Convexity of the embedding parameter sets of some analytic function spaces

Proposition 1. If $\Phi$ is strictly increasing, then $\Phi \in \Delta_{\log}^-$ if and only if $\Phi^{-1} \in \Delta_{\log}^+$ .

Proposition 2. Let $\Phi_0 \in \mathcal{U}^{q_0}$ and $\Phi_1 \in \mathcal{U}^{q_1}$ with $q_0 < q_1$ . For $\theta \in [0,1]$ , define $\Phi_{\theta}$ by $\log \Phi_{\theta}(t) = (1 - \theta) \log \Phi_{0}(t) + \theta \log \Phi_{1}(t).$ Then $\Phi_{\theta} \in \mathcal{U}^{q_{\theta}}$ where $q_{\theta} = (1 - \theta)q_0 + \theta q_1$ .

Proposition 3. Let $\Phi_0 \in \mathcal{L}^{p_0}$ and $\Phi_1 \in \mathcal{L}^{p_1}$ with $p_0, p_1 \leq 1$ . For $\theta \in [0, 1]$ , define $\Phi_\theta$ by $\log \Phi_{\theta}(t) = (1 - \theta) \log \Phi_{0}(t) + \theta \log \Phi_{1}(t).$ Then $\Phi_{\theta} \in \mathcal{L}^{p_{\theta}}$ where $p_{\theta} = (1 - \theta)p_0 + \theta p_1$ .

Lemma 1. Let $\Phi \in \mathcal{L} \cup \mathcal{U}$ and $\Psi \in \mathcal{L} \cup \mathcal{U}$ be such $a_{\Psi} \geq b_{\Phi}$ . Then both that $\Psi/\Phi$ and $\Phi^{-1}/\Psi^{-1}$ are nondecreasing.

Proposition 4. Let $\Phi_0, \Phi_1, \Psi_0, \Psi_1$ be growth functions such that $a_{\Psi_i} \geq b_{\Phi_i}$ , i = 0, 1. Define the log-convex interpolations $$\Phi_{\theta}^{-1}(t) = \Phi_{0}^{-1}(t)^{1-\theta}\Phi_{1}^{-1}(t)^{\theta}, \quad \Psi_{\theta}^{-1}(t) = \Psi_{0}^{-1}(t)^{1-\theta}\Psi_{1}^{-1}(t)^{\theta}.$$ Then the ratios $R_{\theta}(t) = \Psi_{\theta}(t)/\Phi_{\theta}(t)$ and $\tilde{R}_{\theta}(t) = \Psi_{\theta}^{-1}(t)/\Phi_{\theta}^{-1}(t)$ are nondecreasing on $\mathbb{R}_+$ for every $\theta \in [0,1]$ .

Lemma 2. Let F(z) be analytic on the open strip 0 < Re z < 1 and continuous on its closure, and suppose $|F(z)| \le M_0$ on the line Re z = 0 and $|F(z)| \le M_1$ on the line Re z = 1. Then for any z with $\text{Re } z = \theta$ ( $0 < \theta < 1$ ), $$|F(z)| \leq M_0^{1-\theta} M_1^{\theta}$$ .

Theorem 1. Let $\Phi_1$ be a growth function such that $\Phi_1^{-1} \in \Delta_{\log}^-$ with constant $C_1 \geq 1$ , and let $\Phi_2$ be a growth function such that $\Phi_2^{-1} \in \Delta_{\log}^+$ with constant $C_2 \geq 1$ . Then the set $$\mathcal{E} = \left\{ (\alpha, \beta) \in \mathbb{R}^2 \mid \alpha, \beta \ge -1, \quad \exists C, K > 0 \ s.t. \ \Phi_1^{-1}(t^{2+\alpha}) \le C \, \Phi_2^{-1}(Kt^{2+\beta}) \ \forall t > 0 \right\}$$ is convex.

