Proof of Theorem 5

We first consider the symmetric case.

(i) Since $\epsilon _{\alpha }(\omega ')X_{\alpha }(\omega )$ and $X_{\alpha }(\omega )$ are identically distributed, provided that $\{\epsilon _\alpha \}_{\alpha \in \mathbb {N}^n}$ is a Rademacher sequence on a probability space $\left (\Omega ',\mathcal {F}',\mathbb {P}'\right )$ and independent to $\{X_\alpha \}_{\alpha \in \mathbb {N}^n}$ . Let

$$ \begin{align*}\mathcal{R}_{\epsilon}\mathcal{R}f(z)=\sum_{\alpha\in\mathbb{N}^n}a_{\alpha}\epsilon_{\alpha}X_{\alpha}z^{\alpha}.\end{align*} $$

For $q\leq p$ , by Hölder’s inequality and Khintchine’s inequality, we have

(3) $$ \begin{align} \mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_{\epsilon}\mathcal{R}f) & \lesssim\left(\int_{\partial \mathbb{B}_n}\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_\alpha X_\alpha \zeta^{\alpha}|^2r^{2m}\right)^{\frac{p}{2}}d\sigma(\zeta)\right)^{\frac{q}{p}} \nonumber\\ & \leq\left(\int_{\partial \mathbb{B}_n}\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_{\alpha}X_{\alpha}\zeta^{\alpha}|^2r^{2m}d\sigma(\zeta)\right)^{\frac{q}{2}} \nonumber\\ & =M_2^q(r,\mathcal{R}f). \end{align} $$

For $q>p$ , by Minkowski’s inequality and Khintchine’s inequality, we have

(4) $$ \begin{align} \left(\mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_{\epsilon}\mathcal{R}f)\right)^{\frac{p}{q}} & \leq\int_{\partial \mathbb{B}_n}\left(\mathbb{E}_{\epsilon}|\mathcal{R}_{\epsilon}\mathcal{R}f|^q\right)^{\frac{p}{q}}d\sigma(\zeta) \nonumber \\ & \approx\int_{\partial \mathbb{B}_n}\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_{\alpha}X_{\alpha}\zeta^{\alpha}|^2r^{2m}\right)^{\frac{p}{2}}d\sigma(\zeta) \nonumber \\ & \leq M_2^p(r,\mathcal{R}f). \end{align} $$

So we have

(5) $$ \begin{align} \mathbb{E}_X\lVert\mathcal{R}f\rVert_{p,q,\varphi}^q & =\mathbb{E}_X\mathbb{E}_\epsilon\|\mathcal{R}_\epsilon\mathcal{R} f\|_{p,q,\varphi}^q \nonumber\\ & \lesssim\int_{0}^{1}\frac{\varphi^q(r)}{1-r}\mathbb{E}_XM_2^q(r,\mathcal{R} f)dr \nonumber\\ & \leq\left(\mathbb{E}|X_0|^\tau\right)^{\frac{q}{\tau}}\lVert f\rVert_{2,q,\varphi}^q, \end{align} $$

where the “ $\lesssim $ ” follows from (3) and (4), the “ $\leq $ ” is obtained by applying Hölder’s inequality for $0<q\leq 2$ and Minkowski’s inequality for $2<q<\infty $ .

(ii) Similarly, we have

$$ \begin{align*} \mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_{\epsilon}\mathcal{R}f) & \lesssim\left(\int_{\partial \mathbb{B}_n}\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_{\alpha}X_{\alpha}\zeta^{\alpha}|^2r^{2m}\right)^{\frac{p}{2}}d\sigma(\zeta)\right)^{\frac{q}{p}} \\ & \leq\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_{\alpha}X_{\alpha}|^2r^{2m}\left(\int_{\partial \mathbb{B}_n}|\zeta^{\alpha}|^pd\sigma(\zeta)\right)^{\frac{2}{p}}\right)^{\frac{q}{2}} \end{align*} $$

by Hölder’s inequality and Khintchine’s inequality when $q\leq p$ and by Minkowski’s inequality and Khintchine’s inequality when $q>p$ . Let

$$ \begin{align*}I(p):=\int_{\partial \mathbb{B}_n}|\zeta^{\alpha}|^pd\sigma(\zeta)=\frac{(n-1)!\prod_{j=1}^{n}\Gamma(\frac{p\alpha_j}{2}+1)}{\Gamma(n+\frac{p|\alpha|}{2})},\ 0<p<\infty.\end{align*} $$

Let $t_j:=|\zeta _j|^2$ and $\Delta _{n-1}:=\{t\in \mathbb {R}^n:\ t_j\geq 0$ , $\sum _{j=1}^{n}t_j=1\}$ . Then

$$ \begin{align*}|\zeta^{\alpha}|^p=\prod_{j=1}^{n}|\zeta_j|^{p\alpha_j}=\prod_{j=1}^{n}t_j^{\frac{p}{2}\alpha_j}\end{align*} $$

and by $Dirichlet$ - $\beta $ integral [Reference Andrews, Askey and Roy1, p. 32],

$$ \begin{align*}\int_{\partial \mathbb{B}_n}|\zeta^{\alpha}|^pd\sigma(\zeta)=(n-1)!\int_{\Delta_{n-1}}\prod_{j=1}^{n}t_j^{\frac{p}{2}a_{\alpha}}dt=\frac{(n-1)!\prod_{j=1}^{n}\Gamma(\frac{p}{2}\alpha_j+1)}{\Gamma(\frac{p}{2}|\alpha|+n)}.\end{align*} $$

We have

$$ \begin{align*}\frac{(I(p))^{\frac{2}{p}}}{I(2)}\lesssim|\alpha|^{(n-1)(1-\frac{2}{p})}\end{align*} $$

by Stirling’s formula. Since

$$ \begin{align*}\frac{\Gamma(|\alpha|+\nu+1)}{\Gamma(|\alpha|+1)}\approx|\alpha|^{\nu},\ \ \ |\alpha|\to\infty,\end{align*} $$

and

$$ \begin{align*}M_2^q(r,\mathcal{R} f^{[\nu]})\approx\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|\alpha|^{2\nu}|a_{\alpha}X_\alpha|^2r^{2m}\left(\int_{\partial \mathbb{B}_n}|\zeta^{\alpha}|^2d\sigma(\zeta)\right)\right)^{\frac{q}{2}},\end{align*} $$

we have

(6) $$ \begin{align} \mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_{\epsilon}\mathcal{R}f)\lesssim M_2^q(r,\mathcal{R}f^{[\nu]}), \end{align} $$

where $\nu \geq \frac {n-1}{2}(1-\frac {2}{p})$ . By Lemma 8,

(7) $$ \begin{align} \lVert f\rVert_{2,q,\varphi}\approx\lVert f^{[\nu]}\rVert_{2,q,(1-r)^\nu\varphi}. \end{align} $$

So,

(8) $$ \begin{align} \mathbb{E}\lVert\mathcal{R}f\rVert_{p,q,(1-r)^{\nu}\varphi}^q & \lesssim\mathbb{E}\lVert\mathcal{R}f^{[\nu]}\rVert_{2,q,(1-r)^{\nu}\varphi}^q \nonumber\\ & \lesssim\lVert f^{[\nu]}\rVert_{2,q,(1-r)^{\nu}\varphi}^q \nonumber\\ & \approx\lVert f\rVert_{2,q,\varphi}^q, \end{align} $$

where the first “ $\lesssim $ ” holds by (6), the second one by Minkowski’s inequality and $X_0\in L^{q}(\Omega )$ for $2<q<\infty $ , and by Hölder’s inequality and $X_0\in L^2(\Omega )$ for $0<q\leq 2$ , and the last “ $\approx $ ” due to (7).

