{\varepsilon\rightarrow 0^{+}}\left\langle
f\left(x_{0}+\varepsilon
x\right),\phi\left(x\right)\right\rangle=\gamma_{0}\left(\phi\right)$ exists
whenever $\phi\in\mathcal{D}\left(\mathbb{R}^{d}\right).$ The formula
$\left\langle\gamma_{0},\phi\right\rangle=\gamma_{0}\left(\phi\right),$
defines a distribution
$\gamma_{0}\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right),$ and
$\gamma_{0}$ is homogeneous of degree $0,$ that is,
$\gamma_{0}\left(t\mathbf{x}\right)=\gamma_{0}\left(\mathbf{x}\right),$ $t>0.$
As explained in Section 2, using [5, Thm. 2.6.2] we conclude that $\gamma_{0}$
is obtained from a distribution
$\gamma\in\mathcal{D}^{\prime}\left(\mathbb{S}\right)$ by the formula

(7.5)
$\left\langle\gamma_{0},\phi\right\rangle=\int_{0}^{\infty}\left\langle\alpha\left(\mathbf{w}\right),\phi\left(r\mathbf{w}\right)\right\rangle_{\mathcal{D}^{\prime}\left(\mathbb{S}\right)\times\mathcal{D}\left(\mathbb{S}\right)}r^{d-1}\,\mathrm{d}r\,,$

and that $\alpha$ is the thick distributional value of $f$ at
$\mathbf{x}_{0}.$∎

Let $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$ be a sequence of test
functions. If $T$ is an orthogonal transformation of $\mathbb{R}^{d},$ that
is, with $\left|\det T\right|=1,$ then the sequence
$\left\\{\phi_{n}^{T}\right\\}_{n=1}^{\infty},$ where
$\phi^{T}\left(\mathbf{x}\right)=\phi\left(T\mathbf{x}\right),$ is also a
delta sequence. We have then the following result.

######  Proposition 7.2.

Let $f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right).$ Then the
distributional point value $f\left(\mathbf{x}_{0}\right)$ exists if and only
if for all standard delta sequences generated by a positive normalized test
function of $\mathcal{D}\left(\mathbb{R}^{d}\right),\
\left\\{\phi_{n}\right\\}_{n=1}^{\infty},$ the limit
$\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\right\rangle=\gamma_{\left\\{\phi_{n}\right\\}}\,$exists
and for all orthogonal transformations $T$ of $\mathbb{R}^{d},$
$\gamma_{\left\\{\phi_{n}^{T}\right\\}}=\gamma_{\left\\{\phi_{n}\right\\}}.$

###### Proof.

This follows immediately from Proposition 7.1 if we observe that a homogeneous
function or distribution of degree $0$ is a constant if and only if it is
invariant with respect to orthogonal transformations.∎

Notice that in one variable, Proposition 7.1 says that
$\lim_{n\rightarrow\infty}\left\langle
f\left(x_{0}+x\right),\phi_{n}\left(x\right)\right\rangle=\gamma_{\left\\{\phi_{n}\right\\}}$
exists for all standard delta sequences generated by a positive normalized
test function if and only if there are constants $\gamma_{+}$ and $\gamma_{-}$
such that

(7.6) $\lim_{\varepsilon\rightarrow 0^{+}}\left\langle
f\left(x_{0}+\varepsilon
x\right),\psi\left(x\right)\right\rangle=\gamma_{-}\int_{-\infty}^{0}\psi\left(x\right)\,\mathrm{d}x+\gamma_{+}\int_{0}^{\infty}\psi\left(x\right)\,\mathrm{d}x\,,$

