The three point Pick problem on the Bidisk

Jim Agler ^{∗} U.C. San Diego La Jolla, California 92093

John E. McCarthy Washington University St. Louis, Missouri 63130

December 11 1999

Abstract

0 Introduction

The original Pick problem is to determine, given N points λ_{1}, …, λ_{N} in the unit disk 𝔻 and N complex numbers w_{1}, …, w_{N}, whether there exist a function ϕ in the closed unit ball of H^{∞}(𝔻) (the space of bounded analytic functions on 𝔻) that maps each point λ_{i} to the corresponding value w_{i}. This problem was solved by G. Pick in 1916 [9], who showed that a necessary and sufficient condition is that the Pick matrix

(\frac{1 − w̄_{i}w_{j}}{1 − λ̄_{i}λ_{j}} )_{i,j=1}^{N}

be positive semi-definite.

It is well-known that if the problem is extremal, i.e. the problem can be solved with a function of norm one but not with a function of any smaller norm, then the Pick matrix is singular, and the corresponding solution is a unique Blaschke product, whose degree equals the rank of the Pick matrix [6, 5].

In [1], the first author extended Pick’s theorem to the space H^{∞}(𝔻^{2}), the bounded analytic functions on the bidisk; see also [4, 3, 2]. It was shown in [2] that if the problem has a solution, then it has a solution that is a rational inner function; however the qualitative properties of general solutions are not fully understood. The example λ_{1} = (0, 0), λ_{2} = (\frac{1}{2}, \frac{1}{2}), w_{1} = 0, w_{2} = \frac{1}{2} shows that even extremal problems do not always have unique solutions.

∗Partially supported by the National Science Foundation

The two point Pick problem on the bidisk is easily analyzed. It can be solved if and only if the Kobayashi distance between λ_{1} and λ_{2} is greater than or equal to the hyperbolic distance between w_{1} and w_{2}. On the bidisk, the Kobayashi distance is just the maximum of the hyperbolic distance between the first coordinates, and the hyperbolic distance between the second coordinates. A pair of points in 𝔻^{2} is called balanced if the hyperbolic distance between their first coordinates equals the hyperbolic distance between their second coordinates.

The two point Pick problem has a unique solution if and only if the Kobayashi distance between λ_{1} and λ_{2} exactly equals the hyperbolic distance between w_{1} and w_{2}, and moreover (λ_{1}, λ_{2}) is not balanced. In this case the solution is a Möbius map in the coordinate function in which the Kobayashi distance is attained. If the distance between λ_{1} and λ_{2} equals the distance between w_{1} and w_{2}, but the pair (λ_{1}, λ_{2}) is balanced, then the function ϕ will be uniquely determined on the geodesic disk passing through λ_{1} and λ_{2}, but will not be unique off this disk. (For the example λ_{1} = (0, 0), λ_{2} = (\frac{1}{2}, \frac{1}{2}), w_{1} = 0, w_{2} = \frac{1}{2}, on the diagonal {(z, z)} we must have ϕ(z, z) = z; but off the diagonal any convex combination of the two coordinate functions z^{1} and z^{2} will work).

It is the purpose of this article to examine the three point Pick problem on the bidisk. Our main result is the following:

Theorem 0.1 The solution to an extremal non-degenerate three point problem on the bidisk is unique. The solution is given by a rational inner function of degree 2. There is a formula for the solution in terms of two uniquely determined rank one matrices.

In the next section, we shall define precisely the terms “extremal” and “non-degenerate”, but roughly it means that the problem is genuinely two-dimensional, is really a 3 point problem not a 2 point problem, and the minimal norm of a solution is 1.

1 Notation and Preliminaries

We wish to consider the N point Pick interpolation problem

ϕ(λ_{i}) = w_{i},  i = 1, …, N
and  (1.1)
∥ϕ∥_{H^{∞}(𝔻^{2})} ≤ 1.

We shall say that a solution ϕ to (1.1) is an extremal solution if ∥ϕ∥ = 1, and no solution has a smaller norm.

For a point λ in 𝔻^{2}, we shall use superscripts to denote coordinates:

λ = (λ^{1}, λ^{2}).

Let W, Λ^{1} and Λ^{2} denote the N-by-N matrices

W = (1 − w̄_{i}w_{j})_{i,j=1}^{N}
Λ^{1} = (1 − λ̄_{i}^{1}λ_{j}^{1})_{i,j=1}^{N}
Λ^{2} = (1 − λ̄_{i}^{2}λ_{j}^{2})_{i,j=1}^{N} .

