2 Preliminaries

Let us agree that the norm in $L^p({\mathbb D}, {\mathrm {dH}})$ will be denoted by $\|\cdot \|_p.$ Let $a\in \mathbb {B}^n, P_a$ is the orthogonal projection from $\mathbb {C}^n$ onto the one-dimensional subspace $[a]$ generated by $a,$ and $Q_a$ is the orthogonal projection from $\mathbb {C}^n$ onto $\mathbb {C}^n\ominus [a].$ It is known that

$$ \begin{align*} P_a(z)&=\frac{\langle z,a\rangle}{|a|^2}a,\ \ \ z\in\mathbb{C}^n,\\ Q_a(z)&=z-P_a(z)=z-\frac{\langle z,a\rangle}{|a|^2}a,\ \ \ z\in\mathbb{C}^n. \end{align*} $$

The following map

(5) $$ \begin{align} \varphi_a(z)=\frac{a-P_a(z)-s_aQ_a(z)}{1-\langle z,a\rangle},\quad a,z\in\mathbb{B}^n \end{align} $$

defines the class of automorphisms of the unit ball, which is usually called symmetries or involutive automorphisms, i.e., involutions:

$$ \begin{align*}\varphi_a\circ\varphi_a(z)=z,\ \ \ z\in\mathbb{B}^n.\end{align*} $$

Note that for $a=0$ , we assume $\varphi _0(z)=-z.$ We will need the following lemma proven in [Reference Zhu23] which is valid for a bounded symmetric domains $\Omega \in \mathbb {C}^n,$ group of automorphisms $\mathrm {Aut}(\Omega )$ with $G_0$ be a subgroup $\{\psi \in \mathrm {Aut}(\Omega ): \psi (0)=0 \}.$

Lemma 1 [Reference Zhu23, Lemma 2]

For any $a,b \in \Omega $ , there exists a unitary $U\in G_0$ such that

(6) $$ \begin{align} U\varphi_{\varphi_a(b)}=\varphi_b\circ\varphi_a. \end{align} $$

Moreover, the unitary U is given by the formula

(7) $$ \begin{align} U=\varphi_b\circ\varphi_a\circ\varphi_{\varphi_a(b)}. \end{align} $$

Recall that the function

$$ \begin{align*}K(z,w)=\frac{1}{(1-\langle w,z\rangle)^{n+1}}, \qquad z, w\in{\mathbb B}^n,\end{align*} $$

is the Bergman reproducing kernel for the unit ball, and the normalized reproducing kernels $k_z(\cdot ), z\in {\mathbb B}^n$ , are given by the formula

(8) $$ \begin{align} k_z(w)=\frac{K(z,w)}{\|K(z,\cdot)\|_{L^2({\mathbb B}^n,{\mathrm{dA}})}} =\frac{(1-|z|^2)^{\frac{n+1}{2}}}{(1-\langle w,z\rangle)^{n+1}}. \end{align} $$

We will use the following known result (see [Reference Zhu24, Theorem 1.12]).

Lemma 2 [Reference Zhu24].

Suppose $c\in {\mathbb R}$ , $t>-1$ , and

$$ \begin{align*}I_{c,t}(z)=\int_{\mathbb{B}^n}\frac{(1-|w|^2)^t}{|1-\langle z,w\rangle|^{n+1+t+c}}\,{\mathrm{dA}}(w),\ \ z\in{\mathbb B}^n.\end{align*} $$

1. (1) If $c<0,$ then as a function of $z, I_{c,t}(z)$ is bounded from above and bounded from below on ${\mathbb B}^n$ .

2. (2) If $c>0,$ then $I_{c,t}\eqsim (1-|z|^2)^{-c}$ as $|z|\to 1^-$ .

3. (3) If $c=0,$ then $I_{0,t}\eqsim -\ln (1-|z|^2)$ as $|z|\to 1^-$ .

