2 Notations and background

Several spaces of sequences will be of interest.

• • $\mathbb {N} = \{1,2,\dots \}$ .

• • $\ell ^{\infty } = \ell ^{\infty }(\mathbb {N}) = \{ z = (z_j)_{j\in \mathbb {N}} \in \mathbb {C}^{\mathbb {N}}: \|z\|_{\infty } := \sup _j |z_j| <\infty \}.$

• • $\ell ^2 = \ell ^2(\mathbb {N}) = \{ z= (z_j)_{j\in \mathbb {N}} : \sum _{j=1}^{\infty } |z_j|^2 <\infty \}.$

• • $c_0 = c_0(\mathbb {N}) = \{z\in \ell ^{\infty }: \lim _{j\to \infty } z_j = 0\}$ .

• • $c_{00} = c_{00}(\mathbb {N}) = \{ z \in \ell ^{\infty }: \exists N \in \mathbb {N}, z_j = 0 \text { for } j>N\}$ .

• • $\mathrm{Ball}(\ell ^{\infty }) = \{z\in \ell ^{\infty }: \|z\|_{\infty } <1\}$ denotes the open unit ball of $\ell ^{\infty }$ .

• • $\mathbb {D}^{\infty } = \mathbb {D}^{\mathbb {N}} = \{z \in \ell ^{\infty }: \forall j, |z_j| <1\}$ .

• • $\mathrm{Ball}(c_0) = \{z\in c_0: \|z\|_{\infty } <1\}$ .

• • $\mathrm{Ball}(c_{00}) = \{z\in c_{00}: \|z\|_{\infty } < 1\}$ .

• • We identify $\mathbb {D}^{N}$ with $\mathbb {D}^N \times \{(0,0,\dots )\} \subset \mathrm{Ball}(c_{00})$ .

• • The Hilbert multidisk is the set $\mathbb {D}_{2}^{\infty } := \ell ^2 \cap \mathrm{Ball}(\ell ^\infty )$ , namely, the set of sequences $(z_j)_{j\in \mathbb {N}}$ such that $\sup _j |z_j|<1$ and $\sum _{j} |z_j|^2 <\infty $ . Note that for $z,w \in \mathbb {D}_2^{\infty }$ the infinite product

  $$\begin{align*}\prod_{j=1}^{\infty} \frac{1}{1-\bar{w}_j z_j} \end{align*}$$
  converges absolutely.

• • We generally use standard modulus bars $|\cdot |$ for the modulus of complex numbers or vectors (in $\mathbb {C}^N$ or Hilbert space) while double bars $\|\cdot \|$ are reserved for operator norms or other norms as listed above.

Remark 2.1 We use the basics of positive semi-definite functions. Given a set X, a function $A:X\times X \to \mathbb {C}$ is positive semi-definite on X if for every finite subset $Y\subset X$ and every function $a:Y\to \mathbb {C}$ we have

$$\begin{align*}\sum_{z,w \in Y} a(z)\overline{a(w)} A(z,w) \geq 0. \end{align*}$$

We write $A(z,w) \succeq 0$ in this case. More generally, we write $A(z,w) \succeq B(z,w)$ if $A - B \succeq 0$ . We frequently use the Schur product theorem which say that if $A(z,w), B(z,w) \succeq 0$ then $A(z,w)B(z,w) \succeq 0$ . For a function $f:X\to \mathbb {C}$ , we let $f\otimes \bar {f}$ denote the function $(z,w) \mapsto f(z)\overline {f(w)}$ . Also, if $X \subset \mathbb {C}^{\mathbb {N}}$ , we let $Z_j \otimes \bar {Z}_j$ denote the function $(z,w) \mapsto z_j\bar {w}_j$ .

3 Proof of Theorem 1.5

Theorem 1.5 will be proven with three lemmas. The first lemma is our main advance while the other two are standard.

For the first lemma, let $\rho = (\rho _n)_{n\in \mathbb {N}} \in \mathbb {D}_{2}^{\infty }$ be a fixed square summable sequence of positive numbers. Let $\overline {\mathbb {D}}_{\rho } = \{z =(z_n)_{n\in \mathbb {N}}: |z_n|\leq \rho _n\}$ and let $\mathbb {D}_{\rho } = \{z \odot \rho : z \in \mathrm{Ball}(c_0)\}$ where $\rho \odot z = (\rho _jz_j)_{j\in \mathbb {N}}$ . (The notation is not entirely consistent but it is temporary.)

Lemma 3.1 Assume $f: \mathrm{Ball}(c_{00})\to \mathbb {C}$ is in the Agler class, $\mathcal {A}_{\infty }$ .

Then, for $j=0,1,2,\dots $ , there exist positive semi-definite kernels $K_j$ on $\mathbb {D}_{\rho }$ such that

$$\begin{align*}1- f(z)\overline{f(w)} = K_0(z,w) + \sum_{j=1}^{\infty}(1-z_j\bar{w}_j) K_j(z,w) \end{align*}$$

where the sum converges absolutely.