For the non-symmetric case.

(i) We know that $H^{2,q,\varphi }(\mathbb {B}_n)\subset \left (H^{2,q,\varphi }(\mathbb {B}_n)\right )_{\star }$ by the proof of symmetric case. Since $M_p^q(r,f)\leq M_2^q(r,f)$ for $0<p\leq 2$ , we have $\left (H^{2,q,\varphi }\left (\mathbb {B}_n\right )\right )_{\star }\subset \left (H^{p,q,\varphi }\left (\mathbb {B}_n\right )\right )_{\star }$ . So,

$$ \begin{align*}H^{2,q,\varphi}(\mathbb{B}_n)\subset\left(H^{p,q,\varphi}(\mathbb{B}_n)\right)_{\star}.\end{align*} $$

(ii) Let $\{\widetilde {X_\alpha }\}_{\alpha \in \mathbb {N}^n}$ be an independent copy of $\{X_\alpha \}_{\alpha \in \mathbb {N}^n}$ . Then $X_\alpha -\widetilde {X_\alpha }$ is symmetric. Since $\mathbb {E}X_\alpha =0$ , we have

$$ \begin{align*}(\mathcal{R}f)_{\omega}(z)=\int_{\Omega}\left((\mathcal{R}f)_{\omega}(z)-(\widetilde{\mathcal{R}f})_{\omega'}(z)\right)d\mathbb{P}(\omega'),\end{align*} $$

where $(\mathcal {R} f)_{\omega }(z)=(\mathcal {R} f)(z)$ , $(\widetilde {\mathcal {R} f})_{\omega '}(z)=\sum _{\alpha \in \mathbb {N}^n}a_{\alpha }\widetilde {X_\alpha }(\omega ')z^\alpha $ , and

$$ \begin{align*}\lVert(\mathcal{R}f)_{\omega}\rVert_{p,q,(1-r)^{\nu}\varphi}^q=\int_{0}^{1}\frac{\varphi(r)^q}{(1-r)^{q\nu-1}}M_p^q\left(r,\mathbb{E}_{\omega'}\left((\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)dr.\end{align*} $$

Since $2<p<\infty $ , by Minkowski’s inequality, we have

$$ \begin{align*}M_p\left(r,\mathbb{E}_{\omega'}\left((\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)\leq\mathbb{E}_{\omega'}\left(M_p(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'})\right).\end{align*} $$

If $1\leq q<\infty $ , by Jensen’s inequality,

$$ \begin{align*}\left(\mathbb{E}_{\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)\right)^q\leq \mathbb{E}_{\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)^q.\end{align*} $$

Then

(9) $$ \begin{align} \mathbb{E}_\omega\lVert(\mathcal{R}f)_{\omega}\rVert_{p,q,(1-r)^{\nu}\varphi}^q\leq\mathbb{E}_\omega\mathbb{E}_{\omega'}\lVert(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\rVert_{p,q,(1-r)^{\nu}\varphi}^q. \end{align} $$

By the proof of symmetric case, we have $\mathbb {E}_\omega \mathbb {E}_{\omega '}\lVert (\mathcal {R}f)_{\omega }-(\widetilde {\mathcal {R}f})_{\omega '}\rVert _{p,q,(1-r)^{\nu }\varphi }^q<\infty $ . So

$$ \begin{align*}\mathcal{R}f\in H^{p,q,(1-r)^{\nu}\varphi}(\mathbb{B}_n)\ \text{a.s.}\end{align*} $$

Similarly, by Minkowski’s inequality, one has

$$ \begin{align*}M_p\left(r,\mathbb{E}_{\omega'}\left((\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)\leq\mathbb{E}_{\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right).\end{align*} $$

If $0<q<1$ , then

$$ \begin{align*}\lVert\mathcal{R}f\rVert_{p,q,(1-r)^{\nu}\varphi}^q\leq\int_{0}^{1}\frac{{\varphi(r)^q}}{(1-r)^{q\nu-1}}\left(\mathbb{E}_{\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)\right)^qdr.\end{align*} $$

By Hölder’s inequality and Fubini’s theorem,

(10) $$ \begin{align} \mathbb{E}_\omega\lVert(\mathcal{R}f)_{\omega}\rVert_{p,q,(1-r)^{\nu}\varphi}^q & \leq\int_{0}^{1}\frac{\varphi(r)^{q}}{(1-r)^{q\nu-1}}\mathbb{E}_\omega\left(\mathbb{E}_{\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)\right)^qdr \nonumber\\ & \leq\int_{0}^{1}\frac{\varphi(r)^{q}}{(1-r)^{q\nu-1}}\left(\mathbb{E}_{\omega\times\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)\right)^qdr. \end{align} $$

By Hölder’s inequality and (6),

$$ \begin{align*} \mathbb{E}_{\omega\times\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right) & =\mathbb{E}_{\omega\times\omega'}\mathbb{E}_{\epsilon}\left(M_p\left(r,\mathcal{R}_{\epsilon}\left((\mathcal{R} f)_{\omega}-(\widetilde{\mathcal{R} f})_{\omega'})\right)\right)\right) \\ & \lesssim\mathbb{E}_{\omega\times\omega'}\left(M_2\left(r,(\mathcal{R}f^{[\nu]})_{\omega}-(\widetilde{\mathcal{R}f^{[\nu]}})_{\omega'}\right)\right) \\ & \leq(\mathbb{E}_{\omega\times\omega'}|X_0-\widetilde{X_0}|^2)^{\frac{1}{2}}M_2(r,f^{[\nu]}). \end{align*} $$

So, by Lemma 8,

(11) $$ \begin{align} \int_{0}^{1}\frac{\varphi(r)^{q}}{(1-r)^{q\nu-1}}\left(\mathbb{E}_{\omega\times\omega'}\left(M_p\left(r,(\mathcal{R}f)_{\omega}-(\widetilde{\mathcal{R}f})_{\omega'}\right)\right)\right)^qdr\lesssim\lVert f\rVert_{2,q,\varphi}^q<\infty. \end{align} $$

Then $\mathcal {R}f\in H^{p,q,(1-r)^{\nu }\varphi }(\mathbb {B}_n)$ a.s.

Proof of Theorem 6

We first consider the symmetric case.

Since $\epsilon _\alpha (\omega ')X_\alpha (\omega )$ and $X_\alpha (\omega )$ have the same distribution, one has

$$ \begin{align*}\mathbb{P}_X\times\mathbb{P}_\epsilon(\lVert\mathcal{R}_\epsilon\mathcal{R}f)\rVert_{p,q,\varphi}<\infty)=\mathbb{P}_X(\lVert\mathcal{R}f\rVert_{p,q,\varphi}<\infty),\end{align*} $$

where $\{\epsilon _\alpha \}_{\alpha \in \mathbb {N}^n}$ is a Rademacher sequence on a probability space $\left (\Omega ',\mathcal {F}',\mathbb {P}'\right )$ and independent to $\{X_\alpha \}_{\alpha \in \mathbb {N}^n}$ . Since $\mathcal {R}f\in H^{p,q,\varphi }(\mathbb {B}_n)$ a.s., it follows that for a.e. $\omega '\in \Omega '$ ,

$$ \begin{align*}\mathbb{P}_X(\lVert\mathcal{R}_\epsilon\mathcal{R}f\rVert_{p,q,\varphi}<\infty)=1.\end{align*} $$

Then, for a.e. $\omega \in \Omega $ , $\mathbb {P}_\epsilon \left (\mathcal {R}_\epsilon \mathcal {R}f\in H^{p,q,\varphi }(\mathbb {B}_n)\right )=1$ . By Lemma 11,

