Proof The map $f\mapsto T_gf$ is linear. Therefore, it is clear that the topology in $[g:H^q]$ is given by a norm or a quasi-norm, whenever the topology of $H^q$ is. We next investigate the completeness of $[g:H^q]$ . The argument is largely the same as in Theorem 1 of [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis10].

Let $f \in [g:H^q]$ . Let $Z_{Rg}=\{z\in \mathbb {B}_n: Rg(z)=0\}$ . For each point $z \notin Z_{Rg}$ , we have

$$ \begin{align*}|f(z)|\leq C_z \|f\|_{[g:H^q]},\end{align*} $$

which follows from using the boundedness of the functional $h\mapsto Rh(z)$ on $H^q$ . The positive constants $C_z$ depend on $1/|Rg(z)|$ and q but vary continuously on $\mathbb {B}_n\setminus Z_{Rg}$ .

Let now $z \in Z_{Rg}$ . This case needs some extra effort, since the zeros of holomorphic functions of several complex variables need not be isolated inside the domain. We proceed as follows. Since g is not constant, there exists $\zeta \in \mathbb {C}^n$ with $|\zeta |=1$ , and $\varepsilon>0$ , so that the one variable analytic slice $h_\zeta (w)=Rg(z+w\zeta )$ satisfies $|h_\zeta (w)|>\varepsilon $ when $r<|w|<R$ for some $0<r<R$ . We can use subharmonicity of $\log |h_\zeta |$ to arrive at an upper bound for $|f(z)|$ similar to that in [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis10]. This upper bound can be chosen to be uniform in some neighborhood of z (regardless if the new point belongs to $Z_{Rg}$ or not).

From these observations, it follows that Cauchy sequences in $[g:H^q]$ converge to a holomorphic function uniformly on compact subsets. Now that this has been achieved, the rest of the argument in the cited paper carries over.

For the second claim of the proof, the reader can follow the argument of Theorem 4 in [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis10], which also works in the present setting.

Proof One can argue as in [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis10]. Indeed, for any $g \in H^r$ , it is trivially true that $H^p$ is contained in $[g:H^q]$ . If there were a $g \in H^r$ with $H^p=[g:H^q]$ , the identity map would be bounded and onto $H^p\to [g:H^q]$ . The open mapping theorem (which holds also for F-spaces [Reference Rudin31]) then implies that $T_g:H^p\to H^q$ is bounded below.

Since $H^p$ is a subspace of $[g:H^q]$ and embeds there boundedly, if there was a $g \in H^r$ with $H^p=[g:H^q]$ , one would have that $T_g:H^p\to H^q$ is bounded below. It is well-known [Reference Pau27] that for this range of p and q, $T_g:H^p\to H^q$ is bounded if and only if it is compact. A compact operator cannot be bounded below.

Proof We follow the reasoning in [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis10]. Let $ g\in V_{p,q} = H^r$ and $f\in [H^r: H^q]$ . Now, $ T_g f $ belongs to $H^q$ , or equivalently, $S_f g \in H^q$ . It follows from the closed graph theorem (for F-spaces if necessary; see [Reference Rudin31]) that $S_f:H^r\to H^q$ is bounded. The claim now follows from the next theorem.

Proof We note that this result is probably known to experts, at least when $n=1$ . Since we have been unable to find its proof in the literature, one is given here.

We first consider the sufficiency. In fact, it follows easily from the corresponding results for $T_g$ and the multiplication operator $M_g$ . We provide a direct proof, similar to one for $T_g$ in [Reference Pau27], by methods that will be used for the necessity. For $\gamma>1$ and $|\zeta |=1$ , we denote by $\Gamma (\zeta )=\Gamma _{\gamma }(\zeta )$ the approach region

$$ \begin{align*}\Gamma_{\gamma}(\zeta)=\left\lbrace z \in \mathbb{B}_n:|1-\langle z,\zeta\rangle|<\frac{\gamma}{2}(1-|z|^2)\right\rbrace.\end{align*} $$