A pair Γ, Δ of N-by-N positive semi-definite matrices is called permissible if

W = Λ^{1} · Γ + Λ^{2} · Δ.  (1.2)

Here · denotes the Schur or entrywise product:

(A · B)_{ij} := A_{ij}B_{ij}.

The main result of [1] is that the problem (1.1) has a solution if and only if there is a pair Γ, Δ of permissible matrices.

A kernel K on {λ_{1}, …, λ_{N}} × {λ_{1}, …, λ_{N}} is a positive definite N-by-N matrix

K_{ij} = K(λ_{i}, λ_{j}).

We shall call the kernel K admissible if

Λ^{1} · K ≥ 0

and

Λ^{2} · K ≥ 0.

If the problem (1.1) has a solution and K is an admissible kernel, then (1.2) implies that K · W ≥ 0. We shall call the kernel K active if it is admissible and K · W has a non-trivial null-space. Notice that all extremal problems have an active kernel.

If one can find a pair of permissible matrices one of which is 0, then the Pick problem is really a one-dimensional problem because one can find a solution ϕ that depends only on one of the coordinate functions. If this occurs, we shall call the problem degenerate; otherwise we shall call it non-degenerate.

2 The three point problem

We wish to analyze extremal solutions to three point Pick problems. Fix three points λ_{1}, λ_{2}, λ_{3} in 𝔻^{2}, and three numbers w_{1}, w_{2}, w_{3}. Let notation be as in the previous section. We shall make the following assumptions throughout this section:

(a) The function ϕ is an extremal solution to the Pick problem of interpolating λ_{i} to w_{i}, where i ranges from 1 to 3.

(b) The function ϕ is not an extremal solution to any of the three two point Pick problems mapping two of the λ_{i}’s to the corresponding w_{i}’s.

(c) The three point problem is non-degenerate.

Lemma 2.1 If K is admissible, then rank(K · W) > 1.

Proof: Suppose (Γ, Δ) is permissible. By (1.2), we have

K · W = K · Λ^{1} · Γ + K · Λ^{2} · Δ.

If rank(K · W) = 1, then either Γ = 0 (which violates (c)), or there exists t > 0 such that

K · Λ^{1} · Γ = tK · W.

But then (\frac{1}{t}Γ, 0) is permissible, violating assumption (c). □

Lemma 2.2 If K is an admissible kernel with a non-vanishing column, then rank(K ·Λ^{1}) ≥ 2 and rank(K · Λ^{2}) ≥ 2.

Proof: Suppose that rank(K ·Λ^{1}) = 1. As no entry of Λ^{1} can be 0, and some column of K is non-vanishing, there is a column of K · Λ^{1} that is non-vanishing. As K · Λ^{1} is self-adjoint and rank one and has non-zero diagonal entries, the other two columns of K · Λ^{1} must be non-zero multiples of this non-vanishing column. So Q := K · Λ^{1} is a positive rank one matrix with no zero entries, and K has no zero entries.

So

Λ^{2} = (\frac{1}{K} ) · (K · Λ^{2})
= (\frac{1}{Q} · Λ^{1}) · (K · Λ^{2})
= (\frac{1}{Q} · K · Λ^{2}) · Λ^{1},

where by \frac{1}{K} and \frac{1}{Q} is meant the entrywise reciprocal. Now K · Λ^{2} is positive by hypothesis, and \frac{1}{Q} is positive because Q is rank one and non-vanishing; moreover the Schur product of two positive matrices is positive [8, Thm 5.2.1]. Therefore (Γ+Δ· \frac{1}{Q} ·K ·Λ^{2}, 0) is permissible, which violates assumption (c). □

Lemma 2.3 If K is an active kernel, it has a non-vanishing column.

Proof: By assumption (b), we cannot have both K(λ_{1}, λ_{2}) = 0 and K(λ_{1}, λ_{3}) = 0; for then K restricted to {λ_{2}, λ_{3}} × {λ_{2}, λ_{3}} would be an active kernel for the two point problem on λ_{2}, λ_{3}, and so any solution to the two point problem would have norm at least one, so ϕ would be an extremal solution to the two point problem.

