Nonlinear resolvents: distortion and order of starlikeness

Theorem 1.1. Let f have the form f(x) = p(x)x, where $p \in \text{Hol}(\mathbb{B}, \mathbb{C})$ with $\text{Re } p(x) \geq a$ , $x \in \mathbb{B}$ , for some $a \geq 0$ . Let $\{G_{\lambda}\}_{{\lambda}>0}$ be the resolvent family for f. Denote g = p(0). Then for any ${\lambda} > 0$ we have $$||G_{\lambda}(x)|| \le \left(\frac{2}{A + \sqrt{B}}\right)^{\frac{1}{2}} ||x||, \qquad x \in \mathbb{B}, \tag{1.1}$$ where $$A := |1 - \lambda q|^2 + 4\lambda a + 1$$ and $B := (|1 - \lambda q|^2 - 1)^2 + 8\lambda^3 a|q|^2$ . Inter alia, this theorem enables us to show that nonlinear resolvents themselves are holomorphically accretive and to find squeezing ratio of semigroups generated by them (see Theorem 4.5). The second main result establishes order of starlikeness for resolvents. To formulate it we denote $T(r) = \frac{2\alpha r}{(1+\beta)(1-r)^2 + \alpha(1-r^2)}, \, \alpha > 0, \, \beta \geq 0$ , and $\rho^* := \frac{\sqrt{1+\lambda \text{Re }q}}{\sqrt{2\lambda(\text{Re }q-a)} + \sqrt{1+\lambda \text{Re }q}}$ .

Theorem 1.2. Under the above conditions $G_{\lambda}$ is a starlike mapping of order $\frac{1}{2}$ for any $\lambda > 0$ . Moreover, if for some $\lambda > 0$ , $||G_{\lambda}(x)|| \leq \rho$ on $\mathbb{B}$ , then $$\left| \frac{1}{\|x\|^2} \left\langle (G'_{\lambda}(x))^{-1} G_{\lambda}(x), x^* \right\rangle - 1 \right| \le T(\rho) \quad \text{for all} \quad x \in \mathbb{B},$$ where $\alpha = \lambda(\operatorname{Re} q - a)$ and $\beta = \lambda a$ . Consequently, if $\rho \leq \rho^*$ , then $G_{\lambda}$ is starlike of order $\frac{1}{1+T(\rho)}$ and strongly starlike of order $\frac{2 \arcsin T(\rho)}{\pi}$ . In the next section, we introduce required notions including holomorphically accretive mappings and nonlinear resolvents. Section 3 is devoted to the class $\widehat{\text{Hol}}(\mathbb{B},X)$ and its subclasses. In Sections 4–5 we formulate and prove our main results.

Proposition 2.1. A mapping $f \in \text{Hol}(\mathbb{B}, X)$ , f(0) = 0, is holomorphically accretive if and only if $\text{Re } \langle f(x), x^* \rangle \geq 0$ for all $x \in \mathbb{B}$ . Definition 2.4 below presents some refinements connected to this fact. To demonstrate the importance of holomorphically accretive mappings we recall two concepts. One says that $\diamond f \in \operatorname{Hol}(\mathbb{B}, X)$ is an infinitesimal generator if for any $x \in \mathbb{B}$ the Cauchy problem $$\begin{cases} \frac{\partial u(t,x)}{\partial t} + f(u(t,x)) = 0, \\ u(0,x) = x, \end{cases}$$ (2.3) has the unique solution $u(t,x) \in \mathbb{B}$ for all $t \geq 0$ . This solution $\{u(t,\cdot)\}_{t\geq 0} \subset \operatorname{Hol}(\mathbb{B})$ forms the semigroup with respect to composition operator. Now we define non-linear resolvents, the main object of the study in this paper.

Proposition 2.3. Let $f \in \text{Hol}(\mathbb{B}, X)$ . Then - the mapping f is holomorphically accretive if and only if it satisfies the range condition on $\mathbb{B}$ . - If, in addition, f(0) = 0 and f is bounded on each subset strictly inside $\mathbb{B}$ , then f is holomorphically accretive if and only if it is an infinitesimal generator. The set of all holomorphically accretive mappings having an isolated null at the origin is often denoted by $\mathcal{N}_0$ (see, for example, [19]). Hence the following clarification is natural.