For $\beta \in \mathbb {R}$ , we write $dV_\beta (z)=(1-|z|^2)^\beta dV(z)$ , where V is the normalized $2n$ -dimensional Lebesgue measure on $\mathbb {B}_n$ . Now assume $g\in H^r $ with $ r= pq/(p-q)$ . If $f\in H^p $ , then by the area function description of Hardy spaces [Reference Ahern and Bruna1, Reference Pau27], we have

$$ \begin{align*} \|S_g f\|^q_{H^q} &\asymp \int_{\mathbb{S}_n} \left(\int_{\Gamma(\zeta)} |Rf(z) g(z) |^2 dV_{1-n}(z) \right)^{q/2} d\sigma(\zeta)\\ &\lesssim \int_{\mathbb{S}_n}\left( \sup_{z\in\Gamma(\zeta)} |g(z)|\right)^q \left(\int_{\Gamma(\zeta)} |Rf(z)|^2 dV_{1-n}(z)\right)^{q/2}d\sigma(\zeta). \end{align*} $$

We use Hölder’s inequality and get

$$ \begin{align*} \|S_g f\|^q_{H^q} \lesssim &\left( \int_{\mathbb{S}_n}\left( \sup_{z\in\Gamma(\zeta)} |g(z)|\right)^{pq/(p-q)} d\sigma(\zeta)\right)^{1-q/p} \\&\times\left(\int_{\mathbb{S}_n}\left(\int_{\Gamma(\zeta)} |Rf(z)|^2 dV_{1-n}(z)\right)^{p/2}d\sigma(\zeta)\right)^{q/p}. \end{align*} $$

Equivalently, this means

$$ \begin{align*}\|S_g f\|^q_{H^q} \lesssim \|g\|^q_{H^r} \|f\|^q_{H^p},\end{align*} $$

giving the sufficiency.

For the necessity, we proceed with the argument from [Reference Miihkinen, Pau, Perälä and Wang25], based on the Kahane and Khinchine inequalities [Reference Luecking24], atomic decomposition [Reference Coifman and Rochberg13, Reference Pau27], area function description of Hardy spaces [Reference Ahern and Bruna1, Reference Pau27], and factorization of tent spaces [Reference Cohn and Verbitsky11, Reference Coifman, Meyer and Stein12, Reference Miihkinen, Pau, Perälä and Wang25]. The proof follows the outline of either Theorem 7 or Theorem 8 of [Reference Miihkinen, Pau, Perälä and Wang25], except that it is simpler. The proof is somewhat sketchy; an interested reader can take a look at [Reference Miihkinen, Pau, Perälä and Wang25].

Let $Z=\{a_k\}\subseteq \mathbb {B}_n$ be a $\rho $ -lattice in the Bergman metric. Most of the argument below works for any $\rho>0$ , but in the very last stage, we shall assume that $\rho $ is very small. Let $b>n\max (1,2/p)$ . We consider a family of test functions of the form

$$ \begin{align*}F_t(z) = \sum_k \lambda_k r_k(t) \left(\frac{1-|a_k|^2}{1-\langle z,a_k\rangle} \right)^b,\end{align*} $$

where the sequence $ \{\lambda _k\} $ belongs to $ T_2^p$ , a tent space of sequences indexed by the lattice, and $r_k : [0,1]\to \{-1 , +1\} $ are the Rademacher functions. By the comment at the end of Section 1, we may assume that g is already in $H^r$ and focus on establishing $\|g\|_{H^r}\lesssim \|S_g\|$ . The general claim follows from pointwise convergence and a limit argument. For essentially the same reason, we may assume that the sum defining $F_t$ is a finite one. This is a valid simplification, since replacing the elements of $\{\lambda _k\}$ by zeros cannot increase the $T^p_2$ norm, and the sum converges pointwise.

By the area function description of the Hardy spaces, integrating with respect to t, where $ t\in [0,1], $ and using Fubini’s theorem, we have

$$ \begin{align*} \int_{\mathbb{S}_n} \int_{0}^{1} &\left( \int_{\Gamma(\zeta)} \left|g(z) \sum_{k} \lambda_k r_k(t) |\langle z,a_k\rangle|\dfrac{(1-|a_k|^2)^b}{(1-\langle a_k,z\rangle)^{b+1}} \right|{}^2 dV_{1-n}(z)\right)^{q/2} dt \,d\sigma(\zeta)\\ &\lesssim \|S_g\|^q \,\|\lambda\|^q_{T^p_2}. \end{align*} $$

