New York Journal of Mathematics

New York J. Math. 6 (2000) 227–236.

The Three Point Pick Problem on the Bidisk

Jim Agler and John E. McCarthy

Abstract. We prove that a non-degenerate extremal 3 point Pick problem on the bidisk always has a unique solution.

Contents

0. Introduction 227 1. Notation and preliminaries 228 2. The three point problem 229 3. Finding Γ and Δ 235 References 236

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 exists 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

227^{ISSN 1076-9803/00}

Received March 23, 2000.

Mathematics Subject Classification. Primary: 32Q45, 32D15; Secondary: 46J15.

Key words and phrases. bidisk, H^{∞}, Pick interpolation, Nevanlinna-Pick.

The first author was partially supported by the National Science Foundation.

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.

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 “nondegenerate”, 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

(1.1)  ϕ(λ_{i}) = w_{i}, i = 1, …, N, and
∥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 four 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 permissi-

ble, 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} )

+ ( \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 (∗), we get

( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) = sv⃗_{1} + tv⃗_{2} + w_{4}v⃗_{3}. (2.8)

Equation (2.8) 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. (2.9)

Notice that the determinant in (2.9) 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 (2.9) vanished identically. Then there is a set of uniqueness of λ_{4}’s on which Equation (2.8) 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 (2.8) 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 (2.8) has a unique solution, so by Cramer’s rule we get

ϕ(λ_{4}) = w_{4} = \begin{matrix}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} ) \end{matrix}. (2.10)

Equation (2.10) 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}. (2.11)

Realizing both sides of (2.11) 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} ) . (2.12)

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 (2.12) 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 (2.10) 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} )

+ ( \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 (∗∗) 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}:

(3.1)  (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})

Equation (3.2) 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_{3}). 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 (3.1)–(3.3) can be solved. This will produce an inner function ϕ via (2.10); then the function \frac{1}{t}ϕ will be the unique function of minimal norm solving the original problem.

Example 3.4. 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}.

236 Jim Agler and John E. McCarthy

Equations (3.1)–(3.3) 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 (2.10) 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 (2.10) 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.

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[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 (1998), 1–61, MR 2000b:47028, Zbl 914.47020.

[4] B. J. Cole and J. Wermer, Pick interpolation, von Neumann inequalities, and hyperconvex sets, Complex Potential Theory (P. M. Gauthier, ed.), NATO ASI Ser., Ser. C, Math. Phys. Sci., no. 439, Kluwer Acad. Publ., Dordrecht, 1994, pp. 89–129, MR 96m:46092, Zbl 811.30024.

[5] C. Foiaş and A. E. Frazho, The Commutant Lifting Approach to Interpolation Problems, Operator Theory: Advances and Applications, no. 44, Birkhäuser, Basel, 1990, MR 92k:47033, Zbl 718.47010.

[6] John B. Garnett, Bounded Analytic Functions, Pure and Applied Mathematics, no. 96, Academic Press, New York, 1981, MR 83g:30037, Zbl 469.30024.

[7] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1985, MR 87e:15001, Zbl 576.15001.

[8] R. A. Horn and C. R. Johnson, Topics in Matrix Analysis, Cambridge University Press, Cambridge, 1991, MR 92e:15003, Zbl 729.15001.

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

U.C. San Diego, La Jolla, California 92093 jagler@math.ucsd.edu

Washington University, St. Louis, Missouri 63130 mccarthy@math.wustl.edu http://www.math.wustl.edu/˜ mccarthy/

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