If neither of K(λ_{1}, λ_{2}) or K(λ_{1}, λ_{3}) are 0, we are done. So assume without loss of generality that the first is non-zero and the second equals zero. But then K(λ_{2}, λ_{3}) cannot equal zero, for then K restricted to {λ_{1}, λ_{2}}×{λ_{1}, λ_{2}} would be active, violating assumption (b). Thus we can conclude that the second column of K is non-vanishing. □

Lemma 2.4 If (Γ, Δ) is a permissible pair, then rank(Γ) = 1 = rank(Δ).

Proof: Let K be an active kernel. Then K · W is rank 2, and annihilates some vector

γ⃗ = ( \begin{matrix}γ_{1} \\ γ_{2} \\ γ_{3} \end{matrix} ) .

Moreover, by assumption (b), none of the entries of γ⃗ are 0.

Suppose rank(Γ) > 1. We have

K · W = K · Λ^{1} · Γ + K · Λ^{2} · Δ.

As K · Λ^{1} has non-zero diagonal terms, Oppenheim’s theorem [7, Thm 7.8.6] guarantees that rank(K · Λ^{1} · Γ) ≥ rank(Γ). As K · W has rank 2, and K · Λ^{2} · Δ ≥ 0, we must have rank(Γ) = 2. Write

Γ = u⃗ ⊗ u⃗ + v⃗ ⊗ v⃗,

where u⃗ and v⃗ are not collinear; if

u⃗ = ( \begin{matrix}u_{1} \\ u_{2} \\ u_{3} \end{matrix} ) ,

then u⃗ ⊗ u⃗ denotes the matrix

(u⃗ ⊗ u⃗)_{ij} = u_{i}ū_{j}.

Let

K · Λ^{1} = w⃗ ⊗ w⃗ + x⃗ ⊗ x⃗

if K · Λ^{1} is rank two, and

K · Λ^{1} = w⃗ ⊗ w⃗ + x⃗ ⊗ x⃗ + y⃗ ⊗ y⃗

if it is rank three.

Notice that (K · Λ^{1} · Γ) γ⃗ = 0, because K · Λ^{1} · Γ is positive and

⟨(K · Λ^{1} · Γ) γ⃗, γ⃗⟩ = −⟨(K · Λ^{2} · Δ) γ⃗, γ⃗⟩
≤ 0.

Therefore all 4 of (u⃗ ⊗ u⃗) · (w⃗ ⊗ w⃗), (u⃗ ⊗ u⃗) · (x⃗ ⊗ x⃗), (v⃗ ⊗ v⃗) · (w⃗ ⊗ w⃗), (v⃗ ⊗ v⃗) · (x⃗ ⊗ x⃗) annihilate γ⃗. Therefore

∑_{j=1}^{3} ū_{j}w̄_{j}γ_{j} = 0
= ∑_{j=1}^{3} ū_{j}x̄_{j}γ_{j}
= ∑_{j=1}^{3} v̄_{j}w̄_{j}γ_{j}
= ∑_{j=1}^{3} v̄_{j}x̄_{j}γ_{j}

Therefore the vectors

( \begin{matrix}\overline{w_{1}} & γ_{1} \\ \overline{w_{2}} & γ_{2} \\ \overline{w_{3}} & γ_{3} \end{matrix} ) and ( \begin{matrix}\overline{x_{1}} & γ_{1} \\ \overline{x_{2}} & γ_{2} \\ \overline{x_{3}} & γ_{3} \end{matrix} )

are both orthogonal to both u⃗ and v⃗, and therefore are collinear (since u⃗ and v⃗ span a two-dimensional subspace of ℂ^{3}). As none of the entries of γ⃗ are 0, it follows that w⃗ and x⃗ are collinear. Therefore rank(K · Λ^{1}) = 1, contradicting Lemmata (2.2) and (2.3). □

Lemma 2.5 The matrices Γ and Δ are unique.

Proof: If both (Γ_{1}, Δ_{1}) and (Γ_{2}, Δ_{2}) were permissible, then (\frac{1}{2}(Γ_{1} + Γ_{2}), \frac{1}{2}(Δ_{1} + Δ_{2})) would also be permissible. As all permissible matrices are rank one by Lemma 2.4, it follows that Γ_{1} and Γ_{2} are constant multiples of each other, and so are Δ_{1} and Δ_{2}.

So suppose

W = Λ^{1} · Γ + Λ^{2} · Δ

and

W = Λ^{1} · t_{1}Γ + Λ^{2} · t_{2}Δ,

where both t_{1}, t_{2} are positive, one is less than 1, and the other is bigger than 1. Then

(1 − t_{1})Λ^{1} · Γ + (1 − t_{2})Λ^{2} · Δ = 0.