Proposition 2.5. Let f be a holomorphically accretive mapping on $\mathbb{B}$ and $\{u(t,\cdot)\}_{t\geq 0}$ be a semigroup generated by f. (i) If $f \in \mathcal{N}_a$ , a > 0, then $\{u(t, \cdot)\}_{t \geq 0}$ converges to zero uniformly on $\mathbb{B}$ with the squeezing ratio $\kappa = a$ , which means that $$||u(t,x)|| \le e^{-at}||x|| \quad for \ all \ \ x \in \mathbb{B}. \tag{2.7}$$ (ii) If X is a Hilbert space and the semigroup $\{u(t,\cdot)\}_{t\geq 0}$ satisfies inequality (2.7) for some a>0, then $f\in\mathcal{N}_a$ . Assertion (i) follows from [13, Lemma 3.3.2], where we set $\alpha(s) = as$ , while assertion (ii) is proved in [12], see also [14, 30]. 2.2. Starlike mappings. Holomorphically accretive mappings play an essential role not only in dynamical systems, but also in geometric function theory (the reader is referred to [14, 19, 30]). Following [19], see also [13, 23, 26, 33], we recall several notions. A mapping $f \in \text{Hol}(D, D_1)$ is said to be biholomorphic if it is invertible and $f^{-1} \in \text{Hol}(D_1, D)$ .

Proposition 3.2. Let $p \in \text{Hol}(\mathbb{B}, \mathbb{C})$ , q = p(0) and $a \geq 0$ . The following assertions are equivalent: - (i) Re $p(x) \ge a$ for all $x \in \mathbb{B}$ ; - (ii) (the Riesz-Herglotz formula) for every $u \in \partial \mathbb{B}$ there is a probability measure $\mu = \mu_u$ on the unit circle such that for all $z \in \mathbb{D}$ $$p(zu) = \int_{|\zeta|=1} \frac{(1+z\overline{\zeta})(\operatorname{Re} q - a)}{1-z\overline{\zeta}} d\mu(\zeta) + a + i\gamma, \quad \gamma = \operatorname{Im} q;$$ - (iii) for every $x \in \mathbb{B}$ the value p(x) lies in the closed disk centered at the point $c(x) := \frac{p(0) + \|x\|^2 \overline{p(0)} 2a\|x\|^2}{1 \|x\|^2}$ and of radius $r(x) := \frac{2\|x\| (\operatorname{Re} p(0) a)}{1 \|x\|^2};$ - (iv) (Harnack's type inequality) for every $x \in \mathbb{B}$ the following inequality holds $$\frac{(1-\|x\|)\operatorname{Re} p(0)+2a\|x\|}{1+\|x\|} \le \operatorname{Re} p(x) \le \frac{(1+\|x\|)\operatorname{Re} p(0)-2a\|x\|}{1-\|x\|}.$$

Proposition 3.3. Let $f \in \widehat{\mathcal{N}}_0$ , that is, f(x) = p(x)x, and let $\{G_{\lambda}\}_{{\lambda} \geq 0} \in \operatorname{Hol}(\mathbb{B})$ be its resolvent family. Then - (a) $\{G_{\lambda}\}_{{\lambda}>0} \subset \widehat{\operatorname{Hol}}(\mathbb{B})$ , that is, for every ${\lambda} \geq 0$ there is $g_{\lambda} \in \operatorname{Hol}(\mathbb{B},\mathbb{C})$ such that $G_{\lambda}(x) = g_{\lambda}(x)x$ . - (b) for every $u \in \partial \mathbb{B}$ , the function $\widetilde{f}_u \in \operatorname{Hol}(\mathbb{D}, \mathbb{C})$ defined by $\widetilde{f}_u(z) := p(zu)z$ is holomorphically accretive on the open unit disk $\mathbb{D}$ . The resolvent family of $\widetilde{f}_u$ is $\{g_{\lambda}(\cdot u)\cdot\}_{\lambda>0}$ .

Proposition 3.4. Let $f \in \widehat{\text{Hol}}(\mathbb{B}, X)$ , that is, f(x) = p(x)x, where $p \in \text{Hol}(\mathbb{B}, \mathbb{C})$ . Then f is a starlike mapping of order $\gamma \in [0, 1]$ (respectively, strongly starlike of order $\beta \in [0, 1]$ ) if and only if for every $u \in \partial \mathbb{B}$ , the function $h \in \text{Hol}(\mathbb{D}, \mathbb{C})$ defined by h(z) := p(zu)z is starlike of order $\gamma$ (respectively, strongly starlike of order $\beta$ ) in $\mathbb{D}$ .

