Geometric Properties of the Pentablock

In this paper, we give a positive answer to the question raised in Kosi´nski (Com-plex Anal Oper Theory 9(6):1349–1359, 2015) and Zapałowski (J Math Anal Appl 430(1):126–143, 2015), i.e., we show that the pentablock P is a C -convex domain. 2 Pentablock Pentablock a Hartogs

the study of a long-standing open problem whether Lempert's theorem still holds for C-convex domain. However, as far as we know, the answer is positive for C-convex domain with C 2 boundary (see [10]).
In 2015, Agler, Lykova and Young [1] introduced a new bounded domain P by denotes the open unit ball in the space C 2×2 with the usual operator norm. They called this domain the pentablock as P ∩ R 3 is a convex body bounded by five faces, three of which are flat and two are curved (see [1]). The pentablock P is polynomially convex and starlike about the origin, but neither circled nor convex. Moreover, it does not have a C 1 boundary (see [1]). This new domain is also arising from the μ-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 C-convexity of 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 C-convexity of P. More precisely, we obtain the following theorem.

Theorem 1.1 The pntablock P is a C-convex domain.
Throughout this paper, D denotes the open unit disc in the complex plane ,while T denotes the unit circle. And other basic notions, definitions, and properties from the theory of invariant functions, linearly convex and C-convex domains that we shall use in the paper may be found in [3,9,11].

Pentablock
We first recall the definition of the pentablock P.

Pentablock as a Hartogs Domain
Following the description of the pentablock P, we can learn that the pentablock P is closely related to the symmetrized bidisc G 2 , which is defined by In fact, the pentablock P can be seen as a Hartogs domain in C 3 over the symmetrized bidisc G 2 (see [1]), that is, (s, p) ∈ G 2 and β = s−s p 1−| p| 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 P as a Hartogs domain will be great helpful for us to study the convexity of the pentablock P.

Some Useful Results
In this subsection, we will give some useful results on the symmetrized bidisc G 2 and the pentablock P. In order to study the pentablock P, it is sufficient to learn the C-convexity of G 2 .
Through the study of the boundary of 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 P to simplify the problem. In order to study the C-convexity of P in some simple way, we give the whole holomorphic automorphism group Aut(P) as follows.

The Set of All Tangent Hyperplanes to P at the Non-smooth Part
In this section, we will give a full description of the tangent hyperplanes to the pentablock P at the non-smooth boundary part. Set P 0 = (a 0 , s 0 , p 0 ) be a nonsmooth boundary point of the pentablock P, and let P (P 0 ) denote the set of all tangent hyperplanes to P at the boundary point P 0 . Now assume the hyperplane in C 3 that Together with automorphisms of P by Theorem 2.4, we can only need to consider some special boundary points P 0 : Actually we can assume that (s 0 , p 0 ) = (1, 0), and then we have |a 0 | ≤ 1 2 . Now we consider the hyperplane in C 3 passing through the boundary point P 0 , Next suppose that L ∈ P (P 0 ), so we get L ∩ P = ∅. This means that for any (s, p) ∈ G 2 , we have Now set p = 0, we can learn that for any s ∈ D, Together with |a 0 | ≤ 1 2 , we obtain that the following inequality Hence, taking s tends to 1, we can conclude that such k 2 does not exist. It follows that if the hyperplane L in (3.1) belongs to P (P 0 ), then k 1 = 0.
Moreover, if we have a hyperplane L ∈ P (P 0 ), then consider the following hyperplane in C 2 : Easily, we can see that
By the same way, consider the hyperplane passing through the boundary point P 0 with the assumption k 1 = 0, namely, Then suppose that L ∈ P (P 0 ), so this leads to L ∩ P = ∅. Hence, for any (s, p) ∈ G 2 , we have Now set s = 0, we obtain that for any p ∈ D, This means that the following inequality holds for any p ∈ 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 By the assumption, we can also see that s 2 0 = 4 p 0 . So we can assume (a 0 , s 0 , p 0 ) = (1, 0, −1). Then consider the hyperplane and suppose that L ∈ P (P 0 ). This implies that L ∩ P = ∅. Thus for any (s, p) ∈ G 2 , we have Let s = 0, then for any p ∈ D, we have Now we give the following lemma to help us.
Proof Set f (z) = 1 + k + kz, and then we directly learn that f (z) is a holomorphic function on D. Since | f (z)| ≥ 1, it means that f (z) has no zero in D. Hence, by the maximal principle, | f (z)| ≥ 1 holds for all z ∈ D. Now let |z| = 1, we can obtain This means that k is real. Now if k < 0, then there exists M > 0 such that and we can choose z 0 with −1 < z 0 < 0 such that Hence, we have This leads to a contradiction. On the other hand, for k ≥ 0, the inequality is evidently valid. Therefore, we conclude that k ≥ 0.
By Lemma 3.1, we have k 3 ≥ 0. Now we want to prove the inequality (3.2) on the whole G 2 .

So together with (3.3), by the case (1) and case
Back to the inequality (3.2), since (s, p) ∈ G 2 , we can set s = λ 1 + λ 2 and p = λ 1 λ 2 . So for any (λ 1 , λ 2 ) ∈ ∂(D × D), we have Now let λ 1 = 1, then for any λ 2 ∈ ∂D, we can obtain This implies that k is real . On the other hand, we can learn that if k is real , then we have Thus we can obtain that for any λ 2 ∈ ∂D, Since k 3 ≥ 0, k 2 is real. Following this result, we want to omit the assumption λ 1 = 1. Thus we give the following lemma.

Lemma 3.2 For any
holding is equivalent to k 2 3 + k 3 ≥ k 2 2 . Proof By direct calculation, we have Now set a = 1 2 + 2k 2 2 , b = 2k 3 + 2k 2 3 and c = 2k 2 + 4k 2 k 3 . Thus we see that if Notice that a > 0 and b ≥ 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 However, it is not hard to see a + b ≥ |c|. (3.5) So together with a + b ≥ 0, we obtain that the inequality (3.5) is equivalent to In fact, (a + b) 2 − c 2 ≥ (a + b) 2 − 4ab ≥ 0. Hence, this must be an equivalence condition.
Fixed any β ∈ D, define and then we can see that f β ( p) is a well-defined holomorphic function on D. So if we want to show (3.10), we only need to prove that for any fixed β ∈ D, With the maximal principle, we just need to show | f β ( p)| ≤ 1, ∀ p ∈ ∂D with any fixed β ∈ D.
In summary, we can give a full description of P (P 0 ) as follows.
Theorem 3. 3 We can give the description of the tangent hyperplanes to the pentablock P at four different boundary points.
Then Theorems 3.3 and 2.2 imply that 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 P is C-convex. This finishes the proof.