Now, using Kahane’s inequality and Fubini’s theorem, this gives

$$ \begin{align*} \int_{\mathbb{S}_n} &\left( \int_{\Gamma(\zeta)}\int_{0}^{1} \left|g(z) \langle z,a_k\rangle \sum_{k} \lambda_k r_k(t) \dfrac{(1-|a_k|^2)^b}{(1-\langle z,a_k\rangle)^{b+1}} \right|{}^2 dt \,dV_{1-n}(z)\right)^{q/2} \,d\sigma(\zeta)\\ &\lesssim \|S_g\|^q \,\|\lambda\|^q_{T^p_2}. \end{align*} $$

We use Khinchine’s inequality to obtain

$$ \begin{align*} \int_{\mathbb{S}_n} &\left( \int_{\Gamma(\zeta)} \sum_{k} |\lambda_k|^2 |\langle z,a_k\rangle|^2 |g(z)|^2 \dfrac{(1-|a_k|^2)^{2b}}{|1-\langle z,a_k\rangle|^{2b+2}} \,dV_{1-n}(z)\right)^{q/2} \,d\sigma(\zeta)\\ &\lesssim \|S_g\|^q \,\|\lambda\|^q_{T^p_2}. \end{align*} $$

Replacing the integral over $ \Gamma (\zeta ) $ by an integral over Bergman metric balls $B(a_k, \rho )$ , and switching to an approach region with a smaller aperture, also denoted by $\Gamma (\zeta )$ ,

$$ \begin{align*} \int_{\mathbb{S}_n} &\left( \sum_{{a_k}\in \Gamma(\zeta)} |\lambda_k|^2 \int_{B(a_k, r)}|\langle z,a_k\rangle |^2 |g(z)|^2 \dfrac{(1-|a_k|^2)^{2b}}{|1-\langle z,a_k\rangle|^{2b+2}} \,dV_{1-n}(z)\right)^{q/2} \,d\sigma(\zeta)\\ &\lesssim \|S_g\|^q \,\|\lambda\|^q_{T^p_2}. \end{align*} $$

We proceed by using subharmonicity and $|1-\langle z,a_k\rangle | = 1-|a_k|^2 $ for every $ z\in B(a_k, r)$ , arriving at the estimate

(2.1) $$ \begin{align} \int_{\mathbb{S}_n} \left( \sum_{{a_k}\in \Gamma(\zeta)} |\lambda_k|^2 |a_k|^4 |g(a_k)|^2 \right)^{q/2} \,d\sigma(\zeta) \lesssim \|S_g\|^q \,\|\lambda\|^q_{T^p_2}. \end{align} $$

Denoting $\nu _k = |a_k|^2 |g(a_k)|,$ we want to prove that $ \{\nu _k\} \in T^r_\infty $ . This would lead to a discrete way to express that $g\in H^r$ .

Let $s>0$ be big enough so that $2s>1$ and $qs>1$ hold. We are done if we can prove that $\{\nu _k^{1/s}\}\in T^{rs}_\infty $ . By duality of tent spaces (see [Reference Arsenović8, Reference Jevtić21]), since $rs>1$ , this can be shown if we prove

$$ \begin{align*}\left|\sum_k \mu_k \nu_k^{1/s}(1-|a_k|^2)^n\right|\lesssim \|\mu\|_{T^{(rs)'}_1}.\end{align*} $$

To this end, we assume that $\lambda _k$ and $\tau _k$ are positive. Indeed, since $\nu _k$ are positive, the above dual pairing is clearly at its biggest for a positive sequence $\{\mu _k\}$ , and the sign change on individual elements does not affect the norm of the corresponding tent space. We get

$$ \begin{align*}\sum_k \tau_k \lambda_k^{1/s} \nu_k^{1/s} ( 1- |a_k|^2)^n \asymp \int_{\mathbb{S}_n} \left( \sum_{{a_k}\in \Gamma(\zeta)} \tau_k \lambda_k^{1/s} \nu_k^{1/s} \right) \,d\sigma(\zeta).\end{align*} $$

