Gleason’s Problem and Homogeneous Interpolation in Hardy and Dirichlet-Type

Spaces of the Ball

Daniel Alpay^{1} Department of Mathematics Ben-Gurion University of the Negev Beer-Sheva 84105, Israel E-mail: dany@math.bgu.ac.il

and

H. Turgay Kaptanoğlu^{2} Mathematics Department Middle East Technical University Ankara 06531, Turkey E-mail: kaptan@math.metu.edu.tr URL: http://www.math.metu.edu.tr/∼kaptan/

Version: 1 August 2002

We solve Gleason’s problem in the reproducing kernel Hilbert spaces with reproducing kernels 1/(1 − ∑_{1}^{N} z_{j} \overline{w}_{j})^{r} for real r > 0, and their counterparts for r ≤ 0, and study the homogeneous interpolation problem in these spaces.

Key Words: Reproducing kernel Hilbert space, Dirichlet-type space, weighted Hardy space, Schur multiplier, Gleason’s problem, shift and resolvent, homogeneous interpolation.

2000 Mathematics Subject Classification: Primary: 47A57; Secondary: 32A70, 46E22, 47B32.

1. INTRODUCTION

A complete Nevanlinna–Pick kernel is a function K(z, w) that is positive in a set Ω and such that 1/K(z, w) has one positive square in Ω. A typical example is given by

K_{1}(z, w) = \frac{1}{1 − ⟨z, w⟩},  (1)

1The research of this author was supported by the Israel Science Foundation (Grant no. 322/00).

2This author thanks Ben-Gurion University of the Negev for supporting a visit to Beer-Sheva.

where z = (z_{1}, z_{2}, …, z_{N}) and w = (w_{1}, w_{2}, …, w_{N}) are in the unit ball

𝔹_{N} = {z = (z_{1}, z_{2}, …, z_{N}) ∈ ℂ^{N} | |z|^{2} = |z_{1}|^{2} + |z_{2}|^{2} + ⋯ + |z_{N}|^{2} < 1}

of ℂ^{N} and ⟨z, w⟩ = z_{1}\overline{w}_{1} + z_{2}\overline{w}_{2} + ⋯ + z_{N} \overline{w}_{N}. We will denote by H_{1} the reproducing kernel Hilbert space with reproducing kernel (1).

Complete Nevanlinna–Pick kernels originate with the work of Agler (see Quiggin’s paper [9]) and play a central role in the extension of the Nevanlinna–Pick interpolation problem from the case of the disk to more general settings such as the ball. In [8] McCullough and Trent extended Beurling’s theorem to reproducing kernel Hilbert spaces with complete Pick kernels. The theorem they proved reads as follows for H_{1}^{n}, the Hilbert space whose elements are n–vectors and with reproducing kernel (1).

Theorem 1.1. Let M be a closed subspace of H_{1}^{n} invariant under the operators of multiplication by z_{j} (j = 1, …, N). Then there is an n × m– valued function Φ (with m possibly infinite) that takes contractive values in 𝔹_{N} and whose radial limits satisfy Φ(z)Φ(z)^{∗} = I_{n} on the sphere 𝕊 = ∂𝔹_{N} such that M = ΦH_{1}^{m}.

Motivated by a problem in interpolation of analytic functions, we proved in [4] and [2] a particular case of Theorem 1.1 for a class of spaces M of finite codimension. The last hypothesis allowed us to get explicit formulas and to have an estimate on the size of Φ.

In this paper we prove the counterpart of our results of [4] for all reproducing kernel Hilbert spaces H_{r} with reproducing kernels

K_{r}(z, w) = \frac{1}{(1 − ⟨z, w⟩)^{r}}

for r real and positive and

K_{0}(z, w) = log \frac{1}{1 − ⟨z, w⟩}.

The principal value of the logarithm is used in all kernels. Except for 0 ≤ r ≤ 1, these kernels are not complete Nevanlinna–Pick kernels. However, the spaces include in particular the Dirichlet space (r = 0), the Hardy space (r = N), and the weighted Bergman spaces (r > N) of the ball. They are also called Dirichlet-type spaces. Denoting the inner product and norm of H_{r} by (·, ·)_{H_{r}} and ∥ · ∥_{H_{r}}, the reproducing property of the kernels is that

(f(z), K_{r}(z, w))_{H_{r}} = f(w)

for any f ∈ H_{r} and any w ∈ 𝔹_{N}.

Our computations require frequent use of the gamma function Γ with domain all real numbers except for nonpositive integers. The gamma function satisfies Γ(x + 1) = xΓ(x) for such real numbers x and n! = Γ(n + 1) for nonnegative integers n. Stirling’s formula for the asymptotic behavior of the gamma function and its counterpart for the factorial are

Γ(x) ∼ \frac{x^{x}}{e^{x}\sqrt{x}} (x → ∞)  and  m! ∼ \frac{m^{m}\sqrt{m}}{e^{m}} (m → ∞), (2)

where A ∼ B means that |A/B| is bounded above and below by two positive constants. From this it follows that

\frac{Γ(a + b)}{Γ(a + c)} ∼ a^{b−c}  (a → ∞)  (3)

for fixed b and c.

We also use the multi-index notation in which α = (α_{1}, …, α_{N}) is an N-tuple of nonnegative integers, |α| = α_{1} + ⋯ + α_{N}, α! = α_{1}! ⋯ α_{N}!, z^{α} = z_{1}^{α_{1}} ⋯ z_{N}^{α_{N}}, and 0^{0} = 1.

Recall that a ℂ^{n×m}-valued function Φ defined in 𝔹_{N} is called a Schur multiplier from H_{r}^{m×1} into H_{r}^{n×1} if the operator of multiplication by Φ is a contraction from H_{r}^{m×1} into H_{r}^{r×1}, or, equivalently, if the kernel

\frac{I_{n} − Φ(z)Φ(w)^{∗}}{(1 − ⟨z, w⟩)^{r}}

is positive in 𝔹_{N}. The function Φ is in particular holomorphic in 𝔹_{N} and takes contractive values there. There are functions that are holomorphic and contractive in 𝔹_{N} and are not Schur multipliers. Examples are presented in [4] for r = 1. Another family of examples is presented in Proposition 2.5 below.

