1 Introduction

Recently, many authors showed great interest in two domains: the symmetrized bidisc and the tetrablock, arising from the \(\mu \)-synthesis, from the aspect of geometric function theory. Actually, both domains are \({\mathbb {C}}\)-convex but non-convex, and they cannot be exhausted by domains biholomorphic to convex ones, with the Lempert’s theorem (see Lempert [13, 14]) holding on these two domains, i.e., the Lempert function and the Carathéodory distance coincide on them (see [2, 6,7,8, 20]). So from the point of view of the Lempert’s theorem holding, these two domains play an important role in the study of a long-standing open problem whether Lempert’s theorem still holds for \({\mathbb {C}}\)-convex domain. However, as far as we know, the answer is positive for \({\mathbb {C}}\)-convex domain with \({\mathcal {C}}^2\) boundary (see [10]).

In 2015, Agler, Lykova and Young [1] introduced a new bounded domain \({\mathcal {P}}\) by

$$\begin{aligned} {\mathcal {P}}:=\left\{ (a_{21}, \mathrm{tr}\,{A}, \det {A}): A=[a_{ij}]_{i, j=1}^{2}\in {\mathbb {B}} \right\} , \end{aligned}$$

where

$$\begin{aligned} {\mathbb {B}}:= \left\{ A\in {\mathbb {C}}^{2\times 2}:\; ||A||<1 \right\} \end{aligned}$$

denotes the open unit ball in the space \({\mathbb {C}}^{2\times 2}\) with the usual operator norm. They called this domain the pentablock as \({\mathcal {P}}\cap {\mathbb {R}}^{3}\) is a convex body bounded by five faces, three of which are flat and two are curved (see [1]).

The pentablock \({\mathcal {P}}\) is polynomially convex and starlike about the origin, but neither circled nor convex. Moreover, it does not have a \({\mathcal {C}}^{1}\) boundary (see [1]). This new domain is also arising from the \(\mu \)-synthesis, just like the symmetrized bidisc and the tetrablock. So it is naturally to consider analogous properties of the pentablock, such as the question about \({\mathbb {C}}\)-convexity of \({\mathcal {P}}\), and Lemperts theorem on the equality of holomorphically invariant functions and metrics for the pentablock (see [1, 12, 19]). In this paper, we give a positive answer to the \({\mathbb {C}}\)-convexity of \({\mathcal {P}}\). More precisely, we obtain the following theorem.

Theorem 1.1

The pntablock \({\mathcal {P}}\) is a \({\mathbb {C}}\)-convex domain.

Throughout this paper, \({\mathbb {D}}\) denotes the open unit disc in the complex plane ,while \({\mathbb {T}}\) denotes the unit circle. And other basic notions, definitions, and properties from the theory of invariant functions, linearly convex and \({\mathbb {C}}\)-convex domains that we shall use in the paper may be found in [3, 9, 11].

2 Preliminary Results

2.1 Pentablock

We first recall the definition of the pentablock \({\mathcal {P}}\).

Theorem 2.1

[1, Theorem 1.1 and Theorem 5.2] Let

$$\begin{aligned} (s,p)=(\lambda _1+\lambda _2, \lambda _1\lambda _2), \end{aligned}$$

where \(\lambda _1, \lambda _2\in {\mathbb {D}}\). Let \(a\in {\mathbb {C}}\) and

$$\begin{aligned} \beta =\frac{s-\bar{s}p}{1-\vert p\vert ^2}. \end{aligned}$$

The following statements are equivalent:

  1. (1)

    \((a,s,p)\in {\mathcal {P}}\),

  2. (2)

    \(\vert a\vert <\left| 1-\frac{\frac{1}{2}s\bar{\beta }}{1+\sqrt{1-\vert \beta \vert ^2}}\right| \),

  3. (3)

    \(\vert a\vert <\frac{1}{2}\vert 1-\bar{\lambda }_1\lambda _2\vert +\frac{1}{2}(1-\vert \lambda _1\vert ^2)^{\frac{1}{2}} (1-\vert \lambda _2\vert ^2)^{\frac{1}{2}}\),

  4. (4)

    \(\sup _{z\in {\mathbb {D}}}\vert \Psi _{z}(a,s,p)\vert <1\), where \(\Psi _z\) is the linear fractional map

    $$\begin{aligned} \Psi _z(a,s,p)=\frac{a(1-\vert z\vert ^2)}{1-sz+pz^2}. \end{aligned}$$

2.2 Pentablock as a Hartogs Domain

Following the description of the pentablock \({\mathcal {P}}\), we can learn that the pentablock \({\mathcal {P}}\) is closely related to the symmetrized bidisc \({\mathbb {G}}_{2}\), which is defined by

$$\begin{aligned} {\mathbb {G}}_2=\left\{ (s,p)\in {\mathbb {C}}^2: \vert s-\bar{s} p\vert +\vert p\vert ^2<1\right\} . \end{aligned}$$

In fact, the pentablock \({\mathcal {P}}\) can be seen as a Hartogs domain in \({\mathbb {C}}^3\) over the symmetrized bidisc \({\mathbb {G}}_2\) (see [1]), that is,

$$\begin{aligned} {\mathcal {P}}=\left\{ (a,s,p)\in {\mathbb {D}}\times {\mathbb {G}}_2:\vert a\vert ^2<e^{-\varphi (s,p)}\right\} , \end{aligned}$$

where

$$\begin{aligned} \varphi (s,p)=-2\log \left| 1-\frac{\frac{1}{2}s\bar{\beta }}{1+\sqrt{1-\vert \beta \vert ^2}} \right| , \end{aligned}$$

\((s,p)\in {\mathbb {G}}_2\) and \(\beta =\frac{s-\bar{s} p}{1-\vert p\vert ^2}\).