for all $\psi\in\mathcal{D}\left(\mathbb{R}\right).$ On the other hand, since
the only orthogonal transformations in dimension one are the identity and
$x\rightsquigarrow-x,$ Proposition 7.2 says that the distributional point
value $f\left(x_{0}\right)$ exists if and only if for all standard delta
sequences generated by a positive normalized test function of
$\mathcal{D}\left(\mathbb{R}\right),\
\left\\{\phi_{n}\right\\}_{n=1}^{\infty},$ the limit
$\lim_{n\rightarrow\infty}\left\langle
f\left(x_{0}+x\right),\phi_{n}\left(x\right)\right\rangle=\gamma_{\left\\{\phi_{n}\right\\}}\,$exists
and
$\gamma_{\left\\{\phi_{n}\left(-x\right)\right\\}}=\gamma_{\left\\{\phi_{n}\left(x\right)\right\\}}.$

Our results also give the ensuing equivalence.

######  Proposition 7.3.

Let $f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right).$ Then the
distributional point value $f\left(\mathbf{x}_{0}\right)$ exists and equals
$\gamma$ if and only if for $\mathfrak{F}\ $the family of standard delta
sequences generated by a positive normalized test function

(7.7) $f\left(\mathbf{x}_{0}\right)=\gamma\ \left(\mathfrak{F}\right)\,.$

###  7.2. Standard delta sequences generated by an even positive normalized
test function

We now consider the case of symmetric standard delta sequences, the family
considered by Sasane [18].

We first need to explain the idea of symmetric point values. Let
$f\in\mathcal{D}^{\prime}\left(\mathbb{R}\right)$ and $x_{0}\in\mathbb{R}.$
The symmetric distributional point value of $f$ exists at $x_{0}$ and equals
$\gamma$ if

(7.8) $\lim_{\varepsilon\rightarrow 0}\frac{f\left(x_{0}+\varepsilon
x\right)+f\left(x_{0}-\varepsilon x\right)}{2}=\gamma\,,$

in $\mathcal{D}^{\prime}\left(\mathbb{R}\right).$ Each distribution can be
written as the sum of an even one and an odd one,

(7.9) $g=g_{\mathrm{e}}+g_{\mathrm{o}}\,,$

where

(7.10)
$g_{\mathrm{e}}\left(x\right)=\frac{g\left(x\right)+g\left(-x\right)}{2}\,,\ \
\ \
g_{\mathrm{o}}\left(x\right)=\frac{g\left(x\right)-g\left(-x\right)}{2}\,.$

Applying this to $g\left(x\right)=f\left(x_{0}+x\right),$ we see that the
distributional symmetric value $f\left(x_{0}\right)$ exists and equals
$\gamma$ if and only if the distributional value
$g_{\mathrm{e}}\left(0\right)$ exists and equals $\gamma.$

Notice also that if $\phi$ is a test function and we write
$\phi=\phi_{\mathrm{e}}+\phi_{\mathrm{o}},$ then

(7.11) $\left\langle g,\phi\right\rangle=\left\langle
g_{\mathrm{e}},\phi_{\mathrm{e}}\right\rangle+\left\langle
g_{\mathrm{o}},\phi_{\mathrm{o}}\right\rangle\,.$

Therefore we have the following result.

######  Lemma 7.4.

A distribution $f\in\mathcal{D}^{\prime}\left(\mathbb{R}\right)$ has a
symmetric distributional value $\gamma$ at $x_{0}$ if and only if

(7.12) $\lim_{\varepsilon\rightarrow 0}\left\langle f\left(x_{0}+\varepsilon
x\right),\phi_{\mathrm{e}}\left(x\right)\right\rangle=\gamma\int_{-\infty}^{\infty}\phi_{\mathrm{e}}\left(x\right)\,\mathrm{d}x\,,$