The theorem below is the main result of this paper.

Theorem 1.2. Let a_{1}, …, a_{m} ∈ 𝔹_{N} and c_{1}, …, c_{m} ∈ ℂ^{n×1}. The set M of functions f in H_{r}^{n×1} such that

c_{j}^{∗}f(a_{j}) = 0  (j = 1, …, m)  (4)

is of the form ΦM_{r}^{(n+m′(N−1))×1}, where m′ ≤ m independently of r, and Φ is a ℂ^{n×(n+m′(N−1))}–valued holomorphic function in the ball taking coisometric radial limits almost everywhere on the sphere 𝕊. The operator of multiplication by Φ is bounded in all the spaces H_{r} and is a contraction for r ≥ 1.

For related results on the Dirichlet and Bergman spaces when N = 1, we refer to [1].

2. THE SPACES H_{r}

We gain much insight on the structure of the spaces H_{r} by considering the expansion

\frac{1}{(1 − ⟨z, w⟩)^{r}} = ∑_{j=0}^{∞} \frac{Γ(r + j)}{Γ(r) j!}⟨z, w⟩^{j} = ∑_{α} \frac{Γ(r + |α|)}{Γ(r) α!}z^{α}\overline{w}^{α},  (5)

that is valid in the set that Γ(r) makes sense, which includes r > 0. Expansion (5) gives

(z^{α}, z^{β})_{H_{r}} = 0 (α ≠ β),  ∥z^{α}∥_{H_{r}}^{2} = \frac{Γ(r) α!}{Γ(r + |α|)},  (6)

and that H_{r} is the space of holomorphic functions f(z) = ∑_{α} f_{α}z^{α} in the ball for which

∥f∥_{H_{r}}^{2} = ∑_{α} |f_{α}|^{2} \frac{Γ(r) α!}{Γ(r + |α|)} < ∞.  (7)

When r = 0, we have the expansion

log \frac{1}{1 − ⟨z, w⟩} = ∑_{j=1}^{∞} \frac{⟨z, w⟩^{j}}{j} = ∑_{|α|>0} \frac{|α|!}{|α| α!}z^{α}\overline{w}^{α},  (8)

which yields the norms

∥z^{α}∥_{H_{0}}^{2} = \frac{|α| α!}{|α|!}  and  ∥f∥_{H_{0}}^{2} = ∑_{|α|>0} |f_{α}|^{2} \frac{|α| α!}{|α|!}.  (9)

In other words, the monomials {z^{α}} form a complete orthogonal set in each H_{r} (for |α| > 0 when r = 0). By the reproducing property of K_{r} applied on itself, we have ∥K_{r}(·, w)∥_{H_{r}}^{2} = K_{r}(w, w) in each H_{r}.

We denote by k_{r,j} for any r ≥ 0 the coefficients in the expansion of K_{r}(z, w) in powers of ⟨z, w⟩ given in (5) and (8). So

k_{r,j} = \frac{Γ(r + j)}{Γ(r) j!} (r > 0, j = 0, 1, …) and k_{0,j} = \frac{1}{j} (j = 1, 2, …). (10)

The two expressions in (10) are quite similar: by (3), if r > 0, then k_{r,j} ∼ 1/j^{1−r} for large j, which tends to 1/j as r → 0. All k_{r,j} are positive.

We next give integral formulas for the inner products of the spaces H_{r}. Those corresponding to r = 0 and r = 1 are in [7] and [3].

Proposition 2.1. Consider the pairing

L_{s}(f, g) = \frac{(s − N) ⋯ (s − 2)(s − 1)}{N!} ∫_{𝔹_{N}} f(z)\overline{g(z)}(1 − |z|^{2})^{s−N−1} dν(z),

where ν is the volume measure on 𝔹_{N} with total mass 1. Then (f, g)_{H_{r}} = lim_{s→r} L_{s}(f, g) for r > 0. Also

(f, g)_{H_{0}} = lim_{s→0} \frac{(s − N) ⋯ (s − 1)s}{N!} ∫_{𝔹_{N}} f(z)\overline{g(z)}(1 − |z|^{2})^{s−N−1} dν(z).

Proof. We prove only the formula for r > 0. It suffices to obtain the result on the monomials since they form a complete orthogonal set in each H_{r}. By Lemma 1 of [3], L_{s}(z^{α}, z^{β}) = 0 if α ≠ β. By the same lemma, for s > N we have

L_{s}(z^{α}, z^{α}) = \frac{α!}{s(s + 1) ⋯ (s − 1 + |α|)}

The function s ↦ L_{s} continues analytically at least to the region {Re s > 0} and

|||z^{α}|||_{r}^{2} =^{def} lim_{s→r} L_{s}(z^{α}, z^{α}) = \frac{Γ(r) α!}{Γ(r + |α|)} = ∥z^{α}∥_{H_{r}}.

The Hilbert space completion of the polynomials in the norm |||·|||_{r} coincides with H_{r}.

Remark 2.2. When r = N + 1 and r = N, the formulas of Proposition 2.1 actually reduce to the classical forms of the inner products of the Bergman and Hardy spaces of the ball. We consider the measures

dμ_{s}(ρ) = \frac{(s − N) ⋯ (s − 2)(s − 1)}{N!}2Nρ^{2N+1}(1 − ρ^{2})^{s−N−1} dρ

on the interval [0, 1]. Integration in polar coordinates (§1.4.3 of [10]) shows that

|||f|||_{r}^{2} = lim_{s→r} ∫_{0}^{1} dμ_{s}(ρ) ∫_{𝕊} |f(ρζ)|^{2} dσ(ζ),

where σ is the area measure on 𝕊 with total mass 1. The measure μ_{s} is finite if and only if s > N, and then a computation with the beta function gives μ_{s}([0, 1]) = 1. The distribution function of μ_{s} is defined as F_{s}(x) = μ_{s}([0, x]) and given by

F_{s}(x) = \frac{(s − N) ⋯ (s − 2)(s − 1)}{N!} 2N ∫_{0}^{x} ρ^{2N+1}(1 − ρ^{2})^{s−N−1} dρ

= ∑_{j=0}^{N−1} \frac{(−1)^{j}(s − N) ⋯ (s − 2)(s − 1)}{(N − 1 − j)! (s − N + j) j!}[1 − (1 − x^{2})^{s−N+j}]

for 0 ≤ x < 1; also F_{s}(1) = 1 as stated above. By Theorem 2.3 of [5], for all x ∈ [0, 1],

F^{N+1}(x) = lim_{s→N+1} F_{s}(x) = x^{2N}

is the distribution function of the weak* limit μ^{N+1} of μ_{s} as s → N + 1. In other words, dμ^{N+1}(ρ) = dF^{N+1}(ρ) dρ = 2N ρ^{2N−1} dρ for ρ ∈ [0, 1], and

|||f|||_{N+1}^{2} = 2N ∫_{0}^{1} ρ^{2N−1} ∫_{𝕊} |f(ρζ)|^{2} dσ(ζ) = ∫_{𝔹_{N}} |f(z)|^{2} dν(z) = ∥f∥_{H_{N+1}}^{2}.

