Proof of Theorem 1.1

Given a holomorphic function $f(z)=\sum _{n=0}^{\infty } a_nz^n$ mapping $\mathbb {D}$ inside $\mathbb {D}$ , a straightforward application of Theorem A yields

(2.1) $$ \begin{align} |a_0|^p+\left(\sum_{n=1}^{\infty} |a_n|r^n\right)^q\leq \begin{cases} a^p+(1-a^2)^q\left(\frac{r}{1-ar}\right)^q\quad \textrm{for}\,\, r\leq a,\\ a^p+(1-a^2)^{\frac{q}{2}}\left(\frac{r}{\sqrt{1-r^2}}\right)^q\quad \textrm{for}\,\,r\in [0,1). \end{cases} \end{align} $$

Now,

$$ \begin{align*}a^p+(1-a^2)^q\left(\frac{r}{1-ar}\right)^q\leq 1 \end{align*} $$

whenever $r\leq A_{p, q}(a)$ . A little calculation reveals that $A_{p,q}(a)\leq a$ whenever $a^p+a^q\geq 1$ , i.e., whenever $a\geq \widehat {a}$ , $\widehat {a}$ being the root of equation (1.3). Thus, from (2.1), it is clear that

(2.2) $$ \begin{align} |a_0|^p+\left(\sum_{n=1}^{\infty} |a_n|r^n\right)^q\leq 1 \end{align} $$

for $r\leq \inf _{a\in [\widehat {a}, 1)}A_{p, q}(a)$ , provided that $a\geq \widehat {a}$ . On the other hand,

$$ \begin{align*}a^p+(1-a^2)^{\frac{q}{2}}\left(\frac{r}{\sqrt{1-r^2}}\right)^q\leq 1 \end{align*} $$

for $r\leq S_{p, q}(a)$ , i.e., inequality (2.2) remains valid for $r\leq \inf _{a\in [0, \widehat {a}]}S_{p, q}(a)$ , provided that $a\leq \widehat {a}$ . Therefore, we conclude that for any given $p, q\in [1, \infty )$ ,

(2.3) $$ \begin{align} R_{p, q}(\mathbb{C})\geq \min\left\{\inf_{a\in[0, \widehat{a}]}S_{p, q}(a), \inf_{a\in[\widehat{a}, 1)}A_{p, q}(a)\right\}\hspace{-1.5pt}. \end{align} $$

We also record some other facts which we will need to use later. Observe that for all $p, q\in [1, \infty )$ ,

$$ \begin{align*}S_{p, q}(a)=\sqrt{\frac{T(a)}{1+T(a)}}\,\, \mbox{where}\,\,T(a)=\frac{(1-a^p)^{\frac{2}{q}}}{1-a^2}, \end{align*} $$

and therefore

$$ \begin{align*}S_{p, q}^{\prime}(a) =\frac{T^{\prime}(a)}{2\sqrt{T(a)(1+T(a))^3}} \end{align*} $$

for $a\in (0,1)$ , where

(2.4) $$ \begin{align} T^{\prime}(a) =\frac{2a^{p-1}T(a)}{1-a^p}\left(\frac{a^2(1-a^p)}{a^p(1-a^2)}-\frac{p}{q}\right)\hspace{-1.5pt}. \end{align} $$

Setting $y=1/a$ for convenience, we write

$$ \begin{align*}\frac{a^2(1-a^p)}{a^p(1-a^2)} =\frac{y^p-1}{y^2-1}=P(y) \end{align*} $$

defined on $(1,\infty )$ . Note that

(2.5) $$ \begin{align} \frac{d}{da}P(y)=P^{\prime}(y)\frac{dy}{da} =-y^3\frac{py^p-py^{p-2}-2y^p+2}{(y^2-1)^2}, \end{align} $$

and that

(2.6) $$ \begin{align} Q^{\prime}(y)=y^{p-3}(y^2-1)p(p-2), \end{align} $$

where $Q(y)=py^p-py^{p-2}-2y^p+2$ .

Furthermore, observe that for the disk automorphisms $\phi _a(z)=(a-z)/(1-az)$ , $z\in \mathbb {D}, a\in [\widehat {a}, 1)$ , $R_{p, q}(\phi _a, \mathbb {C})=A_{p, q}(a)$ , and hence $R_{p, q}(\mathbb {C})\leq \inf _{a\in [\widehat {a}, 1)}A_{p, q}(a)$ . Moreover, for $\xi (z)=z\phi _{1/\sqrt {2}}(z)$ , $z\in \mathbb {D}$ , we have $R_{p, q}(\xi , \mathbb {C})=1/\sqrt {2}$ . Combining these two facts, we write

(2.7) $$ \begin{align} R_{p, q}(\mathbb{C})\leq\min\left\{(1/\sqrt{2}), \inf_{a\in[\widehat{a}, 1)}A_{p, q}(a)\right\}\hspace{-1.5pt}. \end{align} $$

We now deal with the problem case by case.

$\underline {\text {Case} \ p, q\in [1,2]}$ : Let us start with $p<2$ . From (2.6), it is evident that $Q^{\prime }(y)<0$ for $p<2$ , and hence $Q(y)<Q(1)=0$ for all $y\in (1, \infty )$ . Thus, from (2.5), it is clear that $P(y)$ is strictly increasing in $(0,1)$ with respect to a. Consequently, for all $y\in (1, \infty )$ ,

(2.8) $$ \begin{align} P(y)<\lim_{a\to 1-}P(y)=\frac{p}{2}, \end{align} $$

and using the above estimate in (2.4) gives, for all $a\in (0,1)$ ,

$$ \begin{align*}T^{\prime}(a) <\frac{2a^{p-1}T(a)}{1-a^p}\left(\frac{p}{2}-\frac{p}{q}\right) \leq 0, \end{align*} $$

as $q\leq 2$ . Therefore, $S_{p, q}(a)$ is strictly decreasing in $(0,1)$ , and after some calculations, we have, as a consequence,

$$ \begin{align*}\inf_{a\in[0, \widehat{a}]}S_{p, q}(a)=S_{p, q}(\widehat{a})=A_{p, q}(\widehat{a})\geq \inf_{a\in[\widehat{a}, 1)}A_{p, q}(a). \end{align*} $$

