Proof. Since $h_r$ for every $r\in (0,\,1),$ is analytic on $\overline {\mathbb {D}}$ and convex in $\mathbb {D},$ from (2.1) it follows that for $z\in \mathbb {D}$ and $\zeta \in \mathbb {T},$

\begin{align*} & \operatorname{Re}\left\{ \frac{2\zeta h_r'(\zeta)}{h_r(\zeta)-h_r(z)}-\frac{\zeta+z}{\zeta-z}\right\}\\ & \quad=\operatorname{Re}\left\{ \frac{2r\zeta h'(r\zeta)}{h(r\zeta)-h(rz)}-\frac{r\zeta+rz}{r\zeta-rz}\right\}\\ & \quad=\operatorname{Re}\left\{ \frac{2u h'(u)}{h(u)-h(v)}-\frac{u+v}{u-v}\right\}\ge 0, \end{align*}

where $u:=r\zeta \in \mathbb {D}$ and $v:=rz\in \mathbb {D}.$ Hence and by the fact that $h_r(\zeta )\rightarrow h(\zeta )$ and $h_r'(\zeta )\rightarrow h'(\zeta )$ as $r\rightarrow 1^-,$ we deduce that

\[ \operatorname{Re}\frac{2\zeta h'(\zeta)}{h(\zeta)-h(z)}\ge \operatorname{Re}\frac{\zeta+z}{\zeta-z}=\frac{1-|z|^2}{|\zeta-z|^2}>0,\quad z\in\mathbb{D},\ \zeta\in\mathbb{T}, \]

which shows (2.3).

Proof. Suppose that (2.6) holds. Then by the Lindelöf Principle (e.g., [Reference Goodman3, Vol. I, p. 86]) for every $r\in (0,\,1),$

(2.8)\begin{equation} f_r(\mathbb{D})=f(\mathbb{D}_r)\subset h(\mathbb{D}_r)=h_r(\mathbb{D}). \end{equation}

Since $f_r(0)=f(0)=h(0)=h_r(0)$ and every $h_r$ is univalent, from (1.1) and (2.8) it follows (2.7).

Suppose now that (2.7) holds for every $r\in (0,\,1).$ Then by (1.1) the inclusion (2.8) holds for every $r\in (0,\,1),$ and therefore

\[ f(\mathbb{D})=\bigcup_{r\in(0,1)}f(\mathbb{D}_r)\subset \bigcup_{r\in(0,1)}h(\mathbb{D}_r)=h(\mathbb{D}). \]

Hence and from (1.1) we obtain (2.6).

Proof. Note first that if $\operatorname {Re} \varphi (h(z_0),\,mz_0h'(z_0))=0$ for a certain $z_0\in \mathbb {D},$ then by the minimum principle for harmonic function $\operatorname {Re} \varphi (h(z),\,mzh'(z))=0$ for all $z\in \mathbb {D}$ and hence $\varphi (h(z),\,mzh'(z))=\mathrm {i} a$ for some $a\in \mathbb {R}$ and all $z\in \mathbb {D}.$

Let $p\in \mathcal {H}[\varphi ]$ with $p(0)=h(0).$ Define $\psi :\mathbb {D}\rightarrow \mathbb {C}$ as follows

(2.13)\begin{equation} \psi(z):=p(z)+zp'(z)\varphi(p(z),zp'(z)),\quad z\in\mathbb{D}. \end{equation}

Since $\psi (0)=p(0)=h(0),$ by Theorem 2.3 the condition (2.11) is equivalent to

(2.14)\begin{equation} \psi_r\prec h_r,\quad r\in(0,1). \end{equation}

On the contrary, suppose that $p$ is not subordinate to $h.$ By Theorem 2.3, there exists $r_0\in (0,\,1)$ such that $p_{r_0}$ is not subordinate to $h_{r_0}.$ Since $h_{r_0}$ is analytic in $\overline {\mathbb {D}},$ by lemma 2.2 there exist $z_0\in \mathbb {D}\setminus \{0\},$ $\zeta _0\in \mathbb {T}$ and $m\ge 1$ such that (2.4) and (2.5) hold with $p:=p_{r_0}$ and $h:=h_{r_0},$ i.e.,

