#  $p$-capacity with Bessel convolution

Á. P. Horváth

###### Abstract.

We define and examine nonlinear potential by Bessel convolution with Bessel
kernel. We investigate removable sets with respect to Laplace-Bessel
inequality. By studying the maximal and fractional maximal measure, a Wolff
type inequality is proved. Finally the relation of B-$p$ capacity and
B-Lipschitz mapping, and the B-$p$ capacity and weighted Hausdorff measure and
the B-$p$ capacity of Cantor sets are examined.

###### Key words and phrases:

nonlinear potential, Bessel convolution, Laplace-Bessel equation, Wolff
inequality, weighted Hausdorff measure

###### 2020 Mathematics Subject Classification:

31C45, 26D15, 28A78

##  1\. Introduction

Classical, nonlinear, and Bessel potentials are widespread, have an extensive
literature, and are widely applicable, see e.g. [17], [10], [15] and the
references therein. Below we introduce and examine nonlinear potential defined
by Bessel convolution with Bessel kernel.

Bessel translation was defined by Delsarte [5] and the basic investigation is
due to Levitan, [14]. In a series of works the authors pointed out that Bessel
translation and convolution methods are effective tools to handle Bessel-type
partial differential operators, see e.g. [16], [12], [18], [13]. It also
proved useful for deriving Nikol’skii type inequality, see [4], and for giving
compactness criteria in some Banach spaces, see [11].

This leads to examine nonlinear potential and $p$-capacity with respect to
Bessel convolution. The curiosity of the method is that the underlying space
of Bessel-$p$ capacity is automatically weighted. Weighted nonlinear potential
was studied already in the ’80-s, see e. g. [1], [3]. For logarithmic
potentials with external field see the monograph [21]. In our investigation
the Bessel weighted space is a natural consequence of the definition of
convolution, and so many of the results are very similar to the ones proved in
the unweighted case.

The paper is organized as follows. After the preliminaries, in the third
section, applying recent results on Bessel potential, we investigate removable
sets for Laplace-Bessel equation. In the fourth section a Wolff type
inequality is proved, which is the basis of the study of the last section.
This last section contains some ”metric” results on Lipschitz type mapping and
on capacity of Cantor sets. Since Bessel translation is not a geometric
similarity, moreover the underlying space is weighted, we have to introduce a
special property (B-Lipschitz mapping), and the notion of weighted Hausdorff
measure.

##  2\. Notation, preliminaries

Let $\mathbb{R}^{n}_{+}:=\\{x=(x_{1},\dots,x_{n}):x_{i}\geq 0,\hskip
4.0pti=1,\dots,n.\\}$. $\lambda$ is the $n$-dimensional Lebesgue measure.
$a=a_{1},\dots,a_{n}$ is a multiindex. Let $E\subset\mathbb{R}^{n}_{+}$ and
$\mathcal{M}(E)$ stands for the Radon measures supported on $E$. If
$\mu\in\mathcal{M}(E)$ for some $E$, $d\mu_{a}(x):=x^{a}d\mu(x)$ Define the
Banach space $L^{p}_{a}$ as follows.

$\|f\|_{p,a}^{p}=\int_{\mathbb{R}^{n}_{+}}|f(x)|^{p}d\lambda_{a}(x),$

and as usual

$L^{p}_{a}:=L^{p}_{a}(\mathbb{R}^{n}_{+})=\\{f:\|f\|_{p,a}<\infty\\},\hskip
4.0pt\hskip 4.0ptL^{p+}_{a}:=\\{f\in L^{p}_{a}:f\geq 0\\}.$

The dual index $p^{\prime}$ is defined by
$\frac{1}{p}+\frac{1}{p^{\prime}}=1$.

###  2.1. Bessel translation

Let

$a:=2\alpha_{1}+1,\dots,2\alpha_{n}+1,\hskip 4.0pt\hskip 4.0pt\hskip
4.0pt\alpha_{i}>-\frac{1}{2},\hskip 4.0pti=1,\dots,n,\hskip 4.0pt\hskip
4.0pt|a|=\sum_{i=1}^{n}(2\alpha_{i}+1).$

The Bessel translation of a function, $f$ (see e.g. [14], [16], [20]) is

$T_{a}^{t}f(x)=T_{a_{n}}^{t_{n}}\dots T_{a_{1}}^{t_{1}}f(x_{1},\dots,x_{n}),$

where

$T_{a_{i}}^{t_{i}}f(x_{1},\dots,x_{n})$ (1)
$=\frac{\Gamma(\alpha_{i}+1)}{\sqrt{\pi}\Gamma\left(\alpha_{i}+\frac{1}{2}\right)}\int_{0}^{\pi}f(x_{1},\dots,\sqrt{x_{i}^{2}+t_{i}^{2}-2x_{i}t_{i}\cos\vartheta_{i}},x_{i+1},\dots,x_{n})\sin^{2\alpha_{i}}\vartheta_{i}d\vartheta_{i}.$

The translation can also be expressed as an integral with respect to a kernel
function:

(2)
$T_{a_{i}}^{t_{i}}f(x_{1},\dots,x_{n})=\int_{0}^{\infty}K(x_{i},t_{i},z_{i})f(z_{1},\dots,z_{n})d\lambda_{a_{i}}(z_{i}),$

where

(3)
$K(x,t,z)=\left\\{\begin{array}[]{ll}\frac{\pi^{\alpha+\frac{1}{2}}\Gamma(\alpha+1)}{2^{2\alpha-1}\Gamma\left(\alpha+\frac{1}{2}\right)}\frac{[(x+t)^{2}-z^{2})(z^{2}-(x-t)^{2})]^{\alpha-\frac{1}{2}}}{(xtz)^{2\alpha}},\hskip
4.0pt\hskip 4.0pt|x-t|<z<x+t\\\ 0,\hskip 4.0pt\hskip
4.0pt\mbox{otherwise}.\end{array}\right.$

Obviously

$T_{a}^{t}f(x)=T_{a}^{x}f(t).$

$T_{a}$ is a positive operator, and

(4) $\|T_{a,x}^{t}f(x)\|_{p,a}\leq\|f\|_{p,a},\hskip 4.0pt\hskip 4.0pt1\leq
p\leq\infty,$

see e.g. [14].

The generalized convolution with respect to the Bessel translation is

$f*_{a}g=\int_{\mathbb{R}^{n}_{+}}T_{a,x}^{t}f(x)g(x)d\lambda_{a}(x).$

We have

$f*_{a}g=g*_{a}f,$

and Young’s inequality fulfils i. e. if $1\leq p,q,r\leq\infty$ with
$\frac{1}{r}=\frac{1}{p}+\frac{1}{q}-1$; if $f\in L^{p}_{\alpha}$ and $g\in
L^{q}_{\alpha}$, then

(5) $\|f*_{a}g\|_{r,a}\leq\|f\|_{p,a}\|g\|_{q,a},$

see [20, (3.178)].

Subsequently if it does not cause any confusion, $T^{t}f(x)$ stands for
$T_{a}^{t}f(x)$. For any set $H\subset\mathbb{R}^{n}$ we denote by
$H_{+}:=H\cap\mathbb{R}^{n}_{+}$. The next technical lemma will be useful in
the following sections.

######  Lemma 1.

$\mathrm{supp}T^{t}\chi_{B_{+}(0,r)}(x)=\overline{B_{+}(x,r)}$,
$\mathrm{supp}T^{t}\chi_{[0,r)^{n}}(x)=\times_{i=1}^{n}[x_{i}-r,x_{i}+r]_{+}=:T_{+}(x,r)$.
There is a $c>0$ such that for all $x\in\mathbb{R}^{n}_{+}$, $t\in B_{+}(x,r)$

(6) $T^{t}\chi_{B_{+}(0,r)}(x)\leq
c\prod_{i=1}^{n}\min\left\\{1,\left(\frac{r}{x_{i}}\right)^{a_{i}}\right\\}.$

There is a $c>0$ such that for all $x\in\mathbb{R}^{n}_{+}$, $t\in
T_{+}\left(x,\frac{r}{2}\right)$

(7) $T^{t}\chi_{[0,r)^{n}}(x)\geq
c\prod_{i=1}^{n}\min\left\\{1,\left(\frac{r}{x_{i}}\right)^{a_{i}}\right\\}.$

###### Proof.

The first two statements are direct consequences of the definition, for (6)
see [8, p. 321].
(7): Since $\sqrt{x_{i}^{2}+t_{i}^{2}-2x_{i}t_{i}\cos\vartheta_{i}}\leq
x_{i}+t_{i}$ if $x_{i}+t_{i}\leq r$, then

$\int_{\\{\vartheta\in[0\pi):\sqrt{x_{i}^{2}+t_{i}^{2}-2x_{i}t_{i}\cos\vartheta_{i}}\leq
r\\}}1d\sigma_{i}\vartheta=1,$

where
$d\sigma_{i}:=\frac{\Gamma(\alpha_{i}+1)}{\sqrt{\pi}\Gamma\left(\alpha_{i}+\frac{1}{2}\right)}\sin\vartheta^{2\alpha_{i}}d\vartheta_{i}$
is a probability measure on $[0,\pi]$.
If $r<x_{i}+t_{i}\leq 2r$, using (2) and recalling that
$|x_{i}-t_{i}|\leq\frac{r}{2}$ we have

$I_{i}=\frac{c(a_{i})}{(x_{i}t_{i})^{2\alpha_{i}}}\int_{|x_{i}-t_{i}|}^{r}\left[(z_{i}^{2}-(x_{i}-t_{i})^{2})((x_{i}+t_{i})^{2}-z_{i}^{2})\right]^{\alpha_{i}-\frac{1}{2}}z_{i}dz_{i}$
$\geq
c(a_{i})\frac{1}{(x_{i}+t_{i})^{4\alpha_{i}}}\int_{\frac{5}{8}r}^{\frac{7}{8}r}(\cdot)dz_{i}=c\frac{1}{r^{4\alpha_{i}}}r^{4\alpha_{i}-2}rr\geq
c(a_{i}).$

If $x_{i}+t_{i}>2r$, then $x_{i}\sim t_{i}$ ($f\sim g$ if there are positive
constants $A$ and $B$ such that $Af<g<Bf$) and we have

$I_{i}\geq
c\int_{\frac{5}{8}r}^{r}r^{2\alpha_{i}-1}\frac{(x_{i}+t_{i})^{2\alpha_{i}-1}}{(x_{i}t_{i})^{2\alpha_{i}}}z_{i}dz_{i}=c\left(\frac{r}{x_{i}}\right)^{2\alpha_{i}+1}.$

######  Remark 1.

For any $c\geq 2\sqrt{n}$

$T^{t}\chi_{B_{+}(0,r)}(x)\sim T^{t}\chi_{B_{+}(0,cr)}(x),$

for all $x\in\mathbb{R}^{n}_{+}$, $t\in B_{+}(x,r)$.
Indeed, in view of (2) if $H\subset S\subset\mathbb{R}^{n}_{+}$, then
$T^{t}\chi_{H}(x)\leq T^{t}\chi_{S}(x)$ for any $x,t$. Together with Lemma 1
it implies that $c_{1}T^{t}\chi_{B_{+}(0,r)}(x)\leq
T^{t}\chi_{B_{+}(0,cr)}(x)\leq c_{2}T^{t}\chi_{B_{+}(0,r)}(x)$.

###  2.2. Radially decreasing kernels and B-$p$ capacity

######  Definition 1.

Let $g$ be a non-negative lower semi-continuous, non-increasing function on
$\mathbb{R}_{+}$ for which

(8) $\int_{0}^{1}g(t)t^{n+|a|-1}dt<\infty.$

Then $\kappa:=g(|x|)$ is a radially decreasing kernel on $\mathbb{R}^{n}$.

The B-$p$ capacity with respect to $\kappa$ is as follows.

######  Definition 2.

Let $E\subset\mathbb{R}^{n}_{+}$, $1\leq p<\infty$.

$C_{p,\kappa}(E):=\inf\\{\|f\|_{p,a}^{p}:f\in L^{p+}_{a},\hskip
4.0pt\kappa*_{a}f(x)\geq 1,\hskip 4.0pt\forall x\in E\\}.$

######  Remark 2.

(1) Definition 1 is a special case of [2, Definition 2.3.3]. Thus all the
standard properties proved in [2, Chapter 2.3] are valid.
(2) Notice that by the definition of Bessel translation if $\kappa$ is a
radially decreasing kernel, then $T^{t}\kappa(x)\leq g(|x-t|)$. Thus if $f\geq
0$, $\kappa*_{a}f(x)\leq\kappa*h(x)$, where
$h(x)=f(x)x^{a}\chi_{\mathbb{R}^{n}_{+}}(x)$ and $*$ stands for the standard
convolution.
(3) Let $K\subset\mathbb{R}^{n}_{+}$, $1<p<\infty$. An equivalent form of
Definition 2 is

$C_{p,\kappa}^{\frac{1}{p}}(K)=\sup\\{\mu_{a}(K):\mu\in\mathcal{M}(K),\hskip
4.0pt\|\kappa*_{a}\mu\|_{p^{\prime},a}\leq 1\\},$

where $\mathcal{M}(K)$ is the set of (positive) measures on $K$, see [2,
Theorem 2.5.1].
(4) As usual, the definitions above can be extended to any subsets of
$\mathbb{R}^{n}_{+}$ as it follows. If $O\subset\mathbb{R}^{n}_{+}$ is open,
then $C_{p,\kappa}(O):=\sup\\{C_{p,\kappa}(K):K\subset O,\hskip 4.0ptK\hskip
4.0pt\mbox{is compact}\\}$ and if $E\subset\mathbb{R}^{n}_{+}$ is arbitrary,
then $C_{p,\kappa}(E):=\inf\\{C_{p,\kappa}(O):E\subset O,\hskip 4.0ptO\hskip
4.0pt\mbox{is open}\\}$.
(5) $C_{p,\kappa}$ is monotone and $\sigma$-subadditive (cf. [2, Propositions
2.3.4 and 2.3.6]).