Theorem 4.1. Let $f \in \widehat{\mathcal{N}}_a$ for some $a \geq 0$ , that is, f(x) = p(x)x, where $\operatorname{Re} p(x) \geq a$ , $x \in \mathbb{B}$ . Let $\{G_{\lambda}\}_{{\lambda}>0}$ be the resolvent family for f. Denote q = p(0). Then for any ${\lambda} > 0$ we have $$||G_{\lambda}(x)|| \le \left(\frac{2}{A + \sqrt{B}}\right)^{\frac{1}{2}} ||x||, \qquad x \in \mathbb{B},\tag{4.1}$$ where $A := |1 - \lambda q|^2 + 4\lambda a + 1$ and $B := (|1 - \lambda q|^2 - 1)^2 + 8\lambda^3 a|q|^2$ .

Corollary 4.3. If in hypotheses of Theorem 4.1, $f \in \widehat{\mathcal{N}}_0$ , then $$||G_{\lambda}(x)|| < \begin{cases} ||x||, & as \ \lambda \le \frac{2\operatorname{Re} q}{|q|^2} \\ \frac{||x||}{|1-\lambda q|}, & as \ \lambda > \frac{2\operatorname{Re} q}{|q|^2} \end{cases}, \quad z \in \mathbb{D}.$$ In particular, the mapping $G_{\lambda}$ , $\lambda > \frac{2\operatorname{Re}q}{|q|^2}$ , has no boundary fixed points. In Fig. 1, the graph of the function $\lambda \mapsto \sqrt{\frac{2}{A+\sqrt{B}}}$ is presented for the values of the parameters $q=1,\ a=0$ and $q=1,\ a=\frac{1}{40}$ , respectively. To proceed, recall that by Definition 2.2, $$G_{\lambda}(x) + \lambda f(G_{\lambda}(x)) = x,$$ hence $$\lambda \langle f(G_{\lambda}(x)), x^ \rangle = ||x||^2 - \langle G_{\lambda}(x), x^ \rangle.$$ Therefore, estimate (4.1) implies that $$\lambda \operatorname{Re} \langle f(G_{\lambda}(x)), x^* \rangle \ge ||x||^2 \left( 1 - \left( \frac{2}{A + \sqrt{B}} \right)^{\frac{1}{2}} \right).$$ So, we conclude the following:

Corollary 4.4. Let $f \in \widehat{\mathcal{N}}_a$ for some $a \geq 0$ and $\{G_{\lambda}\}_{{\lambda}>0}$ be the resolvent family for f. Then $$f \circ G_{\lambda} \in \mathcal{N}_{a_{\lambda}}, \quad where \quad a_{\lambda} = \frac{1}{\lambda} \left( 1 - \left( \frac{2}{A + \sqrt{B}} \right)^{\frac{1}{2}} \right).$$ Thus $f \circ G_{\lambda}$ is a holomorphically accretive mapping that generate the semigroup with squeezing ratio $a_{\lambda}$ . It turns out that resolvents themselves are holomorphically accretive. In the next theorem we prove this fact along estimation of squeezing ratio of generated semigroups.

Theorem 4.5. Let $f \in \widehat{\mathcal{N}}_a$ , f(x) = p(x)x, q = p(0) and $G = G_{\lambda}$ is the resolvent of f corresponding to the parameter $\lambda > 0$ . Denote $$\phi(t) := \frac{1 + \lambda \operatorname{Re} q - t (1 - \lambda (\operatorname{Re} q - 2a))}{|1 + \lambda q|^2 - t |1 - \lambda (q - 2a)|^2}.$$ (4.3) Then $G \in \widehat{\mathcal{N}}_{d_{\lambda}}$ , $d_{\lambda} = \min\left\{\frac{1+\lambda \operatorname{Re} q}{|1+\lambda q|^2}, \phi\left(\frac{2}{A+\sqrt{B}}\right)\right\}$ , where $A = |1-\lambda q|^2 + \frac{1}{2}$ $4\lambda a + 1$ and $B = (|1 - \lambda q|^2 - 1)^2 + 8\lambda^3 a |q|^2$ as above.