(12) $$ \begin{align} \mathbb{P}_\epsilon\left(\mathcal{R}_\epsilon\mathcal{R}f\in H^{p,q,\varphi}(\mathbb{B}_n)\right)=1\Leftrightarrow\mathbb{E}_\epsilon\lVert\mathcal{R}_\epsilon\mathcal{R}f\rVert_{p,q,\varphi}^{q}<\infty. \end{align} $$

(i) For $2\leq p<\infty $ and $p\leq q$ , by Hölder’s inequality and Khintchine’s inequality,

(13) $$ \begin{align} \mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_{\epsilon}\mathcal{R}f) & \geq\left(\mathbb{E}_{\epsilon}M_p^p\left(r,\mathcal{R}_{\epsilon}\mathcal{R}f\right)\right)^{\frac{q}{p}} \nonumber \\ & \approx\left(\int_{\partial \mathbb{B}_n}\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_\alpha X_\alpha \zeta^{\alpha}|^2r^{2m}\right)^{\frac{p}{2}}d\sigma(\zeta)\right)^{\frac{q}{p}} \\ & \geq \left(\int_{\partial \mathbb{B}_n}\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_\alpha X_\alpha \zeta^{\alpha}|^2r^{2m}d\sigma(\zeta)\right)^{\frac{q}{2}} \nonumber \\ & =M_2^q(r,\mathcal{R}f). \nonumber \end{align} $$

For $2\leq p<\infty $ and $p>q$ , by Minkowski’s inequality, Khintchine’s inequality, and Hölder’s inequality, one has

(14) $$ \begin{align} \mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_{\epsilon}\mathcal{R}f) & \geq \left(\int_{\partial \mathbb{B}_n}\left(\mathbb{E}_{\epsilon}|\mathcal{R}_{\epsilon}\mathcal{R}f|^q\right)^{\frac{p}{q}}d\sigma(\zeta)\right)^{\frac{q}{p}} \nonumber \\ & \approx\left(\int_{\partial \mathbb{B}_n}\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_\alpha X_\alpha \zeta^{\alpha}|^2r^{2m}\right)^{\frac{p}{2}}d\sigma(\zeta)\right)^{\frac{q}{p}} \\ & \geq \left(\int_{\partial \mathbb{B}_n}\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_\alpha X_\alpha \zeta^{\alpha}|^2r^{2m}\right)d\sigma(\zeta)\right)^{\frac{q}{2}} \nonumber \\ & =M_2^q(r,\mathcal{R}f). \nonumber \end{align} $$

Then for a.e. $\omega \in \Omega $ ,

(15) $$ \begin{align} \mathbb{E}_\epsilon\lVert \mathcal{R}_\epsilon\mathcal{R}f\rVert_{p,q,\varphi}^q<\infty\ \Rightarrow\lVert\mathcal{R}f\rVert_{2,q,\varphi}<\infty. \end{align} $$

Since

$$ \begin{align*}\lVert f\rVert_{2,q,\varphi}^q=\int_{0}^{1}\frac{\varphi(r)^q}{1-r}\left(\sum_{m=0}^{\infty}\left(\sum_{|\alpha|=m}|a_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)\right)r^{2m}\right)^{\frac{q}{2}}dr\end{align*} $$

and

$$ \begin{align*}G_k=\sum_{m\in I_k}\lVert f_m(z)\rVert_{L^2(\partial \mathbb{B}_n)}^2=\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta),\end{align*} $$

by Lemma 10,

(16) $$ \begin{align} \lVert f\rVert_{2,q,\varphi}^q<\infty\Leftrightarrow\sum_{k=0}^{\infty}\varphi(r_k)^q\left(\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)\right)^{\frac{q}{2}}<\infty, \end{align} $$

where $r_k=1-2^{-k}$ . We argue by contradiction. Suppose that

$$ \begin{align*}\sum_{k=0}^{\infty}\varphi(r_k)^q\left(\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)\right)^{\frac{q}{2}}=\infty.\end{align*} $$

By (16), one has $f\notin H^{2,q,\varphi }(\mathbb {B}_n)$ . Combining this with (12) and (15), it suffices to show that

$$ \begin{align*}\sum_{k=0}^{\infty}\varphi(r_k)^q\left(\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2|X_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)\right)^{\frac{q}{2}}=\infty\ \text{a.s.}\end{align*} $$

For $\{(\alpha _1,\ldots ,\alpha _n)\}$ , an order can be introduce by the following method. If

$$ \begin{align*}\alpha=(\alpha_1,\ldots,\alpha_n),\ \beta=(\beta_1,\ldots,\beta_n)\end{align*} $$

satisfy either of the following two conditions, then $\alpha <\beta :$

(a) $|\alpha |<|\beta |$ ;

(b) if $|\alpha |=|\beta |$ , exists some $k$ , $1\leq k\leq n$ , such that $\alpha _1=\beta _1,\ldots ,\alpha _{k-1}=\beta _{k-1}$ , $\alpha _k<\beta _k.$

The single-index sequences formed by arranging

$$ \begin{align*}\{a_\alpha\}_{|\alpha|\in I_k},\ \ \{X_\alpha\}_{|\alpha|\in I_k},\ \ and\ \ \left\{\int_{\partial \mathbb{B}_n}|\zeta^{\alpha}|^2d\sigma(\zeta)\right\}_{|\alpha|\in I_k}\end{align*} $$

in ascending order following the aforementioned order are denoted as $\{b_l^{(k)}\}$ , $\{c_l^{(k)}\},$ and $\{t_l^{(k)}\}$ , respectively. Let

$$ \begin{align*}d_k:=\displaystyle\sum_{m\in I_k}\binom{n+m-1}{m}=\binom{n+2^{k+1}-1}{n}-\binom{n+2^k-1}{n}.\end{align*} $$

Write

$$ \begin{align*}B_k:=\sum_{l=1}^{d_k}|b_l^{(k)}|^2t_l^{(k)},\ \ Z_k:=\sum_{l=1}^{d_k}|b_l^{(k)}|^2|c_l^{(k)}|^2t_l^{(k)}\end{align*} $$

and

$$ \begin{align*}W_k:=\varphi(r_k)^qZ_k^{\frac{q}{2}},\ \ A_k:=\varphi(r_k)^qB_k^{\frac{q}{2}}.\end{align*} $$

Then

(17) $$ \begin{align} \sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)=\sum_{l=1}^{d_k}|b_l^{(k)}|^2t_l^{(k)} \end{align} $$

and

(18) $$ \begin{align} \sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2|X_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)=\sum_{l=1}^{d_k}|b_l^{(k)}|^2|c_l^{(k)}|^2t_l^{(k)}. \end{align} $$

Since $\mathbb {P}(|X_0|>0)>0$ , write $\mathbb {E}|X_0|^2:=\kappa>0$ and $\mathbb {E}|X_0|^4:=\lambda>0$ . We have $\mathbb {E}Z_k=\displaystyle \sum _{l=1}^{d_k}|b_l^{(k)}|^2t_l^{(k)}\mathbb {E}|X_0|^2=\kappa B_k$ and

$$ \begin{align*} \mathbb{E}Z_k^2 & =\sum_{l=1}^{d_k}|b_l^{(k)}|^4(t_l^{(k)})^2\mathbb{E}|X_0|^4+\sum_{\substack{i,j\in [1,d_k]\\i\neq j}}|b_i^{(k)}|^2t_i^{(k)}|b_j^{(k)}|^2t_j^{(k)}\kappa^2 \\ & =\lambda\sum_{l=1}^{d_k}|b_l^{(k)}|^4(t_l^{(k)})^2+\kappa^2\left(\left(\sum_{l=1}^{d_k}|b_l^{(k)}|^2t_l^{(k)}\right)^2-\sum_{l=1}^{d_k}|b_l|^4\left(t_l^{(k)}\right)^2\right) \\ & =\kappa^2B_k^2+(\lambda-\kappa^2)\sum_{l=1}^{d_k}|b_l^{(k)}|^4(t_l^{(k)})^2\\ & \leq\kappa^2B_k^2+(\lambda-\kappa^2)B_k^2 \\ & =\lambda B_k^2. \end{align*} $$