Assume without loss of generality that t_{1} < 1 < t_{2}. Then (\frac{t_{2} − t_{1}}{1 − t_{1}} Γ, 0) is permissible, which contradicts Assumption (c). □

Theorem 2.6 The solution to an extremal non-degenerate three point problem satisfying Assumptions (a)-(c) is unique. It is given by a rational inner function of degree 2, and there is a formula in terms of Γ and Δ.

Proof: We have

W = Λ^{1} · Γ + Λ^{2} · Δ.  (2.7)

Choose vectors a⃗ and b⃗ so that Γ = a⃗ ⊗ a⃗ and Δ = b⃗ ⊗ b⃗.

Choose some point λ_{4} in 𝔻^{2}, distinct from the first three points. Let w_{4} be the value attained at λ_{4} by some solution ϕ of the three point problem (1.1). Then the four point problem, interpolating λ_{i} to w_{i} for i = 1, …, 4 has a solution, so we can find a pair of 4-by-4 permissible matrices Γ̃ and Δ̃ satisfying (1.2). As the restriction of these matrices to the first three points satisfy (2.7), and Γ and Δ are unique by Lemma 2.5, we get that Γ̃ and Δ̃ are extensions of Γ and Δ. Therefore we have

( \begin{matrix}& & & 1 − w̄_{1}w_{4} \\ & W & & 1 − w̄_{2}w_{4} \\ & & & 1 − w̄_{3}w_{4} \\ ∗ & ∗ & ∗ & 1 − |w_{4}|^{2} \end{matrix} ) = ( \begin{matrix}& & & g_{1} \\ & Γ & & g_{2} \\ & & & g_{3} \\ ḡ_{1} & ḡ_{2} & ḡ_{3} & g_{4} \end{matrix} ) · ( \begin{matrix}& & & 1 − λ̄_{1}^{1}λ_{4}^{1} \\ & Λ^{1} & & 1 − λ̄_{2}^{1}λ_{4}^{1} \\ & & & 1 − λ̄_{3}^{1}λ_{4}^{1} \\ ∗ & ∗ & ∗ & 1 − |λ_{4}^{1}|^{2} \end{matrix} )

(2.8)

+ ( \begin{matrix}& & & d_{1} \\ & Δ & & d_{2} \\ & & & d_{3} \\ d̄_{1} & d̄_{2} & d̄_{3} & d_{4} \end{matrix} ) · ( \begin{matrix}& & & 1 − λ̄_{1}^{2}λ_{4}^{2} \\ & Λ^{2} & & 1 − λ̄_{2}^{2}λ_{4}^{2} \\ & & & 1 − λ̄_{3}^{2}λ_{4}^{2} \\ ∗ & ∗ & ∗ & 1 − |λ_{4}^{2}|^{2} \end{matrix} )

As Γ̃ is positive, it must be that ( \begin{matrix}g_{1} \\ g_{2} \\ g_{3} \end{matrix} ) is in the range of Γ, so

( \begin{matrix}g_{1} \\ g_{2} \\ g_{3} \end{matrix} ) = sa⃗ = s ( \begin{matrix}a_{1} \\ a_{2} \\ a_{3} \end{matrix} )

for some constant s. Similarly,

( \begin{matrix}d_{1} \\ d_{2} \\ d_{3} \end{matrix} ) = tb⃗ = t ( \begin{matrix}b_{1} \\ b_{2} \\ b_{3} \end{matrix} )

for some t.

Let

v⃗_{1} = ( \begin{matrix}(1 − λ̄_{1}^{1}λ_{4}^{1})a_{1} \\ (1 − λ̄_{2}^{1}λ_{4}^{1})a_{2} \\ (1 − λ̄_{3}^{1}λ_{4}^{1})a_{3} \end{matrix} )

v⃗_{2} = ( \begin{matrix}(1 − λ̄_{1}^{2}λ_{4}^{2})b_{1} \\ (1 − λ̄_{2}^{2}λ_{4}^{2})b_{2} \\ (1 − λ̄_{3}^{2}λ_{4}^{2})b_{3} \end{matrix} )

v⃗_{3} = ( w̄_{1} ; w̄_{2} ; w̄_{3} ) .