Hartogs domain is a one of important research object in several complex variable. For the studies on Hartogs domain, please refer to [4, 5, 16,17,18]. So considering the pentablock \({\mathcal {P}}\) as a Hartogs domain will be great helpful for us to study the convexity of the pentablock \({\mathcal {P}}\).

2.3 Some Useful Results

In this subsection, we will give some useful results on the symmetrized bidisc \({\mathbb {G}}_{2}\) and the pentablock \({\mathcal {P}}\). In order to study the pentablock \({\mathcal {P}}\), it is sufficient to learn the \({\mathbb {C}}\)-convexity of \({\mathbb {G}}_2\).

Theorem 2.2

[15] The symmetrized bidisc \({\mathbb {G}}_{2}\) is \({\mathbb {C}}\)-convex.

Through the study of the boundary of \({\mathcal {P}}\), we can learn that there are two main part of the boundary, i.e., the smooth part and the non-smooth part. So it is necessary to study some basic convexity property of \({\mathcal {P}}\) to simplify the problem.

Theorem 2.3

[12, Proposition 9] The pentablock \({\mathcal {P}}\) is linearly convex.

In order to study the \({\mathbb {C}}\)-convexity of \({\mathcal {P}}\) in some simple way, we give the whole holomorphic automorphism group \(\text{ Aut }({\mathcal {P}})\) as follows.

Theorem 2.4

[12, Theorem 15] All mappings of the form

$$\begin{aligned} f_{\omega ,\nu }(a, \lambda _{1}+\lambda _{2}, \lambda _{1}\lambda _{2}) =\left( \frac{\omega (1{-}|\alpha |^{2})a}{1{-}\bar{\alpha }(\lambda _{1} {+}\lambda _{2})+\bar{\alpha }^{2}\lambda _{1}\lambda _{2}}, \nu (\lambda _{1})+\nu (\lambda _{2}), \nu (\lambda _{1})\nu (\lambda _{2})\right) , \end{aligned}$$

where \((a, \lambda _{1}+\lambda _{2}, \lambda _{1}\lambda _{2})\in {\mathcal {P}},\)\(\lambda _{1}, \lambda _{2}\in {\mathbb {D}}\), \(\nu \) is a \(M\ddot{o}bius\) function of the form \(\nu (\lambda )=\eta \frac{\lambda -\alpha }{1-\bar{\alpha }\lambda }\), and where \(\omega , \eta \in {\mathbb {T}}\), \(\alpha \in {\mathbb {D}}\), form the whole group \(\text{ Aut }({\mathcal {P}})\) of holomorphic automorphisms of the pentablock \({\mathcal {P}}\).

2.4 \({\mathbb {C}}\)-convex Domain

A domain \(D\subset {\mathbb {C}}^n\) is called \({\mathbb {C}}\)-convex if for any affine complex line \(\ell \) such that \(\ell \cap D\ne \emptyset \), and the set \(\ell \cap D\) is connected and simply connected. For a domain \(D\subset {\mathbb {C}}^n\) and a point \(a\in {\mathbb {C}}^n\), we denote by \(\Gamma _D(a)\) the set of all complex hyperplanes L such that \((a+L)\cap D=\emptyset \). Then we have the basic criterion on \({\mathbb {C}}\)-convexity.

Theorem 2.5

[3, Theorem 2.5.2] The bounded domain \(D\subset {\mathbb {C}}^n\), \(n>1\), is \({\mathbb {C}}\)-convex iff for any boundary point \(x\in \partial D\), the set \(\Gamma _D(x)\) is non-empty and connected.

Remark 2.6

By Theorem 2.5, we only need to give a full description of the tangent hyperplanes to the pentablock \({\mathcal {P}}\). And together with Theorem 2.3, we can only need to consider the non-smooth part of the boundary. Furthermore, through Theorem 2.4 we can simplify the situation into just four different types, i.e., (1) (a, 1, 0) with \(\vert a\vert \le \frac{1}{2}\); (2) \((a,0,-1)\) with \(\vert a\vert <1\); (3) \((1,0,-1)\); (4) (0, 2, 1).