Note that we have introduced the term $K_0$ which is for convenience in the proof. This term can be absorbed into any of the other terms for instance as

$$\begin{align*}(1 - z_1 \bar{w}_1)\left( \frac{K_0(z,w)}{1-z_1\bar{w}_1} + K_1(z,w)\right). \end{align*}$$

Since $K_0$ will be positive semi-definite and since $\frac {1}{1-z_1\bar {w}_1}$ is positive semi-definite, the product is positive semi-definite.

Proof of Lemma 3.1

For each N, let $f_N(z_1,z_2,\dots ) = f(z_1,\dots ,z_N, 0,\dots )$ . Since f restricted to $\mathbb {D}^N$ has an Agler decomposition, we can write

$$\begin{align*}1-f_N(z)\overline{f_N(w)} = \sum_{j=1}^{N}(1-\bar{w}_j z_j) K^N_j(z,w) \end{align*}$$

for positive semi-definite kernels $K_j^N$ that only depend on the first N variables $z_1,\dots , z_N,w_1,\dots , w_N$ , are analytic in z, and are anti-analytic in w.

Note that for $z,w\in \mathbb {D}_{2}^{\infty }$ , the product

$$\begin{align*}S(z,w) = \prod_{j=1}^{\infty} \frac{1}{1-\bar{w}_j z_j} \end{align*}$$

converges and is positive semi-definite. Note that

$$\begin{align*}S_n(z,w) = (1-\bar{w}_n z_n) S(z,w) = \prod_{j\ne n} \frac{1}{1-\bar{w}_j z_j} \end{align*}$$

is also positive semi-definite and $S, S_n \succeq 1$ using the partial order from Remark 2.1. Here ‘ $1$ ’ is the identically $1$ function.

Let $f_N\otimes \overline {f_N}$ denote the function $(z,w) \mapsto f_N(z)\overline {f_N(w)}$ . We have the positive semi-definite function inequality

(3.1) $$ \begin{align} S \succeq (1-f_N\otimes \overline{f_N})S = \sum_{j=1}^{N} K_j^{N} S_j \succeq \sum_{j=1}^{N} K_j^{N} \succeq \sum_{j=1}^{M} K_j^N \end{align} $$

for $M<N$ and by Cauchy–Schwarz we have for $z,w\in \mathbb {D}_{2}^{\infty }$

$$\begin{align*}S(z,z)^{1/2}S(w,w)^{1/2} \geq |K_j^N(z,w)|. \end{align*}$$

Recall $\rho \odot z = (\rho _jz_j)_{j\in \mathbb {N}}$ . For $z,w \in \mathrm{Ball}(c_0)$ , $K_j^N(\rho \odot z, \rho \odot \bar {w})$ is bounded and analytic in $\mathrm{Ball}(c_0)\times \mathrm{Ball}(c_0)$ with supremum norm bounded by $S(\rho ,\rho )$ . By the Montel theorem (see Remark 3.2), for each $j=1,2,\dots $ in succession there is a subsequence of $N\in \mathbb {N}$ such that

(3.2) $$ \begin{align} K_j^N(\rho\odot z, \rho \odot \bar{w}) \overset{N\to\infty}{\to} K_j(\rho\odot z, \rho \odot \bar{w}) \text{ uniformly on compact subsets of } \mathrm{Ball}(c_0). \end{align} $$

The limiting functions, $K_j$ , are necessarily positive semi-definite because this property is preserved under limits. By a standard diagonal argument, we can find a common subsequence of N such that for all j, (3.2) holds. Recall $\mathbb {D}_{\rho } = \{z \odot \rho : z \in \mathrm{Ball}(c_0)\}$ . By (3.1), for $z,w\in \mathbb {D}_{\rho }$ , $ S\succeq \sum _{j=1}^{M} K_j^N $ for $M<N$ and sending $N \to \infty $ we have $S \succeq \sum _{j=1}^{M} K_j$ . Finally, we can send $M\to \infty $ to obtain $ S \succeq \sum _{j=1}^{\infty } K_j $ with absolute convergence. Absolute convergence can be proven by looking at the diagonal $z=w$ first and then applying Cauchy–Schwarz. To finish the proof, we show

$$\begin{align*}K_0(z,w) := 1-f(z)\overline{f(w)} - \sum_{j=1}^{\infty}(1-z_j \bar{w}_j)K_j(z,w) \end{align*}$$

is positive semi-definite. Here we emphasize that f on $\mathbb {D}_{\rho }$ is defined via f’s holomorphic extension from $\mathrm{Ball}(c_{00})$ to $\mathrm{Ball}(c_0)$ as in the traditional Schur class of infinitely many variables.