Corollary 1. The set $\mathcal{E}$ is an epigraph. More precisely, there exists a convex, nondecreasing function $\beta^* : [-1, \infty) \to [-1, \infty]$ such that $$\mathcal{E} = \{ (\alpha, \beta) : \alpha \ge -1, \ \beta \ge \beta^*(\alpha) \}.$$ Moreover, $\beta^*$ is given by $$\beta^*(\alpha) = \inf\left\{\beta \ge -1 \ : \ \exists C, K > 0, \ \Phi_1^{-1}(t^{2+\alpha}) \le C \, \Phi_2^{-1}(Kt^{2+\beta}) \ \forall t > 0\right\}.$$ Proof. We first note that if $(\alpha, \beta_0) \in \mathcal{E}$ and $\beta_1 \geq \beta_0$ , then $(\alpha, \beta_1) \in \mathcal{E}$ . Indeed, for t > 0, $t^{2+\beta_1} \geq t^{2+\beta_0}$ (since $2 + \beta_1 \geq 2 + \beta_0$ ). Because $\Phi_2^{-1}$ is increasing, we have $\Phi_2^{-1}(Kt^{2+\beta_1}) \geq \Phi_2^{-1}(Kt^{2+\beta_0})$ . Hence the inequality $\Phi_1^{-1}(t^{2+\alpha}) \leq C\Phi_2^{-1}(Kt^{2+\beta_0})$ implies the same with $\beta_1$ in place of $\beta_0$ (using the same C). Therefore $\mathcal{E}$ is an upward-closed set in the $\beta$ direction. This allows us to define the boundary function $\beta^*(\alpha)$ as the infimum over all $\beta$ such that $(\alpha, \beta) \in \mathcal{E}$ . The convexity of $\mathcal{E}$ (Theorem 1) implies that $\beta$ is convex. Indeed, for any $\alpha_0, \alpha_1$ and $\theta \in [0,1]$ , take any $\beta_0 \geq \beta^(\alpha_0)$ and $\beta_1 \geq \beta^(\alpha_1)$ . Then $(\alpha_0, \beta_0), (\alpha_1, \beta_1) \in \mathcal{E}$ , so by convexity $(\alpha_\theta, \beta_\theta) \in \mathcal{E}$ with $\beta_\theta = (1-\theta)\beta_0 + \theta\beta_1$ . Taking infimum over $\beta_0, \beta_1$ yields $\beta^(\alpha_\theta) \leq (1-\theta)\beta^(\alpha_0) + \theta\beta^(\alpha_1)$ , i.e., $\beta^*$ is convex. Let us check that $\beta$ is nondecreasing. If $\alpha_1 \geq \alpha_0$ and $(\alpha_0, \beta) \in \mathcal{E}$ , then for t > 0, $t^{2+\alpha_1} \geq t^{2+\alpha_0}$ , so $\Phi_1^{-1}(t^{2+\alpha_1}) \geq \Phi_1^{-1}(t^{2+\alpha_0})$ . This implies for any $\beta_1$ such that $\Phi_1^{-1}(t^{2+\alpha_1}) \leq C\Phi_2^{-1}(Kt^{2+\beta_1})$ , we also have $\Phi_1^{-1}(t^{2+\alpha_0}) \leq C\Phi_2^{-1}(Kt^{2+\beta_1})$ . Hence $\beta^(\alpha_0) \leq \beta_1$ . This implies $\beta^(\alpha_1) \geq \beta^(\alpha_0)$ . Thus $\beta^*$ is nondecreasing. The proof is complete. 3.2. Log-Convexity of $\mathcal{F}$ and interpolation of Bergman-Orlicz space embeddings. We now consider the dual problem: fix the weights $\alpha, \beta \geq -1$ and consider the set of growth function pairs.