So, by the Paley–Zygmund inequality [Reference Li and Queffélec18, Proposition III.3, p. 23], one has

$$ \begin{align*}\mathbb{P}(Z_k\geq\frac{1}{2}\mathbb{E}Z_k)\geq\frac{(\frac{1}{2})^2(\mathbb{E}Z_k)^2}{\mathbb{E}Z_k^2}\geq\frac{\frac{1}{4}\kappa^2B_k^2}{\lambda B_k^2}=\frac{\kappa^2}{4\lambda}:=c^{\prime}_0.\end{align*} $$

Let $\displaystyle (\frac {\kappa }{2})^{\frac {q}{2}}=c^{\prime }_1$ . Then

$$ \begin{align*}\mathbb{P}(W_k\geq c^{\prime}_1A_k)=\mathbb{P}(Z_k\geq \frac{1}{2}\mathbb{E}Z_k)\geq c^{\prime}_0.\end{align*} $$

If $f\notin H^{2,q,\varphi }(\mathbb {B}_n)$ , then $\sum _{k=1}^{\infty }A_k=\infty $ . Let

$$ \begin{align*}S_l:=\sum_{k\in J_l}A_k\geq l,\ \ Y_l:=\sum_{k\in J_l}A_k\mathcal{X}_{\{W_k\geq c^{\prime}_1A_k\}},\end{align*} $$

where $J_1=\{1,\ldots ,t_1\}$ and $J_l=\{t_{l-1}+1,\ldots ,t_l\}$ for $l>1$ , with the indices $\{t_l\}_{l\geq 1}$ chosen so that $\displaystyle \sum _{k\in J_l}A_k\geq l$ for all $l$ . We have

$$ \begin{align*}0\leq Y_l\leq \sum_{k\in J_l}A_k=S_l \quad \text{and} \quad \mathbb{E}Y_l^2\leq S_l^2<\infty.\end{align*} $$

So,

$$ \begin{align*}\mathbb{E}(Y_l)=\sum_{k\in J_l}A_k\mathbb{P}(W_k\geq c^{\prime}_1A_k)\geq c^{\prime}_0S_l\end{align*} $$

and

$$ \begin{align*}Var(Y_l)=\sum_{k\in J_l}A_k^2Var(\mathcal{X}_{\{W_k\geq c^{\prime}_1A_k\}})\leq\sum_{k\in J_l}A_k^2\leq(\max_{k\in J_l}A_k)S_l\leq S_l^2<\infty.\end{align*} $$

By the Paley–Zygmund inequality again,

$$ \begin{align*}\mathbb{P}(Y_l\geq \frac{1}{2}\mathbb{E}Y_l)\geq\frac{\frac{1}{4}(\mathbb{E}Y_l)^2}{Var(Y_l)+(\mathbb{E}Y_l)^2}\geq\frac{\frac{1}{4}(c^{\prime}_0S_l)^2}{S_l^2+(c^{\prime}_0S_l)^2}=\frac{(c^{\prime}_0)^2}{4\left(1+(c^{\prime}_0)^2\right)}:=\delta>0.\end{align*} $$

Since $\sum _{l=1}^{\infty }\mathbb {P}\left (\{Y_l\geq \frac {1}{2}\mathbb {E}Y_l\}\right )=\infty $ , by the Borel–Cantelli lemma,

$$ \begin{align*}\mathbb{P}\left(\{Y_l\geq \frac{1}{2}\mathbb{E}Y_l\}\ i.o.\right)=1.\end{align*} $$

And for a.e. $\omega \in \Omega $ ,

$$ \begin{align*}\sum_{k\in J_l}W_k\geq c^{\prime}_1Y_l\geq\frac{1}{2}c^{\prime}_0c^{\prime}_1S_l\geq\frac{1}{2}c^{\prime}_0c^{\prime}_1l\end{align*} $$

holds for infinitely many $J_l$ . Then

$$ \begin{align*}\sum_{k=1}^{\infty}W_k=\sum_{l=1}^{\infty}\sum_{k\in J_l}W_k=\infty\ \text{a.s.}\end{align*} $$

So $\mathcal {R}f\notin H^{2,q,\varphi }(\mathbb {B}_n)$ a.s., and one has $\mathcal {R}f\notin H^{p,q,\varphi }(\mathbb {B}_n)$ a.s.

(ii) For $0<p<2$ , similar to (13) and (14), by Hölder’s inequality, Khintchine’s inequality, and Minkowski’s inequality,

$$ \begin{align*} & \mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_\epsilon\mathcal{R}f^{[\nu]})\\[-2pt] &\gtrsim \left(\int_{\partial \mathbb{B}_n}\left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_\alpha X_\alpha \zeta^\alpha|^2r^{2m}\left(\frac{\Gamma(|\alpha|+\nu+1)}{\Gamma(|\alpha|+1)}\right)^2\right)^{\frac{p}{2}}d\sigma(\zeta)\right)^{\frac{q}{p}} \\[-2pt] &\geq \left(\sum_{m=0}^{\infty}\sum_{|\alpha|=m}|a_\alpha X_\alpha \zeta^{\alpha}|^2r^{2m}\left(\frac{\Gamma(|\alpha|+\nu+1)}{\Gamma(|\alpha|+1)}\right)^2\left(\int_{\partial \mathbb{B}_n}|\zeta^{\alpha}|^pd\sigma(\zeta)\right)^{\frac{2}{p}}\right)^{\frac{q}{2}}. \end{align*} $$

And

$$ \begin{align*}\frac{I(2)}{I(p)^{\frac{2}{p}}}\lesssim |\alpha|^{(n-1)(\frac{2}{p}-1)}\end{align*} $$

by Stirling’s formula. Then

$$ \begin{align*}\mathbb{E}_{\epsilon}M_p^q(r,\mathcal{R}_{\epsilon}\mathcal{R}f^{[\nu]})\gtrsim M_2^q(r,\mathcal{R}f),\end{align*} $$

where $\nu \geq \frac {n-1}{2}(\frac {2}{p}-1)$ . So, by Lemma 8,

$$ \begin{align*} \mathbb{E}_{\epsilon}\lVert\mathcal{R}_{\epsilon}\mathcal{R}f\rVert_{p,q,\varphi}^q & \approx\mathbb{E}_{\epsilon}\lVert\mathcal{R}_{\epsilon}\mathcal{R}f^{[\nu]}\rVert_{p,q,(1-r)^{\nu}\varphi}^q \\ & \gtrsim\lVert\mathcal{R}f\rVert_{2,q,(1-r)^{\nu}\varphi}^q. \end{align*} $$

Then

$$ \begin{align*}\mathcal{R}f\in H^{p,q,\varphi}(\mathbb{B}_n)\ \text{a.s.}\Rightarrow\lVert\mathcal{R}f\rVert_{2,q,(1-r)^{\nu}\varphi}<\infty\ \text{a.s.}\end{align*} $$

That is, it suffices to prove that

$$ \begin{align*}\lVert\mathcal{R}f\rVert_{2,q,(1-r)^{\nu}\varphi}=\infty\ \text{a.s.}\end{align*} $$

when $f\notin H^{2,q,(1-r)^{\nu }\varphi }(\mathbb {B}_n)$ . The proof is similar to that of the case $2\leq p<\infty $ and is therefore omitted.

For the non-symmetric case.