3 The Set of All Tangent Hyperplanes to \({\mathcal {P}}\) at the Non-smooth Part

In this section, we will give a full description of the tangent hyperplanes to the pentablock \({\mathcal {P}}\) at the non-smooth boundary part. Set \(P_0=(a_0,s_0,p_0)\) be a non-smooth boundary point of the pentablock \({\mathcal {P}}\), and let \(\Gamma _{{\mathcal {P}}}(P_0)\) denote the set of all tangent hyperplanes to \({\mathcal {P}}\) at the boundary point \(P_0\). Now assume the hyperplane in \({\mathbb {C}}^3\) that

$$\begin{aligned} {\mathcal {L}}:=\{(a,s,p)\in {\mathbb {C}}^3: k_1 a+k_2 s+k_3 p=0\}. \end{aligned}$$

Then

$$\begin{aligned} {\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(P_0)\Longleftrightarrow P_0\in {\mathcal {L}} \ \ \ \text {and} \ \ \ {\mathcal {L}}\cap {\mathcal {P}}=\emptyset . \end{aligned}$$

Together with automorphisms of \({\mathcal {P}}\) by Theorem 2.4, we can only need to consider some special boundary points \(P_0\):

  1. (1)

    If \((s_0, p_0)\in \partial {\mathbb {G}}_2{\setminus }\partial _s {\mathbb {G}}_2\) and \(\vert a_0\vert ^2\le e^{-\varphi (s_0, p_0)}\).

Actually we can assume that \((s_0,p_0)=(1,0)\), and then we have \(\vert a_0\vert \le \frac{1}{2}\). Now we consider the hyperplane in \({\mathbb {C}}^3\) passing through the boundary point \(P_0\),

$$\begin{aligned} {} {\mathcal {L}}:=\{(a,s,p)\in {\mathbb {C}}^3: k_1(a-a_0)+k_2(s-s_0)+k_3(p-p_0)=0\}. \end{aligned}$$
(3.1)

If \(k_1\ne 0\), then \({\mathcal {L}}=\{(a,s,p)\in {\mathbb {C}}^3: a=a_0+k_2(1-s)-k_3 p\}\).

Next suppose that \({\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(P_0)\), so we get \({\mathcal {L}}\cap {\mathcal {P}}=\emptyset \). This means that for any \((s,p)\in {\mathbb {G}}_2\), we have

$$\begin{aligned} \vert a_0+k_2(1-s)-k_3 p\vert ^2\ge e^{-\varphi (s,p)}. \end{aligned}$$

Now set \(p=0\), we can learn that for any \(s\in {\mathbb {D}}\),

$$\begin{aligned} \frac{1}{2}+\frac{1}{2}(1-\vert s\vert ^2)^{\frac{1}{2}}\le \vert a_0 +k_2(1-s)\vert \le \vert a_0\vert +\vert k_2(1-s)\vert . \end{aligned}$$

Together with \(\vert a_0\vert \le \frac{1}{2}\), we obtain that the following inequality

$$\begin{aligned} \vert k_2\vert \ge \frac{\frac{1}{2}(1-\vert s\vert ^2)^{\frac{1}{2}}}{\vert 1-s\vert } \end{aligned}$$

holds for all \(s\in {\mathbb {D}}\).

Hence, taking s tends to 1, we can conclude that such \(k_2\) does not exist. It follows that if the hyperplane \({\mathcal {L}}\) in (3.1) belongs to \(\Gamma _{{\mathcal {P}}}(P_0)\), then \(k_1=0\).

Moreover, if we have a hyperplane \({\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(P_0)\), then consider the following hyperplane in \({\mathbb {C}}^2\):

$$\begin{aligned} {\mathcal {L}}':=\{(s,p)\in {\mathbb {C}}^2: (a,s,p)\in {\mathcal {L}}\}. \end{aligned}$$

Easily, we can see that \((s_0,p_0)=(1,0)\in {\mathcal {L}}'\cap \partial {\mathbb {G}}_2\) and \({\mathcal {L}}'\cap {\mathbb {G}}_2=\emptyset \). So

$$\begin{aligned} {\mathcal {L}}'\in \Gamma _{{\mathbb {G}}_2}(s_0,p_0). \end{aligned}$$

This implies that

$$\begin{aligned} \Gamma _{{\mathcal {P}}}(P_0)\subseteq {\mathbb {C}} \times \Gamma _{{\mathbb {G}}_2}(s_0,p_0). \end{aligned}$$

Clearly the other inclusion holds. Therefore, we have

$$\begin{aligned} \Gamma _{{\mathcal {P}}}(a_0, 1,0)={\mathbb {C}} \times \Gamma _{{\mathbb {G}}_2}(1,0) \ \ \ \ \left( \vert a_0\vert \le \frac{1}{2}\right) . \end{aligned}$$
  1. (2)

    If \((s_0,p_0)\in \partial _s{\mathbb {G}}_2\) and \(\vert a_0\vert ^2<e^{-\varphi (s_0,p_0)}\).

From the assumption, we can learn that \(s_0^2\ne 4p_0\). Otherwise, \(e^{-\varphi (s_0,p_0)}=0\). Hence, we can assume that \((s_0,p_0)=(0,-1)\) and \(\vert a_0\vert <1\).