Next, using Hölder’s inequality, yields

$$ \begin{align*} &\sum_k \tau_k \lambda_k^{1/s} \nu_k^{1/s} ( 1- |a_k|^2)^n \\ &\quad\lesssim \int_{\mathbb{S}_n} \left( \sum_{{a_k}\in \Gamma(\zeta)} \tau_k^{2s/(2s-1)} \right)^{(2s-1)/2s} \left(\sum_{{a_k}\in \Gamma(\zeta)} \lambda_k^2 \nu_k^2 \right)^{1/(2s)} \,d\sigma(\zeta). \end{align*} $$

Consider $ 2s/(2s-1) = (2s)' $ and doing another H $ \ddot{\text{l}} $ der’s inequality for $qs$ and $qs/(qs-1) = (qs)'$ gives us now that the expression is dominated by

$$ \begin{align*}\left( \int_{\mathbb{S}_n} \left( \sum_{{a_k}\in \Gamma(\zeta)} \tau_k^{\frac{2s}{2s-1}} \right)^{\frac{(qs)'}{(2s)'}}\,d\sigma(\zeta)\right)^{\frac{1}{(qs)'}} \left( \int_{\mathbb{S}_n} \left(\sum_{{a_k}\in \Gamma(\zeta)} \lambda_k^2 \nu_k^2 \right)^{\frac{q}{2}} \,d\sigma(\zeta)\right)^{\frac{1}{qs}}.\end{align*} $$

The first part is the norm $\|\tau \|_{T^{(qs)'}_{(2s)'}}.$ By using (2.1) for the second part, to get the upper bound $\|S_g\|^{1/s} \|\lambda \|^{1/s}_{T^p_2} = \|S_g\|^{1/s} \|\lambda \|_{T^{ps}_{2s}}$ . Altogether, we have

$$ \begin{align*}\sum_k \tau_k \lambda_k^{1/s} \nu_k^{1/s} ( 1- |a_k|^2)^n \lesssim \|S_g\|^{1/s} \|\lambda^{1/s}\|_{T^{ps}_{2s}} \|\tau\|_{T^{(qs)'}_{(2s)'}}.\end{align*} $$

By factorization of tent spaces (essentially due to Cohn and Verbitsky [Reference Cohn and Verbitsky11], but the present variant is taken from [Reference Miihkinen, Pau, Perälä and Wang25]) and considering $\mu _k =\tau _k \lambda _k^{1/s}$ as an arbitrary positive element of the space $T^{(rs)'}_1,$ using duality, we have $\{\nu _k^{1/s}\} \in T^{rs}_\infty $ . By the discussion and calculation above, this leads to $\|\nu \|_{T^r_\infty } \lesssim \|S_g\|,$ which is what we wanted to prove.

Now, choosing $\rho>0$ small enough and invoking Lemma 3 of [Reference Miihkinen, Pau, Perälä and Wang25], we get

$$ \begin{align*}\int_{\mathbb{S}_n}\left(\sup_{z\in \Gamma(\zeta)}|z_i|^2 |g(z)|\right)^rd\sigma(\zeta)<\infty\end{align*} $$

for every $i\in \{1,2,\ldots ,n\}$ (since $z_i^2g(z)$ is holomorphic), yielding that

(2.2) $$ \begin{align}\int_{\mathbb{S}_n}\left(\sup_{z\in \Gamma(\zeta)}|z|^2 |g(z)|\right)^rd\sigma(\zeta)<\infty. \end{align} $$

By the admissible maximal function description of Hardy spaces, we see that $H^r$ and the tent space of holomorphic functions defined by (2.2) are in bijective correspondence. By the bounded inverse theorem for F-spaces, we get that $\|g\|_{H^r}\lesssim \|S_g\|$ .

Proof Assume $p<q$ and $g\in \mathcal {B}^{\alpha }$ . We consider two cases, $q>2$ and $q\leq 2$ . For the case $q>2$ , let $s\in (2,q)\cap (p,q)$ . Then we have $2<s<q$ and from Theorem 1(1) of [Reference Miihkinen, Pau, Perälä and Wang25], it follows that $T_g: A^s_{n\frac {s}{p}-n-1} \to H^q$ is bounded. Therefore, $ A^s_{n\frac {s}{p}-n-1} \subseteq [g : H^q]$ .