Theorem 5.1. Let $f \in \widehat{\mathcal{N}}_a$ , $a \geq 0$ , and $G(=G_{\lambda})$ be the resolvent of f corresponding to the parameter $\lambda > 0$ . If $||G(x)|| \leq \rho$ on $\mathbb{B}$ , then $$\left| \frac{1}{\|x\|^2} \left\langle (G'(x))^{-1} G(x), x^* \right\rangle - 1 \right| \le T(\rho) \quad \text{for all} \quad x \in \mathbb{B},$$ where function T is be defined by (5.1) with $\alpha = \lambda(\operatorname{Re} q - a)$ and $\beta = \lambda a$ .

Corollary 5.2. Let G be the resolvent of $f \in \widehat{\mathcal{N}}_a$ , $a \geq 0$ , corresponding to the parameter $\lambda > 0$ . Then G is a starlike mapping of order $\frac{1}{2}$ . Moreover, if $||G(x)|| \leq \rho \leq \rho^*$ on $\mathbb{B}$ , then mapping G is starlike of order $\frac{1}{1+T(\rho)}$ and strongly starlike of order $\frac{2 \arcsin T(\rho)}{\pi}$ . Proof. Applying Proposition 3.3 (i), one concludes that $G \in \widehat{\text{Hol}}(\mathbb{B})$ , that is, G(x) = g(x)x, $g \in \text{Hol}(\mathbb{B}, \mathbb{D})$ . In order to prove that G is starlike of some order, we have to show according to Proposition 3.4 that for every $u \in \partial \mathbb{B}$ the function $z \mapsto g(zu)z$ is starlike of the same order. It follows from Proposition 3.3 (ii) that the last function is a non-linear resolvent of the function p(zu)z, which is holomorphically accretive in the open unit disk $\mathbb{D}$ . Therefore the first assertion of the theorem follows from [18, Corollary 2.6] As for the second assertion, we note that $T(\rho) \leq 1$ , and the disk $|\zeta - 1| \leq T(\rho)$ is a subset of the disk $|\zeta - \frac{1}{2\gamma}| \leq \frac{1}{2\gamma}$ with $\gamma = \frac{1}{1+T(\rho)}$ and of the wedge $|\arg \zeta| \leq \arcsin T(\rho)$ . Thus the conclusion follows from Theorem 5.1 and Definition 2.6. It is natural to apply Theorem 5.1 and Corollary 5.2 together with finding concrete $\rho \leq \rho^*$ for which $||G(x)|| \leq \rho$ , $x \in \mathbb{B}$ . To do this, we will use the results of the previous section. By Theorem 4.1, all values of the function $G_{\lambda}$ lies in the disk of radius $\sqrt{\frac{2}{A+\sqrt{B}}}$ , where $$\begin{cases} A = |1 - \lambda q|^2 + 4\lambda a + 1, \\ B = (|1 - \lambda q|^2 - 1)^2 + 8\lambda^3 a |q|^2. \end{cases}$$ (5.3) Thus, we need conditions that ensure the inequality $\sqrt{\frac{2}{A+\sqrt{B}}} \leq \rho$ , where $\rho$ is defined before Corollary 5.2. This relation is equivalent to $$2\left(\sqrt{\frac{2\lambda(\operatorname{Re}q - a)}{1 + \lambda\operatorname{Re}q}} + 1\right)^{2} \le A + \sqrt{B}.$$ (5.4) (So, in this case G is starlike of order $\frac{1}{1+T(\rho)}$ , $\rho = \sqrt{\frac{2}{A+\sqrt{B}}}$ by Corollary 5.2.) Not knowing how to handle this inequality, we will look at stronger ones. Specifically, we analyze the two cases. In the first one, $|1-\lambda q|^2 \ge 1$ , or equivalently, $\lambda |q|^2 \ge 2 \operatorname{Re} q$ . In this case we decrease A and B replacing |q| in (5.3) by $\operatorname{Re} q$ . In particular, $B \ge ((1-\lambda \operatorname{Re} q)^2-1)^2+8\lambda^3 a \operatorname{Re}^2 q$ . Introduce the parameters x= $2\lambda(\operatorname{Re} q - a)$ and $s = \lambda \operatorname{Re} q$ and consider the inequality $$2\left(\sqrt{\frac{x}{1+s}}+1\right)^2 < s^2+2s+2-2x+s\sqrt{(s+2)^2-4x},$$ which is stronger than (5.4). Solving this inequality we get $x < \frac{4s^2(1+s)}{(2+s)^2}$ . This means that $$\operatorname{Re} q - a < \frac{2\lambda \operatorname{Re}^2 q (1 + \lambda \operatorname{Re} q)}{(2 + \lambda \operatorname{Re} q)^2}.$$ The solution to the last inequality with respect to $\lambda$ is $\lambda > M_1(q, a)$ , where $$M_1(q, a) := \frac{\sqrt{5 \operatorname{Re}^2 q - 4a \operatorname{Re} q + \operatorname{Re} q - 2a}}{(\operatorname{Re} q + a) \operatorname{Re} q}.$$ (5.5) In the opposite case $\lambda |q|^2 < 2 \operatorname{Re} q$ . First, assume that q = p(0)is real. Then $B = \lambda^2 q^2 (2 - \lambda q)^2 + 8\lambda^3 a q^2$ . In this situation inequality (5.4) can be solved relative to a: $$a > \frac{4 + 2\lambda q - (\lambda q)^2}{(2 + \lambda q)^2}.$$ In the general situation (that is, when p(0) is not necessarily real), we cannot omit Im q. Therefore let estimate B differently: $\sqrt{B} \geq$ $1-|1-\lambda q|^2$ . Then $A+\sqrt{B}\geq 4\lambda a+2$ . Thus, we replace inequality (5.4) with more stringent: $$\left(\sqrt{\frac{2\lambda(\operatorname{Re}q-a)}{1+\lambda\operatorname{Re}q}}+1\right)^2<2\lambda a+1.$$ This inequality holds if and only if $a > M_2(q, \lambda)$ , where $$M_2(q,\lambda) := \frac{(s+1)\sqrt{2s^2+4s+1}+s^2+s-1}{\lambda(2+s)^2}, \quad s = \lambda \operatorname{Re} q.$$ (5.6) We summarize these facts in the following theorem.