By the same way, consider the hyperplane passing through the boundary point \(P_0\) with the assumption \(k_1\ne 0\), namely,

$$\begin{aligned} {\mathcal {L}}=\{(a,s,p)\in {\mathbb {C}}^3: a=a_0+k_2s+k_3(1+p)\}. \end{aligned}$$

Then suppose that \({\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(P_0)\), so this leads to \({\mathcal {L}}\cap {\mathcal {P}}=\emptyset \). Hence, for any \((s,p)\in {\mathbb {G}}_2\), we have

$$\begin{aligned} \vert a_0+k_2s+k_3(1+p)\vert ^2\ge e^{-\varphi (s,p)}. \end{aligned}$$

Now set \(s=0\), we obtain that for any \(p\in {\mathbb {D}}\),

$$\begin{aligned} 1\le \vert a_0+k_3(1+p)\vert \le \vert a_0\vert +\vert k_3(1+p)\vert . \end{aligned}$$

This means that the following inequality

$$\begin{aligned} \vert k_3(1+p)\vert \ge 1-\vert a_0\vert >0 \end{aligned}$$

holds for any \(p\in {\mathbb {D}}\). Thus such \(k_3\) does not exist. And it follows that \(k_1\) must be zero. Therefore, following by the same procedure, we obtain

$$\begin{aligned} \Gamma _{{\mathcal {P}}}(a_0,0,-1)={\mathbb {C}} \times \Gamma _{{\mathbb {G}}_2}(0,-1) \ \ \ \ (\vert a_0\vert <1). \end{aligned}$$
  1. (3)

    \((s_0,p_0)\in \partial _s{\mathbb {G}}_2{\setminus }\Sigma \) and \(\vert a_0\vert ^2=e^{-\varphi (s_0,p_0)}\).

By the assumption, we can also see that \(s_0^2\ne 4p_0\). So we can assume \((a_0,s_0,p_0)=(1,0,-1)\). Then consider the hyperplane

$$\begin{aligned} {\mathcal {L}}:=\{(a,s,p)\in {\mathbb {C}}^3: a=1+k_2s+k_3(1+p)\}, \end{aligned}$$

and suppose that \({\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(P_0)\). This implies that \({\mathcal {L}}\cap {\mathcal {P}}=\emptyset \). Thus for any \((s,p)\in {\mathbb {G}}_2\), we have

$$\begin{aligned} \vert 1+k_2 s+k_3(1+p)\vert ^2 \ge e^{-\varphi (s,p)}. \end{aligned}$$
(3.2)

Let \(s=0\), then for any \(p\in {\mathbb {D}}\), we have

$$\begin{aligned} \vert 1+k_3+k_3p\vert \ge 1. \end{aligned}$$

Now we give the following lemma to help us.

Lemma 3.1

For any \(z\in {\mathbb {D}}\), the inequality \(\vert 1+k+kz\vert \ge 1\) holds iff \(k\ge 0\).

Proof

Set \(f(z)=1+k+kz\), and then we directly learn that f(z) is a holomorphic function on \({\mathbb {D}}\). Since \(\vert f(z)\vert \ge 1\), it means that f(z) has no zero in \({\mathbb {D}}\). Hence, by the maximal principle, \(\vert f(z)\vert \ge 1\) holds for all \(z\in \overline{{\mathbb {D}}}\). Now let \(\vert z\vert =1\), we can obtain

$$\begin{aligned}&\vert 1+k+kz\vert \ge 1,\\&\quad \Leftrightarrow \vert 1+k\vert ^2+\vert k\vert ^2-1\ge -2\text {Re}\left( (\vert k\vert ^2 +k)z\right) ,\ \ \ ( \vert z\vert =1)\\&\quad \Leftrightarrow 2\vert k\vert ^2+2\text {Re} k\ge 2\left| k +\vert k\vert ^2\right| , \ \ \ (\text {the inequality holds for all}\ z\in \partial {\mathbb {D}})\\&\quad \Rightarrow (\text {Re} k)^2\ge \vert k\vert ^2. \end{aligned}$$

This means that k is real. Now if \(k<0\), then there exists \(M>0\) such that

$$\begin{aligned} M\vert k\vert <1, \end{aligned}$$

and we can choose \(z_0\) with \(-1<z_0<0\) such that

$$\begin{aligned} 1+z_0<M. \end{aligned}$$

Hence, we have

$$\begin{aligned} 0<1+Mk<1+k(1+z_0)<1. \end{aligned}$$

This leads to a contradiction. On the other hand, for \(k\ge 0\), the inequality is evidently valid. Therefore, we conclude that \(k\ge 0\). \(\square \)

By Lemma 3.1, we have \(k_3\ge 0\). Now we want to prove the inequality (3.2) on the whole \(\overline{{\mathbb {G}}}_2\).

If there exists \((s_1,p_1)\in \partial {\mathbb {G}}_2\) such that the inequality (3.2) does not hold. Then set \(a_1=1+k_2 s_1+k_3 (1+p_1)\), and by the assumption, we have

$$\begin{aligned} \vert a_1\vert ^2<e^{-\varphi (s_1,p_1)}. \end{aligned}$$
(3.3)

Since \((a_1, s_1,p_1)\in {\mathcal {L}}\) and \({\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(a_0,s_0,p_0)\), we can see that

$$\begin{aligned} {\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(a_1,s_1,p_1). \end{aligned}$$

So together with (3.3), by the case (1) and case (2), we know that such \({\mathcal {L}}\) does not exist. Hence, the inequality (3.2) also holds for all \((s,p)\in \overline{{\mathbb {G}}}_2\).