3 Hausdorff–Berezin operators

For any one-variable kernel (a function) K on the unit ball $\mathbb {B}^n$ , the corresponding Hausdorff–Berezin operator is defined by formula

$$ \begin{align*} \mathbb{K}f(z)&=\int_{\mathbb{B}^n} K(w)f(\varphi_z(w))\,{\mathrm{dH}}(w)\\ &=\int_{\mathbb{B}^n} K(\varphi_z(w))f(w)\,{\mathrm{dH}}(w), \end{align*} $$

if thus integral makes sense. Here, ${\mathrm {dH}}$ stands for the invariant Haar measure

$$ \begin{align*}{\mathrm{dH}}(z)=\frac{{\mathrm{dA}}(z)}{(1-|z|^2)^{n+1}},\quad z\in\mathbb{B}^n.\end{align*} $$

We will use ${\mathfrak K}$ to denote the class of all Hausdorff–Berezin operators.

Along with the operator $\mathbb {K}$ , we consider the conjugate operator

$$ \begin{align*} \mathbb{K}^*f(z)&=\int_{\mathbb{B}^n} \overline{K(\varphi_w(z))}f(w)\,{\mathrm{dH}}(w)\\&= \int_{\mathbb{B}^n} \overline{K(\varphi_{\varphi_z(\xi)}(z))}f(\varphi_z(\xi))\,{\mathrm{dH}}(\xi). \end{align*} $$

Lemma 3 The class ${\mathfrak K}$ coincides with the class of operators of the type

(9) $$ \begin{align} \mathcal{K}f(z)\equiv\int_{\mathbb{B}^n}\widetilde{K}(z,w)f(w)\,{\mathrm{dH}}(w), \end{align} $$

where the integral kernel $\widetilde {K}$ is invariant in the sense that

(10) $$ \begin{align} \widetilde{K}(z,w)=\widetilde{K}(0,-\varphi_z(w)),\qquad z,w\in{\mathbb D}. \end{align} $$

Proof If we define the two-variable integral kernel by the rule

$$ \begin{align*}\widetilde K(z,w)=K(\varphi_z(w)),\end{align*} $$

then it satisfies (10), because

$$ \begin{align*}\widetilde K(z,w)=K(\varphi_z(w))=K(\varphi_0\left(-\varphi_z(w)\right))=\widetilde K(0,-\varphi_z(w)).\end{align*} $$

Conversely, every operator $\mathcal {K}$ with the kernel $\widetilde K$ satisfying (10) has the form of the operator $\mathbb {K}$ with

$$ \begin{align*}K(\varphi_z(w))=\widetilde K(0,-\varphi_z(w))\end{align*} $$

by definition.

Remark 1 Note that in one-dimensional complex case, i.e., ${\mathbb B}^n={\mathbb D}$ the more strict invariance condition

$$ \begin{align*} \widetilde{K}(\varphi_a(z),\varphi_a(w))=\widetilde{K}(z,w), \qquad a,z,w\in{\mathbb B}^n, \end{align*} $$

is equivalent to the case that initial one-variable kernel K is radial, see [Reference Karapetyants, Samko and Zhu15]. We state the problem of finding an analog of such a condition as an open question for the case ${\mathbb B}^n, n>1.$

4 Boundedness of Hausdorff–Berezin operators in $L^p({\mathbb B}^n,{\mathrm {dH}})$

In this section, we consider the boundedness of our Hausdorff–Berezin operators on the spaces $L^p({\mathbb B}^n,{\mathrm {dH}})$ . We start with the case $p=1$ . Let

(11) $$ \begin{align} \kappa=\sup_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n}\left|K\left(\varphi_w(z)\right)\right|\,{\mathrm{dH}}(w). \end{align} $$

If K is radial, then the formula (11) reads as

$$ \begin{align*} \kappa=\int_{\mathbb{B}^n}\left|K\left(w\right)\right|\,{\mathrm{dH}}(w)=\int_0^1\frac{|K(r)|}{(1-r^2)^{n+1}}2nr^{2n-1}dr. \end{align*} $$