Looking at the first three entries of the last column of Equation (2.8), we get

( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) = sv⃗_{1} + tv⃗_{2} + w_{4}v⃗_{3}.  (2.9)

Equation (2.9) has a unique solution for s, t and w_{4} unless

det ( \begin{matrix}(1 − λ̄_{1}^{1}λ_{4}^{1})a_{1} & (1 − λ̄_{1}^{2}λ_{4}^{2})b_{1} & w̄_{1} \\ (1 − λ̄_{2}^{1}λ_{4}^{1})a_{2} & (1 − λ̄_{2}^{2}λ_{4}^{2})b_{2} & w̄_{2} \\ (1 − λ̄_{3}^{1}λ_{4}^{1})a_{3} & (1 − λ̄_{3}^{2}λ_{4}^{2})b_{3} & w̄_{3} \end{matrix} ) = 0.  (2.10)

Notice that the determinant in (2.10) is analytic in λ_{4}. So if there is a single point λ_{4} for which the determinant does not vanish, there is an open neighborhood of this point for which the determinant doesn’t vanish. Consequently w_{4} (and hence ϕ) would be determined uniquely on this open set, and hence on all of 𝔻^{2}.

Suppose the determinant in (2.10) vanished identically. Then there is a set of uniqueness of λ_{4}’s on which Equation (2.9) can be solved with either s or t equal to 0. (If both s and t were uniquely determined, then v⃗_{3} would be 0, violating Assumption (a)). Without loss of generality, take t = 0. Moreover, we can also assume without loss of generality that w_{1} and w_{2} do not both vanish.

Then one can use the first component of Equation (2.9) to solve for s, and the second one to get

w_{4} = \frac{(1 − λ̄_{1}^{1}λ_{4}^{1})a_{1} − (1 − λ̄_{2}^{1}λ_{4}^{1})a_{2}}{(1 − λ̄_{1}^{1}λ_{4}^{1})a_{1}w̄_{2} − (1 − λ̄_{2}^{1}λ_{4}^{1})a_{2}w̄_{1}}

Then w_{4} is given uniquely as a rational function of degree 1 of λ_{4}^{1}, violating both Assumptions (b) and (c).

Therefore we can assume that there is an open set on which Equation (2.9) has a unique solution, so by Cramer’s rule we get

ϕ(λ_{4}) = w_{4} = \frac{det ( \begin{matrix}(1 − λ̄_{1}^{1}λ_{4}^{1})a_{1} & (1 − λ̄_{1}^{2}λ_{4}^{2})b_{1} & 1 \\ (1 − λ̄_{2}^{1}λ_{4}^{1})a_{2} & (1 − λ̄_{2}^{2}λ_{4}^{2})b_{2} & 1 \\ (1 − λ̄_{3}^{1}λ_{4}^{1})a_{3} & (1 − λ̄_{3}^{2}λ_{4}^{2})b_{3} & 1 \end{matrix} )}{det ( \begin{matrix}(1 − λ̄_{1}^{1}λ_{4}^{1})a_{1} & (1 − λ̄_{1}^{2}λ_{4}^{2})b_{1} & w̄_{1} \\ (1 − λ̄_{2}^{1}λ_{4}^{1})a_{2} & (1 − λ̄_{2}^{2}λ_{4}^{2})b_{2} & w̄_{2} \\ (1 − λ̄_{3}^{1}λ_{4}^{1})a_{3} & (1 − λ̄_{3}^{2}λ_{4}^{2})b_{3} & w̄_{3} \end{matrix} )}.  (2.11)

Equation (2.11) gives a formula for ϕ that shows that ϕ is a rational function of degree at most 2, whose second order terms only involve the mixed product λ_{4}^{1}λ_{4}^{2}.

To show that ϕ is inner, we follow [2]. We can rewrite (1.2) as

1 + λ̄_{i}^{1}λ_{j}^{1}a_{i}ā_{j} + λ̄_{i}^{2}λ_{j}^{2}b_{i}b̄_{j} = w̄_{i}w_{j} + a_{i}ā_{j} + b_{i}b̄_{j}.  (2.12)

Realizing both sides of (2.12) as Grammians, we get that there exists a 3-by-3 unitary U such that, for j = 1, 2, 3,

U ( \begin{matrix}1 \\ λ_{j}^{1}ā_{j} \\ λ_{j}^{2}b̄_{j} \end{matrix} ) = ( w_{j} ; ā_{j} ; b̄_{j} ) .  (2.13)

Writing

U = \begin{aligned}& ℂ ℂ^{2} \\ \begin{matrix}ℂ \\ ℂ^{2} \end{matrix} & ( \begin{matrix}A & B \\ C & D \end{matrix} ) , \end{aligned}

and letting

E_{λ} = ( \begin{matrix}λ^{1} & 0 \\ 0 & λ^{2} \end{matrix} ) ,

we can solve (2.13) to get

w_{j} = A + BE_{λ_{j}}(1 − DE_{λ_{j}})^{−1}C.