for all even test functions $\phi_{\mathrm{e}}.$

###### Proof.

Indeed, if (7.8) is satisfied, then

$\displaystyle\lim_{\varepsilon\rightarrow 0}\left\langle
f\left(x_{0}+\varepsilon
x\right),\phi_{\mathrm{e}}\left(x\right)\right\rangle$
$\displaystyle=\lim_{\varepsilon\rightarrow 0}\left\langle
f\left(x_{0}+\varepsilon x\right)-g_{\mathrm{o}}\left(\varepsilon
x\right),\phi_{\mathrm{e}}\left(x\right)\right\rangle$
$\displaystyle=\lim_{\varepsilon\rightarrow
0}\left\langle\frac{f\left(x_{0}+\varepsilon x\right)+f\left(x_{0}-\varepsilon
x\right)}{2},\phi_{\mathrm{e}}\left(x\right)\right\rangle$
$\displaystyle=\gamma\int_{-\infty}^{\infty}\phi_{\mathrm{e}}\left(x\right)\,\mathrm{d}x\,.$

Conversely, if (7.12) holds, then for any test function
$\phi=\phi_{\mathrm{e}}+\phi_{\mathrm{o}},$

$\displaystyle\lim_{\varepsilon\rightarrow 0}\left\langle
g_{\mathrm{e}}\left(\varepsilon x\right),\phi\left(x\right)\right\rangle$
$\displaystyle=\lim_{\varepsilon\rightarrow 0}\left\langle
g_{\mathrm{e}}\left(\varepsilon
x\right),\phi_{\mathrm{e}}\left(x\right)\right\rangle$
$\displaystyle=\lim_{\varepsilon\rightarrow 0}\left\langle
f\left(x_{0}+\varepsilon
x\right),\phi_{\mathrm{e}}\left(x\right)\right\rangle$
$\displaystyle=\gamma\int_{-\infty}^{\infty}\phi_{\mathrm{e}}\left(x\right)\,\mathrm{d}x$
$\displaystyle=\gamma\int_{-\infty}^{\infty}\phi\left(x\right)\,\mathrm{d}x\,.$

Hence $g_{\mathrm{e}}\left(0\right)=\gamma,$ so that the symmetric
distributional value of $f$ at $x_{0}$ equals $\gamma.$∎

We can now give an equivalence to the existence of the point value
$f\left(x_{0}\right)=\gamma$ $(\mathfrak{F}_{\mathrm{sy}}),$ where
$\mathfrak{F}_{\mathrm{sy}}$ is the family of standard delta sequences
generated by a positive normalized even test function of
$\mathcal{D}\left(\mathbb{R}\right).$

######  Proposition 7.5.

Let $f\in\mathcal{D}^{\prime}\left(\mathbb{R}\right).$ Then the following are
equivalent:

1\. If $\mathfrak{F}_{\mathrm{sy}}\ $is the family of standard delta sequences
generated by a positive normalized even test function then

(7.13) $f\left(x_{0}\right)=\gamma\ \ \
\left(\mathfrak{F}_{\mathrm{sy}}\right)\,.$

2\. The symmetric distributional point value of $f$ exists at $x_{0}$ and
equals $\gamma.$

###### Proof.

Indeed, if (7.13) holds then

(7.14) $\lim_{n\rightarrow\infty}\left\langle
f\left(x_{0}+x\right),\phi_{n}\left(x\right)\right\rangle=\gamma\,,$

for all standard delta sequences $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$
generated by a positive normalized _even_ test function $\phi_{\mathrm{e}},$
and use of Proposition 3.1 yields that

(7.15) $\lim_{\varepsilon\rightarrow 0}\left\langle f\left(x_{0}+\varepsilon
x\right),\phi_{\mathrm{e}}\left(x\right)\right\rangle=\gamma\,,$

for such normalized even test functions $\phi_{\mathrm{e}}.$ This last
statement is equivalent to the fact that (7.12) holds for _all_ even test
functions because of Lemma 6.1, and Lemma 7.4 yields that in turn this is
equivalent to the symmetric distributional point value being equal to
$\gamma.$∎

Actually, using the same ideas as in the proof of this Proposition we see that
the limit $\lim_{n\rightarrow\infty}\left\langle
f\left(x_{0}+x\right),\phi_{n}\left(x\right)\right\rangle=\gamma_{\left\\{\phi_{n}\right\\}}$
exists for all standard delta sequences
$\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$ generated by a positive normalized
even test function $\phi_{\mathrm{e}}$ if and only if this limit is a constant
$\gamma$ and (7.13) is satisfied.