Let us next assume that $q\leq 2 $ . Here, we could also proceed by taking $s\in (p,q)$ , in which case $s\leq \min \{2,q\}$ . But in fact, we can let $s=q$ , obtaining a larger space. According to case (1) of Theorem 1 in [Reference Miihkinen, Pau, Perälä and Wang25] (in fact, in the unit disk, this case was obtained earlier by Wu [Reference Wu33]), the boundedness of $T_g : A^q_{n\frac {q}{p}-n-1} \to H^q$ is equivalent to $ g\in \mathcal {B}^{\alpha }$ .

Now, by using (3.1), the space $H^p$ embeds into $A^s_{n\frac {s}{p}-n-1}$ or $A^q_{n\frac {q}{p}-n-1}$ , whichever we want. The inclusion is also strict as Bergman functions need not have boundary values almost everywhere. Therefore, the proof is complete.

Proof We first prove that $AT^p_{2,1-2\alpha }$ is contained in $[\mathcal {B}^\alpha :H^q]$ . Using the area description of $H^q$ , one obtains

$$ \begin{align*}\|T_gf\|_{H^q}^q\asymp\int_{\mathbb{S}_n}\left(\int_{\Gamma(\zeta)}|f(z)|^2|Rg(z)|^2 dV_{1-n}(z)\right)^{q/2}d\sigma(\zeta).\end{align*} $$

Bearing in mind $g \in \mathcal {B}^\alpha $ , we use the pointwise estimate

$$ \begin{align*}|Rg(z)|\leq (1-|z|^2)^{-\alpha}\|g\|_{\mathcal{B}^\alpha}.\end{align*} $$

We get

$$ \begin{align*}\|T_gf\|_{H^q}^q \lesssim \|g\|_{\mathcal{B}^\alpha}^q\int_{\mathbb{S}_n}\left(\int_{\Gamma(\zeta)}|f(z)|^2 (1-|z|^2)^{-2\alpha}dV_{1-n}(z)\right)^{q/2}d\sigma(\zeta).\end{align*} $$

This gives exactly that $T_g:AT^q_{2,1-2\alpha }\to H^q$ is bounded. Observe that $\alpha <1$ , so $1-2\alpha>-1$ , and the space is nontrivial. We see that $AT^p_{2,1-2\alpha }$ is contained in $[\mathcal {B}^\alpha :H^q]$ .

We prove the other inclusion. It suffices to show that $S_g:\mathcal {B}^\alpha \to H^q$ being bounded implies that $g \in AT^q_{2,1-2\alpha }$ . So let us assume that $S_g$ is bounded. Let us first deal with the case $\alpha =0$ . In that case, we test with $F_i(z)=z_i$ and use the area description of $H^q$ as above. This shows that $z_i g(z)$ belongs to $AT^q_{2,1}$ for every $i=1,2,\ldots ,n$ . As in the end of the proof of Theorem 2.3, this leads to $g \in AT^q_{2,1}$ .

We now turn to the case $0<\alpha <1$ . The idea is similar to that of Theorem 2.3, and to this end, we use the atomic decomposition of $\mathcal {B}^\alpha $ . This result can be found in [Reference Zhao and Zhu34] (if $0<\alpha <1$ , the space $\mathcal {B}^\alpha $ is the same as the Lipschitz space $\Lambda _{1-\alpha }$ in the notation and terminology of the reference), where it is explained how it directly follows from the atomic decomposition of the Bloch space [Reference Zhu35]. Let $\{a_k\}$ be a $\rho $ -lattice in the Bergman metric. For every $\{\lambda _k\}\in \ell ^\infty $ , the functions

$$ \begin{align*}F_t(z)=\sum_{k}\lambda_k r_k(t)\frac{(1-|a_k|^2)^b}{(1-\langle z,a_k\rangle)^{b+\alpha-1}}\end{align*} $$

belong to $\mathcal {B}^\alpha $ , when $b>n$ . The functions $r_k$ are the Rademacher functions like in Theorem 2.3.