(i) Since $X_\alpha \in L^{4}(\Omega )$ , by the symmetric case,

$$ \begin{align*}\left(H^{2,q,\varphi}(\mathbb{B}_n)\right)_{\star}\subset H^{2,q,\varphi}(\mathbb{B}_n).\end{align*} $$

For $2\leq p<\infty $ , since $M_2(r,\mathcal {R}f)^q\leq M_p(r,\mathcal {R}f)^q$ , one has

$$ \begin{align*}\left(H^{p,q,\varphi}(\mathbb{B}_n)\right)_{\star}\subset\left(H^{2,q,\varphi}(\mathbb{B}_n)\right)_{\star}\subset H^{2,q,\varphi}(\mathbb{B}_n).\end{align*} $$

(ii) For $0<p<2$ , if $f\in (H^{p,q,\varphi }(\mathbb {B}_n))_{\star }$ , then $\mathcal {R}f\in H^{p,q,\varphi }(\mathbb {B}_n)$ a.s. and $\widetilde {\mathcal {R}f}\in H^{p,q,\varphi }(\mathbb {B}_n)$ a.s.

If $1\leq p<2$ and $0<q<1$ , then

$$ \begin{align*}M_p(r,\mathcal{R}f-\widetilde{\mathcal{R}f})\leq M_p(r,\mathcal{R}f)+M_p(r,\widetilde{\mathcal{R}f})\end{align*} $$

and

$$ \begin{align*}M_p^q(r,\mathcal{R}f-\widetilde{\mathcal{R}f})\leq M_p^q(r,\mathcal{R}f)+M_p^q(r,\widetilde{\mathcal{R}f}).\end{align*} $$

This implies that

$$ \begin{align*}\lVert\mathcal{R}f-\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}^q\leq\lVert\mathcal{R}f\rVert_{p,q,\varphi}^q+\lVert\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}^q\end{align*} $$

and $\mathcal {R}f-\widetilde {\mathcal {R}f}\in H^{p,q,\varphi }(\mathbb {B}_n)$ a.s. So $f\in H^{2,q,(1-r)^{\nu }\varphi }(\mathbb {B}_n)$ by the symmetric case.

If $1\leq p<2$ and $1\leq q<\infty $ , then

$$ \begin{align*}\lVert\mathcal{R}f-\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}\leq\lVert\mathcal{R}f\rVert_{p,q,\varphi}+\lVert\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}\end{align*} $$

and $\mathcal {R}f-\widetilde {\mathcal {R}f}\in H^{p,q,\varphi }(\mathbb {B}_n)$ a.s. Similarly, $f\in H^{2,q,(1-r)^{\nu }\varphi }(\mathbb {B}_n)$ .

If $0<p,q<1$ , then $H^{p,q,\varphi }(\mathbb {B}_n)$ is an $s$ -Banach space, where $s=\min \{p,q\}$ . So,

$$ \begin{align*}\lVert\mathcal{R}f-\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}^s\leq\lVert\mathcal{R}f\rVert_{p,q,\varphi}^s+\lVert\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}^s\end{align*} $$

and $\mathcal {R}f-\widetilde {\mathcal {R}f}\in H^{p,q,\varphi }(\mathbb {B}_n)$ a.s. Hence, $f\in H^{2,q,(1-r)^{\nu }\varphi }(\mathbb {B}_n)$ .

If $0<p<1$ and $1\leq q<\infty $ , then

$$ \begin{align*}M_p^q(r,\mathcal{R}f-\widetilde{\mathcal{R}f})\leq M_p^q(r,\mathcal{R}f)+M_p^q(r,\widetilde{\mathcal{R}f})\end{align*} $$

and

$$ \begin{align*}M_p^q(r,\mathcal{R}f-\widetilde{\mathcal{R}f})\leq 2^{\frac{q}{p}-1}\left(M_p^q(r,\mathcal{R}f)+M_p^q(r,\widetilde{\mathcal{R}f})\right).\end{align*} $$

This implies that

$$ \begin{align*}\lVert\mathcal{R}f-\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}^q\leq 2^{\frac{q}{p}-1}(\lVert\mathcal{R}f\rVert_{p,q,\varphi}^q+\lVert\widetilde{\mathcal{R}f}\rVert_{p,q,\varphi}^q)\end{align*} $$

and $\mathcal {R}f-\widetilde {\mathcal {R}f}\in H^{p,q,\varphi }(\mathbb {B}_n)$ a.s. Hence, $f\in H^{2,q,(1-r)^{\nu }\varphi }(\mathbb {B}_n)$ .

Proof By Lemma 10, we only prove

$$ \begin{align*}\sum_{k=0}^{\infty}\varphi(r_k)^q\left(\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2|X_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)\right)^{\frac{q}{2}}=\infty\ \text{a.s.}\end{align*} $$

when

$$ \begin{align*}\sum_{n=0}^{\infty}\varphi(r_k)^q\left(\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta)\right)^{\frac{q}{2}}=\infty.\end{align*} $$

Let $K:=\displaystyle \sup _{\alpha \in \mathbb {N}^n}\frac {\mathbb {E}X_\alpha ^4}{(\mathbb {E}X_\alpha ^2)^2}$ . Then $\mathbb {E}|X_\alpha |^4\leq K\left (\mathbb {E}|X_\alpha |^2\right )^2$ . The choice of $\{b_l^{(k)}\}$ , $\{c_l^{(k)}\}$ , $\{t_l^{(k)}\}$ , $B_k$ , $Z_k$ , $W_k$ , and $A_k$ is the same as that in Theorem 6. Then

$$ \begin{align*} & \mathbb{E}\left(\sum_{l=1}^{d_k}|b_l^{(k)}|^2|c_l^{(k)}|^2t_l^{(k)}\right)^2 \\ &\leq K\sum_{l=1}^{d_k}|b_l^{(k)}|^4\left(\mathbb{E}|c_l^{(k)}|^2\right)^2(t_l^{(k)})^2+\left(\sum_{l=1}^{d_k}|b_l^{(k)}|^2\mathbb{E}|c_l^{(k)}|^2t_l^{(k)}\right)^2 \\ & -\sum_{l=1}^{d_k} |b_l^{(k)}|^4\left(\mathbb{E}|c_l^{(k)}|^2\right)^2\left(t_l^{(k)}\right)^2\\ &\leq K\left(\sum_{l=1}^{d_k}|b_l^{(k)}|^2\mathbb{E}|c_l^{(k)}|^2t_l^{(k)}\right)^2. \end{align*} $$

So, by the Paley–Zygmund inequality [Reference Li and Queffélec18, Proposition III.3, p. 23],

$$ \begin{align*}\mathbb{P}\left(\sum_{l=1}^{d_k}|b_l^{(k)}|^2|c_l^{(k)}|^2t_l^{(k)}>\frac{1}{2}\mathbb{E}\sum_{l=1}^{d_k}|b_l^{(k)}|^2|c_l^{(k)}|^2t_l^{(k)}\right)\geq\frac{1}{4K}:=\tilde{c}_0.\end{align*} $$

By (17) and (18),

$$ \begin{align*} \sum_{l=1}^{d_k}|b_l^{(k)}|^2|c_l^{(k)}|^2t_l^{(k)} & = \mathbb{E}\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2|X_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta) \\ & \geq m_0\sum_{m\in I_k}\sum_{|\alpha|=m}|a_\alpha|^2\int_{\partial \mathbb{B}_n}|\zeta^\alpha|^2d\sigma(\zeta) \\ & = m_0\sum_{l=1}^{d_k}|b_l^{(k)}|^2t_l^{(k)}. \end{align*} $$