Theorem 2. (Inverse log-convexity of $\mathcal{F}$ ). Let $\alpha, \beta \geq -1$ be fixed. Define $$\mathcal{F} = \left\{ (\Phi_1, \Phi_2) \in \mathcal{G}^2 \mid \exists C > 0 \text{ such that } \Phi_1^{-1}(t^{2+\alpha}) \le C \Phi_2^{-1}(t^{2+\beta}) \quad \forall t > 0 \right\}.$$ Then $\mathcal{F}$ is inverse log-convex. That is, if $(\Phi_0, \Psi_0), (\Phi_1, \Psi_1) \in \mathcal{F}$ , then for any $\theta \in (0, 1)$ the pair $(\Phi_{\theta}, \Psi_{\theta})$ defined by $$\Phi_{\theta}^{-1}(t) = \Phi_{0}^{-1}(t)^{1-\theta} \, \Phi_{1}^{-1}(t)^{\theta}, \qquad \Psi_{\theta}^{-1}(t) = \Psi_{0}^{-1}(t)^{1-\theta} \, \Psi_{1}^{-1}(t)^{\theta}$$ also belongs to $\mathcal{F}$ . First proof: through the three-lines lemma. Let t > 0 be arbitrary. For each fixed t, define the functions $$u_0(t) = \log \Phi_0^{-1}(t^{2+\alpha}) - \log \Psi_0^{-1}(t^{2+\beta}),$$ $$u_1(t) = \log \Phi_1^{-1}(t^{2+\alpha}) - \log \Psi_1^{-1}(t^{2+\beta}).$$ The hypothesis $(\Phi_0, \Psi_0) \in \mathcal{F}$ implies that there exists a constant $C_0 > 0$ such that for all t > 0, $$(9) u_0(t) \le \log C_0.$$ Similarly, there exists $C_1 > 0$ such that for all t > 0, $$(10) u_1(t) \le \log C_1.$$ For $z \in \mathbb{C}$ with $0 < \operatorname{Re} z < 1$ , define $$\Phi_z^{-1}(t) = \Phi_0^{-1}(t)^{1-z} \, \Phi_1^{-1}(t)^z, \qquad \Psi_z^{-1}(t) = \Psi_0^{-1}(t)^{1-z} \, \Psi_1^{-1}(t)^z.$$ For each fixed t > 0, the functions $$\log \Phi_z^{-1}(t) = (1-z)\log \Phi_0^{-1}(t) + z\log \Phi_1^{-1}(t),$$ $$\log \Psi_z^{-1}(t) = (1-z) \log \Psi_0^{-1}(t) + z \log \Psi_1^{-1}(t)$$ are affine in z. Consequently, the function $$G_t(z) = \exp\left(\log \Phi_z^{-1}(t^{2+\alpha}) - \log \Psi_z^{-1}(t^{2+\beta})\right) = \Phi_z^{-1}(t^{2+\alpha}) \cdot \Psi_z^{-1}(t^{2+\beta})^{-1}$$ is an exponential of an affine function in z, hence analytic in z. On the line z = it, we have $$\log \Phi_{it}^{-1}(t^{2+\alpha}) = (1-it)\log \Phi_0^{-1}(t^{2+\alpha}) + it\log \Phi_1^{-1}(t^{2+\alpha}).$$ The real part is $\log \Phi_0^{-1}(t^{2+\alpha})$ . Similarly, the real part of $\log \Psi_{it}^{-1}(t^{2+\beta})$ is $\log \Psi_0^{-1}(t^{2+\beta})$ . Therefore, $$|G_t(it)| = \exp\left(\log \Phi_0^{-1}(t^{2+\alpha}) - \log \Psi_0^{-1}(t^{2+\beta})\right) = \frac{\Phi_0^{-1}(t^{2+\alpha})}{\Psi_0^{-1}(t^{2+\beta})} \le C_0,$$ by (9). On the line z = 1 + it, $$|G_t(1+it)| = \frac{\Phi_1^{-1}(t^{2+\alpha})}{\Psi_1^{-1}(t^{2+\beta})} \le C_1,$$ by (10). For each fixed t > 0, $G_t(z)$ is analytic on the strip $0 \le \Re z \le 1$ and uniformly bounded (the bound depends on t but is finite pointwise). The three-lines lemma gives, for $z = \theta$ , $$|G_t(\theta)| \le C_0^{1-\theta} C_1^{\theta}.$$ But $G_t(\theta)$ is precisely $$G_t(\theta) = \Phi_{\theta}^{-1}(t^{2+\alpha}) \cdot \Psi_{\theta}^{-1}(t^{2+\beta})^{-1},$$ where $\Phi_{\theta}^{-1}$ and $\Psi_{\theta}^{-1}$ are defined as in the theorem statement. Thus, for all t > 0, $$\Phi_{\theta}^{-1}(t^{2+\alpha}) \le C_{\theta} \Psi_{\theta}^{-1}(t^{2+\beta}),$$ with $C_{\theta} = C_0^{1-\theta} C_1^{\theta}$ . This is exactly the condition that $(\Phi_{\theta}, \Psi_{\theta}) \in \mathcal{F}$ . Therefore, $\mathcal{F}$ is inverse log-convex. $\square$ Second proof: an algebraic approach. This proof is straightforward and uses only elementary algebra. Since $(\Phi_0, \Psi_0) \in \mathcal{F}$ , there exists $C_0 > 0$ such that for all t > 0, (11) $$\Phi_0^{-1}(t^{2+\alpha}) \le C_0 \,\Psi_0^{-1}(t^{2+\beta}).$$ Since $(\Phi_1, \Psi_1) \in \mathcal{F}$ , there exists $C_1 > 0$ such that for all t > 0, (12) $$\Phi_1^{-1}(t^{2+\alpha}) \le C_1 \Psi_1^{-1}(t^{2+\beta}).$$ Now fix $\theta \in (0,1)$ . Raise inequality (11) to the power $1-\theta$ and inequality (12) to the power $\theta$ (all quantities are positive, so this preserves the inequalities): $$\Phi_0^{-1} (t^{2+\alpha})^{1-\theta} \le C_0^{1-\theta} \, \Psi_0^{-1} (t^{2+\beta})^{1-\theta} ,$$ $$\Phi_1^{-1} (t^{2+\alpha})^{\theta} \le C_1^{\theta} \, \Psi_1^{-1} (t^{2+\beta})^{\theta} .$$ Multiply the two inequalities, we obtain $$\Phi_0^{-1} (t^{2+\alpha})^{1-\theta} \Phi_1^{-1} (t^{2+\alpha})^{\theta} \le C_0^{1-\theta} C_1^{\theta} \Psi_0^{-1} (t^{2+\beta})^{1-\theta} \Psi_1^{-1} (t^{2+\beta})^{\theta}.$$ But by definition of the interpolated inverses $$\Phi_{\theta}^{-1}(t^{2+\alpha}) = \Phi_{0}^{-1}(t^{2+\alpha})^{1-\theta} \Phi_{1}^{-1}(t^{2+\alpha})^{\theta},$$ $$\Psi_{\theta}^{-1}(t^{2+\beta}) = \Psi_{0}^{-1}(t^{2+\beta})^{1-\theta} \Psi_{1}^{-1}(t^{2+\beta})^{\theta}.$$ Therefore, $$\Phi_{\theta}^{-1}(t^{2+\alpha}) \le C_{\theta} \Psi_{\theta}^{-1}(t^{2+\beta}),$$ where $C_{\theta} = C_0^{1-\theta} C_1^{\theta}$ . Thus $(\Phi_{\theta}, \Psi_{\theta}) \in \mathcal{F}$ . The proof is complete. We observe the following about the interpolation constant in the above result.