######  Proposition 1.

Let $1<p<\infty$. If $\kappa$ is a radially decreasing kernel, then
(1) if $\|\kappa\|_{p^{\prime},a}<\infty$, then $C_{p,\kappa}(\\{y\\})>0$ for
all $y\in\mathrm{int}\mathbb{R}^{n}_{+}$,
(2) if $\int_{\mathbb{R}^{n}_{+}\setminus
B(0,1)}\kappa^{p^{\prime}}d\lambda_{a}=\infty$, then $C_{p,\kappa}(E)=0$ for
all $E\subset\mathbb{R}^{n}_{+}$,
(3) if $\int_{\mathbb{R}^{n}_{+}\setminus
B(0,1)}\kappa^{p^{\prime}}d\lambda_{a}<\infty$, $E$ is measurable and
$C_{p,\kappa}(E)=0$, then $\lambda_{a}(E)=0$.

###### Proof.

(1) Let $\delta_{y}$ be the Dirac measure concentrated at $y$. According to
Remark 2 and (5),

$C_{p,\kappa}^{\frac{1}{p}}(\\{y\\})=\sup\left\\{\frac{\mu_{a}(\\{y\\})}{\|\kappa*_{a}\delta_{y}\|_{p^{\prime},a}}:\mu\in\mathcal{M}(\\{y\\})\right\\}\geq\frac{y^{a}}{\|\kappa*_{a}\delta_{y}\|_{p^{\prime},a}}\geq\frac{1}{\|\kappa\|_{p^{\prime},a}}>0.$

(2) It is enough to show that $C_{p,\kappa}(B_{+}(0,r))=0$ for all $r>0$. Let
$\mu\in\mathcal{M}(B_{+}(0,r))$. In view of (2.1) $T^{t}\kappa(x)\geq
g(|x+t|)$, thus

$\|\kappa*_{a}\mu\|_{p^{\prime},a}\geq\mu(B_{+}(0,r))\left(\int_{\mathbb{R}^{n}_{+}}g(r+|t|)^{p^{\prime}}t^{a}dt\right)^{\frac{1}{p^{\prime}}}$
$\geq c\left(\int_{\mathbb{R}^{n}_{+}\setminus
B(0,2r)}g(|t|)^{p^{\prime}}t^{a}dt\right)^{\frac{1}{p^{\prime}}}=\infty.$

The last inequality is equivalent with the assumption, and according to Remark
2, it proves the statement.
(3) It is enough to show that $\lambda_{a}(E\cap B_{+}(0,r))=0$ for all $r>0$.
Let $F=E\cap B_{+}(0,r)$ and $f\in L^{p+}_{a}$ such that $\kappa*_{a}f(x)\geq
1$ on $F$. Then by Fubini’s theorem

$\lambda_{a}(F)\leq\int_{F}\kappa*_{a}f(x)x^{a}dx=\int_{\mathbb{R}^{n}_{+}}\chi_{F}(x)\kappa*_{a}f(x)x^{a}dx=\int_{\mathbb{R}^{n}_{+}}\kappa*_{a}\chi_{F}(t)f(t)t^{a}dt$
$\leq\|f\|_{p,a}\|\kappa*_{a}\chi_{F}\|_{p^{\prime},a}\leq\|f\|_{p,a}\|\kappa*_{a}\chi_{B_{+}(0,r)}\|_{p^{\prime},a}.$

We estimate the second factor.

$\|\kappa*_{a}\chi_{B_{+}(0,r)}\|_{p^{\prime},a}$
$\leq\left(\int_{B_{+}(0,2r)}(\kappa*_{a}\chi_{B_{+}(0,r)}(t))^{p^{\prime}}t^{a}dt\right)^{\frac{1}{p^{\prime}}}+\left(\int_{\mathbb{R}^{n}_{+}\setminus
B_{+}(0,2r)}(\cdot)\right)^{\frac{1}{p^{\prime}}}=I+II.$

If $|t|>2r$, $\kappa(x)\leq\kappa\left(\frac{t}{2}\right)$ while by Lemma 1
$T^{t}\chi_{B_{+}(0,r)}(x)\leq cr^{|a|}\frac{1}{x^{a}}$ on $|x-t|<r$. Thus by
the assumption we have

$II=\left(\int_{\mathbb{R}^{n}_{+}\setminus
B_{+}(0,2r)}\left(\int_{\mathbb{R}^{n}_{+}}T^{t}\chi_{B_{+}(0,r)}(x)\kappa(x)x^{a}dx\right)^{p^{\prime}}t^{a}dt\right)^{\frac{1}{p^{\prime}}}$
$\leq c\left(\int_{\mathbb{R}^{n}_{+}\setminus
B_{+}(0,2r)}\kappa^{p^{\prime}}\left(\frac{t}{2}\right)t^{a}\right)^{\frac{1}{p^{\prime}}}<c.$

In the first integral $|t|<2r$ and $|x-t|<r$, so the convolution can be
estimated as

$\kappa*_{a}\chi_{B_{+}(0,r)}(t)\leq\int_{B_{(}0,3r)}(g|x|)x^{a}dx\leq c,$

where in spherical coordinates the last inequality is just (8). Thus, $I$ is
also bounded by a constant. Taking infimum over appropriate functions $f$, we
have that $\lambda_{a}(F)\leq cC_{p,\kappa}(F)$, which implies the statement.

######  Remark 3.

Of course, the nonlinear potential with Bessel convolution is
$V_{\kappa,p}^{\mu}=\kappa*_{a}(\kappa*_{a}\mu)^{p^{\prime}-1}$. Subsequently,
we focus on capacity.

###  2.3. Bessel and Riesz kernels

The modified Bessel function of the second kind, $K_{\alpha}$ is defined as
follows.

$i^{-\alpha}J_{\alpha}(ix)=\sum_{k=0}^{\infty}\frac{1}{k!\Gamma(k+\alpha+1)}\left(\frac{x}{2}\right)^{2k+\alpha},$

where $J_{\alpha}$ is the Bessel function, and

$K_{\alpha}(x)=\frac{\pi}{2}\frac{i^{\alpha}J_{-\alpha}(ix)-i^{-\alpha}J_{\alpha}(ix)}{\sin\alpha\pi}.$

Considering $r>0$, around zero

(9) $K_{\alpha}(r)\sim\left\\{\begin{array}[]{ll}-\ln\frac{r}{2}-c,\hskip
4.0pt\mbox{if}\hskip 4.0pt\alpha=0\\\ C(\alpha)r^{-\alpha},\hskip
4.0pt\mbox{if}\hskip 4.0pt\alpha>0,\end{array}\right.$

and around infinity

(10) $K_{\alpha}(r)\sim\frac{c}{\sqrt{r}}e^{-r}.$

The Bessel kernel is

(11)
$G_{a,\nu}(x):=\frac{2^{\frac{n-a-\nu}{2}+1}}{\Gamma\left(\frac{\nu}{2}\right)\prod_{i=1}^{n}\Gamma(\alpha_{i}+1)}\frac{K_{\frac{n+|a|-\nu}{2}}(|x|)}{|x|^{\frac{n+|a|-\nu}{2}}}.$

Below we also need the Riesz kernel:

(12) $I_{\beta}(x)=\frac{c(\beta)}{|x|^{n-\beta}},\hskip 4.0pt\hskip
4.0ptx\in\mathbb{R}^{n}.$

In the last section we use Bessel kernel rather than the Riesz kernel, because
its behavior at infinity allows wider function classes. On the other hand,
around the origin the Riesz kernel, $I_{\nu-|a|}(x)$, behaves similarly to the
Bessel kernel and is simpler, thus it proved to be a useful tool for
computations.

Below we examine B-$p$ capacity, which is defined by generalized convolution
referring to the Bessel kernel: $C_{p,a,\nu}(E):=C_{p,G_{a,\nu}}(E)$. In view
of (9), (10) and (11) $G_{a,\nu}(x)\in L^{1}_{a}$ if and only if $\nu>0$. On
the other hand according to Proposition 1, B-$p$ capacity is non trivial if
and only if $1<p<\frac{n+|a|}{\nu}$ or $1=p=\frac{n+|a|}{\nu}$. Thus
subsequently we investigate $C_{p,a,\nu}$ if $1<p<\infty$, that is

(13) $0<\nu<\frac{n+|a|}{p}.$

##  3\. The Laplace-Bessel operator

B-elliptic equations are investigated by several authors. For instance
fundamental solutions are given, see e.g.[12] and [13]. Harmonic analysis
associated with Bessel operator is examined, see e.g. [16] and mean-value
theorems are proved, see [19]. Here we give a simple application of B-$p$
capacity.

We begin this section by introducing some additional notation. According to
(13), (4) if $g\in L^{p}_{a}$, $G_{a,\nu}*_{a}g\in L^{p}_{a}$, moreover by
[18, Lemma 4.3 (3)]

(14) $\|G_{a,\nu}*_{a}g\|_{p,a}\leq\|g\|_{p,a}.$

Thus we define the next Banach space.

$L^{p}_{a,\nu}:=L^{p}_{a,\nu}(\mathbb{R}^{n}_{+})=\\{f:f=G_{a,\nu}*_{a}g;\hskip
4.0ptg\in L^{p}_{a}\\},\hskip 4.0pt\hskip 4.0pt\hskip
4.0pt\|f\|_{p,a,\nu}:=\|g\|_{p,a}.$

Let

$B_{\alpha,x}:=\frac{\partial^{2}}{\partial
x^{2}}+\frac{2\alpha+1}{x}\frac{\partial}{\partial x}$

be The Bessel operator. The Laplace-Bessel operator is defined as

$\Delta_{a}=\sum_{i=1}^{n}B_{\alpha_{i},x_{i}}.$

With this notation we define the Sobolev space $W^{m}_{p,a}$ with
$m\in\mathbb{N}$ as it follows.

$W^{m}_{p,a}:=\\{f\in L^{p}_{a}:\Delta_{a}^{k}f\in L^{p}_{a},\hskip
4.0ptk=1,\dots,m\\},\hskip 4.0pt\hskip 4.0pt\hskip
4.0pt\|f\|_{W^{m}_{p,a}}=\sum_{k=0}^{m}\|\Delta_{a}^{k}f\|_{p,a}.$

Notation. We need the ”even” functions from the Schwartz class in
$\mathbb{R}^{n}_{+}$.

$\mathcal{S}_{e}:=\left\\{f\in
C^{\infty}(\mathbb{R}^{n}_{+}):\left.\frac{\partial^{2k+1}f}{\partial
x_{i}^{2k+1}}\right|_{x_{i}=0}=0,\hskip 4.0ptk\in\mathbb{N};\right.$
$\left.\sup_{x\in\mathbb{R}^{n}_{+}}\left|x^{\alpha}D^{\beta}f(x)\right|<\infty\hskip
4.0pt\forall\alpha,\beta\in\mathbb{N}^{n}\right\\}.$

The next lemma describes the relation of Bessel potential and Sobolev spaces
above.

######  Lemma 2.

Let $m$ be a positive integer. Then

$W^{m}_{p,a}=L^{p}_{a,2m}.$

###### Proof.

Let us notice first that if we define
$\|f\|_{\tilde{W}^{m}_{p,a}}=\sum_{k=0}^{m}\|(I-\Delta_{a})^{k}f\|_{p,a}$,
then $\|f\|_{\tilde{W}^{m}_{p,a}}\sim\|f\|_{W^{m}_{p,a}}$. According to [18,
Lemma 4.3] if $g\in L^{p}_{a}$, $1\leq p\leq\infty$ and $k\in\mathbb{N}$,

(15) $(I-\Delta_{a})^{k}(G_{a,\nu}*_{a}g)=G_{a,\nu-2k}*_{a}g,\hskip
4.0pt\hskip 4.0pt\mbox{and}\hskip 4.0ptG_{a,0}*_{a}g=g.$

Comparing this with (14) we have if $f\in L^{p}_{a,2m}$ and $k\leq 2m$, then

$\|(I-\Delta_{a})^{k}f\|_{p,a}\leq\|G_{a,2m-2k}*_{a}g\|_{p,a}\leq\|g\|_{p,a}=\|f\|_{p,a,2m}.$

On the other hand taking into consideration that $S_{e}$ is a dense subset in
$W^{m}_{p,a}$, let $f\in S_{e}$. According to [18, Theorem 4.5] for $1\leq
p\leq\infty$, $k\in\mathbb{N}$

(16) $G_{a,\nu+2k}*_{a}(I-\Delta_{a})^{k}f=G_{a,\nu}*_{a}f.$

Thus by (15) and (16)

$f=G_{a,0}*_{a}f=G_{a,2m}*_{a}(I-\Delta_{a})^{m}f=G_{a,2m}*_{a}g,$

where $g=(I-\Delta_{a})^{m}f\in L^{p}_{a}$. So $f\in L^{p}_{a,2m}$, and

$\|f\|_{p,a,2m}\leq\|f\|_{W^{m}_{p,a}}.$

######  Definition 3.

Let $K\subset\mathbb{R}^{n}_{+}$ compact. $\mathcal{S}$ is the Schwartz class
restricted to $\mathbb{R}^{n}_{+}$.