Hence, from (2.3), we have $R_{p,q}(\mathbb {C})\geq \inf _{a\in [\widehat {a}, 1)}A_{p, q}(a)$ . For $p=2$ , if $q<2$ , then $T^{\prime }(a)<0$ for all $a\in (0,1)$ , which (as in the case $p<2$ ) again gives $R_{2,q}(\mathbb {C})\geq \inf _{a\in [\widehat {a}, 1)}A_{2, q}(a)$ . Otherwise, if $p=q=2$ , then $\widehat {a}=1/\sqrt {2}$ , and for all $a\in [0, 1)$ , we get

$$ \begin{align*}S_{2, 2}(a)=1/\sqrt{2}=\inf_{a\in[\widehat{a}, 1)}A_{2, 2}(a). \end{align*} $$

Therefore, for all $p, q\in [1,2]$ , we have $R_{p,q}(\mathbb {C})\geq \inf _{a\in [\widehat {a}, 1)}A_{p, q}(a)$ , and from (2.7), it is known that $R_{p, q}(\mathbb {C})\leq \inf _{a\in [\widehat {a}, 1)}A_{p, q}(a)$ . This completes the proof for this case.

$\underline {\text {Case} \ p\in (2, \infty ), q\in [1,2]}$ : From (2.6), it is clear that $Q^{\prime }(y)>0$ for $p>2$ , and therefore $Q(y)>Q(1)=0$ for all $y\in (1, \infty )$ . It follows from (2.5) that $P(y)$ is strictly decreasing in $(0,1)$ with respect to a. Thus, for $q<2$ , the value of the quantity

$$ \begin{align*}P(y)-\frac{p}{q}=\frac{a^2(1-a^p)}{a^p(1-a^2)}-\frac{p}{q} \end{align*} $$

decreases from

$$ \begin{align*}\lim_{a\to 0+}(P(y)-(p/q))=+\infty\,\, \mbox{to}\,\, \lim_{a\to 1-}(P(y)-(p/q))=p((1/2)-(1/q))<0, \end{align*} $$

i.e., $P(y)-(p/q)>0$ in $(0, b_1)$ and $P(y)-(p/q)<0$ in $(b_1, 1)$ for some $b_1\in (0,1)$ , where $P(b_1)=(p/q)$ . As a consequence, $T^{\prime }(a)=0$ only for $a=0, b_1$ , and $T^{\prime }(a)>0$ in $(0, b_1)$ , $T^{\prime }(a)<0$ in $(b_1, 1)$ . Hence, $S_{p, q}(a)$ strictly increases in $(0, b_1)$ , and then strictly decreases in $(b_1, 1)$ , which implies that

$$ \begin{align*}\inf_{a\in[0, \widehat{a}]} S_{p, q}(a)=\min\left\{S_{p, q}(0), S_{p, q}(\widehat{a})\right\} =\min\left\{(1/\sqrt{2}), A_{p, q}(\widehat{a})\right\}\hspace{-1.5pt}. \end{align*} $$

Moreover, from the proof of the case $p, q\in [2, \infty )$ , we have $R_{p, 2}(\mathbb {C})=1/\sqrt {2}$ . These two facts combined with (2.3) readily yield

$$ \begin{align*}R_{p, q}(\mathbb{C})\geq \min\left\{(1/\sqrt{2}), \inf\limits_{a\in[\widehat{a}, 1)}A_{p, q}(a)\right\}, \end{align*} $$

and making use of (2.7), we arrive at our desired conclusion.

$\underline {\text {Case} \ p, q\in [2, \infty )}$ : Applying (2.7) of this paper, $(1.9)$ of [Reference Blasco9], and [Reference Blasco10, Remark 1.2] together, the proof follows immediately from the observation:

$$ \begin{align*}(1/\sqrt{2})\geq R_{p, q}(\mathbb{C})\geq R_{2, 2}(\mathbb{C})\geq (1/\sqrt{2})r_2(\mathbb{C})=1/\sqrt{2}. \end{align*} $$

$\underline {\text {Case} \ p\in [1,2], q\in (2, \infty )}$ : The fact that $R_{2, q}(\mathbb {C})=1/\sqrt {2}$ is evident from the proof of the case $p, q\in [2, \infty )$ . Furthermore, as we have already seen, from (2.1) it is clear that inequality (2.2) holds for $r\leq S_{p, q}(a), a\in [0,1)$ , and therefore for $r\leq \inf _{a\in [0,1)}S_{p ,q}(a)$ . From this and (2.7), we have (1.5) as an immediate consequence. The assertion $\inf _{a\in [0,1)}S_{p, q}(a)>0$ is validated from the fact that $S_{p, q}(a)\neq 0$ for all $a\in [0,1)$ and that $\lim _{a\to 1-}S_{p, q}(a)=1$ . Now, we will show that the imposition of the additional condition (1.4) gives an optimal value for $R_{p, q}(\mathbb {C})$ . We know that for $p<2$ , $P(y)$ is strictly increasing in $(0,1)$ with respect to a, and as a result, $P(y)-(p/q)$ increases from

$$ \begin{align*}\lim_{a\to 0+}(P(y)-(p/q))=-p/q\,\, \mbox{to}\,\, \lim_{a\to 1-}(P(y)-(p/q))=p((1/2)-(1/q))>0, \end{align*} $$