Now, for $N<M$

$$\begin{align*}1-f_M(z)\overline{f_M(w)} - \sum_{j=1}^{N}(1-z_j \bar{w}_j)K_j^{M}(z,w)= \sum_{j=N+1}^{M}(1-z_j\bar{w}_j)K_j^{M}(z,w). \end{align*}$$

Let $Z_j \otimes \overline {Z_j}$ denote the function $(z,w) \mapsto z_j\bar {w}_j$ . Since $S \succeq K_j^{M}$ we see that

$$\begin{align*}1 - f_M\otimes \overline{f_M} - \sum_{j=1}^{N} (1 - Z_j \otimes \overline{Z_j}) K_j^M \succeq - \sum_{j=N+1}^{M} (Z_j \otimes \overline{Z_j}) S. \end{align*}$$

This inequality means that for any finite subset $Y \subset \mathbb {D}_{\rho }$ and any function $a:Y \to \mathbb {C}$

$$\begin{align*}\sum_{z,w \in Y} (1-f_M(z)\overline{f_M(w)} - \sum_{j=1}^{N}(1-z_j \bar{w}_j)K_j^{M}(z,w))a(z) \overline{a(w)} \geq -\sum_{z,w\in Y} \sum_{j=N+1}^{M} z_j\bar{w}_j S(z,w) a(z) \overline{a(w)}. \end{align*}$$

(See Remark 2.1.) Now,

$$\begin{align*}\sum_{z,w\in Y} \sum_{j=N+1}^{M} z_j\bar{w}_j S(z,w) a(z) \overline{a(w)} \leq S(\rho,\rho) \sum_{j=N+1}^{M}\rho_j^2 \left(\sum_{z\in Y} |a(z)|\right)^2 \leq S(\rho,\rho) \sum_{j=N+1}^{\infty}\rho_j^2 \left(\sum_{z\in Y} |a(z)|\right)^2. \end{align*}$$

Sending $M\to \infty $

$$\begin{align*}\sum_{z,w \in Y} (1-f(z)\overline{f(w)} - \sum_{j=1}^{N}(1-z_j \bar{w}_j)K_j(z,w))a(z) \overline{a(w)} \geq -S(\rho,\rho) \sum_{j=N+1}^{\infty}\rho_j^2 \left(\sum_{z\in Y} |a(z)|\right)^2 \end{align*}$$

and then sending $N\to \infty $

$$\begin{align*}\sum_{z,w \in Y} (1-f(z)\overline{f(w)} - \sum_{j=1}^{\infty}(1-z_j \bar{w}_j)K_j(z,w))a(z) \overline{a(w)} \geq 0 \end{align*}$$

which proves $K_0(z,w)$ is positive semi-definite.

Lemma 3.3 Let $X \subset \mathrm{Ball}(\ell ^{\infty })$ . Assume $f: X\to \mathbb {C}$ is a function such that for $j=0,1,2,\dots $ , there exist positive semi-definite kernels $K_j:X\times X \to \mathbb {C}$ on X such that

$$\begin{align*}1- f(z)\overline{f(w)} = \sum_{j=1}^{\infty}(1-z_j\bar{w}_j) K_j(z,w), \end{align*}$$

where the sum converges absolutely. Then, there exists a contractive operator V acting on $\mathbb {C}\oplus \bigoplus _{j=1}^{\infty } \mathcal {H}_j$ where $\mathcal {H}_1,\mathcal {H}_2, \dots $ are Hilbert spaces such that when we write V in block form $V = \begin {pmatrix} A & B \\ C & D\end {pmatrix}$ we have

$$\begin{align*}f(z) &= A + B\Delta(z) (1-D\Delta(z))^{-1} C\\K_j(z,w) &= C^*(I-\Delta(w)^* D^*)^{-1}P_j(I-D\Delta(z))^{-1} C, \end{align*}$$

where $\Delta (z) = \sum _{j=1}^{\infty } z_j P_j$ and each $P_j$ represents projection onto $\mathcal {H}_j$ within the direct sum $\bigoplus _{k=1}^{\infty } \mathcal {H}_k$ . With these formulas, f and $K_j$ extend to $\mathrm{Ball}(\ell ^{\infty })$ and there exists a positive semi-definite kernel $K_0(z,w)$ such that

$$\begin{align*}1- f(z)\overline{f(w)} = K_0(z,w) + \sum_{j=1}^{\infty}(1-z_j\bar{w}_j) K_j(z,w) \end{align*}$$

holds on $\mathrm{Ball}(\ell ^{\infty })\times \mathrm{Ball}(\ell ^{\infty })$ .

The proof is a standard lurking isometry argument (for those who know what that is) that we include for completeness (for those who do not).

Proof By the Moore–Aronszajn theorem on reproducing kernel Hilbert space, we can factor $K_j(z,w) = K_{j,z}^* K_{j,w}$ for $K_{j,z}$ an element of some Hilbert space $\mathcal {H}_j$ . We write $K_{j,z}^* K_{j,w}$ instead of $\langle K_{j,w}, K_{j,z} \rangle $ and view $K_{j,z}^*$ as an element of the dual $\mathcal {H}_j^*$ of $\mathcal {H}_j$ . The following map:

$$\begin{align*}\begin{pmatrix} 1 \\ z_1(K_{1,z})^* \\ z_2 (K_{2,z})^* \\ \vdots \end{pmatrix} \mapsto \begin{pmatrix} f(z) \\ (K_{1,z})^* \\ (K_{2,z})^* \\ \vdots \end{pmatrix} \end{align*}$$