Similarly, for 0 ≤ x < 1,

F^{N}(x) = lim_{s→N} F_{s}(x) = 0

is the distribution function of the weak* limit μ^{N} of μ_{s} as s → N. But F^{N}(1) = 1. That means μ^{N} is the unit point mass at ρ = 1, and

|||f|||_{N}^{2} = lim_{ρ→1^{−}} ∫_{𝕊} |f(ρζ)|^{2} dσ(ζ) = sup_{0≤ρ<1} ∫_{𝕊} |f(ρζ)|^{2} dσ(ζ) = ∥f∥_{H_{r}}^{2}

since ∫_{𝕊} |f(ρζ)|^{2} dσ(ζ) is an increasing function of ρ.

Remark 2.3. The spaces H_{r} are ordered by inclusion and the inclusion maps between some are contractions. If 0 < r < s, since

\frac{Γ(s) α!/Γ(s + |α|)}{Γ(r) α!/Γ(r + |α|)} = \frac{(|α| − 1 + r) ⋯ (r + 1)r}{(|α| − 1 + s) ⋯ (s + 1)s} < 1  (11)

for any α, we have ∥f∥_{H_{s}} < ∥f∥_{H_{r}} by (7), and thus H_{r} ⊂ H_{s} contractively. When 0 = r < s, then

\frac{Γ(s) α!/Γ(s + |α|)}{|α| α!/|α|!} = (\frac{|α| − 1}{|α| − 1 + s}) ⋯ (\frac{2}{s + 2})(\frac{1}{s + 1})\frac{1}{s}.  (12)

This ratio is < 1 for s ≥ 1, and thus H_{0} ⊂ H_{s} contractively. For 0 < s < 1, this ratio is bounded above but not necessarily by 1. Thus H_{0} ⊂ H_{s}, but we cannot say whether the inclusion is a contraction or not. However, all the inclusions are proper as we show now. The case r = 1 and s = N is in [4].

Example 2.4. For 0 ≤ r < s, let f(z_{1}, …, z_{N}) = ∑_{j=1}^{∞} j^{(r+s−4)/4} z_{1}^{j}.

Then by (3) as j → ∞

∥f∥_{H_{r}}^{2} ∼ ∑_{j=1}^{∞} j^{(r+s−4)/2} j^{1−r} = ∑_{j=1}^{∞} \frac{1}{j^{1−(s−r)/2}} = ∞,

whereas

∥f∥_{H_{s}}^{2} ∼ ∑_{j=1}^{∞} j^{(s+r−4)/2} j^{1−s} = ∑_{j=1}^{∞} \frac{1}{j^{1+(s−r)/2}} < ∞,

giving us a function f in H_{s} that is not in H_{r}.

Proposition 2.5. When N > 1, a nonconstant inner function is not a Schur multiplier of H_{r} for 0 ≤ r < N. In contrast, every function in the unit ball of H^{∞}(𝔹_{N}), hence every inner function, is a Schur multiplier for the Hardy space H_{N}.

Proof. Let 0 ≤ r < N and f(z) = ∑_{α} f_{α}z^{α} be an inner function in 𝔹_{N}. Since ∥1∥_{H_{r}} = 1, the norm of the operator of multiplication by f is at least ∥f∥_{H_{r}}. But the inner functions are precisely those functions f ∈ H_{N} that have ∥f∥_{H_{N}} = 1 ([11, p. ix]). With s = N ≥ 1, the ratios in (11) and (12) are both < 1 for the values of r considered. Hence 1 = ∥f∥_{H_{N}} < ∥f∥_{H_{r}} for the same r, proving the first claim.

For r = N, H_{N} is the Hardy space of the ball and the square of its norm has the traditional form of the integral of the square of the boundary values. So if f is a bounded holomorphic function with ∥f∥_{H^{∞}(𝔹_{N})} ≤ 1 and g ∈ H_{N}, then ∥fg∥_{H_{N}} ≤ ∥g∥_{H_{N}}. Thus the operator of multiplication by f on H_{N} has norm ≤ 1.

We now identify the spaces that result from restricting the spaces H_{r} to slices of the ball. A slice is the intersection of a complex one-dimensional subspace of ℂ^{N} (a complex line) with the ball 𝔹_{N} and is completely determined by a point ζ on 𝕊. Given a function f in the ball and ζ ∈ 𝕊, the slice function f_{ζ} is defined as f_{ζ}(λ) = f(λζ) for λ in the disc 𝔹_{1}. If f is holomorphic in the ball, then each f_{ζ} is holomorphic in the disc. We work out the restrictions from the point of view of weighted Hardy spaces.

A weighted Hardy space H in the ball is a Hilbert space of holomorphic functions in which the monomials {z^{α}} form a complete orthogonal set with

\frac{∥z^{α_{1}}∥_{H}}{∥z^{α_{1}}∥_{H_{N}}} = \frac{∥z^{α_{2}}∥_{H}}{∥z^{α_{2}}∥_{H_{N}}}

whenever |α_{1}| = |α_{2}|; see [6, p. 23]. A glance at (6) and (9) shows that each of the spaces H_{r} including H_{0} is a weighted Hardy space in the ball.