i.e., $P(y)-(p/q)<0$ in $(0, b_2)$ and $P(y)-(p/q)>0$ in $(b_2, 1)$ for some $b_2\in (0,1)$ , where $P(b_2)=(p/q)$ . As a consequence, $T^{\prime }(a)=0$ only for $a=0, b_2$ , and $T^{\prime }(a)<0$ in $(0, b_2)$ , $T^{\prime }(a)>0$ in $(b_2, 1)$ . Hence, $S_{p, q}(a)$ strictly decreases in $(0, b_2)$ , and then strictly increases in $(b_2, 1)$ . Now, if we assume the condition (1.4) in addition, it is equivalent to saying that $T^{\prime }(\widehat {a})\leq 0$ , i.e., $\widehat {a}\leq b_2$ . Thus, $\inf _{a\in [0, \widehat {a}]}S_{p, q}(a)=S_{p, q}(\widehat {a})=A_{p, q}(\widehat {a})$ . Consequently, from (2.3), we get $R_{p, q}(\mathbb {C})\geq \inf _{a\in [\widehat {a}, 1)}A_{p, q}(a)$ , which completes our proof for this case.

Proof of Proposition 1.3

As any holomorphic function $f:\mathbb {D}\to X$ can also be considered as a holomorphic function from $\mathbb {D}^n$ to X, it immediately follows that $R_{p, q}^n(X)>0$ for any $n\in \mathbb {N}$ implies that $R_{p, q}(X)>0$ . Thus, we only need to establish the converse. Any holomorphic $f:\mathbb {D}^n\to X$ with an expansion (1.1) can be written as

(2.9) $$ \begin{align} f(z)=x_0+\sum_{k=1}^{\infty} P_k(z), z\in\mathbb{D}^n, \end{align} $$

where $P_k(z):=\sum _{|\alpha |=k}x_{\alpha } z^{\alpha }$ . Thus, for any fixed $z_0\in \mathbb {T}^n$ (the n-dimensional torus), we have

(2.10) $$ \begin{align} g(u):=f(uz_0)=x_0+\sum_{k=1}^{\infty} P_k(z_0)u^k:\mathbb{D}\to X \end{align} $$

is holomorphic, and if $\|f\|_{H^{\infty }(\mathbb {D}^n, X)}\leq 1$ , then $\|g\|_{H^{\infty }(\mathbb {D}, X)}\leq 1$ . Hence, starting with the assumption $R_{p, q}(X)=R>0$ , we have $\|P_k(z_0)\|\leq (1/R^{k})(1-\|x_0\|^p)^{1/q}$ , and since $z_0$ is arbitrary, we conclude that $\sup _{z\in \mathbb {T}^n}\|P_k(z)\| \leq (1/R^{k})(1-\|x_0\|^p)^{1/q}$ for any $k\in \mathbb {N}$ . Therefore, for a given $k\in \mathbb {N}$ and for any $\alpha $ with $|\alpha |=k$ , we have

$$ \begin{align*} \|x_{\alpha}\|&=\left\|\frac{1}{(2\pi i)^n}\int_{|z_1|=1}\int_{|z_2|=1}\cdots\int_{|z_n|=1}\frac{P_k(z)}{z^{\alpha+1}}dz_ndz_{n-1}\cdots dz_1\right\|\\ &\leq\sup_{z\in\mathbb{T}^n}\|P_k(z)\|\leq\frac{1}{R^{k}}(1-\|x_0\|^p)^{\frac{1}{q}}. \end{align*} $$

As a result, we have, for all $r<R$ ,

$$ \begin{align*}\|x_0\|^p+\left(\sum_{k=1}^{\infty} r^k\sum_{|\alpha|=k}\|x_{\alpha}\|\right)^q \leq \|x_0\|^p+(1-\|x_0\|^p)\left(\left(\frac{R}{R-r}\right)^n-1\right)^q, \end{align*} $$

which is less than or equal to 1 whenever $r\leq R\left (1-\left (1/2\right )^{1/n}\right )$ , thereby asserting that $R_{p, q}^n(X)>0$ .

Proof of Theorem 1.4

(i) Before we start proving the first part of this theorem, note that the choice of $q\in [2, \infty )$ is again justified due to Proposition 1.3 and [Reference Bhowmik and Das6, Corollary 4]. Now, given a holomorphic $f:\mathbb {D}^n\to \mathcal {H}$ with an expansion (1.1) and with $\|f(z)\|\leq 1$ for all $z\in \mathbb {D}^n$ , we have, for any fixed $R\in (0, 1)$ ,

$$ \begin{align*}\left(2\pi\right)^{-n}\int_{\theta_1=0}^{2\pi}\int_{\theta_2=0}^{2\pi}\cdots\int_{\theta_n=0}^{2\pi}\left\|f\left(Re^{i\theta_1}, Re^{i\theta_2},\ldots, Re^{i\theta_n}\right)\right\|^2d\theta_n d\theta_{n-1}\cdots d\theta_1\leq 1, \end{align*} $$

which is the same as saying that

$$ \begin{align*}\|x_0\|^2+\sum_{|\alpha|\in\mathbb{N}}\|x_{\alpha}\|^2R^{2|\alpha|}+\left(2\pi\right)^{-n}MR^{|\alpha|+|\beta|}\leq 1 \end{align*} $$

with $M:=\sum _{\alpha \neq \beta }\langle x_{\alpha }, x_{\beta }\rangle \int _{\theta _1=0}^{2\pi }\int _{\theta _2=0}^{2\pi }\cdots \int _{\theta _n=0}^{2\pi }e^{i(\theta _1(\alpha _1-\beta _1)+\cdots +\theta _n(\alpha _n-\beta _n))}d\theta _n d\theta _{n-1}\cdots d\theta _1. $ Here, $\langle .,.\rangle $ is the inner product of $\mathcal {H}$ , $\alpha $ and $\beta $ denote as usual n-tuples $(\alpha _1, \alpha _2,\ldots , \alpha _n)$ and $(\beta _1, \beta _2,\ldots , \beta _n)$ of nonnegative integers, respectively. As we know $\int _{0}^{2\pi }e^{ik\theta }d\theta =0$ for any $k\in \mathbb {Z}\setminus \{0\}$ , $M=0$ . Letting $R\to 1-$ in the above inequality, we therefore get $\|x_0\|^2+\sum _{k=1}^{\infty }\sum _{|\alpha |=k}\|x_{\alpha }\|^2\leq 1$ . Taking $z\in r\mathbb {D}^n$ and using this inequality, we obtain