###  7.3. The family of standard delta sequences generated by a radial
positive normalized test function

We now consider the family $\mathfrak{F}_{\mathrm{rad}}$ of standard sequences
generated by a radial positive normalized test function.

Let us start with some notation. We denote $r=\left|\mathbf{x}\right|$ the
radial variable in $\mathbb{R}^{d}.$ A test function
$\phi\in\mathcal{D}\left(\mathbb{R}^{d}\right)$ is called _radial_ if it is a
function of $r,$ $\phi\left(\mathbf{x}\right)=\varphi\left(r\right),$ for some
even function $\varphi\in\mathcal{D}\left(\mathbb{R}\right);$ the space of all
radial test functions of $\mathcal{D}\left(\mathbb{R}^{d}\right)$ is denoted
as $\mathcal{D}_{\mathrm{rad}}\left(\mathbb{R}^{d}\right).$ Similarly, we
denote as $\mathcal{D}_{\mathrm{rad}}^{\prime}\left(\mathbb{R}^{d}\right)$ the
space of all radial distributions; a distribution
$f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$ is radial if
$f\left(T\mathbf{x}\right)=f\left(\mathbf{x}\right)$ for any orthogonal
transformation of $\mathbb{R}^{d},$ and this actually means [4, 7] that
$f\left(\mathbf{x}\right)=f_{1}\left(r\right)$ for some distribution of one
variable $f_{1}.$ Notice, however, that while $\varphi$ is uniquely determined
by $\phi,$ for a given $f$ there are several possible distributions $f_{1}.$

When $d=1$ then $\mathcal{D}_{\mathrm{rad}}\left(\mathbb{R}\right)$ and
$\mathcal{D}_{\mathrm{rad}}^{\prime}\left(\mathbb{R}\right)$ become the spaces
of even test functions and distributions, respectively, and are also denoted
as $\mathcal{D}_{\mathrm{even}}\left(\mathbb{R}\right)$ and
$\mathcal{D}_{\mathrm{even}}^{\prime}\left(\mathbb{R}\right).$ This was the
situation considered in the previous subsection.

Observe that the space
$\mathcal{D}_{\mathrm{rad}}^{\prime}\left(\mathbb{R}^{d}\right)$ is naturally
isomorphic to the dual space
$\left(\mathcal{D}_{\mathrm{rad}}\left(\mathbb{R}^{d}\right)\right)^{\prime},$
that is to say, if the action of a radial distribution is known in all radial
test functions, then it can be obtained for arbitrary test functions. Indeed,
if $f\in\mathcal{D}_{\mathrm{rad}}^{\prime}\left(\mathbb{R}^{d}\right)$ and
$\phi\in\mathcal{D}\left(\mathbb{R}^{d}\right),$ then

(7.16) $\left\langle
f\left(\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle=\left\langle
f\left(\mathbf{x}\right),\widetilde{\phi}\left(\mathbf{x}\right)\right\rangle\,,$

where $\widetilde{\phi}\in\mathcal{D}_{\mathrm{rad}}\left(\mathbb{R}\right)$
is given as

(7.17)
$\widetilde{\phi}\left(\mathbf{x}\right)=\phi^{o}\left(\left|\mathbf{x}\right|\right)\,,$

$\phi^{o}\in\mathcal{D}_{\mathrm{even}}\left(\mathbb{R}\right)$ being defined
as

(7.18)
$\phi^{o}\left(r\right)=\frac{1}{\omega}\int_{\mathbb{S}}\phi\left(r\mathbf{\theta}\right)\,\mathrm{d}\sigma\left(\mathbf{\theta}\right)\,.$

Here we denote by $\mathbb{S}$ the unit sphere of $\mathbb{R}^{d},$
$\mathrm{d}\sigma$ is the Lebesgue measure in $\mathbb{S}$ and
$\omega=2\pi^{d/2}/\Gamma\left(d/2\right)$ is the surface area of the sphere.

