A generalized Schwarz lemma for two domains related to $\mu$-synthesis

We present a set of necessary and sufficient conditions that provides a Schwarz lemma for the tetrablock $\mathbb E$. As an application of this result, we obtain a Schwarz lemma for the symmetrized bidisc $\mathbb G_2$. In either case, our results generalize all previous results in this direction for $\mathbb E$ and $\mathbb G_2$.


Introduction
The aim of this article is to prove an explicit Schwarz lemma for two polynomially convex domains related to µ-synthesis, the symmetrized bidisc G 2 and the tetrablock E, defined as The classical Schwarz lemma gives a necessary and sufficient condition for the solvability of a two-point interpolation problem for analytic functions from the open unit disc D to itself, and describes the extremal functions. It has substantial generalizations in which the two copies of D are replaced by various other domains [9], typically either homogeneous or convex. In this paper, we prove a sharp Schwarz lemma for two domains which are neither homogeneous nor convex. We believe that our results will throw new lights on the spectral NevanlinnaPick problem, which is to interpolate from the unit disc to the set of k × k matrices of spectral radius no greater than 1 by analytic matrix functions.
We first produce several independent necessary and sufficient conditions under which there exists an analytic function from D to E that solves a two-point Nevanlinna-Pick interpolation problem for the tetrablock. This, in a way, generalizes the previous result of Abouhajar, White and Young in this direction, [1]. We mention here that the domain tetrablock was introduced by these three mathematicians in [1] and that this domain has deep connection with the most appealing and difficult problem of µ-synthesis (see [5] for further details). Since then the tetrablock has attracted considerable attention [6,11,10,15,16,5,17,12]. So, our first main result is the following Schwarz lemma for the tetrablock. (1 ′ ) there exists an analytic function ϕ : D → E such that ϕ(0) = (0, 0, 0) and ϕ(λ 0 ) = x; (3) either |b| ≤ |a| and |a −bp| + |ab − p| 1 − |b| 2 ≤ |λ 0 | or |a| ≤ |b| and |b −āp| + |ab − p| 1 − |a| 2 ≤ |λ 0 |; (4) there exists a 2 × 2 function F in the Schur class such that where x = (a 11 , a 22 , det A).
The proof of the theorem is given in Section 3. The functions Ψ and Υ play major role here. These two functions were introduced in [1] and we shall briefly describe them in Section 2. Being armed with the Schwarz lemma for the tetrablock, we apply the result to the symmetrized bidisc to obtain a generalized Schwarz lemma for G 2 , which is another main result of this paper and is presented as Theorem 4.4 in Section 4. This is possible because of the underlying relationship between the two domains E and G 2 which has been described in Lemma 4.1 and Theorem 4.3 in Section 4. This result generalizes the Schwarz lemma obtained by Agler, Young in [3] and also independently by Nokrane and Ransford in [13].

Background material
We recall from [1] the following two rational functions Ψ, Υ which play central role in the study of complex geometry of E.
Also we define We interpret Ψ(., x) to be the constant function equal to x 1 in the event that Indeed, for x 2 ∈ D, the linear fractional function Ψ(., x) maps D to the open disc with centre and radius respectively. Hence Similarly, if x 1 ∈ D, Υ(., x) maps D to the open disc with centre and radius respectively. It was shown in [1] that the closure E of the tetrablock is the following set In [1], the points in the sets E and its closure E are characterized in the following way.

Proof of Theorem 1.1
The equivalence of (1) − (4) was established in Theorem 1.2 of [1]. We shall prove here the rest parts.
(3) ⇔ (5). Condition (3) can be written as we have that condition (3) is equivalent to condition (5). Now for all z ∈ T, We obtained the last step by using the fact that |x| < k ⇔ Re(zx) < k, for all z ∈ T.

Application to the symmetrized bidisc
The symmetrized bidisc G 2 was introduced by a group of control theorists and later have been extensively studied by the complex analysists, geometers and operator theorists during past three decades. We list here a few of the numerous references about the symmetrized bidisc for the readers, [3,4,7,8,14,15,18]. This is another domain which has root in the µ-synthesis problem. The following result provides a beautiful connection between the two domains E and G 2 .
Among the various characterizations of the points of G 2 (see Theorem 1.1, [4]), the following is very elegant.
The following result will play central role in determining the interpolating function in the Schwarz lemma for the symmetrized bidisc.
Conversely, the map maps the tetrablock analytically onto the symmetrized bidisc and when restricted to f (G 2 ), becomes the inverse of the map f . Moreover, both f and g map the origin to the origin.
Proof. The map is clearly a one-one and holomorphic map. All we need to show is that it maps G 2 into the tetrablock. Conversely, the function g maps E to G 2 by Lemma 4.1 and is evidently holomorphic. Now Because, if (a, a, p) ∈ E, then by Lemma 4.1, (2a, p) ∈ G 2 and by the previous part (a, a, p) = f (2a, p). This also proves that g| f (G 2 ) is the inverse of f . Needless to prove that both f and g map the origin to the origin.
Proof. The proof follows easily from Theorem 4.3 by applying the functions f and g. For (s, p) ∈ G 2 , we have f ((s, p)) = ( s 2 , s 2 , p) ∈ E and Theorem 1.1 guarantees the existence of an analytic function φ from D to E or to E that maps 0 to 0 and λ 0 to ( s 2 , s 2 , p), when the conditions (2) − (9) in this theorem. Hence under the conditions (2) − (9), we obtain the analytic function ψ = g • φ that maps 0 to 0 and λ 0 to (s, p).
The converse is also true. That is, if there is a function φ from D to G 2 such that φ(0) = 0 and φ(λ 0 ) = (s, p), then f •φ : D → E maps 0 to 0 and λ 0 to ( s 2 , s 2 , p). So, the conditions (2) − (10) of Theorem 1.1 are satisfied with a = b = s 2 , which is same as saying that the conditions (2) − (9) hold. Hence the proof is complete.