We use $\|\mathbb {K}\|$ to define the operator norm of the operator $\mathbb {K}$ acting in $L^p({\mathbb B}^n,{\mathrm {dH}}).$

Proof The proof is immediate by Minkowski’s inequality. In fact,

$$ \begin{align*} \|\mathbb{K}f\|_1&=\int_{\mathbb{B}^n}\left|\int_{\mathbb{B}^n} K(\varphi_z(w))f(w)\,{\mathrm{dH}}(w)\right|\,{\mathrm{dH}}(z)\\ &\leqslant\int_{\mathbb{B}^n}|f(w)|\,{\mathrm{dH}}(w)\int_{\mathbb{B}^n}|K(\varphi_z(w))|\,{\mathrm{dH}}(z)\leqslant\kappa\|f\|_1 \end{align*} $$

as desired.

Let us consider the case $1<p<\infty $ , where $\frac {1}{p}+\frac {1}{q}=1$ . Given $\sigma \in {\mathbb R}$ , let us write

(12) $$ \begin{align} \kappa_1(p,\sigma)=\sup_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n}|k_\zeta(z)|^{\frac{\sigma}{p}} |K(\zeta)|\,{\mathrm{dH}}(\zeta), \end{align} $$

and

(13) $$ \begin{align} \kappa_2(q,\sigma)= \sup_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n}|k_\zeta(z)|^{\frac{\sigma}{q}} \left|K\left(\varphi_{\varphi_z(\zeta)}(z)\right)\right|\,{\mathrm{dH}}(\zeta). \end{align} $$

If K is radial, then due to (7), the formula in (13) reads as

$$ \begin{align*} \kappa_2(q,\sigma)=\sup_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n}|k_\zeta(z)|^{\frac{\sigma}{q}} \left|K\left(\zeta\right)\right|\,{\mathrm{dH}}(\zeta). \end{align*} $$

Theorem 5 Let $1< p<\infty $ . If there exists $\sigma _0\in {\mathbb R}$ such that

(14) $$ \begin{align} \kappa_1(p,\sigma_0)<\infty,\qquad \kappa_2(q,\sigma_0)<\infty, \end{align} $$

then the operator $\mathbb {K}$ is bounded in $L^p({\mathbb B}^n,{\mathrm {dH}})$ and its operator norm on $L^p({\mathbb B}^n,{\mathrm {dH}})$ satisfies the estimate

$$ \begin{align*}\|\mathbb{K}\|\leqslant\inf\left\{\kappa_1(p,\sigma)^{\frac{1}{q}} \kappa_2(q,\sigma)^{\frac{1}{p}}\right\},\end{align*} $$

where infimum is taken with respect to all those $\sigma =\sigma _0$ for which (14) holds.

Proof Denote $\tau _\sigma (z)=(1-|z|^2)^\sigma , \sigma \in {\mathbb R}.$ By Hölder’s inequality, we obtain

$$ \begin{align*} |\mathbb{K}f(z)|&=\left|\int_{\mathbb{B}^n}\tau_\sigma(w)^{\frac{1}{pq}} \tau_\sigma(w)^{-\frac{1}{pq}}K(\varphi_z(w))f(w)\,{\mathrm{dH}}(w)\right|\\&\leqslant\left(\int_{\mathbb{B}^n}\tau_\sigma(w)^{\frac{1}{p}}|K(\varphi_z(w))| \,{\mathrm{dH}}(w)\right)^{\frac{1}{q}}\\& \quad \times\left(\int_{\mathbb{B}^n}\tau_\sigma(w)^{-\frac{1}{q}}|K(\varphi_z(w))||f(w)|^p\, {\mathrm{dH}}(w)\right)^{\frac{1}{p}}. \end{align*} $$