Theorem 5.3. Let $f \in \widehat{\mathcal{N}}_a$ , $a \geq 0$ , f(x) = p(x)x, q = p(0), and G be the resolvent of f corresponding to the parameter $\lambda > 0$ . Let A, B, $M_1(q, a)$ and $M_2(q, \lambda)$ be defined by (5.3), (5.5) and (5.6). Assume that one of the following conditions holds: - (i) $\lambda |q|^2 \ge 2 \operatorname{Re} q \text{ and } \lambda > M_1(q, a);$ (ii) $\lambda |q|^2 < 2 \operatorname{Re} q \text{ and } a > M_2(q, \lambda).$ Then G is a starlike mapping of order $\frac{1}{1+T(\sqrt{\frac{2}{A^2-Q^2}})}$ .  FIGURE 2. Range of the parameters s and t

Corollary 5.4. If q is real, $a \in [0,q)$ , then $G_{\lambda}$ is starlike of order $\frac{1}{1+T\left(\sqrt{\frac{2}{A+\sqrt{B}}}\right)} \text{ whenever } \lambda > \frac{\sqrt{5p^2(0)-4aq+q-2a}}{(q+a)q} \text{ (cf. (5.5))}.$ It is interesting to determine a set of parameters $(q, a, \lambda)$ for which we found the order of starlikeness. For simplicity, let us take the case $q \in \mathbb{R}$ and depict this set by passing to the parameters $s = \lambda q > 0$ and $t = \frac{a}{q} \in [0, 1)$ . In these new parameters, the desired set can be described as $\left\{(s, t): s > 0, \frac{4+2s-s^2}{(2+s)^2} < t < 1\right\}$ , see Fig. 2. In the one-dimensional case, the order of starlikeness was earlier found for t = 0 and $s \geq r_0 \approx 5.9$ in [11].

Definition 2.2. Let $f \in \text{Hol}(\mathbb{B}, X)$ . One says that f satisfies the range condition on $\mathbb{B}$ if $(\text{Id} + \lambda f)(\mathbb{B}) \supset \mathbb{B}$ for each $\lambda > 0$ and the so-called resolvent equation $$w + \lambda f(w) = x \tag{2.4}$$ has a unique solution $$w = G_{\lambda}(x) \left( = \left( \operatorname{Id} + \lambda f \right)^{-1} (x) \right)$$ (2.5) holomorphic in $\mathbb{B}$ . If it is the case, every mapping $G_{\lambda}$ , $\lambda > 0$ , is called the non-linear resolvent and the family $\{G_{\lambda}\}_{{\lambda} \geq 0} \in \operatorname{Hol}(\mathbb{B})$ is called the resolvent family of f on $\mathbb{B}$ . The following statement combines Theorems 6.11 and 7.5 in [30] with another result obtained at first in [29] (see also the recent books [30, 14]).