(2.15)\begin{equation} p_{r_0}(z_0)=h_{r_0}(\zeta_0) \end{equation}

and

(2.16)\begin{equation} z_0p_{r_0}'(z_0)=m\zeta_0h_{r_0}'(\zeta_0). \end{equation}

Hence

(2.17)\begin{equation} \begin{aligned} \psi_{r_0}(z_0) & =\psi(r_0z_0)\\ & =p(r_0z_0)+r_0z_0p'(r_0z_0)\varphi(p(r_0z_0),r_0z_0p'(r_0z_0))\\ & =p_{r_0}(z_0)+z_0p_{r_0}'(z_0)\varphi(p_{r_0}(z_0),z_0p_{r_0}'(z_0))\\ & =h_{r_0}(\zeta_0)+m\zeta_0h_{r_0}'(\zeta_0)\varphi(h_{r_0}(\zeta_0),m\zeta_0h_{r_0}'(\zeta_0)). \end{aligned} \end{equation}

Moreover by (2.10),

(2.18)\begin{equation} \begin{aligned} \operatorname{Re} \varphi(h_{r_0}(\zeta_0),m\zeta_0h_{r_0}'(\zeta_0)) & =\operatorname{Re} \varphi(h(r_0\zeta_0),mr_0\zeta_0h'(r_0\zeta_0))\\ & =\operatorname{Re} \varphi(h(u_0),mu_0h'(u_0))\ge 0, \end{aligned} \end{equation}

where $u_0:=r_0\zeta _0\in \mathbb {D}.$ In view of (2.14), $\psi _{r_0}\prec h_{r_0}$, so $\psi _{r_0}(\mathbb {D})\subset h_{r_0}(\mathbb {D}).$ Thus $\psi _{r_0}(z_0)\in h_{r_0}(\mathbb {D})$ and therefore $\psi _{r_0}(z_0)=h_{r_0}(z_1)$ for some $z_1\in h_{r_0}(\mathbb {D}).$ Hence from (2.17) and (2.18) we get

\begin{align*} \operatorname{Re}\frac{h_{r_0}(z_1)-h_{r_0}(\zeta_0)}{\zeta_0h_{r_0}'(\zeta_0)}& =\operatorname{Re}\frac{\psi_{r_0}(z_0)-h_{r_0}(\zeta_0)}{\zeta_0h_{r_0}'(\zeta_0)}\\ & =m\operatorname{Re}\varphi(h_{r_0}(\zeta_0),m\zeta_0h_{r_0}'(\zeta_0))\ge 0. \end{align*}

Since $h_{r_0}\in \mathcal {Q},$ it follows that the above inequality contradicts (2.3) with $h:=h_{r_0},$ $z:=z_1$ and $\zeta :=\zeta _0.$ Thus we conclude that $\psi _{r_0}$ is not subordinate to $h_{r_0},$ which contradicts (2.14) and completes the proof.

Proof. Let $\psi$ be defined as in (2.11). By Theorem 2.3 the condition (2.20) is equivalent to (2.14). On the contrary, suppose that $p$ is not subordinate to $h.$ As in the proof of Theorem 2.4, there exist $z_0\in \mathbb {D}\setminus \{0\},$ $\zeta _0\in \mathbb {T}$ and $m\ge 1$ such that (2.15) and (2.16) hold. Thus

(2.22)\begin{equation} \begin{aligned} \psi_{r_0}(z_0) & =\psi(r_0z_0)\\ & =p(r_0z_0)+r_0z_0p'(r_0z_0)\varphi(p(r_0z_0),r_0z_0p'(r_0z_0))\\ & =p_{r_0}(z_0)+z_0p_{r_0}'(z_0)\varphi(p(r_0z_0),r_0z_0p'(r_0z_0))\\ & =h_{r_0}(\zeta_0)+m\zeta_0h_{r_0}'(\zeta_0)\varphi(p(r_0z_0),r_0z_0p'(r_0z_0))\\ & =h_{r_0}(\zeta_0)+m\zeta_0h_{r_0}'(\zeta_0)\varphi(p(u_0),u_0p'(u_0)), \end{aligned} \end{equation}

where $u_0:=r_0z_0\in \mathbb {D}.$ In view of (2.14), $\psi _{r_0}\prec h_{r_0}$, so $\psi _{r_0}(\mathbb {D})\subset h_{r_0}(\mathbb {D}).$ Thus $\psi _{r_0}(z_0)\in h_{r_0}(\mathbb {D})$ and therefore $\psi _{r_0}(z_0)=h_{r_0}(z_1)$ for some $z_1\in\mathbb{D}.$ Hence from (2.22) and (2.19) it follows that