By using the set of techniques demonstrated in the proof of Theorem 2.3, and the calculations leading up to (2.1) to control $S_gF_t$ , one arrives at the estimate

(3.2) $$ \begin{align} \int_{\mathbb{S}_n}\left(\sum_{a_k \in \Gamma(\zeta)} |\lambda_k|^2 |a_k|^4 |g(a_k)|^{2} (1-|a_k|^2)^{-2\alpha+2}\right)^{q/2}d\sigma(\zeta)\lesssim \|S_g\|^q \,\|\lambda\|^q_{\ell^\infty}. \end{align} $$

If we select $\lambda _k=1$ for every k, we get the discrete variant of what we want. Assuming $\rho>0$ is small enough, we can use Lemma 3 of [Reference Miihkinen, Pau, Perälä and Wang25] and reason as in the proof of Theorem 2.3 to get that $ g \in AT^q_{2,1-2\alpha }$ .

The latter claim is clear: If $\alpha <0$ , then $\mathcal {B}^\alpha $ consists of the constant functions only, and $T_g$ is the zero operator.

Proof By the same reasoning as in Theorem 2 of [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis10] – see also the proof of Theorem 2.1 in the present article – it suffices to show that $T_g:H^p\to H^p$ is not bounded below.

We first consider the case $p=2$ . If $f \in H^2$ , then the $H^2$ norm of $T_gf$ is comparable to

(4.1) $$ \begin{align} \int_{\mathbb{B}_n}|f(z)|^2 |Rg(z)|^2 (1-|z|^2)dV(z). \end{align} $$

Note that $g \in BMOA$ if and only if the measure $d\mu _g(z)=|Rg(z)|^2(1-|z|^2)dV(z)$ is a Carleson measure, meaning that $H^2$ embeds boundedly into $L^2_{\mu _g}$ (see Theorem 5.14 of [Reference Zhu35]). Now, let $\varphi $ be an inner function on the unit ball, and consider $f_k=\varphi ^k$ . With these choices, one has $|\varphi (z)|< 1$ for every point $z \in \mathbb {B}_n$ and $\|f_k\|_{H^2}=1$ for every k. By the dominated convergence theorem, the equivalent norms (4.1) tend to zero as $k\to \infty $ . This shows that $T_g:H^2\to H^2$ is not bounded below.

Now, if $p\neq 2$ , the argument of Lemma 1 of [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis10] shows how to reduce this case to the case $p=2$ above.

Proof We only need to show that each element in $[BMOA:H^p]$ is an $H^p$ function. To show this, it is sufficient to show that the boundedness of $S_g:BMOA\to H^p$ implies $g \in H^p$ .

Again, we invoke the machinery used in Theorem 2.3. Let again $Z=\{a_k\}$ be a $\rho $ -lattice in the Bergman metric. By Theorem 5.29 of [Reference Zhu35], if $b>n$ , then the function

$$ \begin{align*}F_t(z)=\sum_k \lambda_k r_k(t)\left(\frac{1-|a_k|^2}{1-\langle z,a_k\rangle}\right)^b\end{align*} $$

belongs to $BMOA$ , if the sequence $\{\lambda _k\}$ is chosen so that

$$ \begin{align*}d\mu_{\lambda}=\sum_k |\lambda_k|^2 (1-|a_k|^2)^n d\delta_{a_k}\end{align*} $$

is a Carleson measure. In the language of tent spaces, this means that $\{\lambda _k\} \in T^{\infty }_2$ or $\{\lambda _k^2\} \in T^\infty _1$ . This is a well-known fact about tent spaces (see Section 2.6 of [Reference Miihkinen, Pau, Perälä and Wang25] for the formulation on the unit ball).

We assume boundedness of $S_g:BMOA\to H^p$ , test with $S_gF_t$ , and estimate as we did leading up to (2.1), to obtain

(4.2) $$ \begin{align} \int_{\mathbb{S}_n} \left( \sum_{{a_k}\in \Gamma(\zeta)} |\lambda_k|^2 |a_k|^4 |g(a_k)|^2 \right)^{p/2} \,d\sigma(\zeta) \lesssim \|S_g\|^p \,\|\lambda\|^p_{T^\infty_2}. \end{align} $$