$$ \begin{align*} \|x_0\|^p+\left(\sum_{k=1}^{\infty} \sum_{|\alpha|=k}\|x_{\alpha} z^{\alpha}\|\right)^q &\leq\|x_0\|^p+\left(\sum_{k=1}^{\infty} \sum_{|\alpha|=k}\|x_{\alpha}\|^2\right)^{\frac{q}{2}}\left(\sum_{k=1}^{\infty} \sum_{|\alpha|=k}|z^{\alpha}|^2\right)^{\frac{q}{2}}\\ &\leq\|x_0\|^p+(1-\|x_0\|^2)^{\frac{q}{2}}\left(\sum_{k=1}^{\infty} {n+k-1 \choose k} r^{2k}\right)^{\frac{q}{2}}\\ &=\|x_0\|^p+(1-\|x_0\|^2)^{\frac{q}{2}}\left(\frac{1}{(1-r^2)^n}-1\right)^{\frac{q}{2}}, \end{align*} $$

which is less than or equal to 1 if

(2.11) $$ \begin{align} r\leq\left(1-(1-(S_{p,q}(\|x_0\|))^2)^{\frac{1}{n}}\right)^{\frac{1}{2}}\hspace{-2pt}, \end{align} $$

and therefore

(2.12) $$ \begin{align} R_{p, q}^n(\mathcal{H})\geq \inf_{a\in[0,1)}\left(1-(1-(S_{p,q}(a))^2)^{\frac{1}{n}}\right)^{\frac{1}{2}}\hspace{-2pt}. \end{align} $$

As the quantity on the right-hand side of inequality (2.11) becomes $\sqrt {1-(1/2)^{1/n}}$ at $x_0=0$ and converges to $1$ as $\|x_0\|\to 1-$ , we conclude that the infimum in inequality (2.12) is attained at some $b_3\in [0, 1)$ . Since every Hilbert space $\mathcal {H}$ has an orthonormal basis and, in our case, $\mbox {dim}(\mathcal {H})=\infty $ , we can choose a countably infinite set $\{e_{\alpha }\}_{|\alpha |\in \mathbb {N}\cup \{0\}}$ of orthonormal vectors in $\mathcal {H}$ . Setting $r_3=(1-(1-(S_{p,q}(b_3))^2)^{\frac {1}{n}})^{\frac {1}{2}}$ , we construct

$$ \begin{align*}\chi(z):=b_3e_0+\frac{1-b_3^2}{(1-b_3^p)^{\frac{1}{q}}}\sum_{k=1}^{\infty} r_3^k\left(\sum_{|\alpha|=k}z^{\alpha} e_{\alpha}\right):\mathbb{D}^n\to\mathcal{H}, \end{align*} $$

which satisfies $\|\chi (z)\|\leq 1$ for all $z\in \mathbb {D}^n$ , and $r_3=R_{p, q}^n(\chi , \mathcal {H})\geq R_{p, q}^n(\mathcal {H})$ . This completes the proof for the first part of this theorem.

(ii) The proof for this part is rather lengthy, so we break it into a couple of steps. Prior to each step, we will provide some auxiliary information whenever needed.

$\underline{\mathrm{Background\ for\ Step\ 1:}}$ If $R_{p, q}(X)>0$ , we have

$$ \begin{align*} \Omega_X(\delta)\leq C\left((1+\delta)^q-(1+\delta)^{q-p}\right)^{1/q},\,\delta\geq 0 \end{align*} $$

for some constant C (see (1.6) in the introduction). Given any such X, and given any holomorphic function $G(u)=\sum _{n=0}^{\infty } y_nu^n:\mathbb {D}\to X$ with $\|G(u)\|\leq 1$ in $\mathbb {D}$ , it is known from the proof of [Reference Bhowmik and Das6, Theorem 1] that

(2.13) $$ \begin{align} \|y_k\|\leq 2\Omega_X(1-\|y_0\|)\leq 2C\left((2-\|y_0\|)^q-(2-\|y_0\|)^{q-p}\right)^{1/q} \end{align} $$

for all $k\geq 1$ .

$\underline{\mathrm{Step\ 1:}}$ In our context, for any given holomorphic $f:\mathbb {D}^n\to X$ with an expansion (1.1) and with $\|f\|_{H^{\infty }(\mathbb {D}^n, X)}\leq 1$ , we define the holomorphic function $g(u)=x_0+\sum _{k=1}^{\infty } P_k(z_0)u^k:\mathbb {D}\to X$ as in (2.10), which satisfies $\|g(u)\|\leq 1$ for all $u\in \mathbb {D}$ , $z_0$ being any chosen point on $\mathbb {T}^n$ . Since $R_{p, q}(X)>0$ , making use of inequality (2.13), we conclude that for any $k\geq 1$ ,

$$ \begin{align*} \|P_k(z_0)\|\leq 2C\left((2-\|x_0\|)^q-(2-\|x_0\|)^{q-p}\right)^{1/q} \end{align*} $$

for any $z_0\in \mathbb {T}^n$ . Therefore,

(2.14) $$ \begin{align} \sup_{z\in\mathbb{T}^n}\|P_k(z)\|\leq 2C\left((2-\|x_0\|)^q-(2-\|x_0\|)^{q-p}\right)^{1/q} \end{align} $$

for any $k\in \mathbb {N}$ , C being the constant for which (1.6) is satisfied.