Corollary 2. For fixed $(\alpha, \beta)$ and fixed endpoint pairs $(\Phi_0, \Psi_0), (\Phi_1, \Psi_1) \in \mathcal{F}$ , the minimal constant $$C_{\min}(\Phi, \Psi) = \sup_{t>0} \frac{\Phi^{-1}(t^{2+\alpha})}{\Psi^{-1}(t^{2+\beta})}$$ satisfies $$C_{\min}(\Phi_{\theta}, \Psi_{\theta}) \le C_{\min}(\Phi_{0}, \Psi_{0})^{1-\theta} C_{\min}(\Phi_{1}, \Psi_{1})^{\theta}.$$ Thus $\log C_{\min}$ is a convex function along log-convex interpolations.

Theorem 3. Let $\alpha \geq -1$ , and $\beta > -1$ be fixed. Suppose that for two pairs of growth functions $(\Phi_0, \Psi_0) \in \mathcal{L} \cup \mathcal{U}$ and $(\Phi_1, \Psi_1) \in \mathcal{L} \cup \mathcal{U}$ we have $a_{\Psi_i} \geq b_{\Phi_i}$ , i = 0, 1. Then for any $\theta \in (0,1)$ , the interpolated growth functions defined by $$\Phi_{\theta}^{-1}(t) = \Phi_{0}^{-1}(t)^{1-\theta}\Phi_{1}^{-1}(t)^{\theta}, \quad \Psi_{\theta}^{-1}(t) = \Psi_{0}^{-1}(t)^{1-\theta}\Psi_{1}^{-1}(t)^{\theta}$$ satisfy - The ratio Ψ_θ(t)/Φ_θ(t) is nondecreasing, The embedding A^Φ_θ_α(Ω) ⇔ A^Ψ_θ_β(Ω) holds.

