1 Introduction

Ando [1] proved that the Banach space \(H^\infty ({\mathbb {D}})\) of bounded holomorphic functions on the unit disc \({\mathbb {D}}\) has a unique isometric predual. Let us denote it by \(G^\infty ({\mathbb {D}})\). By the Bishop-Phelps theorem, the set \(NA(G^\infty ({\mathbb {D}}))\) of functions \(f \in H^\infty ({\mathbb {D}})\) which attain their norm as elements of the dual of \(G^\infty ({\mathbb {D}})\) is a norm-dense subset of \(H^\infty ({\mathbb {D}})\). Fisher [6] showed that \(f \in H^\infty ({\mathbb {D}}), ||f||=1,\) attains its norm as an element of the dual of \(G^\infty ({\mathbb {D}})\) if and only if the radial limits \(f^*(w)\) of f in the torus \({\mathbb {T}}\) satisfy that the set \(\{w\in {\mathbb {T}}:\, |f^*(w)|=1\}\) has positive Lebesgue measure on \({\mathbb {T}}\). The aim of this article is to investigate versions of Fisher’s result for the Banach space of bounded holomorphic functions on the N-dimensional ball and the N-dimensional polydisc. Our main results are Theorems 5 and 8 and Propositions 6 and 7 in the case of the ball. The case of the polydisc is treated in Sect. 3. The final section deals with the Banach space of bounded Dirichlet series.

Let X be a complex Banach space. Its open unit ball is denoted by \(B_X\) and its closed unit ball by \(U_X\). The space of all holomorphic functions on \(B_X\) (i.e. the \({\mathbb {C}}-\)Fréchet differentiable functions \(f:B_X \rightarrow {\mathbb {C}}\)) will be denoted \(H(B_X).\) The Banach space \(H^\infty (B_X)\) of all bounded holomorphic functions f in \(H(B_X)\) is endowed with the supremum norm \(\Vert f\Vert _\infty =\sup _{x\in B_X}|f(x)|\). We denote by \(\tau _0\) the compact-open topology on \(H^\infty (B_X),\) that is, the topology of uniform convergence on compact subsets of \(B_X.\) Recall that \(\tau _0\) is Hausdorff and coarser than the norm topology. Let \(U_{H^\infty (B_X)}\) denote the closed unit ball of \(H^\infty (B_X)\). The vector space \(G^\infty (B_X),\) given by

$$\begin{aligned} G^\infty (B_X):=\{ \varphi \in H^\infty (B_X)^*\,:\, \varphi _{|U_{H^\infty (B_X)}} \text { is}\ \tau _0\text {-continuous}\} \end{aligned}$$

is a Banach space when endowed with the dual norm. By using the Ng-Dixmier Theorem [12], Mujica [11], proved that the topological dual of \(G^\infty (B_X)\) is isometrically isomorphic to \(H^\infty (B_X)\). We abbreviate this fact by

$$\begin{aligned} G^\infty (B_X)^*{\mathop {=}\limits ^{1}}H^\infty (B_X). \end{aligned}$$

For each \(x \in B_X\) we denote by \(\delta _x: H^\infty (B_X) \rightarrow {\mathbb {C}}\) the evaluation \(\delta _x(f):=f(x)\) at the point x. Clearly \(\delta _x\) is \(\tau _0\) continuous. Moreover, the vector space \(\text {span}\{\delta _x:\ x\in B_X\}\) is a norm-dense subset in \(G^\infty (B_X)\). Indeed, \(\{\delta _x:\ x\in B_X\}\) separates points of \(H^\infty (B_X)\). Hence \(\text {span}\{\delta _x\,:\ x\in B_X\}\) is a subspace of \(G^\infty (B_X)\) that is \(w(G^\infty (B_X), H^\infty (B_X))\)-dense in \(G^\infty (B_X)\). Thus it is is also norm-dense subset of \(G^\infty (B_X)\). We collect the following consequence for reference later in the paper.

Lemma 1

If \({\mathcal {F}}\) is a closed subspace of \(G^\infty (B_X)\) containing \(\{\delta _x:\ x\in B_X\}\), then \({\mathcal {F}}= G^\infty (B_X)\).