Equations (7.17) and (7.18) define the _radial component of a test function._
We can also define the radial component of a distribution $f,$
$\widetilde{f}\in\mathcal{D}_{\mathrm{rad}}^{\prime}\left(\mathbb{R}^{d}\right),$
as

(7.19)
$\left\langle\widetilde{f}\left(\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle=\left\langle
f\left(\mathbf{x}\right),\widetilde{\phi}\left(\mathbf{x}\right)\right\rangle\,.$

The distributional analog of (7.18) is not well defined, however [4, 7].

We say that a distribution $f$ has a _radial distributional point value_ at
$\mathbf{x}_{0}$ equal to $\gamma$ if

(7.20) $\widetilde{g}\left(\mathbf{0}\right)=\gamma\,,$

where $\widetilde{g}$ is the radial component of
$g\left(\mathbf{x}\right)=f\left(\mathbf{x}_{0}\mathbf{+x}\right).$ Similar to
Lemma 7.4, we have the following characterization.

######  Lemma 7.6.

A distribution $f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$ has a
radial distributional value $\gamma$ at $\mathbf{x}_{0}$ if and only if

(7.21) $\lim_{\varepsilon\rightarrow 0}\left\langle
f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right),\phi_{\mathrm{rad}}\left(\mathbf{x}\right)\right\rangle=\gamma\int_{\mathbb{R}^{d}}\phi_{\mathrm{rad}}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x\,},$

for all radial test functions $\phi_{\mathrm{rad}}.$

###### Proof.

If $\widetilde{g}\left(\mathbf{0}\right)=\gamma,$ then

$\displaystyle\lim_{\varepsilon\rightarrow 0}\left\langle
f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right),\phi_{\mathrm{rad}}\left(\mathbf{x}\right)\right\rangle$
$\displaystyle=\lim_{\varepsilon\rightarrow
0}\left\langle\widetilde{g}\left(\varepsilon\mathbf{x}\right),\phi_{\mathrm{rad}}\left(\mathbf{x}\right)\right\rangle$
$\displaystyle=\gamma\int_{\mathbb{R}^{d}}\phi_{\mathrm{rad}}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\,.$

On the other hand, if (7.21) holds, then for any test function $\phi,$

$\displaystyle\lim_{\varepsilon\rightarrow
0}\left\langle\widetilde{g}\left(\varepsilon\mathbf{x}\right),\phi\left(\mathbf{x}\right)\right\rangle$
$\displaystyle=\lim_{\varepsilon\rightarrow
0}\left\langle\widetilde{g}\left(\varepsilon\mathbf{x}\right),\widetilde{\phi}\left(\mathbf{x}\right)\right\rangle$
$\displaystyle=\lim_{\varepsilon\rightarrow 0}\left\langle
f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right),\widetilde{\phi}\left(\mathbf{x}\right)\right\rangle$
$\displaystyle=\gamma\int_{\mathbb{R}^{d}}\widetilde{\phi}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}$
$\displaystyle=\gamma\int_{\mathbb{R}^{d}}\phi\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\,.$

Hence $\widetilde{g}\left(\mathbf{0}\right)=\gamma,$ that is, the radial
distributional value of $f$ at $\mathbf{x}_{0}$ equals $\gamma.$∎

Therefore, we obtain the ensuing equivalence for the fact that
$f\left(\mathbf{x}_{0}\right)=\gamma$ $(\mathfrak{F}_{\mathrm{rad}}).$

######  Proposition 7.7.

Let $f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right).$ Then the following
are equivalent:

1\. If $\mathfrak{F}_{\mathrm{rad}}\ $is the family of standard delta
sequences generated by a positive normalized radial test function then