initially defined for vectors indexed by $z\in X$ extends linearly and in a well-defined way to a contractive operator V from $\mathbb {C} \oplus \bigoplus _{j=1}^{\infty } \mathcal {H}_j^*$ to $\mathbb {C} \oplus \bigoplus _{j=1}^{\infty } \mathcal {H}_j^* $ . We write V in block form $ \begin {pmatrix} A & B \\ C & D \\ \end {pmatrix} $ with $A \in \mathbb {C} \cong \mathcal {B}(\mathbb {C},\mathbb {C})$ , $B \in \mathcal {B}(\bigoplus _{j=1}^{\infty } \mathcal {H}_j^*, \mathbb {C})$ , $C \in \mathcal {B}(\mathbb {C}, \bigoplus _{j=1}^{\infty } \mathcal {H}_j^*)$ , $D \in \mathcal {B}(\bigoplus _{j=1}^{\infty } \mathcal {H}_j^*)$ , using the notation $\mathcal {B}(X,Y)$ to denote the bounded linear operators from X to Y (as well as $\mathcal {B}(X)= \mathcal {B}(X,X)$ ). Let $\Delta (z) \in \mathcal {B}(\bigoplus _{j=1}^{\infty } \mathcal {H}_j^*)$ be the diagonal operator sending

$$\begin{align*}(h_j)_{j\in \mathbb{N}} \in \bigoplus_{j=1}^{\infty} \mathcal{H}_j^* \mapsto (z_j h_j)_{j\in\mathbb{N}} \in \bigoplus_{j=1}^{\infty} \mathcal{H}_j^*. \end{align*}$$

Let $F(z) := (K_{j,z}^*)_{j\in \mathbb {N}} \in \bigoplus _{j=1}^{\infty } \mathcal {H}_j^*$ . Then,

$$\begin{align*}V \begin{pmatrix} 1 \\ \Delta(z) F(z) \end{pmatrix} = \begin{pmatrix} f(z) \\ F(z) \end{pmatrix} \end{align*}$$

which implies

(3.3) $$ \begin{align} A + B \Delta(z) F(z) &= f(z) \end{align} $$

(3.4) $$ \begin{align} C+ D \Delta(z) F(z) &= F(z). \end{align} $$

Solving for $F(z)$ and then $f(z)$ we obtain

$$\begin{align*}\begin{aligned} F(z) &= (I-D\Delta(z))^{-1} C \\ f(z) &= A + B \Delta(z) (I-D \Delta(z))^{-1} C. \end{aligned} \end{align*}$$

Note the expressions on the right are defined as written for $z \in \mathrm{Ball}(\ell ^{\infty })$ . Evidently, $F(w)^*P_jF(z)$ extends $K_j(z,w)$ . Since V is contractive, we can factor ${I-V^*V = W^*W}$ for some operator W. Let

$$\begin{align*}G(z) = W \begin{pmatrix} 1 \\ \Delta(z) F(z) \end{pmatrix} \end{align*}$$

so that

$$\begin{align*}\begin{pmatrix} 1 \\ \Delta(w) F(w) \end{pmatrix}^* \begin{pmatrix} 1 \\ \Delta(z) F(z) \end{pmatrix} = \begin{pmatrix} f(w) \\ F(w) \end{pmatrix}^* \begin{pmatrix} f(z) \\ F(z) \end{pmatrix} + G(w)^*G(z). \end{align*}$$

This rearranges into

$$\begin{align*}1- f(z)\overline{f(w)} = F(w)^*(I-\Delta(w)^*\Delta(z))F(z) + G(w)^*G(z) \end{align*}$$

and since

$$\begin{align*}F(w)^*(I-\Delta(w)^*\Delta(z))F(z) = \sum_{j=1}^{\infty} (1-\bar{w}_jz_j) F(w)^*P_j F(z) \end{align*}$$

we have the desired extension of the Agler decomposition using $K_0(z,w) = G(w)^*G(z)$ .

Lemmas 3.1 and 3.3 establish most of Theorem 1.5. For the remaining part, we need a basic estimate on transfer function formulas in order to establish the full von Neumann inequality for the Agler class.

Lemma 3.4 Suppose V is a contractive operator acting on $\mathbb {C}\oplus \bigoplus _{j=1}^{\infty } \mathcal {H}_j$ , where $\mathcal {H}_1,\mathcal {H}_2, \dots $ are Hilbert spaces. Writing V in block form $V = \begin {pmatrix} A & B \\ C & D\end {pmatrix}$ we define for $z \in \mathrm{Ball}(\ell ^{\infty })$

$$\begin{align*}f(z) = A + B\Delta(z) (1-D\Delta(z))^{-1} C \end{align*}$$

where $\Delta (z) = \sum _{j=1}^{\infty } z_j P_j$ and each $P_j$ represents projection onto $\mathcal {H}_j$ within the direct sum $\bigoplus _{k=1}^{\infty } \mathcal {H}_k$ . Then, for $z,w \in \mathrm{Ball}(\ell ^{\infty })$

$$\begin{align*}f(z) - f(w) = B(1-\Delta(z)D)^{-1}\Delta(z-w)(1-D\Delta(w))^{-1}C. \end{align*}$$