Back to the inequality (3.2), since \((s,p)\in \overline{{\mathbb {G}}}_2\), we can set \(s=\lambda _1+\lambda _2\) and \(p=\lambda _1\lambda _2\). So for any \((\lambda _1,\lambda _2)\in \partial ({{\mathbb {D}}}\times {{\mathbb {D}}})\), we have

$$\begin{aligned} \vert 1+k_2\lambda _1+k_3+\lambda _2(k_2+k_3\lambda _1)\vert \ge \frac{1}{2}\vert 1-\bar{\lambda }_1\lambda _2\vert . \end{aligned}$$

Now let \(\lambda _1=1\), then for any \(\lambda _2\in \partial {\mathbb {D}}\), we can obtain

$$\begin{aligned}&\vert 1+k_2+k_3+\lambda _2(k_2+k_3)\vert \ge \frac{1}{2}\vert 1-\lambda _2\vert ,\\&\quad \Leftrightarrow \vert 1+k\vert ^2+\vert k\vert ^2-\frac{1}{2}\ge -\text {Re}\left( \left( 2k+2\vert k\vert ^2+\frac{1}{2}\right) \lambda _2\right) ,\ \ \ (k=k_2+k_3)\\&\quad \Leftrightarrow 2\vert k\vert ^2{+}2\text {Re}k{+}\frac{1}{2} {\ge }\left| 2\vert k\vert ^2{+}2k{+}\frac{1}{2}\right| {,} \ \ \ (\text {the inequality holds for all}\ \lambda _2\in \partial {\mathbb {D}})\\&\quad \Rightarrow (\text {Re}k)^2\ge \vert k\vert ^2. \end{aligned}$$

This implies that k is real . On the other hand, we can learn that if k is real , then we have

$$\begin{aligned} 2\vert k\vert ^2+2\text {Re}k+\frac{1}{2}&=2k^2+2k+\frac{1}{2}\\&=\frac{1}{2}(2k+1)^2\ge 0. \end{aligned}$$

Thus we can obtain that for any \(\lambda _2\in \partial {\mathbb {D}}\),

$$\begin{aligned} \vert 1+k_2+k_3+\lambda _2(k_2+k_3)\vert \ge \frac{1}{2}\vert 1-\lambda _2\vert \Longleftrightarrow k=k_2+k_3\ \ \text {is real }. \end{aligned}$$

Since \(k_3\ge 0\), \(k_2\) is real. Following this result, we want to omit the assumption \(\lambda _1=1\). Thus we give the following lemma.

Lemma 3.2

For any \((\lambda _1,\lambda _2)\in \partial {{\mathbb {D}}}\times \partial {{\mathbb {D}}}\), the inequality

$$\begin{aligned} \vert 1+k_2\lambda _1+k_3+\lambda _2(k_2+k_3\lambda _1)\vert \ge \frac{1}{2}\vert \lambda _1 -\lambda _2\vert \end{aligned}$$

holding is equivalent to \(k_3^2+k_3\ge k_2^2\).

Proof

By direct calculation, we have

$$\begin{aligned}&\vert 1+k_2\lambda _1+k_3+\lambda _2(k_2+k_3\lambda _1)\vert \ge \frac{1}{2}\vert \lambda _1 -\lambda _2\vert ,\\&\quad \Leftrightarrow \vert 1+k_2\lambda _1+k_3\vert ^2 +\vert k_2+k_3\lambda _1\vert ^2-\frac{1}{2}\\&\quad \ge -\text {Re}\left( \left( 2(k_2+k_3\lambda _1) (1+k_2\bar{\lambda }_1+k_3)+\frac{1}{2}\bar{\lambda }_1\right) \lambda _2\right) ,\\&\quad \Leftrightarrow \left( \frac{1}{2}+2k_2^2\right) +(2k_3+2k_3^2)+(2k_2+4k_2k_3) \text {Re}\lambda _1\\&\quad \ge \left| \left( \frac{1}{2}+2k_2^2\right) \bar{\lambda }_1+(2k_3+2k_3^2) \lambda _1+(2k_2+4k_2k_3)\right| . \end{aligned}$$

Now set \(a=\frac{1}{2}+2k_2^2\), \(b=2k_3+2k_3^2\) and \(c=2k_2+4k_2k_3\). Thus we see that if

$$\begin{aligned} a+b+c\text {Re}\lambda _1\ge \left| a\bar{\lambda }_1+b\lambda _1+c\right| , \end{aligned}$$
(3.4)

then we have

$$\begin{aligned} (1-(\text {Re}\lambda _1)^2)(4ab-c^2)\ge 0. \end{aligned}$$

It follows

$$\begin{aligned} k_3^2+k_3\ge k_2^2. \end{aligned}$$

Notice that \(a>0\) and \(b\ge 0\). So if we want to get the equivalence condition for the inequality (3.4), we only need to consider the following inequality holding for all \(\lambda _1\in \partial {\mathbb {D}}\),

$$\begin{aligned} a+b+c\text {Re}\lambda _1\ge 0. \end{aligned}$$

However, it is not hard to see

$$\begin{aligned} a+b\ge \vert c\vert . \end{aligned}$$
(3.5)

So together with \(a+b\ge 0\), we obtain that the inequality (3.5) is equivalent to

$$\begin{aligned} (a+b)^2\ge c^2. \end{aligned}$$

In fact, \((a+b)^2-c^2\ge (a+b)^2-4ab\ge 0\). Hence, this must be an equivalence condition. \(\square \)

Therefore, we have

$$\begin{aligned} \Gamma _{{\mathcal {P}}}(1,0,-1)\subseteq ({\mathbb {C}} \times \Gamma _{{\mathbb {G}}_2}(0,-1))\cup \{(a,s,p)\in {\mathbb {C}}^3:a=1+k_2s+k_3(1+p)\}, \end{aligned}$$

where \(k_2\) is real, \(k_3\ge 0\) and \(k_3^2+k_3\ge k_2^2\).