Write $\tilde {c}_1:=(\frac {m_0}{2})^{\frac {q}{2}}$ . We have $\mathbb {P}(W_k\geq \tilde {c}_1A_k)\geq \tilde {c}_0$ . Assume that

$$ \begin{align*}S_l:=\sum_{k\in J_l}A_k\geq l,\ \ \ Y_l:=\sum_{k\in J_l}A_k\mathcal{X}_{\{W_k\geq \tilde{c}_1A_k\}},\end{align*} $$

where $J_l$ is the same as $J_l$ defined in Theorem 6. Then $\mathbb {E}Y_l\geq \tilde {c}_0\sum _{k\in J_l}A_k$ and $\mathbb {E}Y_l^2\leq (S_l)^2$ . So, by the Paley–Zygmund inequality [Reference Li and Queffélec18, Proposition III.3, p. 23],

$$ \begin{align*}\mathbb{P}(Y_l\geq\frac{1}{2}\mathbb{E}Y_l)\geq\frac{\tilde{c}_0}{4}\end{align*} $$

and

$$ \begin{align*}\sum_{l=1}^{\infty}\mathbb{P}\left(\{Y_l\geq\frac{1}{2}\mathbb{E}Y_l\}\right)=\infty.\end{align*} $$

Then $\mathbb {P}(\{Y_l\geq \frac {1}{2}\mathbb {E}Y_l\}\ i.o.)=1$ by the Borel–Cantelli lemma. So, we have that for a.e. $\omega \in \Omega $ ,

$$ \begin{align*}\sum_{k\in J_l}W_k\geq \tilde{c}_1Y_l\geq \frac{1}{2}\tilde{c}_1\tilde{c}_0S_l\geq \frac{1}{2}\tilde{c}_1\tilde{c}_0l\end{align*} $$

holds for infinitely many $J_l$ . Then $\sum _{k=1}^{\infty }W_k=\sum _{l=1}^{\infty }\sum _{k\in J_l}W_k=\infty $ a.s. That is,

$$ \begin{align*}\mathcal{R}f\notin H^{2,q,\varphi}(\mathbb{B}_n)\ \text{a.s.}\end{align*} $$

Proof of Theorem 7

(i) For $0<p<2$ , assume that

(19) $$ \begin{align} \mathbb{E}\lVert\mathcal{R}f\rVert_{p,q,(1-r)^\nu\varphi}^q=\infty. \end{align} $$

Since $M_p^q(r,\mathcal {R}f)\leq M_2^q(r,\mathcal {R}f),$ one has

(20) $$ \begin{align} \mathbb{E}\lVert\mathcal{R}f\rVert_{p,q,(1-r)^\nu\varphi}^q\leq\mathbb{E}\lVert\mathcal{R}f\rVert_{2,q,(1-r)^\nu\varphi}^q. \end{align} $$

By Jensen’s inequality for $0<q\leq 2$ and Minkowski’s inequality for $2<q<\infty ,$

(21) $$ \begin{align} \mathbb{E}\lVert\mathcal{R}f\rVert_{2,q,(1-r)^\nu\varphi}^q\lesssim \lVert f\rVert_{2,q,(1-r)^\nu\varphi}^q. \end{align} $$

By (19)–(21), $\lVert f\rVert _{2,q,(1-r)^\nu \varphi }=\infty $ . Therefore, by Theorem 6 and the Kolmogorov zero-one law, for $0<p<2$ ,

$$ \begin{align*}\mathcal{R}f\notin H^{p,q,\varphi}(\mathbb{B}_n) \ \text{a.s.,}\end{align*} $$

which contradicts the assumption $\|\mathcal {R}f\|_{p,q,\varphi }<\infty $ a.s.

(ii) For $p=2$ , the proof is similar to the case $0<p<2$ .

(iii) For $2<p<\infty $ and $1\leq q<\infty ,$ assume that $\mathbb {E}\lVert \mathcal {R}f\rVert _{p,q,(1-r)^\nu \varphi }^q=\infty $ . By (8) of Theorem 5,

$$ \begin{align*}\mathbb{E}_{\omega}\lVert\mathcal{R}f\rVert_{p,q,(1-r)^\nu\varphi}^q\leq\mathbb{E}_{\omega\times\omega'}\lVert\mathcal{R}f-\widetilde{\mathcal{R}f}\rVert_{p,q,(1-r)^\nu\varphi}^q.\end{align*} $$

Then

$$ \begin{align*}\mathbb{E}_{\omega\times\omega'}\lVert\mathcal{R}f-\widetilde{\mathcal{R}f}\rVert_{p,q,(1-r)^\nu\varphi}^q=\infty.\end{align*} $$

So, by (8) of Theorem 5, $\lVert f\rVert _{2,q,\varphi }=\infty $ . It follows from Theorem 6 that $\mathcal {R}f\notin H^{p,q,\varphi }(\mathbb {B}_n)$ a.s., which contradicts the assumption $\|\mathcal {R}f\|_{p,q,\varphi }<\infty $ a.s.

For $2<p<\infty $ and $0<q<1$ , by (10) and (11) of Theorem 5, one has

$$ \begin{align*}\mathbb{E}\|\mathcal{R} f\|_{p,q,(1-r)^\nu\varphi}^q\lesssim\|f\|_{2,q,\varphi}^q.\end{align*} $$

The rest of the proof is analogous to that for the case $2<p<\infty $ and $1\leq q<\infty $ .

Proof We first consider the symmetric case. Since $\mathbb {E} X_0=0$ , the proof of (ii) is analogous to that of (i). Therefore, it suffices to prove (i).

(i) Since $\mathcal {R} f(z)=\sum _{n=0}^{\infty }a_n X_n z^n$ and $\mathcal {R}_\epsilon \mathcal {R} f(z)=\sum _{n=0}^{\infty }a_n \epsilon _n X_n z^n$ are identically distributed, where $\{\epsilon _n\}_{n\geq 0}$ is a Rademacher sequence on a probability space $(\Omega ',\mathcal {F}',\mathbb {P}')$ and independent to $\{X_n\}_{n\geq 0}$ . Then

(22) $$ \begin{align} \mathbb{E}_XM_p^q(r,\mathcal{R} f)=\mathbb{E}_X\mathbb{E}_\epsilon\left(M_p^q(r,\mathcal{R}_\epsilon\mathcal{R} f)\right). \end{align} $$

Furthermore, one has

(23) $$ \begin{align} \mathbb{E}_X\mathbb{E}_\epsilon\left(M_p^q(r,\mathcal{R}_\epsilon\mathcal{R} f)\right)\approx\mathbb{E}_X(\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n})^{\frac{q}{2}}=\mathbb{E}_XM_2^q(r,\mathcal{R} f). \end{align} $$

Now, we will prove (23). For $p\geq q,$ on the one hand, by Hölder’s inequality and Khintchine’s inequality,

$$ \begin{align*}\mathbb{E}_\epsilon\left(M_p^q(r,\mathcal{R}_\epsilon \mathcal{R} f)\right)\leq\left(\mathbb{E}_\epsilon M_p^p(r,\mathcal{R}_ \epsilon \mathcal{R} f)\right)^{\frac{q}{p}}\approx\left(\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n}\right)^{\frac{q}{2}}.\end{align*} $$

On the other hand, by Jensen’s inequality and Khintchine’s inequality,

$$ \begin{align*}\mathbb{E}_\epsilon\left(M_p^q(r,\mathcal{R}_\epsilon \mathcal{R} f)\right)\geq\mathbb{E}_\epsilon\left(M_q^q(r,\mathcal{R}_\epsilon \mathcal{R} f)\right)\approx\left(\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n}\right)^{\frac{q}{2}}.\end{align*} $$

So, (23) is established for $p\geq q.$ For $p<q$ , the proof is similar to that of $p\geq q.$