So the function

ψ(λ) = A + BE_{λ}(1 − DE_{λ})^{−1}C

interpolates the original data. Moreover ψ is inner, because a calculation shows that

1 − \overline{ψ(λ)}ψ(λ) = ((1 − DE_{λ})^{−1}C)^{∗}(1 − E_{λ}^{∗}E_{λ})((1 − DE_{λ})^{−1}C),

so |ψ| is less than 1 on 𝔻^{2} and equals 1 on the distinguished boundary. By uniqueness, we must have ψ = ϕ, and hence ϕ is inner.

Finally, we must show that the degree of ϕ is exactly two. This is because an easy calculation shows that a rational function of degree one

\frac{c_{1} + c_{2}z^{1} + c_{3}z^{2}}{c_{4} + c_{5}z^{1} + c_{6}z^{2}}

is inner only if it is a function of either just z^{1} or just z^{2}, i.e. either both c_{2} and c_{5} or both c_{3} and c_{6} can be chosen to be zero. This would violate Assumption (c). □

3 Finding Γ and Δ

Formula (2.11) works fine, provided one knows Γ and Δ (or, equivalently, a_{1}, a_{2}, a_{3} and b_{1}, b_{2}, b_{3}). Lemma 2.5 assures us that Γ and Δ are unique; how does one find them?

First, let us make a simplifying normalization. One can pre-compose ϕ with an automorphism of 𝔻^{2}, and post-compose it with an automorphism of 𝔻; so one can assume that λ_{1} = (0, 0) and w_{1} = 0. Write λ_{2} = (α_{2}, β_{2}) and λ_{3} = (α_{3}, β_{3}). Moreover, as Γ_{ij} = a_{i}ā_{j} and Δ_{ij} = b_{i}b̄_{j}, we can choose a_{1} ≥ 0 and b_{1} ≥ 0; again without loss of generality we can assume that b_{1} > 0. Thus we have

( \begin{matrix}1 & 1 & 1 \\ 1 & 1 − |w_{2}|^{2} & 1 − w̄_{2}w_{3} \\ 1 & 1 − w_{2}w̄_{3} & 1 − |w_{3}|^{2} \end{matrix} ) = ( \begin{matrix}a_{1}^{2} & a_{1}ā_{2} & a_{1}ā_{3} \\ a_{1}a_{2} & |a_{2}|^{2} & a_{2}ā_{3} \\ a_{1}a_{3} & ā_{2}a_{3} & |a_{3}|^{2} \end{matrix} ) · ( \begin{matrix}1 & 1 & 1 \\ 1 & 1 − |α_{2}|^{2} & 1 − ᾱ_{2}α_{3} \\ 1 & 1 − α_{2}ᾱ_{3} & 1 − |α_{3}|^{2} \end{matrix} )

(3.1)

+ ( \begin{matrix}b_{1}^{2} & b_{1}b̄_{2} & b_{1}b̄_{3} \\ b_{1}b_{2} & |b_{2}|^{2} & b_{2}b̄_{3} \\ b_{1}b_{3} & b̄_{2}b_{3} & |b_{3}|^{2} \end{matrix} ) · ( \begin{matrix}1 & 1 & 1 \\ 1 & 1 − |β_{2}|^{2} & 1 − β̄_{2}β_{3} \\ 1 & 1 − β_{2}β̄_{3} & 1 − |β_{3}|^{2} \end{matrix} )

Looking at the first column of (3.1) we get

b_{1} = \frac{1 − a_{1}^{2}}{\sqrt{1 − a_{1}^{2}}}
b_{2} = \frac{1 − a_{1}a_{2}}{\sqrt{1 − a_{1}^{2}}}
b_{3} = \frac{1 − a_{1}a_{3}}{\sqrt{1 − a_{1}^{2}}}

Thus we have three equations that uniquely determine a_{1}, a_{2}, a_{3}:

(1 − a_{1}^{2})(1 − |w_{2}|^{2}) = (1 − a_{1}^{2})|a_{2}|^{2}(1 − |α_{2}|^{2}) + |1 − a_{1}a_{2}|^{2}(1 − |β_{2}|^{2})  (3.2)
(1 − a_{1}^{2})(1 − w̄_{2}w_{3}) = (1 − a_{1}^{2})a_{2}ā_{3}(1 − ᾱ_{2}α_{3}) + (1 − a_{1}a_{2})(1 − a_{1}ā_{3})(1 − β̄_{2}β_{3}) (3.3)
(1 − a_{1}^{2})(1 − |w_{3}|^{2}) = (1 − a_{1}^{2})|a_{3}|^{2}(1 − |α_{3}|^{2}) + |1 − a_{1}a_{3}|^{2}(1 − |β_{3}|^{2})  (3.4)

Equation (3.3) can be used to solve for a_{3} as a rational function of a_{1} and ā_{2}; then one is left with two real algebraic equations in three real variables, a_{1}, ℜ(a_{2}) and ℑ(a_{2}). Provided the original data is really extremal, this system of two equations will have a unique solution with a_{1} ≥ 0. If the original data is not extremal, multiply w_{2} and w_{3} by a positive real number t, and choose the largest t for which equations (3.2)–(3.4) can be solved. This will produce an inner function ϕ via (2.11); then the function \frac{1}{t}ϕ will be the unique function of minimal norm solving the original problem.

Example 3.5 Let us consider a very symmetric special case. Let λ_{1} = (0, 0), λ_{2} = (r, 0), λ_{3} = (0, r), w_{1} = 0, w_{2} = t and w_{3} = t, where t is to be chosen as large as possible and r is a fixed positive number. Then by symmetry, we can assume that a_{1} = b_{1} = \frac{1}{\sqrt{2}} and a_{2} = b_{3} = ā_{2}.

Equations (3.2)–(3.4) then reduce to:

\frac{1}{2}(1 − t^{2}) = \frac{1}{2}a_{2}^{2}(1 − r^{2}) + (1 − \frac{1}{\sqrt{2}}a_{2})^{2}
\frac{1}{2}(1 − t^{2}) = a_{2}(\sqrt{2} − a_{2})

Solving, one gets two solutions. One solution is

t = \frac{r}{2 − r}, a_{2} = \frac{\sqrt{2}}{2 − r};

the other is

t = \frac{r}{2 + r}, a_{2} = \frac{\sqrt{2}}{2 + r}.

The first of these is clearly the extremal solution, and formula (2.11) then gives

ϕ(z) = \frac{z^{1} + z^{2} − 2z^{1}z^{2}}{2 − z^{1} − z^{2}}

as the extremal solution.

The second solution also corresponds to a pair of rank one matrices Γ and Δ that satisfy (2.7), even though the problem is non-extremal. If one plugs in to (2.11) one gets the inner function

ϕ_{2}(z) = \frac{z^{1} + z^{2} + 2z^{1}z^{2}}{2 + z^{1} + z^{2}} .

References

[1] J. Agler. Some interpolation theorems of Nevanlinna-Pick type. Preprint, 1988.

[2] J. Agler and J.E. McCarthy. Nevanlinna-Pick interpolation on the bidisk. J. Reine Angew. Math., 506:191–204, 1999.

[3] J.A. Ball and T.T. Trent. Unitary colligations, reproducing kernel Hilbert spaces, and Nevanlinna-Pick interpolation in several variables. J. Funct. Anal., 197:1–61, 1998.

[4] B.J. Cole and J. Wermer. Pick interpolation, von Neumann inequalities, and hyperconvex sets. In Complex Potential Theory, pages 89–129. Kluwer Acad. Publ., Dordrecht, 1994.

[5] C. Foiaş and A.E. Frazho. The commutant lifting approach to interpolation problems. Birkhäuser, Basel, 1990.

[6] John B. Garnett. Bounded Analytic Functions. Academic Press, New York, 1981.

[7] R.A. Horn and C.R. Johnson. Matrix Analysis. Cambridge University Press, Cambridge, 1985.

[8] R.A. Horn and C.R. Johnson. Topics in Matrix Analysis. Cambridge University Press, Cambridge, 1991.

[9] G. Pick. Über die Beschränkungen analytischer Funktionen, welche durch vorgegebene Funktionswerte bewirkt werden. Math. Ann., 77:7–23, 1916.