Now we want to show the other inclusion. Let \(k_2\) be a real number, and \(k_3\ge 0\) with \(k_3^2+k_3\ge k_2^2\). Then consider the hyperpalne

$$\begin{aligned} {\mathcal {L}}:=\{(a,s,p)\in {\mathbb {C}}^3: a=1+k_2s+k_3(1+p)\}. \end{aligned}$$

In order to prove \({\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(1,0,-1)\), we only need to show the following inequality

$$\begin{aligned} \vert 1+k_2s+k_3(1+p)\vert ^2\ge e^{-\varphi (s,p)} \end{aligned}$$
(3.6)

holding for any \((s,p)\in {\mathbb {G}}_2\).

Define

$$\begin{aligned} h(s,p)=1+k_2s+k_3(1+p). \end{aligned}$$

If \(k_2\ne 0\) and \(k_3\ne 0\), then h(s, p) is a well-defined holomorphic function on \({\mathbb {C}}^2\). By Lemma 3.2, for any \((s,p)\in \partial _s{\mathbb {G}}_2\), we have

$$\begin{aligned} \vert h(s,p)\vert ^2\ge e^{-\varphi (s,p)}. \end{aligned}$$

Now set \(s=\lambda _1+\lambda _2\) and \(p=\lambda _1\lambda _2\), and then for any \((\lambda _1,\lambda _2)\in \partial {\mathbb {D}}\times \partial {\mathbb {D}}\), we have

$$\begin{aligned} \vert 1+k_2(\lambda _1+\lambda _2)+k_3(1+\lambda _1\lambda _2)\vert \ge \frac{1}{2}\vert \lambda _1-\lambda _2\vert . \end{aligned}$$
(3.7)

Thus, if there exists \((s_1,p_1)\in \partial _s{\mathbb {G}}_2\) such that \(h(s_1,p_1)=0\), then we must have \(s_1^2=4p_1\). Hence, we can assume that \(s_1=2\lambda _0\) and \(p_1=\lambda _0^2\) for some \(\lambda _0\in \partial {\mathbb {D}}\). So for \(h(s_1,p_1)=0\), we obtain

$$\begin{aligned} 1+2k_2\lambda _0+k_3(1+\lambda _0^2)=0. \end{aligned}$$
(3.8)

Notice that \(\vert \lambda _0\vert =1\), and then assume \(\lambda _0=x_0+y_0i\). Thus we have \(x_0^2+y_0^2=1\). From (3.8), we see that

$$\begin{aligned} 1+2k_2(x_0+y_0i)+k_3(2x_0^2+2x_0y_0i)=0. \end{aligned}$$

Thus we get

$$\begin{aligned} \left\{ \begin{array}{l} k_2y_0+k_3x_0y_0=0,\\ 1+2k_2x_0+2k_3x_0^2=0. \end{array}\right. \end{aligned}$$

If \(y_0=0\), then \(2\vert k_2\vert =2k_3+1\), which contradicts to \(k^2_3+k_3\ge k^2_2\); if \(y_0\ne 0\), then \(k_2=-k_3x_0\). It follows that \(1+2k_2x_0+2k_3x_0^2=1\ne 0\). Hence, such \(\lambda _0\) does not exist. This means that for any \((s,p)\in \partial _s{\mathbb {G}}_2\), we have

$$\begin{aligned} h(s,p)\ne 0. \end{aligned}$$
(3.9)

Next we want to prove that the inequality (3.7) still holds for any \((\lambda _1,\lambda _2)\in \partial {\mathbb {D}}\times {\mathbb {D}}\), i.e.,

$$\begin{aligned} \vert 1+k_2(\lambda _1+\lambda _2)+k_3(1+\lambda _1\lambda _2)\vert \ge \frac{1}{2}\vert \lambda _1-\lambda _2\vert , \ \ \ \ (\lambda _1,\lambda _2)\in \partial {\mathbb {D}}\times {\mathbb {D}}. \end{aligned}$$

If there exists \((\lambda _1^0,\lambda _2^0) \in \partial {\mathbb {D}}\times {\mathbb {D}}\) such that

$$\begin{aligned} 1+k_2(\lambda _1^0+\lambda _2^0)+k_3(1+\lambda _1^0\lambda _2^0)=0, \end{aligned}$$

then we have

$$\begin{aligned} 1+k_2\lambda _1^0+k_3=-(k_2+k_3\lambda _1^0)\lambda _2^0. \end{aligned}$$