Hence, for $0<q\leq 2$ , by (23) and Hölder’s inequality,

$$ \begin{align*} \mathbb{E}_{X}\int_{0}^{1}(1-r)^{qu-1}M_p(r,\mathcal{R}f)^qdr & =\mathbb{E}_X\int_{0}^{1}(1-r)^{qu-1}\mathbb{E}_\epsilon M_p(r,\mathcal{R}_\epsilon\mathcal{R} f)^qdr \\ & \approx\mathbb{E}_X\int_{0}^{1}(1-r)^{qu-1}\left(\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n}\right)^{\frac{q}{2}}dr \\ & \leq\int_{0}^{1}(1-r)^{qu-1}\left(\mathbb{E}_X\left(\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n}\right)\right)^{\frac{q}{2}}dr \\ & =(\mathbb{E}_X|X_0|^2)^{\frac{q}{2}}\int_{0}^{1}(1-r)^{qu-1}\left(\sum_{n=0}^{\infty}|a_n|^2r^{2n}\right)^{\frac{q}{2}}dr \\ & =(\mathbb{E}_X|X_0|^2)^{\frac{q}{2}}\lVert f\rVert_{2,q,u}^q. \end{align*} $$

For $2<q<\infty $ , by Minkowski’s inequality, one has

$$ \begin{align*} \left(\mathbb{E}_X\left(\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n}\right)^{\frac{q}{2}}\right)^{\frac{2}{q}} & =\left\|\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n}\right\|_{L^{\frac{q}{2}}(\Omega)} \\[-2pt] & \leq\sum_{n=0}^{\infty}\left\lVert|a_n|^2|X_n|^2r^{2n}\right\rVert_{L^{\frac{q}{2}}(\Omega)} \\[-2pt] & =\left(\mathbb{E}_X|X_0|^q\right)^{\frac{2}{q}}\sum_{n=0}^{\infty}|a_n|^2r^{2n}. \end{align*} $$

Then $\mathbb {E}\lVert \mathcal {R}f\rVert _{p,q,u}^q\leq \mathbb {E}|X_0|^\tau \lVert f\rVert _{2,q,u}^q<\infty .$

For the non-symmetric case. From (22), (23), and the result of the symmetric case, the proof for the non-symmetric case is similar to that of the non-symmetric case in Theorem 5 and is therefore omitted.

Proof The proof of (ii) is identical to that of (i), so we only prove (i).

(i) Assume that $f(z)=\sum _{n=0}^{\infty }a_n z^{k(n)}\in H^{2,q,u}(\mathbb {D})$ . Since

$$ \begin{align*}H^{2,q,u}(\mathbb{D})\subset(H^{p,q,u}(\mathbb{D}))_{\star},\end{align*} $$

by Lemma 15,

(24) $$ \begin{align} \sum_{n=0}^{\infty}|a_n2^{-nu}X_n|^q<\infty \quad \text{a.s.} \end{align} $$

Write $c_n:=a_n 2^{-nu}$ . Then $\{c_n\}_{n\geq 0}\in l^q$ and (24) can be rewritten as

(25) $$ \begin{align} \sum_{n=0}^{\infty}|c_n|^q|X_n|^q<\infty\ \ \text{a.s.} \end{align} $$

Assume $X_0\notin L^{q}(\Omega )$ . Let $K_0:=|X_0|^q$ and $K_n:=|X_n|^q$ . Then

$$ \begin{align*}\mathbb{E} K_0=\int_{0}^{\infty}\mathbb{P}(K_0>t)dt=\infty.\end{align*} $$

Next, we will construct a sequence $\{k_n\}_{n\geq 0}\in l^q$ that violates (25), thereby contradicting (24). Let $\{s_m\}_{m\geq 0}$ be a sequence such that $s_m\geq m^4$ and

$$ \begin{align*}\int_{s_m}^{2s_m}\mathbb{P}(K_0>t)dt\geq m^4.\end{align*} $$

Then

$$ \begin{align*}p_m:=\mathbb{P}(K_0>s_m)\geq\frac{1}{s_m}\int_{s_m}^{2s_m}\mathbb{P}(K_0>t)dt\geq\frac{m^4}{s_m}.\end{align*} $$

Partition $\mathbb {N}$ into a sequence of consecutive finite blocks $\{H_m\}_{m=1}^\infty $ , which satisfies $|H_m|=N_m=\displaystyle \left [\frac {m^2}{p_m}\right ]$ . We denote $k_n:=\displaystyle \left (\frac {1}{s_m}\right )^{\frac {1}{q}},$ where $n\in H_m$ . Since $s_mp_m\geq m^4$ , we have

$$ \begin{align*}\sum_{n=0}^{\infty}k_n^q =\sum_{m=1}^{\infty}\sum_{n\in H_m}\frac{1}{s_m}=\sum_{m=1}^{\infty}\frac{N_m}{s_m} \leq\sum_{m=1}^{\infty}\left(\frac{m^2}{p_ms_m}+\frac{1}{s_m}\right) <\infty.\end{align*} $$

So, $\{k_n\}_{n\geq 1}\in l^q$ . Let $E_m:=\{\omega :\ \exists \ n\in H_m,\ K_n(\omega )>s_m\}$ . Then

$$ \begin{align*}\mathbb{P}(E_m^{c})=(1-p_m)^{N_m}\leq e^{-N_mp_m}\leq e^{-m^2}.\end{align*} $$

So, $\sum _{m=1}^{\infty }\mathbb {P}(E_m^{c})<\infty $ . By the Borel–Cantelli lemma, $\mathbb {P}(E_m^{c}\ i.o.)=0$ . That is, $\mathbb {P}(\liminf _{m\to \infty }E_m)=1$ . Then

$$ \begin{align*}\sum_{n\in H_m}k_n^q K_n\geq \frac{1}{s_m}K_n\geq 1\ \text{a.s.},\end{align*} $$

where $m$ is large enough. So,

$$ \begin{align*}\sum_{n=0}^{\infty}k_n^q K_n=\sum_{m=1}^{\infty}\sum_{n\in H_m}k_n^q K_n=\infty\ \text{a.s.}\end{align*} $$

Contradiction.

(iii) The proof for $X_0\in L^q(\Omega )$ is similar to that of (i). It is sufficient to prove that $\mathbb {E} X_0=0$ . If $\mathbb {E} X_0\neq 0$ , we take $X_n^{'}=X_n-\mathbb {E} X_0$ , then $\mathbb {E} X_n^{'}=0$ . Since

(26) $$ \begin{align} H^{2,q,u}(\mathbb{D})\subset\left(H^{p,q,u}(\mathbb{D})\right)_{\star}, \end{align} $$

we pick any $\sum _{n=0}^{\infty }c_n z^n\in H^{2,q,u}(\mathbb {D})\backslash H^{p,q,u}(\mathbb {D})$ with $p>2$ . By Theorem 14, one has

$$ \begin{align*}\sum_{n=0}^{\infty}c_n X_n z^n=\sum_{n=0}^{\infty}c_n (X_n^{'}+\mathbb{E} X_0)z^n\notin H^{p,q,u}(\mathbb{D})\ \text{a.s.},\end{align*} $$

which contradicts (26).

Proof We first consider the symmetric case.