Note that

$$ \begin{align*} \int_{\mathbb{B}^n}\tau_\sigma(w)^{\frac{1}{p}}|K(\varphi_z(w))|\,{\mathrm{dH}}(w)&= \int_{\mathbb{B}^n}\tau_\sigma(\varphi_z(\zeta))^{\frac{1}{p}}|K(\zeta)|\,{\mathrm{dH}}(\zeta)\\ &=(1-|z|^2)^{\frac{\sigma}{p}}\int_{\mathbb{B}^n}|k_\zeta(z)|^{\frac{2\sigma}{(n+1)p}}|K(\zeta)|\,{\mathrm{dH}}(\zeta)\\ &\leqslant\kappa_1(p,2\sigma/(n+1))\, (1-|z|^2)^{\frac{\sigma}{p}}. \end{align*} $$

Therefore, we have

$$ \begin{align*} &\kappa_1(p,2\sigma/(n+1))^{-\frac{p}{q}}\|\mathbb{K}f\|^p_p\\& \quad \le\int_{\mathbb{B}^n}(1-|z|^2)^{\frac{\sigma}{q}}\,{\mathrm{dH}}(z)\int_{\mathbb{B}^n}\tau_\sigma(w)^{-\frac{1}{q}} |K(\varphi_z(w))||f(w)|^p\,{\mathrm{dH}}(w)\\& \quad =\int_{\mathbb{B}^n}|f(w)|^p\,{\mathrm{dH}}(w)\int_{\mathbb{B}^n}(1-|z|^2)^{\frac{\sigma}{q}} \tau_\sigma(w)^{-\frac{1}{q}}|K(\varphi_z(w))|\,{\mathrm{dH}}(z)\\& \quad =\int_{\mathbb{B}^n}|f(w)|^p\,{\mathrm{dH}}(w) \int_{\mathbb{B}^n}(1-|\varphi_w(\zeta)|^2)^{\frac{\sigma}{q}}\tau_\sigma(w)^{-\frac{1}{q}} \left|K\left(\varphi_{\varphi_w(\zeta)}(w)\right)\right|\,{\mathrm{dH}}(\zeta)\\& \quad =\int_{\mathbb{B}^n}|f(w)|^p\,{\mathrm{dH}}(w)\int_{\mathbb{B}^n}|k_\zeta(w)|^{\frac{2\sigma}{(n+1)q}}\left|K\left(\varphi_{\varphi_w(\zeta)}(w)\right)\right|\,{\mathrm{dH}}(\zeta)\\& \quad \leqslant\kappa_2(q,2\sigma/(n+1))\|f\|_p^p. \end{align*} $$

Here, to justify the change of the order of integration, we used Fubini’s theorem. Finally, collecting the above estimates, we obtain

$$ \begin{align*}\|{\mathbb K}f\|_p\le\kappa_1(p,2\sigma/(n+1))^{\frac1q} \kappa_2(q,2\sigma/(n+1))^{\frac1p}\|f\|_p.\end{align*} $$

This finishes the proof.

As a corollary, we formulate below the corresponding boundedness result for the conjugate operator $\mathbb {K}^*$ (we will need it to prove Theorem 7).

Corollary 6 Let $1<q<\infty $ . If there exist $\sigma \in {\mathbb R}$ such that (14) holds, then the operator $\mathbb {K}^*$ is bounded on $L^q({\mathbb B}^n,{\mathrm {dH}})$ and its norm on $L^q({\mathbb B}^n,{\mathrm {dH}})$ satisfies

$$ \begin{align*}\|\mathbb{K}^*\|\leqslant \inf \{\kappa_{1}(p,\sigma)^{\frac{1}{q}} \kappa_{2}(q,\sigma)^{\frac{1}{p}}\},\end{align*} $$

where infimum is taken with respect to all those $\sigma $ for which (14) holds.