Proof

$$\begin{align*}\begin{aligned} f(z) - f(w) &= B(\Delta(z)(1-D\Delta(z))^{-1} - \Delta(w)(1-D\Delta(w))^{-1})C \\ &= B( \Delta(z)((1-D\Delta(z))^{-1} - (1-D\Delta(w))^{-1}) + \Delta(z-w) (1-D\Delta(w))^{-1})C\\ &=B(\Delta(z)(1-D\Delta(z))^{-1})(D\Delta(z-w))(1-D\Delta(w))^{-1} + \Delta(z-w)(1-D\Delta(w))^{-1})C\\ &= B(\Delta(z)(1-D\Delta(z))^{-1})D + 1)\Delta(z-w)(1-D\Delta(w))^{-1})C\\ &= B(1-\Delta(z)D)^{-1}\Delta(z-w)(1-D\Delta(w))^{-1}C. \end{aligned}\\[-34pt] \end{align*}$$

Now we finish the proof of Theorem 1.5. As written in Theorem 1.5, for an infinite tuple $T = (T_1,T_2,\dots )$ of operators on a common Hilbert space $\mathcal {H}$ satisfying

$$\begin{align*}\|T\|_{\infty} := \sup_j \|T_j\| < 1 \end{align*}$$

we define

$$\begin{align*}f(T) := (A\otimes I) + (B\otimes I) \Delta(T) (1-(D\otimes I)\Delta(T))^{-1} (C\otimes I) \end{align*}$$

where $\Delta (T) := \sum _{j=1}^{\infty } P_j \otimes T_j$ . Note that this definition does not require the operators $(T_j)_j$ to pairwise commute but if they do then each $T_j$ commutes with $f(T)$ since each $I\otimes T_j$ commutes with $\Delta (T)$ .

Since $(D\otimes I)\Delta (T)$ is strictly contractive, $f(T)$ equals the absolutely convergent sum

$$\begin{align*}(A\otimes I) + (B\otimes I) \Delta(T)\sum_{j=0}^{\infty} ((D\otimes I)\Delta(T))^{j} (C\otimes I) \end{align*}$$

and substituting $z\in \mathrm{Ball}(\ell ^{\infty })$ we obtain an absolutely convergent homogeneous expansion for $f(z)$

(3.5) $$ \begin{align} f(z) = A + B\Delta(z)\sum_{j=0}^{\infty} ((D\Delta(z))^{j} C. \end{align} $$

Before we finish the proof of Theorem 1.5 we make some clarifications about our functional calculus.

The proof of Lemma 3.4 extends directly to prove that for another such tuple S acting on the same Hilbert space $\mathcal {H}$ as T we have

$$\begin{align*}f(T) - f(S) = (B\otimes I)(1-\Delta(T)(D\otimes I))^{-1}\Delta(T-S)(1-(D\otimes I)\Delta(S))^{-1}(C\otimes I). \end{align*}$$

This implies the estimate that for $x \in \mathcal {H}$

$$\begin{align*}|(f(T)-f(S))x|^2 \leq \|(B\otimes I)(1-\Delta(T)(D\otimes I))^{-1}\|^2 |\Delta(T-S)(1-(D\otimes I)\Delta(S))^{-1}(C\otimes x)|^2. \end{align*}$$

Letting $T^{(N)} = (T_1,\dots , T_N,0,\dots )$ we have $f(T^{(N)}) = f_N(T)$ and

$$\begin{align*}\begin{aligned} |(f_N(T)-f(T))x|^2 &\leq (1-\|T\|_{\infty})^{-2} |\Delta(T^{(N)}-T)(1-(D\otimes I)\Delta(T))^{-1}(C\otimes x)|^2\\ &= (1-\|T\|_{\infty})^{-2} \sum_{j=N+1}^{\infty} |(P_j\otimes T_j)(1-(D\otimes I)\Delta(T))^{-1}(C\otimes x)|^2 \end{aligned} \end{align*}$$

and this goes to $0$ as $N\to \infty $ . Thus, $f_N(T) \to f(T)$ in the strong operator topology. This concludes the proof of Theorem 1.5.

4 Proof of Theorem 1.7

That (1) implies (2) follows from Lemmas 3.1 and 3.3. Proving (2) implies (1) is a standard lurking isometry argument that is somewhat similar to our proof of Theorem 1.5. Proving (1) implies (3) consists mostly of technicalities that we discuss next. The proof of (3) implies (1) is the main contribution below.

Regarding (1) implies (3), by definition, we can make sense of $\tilde {f}(T)$ for $\tilde {f} \in \mathcal {A}_\infty $ when $(\|T_j\|)_{j\in \mathbb {N}} \in \mathrm{Ball}(c_{00})$ and Theorem 1.5 lets us make sense of it when ${(\|T_j\|)_{j\in \mathbb {N}} \in \mathrm{Ball}(\ell ^{\infty })}$ . To prove $\|\tilde {f}(T)\|\leq 1$ when T consists of matrices that are commuting, contractive, simultaneously diagonalizable with joint spectrum $\sigma (T) \subset X$ and eigenspaces of dimension at most $1$ , we can give a continuity argument. First, $\|\tilde {f}(rT)\|\leq 1$ holds for $r<1$ because we will have