Since $\epsilon _n(\omega ')X_n(\omega )$ is identical distribution with $X_n(\omega )$ , we have

$$ \begin{align*}\mathbb{P}_X\times\mathbb{P}_\epsilon(\lVert\mathcal{R}_\epsilon\mathcal{R}f)\rVert_{p,q,u}<\infty)=\mathbb{P}_X(\lVert\mathcal{R}f\rVert_{p,q,u}<\infty),\end{align*} $$

where $\{\epsilon _n\}_{n\geq 0}$ is a Rademacher sequence on a probability space $(\Omega ',\mathcal {F}',\mathbb {P}')$ and independent to $\{X_n\}_{n\geq 0}$ . Since $\mathcal {R}f\in H^{p,q,u}(\mathbb {D})$ a.s. and for a.e. $\omega '\in \Omega '$ ,

$$ \begin{align*}\mathbb{P}_X(\lVert\mathcal{R}_\epsilon\mathcal{R}f\rVert_{p,q,u}<\infty)=1,\end{align*} $$

by [Reference Li and Queffélec18, Proposition III.7], we have that for a.e. $\omega \in \Omega $

$$ \begin{align*}\mathbb{P}_\epsilon(\mathcal{R}_\epsilon\mathcal{R}f\in H^{p,q,u}(\mathbb{D}))=1.\end{align*} $$

By [Reference Cheng, Fang and Liu7, Theorem 8],

$$ \begin{align*}\mathbb{P}_\epsilon\left(\mathcal{R}_\epsilon\mathcal{R}f\in H^{p,q,u}(\mathbb{D})\right)=1\Leftrightarrow\mathbb{E}_\epsilon\lVert\mathcal{R}_\epsilon\mathcal{R}f\rVert_{p,q,u}^{q}<\infty.\end{align*} $$

Moreover, since

$$ \begin{align*}\mathbb{E}_\epsilon\lVert\mathcal{R}_\epsilon\mathcal{R}f\rVert_{p,q,u}^q\approx\int_{0}^{1}(1-r)^{qu-1}\left(\sum_{n=0}^{\infty}|a_n|^2|X_n|^2r^{2n}\right)^{\frac{q}{2}}dr=\lVert \mathcal{R} f\rVert_{2,q,u}^q,\end{align*} $$

it follows that $\mathcal {R}f\in H^{2,q,u}(\mathbb {D})$ a.s. That is,

$$ \begin{align*}\mathcal{R} f\notin H^{2,q,u}(\mathbb{D}) \quad \text{a.s.} \quad \Rightarrow \quad \mathcal{R}f\notin H^{p,q,u}(\mathbb{D}) \quad \text{a.s.} \end{align*} $$

Suppose that $f\notin H^{2,q,u}(\mathbb {D})$ . It suffices to prove that $\mathcal {R}f\notin H^{2,q,u}(\mathbb {D})$ a.s. By Lemma 17, that is, if

$$ \begin{align*}\sum_{k=0}^{\infty}\left(\sum_{n\in I_k}|n^{-u}a_n|^2\right)^{\frac{q}{2}}=\infty,\end{align*} $$

it suffices to show that

$$ \begin{align*}\sum_{k=0}^{\infty}\left(\sum_{n\in I_k}|n^{-u}a_nX_n|^2\right)^{\frac{q}{2}}=\infty\ \ \text{a.s.}\end{align*} $$

The remainder of the proof of is similar to that of the symmetric case in Theorem 6 and is therefore omitted.

For the non-symmetric case. Making use of the result from the symmetric case, the proof for the non-symmetric case is similar to that of the non-symmetric case in Theorem 6 and is therefore omitted.

Proof of Theorem 4

It is sufficient to prove that

$$ \begin{align*}\mathcal{R}f\in H^{p,q,u}(\mathbb{D})\ \text{a.s.}\Rightarrow\ \mathbb{E}\lVert\mathcal{R}f\rVert_{p,q,u}^{q}<\infty.\end{align*} $$

(i) Suppose that $\{X_n\}_{n\geq 0}$ is symmetric and $\mathbb {E}\lVert \mathcal {R}f\rVert _{p,q,u}^q=\infty $ . Combining (22) and (23), one has

$$ \begin{align*}\mathbb{E}M_p^q(r,\mathcal{R} f)\approx\mathbb{E}M_2^q(r,\mathcal{R} f).\end{align*} $$

Then $\mathbb {E}\lVert \mathcal {R}f\rVert _{2,q,u}^q=\infty .$ If $0<q\leq 2$ , then

$$ \begin{align*}\mathbb{E}\lVert\mathcal{R}f\rVert_{2,q,u}^q\lesssim(\mathbb{E}|X_0|^2)^{\frac{q}{2}}\lVert f\rVert_{2,q,u}^q.\end{align*} $$

So, $\lVert f\rVert _{2,q,u}=\infty $ . By Theorem 18, $\mathbb {P}(\lVert \mathcal {R}f\rVert _{p,q,u}=\infty )>0$ , a contradiction.

If $2<q<\infty $ , similarly, assume $\mathbb {E}\lVert \mathcal {R}f\rVert _{2,q,u}^q=\infty $ . Since

$$ \begin{align*} & \mathbb{E}\int_0^1 (1 - r)^{qu - 1}\left( \sum_{n=0}^{\infty} |a_n|^2 |X_n|^2 r^{2n} \right)^{\frac{q}{2}} dr \\ &= \mathbb{E} \int_0^1 (1 - r)^{qu - 1}\left( \sum_{n=0}^{\infty} |a_n|^2 r^{2n} \right)^{\frac{q}{2}}\left( \sum_{n=0}^{\infty} \frac{|a_n|^2 r^{2n}}{\sum_{n=0}^{\infty} |a_n|^2 r^{2n}} |X_n|^2 \right)^{\frac{q}{2}} dr \\ &\leq \mathbb{E} \int_0^1 (1 - r)^{qu - 1}\left( \sum_{n=0}^{\infty} |a_n|^2 r^{2n} \right)^{\frac{q}{2} - 1} \left( \sum_{n=0}^{\infty} |a_n|^2 r^{2n} |X_n|^q \right) dr \\ &= \mathbb{E}|X_0|^q||f||^q_{2,q,u}, \end{align*} $$

we have $\lVert f\rVert _{2,q,u}^q=\infty $ . By Theorem 18, $\mathbb {P}(\lVert \mathcal {R}f\rVert _{p,q,u}=\infty )>0$ , a contradiction.

Assume that $\{X_n\}_{n\geq 0}$ is non-symmetric. For $0<p\leq 2$ , if $\mathbb {E}\lVert \mathcal {R}f\rVert _{2,q,u}^q=\infty $ , then $\lVert f\rVert _{2,q,u}=\infty $ . Since $M_p^q(r,\mathcal {R}f)\leq M_2^q(r,\mathcal {R}f)$ , one has $\mathbb {E}\lVert \mathcal {R}f\rVert _{p,q,u}^q\leq \mathbb {E}\lVert \mathcal {R}f\rVert _{2,q,u}^q$ . If $\mathbb {E}\lVert \mathcal {R} f\rVert _{p,q,u}^q=\infty ,$ then $\lVert f\rVert _{2,q,u}=\infty .$ So, by Theorem 18, $\mathbb {P}(\lVert \mathcal {R}f\rVert _{p,q,u}=\infty )>0$ , a contradiction.

(ii) For $2<p<\infty $ and $1\leq q<\infty $ , suppose that $\mathbb {E}\lVert \mathcal {R}f\rVert _{p,q,u}^q=\infty $ . By (9), $\mathbb {E}\lVert \mathcal {R}f\rVert _{p,q,u}^q\leq \mathbb {E}\lVert \mathcal {R}f-\widetilde {\mathcal {R}f}\rVert _{p,q,u}^q.$ From the proof of (i), we have $\lVert f\rVert _{2,q,u}=\infty $ . So, by Theorem 18, $\mathbb {P}(\lVert \mathcal {R}f\rVert _{p,q,u}=\infty )>0$ . This leads to a contradiction. For $2 < p <\infty $ and $0 < q < 1$ , the proof is similar to the case $2 < p < \infty $ and $1 \leq q < \infty ,$ and is therefore omitted.