$N_{p,a,\nu}(K):=\inf\\{\|f\|_{p,a,\nu}^{p}:f=G_{a,\nu}*_{a}g\in\mathcal{S},\hskip
4.0ptf\equiv 1\hskip 4.0pt\mbox{in a neigborhood of}\hskip 4.0ptK\\}.$

######  Remark 4.

(1) If $p=2$ with standard convolution, $N$ is the spectral measure defined by
Deny, see [6].
(2)Of course, $N_{p,a,\nu}$ can be extended as above and $C_{p,a,\nu}(E)\leq
N_{p,a,\nu}(E)$.

Notation. Let us introduce the inner product for measurable functions

$\langle f,g\rangle_{a}:=\int_{\mathbb{R}^{n}_{+}}fgd\lambda_{a}.$

We denote in the same way the effect of a distribution.

######  Theorem 1.

Let $1<p<\infty$, $K\subset\mathrm{int}\mathbb{R}^{n}_{+}$ compact, and
$L=\sum_{k=0}^{m}a_{k}(I-\Delta_{a})^{k}$, $a_{k}\in\mathbb{R}$, $k=1,\dots,m$
be defined in a bounded open neighborhood $O$ of $K$ such that
$\overline{O}\subset\mathrm{int}\mathbb{R}^{n}_{+}$ . Let $u\in
L^{p}_{a}(O\setminus K)$ is a solution to $Lu=0$ in $O\setminus K$. If
$N_{p^{\prime},a,2m}(K)=0$, $u$ can be extended to $\tilde{u}\in L^{p}_{a}(O)$
such that $L\tilde{u}=0$ in $O$ in weak sense.

######  Remark 5.

(1) Let $1<p<\infty$, $\nu>0$. If $N_{p,a,\nu}(K)=0$, then $\lambda(K)=0$.
Indeed, let $\varepsilon>0$ be arbitrary and $f=G_{a,\nu}*_{a}g$ such that
$\|g\|_{p^{\prime},a}\leq\varepsilon$, $f\equiv 1$ on $K$. By (14)

$\varepsilon^{p}\geq\|G_{a,\nu}*_{a}g\|_{p^{\prime},a}^{p}\geq\int_{K}|G_{a,\nu}*_{a}g|^{p}d\mu_{a}=\mu_{a}(K).$

Thus $\lambda_{a}(K)=0$ and so $\lambda(K)=0$.

(2) If $f\in C_{0}^{\infty}(\mathrm{int}\mathbb{R}^{n}_{+})$ and $g\in
C_{0}^{\infty}(\mathrm{int}\mathbb{R}^{n}_{+})^{\prime}$, that is in the dual
space, then

(17) $\langle Lf,g\rangle_{a}=\langle f,Lg\rangle_{a}.$

In one dimension for $L=B_{\alpha}$ it is [16, (2.4)]. Similarly to this case
if $f\in C_{0}^{\infty}(\mathrm{int}\mathbb{R}^{n}_{+})$ and $g$ is smooth
enough, then integration by parts implies the result. For general elements of
the dual space we extend $Lg$ by the formula above, that is $Lgf:=\langle
Lf,g\rangle_{a}$.

###### Proof.

(of Theorem 1) Let $O$ be a bounded open neighborhood of $K$ in
$\mathbb{R}^{n}_{+}$ and $u\in L^{p}_{a}(O\setminus K)$ for which $Lu=0$ in
$O\setminus K$. Since $N_{p^{\prime},a,2m}(K)=0$, for all $\varepsilon>0$
there is a $\varphi=G_{a,2m}*_{a}f\in\mathcal{S}$ such that $\varphi\equiv 1$
on a neighborhhood $U\subset O$ of $K$ and
$\|f\|_{p^{\prime},a}\leq\varepsilon$. Let $g\in C_{0}^{\infty}(O)$. Then
$(1-\varphi)g\in C_{0}^{\infty}(O\setminus K)$. In view of (1) of Remark 5 $u$
is a.e. defined, so it can be handled as a distribution. Since
$(1-\varphi)g\in C_{0}^{\infty}(O)$, by the assumption

$\langle u,L(1-\varphi)g\rangle_{a}=0.$

This implies that

$\left|\langle u,Lg\rangle_{a}\right|=\left|\langle u,L\varphi
g\rangle_{a}\right|\leq\|u\|_{p,a,O}\|L\varphi g\|_{p^{\prime},a}\leq
c\|u\|_{p,a,O}\|\varphi g\|_{W^{m}_{p^{\prime},a}},$

where $c=c(L)$ depends on the coefficients of $L$.
Applying Lemma 2 and the inversion of Bessel potential (for formulae see e.g.
[7, Theorem 1]) we have

$\left|\langle u,Lg\rangle_{a}\right|\leq c\|u\|_{p,a,O}\|\varphi
g\|_{p^{\prime},a,\nu}\leq c\|\varphi g\|_{p^{\prime},a}\leq
c\|g\|_{\infty}\|f\|_{p^{\prime},a},$

where $c$ depends on $u$ and $L$. Since $\varepsilon$ was arbitrary, for all
$g\in C_{0}^{\infty}(O)$ $\langle u,Lg\rangle_{a}=0$, so in view of (17) $u$
is a weak solution on $O$.

The fundamental solution for the Laplace-Bessel operator, that is

$\Delta_{a}E=\delta_{a},\hskip 4.0pt\hskip 4.0pt\mbox{where}\hskip
4.0pt\langle\delta_{a},\varphi\rangle_{a}=\varphi(0),\hskip 4.0pt\hskip
4.0pt\varphi\in\mathcal{S}_{e},$

is

(18) $E(x)=\left\\{\begin{array}[]{ll}c(n,a)\ln x,n+|a|=2\\\
c(n,a)|x|^{2-n-|a|},n+|a|>2,\end{array}\right.$

see e.g. [20, Theorem 93 (page 324)].

######  Corollary 1.

With the notation above, let $L=\Delta_{a}$, that is $\Delta_{a}u=0$ on
$O\setminus K$, and $u\in L^{p}_{a}(O\setminus K)$. Let
$2\leq\nu<\frac{n+|a|}{p^{\prime}}$. Then $u$ can be extended to $\tilde{u}\in
L^{p}_{a}(O)$ such that $L\tilde{u}=0$ in $O$ in weak sense if and only if
$C_{p^{\prime},a,\nu}(K)=0$.

###### Proof.

If $C_{p^{\prime},a,\nu}(K)>0$, there is a nonzero measure
$\mu\in\mathcal{M}(K)$ such that $G_{a.\nu}*_{a}\mu\in
L^{p}_{a}(\mathbb{R}^{n}_{+})$. In wiev of (18), $E*_{a}\mu\in
L^{p}_{a,loc}(\mathbb{R}^{n}_{+})$. Since $E*_{a}\mu$ is a solution to
$\Delta_{a}u=0$ in $O\setminus K$, $K$ is not removable.

On the other hand we prove that if $C_{p^{\prime},a,\nu}(K)=0$, then
$N_{p^{\prime},a,2m}(K)=0$ too. According to Definition 2 for an
$\varepsilon>0$ there is a nonnegative function $f\in
L^{p}_{a}(\mathbb{R}^{n}_{+})$ such that $G_{a,\nu}*_{a}f\geq 1$ on a
neighborhood of $K$ and $\|f\|_{p,a}^{p}\leq\varepsilon$.
Define a function $h\in C^{\infty}(\mathbb{R}_{+})$, $0\leq h\leq 1$ and
$h(t)=0$ on $t\in\left[0,\frac{1}{2}\right]$; $h(t)=1$ if $t\geq 1$. Taking
into consideration that $G_{a,\nu}*_{a}f\geq 0$,
$\varepsilon\geq\int_{\mathbb{R}_{+}}(G_{a,\nu}*_{a}f)^{p}x^{a}dx\geq\int_{G_{a,\nu}*_{a}f\geq\frac{1}{2}}\frac{1}{2^{p}}x^{a}dx$.
Thus
$\int_{\mathbb{R}_{+}}(h(G_{a,\nu}*_{a}f))^{p}x^{a}dx\leq\int_{G_{a,\nu}*_{a}f\geq\frac{1}{2}}x^{a}dx\leq
2^{p}\varepsilon$.
Noticing that $h(G_{a,\nu}*_{a}f)$ fulfils the requirements of Definition 3,
since $\varepsilon$ is arbitrary, the statement is proved.

##  4\. Maximal measure and a Wolff type inequality

Bessel maximal function was introduced and examined e.g. in cf. e.g. [8], see
also the references therein. The boundedness of the maximal operator in some
Morrey spaces is studied and applied to prove a Hardy-Littlewood-Sobolev type
theorem in [9]. The maximal measure presented below has proved useful in
formulating a Wolff type inequality which is the main tool of the next
section. Wolff type inequalities can be applied in different situations, for
instance in martingale theory, see [3] or deducing trace inequalities or
characterize the trace measures via Wolff’s inequality, see e.g. [24], [25],
[23] and the references therein.

Below we define the maximal measure with respect to Bessel convolution.

######  Definition 4.

$M_{a}\mu(x):=\sup_{r>0}\frac{1}{\lambda_{a}(B_{+}(0,r))}\chi_{B_{+}(0,r)}*_{a}\mu(x).$

Since

$\lambda_{a}(B_{+}(0,r))=cr^{n+|a|},$

we define the fractional maximal measure as

(19)
$M_{a,d}\mu(x):=\sup_{r>0}\frac{\chi_{B_{+}(0,r)}*_{a}\mu(x)}{r^{n+|a|-d}},$

and the truncated one as

(20) $M_{a,d,b}\mu(x):=\sup_{0<r\leq
b}\frac{\chi_{B_{+}(0,r)}*_{a}\mu(x)}{r^{n+|a|-d}}.$

######  Lemma 3.

$I_{\beta}*_{a}\mu(x)=c\int_{0}^{\infty}\frac{\chi_{B_{+}(0,r)}*_{a}\mu(x)}{r^{n-\beta}}\frac{dr}{r}.$
$I_{\beta}\chi_{B_{+}(0,\delta)}*_{a}\mu(x)=c\int_{0}^{\delta}\frac{\chi_{B_{+}(0,r)}*_{a}\mu(x)}{r^{n-\beta}}\frac{dr}{r}+c\frac{\chi_{B_{+}(0,\delta)}*_{a}\mu(x)}{\delta^{n-\beta}}.$

###### Proof.

Let $d\Theta_{x,a}(z)=T^{x}\mu(z)z^{a}dz$. Changing the order of integration
we get

$\int_{0}^{\delta}\frac{\chi_{B_{+}(0,r)}*_{a}\mu(x)}{r^{n-\beta}}\frac{dr}{r}=\int_{0}^{\delta}\frac{1}{r^{n-\beta+1}}\int_{B_{+}(0,r)}1d\Theta_{x,a}(z)dr$
$=\int_{B_{+}(0,\delta)}\int_{|z|}^{\delta}\frac{1}{r^{n-\beta+1}}drd\Theta_{x,a}(z)=c\int_{B_{+}(0,\delta)}I_{\beta}(z)d\Theta_{x,a}(z)-c\int_{B_{+}(0,\delta)}\frac{1}{\delta^{n-\beta}}d\Theta_{x,a}(z)$
$=cI_{\beta}\chi_{B_{+}(0,\delta)}*_{a}\mu(x)-c\frac{\chi_{B_{+}(0,\delta)}*_{a}\mu(x)}{\delta^{n-\beta}}.$

Notation.

$I_{\beta}^{\delta}*_{a}\mu(x):=c\int_{0}^{\delta}\frac{\chi_{B_{+}(0,r)}*_{a}\mu(x)}{r^{n-\beta}}\frac{dr}{r}.$

######  Theorem 2.

Let $1\leq p<\infty$, $0<\nu<n+|a|$, $\delta>0$ arbitrary. Then for all
positive measure $\mu$, there are constants $c=c(n,a,\delta)$ (are not the
same at each occurrence) such that

(21) $\|I_{\nu-|a|}*_{a}\mu\|_{p,a}\leq c\|M_{a,\nu}\mu\|_{p,a}$

and

(22) $\|I_{\nu-|a|}^{\delta}*_{a}\mu\|_{p,a}\leq
c\|M_{a,\nu,\delta}\mu\|_{p,a}.$

Notation.