We define our extension and restriction operators E and R only with respect to the slice obtained from ζ = (1, 0, …, 0) since any slice can be transformed to any other slice by a unitary rotation, and such a rotation carries holomorphic functions to holomorphic functions. For f holomorhic in the disc 𝔻 = 𝔹_{1} and F holomorphic in the ball 𝔹_{N}, we set E(f)(z_{1}, z_{2}, …, z_{N}) = f(z_{1}) and R(F)(λ) = F (λ, 0, …, 0). So R(F)(λ) = F_{(1,0,... ,0)}(λ).

Proposition 2.6. (Proposition 2.21 of [6]) The map E is an isometry from H_{r}(𝔻) into H_{r}(𝔹_{N}), and R is a norm-decreasing map from H_{r}(𝔹_{N}) onto H_{r}(𝔻), so that RE = I, the identity operator.

Therefore the restriction of the Dirichlet space of the ball to a slice is the Dirichlet space of the disc, the restriction of the space H_{1} of the ball to a slice is the Hardy space of the disc, the restriction of the space H_{2} of the ball to a slice is the Bergman space of the disc, the restriction of the Hardy space of the ball to a slice is the weighted Bergman space A_{N−2}^{2} of the disc, the restriction of the Bergman space of the ball to a slice is the weighted Bergman space A_{N−1}^{2} of the disc, and similar results hold for intermediate spaces and weighted Bergman spaces of the ball too.

There is also the converse problem, which is the question of whether a function is in H_{r}(𝔹_{N}) for N > 1 if all its slice functions are in H_{r}(𝔻). This question is answered in the negative in Example 7.2.11 of [10] for the case r = N = 2. That example can be modified to answer the same question in the negative for all N > 1 and r ≥ 0.

Example 2.7. For N > 1 and j = 1, 2, …, let F_{j}(z) = N^{Nj}(z_{1} ⋯ z_{N})^{2j}. So F_{j}(z) = N^{Nj}z^{α} with α = (2j, …, 2j) and |α| = 2Nj. Then F_{j_{ζ}}(λ) = F_{j}(ζ)λ^{2Nj}, and |F_{j}(ζ)| ≤ 1 for any ζ ∈ 𝕊 by Proposition 7.2.8 (i) of [10]. This last fact along with (6), (9), and (3) give

∥F_{j_{ζ}}∥_{H_{r}(𝔻)}^{2} ∼ \frac{1}{j^{r−1}}

for all r ≥ 0. On the other hand, (6), (9), and (2) together imply

∥F_{j}∥_{H_{r}(𝔹_{N})}^{2} ∼ \frac{1}{j^{r−(N+1)/2}}

for all r ≥ 0. Now pick c_{j} ∈ ℂ so that

∑_{k=1}^{∞} \frac{|c_{j}|^{2}}{j^{r−1}} < ∞  but  ∑_{j=1}^{∞} \frac{|c_{j}|^{2}}{j^{r−(N+1)/2}} = ∞;

for example, pick c_{j} = j^{r/2−1−ε} with 0 < ε ≤ (N − 1)/4. Then F (z) = ∑_{j=1}^{∞} c_{j}F_{j}(z) is not in H_{r}(𝔹_{N}) although all its slice functions F_{ζ} are in H_{r}(𝔻).

3. GLEASON’S PROBLEM AND RESOLVENT OPERATORS

In this section we solve Gleason’s problem in the spaces H_{r}. Recall that Gleason’s problem consists of finding functions g_{1}, …, g_{N} in H_{r} satisfying

f(z) − f(a) = ∑_{k=1}^{N}(z_{k} − a_{k})g_{k}(z, a)

for all z ∈ 𝔹_{N} for given f in H_{r} and a ∈ 𝔹_{N}. Said differently, it is the problem of determining if the coordinate functions shifted to a generate the maximal ideal consisting of all f ∈ H_{r} with f(a) = 0; see [10, §6.6.1]. The case r = 1 was considered in [4], where g_{k} turned out to be the k-th backward shift (resolvent) operator in H_{1}. Here the situation is reversed, and we define the backward shift operators by solving Gleason’s problem first. We obtain our results in two stages, first for the kernel functions K_{r}(·, w), and then for arbitrary functions in H_{r}. The cases r > 0 and r = 0 are treated the same way. So let first f(z) = K_{r}(z, w) for some arbitrary but fixed w ∈ 𝔹_{N}. Following §6.6.2 of [10], we see that the functions

g_{k}(z, a) = \frac{\overline{w}_{k}}{⟨z − a, w⟩}(K_{r}(z, w) − K_{r}(a, w))  (13)

do the job provided they lie in H_{r}. To prove they do, we use (5) and (6). Then

g_{k}(z, a) = \frac{\overline{w}_{k}}{⟨z − a, w⟩} ∑_{j=1}^{∞} k_{r,j}(⟨z, w⟩^{j} − ⟨a, w⟩^{j})

= \frac{\overline{w}_{k}(⟨z, w⟩ − ⟨a, w⟩)}{⟨z − a, w⟩} ∑_{j=1}^{∞} k_{r,j} ∑_{l=0}^{j−1}⟨a, w⟩^{l}⟨z, w⟩^{j−1−l}

= \overline{w}_{k} ∑_{l=0}^{∞}⟨a, w⟩^{l} ∑_{j=l+1}^{∞} k_{r,j}⟨z, w⟩^{j−1−l}

= \overline{w}_{k} ∑_{l=0}^{∞}⟨a, w⟩^{l} ∑_{j=0}^{∞} k_{r,j+l+1}⟨z, w⟩^{j},

which converges absolutely. Call the inner sum K_{r}^{l+1}(z, w). Then

K_{r}^{l+1}(z, w) = \frac{1}{⟨z, w⟩^{l+1}} ∑_{j=l+1}^{∞} k_{r,j}⟨z, w⟩^{j}

= \frac{1}{⟨z, w⟩^{l+1}} (K_{r}(z, w) − ∑_{j=0}^{l} k_{r,j}⟨z, w⟩^{j})
= k_{r,l+1} + k_{r,l+2}⟨z, w⟩ + k_{r,l+3}⟨z, w⟩^{2} + ⋯,

where k_{r,j} are defined by (10). So by (7), ∥K_{r}^{l+1}(·, w)∥_{H_{r}} ≤ ∥K(·, w)∥_{H_{r}} for any w. Hence the linear operator A_{l} defined on the span of the functions K_{r}(·, w) that takes K_{r}(z, w) to K_{r}^{l+1}(z, w) is bounded with ∥A_{l}∥ ≤ 1, and K_{r}^{l+1}(z, w) belongs to H_{r}. Then also

∥g_{k}∥_{H_{r}} ≤ ∑_{l=0}^{∞} |⟨a, w⟩|^{l}∥K(·, w)∥_{H_{r}} ≤ K_{r}(w, w) ∑_{l=0}^{∞} |a|^{l} = \frac{K_{r}(w, w)}{1 − |a|} < ∞.