(7.22) $f\left(\mathbf{x}_{0}\right)=\gamma\ \ \
\left(\mathfrak{F}_{\mathrm{rad}}\right)\,.$

2\. The radial distributional point value of $f$ exists at $\mathbf{x}_{0}$
and equals $\gamma.$

###### Proof.

Indeed, if (7.22) holds then

(7.23) $\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\right\rangle=\gamma\,,$

for all standard delta sequences $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$
generated by a positive normalized _radial_ test function
$\phi_{\mathrm{rad}}.$ Use of Proposition 3.1 yields that

(7.24) $\lim_{\varepsilon\rightarrow 0}\left\langle
f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right),\phi_{\mathrm{rad}}\left(\mathbf{x}\right)\right\rangle=\gamma\,,$

for such normalized radial test functions $\phi_{\mathrm{rad}}.$ This last
statement is equivalent to the fact that (7.12) holds for _all_ radial test
functions because of Lemma 6.1, and Lemma 7.6 yields that, in turn, this is
equivalent to the radial distributional point value being equal to $\gamma.$∎

We also have the next result, that is obtained from Lemma 6.1.

######  Proposition 7.8.

The limit $\lim_{n\rightarrow\infty}\left\langle
f\left(\mathbf{x}_{0}+\varepsilon\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\right\rangle=\gamma_{\left\\{\phi_{n}\right\\}}$
exists for all standard delta sequences
$\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$ generated by a positive normalized
radial test function $\phi_{\mathrm{rad}}$ if and only if this limit is a
constant $\gamma$ and $f\left(\mathbf{x}_{0}\right)=\gamma\ \
\left(\mathfrak{F}_{\mathrm{rad}}\right).$

###  7.4. The family of all positive normalized test functions

We saw in Subsection 7.2 that Sasane’s notion of point values was not
equivalent to the standard definition, nor, in the next subsection, is the
notion based on the family $\mathfrak{F}_{\mathrm{rad}}$ of standard delta
sequences generated by a positive normalized radial test function.
Nevertheless, for the family $\mathfrak{F}$ of standard delta sequences
generated by a positive normalized test function the point value definition is
in fact equivalent to the standard Łojasiewicz definition. Of course, Sasane
was considering the family of standard delta sequences generated by an _even_
positive normalized test function $\mathfrak{F}_{\mathrm{sy}}.$ Both
$\mathfrak{F}_{\mathrm{sy}}$ and $\mathfrak{F}_{\mathrm{rad}}$ are subfamilies
of $\mathfrak{F}.$ We can also consider families larger than $\mathfrak{F}.$
For instance, we can consider the family $\mathfrak{F}_{\mathrm{all}}$ of
_all_ delta sequences formed with positive normalized test functions. In this
next example, we will see that the distributional point value
$f\left(\mathbf{x}_{0}\right)=\gamma$ is not equivalent to
$f\left(\mathbf{x}_{0}\right)=\gamma\ \
\left(\mathfrak{F}_{\mathrm{all}}\right).$ Later on we shall find an
equivalent formulation to $f\left(\mathbf{x}_{0}\right)=\gamma\ \
\left(\mathfrak{F}_{\mathrm{all}}\right).$

######  Example 7.9.

Let $f$ be the regular distribution given by $f(x)=\sin(1/x)$. Then $f(0)=0$
distributionally [11]. Let $a_{n}$ be a positive sequence with
$a_{n}\rightarrow 0$ and $f(a_{n})=C>0.$ For instance, we could take
$a_{n}=1/(2\pi n+\pi/6).$ For a fixed $n,$ let
$\left\\{\psi_{n,m}\right\\}_{m=1}^{\infty}$ be a sequence of positive test
functions such that $\psi_{n,m}\rightarrow\delta(x-a_{n})$ as
$m\rightarrow\infty.$ Then as $n\rightarrow\infty$, we obtain a sequence
$\delta_{n}(x)=\delta(x-a_{n})$ that converges to $\delta(x).$ For each $n,$
let $m_{n}$ be large enough so that