For a nonnegative kernel (function) K, let us write

$$ \begin{align*} \varkappa&=\inf_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n} K\left(\varphi_\zeta(z)\right)\,{\mathrm{dH}}(\zeta),\\ \varkappa_1(p,\sigma)&=\inf_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n}|k_\zeta(z)|^{\frac{\sigma}{p}}K(\zeta)\,{\mathrm{dH}}(\zeta),\\ \varkappa_2(q,\sigma)&=\inf_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n}|k_\zeta(z)|^{\frac{\sigma}{q}} K\left( \varphi_{\varphi_z(\zeta)}(z)\right)\,{\mathrm{dH}}(\zeta). \end{align*} $$

If K is also radial, then

$$ \begin{align*}\varkappa_2(q,\sigma)=\inf_{z\in{\mathbb B}^n}\int_{\mathbb{B}^n}|k_\zeta(z)|^{\frac{\sigma}{q}} K\left(\zeta\right)\,{\mathrm{dH}}(\zeta).\end{align*} $$

Proof First, suppose that the operator $\mathbb {K}$ is bounded in $L^1({\mathbb D},{\mathrm {dH}})$ . Then for $\phi (z)=(1-|z|^2)^{n+1}$ , we have

$$ \begin{align*} \|\mathbb{K}\phi\|_1&=\int_{\mathbb{B}^n}\int_{\mathbb{B}^n} K(\varphi_z(w))\,{\mathrm{dA}}(w)\,{\mathrm{dH}}(z)\\ &=\int_{\mathbb{B}^n}\,{\mathrm{dA}}(w)\int_{\mathbb{B}^n} K(\varphi_z(w))\,{\mathrm{dH}}(z)\geqslant\varkappa. \end{align*} $$

Suppose that the operator $\mathbb {K}$ is bounded in $L^p({\mathbb B}^n,{\mathrm {dH}})$ for some $1<p<\infty $ . Then for all $\phi \in L^p({\mathbb B}^n, {\mathrm {dH}})$ and $\psi \in L^q({\mathbb D}, {\mathrm {dH}})$ , we have

$$ \begin{align*}\left|\int_{\mathbb{B}^n}(\mathbb{K}\phi)(z)\,\psi(z)\,{\mathrm{dH}}(z)\right|\leqslant \|\mathbb{K}\|\|\phi\|_p\|\psi\|_q.\end{align*} $$

Let

$$ \begin{align*}\phi(z)=(1-|z|^2)^{\frac{\sigma}{p}},\quad \psi(z)=(1-|z|^2)^{\frac{\sigma}{q}},\quad \sigma>n.\end{align*} $$

We obtain

$$ \begin{align*} &\int_{\mathbb{B}^n}(\mathbb{K}\phi)(z)\,\psi(z)\,{\mathrm{dH}}(z)\\& \quad = \int_{\mathbb{B}^n}(1-|z|^2)^{\frac{\sigma}{q}}\,{\mathrm{dH}}(z) \int_{\mathbb{B}^n}(1-|\varphi_z(w)|)^{\frac{\sigma}{p}}K(w)\,{\mathrm{dH}}(w)\\& \quad = \int_{\mathbb{B}^n}(1-|z|^2)^{\sigma}\,{\mathrm{dH}}(z) \int_{\mathbb{B}^n}|k_w(z)|^{\frac{2\sigma}{(n+1)p}}K(w)\,{\mathrm{dH}}(w)\\& \quad \geqslant\varkappa_1(p,2\sigma/(n+1))\int_{\mathbb{B}^n}(1-|z|^2)^{\sigma}\,{\mathrm{dH}}(z). \end{align*} $$

The integral $\int _{\mathbb {B}^n}(1-|z|^2)^{\sigma }\,{\mathrm {dH}}(z)$ is finite if and only if $\sigma>n.$ This implies that $\varkappa _1(p,\sigma )<\infty $ for any $\sigma>\frac {2n}{n+1}.$ By the same arguments applied to the conjugate operator $\mathbb {K}^*$ , we obtain that $\varkappa _2(q,\sigma )<\infty $ for any $\sigma>\frac {2n}{n+1}.$

At the conclusion of this section, let us consider the important example of the kernel K given by the formula $K(z)=(1-|z|^2)^{\alpha }$ . First, we prove the following technical lemma.