Thus the g_{k} belong to H_{r} and the solution to Gleason’s problem on the span of the functions K_{r}(·, w) is complete.

Another consequence we derive from a computation like the last one is that the operator

T_{k}^{r}(a)^{∗} = \overline{w}_{k} ∑_{l=0}^{∞}⟨a, w⟩^{l}A_{l}

defined on the span of the functions K_{r}(·, w) is bounded with

∥T_{k}^{r}(a)^{∗}∥ ≤ \frac{1}{1 − |a|},  (14)

which does not depend on w. This fact allows us to extend the operator T_{k}^{r}(a)^{∗} as a bounded operator with the bound in (14) to all of H_{r} since it contains the span of the functions K_{r}(·, w) densely. We conclude that Gleason’s problem is solvable for f ∈ H_{r} with g_{k} = T_{k}^{r}(a)^{∗}(f). We call the operator T_{k}^{r}(a)^{∗} the backward shift (resolvent) operator on H_{r} in analogy with the results in [4].

The adjoint T_{k}^{r}(a) of each T_{k}^{r}(a)^{∗} is also bounded on H_{r}, and we call it the forward shift operator on H_{r}. The computation

T_{k}^{r}(a)(K_{r}(z, w))(v) = (T_{k}^{r}(a)(K_{r}(z, w)), K_{r}(z, v))_{H_{r}}
= (K_{r}(z, w), T_{k}^{r}(a)^{∗}(K_{r}(z, v)))_{H_{r}}
= (K_{r}(z, w), \frac{\overline{v}_{k}}{⟨z − a, v⟩}(K_{r}(z, v) − K_{r}(a, v)))_{H_{r}}

= \overline{\frac{\overline{v}_{k}}{⟨w − a, v⟩}(K_{r}(w, v) − K_{r}(a, v))}

which uses the reproducing property of K_{r}(z, w) yields

T_{k}^{r}(a)(K_{r}(z, w)) = \frac{z_{k}}{⟨z, w − a⟩}(K_{r}(z, w) − K_{r}(z, a)).

In terms of series, this is

T_{k}^{r}(a)(K_{r}(z, w)) = z_{k} ∑_{l=0}^{∞}⟨z, a⟩^{l} ∑_{j=l+1}^{∞} k_{r,j}⟨z, w⟩^{j−1−l}

= z_{k} ∑_{l=0}^{∞}⟨z, a⟩^{l}K_{r}^{l+1}(z, w).

Setting w = 0 when r > 0 gives

T_{k}^{r}(a)^{∗}(1) = 0  and  T_{k}^{r}(a)(1) = \frac{z_{k}}{⟨z, a⟩}(K_{r}(z, a) − 1).

Remark 3.1. A careful examination of the above shows that our procedure solves Gleason’s problem and defines shift operators in any reproducing kernel Hilbert space on the ball whose reproducing kernel is a holomorphic function of ⟨z, w⟩.

It is possible to write down an explicit formula for the action of T_{k}^{r}(a) on an arbitrary f ∈ H_{r}. By the reproducing property of K_{r} and using (6), (9), and (10), we obtain

T_{k}^{r}(a)(f)(w) = (T_{k}^{r}(a)(f(z)), K_{r}(z, w))_{H_{r}} = (f(z), T_{k}^{r}(a)^{∗}(K_{r}(z, w)))_{H_{r}}
= (∑_{α} f_{α}z^{α}, \overline{w}_{k} ∑_{l=0}^{∞}⟨a, w⟩^{l} ∑_{j=0}^{∞} k_{r,j+l+1}⟨z, w⟩^{j})_{H_{r}}

= (∑_{α} f_{α}z^{α}, \overline{w}_{k} ∑_{l=0}^{∞}⟨a, w⟩^{l} ∑_{β} k_{r,|β|+l+1} \frac{|β|!}{β!}z^{β}\overline{w}^{β})_{H_{r}}

= w_{k} ∑_{l=0}^{∞}⟨w, a⟩^{l} ∑_{α} k_{r,|α|+l+1} \frac{|α|!}{α!}∥z^{α}∥_{H_{r}}^{2}f_{α}w^{α}

= w_{k} ∑_{l=0}^{∞}⟨w, a⟩^{l} ∑_{α} \frac{k_{r,|α|+l+1}}{k_{r,|α|}} f_{α}w^{α},

which yields

T_{k}^{r}(a)(f(z)) = z_{k} ∑_{l=0}^{∞}⟨z, a⟩^{l} ∑_{j=0}^{∞} \frac{k_{r,j+l+1}}{k_{r,j}} ∑_{|α|=j} f_{α}z^{α}.

An application of (3) gives k_{r,j+l+1}/k_{r,j} ∼ 1 as j → ∞ for each l, and this implies that T_{k}^{r}(a)(f) is in H_{r} since f is. In particular, when f(z) = z^{α}, we have

T_{k}^{r}(a)(z^{α}) = \frac{z_{k}z^{α}}{k_{r,|α|}} ∑_{l=0}^{∞} k_{r,|α|+l+1}⟨z, a⟩^{l} = \frac{z_{k}z^{α}}{k_{r,|α|}} K_{r}^{|α|+1}(z, a).

A similar computation shows that T_{k}^{0}(a)(1) = 0 since ∥1∥_{H_{0}} = 0. From this it follows by the orthogonality of {z^{α}} in H_{0} that T_{k}^{0}(a)^{∗}(1) = 0 too.