$\left|\int_{B_{1/n}\left(a_{n}\right)}f\left(x\right)\psi_{n,m}\left(x\right)\,\mathrm{d}x-f\left(a_{n}\right)\right|<C/2$

and $\operatorname*{supp}\psi_{n,m}\subset B_{1/n}\left(a_{n}\right)$ for
$m\geq m_{n}$. Then we can define the sequence
$\phi_{n}\left(x\right)=\psi_{n,m_{n}}\left(x\right)$. By Lemma 5.1, this is a
delta sequence and we have $\langle f(x),\phi_{n}\left(x\right)\rangle>C/2$
for all $n$ and so $\lim\limits_{n\rightarrow\infty}\langle
f(x),\phi_{n}\left(x\right)\rangle$ cannot be equal to $0.$

The next lemma will be useful momentarily.

######  Lemma 7.10.

If $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$ is a delta sequence of positive
test functions then

(7.25)
$\lim_{n\rightarrow\infty}\left\|\phi_{n}\right\|_{L^{1}\left(B\setminus
U\right)}=0\,,$

where $B$ and $U$ are both neighborhoods of the origin.

###### Proof.

Choose $\psi\in\mathcal{D}\left(\mathbb{R}^{d}\right)$ such that $\psi\geq 0,$

(7.26) $\psi\left(\mathbf{x}\right)=1\,,\ \ \ \mathbf{x}\in B\setminus U\,,\ \
\ \psi\left(\mathbf{0}\right)=0\,,$

which is possible because $\mathbf{0}\notin\overline{B\setminus U}.$ Then

$\displaystyle\left\|\phi_{n}\right\|_{L^{1}\left(B\setminus U\right)}$
$\displaystyle=\int_{B\setminus
U}\left|\phi_{n}\left(\mathbf{x}\right)\right|\,\mathrm{d}\mathbf{x}=\int_{B\setminus
U}\phi_{n}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}$
$\displaystyle=\int_{B\setminus
U}\psi\left(\mathbf{x}\right)\phi_{n}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}$
$\displaystyle\leq\int_{B}\psi\left(\mathbf{x}\right)\phi_{n}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\rightarrow\psi\left(\mathbf{0}\right)=0\,,$

as $n\rightarrow\infty.$∎

We are now ready to prove the main result of this section.

######  Proposition 7.11.

Suppose $f\in\mathcal{D}^{\prime}\left(\mathbb{R}^{d}\right)$ and
$\mathbf{x}_{0}\in\mathbb{R}^{d}.$ If

(7.27) $\lim\limits_{n\rightarrow\infty}\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\rangle=\gamma\,,$

for all positive delta sequences $\left\\{\phi_{n}\right\\}$, then the
following two conditions hold:

  1. (1)

There is an $r^{\ast}>0$ such that
$\left.f\right|_{B_{r^{\ast}}\left(\mathbf{x}_{0}\right)}\in
L^{\infty}\left(B_{r^{\ast}}\left(\mathbf{x}_{0}\right)\right)$.

  2. (2)

$\lim\limits_{r\rightarrow
0}\left\|\left.f\right|_{B_{r}\left(\mathbf{x}_{0}\right)}-\gamma\chi_{B_{r}\left(\mathbf{x}_{0}\right)}\right\|_{\infty}=0.$

Here $\chi_{B_{r}\left(\mathbf{x}_{0}\right)}$ is the characteristic function
of the ball $B_{r}\left(\mathbf{x}_{0}\right).$ Conversely, if (1) and (2) are
satisfied, then (7.27) holds for all positive delta sequences with support
contained in $B_{r^{\ast}}\left(\mathbf{0}\right).$

###### Proof.

Suppose that (7.27) holds. Notice that (1) follows from Lemma 6.2. To see that
(2) is true, suppose instead that