Thus, if \(k_2+k_3\lambda _1^0=0\), then \(\lambda _1^0\) is real. So we have

$$\begin{aligned} 0=1+k_2\lambda _1^0+k_3=1-k_3(\lambda _1^0)^2+k_3=1. \end{aligned}$$

This leads to a contradiction. Hence, since \(\vert \lambda _2^0\vert <1\), we can see

$$\begin{aligned}&\vert 1+k_2\lambda _1^0+k_3\vert<\vert k_2+k_3\lambda _1^0\vert ,\\&\quad \Leftrightarrow 1+2k_3<-2k_2\text {Re}\lambda _1^0, \ \ \ \ \ \ \ \ (\vert \lambda _1^0\vert =1)\\&\quad \Leftrightarrow 1+4k_3+4k_3^2<4k_2^2(\text {Re}\lambda _1^0)^2 \le 4k_2^2\le 4k_3+4k_3^2. \end{aligned}$$

This leads to another contradiction. Thus, define

$$\begin{aligned} g_{\lambda _1}(\lambda _2):=\frac{\frac{1}{2}(\lambda _1-\lambda _2)}{1+k_2(\lambda _1+\lambda _2)+k_3(1+\lambda _1\lambda _2)}, \end{aligned}$$

and then we get that for any fixed \(\lambda _1\in \partial {\mathbb {D}}\), \(g_{\lambda _1}(\lambda _2)\) is a well-defined holomorphic function on \(\overline{{\mathbb {D}}}\).Thus, by the maximal principle, together with (3.7), we have

$$\begin{aligned} \vert g_{\lambda _1}(\lambda _2)\vert \le 1 ,\ \ \ \ \ \ \ \ \ \ \ \forall \lambda _2\in \overline{{\mathbb {D}}}. \end{aligned}$$

So, we can obtain that the inequality

$$\begin{aligned} \vert h(\lambda _1, \lambda _2)\vert \ge \frac{1}{2}\vert \lambda _1-\lambda _2\vert \end{aligned}$$

holds for all \((\lambda _1,\lambda _2)\in \partial {\mathbb {D}}\times {\mathbb {D}}\), and also for all \((\lambda _1,\lambda _2)\in {\mathbb {D}}\times \partial {\mathbb {D}}\). Thus, together with (3.9), we can conclude that

$$\begin{aligned} h(s,p)\ne 0\ \ \ \ \ \ \ \ \ (s,p)\in \partial {\mathbb {G}}_2. \end{aligned}$$

Notice that \(k_2\ne 0\) and \(k_3\ne 0\), so by the Hartogs theorem we have

$$\begin{aligned} h(s,p)\ne 0\ \ \ \ \ \ \ \ \ (s,p)\in \overline{{\mathbb {G}}}_2. \end{aligned}$$

Now set \(s=\beta +\bar{\beta }p\), then if we want to show (3.6), we only need to prove the following inequality

$$\begin{aligned} \vert 1+k_2\beta +k_3+(k_2\bar{\beta }+k_3)p\vert \ge \left| 1-\frac{\frac{1}{2}(\vert \beta \vert ^2 +\bar{\beta }^2p)}{1+\sqrt{1-\vert \beta \vert ^2}}\right| \end{aligned}$$
(3.10)

holds for any \((\beta ,p)\in {\mathbb {D}}\times {\mathbb {D}}\).

Fixed any \(\beta \in {\mathbb {D}}\), define

$$\begin{aligned} f_\beta (p)=\frac{1-\frac{\frac{1}{2}(\vert \beta \vert ^2 +\bar{\beta }^2p)}{1+\sqrt{1-\vert \beta \vert ^2}}}{1+k_2\beta +k_3+(k_2\bar{\beta }+k_3)p}, \end{aligned}$$

and then we can see that \(f_\beta (p)\) is a well-defined holomorphic function on \(\overline{{\mathbb {D}}}\). So if we want to show (3.10), we only need to prove that for any fixed \(\beta \in {\mathbb {D}}\),

$$\begin{aligned} \vert f_\beta (p)\vert \le 1, \ \ \ \ \ \ \ \ \forall p\in {\mathbb {D}}. \end{aligned}$$

With the maximal principle, we just need to show

$$\begin{aligned} \vert f_\beta (p)\vert \le 1, \ \ \ \ \forall p\in \partial {\mathbb {D}} \ \text {with any fixed}\ \beta \in {\mathbb {D}}. \end{aligned}$$

However, for any fixed \(\beta \in {\mathbb {D}}\) and \(p\in \partial {\mathbb {D}}\), there exist \(\lambda _1\), \(\lambda _2\) such that

$$\begin{aligned} \vert \lambda _1\vert =\vert \lambda _2\vert =1,\ \beta =\frac{\lambda _1+\lambda _2}{2}\ \text {and}\ p=\lambda _1\lambda _2. \end{aligned}$$

So it suffices to show the following inequality

$$\begin{aligned} \vert 1+k_2(\lambda _1+\lambda _2)+k_3(1+\lambda _1\lambda _2)\vert \ge \frac{1}{2}\vert \lambda _1-\lambda _2\vert \end{aligned}$$

holding for all \((\lambda _1,\lambda _2)\in \partial {\mathbb {D}}\times \partial {\mathbb {D}}\). By Lemma 3.2, we can conclude that

$$\begin{aligned} {\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(1,0,-1). \end{aligned}$$