$\underline{\mathrm{Background\ for\ Step\ 2:}}$ For $1\leq p<\infty $ and for a linear operator $U:X_0\to Y_0$ between the complex Banach spaces $X_0$ and $Y_0$ , we say that U is p-summing if there exists a constant $c\geq 0$ such that regardless of the natural number m and regardless of the choice of $f_1, f_2,\ldots , f_m$ in $X_0$ , we have

$$ \begin{align*} \left(\sum_{i=1}^m\|U(f_i)\|^p\right)^{1/p}\leq c\sup_{\phi\in B_{X_0^*}}\left(\sum_{i=1}^m|\phi(f_i)|^p\right)^{1/p}, \end{align*} $$

where $B_{X_0^*}$ is the open unit ball in the dual space $X_0^*$ . The least c for which the above inequality always holds is denoted by $\pi _p(U)$ , and the set of all p-summing operators from $X_0$ into $Y_0$ is denoted by $\Pi _p(X_0, Y_0)$ . Now, from [Reference Diestel, Jarchow and Tonge18, Proposition 2.3], we know that:

Fact I. If $U:X_0\to Y_0$ is a bounded linear operator and $\mbox {dim}(U(X_0))<\infty $ , then U is p-summing for every $p\in [1, \infty )$ .

Moreover, [Reference Diestel, Jarchow and Tonge18, Theorem 2.8] states that:

Fact II. If $1\leq p<q<\infty $ , then $\Pi _p(X_0, Y_0)\subset \Pi _q(X_0, Y_0)$ . Moreover, for $U\in \Pi _p(X_0, Y_0)$ , we have $\pi _q(U)\leq \pi _p(U)$ .

$\underline{\mathrm{Step\ 2:}}$ Coming back to our proof now, we set $X_0=Y_0=X$ and $U=I$ —the identity operator on X. As X is finite-dimensional, $\mbox {dim}(I(X))<\infty $ in this case and thus using Fact I, we have $I\in \Pi _p(X, X)$ for all $p\geq 1$ . Therefore,

$$ \begin{align*}\left(\sum_{|\alpha|=k}\|x_{\alpha}\|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}}\leq \pi_{\frac{2k}{k+1}}(I)\sup_{\phi\in B_{X^*}}\left(\sum_{|\alpha|=k}|\phi(x_{\alpha})|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}} \end{align*} $$

for all $k\in \mathbb {N}$ . Since $2k/(k+1)>1$ for all $k\geq 2$ , Fact II asserts that $\pi _{\frac {2k}{k+1}}(I)\leq \pi _1(I)$ . Hence, there exists a constant $D=\pi _1(I)$ (depending only on X) such that

(2.15) $$ \begin{align} \left(\sum_{|\alpha|=k}\|x_{\alpha}\|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}}\leq D\sup_{\phi\in B_{X^*}}\left(\sum_{|\alpha|=k}|\phi(x_{\alpha})|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}} \end{align} $$

for all $k\in \mathbb {N}$ .

$\underline{\mathrm{Background\ for\ Step\ 3:}}$ From [Reference Bayart, Pellegrino and Seoane-Sepúlveda4, Theorem 1.1], we know that for any $\epsilon>0$ , there exists $\mu>0$ such that, for any complex k-homogeneous polynomial ( $k\geq 1$ ) $P(z)=\sum _{|\alpha |=k}c_{\alpha } z^{\alpha }$ ( $c_{\alpha }\in \mathbb {C}$ ), we have

$$ \begin{align*}\left(\sum_{|\alpha|=k}|c_{\alpha}|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}}\leq\mu(1+\epsilon)^k\sup_{z\in\mathbb{D}^n}|P(z)|. \end{align*} $$

$\underline{\mathrm{Step\ 3:}}$ Recall from (2.9) now that $P_k(z)=\sum _{|\alpha |=k}x_{\alpha } z^{\alpha },\, x_{\alpha }\in X$ , and hence $\phi (P_k(z))=\sum _{|\alpha |=k}\phi (x_{\alpha })z^{\alpha }$ for any $\phi \in B_{X^*}$ . Consequently, using the above inequality, we get that for any $\epsilon>0$ , there exists $\mu>0$ such that

$$ \begin{align*}\sup_{\phi\in B_{X^*}}\left(\sum_{|\alpha|=k}|\phi(x_{\alpha})|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}} \leq\mu(1+\epsilon)^k\sup_{\phi\in B_{X^*}}\sup_{z\in\mathbb{D}^n}\left|\phi\left(P_k(z)\right)\right| =\mu(1+\epsilon)^k\sup_{z\in\mathbb{T}^n}\|P_k(z)\| \end{align*} $$

for all $k\geq 1$ . Combining this inequality with inequalities (2.14) and (2.15) appropriately, we get

$$ \begin{align*} \left(\sum_{|\alpha|=k}\|x_{\alpha}\|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}}\leq 2\mu CD(1+\epsilon)^k\left((2-\|x_0\|)^q-(2-\|x_0\|)^{q-p}\right)^{1/q}. \end{align*} $$

It follows that

$$ \begin{align*} \left(\sum_{k=1}^{\infty} r^k\sum_{|\alpha|=k}\|x_{\alpha}\|\right)^q &\leq\left(\sum_{k=1}^{\infty} r^k\left(\sum_{|\alpha|=k}\|x_{\alpha}\|^{\frac{2k}{k+1}}\right)^{\frac{k+1}{2k}} {n+k-1 \choose k}^{\frac{k-1}{2k}}\right)^q\\ &\leq X\left(\sum_{k=1}^{\infty} r^k(1+\epsilon)^k{n+k-1 \choose k}^{\frac{k-1}{2k}}\right)^q, \end{align*} $$

where $X=\mu ^q C_1^q\left ((2-\|x_0\|)^q-(2-\|x_0\|)^{q-p}\right )$ , $C_1=2CD$ . Hence, for $z\in r\mathbb {D}^n$ , the inequality

$$ \begin{align*}\|x_0\|^p+\left(\sum_{k=1}^{\infty}\sum_{|\alpha|=k}\|x_{\alpha} z^{\alpha}\|\right)^q\leq 1 \end{align*} $$

is satisfied if

(2.16) $$ \begin{align} \left(\frac{X}{1-\|x_0\|^p}\right)^{\frac{1}{q}}\left(\sum_{k=1}^{\infty} r^k(1+\epsilon)^k{n+k-1 \choose k}^{\frac{k-1}{2k}}\right)\leq 1. \end{align} $$