$$\begin{align*}\tilde{f}_N(rT) \to \tilde{f}(rT) \end{align*}$$

in the strong operator topology as $N\to \infty $ —as in previous sections $\tilde {f}_N$ refers to the restriction of $\tilde {f}$ to $\mathbb {D}^N$ . Note that $\tilde {f}_N(rT)$ is defined in terms of the absolutely convergent power series of $\tilde {f}_N$ . Next, we use the diagonalizability properties of T; let $\vec {b}(z)$ be the eigenvector associated with joint eigenvalue $z\in \sigma (T)$ . Then, for any function $a: \sigma (T) \to \mathbb {C}$ , we have

$$\begin{align*}| \tilde{f}(rT) \sum_{z\in \sigma(T)} a(z) \vec{b}(z)| = |\sum_{z\in \sigma(T)} \tilde{f}(rz) a(z) \vec{b}(z)| \leq |\sum_{z\in \sigma(T)} a(z) \vec{b}(z)|. \end{align*}$$

We can send $r\nearrow 1$ to conclude $\|\tilde {f}(T)\| \leq 1$ . This proves (1) implies (3).

Our main contribution is the proof of (3) implies (2) assuming the finite set X belongs to $\mathbb {D}_{2}^{\infty }$ . This is a modification of the finite variable cone separation argument; the main difference being Lemma 4.1 below. Consider the following cone of functions on $X\times X$

$$\begin{align*}\begin{aligned} \mathcal{C} = \left\{ (z,w) \right.&\mapsto \sum_{j=1}^{\infty} (1-z_j\bar{w}_j) A_j(z,w): \\ & A_1, A_2,\dots \text{ are positive semi-definite functions on } X; \\ &\left. \forall z\in X, \sum_{j=1}^{\infty} A_j(z,z)<\infty\right\}. \end{aligned} \end{align*}$$

Proof Let

$$\begin{align*}C_n(z,w) = \sum_{j=1}^{\infty} (1-z_j\bar{w}_j) A_{n,j}(z,w) \end{align*}$$

define a sequence of functions in $\mathcal {C}$ that converges to the function $C:X\times X\to \mathbb {C}$ pointwise; in particular, each $A_{n,j}$ is positive semi-definite. We must show $C \in \mathcal {C}$ .

Let $\delta :X\times X \to \mathbb {C}$ denote the function $\delta (z,w) = 1$ if $z=w$ and $\delta (z,w) = 0$ if $z\ne w$ . There necessarily exist constants $c_1, c_2$ such that for all n, $c_1 \delta \succeq C_n \succeq c_2 \delta .$ This looks more familiar when we view our functions on $X\times X$ as matrices. Since $X \subset \mathbb {D}_2^{\infty }$ we can define

$$\begin{align*}S(z,w) = \prod_{j=1}^{\infty} \frac{1}{1-z_j \bar{w}_j} \text{ and } S_j(z,w) = (1-z_j\bar{w}_j)S(z,w) = \prod_{i\ne j} \frac{1}{1-z_i \bar{w}_i}; \end{align*}$$

which are positive semi-definite and satisfy $S, S_j \succeq 1$ , with ‘ $1$ ’ representing the identically $1$ function. Some parts of what follow are similar to the proof of Lemma 3.1. Now, for any $i \in \mathbb {N}$

$$\begin{align*}c_1S \succeq C_n S = \sum_{j=1}^{\infty} S_j A_{n,j} \succeq \sum_{j=1}^{\infty} A_{n,j} \succeq A_{n,i} \succeq 0 \end{align*}$$

which shows the matrices $A_{n,i}$ are uniformly bounded. Let $c_3>0$ satisfy $c_3 \delta \succeq c_1 S$ .

Using a diagonal argument we can select a subsequence such that each $A_{n,i}$ converges to a positive semi-definite matrix $A_i$ as $n\to \infty $ . Also, for each N we have $ C_n S \succeq \sum _{j=1}^{N} S_j A_{n,j} \succeq \sum _{j=1}^{N} A_{n,j}. $ Sending $n \to \infty $ we have $ C S \succeq \sum _{j=1}^{N} S_j A_j \succeq \sum _{j=1}^{N} A_j $ and sending $N\to \infty $ we have $ C S \succeq \sum _{j=1}^{\infty } S_j A_j \succeq \sum _{j=1}^{\infty } A_j $ where the sums converge absolutely. Next, recalling $Z_j\otimes \bar {Z}_j$ denotes the function $(z,w) \mapsto z_j\bar {w}_j$ we have

$$\begin{align*}\begin{aligned} C_n &- \sum_{j=1}^{N} (1-Z_{j}\otimes \bar{Z}_j)A_{n,j} \\ &= \sum_{j=N+1}^{\infty} (1-Z_j\otimes \bar{Z}_j)A_{n,j}\\ &\succeq -c_3 \sum_{j=N+1}^{\infty} (Z_{j}\otimes \bar{Z}_j) \delta\\ &\succeq -c_3 \max\{\sum_{j=N+1}^{\infty} |z_j|^2: z \in X\} \delta. \end{aligned} \end{align*}$$