As a corollary of Lemma 8, we obtain the following result on the boundedness of the operator $\mathbb {K}$ induced by the positive kernel $K(z)=(1-|z|^2)^\alpha $ on $L^p({\mathbb D},{\mathrm {dH}}).$ This result shows that for this kernel, the sufficient condition $\kappa <\infty $ in Theorem 4 and the sufficient conditions (14) with $\sigma =\frac {2n}{n+1}$ stated in Theorem 5 turn out to be necessary.

Theorem 9 Let $1\leqslant p<\infty $ and $\mathbb {K}$ be the Hausdorff–Berezin operator with kernel $K(z)=(1-|z|^2)^{\alpha }$ . Then $\mathbb K$ is bounded in $L^p({\mathbb B}^n,{\mathrm {dH}})$ if and only if

(15) $$ \begin{align} \alpha>n\, \max\left\{\frac{1}{p}, 1-\frac{1}{p}\right\}. \end{align} $$

Proof The sufficiency of (15) follows from Theorems 4 and 5 and Lemma 8 under the choice $\sigma =\frac {2n}{n+1}.$

To prove necessity, we choose the minimizing function

$$ \begin{align*}f_\beta(z)=(1-|z|^2)^{\frac{\beta}{p}},\qquad\beta\in{\mathbb R}.\end{align*} $$

It is obvious that $f_\beta \in L^p({\mathbb B}^n,{\mathrm {dH}})$ if and only if $\beta>n$ , and

$$ \begin{align*} \mathbb{K}f_\beta(z)&=\int_{\mathbb{B}^n}(1-|\varphi_z(w)|^2)^\alpha(1-|w|^2)^{\frac{\beta}{p}} \,{\mathrm{dH}}(w)\\ &=(1-|z|^2)^\alpha\int_{\mathbb{B}^n}\frac{(1-|w|^2)^{\alpha+\frac{\beta}{p}}} {|1-\langle w,z\rangle|^{2\alpha}}\,{\mathrm{dH}}(w). \end{align*} $$

It is clear that there should be $\alpha>n-\frac {\beta }{p}$ . Otherwise, the integral on the right side of the above equality is infinite for any $z\in {\mathbb B}^n$ . From Lemma 2, we see that

$$ \begin{align*}\mathbb{K}f_\beta(z)\eqsim (1-|z|^2)^{\frac{\beta}{p}},\,\,\,\, |z|\to1^- ,\end{align*} $$

if and only if $\alpha>\beta /p$ . Since $\beta>n$ can be chosen arbitrarily close to n, this implies that $\alpha \geqslant n/p$ . The case $\alpha =n/p$ is excluded since in that case from Lemma 2, we obtain

$$ \begin{align*}\mathbb{K}f_\beta(z)\eqsim (1-|z|^2)^{\frac{n}{p}},\,\,\,\,\ |z|\to 1^- .\end{align*} $$

Now the rest of the proof follows by Lemma 8.

5 Compactness of Hausdorff–Berezin operators

Before we formulate the next theorem, recall the definition of a positive-definite kernel. Let X be a nonempty set, and let $\mathbf {K}$ be a function on $X \times X$ such that

$$ \begin{align*}\sum_{j=1}^N\sum_{k=1}^N \mathbf{K}(x_j,x_k)\xi_j\overline{\xi_k}\ge 0 \mbox{ for any } x_1,\dots, x_N \in X, \xi_j\in\mathbb{C}, \mbox{ and any } N. \end{align*} $$

Then $\mathbf {K}$ is called a positive-definite kernel on X. In this case, $\mathbf {K}(x,x)\ge 0$ for all x. Let, in addition, X be a locally compact space which is equipped with regular Borel measure $\mu $ , and let $\mathbf {K}$ be continuous. The well-known sufficient condition for an integral operator B on $L^2(\mu )$ (the space of $\mu $ – measurable quadratically summable functions with respect to the measure $\mu $ ) with a kernel $\mathbf {K}$ to be in a trace class states that this is the case if $\mathbf {K}$ is a positive-definite kernel on X and

$$ \begin{align*}I:=\int_X \mathbf{K}(x,x)d\mu(x)<\infty. \end{align*} $$

In such a case, the trace equals to: $\mathrm {tr}B=I$ (see, e.g., Theorem 2.12 in [Reference Simon22] or arguments before Theorem XI.31 in [Reference Reed and Simon21]).