Now, analyzing the function $f_1(t)=((2-t)^p-1)/(1-t^p), t\in [0,1)$ , we see that $f_1(t)\leq f_1(0)=2^p-1$ for all $t\in [0,1)$ , and hence

$$ \begin{align*}\frac{X}{1-\|x_0\|^p} =\mu^qC_1^q(2-\|x_0\|)^{q-p}f_1(\|x_0\|) \leq \begin{cases} \mu^qC_1^q 2^{q-p}(2^p-1)\,\,\mbox{if}\,\,q\geq p,\\ \mu^qC_1^q(2^p-1)\,\,\mbox{if}\,\,q\leq p. \end{cases} \end{align*} $$

Thus, inequality (2.16) is satisfied if

$$ \begin{align*} C_2\left(\sum_{k=1}^{\infty} r^k(1+\epsilon)^k{n+k-1 \choose k}^{\frac{k-1}{2k}}\right)\leq 1, \end{align*} $$

where $C_2$ is a new constant depending on $\mu , p, q$ and the Banach space X. Using the estimate

$$ \begin{align*}{n+k-1\choose k} \leq\frac{(n+k-1)^k}{k!}<\left(\frac{e}{k}\right)^k(n+k-1)^k <e^k\left(1+\frac{n}{k}\right)^k, \end{align*} $$

we get, by setting $r=(1-2\epsilon )\sqrt {(\log n)/n}$ ,

$$ \begin{align*}\sum_{k=1}^{\infty} r^k(1+\epsilon)^k{n+k-1 \choose k}^{\frac{k-1}{2k}}\leq \sum_{k=1}^{\infty}\left(\sqrt{\frac{\log n}{n}}\sqrt{e}(1-2\epsilon)(1+\epsilon)\right)^k \left(1+\frac{n}{k}\right)^{\frac{k-1}{2}}. \end{align*} $$

Hence, inequality (2.16) is satisfied if

(2.17) $$ \begin{align} C_2\sum_{k=1}^{\infty}\left(\sqrt{\frac{\log n}{n}}\sqrt{e}(1-2\epsilon)(1+\epsilon)\right)^k \left(1+\frac{n}{k}\right)^{\frac{k-1}{2}}\leq 1. \end{align} $$

Starting here, we will follow the similar lines of argument as in [Reference Bayart, Pellegrino and Seoane-Sepúlveda4, pp. 743–744]. For n large enough,

$$ \begin{align*}t_n:=\frac{\sqrt{\log n}}{n^{1/4}}\sqrt{2e}(1-2\epsilon)(1+\epsilon)<1, \end{align*} $$

and for $k>\sqrt {n}$ , observe that

$$ \begin{align*}\left(1+\frac{n}{k}\right)^{\frac{k-1}{2}} <(2\sqrt{n})^{\frac{k}{2}}. \end{align*} $$

Using both the above facts,

$$ \begin{align*} &\sum_{k>\sqrt{n}}\left(\sqrt{\frac{\log n}{n}}\sqrt{e}(1-2\epsilon)(1+\epsilon)\right)^k \left(1+\frac{n}{k}\right)^{\frac{k-1}{2}} &&\\ &\leq\sum_{k>\sqrt{n}}\left(\frac{\sqrt{\log n}}{n^{1/4}}\sqrt{2e}(1-2\epsilon)(1+\epsilon)\right)^k \leq\frac{t_n}{1-t_n}, \end{align*} $$

which goes to $0$ as $n\to \infty $ . For $k\leq \sqrt {n}$ , we start by making n sufficiently large such that $2<k_0\leq \log n$ can be chosen for which the inequalities

$$ \begin{align*}k_0^{\frac{1}{k_0-1}}\leq 1+\frac{\epsilon}{2}\;,\; \sum_{k_0\leq k\leq\sqrt{n}}((1-2\epsilon)(1+\epsilon)^{3/2})^k\leq\frac{1}{2C_2}\;\;\mbox{and}\;\; \left(\frac{1}{n}\right)^{\frac{k_0-2}{2(k_0-1)}}\leq \frac{\epsilon}{2} \end{align*} $$

are satisfied. Observing that $x^{1/(x-1)}$ is decreasing and $(x-2)/2(x-1)$ is increasing in $(1, \infty )$ , we obtain, for $k\geq k_0$ ,

$$ \begin{align*} \left(k^{\frac{k}{k-1}}\left(\frac{1}{n}+\frac{1}{k}\right)\right)^{\frac{k-1}{k}} &\leq\left(\left(\frac{1}{n}\right)^{\frac{k-2}{2(k-1)}}+k^{\frac{1}{k-1}}\right)^{\frac{k-1}{k}} &&\\ &\leq\left(\left(\frac{1}{n}\right)^{\frac{k_0-2}{2(k_0-1)}}+k_0^{\frac{1}{k_0-1}}\right)^{\frac{k-1}{k}} \leq(1+\epsilon)^{\frac{k-1}{k}} \leq 1+\epsilon, \end{align*} $$

which, after a little simplification, gives

$$ \begin{align*}\left(1+\frac{n}{k}\right)^{\frac{k-1}{2}} \leq(1+\epsilon)^{\frac{k}{2}} \frac{n^{\frac{k}{2}}}{n^{\frac{1}{2}}k^{\frac{k}{2}}}. \end{align*} $$