Definition 1. $\Phi \in \mathcal{G}$ has upper type q if there exists C > 0 such that $$\Phi(st) < Cs^q \Phi(t), \quad \forall s > 1, \ \forall t > 0.$$ Let $\mathcal{U}^q$ denote the set of growth functions of upper type q such that $t \mapsto \Phi(t)/t$ is nondecreasing. Then $$\mathcal{U}:=\bigcup_{q\geq 1}\mathcal{U}^q.$$

Definition 2. $\Phi \in \mathcal{G}$ has lower type p if there exists C > 0 such that $$\Phi(st) \le Cs^p \Phi(t), \quad \forall 0 < s < 1, \ \forall t > 0.$$ Let $\mathcal{L}^p$ denote the set of growth functions of lower type p such that $t \mapsto \Phi(t)/t$ is nonincreasing. Then $$\mathcal{L} := \bigcup_{p \leq 1} \mathcal{L}^p.$$ Let us denote by $d\nu$ the Lebesgue measure on the unit ball $\mathbb{B}^n$ of $\mathbb{C}^n$ . For $\alpha > -1$ , we denote by $d\nu_{\alpha}$ the normalized Lebesgue measure $d\nu_{\alpha}(z) = c_{\alpha}(1-|z|^2)^{\alpha}d\nu(z)$ , $c_{\alpha}$ being the normalization constant. In the upper half-plane of $\mathbb{C}$ , we use the notation $dV_{\alpha}(x+iy) = y^{\alpha}dxdy$ . Let $\Omega$ will be either the unit ball of $\mathbb{C}^n$ $(n \in \mathbb{N})$ or the upper half-plane of $\mathbb{C}$ ; we use the notation $d\Omega_{\alpha}$ for either $d\nu_{\alpha}(z) = c_{\alpha}(1-|z|^2)^{\alpha}d\nu(z)$ or $dV_{\alpha}(x+iy)$ . For $\Phi$ a growth function, the weighted Bergman-Orlicz space $A^{\Phi}_{\alpha}(\Omega)$ is the space of all holomorphic functions f such that $$||f||_{\Phi,\alpha} = ||f||_{A^{\Phi}_{\alpha}} := \int_{\Omega} \Phi(|f(z)|) d\Omega_{\alpha}(z) < \infty.$$ We define on $A^{\Phi}_{\alpha}(\Omega)$ the following (quasi)-norm (1) $$||f||_{\Phi,\alpha}^{lux} = ||f||_{A_{\alpha}^{\Phi}}^{lux} := \inf\{\lambda > 0 : \int_{\Omega} \Phi\left(\frac{|f(z)|}{\lambda}\right) d\Omega_{\alpha}(z) \le 1\}.$$ And associated space to the above, is the Hardy-Orlicz space $H^{\Phi}(\Omega)$ that we understand as the limit of $A^{\Phi}_{\alpha}(\Omega)$ when $\alpha \to -1^+$ . For the specific definition, we refer to [4].

Definition 3. A growth function $\Phi$ is said to be log-convex up to a constant $C \geq 1$ (written $\Phi \in \Delta_{\log}^-$ ) if $$\Phi(u^{\theta}v^{1-\theta}) \le C^{\theta(1-\theta)}\Phi(u)^{\theta}\Phi(v)^{1-\theta}, \qquad \forall u, v > 0, \ \theta \in [0, 1].$$ It is log-concave up to a constant $C \geq 1$ (written $\Phi \in \Delta_{\log}^+$ ) if $$\Phi(C^{\theta(1-\theta)}u^{\theta}v^{1-\theta}) \ge \Phi(u)^{\theta}\Phi(v)^{1-\theta}, \qquad \forall u, v > 0, \ \theta \in [0, 1].$$ We observe the following.

Definition 4. For $\Phi \in \mathcal{C}^1(\mathbb{R}_+)$ a growth function, define $$a_{\Phi} := \inf_{t>0} \frac{t\Phi'(t)}{\Phi(t)}, \qquad b_{\Phi} := \sup_{t>0} \frac{t\Phi'(t)}{\Phi(t)}.$$ These are called the Matuszewska-Orlicz indices; $a_{\Phi}$ is the lower indice, while $b_{\Phi}$ is the upper indice. We need the following to prove the next result.