Definition 2.4. A mapping $f \in \text{Hol}(\mathbb{B}, X)$ , f(0) = 0, is said to be holomorphically accretive if $$\operatorname{Re}\langle f(x), x^ \rangle \ge a \|x\|^2 \quad \text{for all} \quad x \in \mathbb{B} \quad \text{and} \quad x^ \in J(x) \quad (2.6)$$ for some $a \geq 0$ . We denote the class of mappings that satisfy (2.6) with a given $a \geq 0$ by $\mathcal{N}_a$ . Note that for any holomorphically accretive mapping f, the number a can be chosen to be zero. Whence a>0 the mapping f is named strongly holomorphically accretive. To explain our interest in the classes $\mathcal{N}_a$ with a>0, recall that if the semigroup generated by $f\in\mathcal{N}_0$ contains neither an elliptic automorphism, nor the identity mapping, then it converges to zero uniformly on any ball $\mathbb{B}_r$ , r<1. It is reasonable to inquire whether this convergence is uniform on the whole ball $\mathbb{B}$ . The following statement answers this question.

Definition 2.6. Let $f \in \text{Hol}(\mathbb{B}, X)$ , f(0) = 0, be a biholomorphic mapping. It is called starlike if Re $$\langle (f'(x))^{-1} f(x), x^* \rangle \ge 0$$ (2.8) for all $x \in \mathbb{B}$ , $x^* \in J(x)$ . Moreover, the mapping f is called starlike of order $\gamma \in (0,1]$ if it satisfies $$\left| \frac{\left\langle (f'(x))^{-1} f(x), x^* \right\rangle}{\|x\|^2} - \frac{1}{2\gamma} \right| \le \frac{1}{2\gamma}, \qquad x \in \mathbb{B} \setminus \{0\}. \tag{2.9}$$ It is called strongly starlike of order $\beta \in [0,1]$ if $$\left|\arg\left\langle \left(f'(x)\right)^{-1}f(x),x^*\right\rangle\right| \le \frac{\pi\beta}{2}, \qquad x \in \mathbb{B}.$$ (2.10) In the one-dimensional case these definitions coincide with classical ones, see, for example, books [9, 16, 19], where more details can be found. Starlike mappings of order $\gamma=0$ (as well as strongly starlike of order $\beta=1$ ) are just starlike, while the only mappings starlike of order $\gamma=1$ (strongly starlike of order $\beta=0$ ) are linear ones. Notice that the normalization $f'(0)=\mathrm{Id}$ is usually imposed. Since the relations (2.8)–(2.10) are invariant under the transformation $f\mapsto Bf$ , where B is an invertible bounded linear operator, we conclude that the normalization is unnecessary. 3. The class $$\widehat{\operatorname{Hol}}(\mathbb{B},X)$$ In this paper we focus on a class of holomorphic mappings which in the one-dimensional case coincides with the class of all holomorphic functions in the open unit disk vanishing at zero.

Definition 3.1. We write $f \in \widehat{\text{Hol}}(\mathbb{B}, X)$ if f has the form f(x) = p(x)x, where $p \in \text{Hol}(\mathbb{B}, \mathbb{C})$ , that is, for every point $x \in \mathbb{B}$ , the vectors x and f(x) are $\mathbb{C}$ -proportional. We also denote $\widehat{\mathcal{N}}_a := \mathcal{N}_a \cap \widehat{\text{Hol}}(\mathbb{B}, X)$ and $\widehat{\text{Hol}}(\mathbb{B}) := \text{Hol}(\mathbb{B}) \cap \widehat{\text{Hol}}(\mathbb{B}, X)$ . Writing $f \in \widehat{\text{Hol}}(\mathbb{B}, X)$ in the above form f(x) = p(x)x, we get $\langle f(x), x^* \rangle = ||x||^2 p(x)$ and f'(0) = p(0) Id. Hence $f \in \widehat{\mathcal{N}}_a$ , see Definition 2.4, if and only if $\text{Re } p(x) \geq a, x \in \mathbb{B}$ . In the next proposition we collect some properties of such functions p. We provide their proof for the sake of completeness. Alternatively, they can be obtained from well-known one-dimensional results (cf. [9, 14, 19]).