Since $x\mapsto n^{1/x}x$ is decreasing up to $x=\log n$ and increasing thereafter, we have $n^{1/k}k\geq e\log n$ . Therefore,

$$ \begin{align*} &\sum_{k_0\leq k\leq\sqrt{n}}\left(\sqrt{\frac{\log n}{n}}\sqrt{e}(1-2\epsilon)(1+\epsilon)\right)^k \left(1+\frac{n}{k}\right)^{\frac{k-1}{2}} &&\\ &\leq \sum_{k_0\leq k\leq\sqrt{n}} \left(\sqrt{e\log n}(1-2\epsilon)(1+\epsilon)^{3/2}\sqrt{\frac{1}{n^{1/k}k}}\right)^k &&\\ &\leq\sum_{k_0\leq k\leq\sqrt{n}}((1-2\epsilon)(1+\epsilon)^{3/2})^k\leq\frac{1}{2C_2}. \end{align*} $$

It remains to analyze the case $1\leq k\leq k_0$ . In this case, we observe that for n large enough,

$$ \begin{align*}\frac{k}{n}+1\leq\frac{k_0}{n}+1\leq\epsilon+1, \end{align*} $$

and hence

$$ \begin{align*}\left(1+\frac{n}{k}\right)^{\frac{k-1}{2}} \leq (1+\epsilon)^{\frac{k}{2}}\left(\frac{n}{k}\right)^{\frac{k-1}{2}}. \end{align*} $$

Making use of the above inequality and the fact that $x\mapsto n^{1/x}x$ is decreasing in $[1, k_0]$ (i.e., $n^{1/k}k\geq n^{1/k_0}k_0$ ), it is easily seen that

$$ \begin{align*} &\sum_{k=1}^{k_0}\left(\sqrt{\frac{\log n}{n}}\sqrt{e}(1-2\epsilon)(1+\epsilon)\right)^k \left(1+\frac{n}{k}\right)^{\frac{k-1}{2}} &&\\ &\leq \sum_{k=1}^{k_0}\left(\sqrt{e\log n}(1-2\epsilon)(1+\epsilon)^{3/2}\frac{k^{1/(2k)}}{k_0^{1/2}n^{1/(2k_0)}}\right)^k, \end{align*} $$

which tends to $0$ as $n\to \infty $ . Combining all the above three estimates, we have

$$ \begin{align*}\sum_{k=1}^{\infty}\left(\sqrt{\frac{\log n}{n}}\sqrt{e}(1-2\epsilon)(1+\epsilon)\right)^k \left(1+\frac{n}{k}\right)^{\frac{k-1}{2}} \leq\frac{1}{2C_2}+o(1) \end{align*} $$

for n large enough. Therefore, inequality (2.17) is satisfied for large enough n. Hence, for any given $\epsilon>0$ , $R_{p, q}^n(X)\geq (1-2\epsilon )\sqrt {\log n}/\sqrt {n}$ for sufficiently large n. This yields the following:

$$ \begin{align*}\liminf_{n\to\infty}R_{p, q}^n(X)\sqrt{n}/\sqrt{\log n}\geq 1. \end{align*} $$

$\underline{\mathrm{Step\ 4:}}$ In view of the above, it is only left to show that

(2.18) $$ \begin{align} \limsup_{n\to\infty}R_{p, q}^n(X)\sqrt{n}/\sqrt{\log n}\leq 1. \end{align} $$

As $R_{p, q}^n(X)\leq R_{p, q}^n(\mathbb {C})$ , it is sufficient to establish this part for $X=\mathbb {C}$ . The proof is exactly the same as the proof for the case $p=q=1$ given in [Reference Boas and Khavinson12, p. 2977], but for the sake of completeness, we reproduce the argument here. From the Kahane–Salem–Zygmund inequality, it is known that there is a constant B such that for every collection of complex numbers $c_{\alpha }$ and every integer $k>1$ , there is a choice of plus and minus signs for which the supremum of the modulus of $\sum _{|\alpha |=k}\pm c_{\alpha } z^{\alpha }$ in $\mathbb {D}^n$ does not exceed $B\left (n\sum _{|\alpha |=k}|c_{\alpha }|^2\log k\right )^{1/2}$ . We choose $c_{\alpha }=k!/\alpha !$ . Then $\sum _{|\alpha |=k}|c_{\alpha }|^2\leq k!n^k$ . By the definition of the generalized Bohr inequality in our context, we get

$$ \begin{align*} \left(\left(R_{p, q}^n(\mathbb{C})\right)^kn^k\right)^q &=\left(\sum_{|\alpha|=k}|c_{\alpha}|\left(R_{p, q}^n(\mathbb{C})\right)^k\right)^q\\ &\leq B^q\left(n\sum_{|\alpha|=k}|c_{\alpha}|^2\log k\right)^{q/2}\leq B^q\left(n^{\frac{k+1}{2}}(k!\log k)^{1/2}\right)^q, \end{align*} $$

or, equivalently,

$$ \begin{align*}R_{p, q}^n(\mathbb{C})\leq B^{1/k}n^{\frac{1-k}{2k}}(k!\log k)^{\frac{1}{2k}}. \end{align*} $$

We use Stirling’s formula $\lim _{k\to \infty }k!(\sqrt {2\pi k}(k/e)^k)^{-1}=1$ to conclude that

$$ \begin{align*}R_{p, q}^n(\mathbb{C})\leq\sqrt{\frac{k}{n}}\left(\frac{B_1^{1/k}n^{\frac{1}{2k}}k^{\frac{1}{4k}}(\log k)^{\frac{1}{2k}}}{\sqrt{e}}\right) \end{align*} $$

for a new constant $B_1$ . Setting $k=\lfloor \log n\rfloor $ ( $\lfloor .\rfloor $ is the floor function), we observe

$$ \begin{align*}\limsup_{n\to\infty}R_{p, q}^n(\mathbb{C})\sqrt{\frac{n}{\log n}} \leq\lim_{n\to\infty}\frac{B_1^{1/\lfloor \log n\rfloor}n^{\frac{1}{2\lfloor \log n\rfloor}}\lfloor \log n\rfloor^{\frac{1}{4\lfloor \log n\rfloor}}(\log \lfloor \log n\rfloor)^{\frac{1}{2\lfloor \log n\rfloor}}}{\sqrt{e}}=1, \end{align*} $$

which implies our desired inequality (2.18). This completes the proof.