Proof To prove statement (i), we are going to apply [Reference Kantorovich and Akilov9, Chapter XI, Theorem 3] to the operator $\mathbb {K}$ in the form

$$ \begin{align*} \mathbb{K}f(z)=\int_{{\mathbb B}^n} K(\varphi_z(w))f(w)\,{\mathrm{dH}}(w)=\int_{{\mathbb B}^n} \frac{K(\varphi_z(w))}{(1-|w|^2)^{n+1}}f(w)\,{\mathrm{dA}}(w). \end{align*} $$

To this end, we identify the space $\mathbb C^n$ with ${\mathbb R}^{2n}$ and ${\mathbb B}^n$ with the unit ball D of ${\mathbb R}^{2n}$ . Then the point $z=s_1+\imath s_2\in {\mathbb B}^n$ corresponds to the point $s=(s_1,s_2)\in {\mathbb R}^{2n}$ and the point $w=t_1+\imath t_2\in {\mathbb B}^n$ corresponds to the point $t=(t_1,t_2)\in {\mathbb R}^{2n}$ . Moreover, let the function $K(\varphi _z(w))/(1-|w|^2)^{n+1}$ corresponds to $\mathbf {K}(s,t)$ in the aforementioned theorem. Further, we put $D'=D$ and $q=p$ in this theorem. Since $dt=|{\mathbb B}^n|{\mathrm {dA}}(w)$ ( $|{\mathbb B}^n|$ stands for the Euclidean volume of ${\mathbb B}^n$ ), the condition (16) implies the validity of the condition 1) from [Reference Kantorovich and Akilov9, Chapter XI, Theorem 1] because

$$ \begin{align*}\int_{{\mathbb B}^n}|\mathbf{K}(s,t)|^rdt=|{\mathbb B}^n|\int_{{\mathbb B}^n}\frac{|K(\varphi_z(w))|^r}{(1-|w|^2)^{r(n+1)}}{\mathrm{dA}}(w)<\infty \ \mbox{ for a.e. } z\in {\mathbb B}^n. \end{align*} $$

Similarly, the condition (17) implies the validity of the condition 2) from [Reference Kantorovich and Akilov9, Chapter XI, Theorem 1]. The remaining condition of this theorem is also satisfied due to our assumptions.

The statement (ii) is a corollary of the well-known result (see, e.g., [Reference Simon22, Theorem 2.11]).

Now, to prove (iii), we use the mentioned above sufficient condition for an integral operator to be in a trace class. Since $\mathbf {K}(s,t)$ is a positive-definite kernel, it suffices to note that the second condition holds, since $\mathbf {K}(s,s)=K(0)=0$ . This completes the proof.

Since ${\mathrm {dH}}$ is invariant, choosing $r=1$ , we get the following corollary.

Proof The condition (16) is valid for $r=1$ , since for all $z\in {\mathbb B}^n,$

$$ \begin{align*} \int_{{\mathbb B}^n}\frac{|K(\varphi_z(w))|}{(1-|w|^2)^{n+1}}{\mathrm{dA}}(w)&=\int_{{\mathbb B}^n}|K(\varphi_z(w))|{\mathrm{dH}}(w)\\ &=\int_{{\mathbb B}^n}|K(w)|{\mathrm{dH}}(w)=:c_1<\infty. \end{align*} $$

It remains to note that in our case $(1-\sigma /p)p'<1$ .

Example 12 The Hausdorff–Berezin operator with kernel

$$ \begin{align*}K(z)=(1-|z|^2)^\alpha\end{align*} $$

is a Hilbert–Schmidt operator in $L^2({\mathbb B}^n,{\mathrm {dA}})$ if and only if $\alpha>n+1/2$ .