$H_{s}^{\mu}:=\\{x:I_{\nu-|a|}*_{a}\mu(x)>s\\},\hskip 4.0pt\hskip
4.0ptK_{s}^{\mu}:=\\{x:M_{a,\nu}\mu(x)>s\\}.$
${}^{1}H_{s}^{\mu}:=\\{x:I_{\nu-|a|}^{1}*_{a}\mu(x)>s\\},\hskip 4.0pt\hskip
4.0pt^{1}K_{s}^{\mu}:=\\{x:M_{a,\nu,1}\mu(x)>s\\}.$

######  Lemma 4.

There is a $\varrho>1$ and a $b>0$, such that for all $s>0$ and
$\varepsilon\in(0,1]$,

(23) $\lambda_{a}(H_{\varrho s}^{\mu})\leq
b\varepsilon^{\frac{n+|a|}{n+|a|-\nu}}\lambda_{a}(H_{s}^{\mu})+\lambda_{a}(K_{\varepsilon
s}^{\mu}).$

Similarly

(24) $\lambda_{a}(^{1}H_{\varrho s}^{\mu})\leq
b\varepsilon^{\frac{n+|a|}{n+|a|-\nu}}\lambda_{a}(^{1}H_{s}^{\mu})+\lambda_{a}(^{1}K_{\varepsilon
s}^{\mu}).$

###### Proof.

By lower semicontinuity we can take Whithney’s decomposition of $H_{s}^{\mu}$,
i.e. $H_{s}^{\mu}=\cup_{i=1}^{\infty}Q_{i}$, where $Q_{i}$-s are dyadic cubes,
$\mathrm{int}Q_{i}\cap\mathrm{int}Q_{j}=\emptyset$ if $i\neq j$ and
$\mathrm{diam}Q_{i}\leq\mathrm{dist}(Q_{i},(H_{s}^{\mu})^{c})\leq
4\mathrm{diam}Q_{i}$. (Dyadic cubes means cubes with side $2^{-k}$,
$k\in\mathbb{Z}$, whose vertices belong to the lattice
$\\{m2^{-k}:m\in\mathbb{Z}^{n}\\}$. For Whithney’s decomposition see [22, page
16, Theorem 3].) In addition, to prove (24) if
$\mathrm{diam}Q_{i}\geq\frac{1}{8}$, then we decompose it to subcubes with
diameter is between $\frac{1}{16}$ and $\frac{1}{8}$, and we consider this new
sequence of cubes.

Let $Q$ be an element of this decomposition. Let $x\in Q$ be arbitrary, denote
the center of $Q$ by $x_{c}$ and let $d:=\mathrm{diam}Q$. Let
$G:=B(x_{c},6d)$, $B=B(x,8d)$, that is $Q\subset G\subset B$. Let
$\mu=\mu_{1}+\mu_{2}$, where $\mu_{1}=\mu|_{G}$.

At first we deal with $I_{\nu-|a|}*_{a}\mu_{2}$. To this we estimate
$T^{t}\chi_{B_{+}(0,r)}(x)$. We can assume, that $r>\frac{11}{2}$, otherwise
$\mathrm{supp}\mu_{2}\cap B_{+}(x,r)=\emptyset$. Let
$x_{1}\in(H_{s}^{\mu})^{c}$ such that $\mathrm{dist}(x_{1},Q)<4d$. Then by (6)
if $t\in B_{+}(x,r)$,

$T^{t}\chi_{B_{+}(0,r)}(x)\leq
c\prod_{i=1}^{n}\min\left\\{1,\left(\frac{r}{x_{i}}\right)^{a_{i}}\right\\}\leq
c\prod_{i=1}^{n}\min\left\\{1,\left(\frac{r}{x_{1,i}}\right)^{a_{i}}\right\\}=:P.$

Indeed, it is obvious if $x_{1,i}\leq x_{i}$. If $x_{1,i},x_{i}<r$, then the
minimum is 1 in both cases. If $x_{i}<r\leq x_{1,i}$, then
$\min\left\\{1,\frac{r}{x_{i}}\right\\}=1$ thus we have to show that $1\leq
c\frac{r}{x_{1,i}}$. Since $x_{1,i}<x_{i}+5d$,
$\frac{r}{x_{1,i}}>\frac{r}{x_{i}+5d}>\frac{r}{r+5d}\geq\frac{\frac{11}{2}d}{\frac{11}{2}d+5d}=\frac{11}{21}$.
If $\frac{11}{2}d<r<x_{i}<x_{1,i}$, then
$\frac{r}{x_{1,i}}>\frac{r}{x_{i}+\frac{20}{11}r}>\frac{11}{21}\frac{r}{x_{i}}$.

In view of (7) if $t\in T_{+}(x_{1},2r)$

$P\leq cT^{t}\chi_{[0,4r)^{n}}(x_{1}).$

Thus

$\chi_{B_{+}(x,r)}*_{a}\mu_{2}\leq c\chi_{[0,4r)^{n}}*_{a}\mu_{2}(x_{1})\leq
c\chi_{B_{+}(0,4\sqrt{n}r)}*_{a}\mu_{2}(x_{1}).$

Recalling that $x_{1}\in(H_{s}^{\mu})^{c}$, we have

$I_{\nu-|a|}*_{a}\mu_{2}(x)\leq
c\int_{\frac{11}{2}d}^{\infty}\frac{\chi_{B_{+}(0,4\sqrt{n}r)}*_{a}\mu_{2}(x_{1})}{(4\sqrt{n}r)^{n+|a|-\nu}}dr=cI_{\nu-|a|}*_{a}\mu_{2}(x_{1})\leq
cs.$

Now we choose $\varrho$ so that $I_{\nu-|a|}*_{a}\mu_{2}(x)\leq\frac{\varrho
s}{2}$, which implies that

(25) $H_{\varrho s}^{\mu}\cap Q\subset H_{\frac{\varrho s}{2}}^{\mu_{1}}\cap
Q.$

If the diameter of $Q$ was originally less then $\frac{1}{8}$, then the whole
construction is contained in a ball of radius less than one, so the same chain
of ideas leads to

(26) $^{1}H_{\varrho s}^{\mu}\cap Q\subset\hskip 4.0pt^{1}H_{\frac{\varrho
s}{2}}^{\mu_{1}}\cap Q.$

If there is no $x_{1}\in(^{1}H_{s}^{\mu})^{c}$ such that
$\mathrm{dist}(Q,x_{1})\leq 4d$, then $\mathrm{diam}Q>\frac{1}{16}$. Let
$x_{0}\in Q\cap(^{1}K_{\varepsilon s})^{c}$. Then, recalling that
$r>\frac{11}{2}d$, we have

$I_{\nu-|a|}^{1}*_{a}\mu_{2}(x_{0})=c\int_{\frac{11}{32}}^{1}\frac{\chi_{B_{+}(0,r)}*_{a}\mu_{2}(x_{0})}{r^{n+|a|-\nu+1}}dr$
$\leq\frac{32}{11}\int_{0}^{1}\frac{\chi_{B_{+}(0,r)}*_{a}\mu_{2}(x_{0})}{r^{n+|a|-\nu}}dr\leq
cM_{a,\nu,1}(x_{0})\leq c\varepsilon s.$

Thus if $Q\cap(^{1}K_{\varepsilon s})^{c}\neq\emptyset$, we can choose
$\varrho$ again so that (26) is satisfied.

Let $x_{0}\in Q\cap(K_{\varepsilon s})^{c}$ again. According to [8, Theorem 2
(c)]

$\lambda_{a}\left(Q\cap H_{\frac{\varrho s}{2}}^{\mu_{1}}\right)\leq
c\left(\frac{1}{\varrho
s}\int_{\mathbb{R}^{n}_{+}}1d\mu_{1,a}(t)\right)^{\frac{n+|a|}{n+|a|-\nu}}=c\left(\frac{1}{\varrho
s}\int_{G}1d\mu_{a}(t)\right)^{\frac{n+|a|}{n+|a|-\nu}}$ $\leq
c\left(\frac{1}{\varrho
s}\int_{B}1d\mu_{a}(t)\right)^{\frac{n+|a|}{n+|a|-\nu}}=(*).$

In view of Lemma 1 $B=\mathrm{supp}T^{t}\chi_{B_{+}(0,8d)(x_{0})}\subset
T_{+}(x_{0},8d)$. Thus by (7)

$(*)\leq c\left(\frac{1}{\varrho
s}\prod_{i=1}^{n}\max\left\\{1,\left(\frac{x_{0,i}}{16d}\right)^{a_{i}}\right\\}\int_{\mathbb{R}^{n}_{+}}T^{t}\chi_{[0,16d)^{n}}(x_{0})d\mu_{a}(t)\right)^{\frac{n+|a|}{n+|a|-\nu}}$
(27) $\leq c\left(\frac{M_{a,\nu}\mu(x_{0})}{\varrho
s}\right)^{\frac{n+|a|}{n+|a|-\nu}}(\prod_{i=1}^{n}\max\\{1,\left(\frac{x_{0,i}}{16d}\right)^{a_{i}}\\}))^{\frac{n+|a|}{n+|a|-\nu}}d^{n+|a|}.$

Taking into consideration that
$\lambda_{a}(Q)\sim\prod_{i=1}^{n}x_{c,i}^{a_{i}}\left(\frac{d}{\sqrt{n}}\right)^{n}\sim\prod_{i=1}^{n}\left(\frac{x_{c,i}}{d}\right)^{a_{i}}d^{n+|a|}$
and recalling that $|x_{0,i}-x_{c,i}|<d$, we have

$\left(\prod_{i=1}^{n}\max\\{1,\left(\frac{x_{0,i}}{16d}\right)^{a_{i}}\\}\right)^{\frac{n+|a|}{n+|a|-\nu}}d^{n+|a|}\leq
c|Q|_{a}\left(\prod_{i=1}^{n}\left(\frac{x_{c,i}}{d}\right)^{a_{i}}\right)^{\frac{\nu}{n+|a|-\nu}}\leq
c\lambda_{a}(Q).$

Finally as $x_{0}\in(K_{\varepsilon s})^{c}$,

(28) $\lambda_{a}\left(Q\cap H_{\frac{\varrho s}{2}}^{\mu_{1}}\right)\leq
b\varepsilon^{\frac{n+|a|}{n+|a|-\nu}}\lambda_{a}(Q).$

Since $\mathrm{diam}Q\leq\frac{1}{8}$, similarly we have

(29) $\lambda_{a}\left(Q\cap\hskip 4.0pt^{1}H_{\frac{\varrho
s}{2}}^{\mu_{1}}\right)\leq
b\varepsilon^{\frac{n+|a|}{n+|a|-\nu}}\lambda_{a}(Q),$

cf. (27).

That is if $Q\cap(K_{\varepsilon s})^{c}\neq\emptyset$ or
$Q\cap(^{1}K_{\varepsilon s})^{c}\neq\emptyset$, then (28) or (29),
respectively, is fulfilled, otherwise $Q\subset K_{\varepsilon s}\mu$.
Recalling (25) or (26) and adding over all $Q\in\\{Q_{i}\\}$, we obtain the
required result.

###### Proof.

(of Theorem 2) (21) and (22) can be derived by the same way from (23) and
(24). Let us see the second one, say. Let $\delta=1$. Integrating (24) and
changing the variables we have

$\frac{1}{\varrho^{p}}\int_{0}^{\varrho
R}\lambda_{a}(^{1}H_{u}^{\mu})u^{p-1}du\leq
b\varepsilon^{\frac{n+|a|}{n+|a|-\nu}}\int_{0}^{R}\lambda_{a}(^{1}H_{s}^{\mu})u^{p-1}du+\frac{1}{\varepsilon^{p}}\int_{0}^{\varepsilon
R}\lambda_{a}(^{1}K_{u}^{\mu})u^{p-1}du.$

Supposing that $\mathrm{supp}\mu$ is bounded, all the integrals are finite. We
choose $\varepsilon$ small enough so that
$b\varepsilon^{\frac{n+|a|}{n+|a|-\nu}}\leq\frac{1}{2\varrho^{p}}$. Then we
have

$\frac{1}{\varrho^{p}}\int_{0}^{\varrho
R}\lambda_{a}(^{1}H_{u}^{\mu})u^{p-1}du\leq\frac{2}{\varepsilon^{p}}\int_{0}^{\varepsilon
R}\lambda_{a}(^{1}K_{u}^{\mu})u^{p-1}du.$

Letting $R\to\infty$,

$\|I_{\nu-|a|}^{1}*_{a}\mu(x)\|_{a,p}\leq\frac{\varrho}{\varepsilon}\|M_{a,\nu,1}\mu(x)\|_{a,p},$

cf. e.g. [20, (1.46)]. If $\mathrm{supp}\mu$ is not compact, then let
$\mu_{n}:=\mu|_{B_{+}(0,n)}$. Since
$\|M_{a,\nu,1}\mu_{n}\|_{a,p}\leq\|M_{a,\nu,1}\mu\|_{a,p}$, we can apply the
monotone convergence theorem.