Proof of Theorem 1.5

(i) Given a complex-valued holomorphic function f with an expansion (1.1) in $\mathbb {D}^n$ (“ $x_{\alpha }$ ’s” are complex numbers in this case) and satisfying $\|f\|_{H^{\infty }(\mathbb {D}^n, \mathbb {C})}\leq 1$ , an application of Hölder’s inequality yields

$$ \begin{align*} |x_0|^p+\sum_{k=1}^{\infty} r^{kp}\sum_{|\alpha|=k}|x_{\alpha}|^p &=\sum_{k=0}^{\infty} \sum_{|\alpha|=k}|x_{\alpha}|^{2-p}r^{kp}|x_{\alpha}|^{2p-2}\\ &\leq\left(\sum_{k=0}^{\infty} r^{\frac{kp}{2-p}}\sum_{|\alpha|=k}|x_{\alpha}|\right)^{2-p}\left(\sum_{k=0}^{\infty} \sum_{|\alpha|=k}|x_{\alpha}|^2\right)^{p-1}\\ &\leq\left(\sum_{k=0}^{\infty} r^{\frac{kp}{2-p}}\sum_{|\alpha|=k}|x_{\alpha}|\right)^{2-p}. \end{align*} $$

Therefore, $r_p^n(\mathbb {C})\geq (r_1^n(\mathbb {C}))^{(2-p)/p}$ . Since $\lim _{n\to \infty }r_1^n(\mathbb {C})\left (\sqrt {n}/\sqrt {\log n}\right )=1$ (cf. [Reference Bayart, Pellegrino and Seoane-Sepúlveda4]), we have

$$ \begin{align*}\liminf_{n\to\infty} r_p^n(\mathbb{C})\left(\frac{n}{\log n}\right)^{\frac{2-p}{2p}}\geq \liminf_{n\to\infty}\left(r_1^n(\mathbb{C})\sqrt{\frac{n}{\log n}}\right)^{\frac{2-p}{p}} =1, \end{align*} $$

and thus $r_p^n(\mathbb {C})\geq C((\log n)/n)^{(2-p)/2p}$ for some constant $C>0$ and for all $n>1$ . The upper bound $r_p^n(\mathbb {C})\leq D\left ((\log n)/n\right )^{(2-p)/2p}$ for some $D>0$ has already been established in [Reference Djakov and Ramanujan20, p. 76]. This completes the proof.

(ii) To handle the second part of this theorem, we first construct $g(u)$ as in (2.10) from a given holomorphic $f:\mathbb {D}^n\to X$ with an expansion (1.1) and satisfying $\|f\|_{H^{\infty }(\mathbb {D}^n, X)}\leq 1$ . Now, since X is p-uniformly $PL$ -convex, from the proof of [Reference Blasco and Pavlović11, Proposition 2.1(ii)], we obtain

$$ \begin{align*}\|P_1(z_0)\|\leq\frac{2}{(I_p(X))^{\frac{1}{p}}}(1-\|x_0\|^p)^{\frac{1}{p}} \end{align*} $$

for any arbitrary $z_0\in \mathbb {T}^n$ . Using a standard averaging trick (see, f.i., [Reference Blasco10, p. 94]), it can be shown that the $P_1(z_0)$ in the above inequality could be replaced by $P_k(z_0)$ for any $k\geq 2$ . Thus, we conclude that

(2.19) $$ \begin{align} \sup_{z\in\mathbb{T}^n}\|P_k(z)\| \leq\frac{2}{(I_p(X))^{\frac{1}{p}}}(1-\|x_0\|^p)^{\frac{1}{p}}. \end{align} $$

Now, from [Reference Defant, García, Maestre and Sevilla-Peris16, Lemma 25.18], it is known that there exists $R>0$ such that

$$ \begin{align*}\left(\sum_{|\alpha|=k}\|x_{\alpha}\|^p\right)R^{kp}\leq\int_{\mathbb{T}^n}\|P_k(z)\|^pdz. \end{align*} $$

Using inequality (2.19) gives

$$ \begin{align*} \sum_{|\alpha|=k}\|x_{\alpha}\|^p\leq\frac{2^p}{I_p(X)R^{kp}}(1-\|x_0\|^p). \end{align*} $$

Assuming $r<R$ , it is easy to see that

$$ \begin{align*} \|x_0\|^p+\sum_{k=1}^{\infty} r^{kp}\sum_{|\alpha|=k}\|x_{\alpha}\|^p &\leq\|x_0\|^p+\frac{2^p}{I_p(X)}(1-\|x_0\|^p)\sum_{k=1}^{\infty}\left(\frac{r}{R}\right)^{kp}\\ &\leq\|x_0\|^p+\frac{2^p}{I_p(X)}(1-\|x_0\|^p)\frac{r^p}{R^p-r^p}, \end{align*} $$

which is less than or equal to $1$ if

$$ \begin{align*}r\leq R\left(\frac{I_p(X)}{2^p+I_p(X)}\right)^{\frac{1}{p}}=\left(\frac{I_p(X)}{2^p+I_p(X)}\right)^{\frac{2}{p}}, \end{align*} $$

as from the arguments in [Reference Defant, García, Maestre and Sevilla-Peris16, p. 627], it is clear that we can take $R^p=I_p(X)/(I_p(X)+2^p)$ . This proves the lower estimate for $r_p^n(X)$ , and the upper estimate is trivial due to the fact that $r_p^n(X)\leq r_p^n(\mathbb {C})=1$ for $p\geq 2$ .