\begin{align*} \operatorname{Re}\frac{h_{r_0}(z_1)-h_{r_0}(\zeta_0)}{\zeta_0h_{r_0}'(\zeta_0)}& =\operatorname{Re}\frac{\psi_{r_0}(z_0)-h_{r_0}(\zeta_0)}{\zeta_0h_{r_0}'(\zeta_0)}\\ & =m\operatorname{Re}\varphi(p(u_0),u_0p'(u_0))\ge 0. \end{align*}

Since $h_{r_0}\in \mathcal {Q},$ it follows that the above inequality contradicts (2.3) with $h:=h_{r_0},$ $z:=z_1$ and $\zeta :=\zeta _0.$ Thus we conclude that $\psi _{r_0}$ is not subordinate to $h_{r_0},$ which contradicts (2.14), so (2.20) and completes the proof.

Proof. Let $\beta \ge 0,$ $h\in \mathcal {S}^c$ and $\phi \in \mathcal {H}(D)$ be such that $h(\mathbb {D})\subset D.$ Define $\varphi :\mathbb {C}\times \mathbb {C}\rightarrow \mathbb {C}$ as

\[ \varphi(u,v):=\phi(u)+\beta\frac{v}{u},\quad (u,v)\in (D\setminus\{0\})\times \mathbb{C}. \]

Since $h(0)=0,$ $h(z)\not =0$ for $z\not =0$ and $h'(0)\not =0,$ it follows that

(3.4)\begin{equation} \lim_{z\to 0}\dfrac{zh'(z)}{h(z)}=1. \end{equation}

Thus the function

\[ \mathbb{D}\setminus\{0\}\ni z\mapsto \varphi(h(z),mzh'(z))=\phi(h(z))+\beta \dfrac{mzh'(z)}{h(z)} \]

has an analytic extension on $\mathbb {D}$ by setting $\phi (0)+\beta m$ at zero. Moreover by (2.2) and (3.1) for every $m\ge 1,$

(3.5)\begin{equation} \begin{aligned} & \operatorname{Re}\varphi(h(z),mzh'(z))\\ & =\operatorname{Re}\phi(h(z))+\beta m\operatorname{Re}\frac{zh'(z)}{h(z)}\ge \frac{\beta}{2}\ge 0,\quad z\in\mathbb{D}. \end{aligned} \end{equation}

Let $p\in \mathcal {H},$ $p(0)=0,$ $p(z)\not =0$ for $z\not =0$ and $p(\mathbb {D})\subset D.$ Because $p(0)=0,$ there exists a positive integer $k$ such that $p(z)=z^kq(z),\,\ z\in \mathbb {D},$ where $q\in \mathcal {H}$ and $q(0)\not =0.$ Hence and by the fact that $p(z)\not =0$ for $z\not =0,$ it follows that $q(z)\not =0$ for $z\in \mathbb {D}.$ Thus the function

\[ \mathbb{D}\ni z\mapsto \varphi(p(z),zp'(z))=\phi(p(z))+\beta \dfrac{zp'(z)}{p(z)} \]

has an analytic extension on $\mathbb {D}$ by setting $\phi (0)+\beta k$ at zero, i.e., $p\in \mathcal {H}[\varphi ].$ At the end note that

\begin{align*} & p(z)+zp'(z)\varphi\left(p(z),zp'(z)\right)\\ & =p(z)+zp'(z)\phi(p(z))+\beta p(z)\left(\frac{zp'(z)}{p(z)}\right)^2,\quad z\in\mathbb{D}. \end{align*}

Thus, the assumptions of Theorem 2.4 are satisfied, which ends the proof of the corollary.

Proof. Take

\[ \phi(w):=\frac{1}{\delta w+\gamma},\quad w\in D:=\mathbb{C}\setminus\{-\gamma/\delta\}. \]

Then by (3.7) the function $\phi \circ h$ is analytic in $\mathbb {D},$ and by (3.8),

\[ \operatorname{Re} \phi(h(z))=\operatorname{Re}\frac{1}{\delta h(z)+\gamma}> 0,\quad z\in\mathbb{D}, \]

and we apply Corollary 3.1.