The corresponding results for T_{k}^{r}(a)^{∗} are more complicated to write because of the presence of z in too many places in T_{k}^{r}(a)(K_{r}(z, w)). Denoting by 1_{k} the multi-index whose coordinates are all 0 except for a 1 in kth position, we have

T_{k}^{r}(a)^{∗}(f(z)) = ∑_{α} ∑_{β+γ+1_{k}=α} \frac{|γ|! |β|!}{|α|!} \frac{α!}{γ! β!} f_{α}a^{γ}w^{β}.

A case with a simple answer is T_{k}^{r}(a)^{∗}(z_{l}) = δ_{kl}, where δ is Kronecker’s delta. This follows from the obvious identity z_{l} − a_{l} = ∑_{k=1}^{N}(z_{k} − a_{k})δ_{kl}.

Let’s investigate further the case that r is a positive integer. Equation (13) then takes the simpler form

T_{k}^{r}(a)^{∗}(K_{r}(z, w)) = \frac{\overline{w}_{k}}{⟨z − a, w⟩} \frac{(1 − ⟨a, w⟩)^{r} − (1 − ⟨z, w⟩)^{r}}{(1 − ⟨z, w⟩)^{r}(1 − ⟨a, w⟩)^{r}}
= \frac{\overline{w}_{k}}{⟨z − a, w⟩} \frac{⟨z, w⟩ − ⟨a, w⟩}{(1 − ⟨z, w⟩)^{r}(1 − ⟨a, w⟩)^{r}} ×

× ∑_{l=0}^{r−1}(1 − ⟨a, w⟩)^{r+1−l}(1 − ⟨z, w⟩)^{l}

= \overline{w}_{k} ∑_{l=1}^{r} \frac{1}{(1 − ⟨a, w⟩)^{r+1−l}(1 − ⟨z, w⟩)^{l}} .

Similarly

T_{k}^{r}(a)(K_{r}(z, w)) = z_{k} ∑_{l=1}^{r} \frac{1}{(1 − ⟨z, a⟩)^{r+1−l}(1 − ⟨z, w⟩)^{l}}

and

T_{k}^{r}(a)(z^{α}) = \binom{r − 1 + |α|}{r − 1}^{−1}z_{k}z^{α} ∑_{l=1}^{r} \binom{l − 1 + |α|}{l − 1}\frac{1}{(1 − ⟨z, a⟩)^{r+1−l}} .

The results for T_{k}^{1}(a) and T_{k}^{1}(a)^{∗} are identical to those given in [4].

4. HOMOGENEOUS INTERPOLATION IN THE SPACES H_{r}

In [4] we proved the following theorem for r = 1.

Theorem 4.1. Let a ∈ 𝔹_{N} and c ∈ ℂ^{n×1}. Then

N =^{def} {f ∈ H_{r}^{n×1} | c^{∗}f(a) = 0} = B_{a}H_{r}^{(n+N−1)×1},

where the equality is as sets and B_{a} is the ℂ^{n×(n+N−1)}–valued function given by

B_{a}(z) = U ( \begin{matrix}b_{a}(z) & 0_{1×(n−1)} \\ 0_{(n−1)×N} & I_{n−1} \end{matrix} ) .

In the above expression, U is any unitary matrix whose first column is c/|c|, and b_{a} is the 1 × N Blaschke factor

b_{a}(z) = \frac{(1 − |a|^{2})^{1/2}}{1 − ⟨z, a⟩} (z − a)(I_{N} − a^{∗}a)^{−1/2}.  (15)

Proof. We first prove that the operator of multiplication by z_{j} is bounded in the spaces H_{r} (r > 0) and in the Dirichlet space. Equivalently, we have to show that there is a positive constant p such that the function (p − ⟨z, w⟩)K_{r}(z, w) is positive in the ball. For r > 0 and p to be determined in the sequel, we have

(p − ⟨z, w⟩)K_{r}(z, w) = \frac{p − ⟨z, w⟩}{(1 − ⟨z, w⟩)^{r}}  (16)
= (p − ⟨z, w⟩)(∑_{j=0}^{∞} \frac{Γ(r + j)}{Γ(r)j!}⟨z, w⟩^{j})

= p + ∑_{j=1}^{∞}(p\frac{Γ(r + j)}{Γ(r)j!} − \frac{Γ(r + j − 1)}{Γ(r)(j − 1)!})⟨z, w⟩^{j}.

Hence the function (16) is positive for p satisfying

p\frac{Γ(r + j)}{Γ(r)j!} − \frac{Γ(r + j − 1)}{Γ(r)(j − 1)!} ≥ 0, (j = 1, 2, …),

which is equivalent to finding a p ≥ j/(r + j − 1) for all j = 1, 2, . . .. So we can pick p = 1/r for the given value of r. When r = 0, we use expansion (8) and see that it suffices to take p = 1.

Next we prove the claim of the theorem for n = 1 and c = 1. By Section 3, f ∈ N if and only if f(z) = (z − a)g(z) with g ∈ H_{r}^{N}. Thus

f(z) = (z − a)g(z) = \frac{1}{1 − ⟨z, a⟩}(z − a)(1 − ⟨z, a⟩)g(z) = b_{a}(z)h(z),

where, in view of the preceding step, the vector–valued function

h(z) = \frac{1}{(1 − |a|^{2})^{1/2}} (1 − ⟨z, a⟩)(I_{N} − a^{∗}a)^{1/2}g(z)

has its entries in H_{r}.

When n > 1, by the previous case we have

c^{∗}f(z) = b_{a}(z)g(z)

for some g ∈ H_{r}^{N×1}. Thus

f(z) = \frac{cc^{∗}}{c^{∗}c}f(z) + (I_{n} − \frac{cc^{∗}}{c^{∗}c}) f(z) = \frac{cb_{a}(z)g(z)}{c^{∗}c} + (I_{n} − \frac{cc^{∗}}{c^{∗}c}) f(z)

= ( \frac{c}{c^{∗}c} (I_{n} − \frac{cc^{∗}}{c^{∗}c}) ) ( \begin{matrix}b_{a}(z)g(z) \\ (I_{n} − \frac{cc^{∗}}{c^{∗}c}) f(z) \end{matrix} ) .