Now if \(k_3=0\), then \(k_2=0\). Easily, we can see that

$$\begin{aligned} {\mathcal {L}}=\{a=1\}\in \Gamma _{{\mathcal {P}}}(1,0,-1); \end{aligned}$$

and if \(k_2=0\), then \({\mathcal {L}}=\{a=1+k_3(1+p)\}\). Notice that

$$\begin{aligned} \vert 1+k_3(1+p)\vert \ge 1+k_3-k_3\vert p\vert \ge 1. \end{aligned}$$

Hence, we can also see that \({\mathcal {L}}\in \Gamma _{{\mathcal {P}}}(1,0,-1)\).

Therefore, we have

$$\begin{aligned} \Gamma _{{\mathcal {P}}}(1,0,-1)=({\mathbb {C}} \times \Gamma _{{\mathbb {G}}_2}(0,-1))\cup \{(a,s,p) \in {\mathbb {C}}^3:a=1+k_2s+k_3(1+p)\}, \end{aligned}$$

where \(k_2\) is real, \(k_3\ge 0\) and \(k_3^2+k_3\ge k_2^2\).

  1. (4)

    \((s_0,p_0)\in \partial {\mathbb {G}}_2\cap \Sigma \) and then \(a_0=0\).

Now suppose that \((a_0,s_0,p_0)=(0,2,1)\), and then by the same way, consider the hyperplane

$$\begin{aligned} {\mathcal {L}}:=\{(a,s,p)\in {\mathbb {C}}^3: a=k_2(2-s) +k_3(1-p)\}\in \Gamma _{{\mathcal {P}}}(0,2,1). \end{aligned}$$

Using the same argument, we obtain that for any \((s,p)\in \overline{{\mathbb {G}}}_2\),

$$\begin{aligned} \vert k_2(2-s)+k_3(1-p)\vert ^2\ge e^{-\varphi (s,p)}. \end{aligned}$$

Now set \(p=1\), then we have \(-2\le s\le 2\). So we obtain

$$\begin{aligned} \vert k_2(2-s)\vert \ge \sqrt{1-\frac{1}{4} s^2}. \end{aligned}$$

Thus, such \(k_2\) does not exist as \(s\rightarrow 2^{-}\). This leads that \(k_1\) must be zero. Therefore, we obtain

$$\begin{aligned} \Gamma _{{\mathcal {P}}}(0,2,1)={\mathbb {C}}\times \Gamma _{{\mathbb {G}}_2}(2,1). \end{aligned}$$

In summary, we can give a full description of \(\Gamma _{{\mathcal {P}}}(P_0)\) as follows.

Theorem 3.3

We can give the description of the tangent hyperplanes to the pentablock \({\mathcal {P}}\) at four different boundary points.

  1. (1)

    \(\Gamma _{{\mathcal {P}}}(a_0, 1,0)={\mathbb {C}}\times \Gamma _{{\mathbb {G}}_2}(1,0)\), \((\vert a_0\vert \le \frac{1}{2})\);

  2. (2)

    \(\Gamma _{{\mathcal {P}}}(a_0,0,-1) ={\mathbb {C}}\times \Gamma _{{\mathbb {G}}_2}(0,-1)\), \((\vert a_0\vert <1)\);

  3. (3)

    \(\Gamma _{{\mathcal {P}}}(1,0,-1)=({\mathbb {C}} \times \Gamma _{{\mathbb {G}}_2}(0,-1))\cup \{(a,s,p)\in {\mathbb {C}}^3:a =1+k_2s+k_3(1+p)\}\), where \(k_2\) is real, \(k_3\ge 0\) and \(k_3^2+k_3\ge k_2^2\);

  4. (4)

    \(\Gamma _{{\mathcal {P}}}(0,2,1)={\mathbb {C}} \times \Gamma _{{\mathbb {G}}_2}(2,1)\).

4 Proof of Theorem 1.1

Proof

Linear convexity of \({\mathcal {P}}\) implies that in the case of a smooth boundary point \(P_0\in \partial {\mathcal {P}}\), the set \(\Gamma _{{\mathcal {P}}}(P_0)\) is a singleton. Consider then the non-smooth point \(P_0\in \partial {\mathcal {P}}\). By Theorem 2.4, it is sufficient to consider the only four different cases

  • \(P_0=(a_0, 1,0)\), \(\vert a_0\vert \le \frac{1}{2}\);

  • \(P_0=(a_0,0,-1)\), \(\vert a_0\vert <1\);

  • \(P_0=(1,0,-1)\);

  • \(P_0=(0,2,1)\).

Then Theorems 3.3 and 2.2 imply that \(\Gamma _{{\mathcal {P}}}(P_0)\) is the union of connected sets whose intersection is non-empty for the non-smooth boundary point \(P_0\), so it is connected. Thus, by Theorem 2.5, we can conclude that the pentablock \({\mathcal {P}}\) is \({\mathbb {C}}\)-convex. This finishes the proof. \(\square \)