######  Corollary 2.

With the assumptions of Theorem 2 we have

(30) $\|I_{\nu-|a|}^{\delta}*_{a}\mu\|_{p,a}\leq
c\|G_{a,\nu}*_{a}\mu\|_{p,a}\leq c\|M_{a,\nu,\delta}\mu\|_{p,a}.$

###### Proof.

(9) and (10) ensure the first inequality and

$\|G_{a,\nu}*_{a}\mu\|_{p,a}\leq\|I_{\nu-|a|}\chi_{B_{+}(0,\delta)}*_{a}\mu\|_{p,a}+\|e^{-\frac{|\cdot|}{2}}*_{a}\mu\|_{p,a}$
$\leq\|I_{\nu-|a|}^{\delta}*_{a}\mu\|_{p,a}+\|M_{a,\nu,\delta}\mu\|_{p,a}+\|e^{-\frac{|\cdot|}{2}}*_{a}\mu\|_{p,a}.$

Observe that

$\frac{e^{-\frac{|\cdot|}{2}}*_{a}\chi_{B_{+}(0,r)}(x)}{r^{n+|a|}}$
$=\frac{1}{r^{n+|a|}}\int_{B_{+}(0,r)}\int_{[0,\pi)^{n}}e^{-\frac{1}{2}\sqrt{\sum_{i=1}^{n}x_{i}^{2}+t_{i}^{2}-2x_{i}t_{i}\cos\vartheta_{i}}}d\sigma^{a}(\vartheta)d\lambda_{a}(t)$
$\geq\frac{1}{r^{n+|a|}}\int_{B_{+}(0,r)}e^{-\frac{1}{2}|x+t|}d\lambda_{a}(t)\geq
ce^{-\frac{r}{2}}e^{-\frac{|x|}{2}}.$

Thus

$e^{-\frac{|x|}{2}}\leq
ce^{\frac{r}{2}}\frac{e^{-\frac{|\cdot|}{2}}*_{a}\chi_{B_{+}(0,r)}(x)}{r^{n+|a|}},$

and so if $r\leq\delta$,

$e^{-\frac{|\cdot|}{2}}*_{a}\mu(x)\leq
c(r)e^{-\frac{|\cdot|}{2}}*_{a}\frac{\chi_{B_{+}(0,r)}(x)}{r^{n+|a|}}*_{a}\mu\leq
c(r,\nu)e^{-\frac{|\cdot|}{2}}*_{a}M_{a,\nu,\delta}\mu.$

According to (5)

$\|e^{-\frac{|\cdot|}{2}}*_{a}\mu\|_{p,a}\leq\|e^{-\frac{|\cdot|}{2}}\|_{1,a}\|M_{a,\nu,\delta}\mu\|_{p,a},$

which, together with Theorem 2, implies the statement.

Notation. Denote by

$b_{k}(x):=2^{k(n+|a|-\nu)}\chi_{B_{+}(0,2^{-k})}*_{a}\mu(x)$

and by

$c_{k}(x):=2^{k(n+|a|-p\nu)}\chi_{B_{+}(0,2^{-k})}*_{a}\mu(x).$

The corresponding Wolff-function is

$W_{a,\nu,p}^{\mu}(x):=\int_{0}^{1}\left(\frac{\chi_{B_{+}(0,r)}*_{a}\mu(x)}{r^{n+|a|-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}.$

######  Remark 6.

In view of Lemma 3 and (20) we can observe that

(31)
$I_{\nu-|a|}*_{a}\mu(x)\sim\|\\{b_{k}(x)\\}_{-\infty}^{\infty}\|_{l^{1}},\hskip
4.0pt\hskip
4.0ptI_{\nu-|a|}^{1}*_{a}\mu(x)\sim\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{1}},$

and

(32) $M_{a,\nu,1}\mu(x)\sim\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{\infty}}.$

We are in position to prove the next Wolff type inequality.

######  Theorem 3.

$1<p,q<\infty$

$\|G_{a,\nu}*_{a}\mu\|_{p^{\prime},a}\sim\|\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{q}}\|_{p^{\prime},a}$
$\sim\left(\int_{\mathbb{R}^{n}_{+}}\|\\{c_{k}(x)\\}_{0}^{\infty}\|_{l^{p^{\prime}-1}}^{p^{\prime}-1}d\mu_{a}(x)\right)^{\frac{1}{p^{\prime}}}\sim\left(\int_{\mathbb{R}^{n}_{+}}W_{a,\nu,p}^{\mu}(x)d\mu_{a}(x)\right)^{\frac{1}{p^{\prime}}}.$

###### Proof.

Corollary 2, (31) and (32) ensure that

$\|G_{a,\nu}*_{a}\mu\|_{p,a}\leq c\|M_{a,\nu,1}\mu\|_{p,a}\leq
c\|\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{\infty}}\|_{p,a}$
$\leq\|\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{p}}\|_{p,a}\leq\|\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{1}}\|_{p,a}\leq
c\|I_{\nu-|a|}^{2}*_{a}\mu\|_{p,a}\leq c\|G_{a,\nu}*_{a}\mu\|_{p,a}.$

To prove the Wolff type inequality we have

$\|\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{p^{\prime}}}\|_{p^{\prime},a}^{p^{\prime}}=\int_{\mathbb{R}^{n}_{+}}\sum_{k=0}^{\infty}b_{k}(t)^{p^{\prime}}d\lambda_{a}(t)$
$=\sum_{k=0}^{\infty}2^{k(n+|a|-\nu)p^{\prime}}\int_{\mathbb{R}^{n}_{+}}\chi_{B_{+}(0,2^{-k})}*_{a}\mu(t)(\chi_{B_{+}(0,2^{-k})}*_{a}\mu)^{p^{\prime}-1}(t)d\lambda_{a}(t)$
$=\sum_{k=0}^{\infty}2^{k(n+|a|-\nu)p^{\prime}}I_{k}.$
$I_{k}=\int_{\mathbb{R}^{n}_{+}}\int_{\mathbb{R}^{n}_{+}}T^{x}\chi_{B_{+}(0,2^{-k})}(t)d\mu_{a}(x)(\chi_{B_{+}(0,2^{-k})}*_{a}\mu)^{p^{\prime}-1}(t)d\lambda_{a}(t)$
$=\int_{\mathbb{R}^{n}_{+}}\int_{\mathbb{R}^{n}_{+}}\int_{\mathbb{R}^{n}_{+}}K(x,t,z)\chi_{B_{+}(0,2^{-k})}(z))d\lambda_{a}(z)d\mu_{a}(x)(\chi_{B_{+}(0,2^{-k})}*_{a}\mu)^{p^{\prime}-1}(t)d\lambda_{a}(t)$
$=\int_{\mathbb{R}^{n}_{+}}\int_{B_{+}(0,\frac{1}{2^{k}})}\int_{\mathbb{R}^{n}_{+}}K(x,t,z)(\chi_{B_{+}(0,2^{-k})}*_{a}\mu)^{p^{\prime}-1}(t)d\lambda_{a}(t)d\lambda_{a}(z))d\mu_{a}(x).$

Since $0<z_{i}<2^{-k}$, recalling that $t_{i}\in(|x_{i}-z_{i}|,x_{i}+z_{i})$,
$(\chi_{B_{+}(0,2^{-k})}*_{a}\mu)(t)\sim(\chi_{B_{+}(0,2^{-k})}*_{a}\mu)(x)$.
As $K$ is a reproducing kernel

$I_{k}\sim\lambda_{a}(B_{+}(0,2^{-k}))\int_{\mathbb{R}^{n}_{+}}(\chi_{B_{+}(0,2^{-k})}*_{a}\mu)^{p^{\prime}-1}(x)d\mu_{a}(x).$

Thus recalling that $\lambda_{a}(B_{+}(0,2^{-k}))\sim 2^{-k(n+|a|)}$ by
Fubini’s theorem again

$\|\|\\{b_{k}(x)\\}_{0}^{\infty}\|_{l^{q}}\|_{p^{\prime},a}^{p^{\prime}}\sim\int_{\mathbb{R}^{n}_{+}}\sum_{k=0}^{\infty}c_{k}(x)^{p^{\prime}-1}d\mu_{a}(x).$

######  Remark 7.

In view of Corollary 2, instead of $M_{a,\nu,1}\mu$ we can consider
$M_{a,\nu,\delta}\mu$ with the corresponding sequence
$\\{b_{k}^{\delta}(x)\\}$.

##  5\. Metric properties

Applying the previous section, below we investigate some ”metric” properties
of B-$p$ capacity. Since the Bessel-translation is not a geometric congruence,
we need a special ”Lipschitz”- condition. It is also necessary to introduce
the notion of ”weighted Hausdorff measure”, to examine Cantor-type sets.

At the beginning of this section let us recall that for $1<p<\infty$ the B-$p$
capacity of $K\subset\mathbb{R}^{n}_{+}$ is

(33)
$C_{a,\nu,p}^{\frac{1}{p}}(K):=\sup\\{\mu_{a}(K):\mu\in\mathcal{M}(K),\hskip
4.0pt\|G_{a,\nu}*_{a}\mu\|_{p,a}\leq 1\\},$

and the B-$p$ capacity is non trivial if $1<p<\frac{n+|a|}{\nu}$.

###  5.1. B-Lipschitz mappings

The next Lipschitz type property is corresponding to the Bessel translation.

######  Definition 5.

Let $\Phi:\mathbb{R}^{n}_{+}\to\mathbb{R}^{n}_{+}$. Consider
$z(x,t,\vartheta)=z_{1}(x,t,\vartheta),\dots,z_{n}(x,t,\vartheta)$, where
$z_{k}(x,t,\vartheta)=x_{k}-t_{k}\cos\vartheta_{k}+it_{k}\sin\vartheta_{k}$.
Let $\Psi:\mathbb{C}^{n}_{+}\to\mathbb{C}^{n}_{+}$ such that
$\Psi(z)_{k}(x,t\vartheta)=\Phi(x)_{k}-\Phi(t)_{k}\cos\vartheta_{k}+i\Phi(t)_{k}\sin\vartheta_{k}$.
We say that $\Phi$ fulfils the B-Lipschitz condition with B-Lipschitz constant
$L$ if for a.e. $\vartheta\in[0\pi)^{n}$

(34) $|\Psi(z)(\vartheta)|\leq L|z(\vartheta)|.$

######  Remark 8.

Of course, linear mappings posses the B-Lipschitz property.
For $\vartheta=0$ (34) gives back the standard Lipschitz condition, and for
$\vartheta_{k}=\pi$ $k=1,\dots,n$ (34) means that $|\Phi(x)+\Phi(t)|\leq
L|x+t|$.

Example. Let $f:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be a Lipschitz function. Let
$K\subset\mathrm{int}\mathbb{R}_{+}^{n}$ compact and
$\Phi:K\to\mathbb{R}_{+}^{n}$; $\Phi(x)=f(|x|)x$. On $K$ $\Phi$ also fulfils
the Lipschitz condition with constant $\tilde{L}(K)$ and
$\frac{\Phi(x)_{k}}{x_{k}}\leq M(K)$. Let

$G(\vartheta):=L^{2}L|z(\vartheta)|^{2}-|\Phi(z)(\vartheta)|^{2}$
$=\sum_{k=1}^{n}L^{2}(x_{k}^{2}+t_{k}^{2})-\Phi(x)_{k}^{2}-\Phi(t)_{k}^{2}-2\sum_{k=1}^{n}\cos\vartheta_{k}(L^{2}x_{k}t_{k}-\Phi(x)_{k}\Phi(t)_{k}).$

If $L>M(K)$, we have

$\frac{\partial G}{\partial\vartheta_{k}}=0\hskip 4.0pt\hskip
4.0pt\mbox{iff}\hskip 4.0pt\vartheta_{k}=0\hskip 4.0pt\mbox{or}\hskip
4.0pt\vartheta_{k}=\pi.$

It can be readily seen, that the Hessian is positive (negative) definite if
$\vartheta=0$ ($\vartheta_{k}=\pi$ $k=1,\dots,n$), respectively. Thus the
Lipschitz property of $\Phi(x)$ implies that the B-Lipschitz condition fulfils
for all $\vartheta\in[0\pi)^{n}$.

######  Theorem 4.

Let $\nu>0$, $1<p<\frac{n+|a|}{\nu}$. Let $E\subset\mathbb{R}^{n}_{+}$ and
$\Phi:E\to\mathbb{R}^{n}_{+}$ is a B-Lipschitz mapping with B-Lipschitz
constant $L$. Then

$C_{p,a,\nu}(\Phi(E))\leq c\hskip 4.0ptC_{p,a,\nu}(E),$

where $c$ depends only on $n,p,a,L$.