To conclude, we first note that c^{∗}f(z) does not depend on

(I_{n} − \frac{cc^{∗}}{c^{∗}c}) f(z) =^{def} g_{2}(z)

and thus this last expression is arbitrary. Also rank (I_{n} − \frac{cc^{∗}}{c^{∗}c}) = n−1, and

the only nonzero eigenvalue of this matrix is 1 since I_{n} − \frac{cc^{∗}}{c^{∗}c} is a projection. So there is a unitary matrix U_{1} such that

I_{n} − \frac{cc^{∗}}{c^{∗}c} = U_{1} (diag (1, 1, …, 1, 0)) U_{1}^{∗}.

Let V_{1} denote the n×(n−1) matrix that consists of the first n−1 columns of U_{1}. We can write

f(z) = ( \frac{c}{c^{∗}c} V_{1} ) ( \begin{matrix}b_{a} & 0_{1×(n−1)} \\ 0_{(n−1)×N} & I_{n−1} \end{matrix} ) ( \begin{matrix}g(z) \\ g_{3}(z) \end{matrix} ) ,

where g_{3} is the ℂ^{n−1}–valued function whose components are the first n − 1 components of U_{1}^{∗}g_{2}(z). The matrix

U = ( \frac{c}{c^{∗}c} V_{1} )

is unitary since the ranges of \frac{c}{c^{∗}c} and V_{1} are orthogonal, and this concludes the proof.

Proof. (of Theorem 1.2) This is obtained from Theorem 4.1 recursively. A function f ∈ H_{r}^{n×1} satisfies the interpolation conditions (4) if and only if it can be written as f(z) = B_{a_{1},c_{1}}(z)g(z), where g ∈ H_{r}^{(n+N−1)×1} satisfies the interpolation conditions

c_{j}^{∗}B_{a_{1},c_{1}}(a_{j})g(a_{j}) = d_{j}^{∗}g(a_{j}) = 0 (j = 2, …, m),

where d_{j}^{∗} = c_{j}^{∗}B_{a_{1},c_{1}}(a_{j}) for j = 2, …, m. We thus find the set of all g ∈ H_{r}^{(n+N−1)×1} for which d_{2}^{∗}g(a_{2}) = 0 first and then iterate on g.

5. THE CASE OF INTEGER r

We now specialize to the case that r is a positive integer. The analysis is then very close to that of [4].

Theorem 5.1. Let r be a positive integer. Then

N =^{def} {f ∈ H_{r}^{n×1} | c^{∗}f(a) = 0} = B_{a}H_{r}^{(n+N−1)×1},

where equality and B_{a} are as in Theorem 4.1, and the norm in N is equivalent to the norm

∥B_{a}f∥ = ∥(I − π)f∥_{H_{r}^{(n+N−1)×1}}

with π denoting the orthogonal projection on the set of functions f ∈ H_{r}^{(n+N−1)×1} satisfying B_{a}(z)f(z) ≡ 0.

Proof. We set N^{⊥} = M to be the one-dimensional subspace of H_{r}^{n×1} spanned by the function c(1 − ⟨z, a⟩)^{−r}. Then M is a reproducing kernel Hilbert space whose reproducing kernel is

K_{M}(z, w) = (\frac{I_{n} − B_{a}(z)B_{a}(w)}{1 − ⟨z, a⟩})^{r} .

The arguments are as in [4] and repeated for completeness. We first consider the case n = 1 and c = 1. Then the reproducing kernel of M is

K_{M}(z, w) = \frac{(1 − ⟨a, a⟩)^{r}}{(1 − ⟨z, a⟩)^{r}(1 − ⟨a, w⟩)^{r}} = (\frac{1 − b_{a}(z)b_{a}(w)}{1 − ⟨z, w⟩})^{r} ,

where we have used the formula (see [10, Theorem 2.2.2] and [4] for a different proof)

\frac{1 − b_{a}(z)b_{a}(w)^{∗}}{1 − ⟨z, w⟩} = \frac{1 − ⟨a, a⟩}{(1 − ⟨z, a⟩)(1 − ⟨a, w⟩)} (z, w ∈ 𝔹_{N})

to get the second identity. Next the choice of c_{0} = ( 1 0 ⋯ 0 )^{∗} leads to

K_{M}(z, w) = ( \begin{matrix}\frac{(1−⟨a,a⟩)^{r}}{(1−⟨z,a⟩)^{r}(1−⟨a,w⟩)^{r}} & 0 & ⋯ & 0 \\ 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 0 \end{matrix} )

= ( \begin{matrix}\frac{(1−⟨a,a⟩)}{(1−⟨z,a⟩)(1−⟨a,w⟩)} & 0 & ⋯ & 0 \\ 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 0 \end{matrix} )^{r}

= ( \begin{matrix}\frac{1−b_{a}(z)b_{a}(w)^{∗}}{1−⟨z,w⟩} & 0 & ⋯ & 0 \\ 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 0 \end{matrix} )^{r} ,

which is of the asserted form with B_{a}(z) = ( \begin{matrix}b_{a}(z) & 0_{1×(n−1)} \\ 0_{(n−1)×N} & I_{n−1} \end{matrix} ). Fi-

nally, let c ≠ 0_{n×1} and U be a unitary matrix such that Uc_{0} = c/|c|. Then K_{M}(z, w) is equal to

\frac{cc^{∗}}{c^{∗}c} (\frac{1 − b_{a}(z)b_{a}(w)^{∗}}{1 − ⟨z, w⟩} )^{r} = Uc_{0}c_{0}^{∗} \frac{1 − b_{a}(z)b_{a}(w)^{∗}}{1 − ⟨z, w⟩} U^{∗}

= U (\frac{I_{n} − ( \begin{matrix}b_{a}(z) & 0_{1×(n−1)} \\ 0_{(n−1)×N} & I_{n−1} \end{matrix} ) ( \begin{matrix}b_{a}(w) & 0_{1×(n−1)} \\ 0_{(n−1)×N} & I_{n−1} \end{matrix} )^{∗}}{1 − ⟨z, w⟩} )^{r}

= (\frac{I_{n} − U ( \begin{matrix}b_{a}(z) & 0_{1×(n−1)} \\ 0_{(n−1)×N} & I_{n−1} \end{matrix} ) ( \begin{matrix}b_{a}(w) & 0_{1×(n−1)} \\ 0_{(n−1)×N} & I_{n−1} \end{matrix} )^{∗}}{1 − ⟨z, w⟩} )^{r} ,

since U is unitary.