The last inequality amounts to the fact that for any function $a:X\to \mathbb {C}$ we have

$$\begin{align*}\begin{aligned} &\sum_{z,w\in X} \sum_{j=N+1}^{\infty} (z_j\bar{w}_j) \delta(z,w) a(z) \overline{a(w)} \\ &\quad = \sum_{z\in X}\sum_{j=N+1}^{\infty} |z_j|^2 |a(z)|^2 \\ &\quad\leq \max\{\sum_{j=N+1}^{\infty} |z_j|^2: z\in X\} \sum_{z\in X}|a(z)|^2. \end{aligned} \end{align*}$$

Setting $M_N = \max \{\sum _{j=N+1}^{\infty } |z_j|^2: z\in X\}$ we have $M_N \to 0$ since $X \subset \ell ^2$ . Sending $n \to \infty $ we have

$$\begin{align*}C - \sum_{j=1}^{N} (1-Z_j\otimes \bar{Z}_j)A_j \succeq -c_3 M_N \delta \end{align*}$$

and finally sending $N\to \infty $ we have

$$\begin{align*}A_0 := C - \sum_{j=1}^{\infty} (1-Z_j\otimes \bar{Z}_j)A_j \succeq 0. \end{align*}$$

Finally, $A_0$ can be absorbed into any other term, say $A_1$ as $\tilde {A}_1 = A_1 + \frac {1}{1-Z_1\otimes \bar {Z}_1} A_0$ to see that

$$\begin{align*}C = (1-Z_1\otimes \bar{Z}_1)\tilde{A}_1+ \sum_{j=2}^{\infty} (1-Z_j\otimes \bar{Z}_j)A_j \end{align*}$$

is of the desired form.

What follows is now standard. Suppose $f:X \to \overline {\mathbb {D}}$ is a function with the assumed property in item (3). Suppose the function on $X\times X$ , $F= 1 - f\otimes \bar {f}$ is not in the cone $\mathcal {C}$ . By the Hahn–Banach hyperplane separation theorem, there exists a function ${B: X\times X \to \mathbb {C}}$ with $B(z,w) = \overline {B(w,z)}$ such that

$$\begin{align*}\sum_{z,w \in X} F(z,w) B(z,w) <0 \text{ and for all } C \in \mathcal{C} \text{ we have } \sum_{z,w\in X} C(z,w) B(z,w) \geq 0. \end{align*}$$

The second condition implies $B \succeq 0$ by setting $C = (1-Z_1\otimes \bar {Z}_1) \frac { a\otimes \bar {a} }{1-Z_1\otimes \bar {Z}_1}$ for an arbitrary function $a:X\to \mathbb {C}$ and observing

$$\begin{align*}\sum_{z,w\in X} C(z,w) B(z,w) = \sum_{z,w} a(z)\overline{a(w)} B(z,w) \geq 0. \end{align*}$$

Next, we factor $B(w,z) = \overline {B(z,w)} = \vec {b}(z)^* \vec {b}(w)$ for vectors $\vec {b}(z) \in \mathbb {C}^r$ where r is the rank of the matrix $(B(z,w))_{z,w\in X}$ .

Choosing now $C_j = (1-Z_j\otimes \bar {Z}_j) (a\otimes \bar {a})$ for an arbitrary $a:X\to \mathbb {C}$

$$\begin{align*}\begin{aligned} 0&\leq \sum_{z,w \in X} (1-z_j\bar{w}_j)a(z) \overline{a(w)}B(z,w)\\ &= \sum_{z,w\in X} (1-z_j\bar{w}_j) a(z) \overline{a(w)} \vec{b}(z)^t \overline{\vec{b}(w)}\\ &= {\left|\sum_{w\in X} a(w) \vec{b}(w)\right|}^2 - {\left|\sum_{w\in X} w_j a(w) \vec{b}(w)\right|}^2 \\ \end{aligned} \end{align*}$$

which proves that maps $T_j: \sum _{w\in X} a(w) \vec {b}(w) \mapsto \sum _{w\in X} w_j a(w) \vec {b}(w)$ are well-defined, contractive, diagonalizable. Setting $T=(T_1,T_2,\dots )$ , we have assumed that $f(T)$ is contractive which means for all functions $a:X\to \mathbb {C}$

$$\begin{align*}{\left|\sum_{w\in X} a(w) \vec{b}(w)\right|}^2 - {\left|\sum_{w\in X} f(w) a(w) \vec{b}(w)\right|}^2 \geq 0 \end{align*}$$

and this means

$$\begin{align*}\sum_{z,w\in X} (1-f(z)\overline{f(w)}) B(z,w) a(z) \overline{a(w)} \geq 0 \end{align*}$$

and this means

$$\begin{align*}FB = (1- f\otimes \bar{f})B \succeq 0 \end{align*}$$

(recall $F = 1-f\otimes f$ ) and in particular

$$\begin{align*}\sum_{z,w\in X} F(z,w) B(z,w) \geq 0 \end{align*}$$

which is a contradiction. This proves $F \in \mathcal {C}$ or more precisely, there exist positive semi-definite matrices $A_1, A_2,\dots $ with rows and columns indexed by S such that

$$\begin{align*}1-f(z)\overline{f(w)} = \sum_{j=1}^{\infty} (1-z_j\bar{w}_j) A_j(z,w) \end{align*}$$

and $\sum _{j=1}^{\infty } A_j(z,z) < \infty $ . This proves that (2) implies (3) as well as the full proof of Theorem 1.7.