###### Proof.

By standard arguments it is enough to prove for any
$K\subset\mathbb{R}^{n}_{+}$, compact. Let $\mu\in\mathcal{M}(\Phi(K))$ Then
by [2, Lemma 5.2.2] there is a $\mu_{\Phi}\in\mathcal{M}(K)$ such that

$\int_{\Phi(K)}W_{a,\nu,p}^{\mu}(y)d\mu_{a}(y)=\int_{K}W_{a,\nu,p}^{\mu}(\Phi(x))\Phi^{a}(x)d\mu_{\Phi}(x)$
$=\int_{K}\int_{0}^{1}\left(\frac{\int_{K}T^{\Phi(u)}\chi_{B_{+}(0,r)}(\Phi(x))\Phi^{a}(u)d\mu_{\Phi}(u)}{r^{n+|a|-\nu
p}}\right)^{p^{\prime}-1}\frac{dr}{r}\Phi^{a}(x)d\mu_{\Phi}(x)$
$\geq\int_{K}\int_{0}^{1}\left(\frac{\chi_{B_{+}\left(0,\frac{r}{L}\right)}*_{a}\mu_{\Phi}(x)}{r^{n+|a|-\nu
p}}\right)^{p^{\prime}-1}\frac{dr}{r}\Phi^{a}(x)d\mu_{\Phi}(x)$
$=c\int_{K}\int_{0}^{\frac{1}{L}}\left(\frac{\chi_{B_{+}\left(0,r\right)}*_{a}\mu_{\Phi}(x)}{r^{n+|a|-\nu
p}}\right)^{p^{\prime}-1}\frac{dr}{r}\Phi^{a}(x)d\mu_{\Phi}(x),$

where $c=c(n,p,a,L)$.
This implies that

$\int_{\Phi(K)}W_{a,\nu,p}^{\mu}(y)d\mu_{a}(y)>c\int_{K}W_{a,\nu,p}^{\mu_{\Phi}}(x)\Phi^{a}(x)d\mu_{\Phi}(x).$

Indeed, if $L\leq 1$, we have immediately the inequality above, if $L>1$ we
have to consider Remark 7 with $\delta=\frac{1}{L}$, which leads again to the
inequality above. According to Theorem 3 it proves the statement, cf. the
definition above.

###  5.2. Coverings

In the next subsections coverings in Bessel-weighted space are introduced.
Since Bessel-convolution lives in a weighted space, B-$p$ capacity of a set
depends also on the location of the set. As capacity is in close connection
with Hausdorff measure, in the next subsection we extend this notion to
weighted spaces as well.

Notation. Let $K\subset\mathbb{R}^{n}_{+}$ compact.

(1) Let $\mathcal{A}(r)$ be the minimal number of balls of radius $r$ required
to cover $K$.

(2)

$m^{a}:=m^{a}(u,r,K)=\max\\{x^{a}:x\in\overline{B(u,r)}\cap K\\}.$
$\mathcal{B}(r):=\inf\\{\sum_{j=1}^{\mathcal{A}(r)}m_{j}^{a}:K\subset\cup_{j=1}^{\mathcal{A}(r)}B(u_{j},r)\hskip
4.0pt\mbox{is a minimal covering}\\}.$

######  Remark 9.

(1) $\mathcal{A}(r)$ is obviously decreasing.
(2) Let $\cup_{j=1}^{\mathcal{A}(r)}B(u_{j},r)$ be a minimal covering of $K$.
Then there is a constant $C_{n}$ such that any point of $K$ belongs to at most
$C_{n}$ balls. Indeed, let $C_{n}=C_{n}(q)$ be the minimal number of balls of
radius $q\leq\frac{1}{2}$ which covers the unite ball. Suppose that there is a
point $x\in K$ which belongs to $C_{n}(q)+1$ balls. Then
$B\left(x,\frac{r}{q}\right)$ contains all these balls and it can be covered
by $C_{n}(q)$ balls of radius $r$ which contradicts with minimality, cf. e.g.
[2, page 145].
(3) Let $r_{k}:=\frac{1}{2^{k}}$, $r_{k+1}\leq r\leq r_{k}$ and
$\frac{1}{4}\leq q:=\frac{1}{2^{k+2}r}\leq\frac{1}{2}$. Let
$\\{B(v_{j},r_{1})\\}_{j=1}^{\mathcal{A}(r)}$ and
$\\{B(u_{i},r_{k+2})\\}_{i=1}^{\mathcal{A}(r_{k+2})}$ be minimal coverings of
$K$ with the corresponding points $\\{m_{j}\\}_{j=1}^{\mathcal{A}(r)}$ and
$\\{M_{i}\\}_{i=1}^{\mathcal{A}(r_{k+2})}$, respectively. Any $m_{j}$ belongs
to a ball, $B(u_{i},r_{k+2})$, and so $m_{j}^{a}\leq M_{i}^{a}$. Since at most
$C_{n}\left(\frac{1}{q+2}\right)$ maximum points ($m_{j}$) can be in the same
ball, $\sum_{j=1}^{\mathcal{A}(r)}m_{j}^{a}\leq
C_{n}\left(\frac{1}{q+2}\right)\sum_{j=1}^{\mathcal{A}(r_{k+2})}M_{j}^{a}$,
thus $\mathcal{B}(r)\leq C_{n}\left(\frac{1}{q+2}\right)\mathcal{B}(r_{k+2})$.
Repeating the chain of ideas with $r_{k-1}$ and $q_{1}=2^{k-1}r$, we have
$\frac{1}{C(n,q_{1})}\mathcal{B}(r_{k-1})\leq\mathcal{B}(r)$.

######  Theorem 5.

As above, let $\nu>0$, $1<p<\frac{n+|a|}{\nu}$. Then

$C_{a,\nu,p}^{\frac{1}{p}}(K)\leq
c\left(\int_{0}^{1}\left(\mathcal{B}(r)r^{n-p\nu}\right)^{1-p^{\prime}}\frac{dr}{r}\right)^{1-p},$

where $c=c(n,a)$.

###### Proof.

Let $\mu\in\mathcal{M}(K)$ and $r_{k}=\frac{1}{2^{k}}$, as above. According to
Corollary 2 and (31) we have

$\|G_{a,\nu}*_{a}\mu\|_{p^{\prime},a}^{p^{\prime}}\geq\|I_{\nu-|a|}^{1}*_{a}\mu\|_{p^{\prime},a}^{p^{\prime}}$
$\geq
c\int_{\mathbb{R}^{n}_{+}}\left(\sum_{k=0}^{\infty}2^{k(n+|a|-\nu)}\chi_{B_{+}\left(0,r_{k}\right)}*_{a}\mu(x)\right)^{p^{\prime}}d\lambda_{a}(x)$
$\geq
c\sum_{k=0}^{\infty}\frac{1}{r_{k}^{p^{\prime}(n+|a|-\nu)}}\int_{\mathbb{R}^{n}_{+}}\left(\chi_{B_{+}\left(0,r_{k}\right)}*_{a}\mu(x)\right)^{p^{\prime}}d\lambda_{a}(x),$

where the last inequality follows from the monotone convergence theorem. Let
$K\subset\cup_{j=1}^{\mathcal{A}(r_{k+1})}B(u_{j},r_{k+1})$ be a minimal
covering. Recalling that
$\chi_{B_{+}\left(0,r_{k}\right)}*_{a}\mu(x)=\int_{\mathbb{R}^{n}_{+}}T^{t}\chi_{B_{+}\left(0,r_{k}\right)}(x)d\mu_{a}(t)$,
if $t\in K$, $x\in U_{k}$, where $U_{k}$ is the neighbourhood of $K$ of radius
$r_{k}$, otherwise $T^{t}\chi_{B_{+}\left(0,r_{k}\right)}(x)=0$. Noticing that
$\cup_{j=1}^{\mathcal{A}(r_{k+1})}B(u_{j},r_{k+1})\subset U_{k}$, in view of
Remark 9 we have

$\int_{\mathbb{R}^{n}_{+}}\left(\chi_{B_{+}\left(0,r_{k}\right)}*_{a}\mu(x)\right)^{p^{\prime}}d\lambda_{a}(x)$
$\geq\frac{1}{C_{n}}\sum_{j=1}^{\mathcal{A}(r_{k+1})}\int_{B(u_{j},r_{k+1})}\left(\int_{\mathbb{R}^{n}_{+}}T^{t}\chi_{B_{+}\left(0,r_{k}\right)}(x)d\mu_{a}(t)\right)^{p^{\prime}}d\lambda_{a}(x)=:(**).$

Considering again the support of $T^{t}\chi_{B_{+}\left(0,r_{k}\right)}$, by
Hölder’s inequality we have

$(**)\geq\frac{1}{C_{n}}\sum_{j=1}^{\mathcal{A}(r_{k+1})}\lambda_{a}(B(u_{j},r_{k+1}))^{-\frac{p^{\prime}}{p}}$
$\times\left(\int_{B(u_{j},r_{k+1})}\int_{B_{+}(x,r_{k})}T^{t}\chi_{B_{+}(0,r_{k})}(x)d\mu_{a}(t)d\lambda_{a}(x)\right)^{p^{\prime}}=:(***).$

Since $x\in B\left(u_{j},r_{k+1}\right)$, $B(x,r_{k})\supset
B(u_{j},r_{k+1})$. As $T^{t}\chi_{B_{+}\left(0,r_{k}\right)}(x)$ is continuous
(actually it belongs to the $\mathrm{Lip}(\frac{1}{2})$ class) on
$B_{+}\left(x,r_{k}\right)$, $T^{t}\chi_{B_{+}\left(0,r_{k}\right)}(x)\geq
cT^{t}\chi_{B\left(0,r_{k}\right)}(u_{j})$. According to Lemma 1, if $t\in
T_{+}\left(u_{j},\frac{r_{k}}{4\sqrt{n}}\right)=:T_{k,j}$, then
$T^{t}\chi_{B\left(0,r_{k}\right)}(u_{j})\geq
c\prod_{i=1}^{n}\min\left\\{1,\left(\frac{r_{k}}{u_{j,i}}\right)^{a_{i}}\right\\}$.
Thus we have

$(***)$
$\geq\frac{c}{C_{n}}\sum_{j=1}^{\mathcal{A}(r_{k+1})}\lambda_{a}(B\left(u_{j},r_{k+1}\right))^{-\frac{p^{\prime}}{p}}\mu_{a}^{p^{\prime}}(B\left(u_{j},r_{k+1}\right))$
$\times\left(\int_{T_{k,j}}\prod_{i=1}^{n}\min\left\\{1,\left(\frac{r_{k}}{u_{j,i}}\right)^{a_{i}}\right\\}d\lambda_{a}(x)\right)^{p^{\prime}}.$

Estimating the last integral we have

$\int_{T_{k,j}}\prod_{i=1}^{n}\min\left\\{1,\left(\frac{r_{k}}{u_{j,i}}\right)^{a_{i}}\right\\}d\lambda_{a}(x)$
$\geq\prod_{i=1}^{n}\int_{u_{j,i}-\frac{r_{k+1}}{4\sqrt{n}}}^{u_{j,i}+\frac{r_{k+1}}{4\sqrt{n}}}\min\left\\{1,\left(\frac{r_{k}}{u_{j,i}}\right)^{a_{i}}\right\\}x_{i}^{a_{i}}dx_{i}=\prod_{i=1}^{n}I_{j,i}.$

If $r_{k}>u_{j,i}$,

$I_{j,i}\geq c\int_{0}^{\frac{r_{k+1}}{4\sqrt{n}}}x_{i}^{a_{i}}dx_{i}\geq
cr_{k+1}^{a_{i}+1}.$

If $r_{k}\leq u_{j,i}$,

$I_{j,i}\geq
c\int_{u_{j,i}-\frac{r_{k+1}}{4\sqrt{n}}}^{u_{j,i}+\frac{r_{k+1}}{4\sqrt{n}}}r_{k+1}^{a_{i}}dx_{i}\geq
cr_{k+1}^{a_{i}+1}.$

So we have

$(***)\geq
c\sum_{j=1}^{\mathcal{A}(r_{k+1})}r_{k}^{n(1-p^{\prime})}m_{j}^{a(1-p^{\prime})}\mu_{a}^{p^{\prime}}(B(u_{j},r_{k+1}))r_{k}^{(n+|a|)p^{\prime}}$
$=cr_{k}^{n+|a|p^{\prime}}\sum_{j=1}^{\mathcal{A}(r_{k+1})}m_{j}^{a(1-p^{\prime})}\mu_{a}^{p^{\prime}}(B(u_{j},r_{k+1})).$

By Hölder’s inequality

$\mu_{a}(K)\leq\left(\sum_{j=1}^{\mathcal{A}(r_{k+1})}\mu_{a}(B(u_{j},r_{k+1}))^{p^{\prime}}m_{j}^{a(1-p^{\prime})}\right)^{\frac{1}{p^{\prime}}}\left(\sum_{j=1}^{\mathcal{A}(r_{k+1})}m_{j}^{a}\right)^{\frac{1}{p}}.$

Thus

$\|G_{a,\nu}*_{a}\mu\|_{p^{\prime},a}^{p^{\prime}}\geq
c\mu_{a}(K)^{p^{\prime}}\sum_{k=0}^{\infty}\left(r_{k}^{n-\nu
p}\mathcal{B}(r_{k+1})\right)^{1-p^{\prime}}.$

Taking into consideration Remark 9 we have

$\|G_{a,\nu}*_{a}\mu\|_{p^{\prime},a}^{p^{\prime}}\geq
c\mu_{a}(K)^{p^{\prime}}\int_{0}^{1}\left(\mathcal{B}(r)r^{n-p\nu}\right)^{1-p^{\prime}}\frac{dr}{r}.$

Comparing it with (33) the proof is finished.

###  5.3. Hausdorff measure with Bessel external field

######  Definition 6.

Let $h$ be an increasing function on $\mathbb{R}_{+}$ with $h(0)=0$. Let
$E\subset\mathbb{R}_{+}^{n}$, $\varrho>0$.