Now, writing $\nu _k=|a_k|^2 |g(a_k)|$ , we aim to show that $\{\nu _k\} \in T^p_\infty $ . This is the same as showing that $\{\nu _k^{2/s}\}\in T^{ps/2}_\infty $ . We choose s big enough so that $ps/2>1$ and $s>1$ hold true. This means that we need to show that $\{\nu _k^{2/s}\}$ belongs to the dual of $T^{(ps/2)'}_1$ . According to the formula in the beginning of the proof of Proposition 6 in [Reference Miihkinen, Pau, Perälä and Wang25], we can factorize

$$ \begin{align*}T^{(ps/2)'}_1=T^{(ps/2)'}_\infty \cdot T^\infty_1.\end{align*} $$

Therefore, we look at

$$ \begin{align*}\sum_k \tau_k \lambda_k^2 \nu_k^{2/s} (1-|a_k|^2)^n\asymp \int_{\mathbb{S}_n}\left(\sum_{a_k \in \Gamma(\zeta)}\tau_k\lambda_k^2 \nu_k^{2/s}\right)d\sigma(\zeta),\end{align*} $$

where $\{\tau _k\} \in T^{(ps/2)'}_\infty $ .

We now make the split $\lambda _k^2=\lambda _k^{2/s'}\lambda _k^{2/s}$ , observe that $\{\lambda _k^{2/s'}\}\in T^\infty _{s'}$ , and conclude that if $\alpha _k=\tau _k \lambda _k^{2/s'}$ , then $\{\alpha _k\}\in T^{(ps/2)'}_{s'}$ .

With these preparations, we do Hölder’s inequality with $s'$ and s and arrive at

$$ \begin{align*}\int_{\mathbb{S}_n}\left(\sum_{a_k \in \Gamma(\zeta)}\tau_k\lambda_k^2 \nu_k^{2/s}\right)d\sigma(\zeta)\leq \int_{\mathbb{S}_n}\left(\sum_{a_k \in \Gamma(\zeta)} \alpha_k^{s'}\right)^{1/s'}\left(\sum_{a_k \in \Gamma(\zeta)} \lambda_k^2 \nu_k^2 \right)^{1/s}d\sigma(\zeta).\end{align*} $$

Finally, we do another Hölder’s inequality with $(ps/2)'$ and $ps/2$ , so that we have proven

$$ \begin{align*}\sum_k \tau_k \lambda_k^2 \nu_k^{2/s} (1-|a_k|^2)^n\leq \|\alpha\|_{T^{(ps/2)'}_{s'}}\left(\int_{\mathbb{S}_n} \left( \sum_{{a_k}\in \Gamma(\zeta)} \lambda_k^2 \nu_k^2 \right)^{p/2} \,d\sigma(\zeta)\right)^{2/ps}.\end{align*} $$

Since

$$ \begin{align*}\|\alpha\|_{T^{(ps/2)'}_{s'}}\lesssim \|\tau\|_{T^{(ps/2)'}_\infty}\cdot \|\lambda^{2/s'}\|_{T^\infty_{s'}},\end{align*} $$

we return to the estimate (4.2) and see that

$$ \begin{align*}\sum_k \tau_k \lambda_k^2 \nu_k^{2/s} (1-|a_k|^2)^n \lesssim \|\tau\|_{T^{(ps/2)'}_\infty}\cdot \|\lambda^{2}\|_{T^\infty_{1}}^{1/s'}\cdot\|\lambda^2\|_{T^\infty_1}^{1/s}\|S_g\|^{2/s}.\end{align*} $$

By factorization, we finally get

$$ \begin{align*}\sum_k \tau_k \lambda_k^2 \nu_k^{2/s} (1-|a_k|^2)^n\lesssim \|\tau \lambda^2\|_{T^{(ps/2)'}_1}\|S_g\|^{2/s}.\end{align*} $$

Therefore, $\{\nu _k^{2/s}\}$ must belong to $T^{ps/2}_\infty $ , with the norm bound $\|\nu ^{2/s}\|_{T^{(ps/2)}_\infty }\lesssim \|S_g\|^{2/s}$ . This is equivalent to the membership of $\{\nu _k\}$ in $T^p_\infty $ and $\|\nu \|_{T^p_\infty }\lesssim \|S_g\|$ . We can use the same reasoning as in the end of the proof of Theorem 2.3 to see how this leads to $g \in H^p$ with the appropriate norm bound.