5 Proof of Theorem 1.9

The equivalence of (1) and (2) is straightforward based on previous results. The implication (1) implies (3) follows from Theorem 1.7.

To prove (3) implies (2), we note that our function $f:\mathbb {C}_+\to \overline {\mathbb {D}}$ gives rise to a function $F: X \to \overline {\mathbb {D}}$ where

$$\begin{align*}X = \{\pi(s) := (p_1^{-s},p_2^{-s},\dots): s\in \mathbb{C}_+\} \subset \mathrm{Ball}(c_0) \end{align*}$$

and $F(\pi (s)) = f(s)$ . We can apply the interpolation result of Theorem 1.7 as follows. Let $Y \subset \{s \in \mathbb {C}: \Re s>1/2\}$ be a countable set with a limit point in $\{s \in \mathbb {C}: \Re s>1/2\}$ . We want a limit point so that it is a determining set for holomorphic functions and we want $\Re s> 1/2$ so that $\pi (s) \in \mathbb {D}_{2}^{\infty }$ . Let $\bigcup _{n=1}^{\infty } Y_n = Y$ be an increasing union of finite sets. Set $X_n = \pi (Y_n), X_{\infty } = \pi (Y)$ .

Fix n and let $T = (T_1, T_2,\dots )$ be an infinite tuple of contractive, commuting matrices with $\sigma (T) \subset X_n$ and one-dimensional joint eigenspaces. The eigenvalues of $T_1$ are of the form $p_1^{-s} = 2^{-s}$ for $s \in Y_n$ .

We can take $M = -\log T_1/\log 2$ using the principal log and then $T_j = p_j^{-M}$ for all j. By assumption, $\|T_j\|= \|p_j^{-M}\| \leq 1$ and this extends to all natural numbers by factoring into primes: $\|n^{-M}\| \leq 1$ for all $n\in \mathbb {N}$ . By assumption (3), $\|f(M)\| = \|F(T)\| \leq 1$ . Therefore, by the implication (3) $\implies $ (1) in Theorem 1.7, there exists $G_n \in \mathcal {A}_{\infty }$ such that $\left .G_n\right |{}_{X_n} = \left .F\right |{}_{X_n}$ . In particular, $g_n(s) := G_n(\pi (s)) \in \mathscr {A}^{\infty }$ and agrees with $f_n$ on $Y_n$ .

By the Montel Theorem of Remark 3.2, since $G_n \in \mathcal {A}_{\infty } \subset \mathcal {S}_{\infty }$ , there is a subsequence of $n\in \mathbb {N}$ such that $G_n \to G \in \mathcal {S}_{\infty }$ uniformly on compact subsets of $\mathrm{Ball}(c_0)$ . The limit G necessarily agrees with F on $X_{\infty }$ . Since membership in $\mathcal {A}_{\infty }$ can be tested by checking whether G restricted to $\mathbb {D}^N$ belongs to $\mathcal {A}_N$ by Theorem 1.5, and since this in turn can be tested by examining G on finite subsets of $\mathbb {D}^N$ , we see that the limit G belongs to $\mathcal {A}_{\infty }$ since each $G_n$ belongs to $\mathcal {A}_{\infty }$ . Finally, $g = G\circ \pi \in \mathscr {A}^{\infty }$ and agrees with f on the set of uniqueness Y, so we see that $g=f \in \mathscr {A}^{\infty }$ .

6 Proof of Theorem 1.10

Given $f \in \mathscr {A}^{\infty }$ , let $F \in \mathcal {A}_{\infty }$ be the Bohr lift of f namely, $f(s) = F(p_1^{-s},p_2^{-s},\dots )$ . Again, forming $F_N(z_1,z_2,\dots ) = F(z_1,\dots , z_N,0,\dots )$ , we see that the functions $f_N(s) := F_N(p_1^{-s},p_2^{-s},\dots )$ converge uniformly to f on the half planes $\{s\in \mathbb {C}: \Re s \geq \epsilon \}$ for each $\epsilon>0$ . The power series for $F_N$ converges absolutely on $\mathbb {D}^N$ and therefore the Dirichlet series for $f_N$ converges to $f_N$ absolutely on $\mathbb {C}_{+}$ . Because of this the holomorphic functional calculus evaluation $f_N(M)$ agrees with $F_N(p_1^{-M},p_2^{-M},\dots )$ . Since $F_N \in \mathcal {A}_{\infty }$ ,

$$\begin{align*}\|f_N(M)\| = \|F_N(p_1^{-M},p_2^{-M},\dots)\| \leq 1. \end{align*}$$

Since $f_N$ converges uniformly to f on a half-plane containing $\sigma (M)$ , $f_N(M) \to f(M)$ in operator norm and therefore $\|f(M)\| \leq 1$ as desired. This completes the proof.