(35)
$\Lambda^{\varrho}_{h,a}(E):=\inf\\{\sum_{i=1}^{\infty}x_{i,r_{i}}^{a}h(r_{i}):E\subset\cup_{i=1}^{\infty}B(x_{i},r_{i}),\hskip
4.0ptx_{i}\in\mathbb{R}_{+}^{n},\hskip 4.0ptr_{i}\leq\varrho\\},$

where $x_{i,r_{i}}^{a}:=\max\\{t^{a}:t\in\overline{B(x_{i},r_{i})}\\}$.
Since $\Lambda^{\varrho}_{h,a}(E)$ is a decreasing function of $\varrho$, we
can define the (finite or infinite) $a$-Hausdorff measure of $E$ as

(36) $\Lambda_{h,a}(E)=\lim_{\varrho\to 0}\Lambda^{\varrho}_{h,a}(E).$

######  Remark 10.

(1) $h(x,r):=x_{r}^{a}h(r)$ is an increasing function of $r$, but it depends
on $x$ as well, that is the $a$-Hausdorff measure of $E$ depends also on the
location of $E$.
(2) If $K\subset\mathrm{int}\mathbb{R}^{n}_{+}$ is compact, then there are
constants $c_{i}=c_{i}(a,K)$, $i=1,2$ such that
$c_{1}\Lambda_{h}(K)\leq\Lambda_{h,a}(K)\leq c_{2}\Lambda_{h}(K)$.
(3) As in the standard case $\Lambda^{\infty}_{h,a}(E)=0$ if and only if
$\Lambda_{h,a}(E)=0$. Of course, for all $\varrho>0$
$\Lambda^{\infty}_{h,a}(E)\leq\Lambda^{\varrho}_{h,a}(E)$, and so
$\Lambda^{\infty}_{h,a}(E)\leq\Lambda_{h,a}(E)$. Oppositely, by standard
arguments if $\Lambda^{\infty}_{h,a}(E)>0$, then there is a constant $c$ such
that $\Lambda^{\infty}_{h,a}(E)>c>0$. Let $\varrho$ be so small such that
$\Lambda^{\varrho}_{h,a}(E)>c$. The for all $\varrho$-covering of $E$,
$\cup_{i=1}^{\infty}B(x_{i},r_{i})$,
$\sum_{i=1}^{\infty}x_{i,r_{i}}^{a}h(r_{i})>c$. If
$\cup_{j=1}^{\infty}B(u_{j},r_{j})$ is not a $\varrho$-covering of $E$, there
exists an $r_{l}>\varrho$, and because $u_{j}\in\mathbb{R}^{n}_{+}$,
$\sum_{j=1}^{\infty}u_{j,r_{j}}^{a}h(r_{j})>u_{l,r_{l}}^{a}h(\varrho)>c(n)\varrho^{|a|}h(\varrho)$.
Thus $\Lambda^{\infty}_{h,a}(E)>\min\\{c,c(n)\varrho^{|a|}h(\varrho)\\}>0$.

Let us denote by $\mathcal{Q}_{k}$ the set of the dyadic cubes in
$\mathbb{R}_{+}^{n}$ with edge length $\frac{1}{2^{k}}$, $k\in\mathbb{Z}$.

######  Theorem 6.

With the notation above, let $h$ be an increasing function,
$E\subset\mathbb{R}_{+}^{n}$ and $\mu\in\mathcal{M}(E)$ such that for all
balls $\mu(B(x,r))\leq h(r)$. Then

$\mu_{a}(E)\leq\Lambda^{\infty}_{h,a}(E).$

Let $h$ be an increasing function, $E\subset Q\in\mathcal{Q}_{k}$. Then there
is a constant $c$ depending only on $n$, $k$ and $a$ and a measure
$\mu\in\mathcal{M}(E)$ satisfying that $\mu(B(x,r))\leq h(r)$ for all balls,
such that

$\Lambda^{\infty}_{h,a}(E)\leq c\mu_{a}(E).$

###### Proof.

Obviously if $E\subset\cup_{i=1}^{\infty}B(x_{i},r_{i})$, then

$\mu_{a}(E)\leq\sum_{i=1}^{\infty}\mu_{a}(B(x_{i},r_{i}))\leq\sum_{i=1}^{\infty}x_{i,r_{i}}^{a}h(r_{i}).$

The first part of the second statement is proved in [2, page 137], namely
there are measures $\mu_{ll}$ such that
$\mathrm{supp}\mu_{ll}=\\{\cup_{j}Q_{j}:Q_{j}\in\mathcal{Q}_{l},\hskip
4.0ptQ_{j}\cap E\neq\emptyset\\}$ and $\mu_{ll}(Q_{i})\leq h(r_{i})$ for all
$Q_{i}\in\mathcal{Q}_{i}$, $i=0,\dots,l$, where $r_{i}=\frac{1}{2^{i}}$.
Moreover $\mu_{ll}$ has constant density on each $Q_{j}\in\mathcal{Q}_{l}$.
Finally $\mu$ is defined as a weak accumulation point of $\\{\mu_{ll}\\}$.
Then $\mathrm{supp}\mu=E$ and $\mu(Q_{k})\leq 3^{n}h(r_{k})$ for all
$Q_{k}\in\mathcal{Q}_{k}$, $k\in\mathbb{N}$. It is also pointed out that $E$
has a disjoint covering with dyadic cubes, $E\subset\cup_{j}Q_{j}$, such that
$Q_{j}\in\mathcal{Q}_{n_{j}}$ and $\mu_{ll}(Q_{j})=h(r_{n_{j}})$,
$j=1,2,\dots$. Thus $\mu_{ll}(Q)=\sum_{j}h(r_{n_{j}})$. Then, with
$m_{n_{j}}^{a}=\max_{x\in\overline{Q}_{j}}x^{a}$, we have

$\mu_{a,ll}(Q)=\int_{Q}x^{a}d\mu_{ll}(x)\geq c\sum_{j,n_{j}\leq
l}m_{n_{j}}^{a}h(r_{n_{j}})\geq c\inf\sum_{i}m_{n_{i}}^{a}h(r_{n_{i}}),$

where $c=c(n,a,k)$ and the infimum is taken over all finite or denumerable
coverings of $E$. So

$\mu_{a}(Q)=\mu_{a}(E)\geq c\inf\sum_{i}m_{n_{i}}^{a}h(r_{n_{i}}).$

Taking into consideration that a $Q_{i}\in\mathcal{Q}_{i}$ can be covered by
$c(n)$ balls of radius $r_{i}$,

$\Lambda^{\infty}_{h,a}(E)\leq c(n)\inf\sum_{i}x_{i,r_{i}}^{a}h(r_{i})\leq
c\inf\sum_{i}m_{n_{i}}^{a}h(r_{n_{i}})\leq c\mu_{a}(E),$

where $c=c(n,a,k)$.

###  5.4. Capacity of Cantor sets with Bessel external field

Let $L:=\\{l_{k}\\}_{k=0}^{\infty}$ be a decreasing sequence such that
$0<2l_{k+1}<l_{k}$ for $k\in\mathbb{N}$. Let $C_{0}$ be a closed interval of
length $l_{0}$. $C_{1}$ is obtained by removing an open interval of length
$l_{0}-2l_{1}$ in the middle of $C_{0}$, etc., $C_{k}$ consists of $2^{k}$
closed intervals of length $l_{k}$. Let $C_{k}^{n}:=C_{k}\times\dots\times
C_{k}$, the Cartesian product of $n$ copies of $C_{k}$. Let
$C_{L}:=\cap_{k=0}^{\infty}C_{k}^{n}$. $C_{L}=C_{L}(n,Q)$, where
$Q=C_{0}\times\dots\times C_{0}$, the cube which contains $C_{L}$.

Notation.

Let $C_{k}^{n}=C_{k}^{n}(Q,L)=\cup_{i=1}^{2^{nk}}q_{k,i}$ as above, where
$q_{k,i}$ are the closed cubes in $C_{k}^{n}$ of edge length $l_{k}$. Let
$v_{k,i}^{a}:=\max_{x\in q_{k,i}}x^{a}$. Let us denote by

(37)
$h_{L}(l_{k}):=h_{Q,L,a}(l_{k})=\frac{1}{\sum_{i=1}^{2^{nk}}v_{k,i}^{a}}.$

Obviously, $h(l_{k})>h(l_{k+1})$. Let $h_{L}(r):=h_{Q,L,a}(r)$ be an
increasing function on $[0,\infty)$, $h_{L}(0)=0$ and $h_{L}(l_{k})$ is given
by (37).

######  Theorem 7.

Let $0<p\nu<n+|a|$, $C_{L}(n,Q)$, $h_{L}=h_{Q,L,a}$ as above. Then
$C_{a,\nu,p}(C_{L}(n,Q))>0$ if and only if

$\int_{0}^{1}\left(\frac{h_{L}(r)}{r^{n-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}<\infty.$

###### Proof.

With the notation above $C_{L}$ can be covered by $2^{kn}$ balls of radius
$l_{k}\frac{\sqrt{n}}{2}$, $\mathcal{A}(r)\leq 2^{kn}$, and if
$l_{k+1}\frac{\sqrt{n}}{2}\leq r\leq l_{k}\frac{\sqrt{n}}{2}$,

$\mathcal{B}(r)\leq\frac{1}{h_{L}(l_{k})}.$

Comparing with Theorem 5 it shows that $C_{a,\nu,p}(C_{L})=0$ if
$\int_{0}^{1}\left(\frac{h_{L}(r)}{r^{n-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}$
diverges.

On the other hand, considering $h_{L}$ let us construct the measure $\mu_{L}$
ensured by Theorem 6. In view of Lemma 1

$\chi_{B_{+}(0,r)}*_{a}\mu_{L}(x)\leq cr^{|a|}\mu_{L}(B(x,r))\leq
cr^{|a|}h_{L}(r).$

According to Theorem 3

$\|G_{a,\nu}*_{a}\mu_{L}\|_{p^{\prime},a}^{p^{\prime}}\leq
c\int_{\mathbb{R}_{+}}\int_{0}^{1}\left(\frac{\chi_{B_{+}(0,r)}*_{a}\mu_{L}(x)}{r^{n+|a|-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}d\mu_{L,a}(x)$
$\leq
c\int_{\mathbb{R}_{+}}\int_{0}^{1}\left(\frac{r^{|a|}h_{L}(r)}{r^{n+|a|-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}d\mu_{L,a}(x).$

In view of (33)

$C_{a,\nu,p}^{\frac{1}{p}}(C_{L})\geq\frac{\mu_{L}(C_{L})}{\|G_{a,\nu}*_{a}\mu_{L}\|_{p^{\prime},a}}\geq
c\frac{\mu_{L}(C_{L})^{1-\frac{1}{p^{\prime}}}}{I(h_{L})^{\frac{1}{p^{\prime}}}}$

which proves the converse statement.

######  Theorem 8.

With the notation above and supposing that $l_{0}=1$ we have that
$C_{a,\nu,p}(C_{L})>0$ if and only if

$\sum_{k=0}^{\infty}\left(l_{k}^{n-p\nu}\sum_{i=1}^{2^{nk}}v_{k,i}^{a}\right)^{1-p^{\prime}}<\infty.$

###### Proof.

First we observe that

(38)
$\frac{1}{h_{L}(l_{k+1})}=\sum_{i=1}^{2^{n(k+1)}}v_{k+1,i}^{a}=\sum_{j=1}^{2^{nk}}\sum_{i:v_{k+1,i}\in
q_{j}}v_{k+1,i}^{a}\leq
2^{n}\sum_{i=1}^{2^{nk}}v_{k,i}^{a}=2^{n}\frac{1}{h_{L}(l_{k})}.$

In view of (38)

$I(h_{L})=\sum_{k=0}^{\infty}\int_{l_{k+1}}^{l_{k}}\left(\frac{h_{L}(r)}{r^{n-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}\leq\sum_{k=0}^{\infty}h_{L}^{p^{\prime}-1}(l_{k})\int_{l_{k+1}}^{l_{k}}\left(\frac{1}{r^{n-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}$
$\leq
c\sum_{k=0}^{\infty}h_{L}^{p^{\prime}-1}(l_{k})l_{k+1}^{(p\nu-n)(p^{\prime}-1)}\leq
c2^{n}\sum_{k=0}^{\infty}h_{L}^{p^{\prime}-1}(l_{k+1})l_{k+1}^{(p\nu-n)(p^{\prime}-1)}$
$\leq
c2^{n}\sum_{k=0}^{\infty}h_{L}^{p^{\prime}-1}(l_{k})l_{k}^{(p\nu-n)(p^{\prime}-1)}.$

On the other hand

$I(h_{L})\geq
c\sum_{k=0}^{\infty}h_{L}^{p^{\prime}-1}(l_{k+1})\left(l_{k+1}^{(p\nu-n)(p^{\prime}-1)}-l_{k}^{(p\nu-n)(p^{\prime}-1)}\right)$
$\geq
c\sum_{k=0}^{\infty}h_{L}^{p^{\prime}-1}(l_{k+1})l_{k}^{(p\nu-n)(p^{\prime}-1)}\geq\frac{c}{2^{n}}\sum_{k=0}^{\infty}h_{L}^{p^{\prime}-1}(l_{k})l_{k}^{(p\nu-n)(p^{\prime}-1)},$

where in th last but one inequality we used that $2l_{k+1}<l_{k}$, and then
(38).