Now we prove the claim of the theorem for n = 1 and c = 1. We write

\frac{1 − (1 − b_{a}(z)b_{a}(w))^{r}}{(1 − ⟨z, w⟩)^{r}} = b_{a}(z)C_{r}(z, w)b_{a}(w)^{∗},

where C_{r}(z, w) is the ℂ^{N×N}-valued positive function defined by

C_{r}(z, w) = ∑_{j=0}^{r−1} (\frac{1 − b_{a}(z)b_{a}(w)}{1 − ⟨z, w⟩})^{j} \frac{I_{N}}{(1 − ⟨z, w⟩)^{r−j}} .

Each of the functions

(\frac{1 − b_{a}(z)b_{a}(w)}{1 − ⟨z, w⟩})^{j} \frac{I_{N}}{(1 − ⟨z, w⟩)^{r−j}} (j = 0, 1, …, r − 1)

is positive in the ball and hence

\frac{I_{N}}{(1 − ⟨z, w⟩)^{r}} ≤ C_{r}(z, w).

Since

\frac{I_{N}}{(1 − ⟨z, w⟩)^{r}} − (\frac{1 − b_{a}(z)b_{a}(w)^{∗}}{1 − ⟨z, w⟩} )^{j} \frac{I_{N}}{(1 − ⟨z, w⟩)^{r−j}} =

= \frac{1 − (1 − b_{a}(z)b_{a}(w)^{∗})^{j}}{(1 − ⟨z, w⟩)^{r}} I_{N}

= \frac{b_{a}(z)b_{a}(w)^{∗}}{(1 − ⟨z, w⟩)^{r−k}} (∑_{ℓ=0}^{j−1} (\frac{1 − b_{a}(z)b_{a}(w)^{∗}}{1 − ⟨z, w⟩} )^{ℓ}) I_{N} ≥ 0,

we have

\frac{I_{N}}{(1 − ⟨z, w⟩)^{r}} ≤ C_{r}(z, w) ≤ m \frac{I_{N}}{(1 − ⟨z, w⟩)^{r}} .

Therefore, as vector spaces,

H(C_{r}) = H (\frac{I_{N}}{(1 − ⟨z, w⟩)^{r}} ) ,

their norms are equivalent (this last fact is also a consequence of Banach’s theorem), and M^{⊥} = b_{a}H_{r}^{N×1}. The last claim on the norm is easily verified. Next for n > 1 we first take

c = c_{0} = ( 1 0 ⋮ 0 ) .

The reproducing kernel of M^{⊥} is then

K_{M^{⊥}}(z, w) = \frac{I_{n}}{(1 − ⟨z, a⟩)^{r}} − \frac{(1 − ⟨a, a⟩)^{r}}{(1 − ⟨z, a⟩)^{r}(1 − ⟨a, w⟩)^{r}} ×

× ( 1 0 ⋮ 0 ) ( 1 0 ⋯ 0 )

= ( \begin{matrix}\frac{1−(1−b_{a}(z)b_{a}(w))^{r}}{(1−⟨z,w⟩)^{r}} & 0_{1×(n−1)} \\ 0_{(n−1)×1} & \frac{I_{n−1}}{(1−⟨z,w⟩)^{r}} \end{matrix} ) .

Thus, with the current choice of c = c_{0} and taking into account the results for n = 1, we have

M^{⊥} = \{F (z) = ( \begin{matrix}b_{a}(z)f(z) \\ g(z) \end{matrix} ) | f ∈ H_{r}^{N×1}, g ∈ H_{r}^{(n−1)×1}\}
= B_{a}H_{r}^{(n+N−1)×1}.

The case of arbitrary c is easily adapted.

Remark 5.2. The case where r = N is of special interest since it concerns the Hardy space of the ball. Assume that B is ℂ^{1×ℓ}-valued. Then for any f ∈ H_{N}^{ℓ×ℓ} we have

∥Bf∥_{H_{N}^{ℓ×ℓ}} = ∫_{𝕊} Tr f(z)^{∗}B(z)^{∗}B(z)f(z) dσ(z)
= ∫_{𝕊} Tr B(z)f(z)f(z)^{∗}B(z)^{∗} dσ(z),

which is equal to 1 for functions f taking unitary values on 𝕊.

Added in proof. The reproducing kernel Hilbert spaces H_{r} for r ≤ 0 with reproducing kernels

K_{r}(z, w) = 1 + ∑_{j=1}^{∞} \frac{Γ(1 − r) (j − 1)!}{Γ(1 − r + j)} ⟨z, w⟩^{j}  (r ≤ 0)  (17)

are studied in [13]. These spaces naturally extend the spaces H_{r} (r ≥ 0) of this paper to all real r, and the two families have many properties in common. In fact, when r = 0, (17) reduces to the K_{0}(z, w) given earlier. The k_{r,j} corresponding to (17) satisfies k_{r,j} ∼ 1/j^{1−r} for r ≤ 0 and large j exactly as it does for r > 0, although

∥z^{α}∥_{H_{r}}^{2} = \frac{Γ(1 − r + |α|) |α| α!}{Γ(1 − r) (|α|!)^{2}} (r ≤ 0)

is ostensibly different.

All the new kernels are complete Nevanlina-Pick kernels by Lemma 7.38 of [12]. Thus the spaces H_{r} have the complete Nevanlinna-Pick property if and only if r ≤ 1. By contrast, a reworking of Example 5.17 of [12] shows that none of the spaces H_{r} for r > 1 has the weaker two-point scalar Pick property.

Continuing, H_{r} ⊂ H_{s} if −∞ < r < s < ∞. In addition to Remark 2.3, the inclusion is a contraction also when r < 0 if s ≥ 1 or s ≤ 0. The first statement of Proposition of 2.5 and Proposition 2.6 hold for H_{r} when r < 0 too, as well as Examples 2.4 and 2.7.

Remark 3.1 already shows that the work of Section 3 carries over to H_{r} for r < 0. Finally, repeating the first part of the proof of Theorem 4.1 yields that the operator of multiplication by z_{j} is a contraction (choosing p = 1 works) on H_{r} when r < 0. Thus the homogeneous interpolation described in Theorem 4.1 is valid in H_{r} for any real r.

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