(7.28) $\limsup_{r\rightarrow
0}\left\|\left.f\right|_{B_{r}\left(\mathbf{x}_{0}\right)}-\gamma\chi_{B_{r}\left(\mathbf{x}_{0}\right)}\right\|_{\infty}=C>0\,.$

Let $r_{n}$ be a decreasing sequence of positive numbers with $r_{n}<r^{\ast}$
and $r_{n}\rightarrow 0.$ For each $n$, there is a positive normalized test
function supported in $B_{r_{n}}\left(\mathbf{0}\right),$ say $\phi_{n},$ such
that

(7.29) $\left|\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\rangle-\gamma\right|>\frac{C}{2}\,.$

By Lemma 5.1, $\left\\{\phi_{n}\right\\}_{n=1}^{\infty}$ forms a delta
sequence and so $\lim\limits_{n\rightarrow\infty}\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\phi_{n}\left(\mathbf{x}\right)\rangle=\gamma,$
which contradicts (7.29).

For the converse, let $\left\\{\psi_{n}\right\\}$ be a delta sequence of
positive normalized test functions supported in
$B_{r^{\ast}}\left(\mathbf{0}\right).$ Since by (1) $f$ is a regular
distribution in $B_{r^{\ast}}\left(\mathbf{0}\right)$ we have

(7.30) $\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\psi_{n}\left(\mathbf{x}\right)\rangle-\gamma=\int_{B_{r^{\ast}}\left(\mathbf{0}\right)}\left(f\left(\mathbf{x}_{0}+\mathbf{x}\right)-\gamma\right)\psi_{n}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\,.$

Let $\varepsilon>0.$ By condition (2), we can find an open neighborhood $V$ of
the origin that is contained in $B_{r^{\ast}}\left(\mathbf{0}\right)$ such
that if $W=\mathbf{x}_{0}+V,$ then
$\left\|f-\gamma\right\|_{L^{\infty}\left(W\right)}<\varepsilon$ and for this
$V$ we can find $n_{0}$ such that
$\left\|\psi_{n}\right\|_{L_{1}\left(B_{r^{\ast}}\left(\mathbf{0}\right)\setminus
V\right)}<\varepsilon$ if $n\geq n_{0}.$ If $M$ is the constant
$\|f-\gamma\|_{L^{\infty}\left(B_{r^{\ast}}\left(\mathbf{x}_{0}\right)\right)},$
then we have

$\displaystyle\left|\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\psi_{n}\left(\mathbf{x}\right)\rangle-\gamma\right|$
$\displaystyle=\left|\int_{W}\left(f\left(\mathbf{x}\right)-\gamma\right)\psi_{n}\left(\mathbf{x-x}_{0}\right)\,\mathrm{d}\mathbf{x}\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \ \
+\int_{B_{r^{\ast}}\left(\mathbf{0}\right)\setminus
V}\left(f\left(\mathbf{x}_{0}+\mathbf{x}\right)-\gamma\right)\psi_{n}\left(\mathbf{x}\right)\,\mathrm{d}\mathbf{x}\right|$
$\displaystyle\leq\left\|f-\gamma\right\|_{L^{\infty}\left(W\right)}\left\|\psi_{n}\right\|_{L^{1}\left(V\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \
+\|f-\gamma\|_{L^{\infty}\left(B_{r^{\ast}}\left(\mathbf{x}_{0}\right)\setminus
W\right)}\left\|\psi_{n}\right\|_{L^{1}\left(B_{r^{\ast}}\left(\mathbf{0}\right)\setminus
V\right)}$ $\displaystyle<\varepsilon+M\varepsilon\,,$

and consequently $\lim\limits_{n\rightarrow\infty}\langle
f\left(\mathbf{x}_{0}+\mathbf{x}\right),\psi_{n}\left(\mathbf{x}\right)\rangle=\gamma.$
∎

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