######  Corollary 3.

With the notation above $C_{a,\nu,p}(C_{L})>0$ if and only if

$\sum_{k=0}^{\infty}\left(l_{k}^{n-p\nu}2^{nk}\right)^{1-p^{\prime}}<\infty.$

###### Proof.

$S_{k}:=\frac{1}{2^{nk}}\sum_{i=1}^{2^{nk}}v_{k,i}^{a}$. If
$C_{L}\subset\mathrm{int}\mathbb{R}^{n}_{+}$, then $S_{k}$ obviously can be
estimated by a constant from above and from below. If $C_{L}\subset[0,1]^{n}$,
then

$1\geq
S_{k}\geq\frac{1}{2^{nk}}\sum_{i:v_{k,i}\in[1-l_{1},1]^{n}}v_{k,i}^{a}\geq\frac{1}{2^{n}}(1-l_{1})^{|a|}.$

The computation is similar if $C_{L}\subset Q\not=[0,1]^{n}$, but
$C_{L}\cap\partial\mathbb{R}^{n}_{+}\not=\emptyset$.

######  Proposition 2.

(1) Let $K\subset\mathbb{R}^{n}_{+}$ be an arbitrary bounded set and
$\varrho>0$.
If $\liminf_{r\to 0}\frac{h(r)}{r^{n}}=0$, then
$\Lambda_{h,a}^{\varrho}(K)=0$.
If $\liminf_{r\to 0}\frac{h(r)}{r^{n}}>0$, then there is a function
$\tilde{h}(r)$, increasing and $\tilde{h}(0)=0$ such that
$\frac{\tilde{h}(r)}{r^{n}}$ is decreasing and
$\Lambda_{h,a}^{\varrho}(K)\sim\Lambda_{\tilde{h},a}^{\varrho}(K)$.
(2) Let $L:=\\{l_{k}\\}$ such that $2l_{k+1}<l_{k}$ and $C_{L}$ be the
corresponding Cantor set. Let $h(r)$ be increasing on $[0,\infty)$, $h(0)=0$.
If $\liminf_{k\to\infty}\frac{h(l_{k})}{h_{L}(l_{k})}>0$, then
$\Lambda_{h,a}(C_{L})>0$.
(3) With the notation above, there is a constant $c=c(a,Q,n)$ such that

$\Lambda_{h,a}(C_{L})\leq c\liminf_{k\to\infty}\frac{h(l_{k})}{h_{L}(l_{k})}.$

###### Proof.

(1) For any bounded set $\Lambda_{h,a}^{\varrho}(K)\leq
c(a,K)\Lambda_{h}^{\varrho}(K)$, thus [2, Proposition 5.1.8 (a)] implies the
first statement. To prove the second statement define $\tilde{h}(r)$ with

$\frac{\tilde{h}(r)}{r^{b}}:=\inf_{0<t\leq r}\frac{h(t)}{t^{b}}.$

If $\liminf_{r\to 0}\frac{h(r)}{r^{b}}>0$, $\tilde{h}(r)>0$ $\forall r>0$.
$\frac{\tilde{h}(r)}{r^{b}}$ decreasing and repeated the chain of ideas of [2]
for an arbitrary $\varepsilon>0$ we choose a $t\in[r,R]$ such that
$\frac{h(t)}{t^{b}}\leq(1+\varepsilon)\frac{\tilde{h}(R)}{R^{b}}\leq(1+\varepsilon)\frac{\tilde{h}(t)}{t^{b}}$
Thus for all $\varepsilon>0$, $\tilde{h}\leq h(r)\leq
h(t)\leq(1+\varepsilon)\tilde{h}(R)\left(\frac{t}{R}\right)^{b}\leq(1+\varepsilon)\tilde{h}(R)$,
because $b$ must be positive. That is $\tilde{h}$ is increasing.
Since $\tilde{h}\leq h(r)$, it is enough to show that $\Lambda_{h,a}(K)\leq
c\Lambda_{\tilde{h},a}(K)$. To prove this we assume that $b=n$. Let
$K\subset\cup_{i}B(x_{i},r_{i})$, $r_{i}\leq\varrho$ is a covering such that
$\sum_{i}x_{i,r_{i}}^{a}\tilde{h}(r_{i})<\Lambda_{\tilde{h},a}^{\varrho}(K)+\varepsilon$,
where $\varepsilon>0$ is arbitrary. All $B(x_{i},r_{i})$ can be covered by
$c(n)\left(\frac{r_{i}}{r}\right)^{n}$ balls of radius $r\leq r_{i}$,
$B(x_{i},r_{i})\subset\cup_{j}B(x_{i,j},r)$. Taking into account that

$c(n)\sum_{i=1}^{\infty}\frac{h(r)}{r^{n}}r_{i}^{n}\frac{1}{c(n)}\left(\frac{r}{r_{i}}\right)^{n}\sum_{j=1}^{c(n)\left(\frac{r_{i}}{r}\right)^{n}}x_{i,j,r}^{a}\leq
c(n)\sum_{i=1}^{\infty}\frac{h(r)}{r^{n}}r_{i}^{n}x_{i,r_{i}}^{a},$

we have

$\Lambda_{h,a}(K)\leq c(n)\sum_{i=1}^{\infty}\inf_{0<r\leq
r_{i}}\frac{h(r)}{r^{n}}r_{i}^{n}x_{i,r_{i}}^{a}\leq
c(n)\sum_{i=1}^{\infty}x_{i,r_{i}}^{a}\tilde{h}(r_{i}),$

which implies the inequality by choice of the covering.
(2) There is a measure $c(n)\mu_{L}$ such that $c(n)\mu_{L}(B(x,r))\leq h(r)$
for all $r\leq r_{0}$, cf. [2, Theorem 5.3.1]. Thus Theorem 6 ensures the
result.
(3) $C_{L}\subset Q$ can be covered by $c(n)2^{kn}$ balls of radius $l_{k}$,
$B(x_{i},l_{k})$. As $x_{i,l_{k}}^{a}\leq cv_{i,k}^{a}$, if $\varrho\geq
l_{k}$ $(c=c(Q,n,a)$,

$\Lambda_{\tilde{h},a}^{\varrho}(C_{L})\leq
cc(n)\sum_{i=1}^{2^{kn}}v_{i,k}^{a}h(l_{k})=c(cn)\frac{h(l_{k})}{h_{L}(l_{k})},$

which implies the statement.

Another corollary of Theorem 8 is given below.

######  Corollary 4.

(a) Let $h(r)$ be increasing on $[0,\infty)$, $h(0)=0$. Let $0<p\nu\leq n$. If

$\int_{0}^{1}\left(\frac{h(r)}{r^{n-p\nu}}\right)^{p^{\prime}-1}\frac{dr}{r}=\infty,$

then there exists a compact set $K\subset\mathbb{R}^{n}_{+}$ such that
$\Lambda_{h,a}(K)>0$ and $C_{a,\nu,p}(K)=0$.
(b) With the notation above, if

(39) $\liminf_{r\to 0}\frac{h(r)}{r^{n-p\nu}}=0,$

then there exists a compact set $K\subset\mathbb{R}^{n}_{+}$ such that
$\Lambda_{h,a}(K)=0$ and $C_{a,\nu,p}(K)>0$.

###### Proof.

Comparing Theorem 8 with [2, Theorem 5.3.2], it can be seen that if
$0<p\nu\leq n$, $C_{a,\nu,p}(C_{L})>0$ if and only if $C_{\nu,p}(C_{L})>0$. If
$C_{L}\subset[0,1]^{n}$,
$\Lambda_{h}(C_{L})\geq\Lambda_{h,a}(C_{L})\geq\Lambda_{h,a}(C_{L}^{1})\geq(1-l_{1})^{|a|}\Lambda_{h}(C_{L}^{1})\geq
c(l_{1},a,n)\Lambda_{h}(C_{L})$, where $C_{L}^{1}$ is the Cantor set
associated with $L^{1}:=\\{l_{i}\\}_{i=1}^{\infty}$ and located in
$[1-l_{1},1]^{n}$. According to [2, Theorem 5.4.2] assumptions of (a) ensures
that there is a Cantor set $C_{L}$ with $\Lambda_{h}(C_{L})>0$ and
$C_{\nu,p}(C_{L})=0$, which example fulfils the requirements of part (a).
According to [2, Theorem 5.4.1] assumptions of part (b) imply that there is a
Cantor set $C_{L}$ with $\Lambda_{h}(C_{L})=0$ and $C_{\nu,p}(C_{L})>0$, which
example fulfils the requirements of part (b).

Construction. Comparing Theorem 7, Proposition 2 and Corollary 4 there is a
sequence $L=\\{l_{k}\\}_{k=0}^{\infty}$ with a corresponding Cantor set such
that $h(l_{k})\sim h_{L}(l_{k})$. If $\frac{h(r)r^{|a|}}{r^{n}}$ is
decreasing, it is not difficult to construct a sequence which generates a
Cantor set and fulfils that $h(l_{k})=h_{L}(l_{k})$.
Indeed, Let $Q=[0,1]^{n}$. We can assume that $h(1)=1=l_{0}$. The right
endpoints of the intervals of $C_{1}$ are $l$, $1$. Thus all the coordinates
of $v_{1},i(l)$ are $l$ or $1$ and so
$u_{1}(l):=\sum_{i=1}^{2^{n}}v_{1},i^{a}(l)$ is increasing in $l$, and
$f_{l}(l):=\frac{1}{u_{1}(l)}$ is decreasing, positive $f_{1}(0)=1$ and
$f_{1}(1)=\frac{1}{2^{n}}$. So $l_{1}$ is defined by $h(l_{1})=f_{1}(l_{1})$.
$u_{2}(l)=u_{1}(l_{1})+s_{2}(l)=\sum_{i=1}^{2^{2n}}v_{2,i}^{a}(l_{1},l)$,
where $s_{2}(l)$ is an increasing function of $l$ since the coordinates of
$v_{2,i}$ contains the right endpoints of $C_{2}=C_{2}(l_{1},l)$.
$u_{2}(l_{1})=2^{n}u_{1}(l_{1})$, because the right endpoints
$l;l_{1};1-l_{1}+l;1$ become $l_{1};l_{1};1;1$. Thus
$f_{2}(l):=\frac{1}{u_{2}(l)}$ is decreasing,
$f_{2}(l_{1})=\frac{1}{2^{n}}f_{1}(l_{1})$, $f_{2}(0)>0$, so $l_{2}$ is
defined by $h(l_{2})=f_{2}(l_{2})$. … $u_{k}(l)=u_{k-1}(l_{k-1})+s_{k}(l)$,
$f_{k}(l):=\frac{1}{u_{k}(l)}$ is decreasing,
$f_{k}(l_{k-1})=\frac{1}{2^{n}}f_{k-1}(l_{k-1})=\frac{1}{2^{n}}h(l_{k-1})$,
$f_{k}(0)>0$, which defines $l_{k}$. $L:=\\{l_{k}\\}$ is obviously decreasing
and $h(l_{k-1})\leq 2^{n}h(l_{k})$. It remains to prove that $L$ defines a
Cantor set. We have to show that

$f_{k}\left(\frac{l_{k-1}}{2}\right)\leq h\left(\frac{l_{k-1}}{2}\right).$

Consider
$u_{k}\left(\frac{l}{2}\right)=u_{k-1}(l_{k-1})+s_{k}\left(\frac{l}{2}\right)$,
and the members of $s_{k}\left(\frac{l}{2}\right)$ consist of products of
terms $\left(d+\frac{l}{2}\right)^{a_{i}}$, where $d\geq 0$. Thus
$u_{k}\left(\frac{l}{2}\right)\geq\frac{1}{2^{|a|}}u_{k}(l)$. So

$\frac{1}{2^{|a|}}f_{k}\left(\frac{l_{k-1}}{2}\right)\leq
f_{k}(l_{k-1})=\frac{1}{2^{n}}f_{k-1}(l_{k-1})=\frac{1}{2^{n}}h(l_{k-1})\leq\frac{1}{2^{|a|}}h\left(\frac{l_{k-1}}{2}\right),$

where in the last inequality we used the assumption.

Notice, that together with Theorem 5 this leads to the construction of a
Cantor-type set in $[0,1]^{n}$ with ”prescribed” B-$p$-capacity. More
precisely, if $0<|a|<p\nu$, let $h(r)=r^{n-b}$, where $0<p\nu-b<p\nu-|a|$.
Then $0<C_{a,\nu,p}(C_{L})<c(n,a)p(p\nu-b)$, where $h(l_{k})=h_{L}(l_{k})$.

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Department of Analysis, Institute of Mathematics,
Budapest University of Technology and Economics
Műegyetem rkp. 3., H-1111 Budapest, Hungary.

Department of Analysis,
Alfréd Rényi Institute of Mathematics,
Reáltanoda street 13-15, H-1053, Budapest, Hungary

<EMAIL_ADDRESS>