Let Y be a Banach space. The set of norm attaining functionals is defined to be the following subset of \(Y^*:\)

$$\begin{aligned} NA(Y):=\{ y^*\in Y^*:\, \text { there exists}\ y\in Y, \Vert y\Vert =1\ \text {such that } \Vert y^*\Vert =y^*(y)\} \end{aligned}$$

The Bishop–Phelps theorem (see, e.g., Theorem 8.11 in [2]) ensures that the set NA(Y) of norm attaining functionals is a norm-dense subset of \(Y^*\). As a consequence, for each non-trivial, complex Banach space X, there exists a norm-dense subset \(NA(G^\infty (B_X))\) of \(H^\infty (B_X)\), such that for every \(f\in NA(G^\infty (B_X))\), there exists an element \(\varphi \in G^\infty (B_X)\) with \(\Vert \varphi \Vert =1\) such that

$$\begin{aligned} \Vert f\Vert _\infty =\varphi (f). \end{aligned}$$

The aim of this paper is to study those functions \(f \in H^\infty (B_X)\) that attain their norm as elements of the dual of \(G^\infty (B_X)\), that is, those \(f \in NA(G^\infty (B_X))\). We mainly concentrate on the case \(X=({\mathbb {C}}^N,\Vert .\Vert _2)\) and hence, \(B_X\) is the N-dimensional Euclidean ball which henceforth will be denoted \(B_N.\)

In the one dimensional case, \(B_N = {\mathbb {D}}\) and its boundary is the torus \({\mathbb {T}}=\{z\in {\mathbb {C}}:\, |z|=1\}\). In this case, by a result by Fatou, there is an isometric isomorphism between \(H^\infty ({\mathbb {D}})\) and

$$\begin{aligned} H^\infty ({\mathbb {T}}):=\left\{ g\in L^\infty ({\mathbb {T}}):\; {\hat{g}}(k)=\int _{{\mathbb {T}}}w^{-k}g(w)dm_1(w)=0, \ k=-1,-2,\ldots \right\} . \end{aligned}$$

The isometric isomorphism \(H^\infty ({\mathbb {D}}) \rightarrow H^\infty ({\mathbb {T}})\) is given by

$$\begin{aligned}{} & {} H^\infty ({\mathbb {D}}) \longrightarrow H^\infty ({\mathbb {T}}) \\{} & {} f\longrightarrow f^* \end{aligned}$$

where the radial limit

$$\begin{aligned} f^*(w):=\lim _{r\rightarrow 1-}f(rw), \end{aligned}$$

exists almost everywhere on \({\mathbb {T}}\) (with respect to the Lebesgue normalized measure on \({\mathbb {T}},\) denoted by \( dm_1(w)=\frac{dt}{2\pi }\), where \(w=e^{it}\).) From this point of view \(H^\infty ({\mathbb {D}}) {\mathop {=}\limits ^{1}} H^\infty ({\mathbb {T}})\) is a closed subspace of \(L^\infty ({\mathbb {T}})\), and hence it is a dual space. In fact, if \(H^1_0({\mathbb {T}})\) is the closed subspace of \(L^1({\mathbb {T}})\) given by

$$\begin{aligned} H_0^1({\mathbb {T}})\!=\!\left\{ f\in L_1({\mathbb {T}}):\, {\hat{f}}(-n)\!=\!\int _{{\mathbb {T}}}f(w)w^{n}dm_1(w)=0,\ \text {for all } n\!=\!0,1,2,\ldots \right\} , \end{aligned}$$

then

$$\begin{aligned} H^\infty ({\mathbb {T}}){\mathop {=}\limits ^{1}}\Big (L^1({\mathbb {T}})/H_0^1({\mathbb {T}})\Big )^*. \end{aligned}$$

Ando in [1] proved that \(H^\infty ({\mathbb {D}})\) has a unique isometric predual. Accordingly, \(L^1({\mathbb {T}})/H_0^1({\mathbb {T}}) {\mathop {=}\limits ^{1}} G^\infty ({\mathbb {D}})\). As far as we know, it is an open question for \(N\ge 2\) whether there is a unique predual of the corresponding \(H^\infty \)-spaces in the case of the N-dimensional ball and the N-polydisc. In this paper, we will introduce another natural predual and show, in Theorems 5 and 10, that it coincides with \(G^\infty (B_X)\).

The characterization of norm attaining elements of \(f \in H^\infty ({\mathbb {D}})\) was obtained by S. Fisher in 1969.

Theorem 2

(Fisher [6, Theorem 2]). Let f be an element of norm one in \(H^\infty ({\mathbb {D}})\). The function f attains its norm as an element of the dual of \(L^1({\mathbb {T}})/H_0^1({\mathbb {T}})=G^\infty ({\mathbb {D}})\) if and only if \(f^*(w)=\lim _{r\rightarrow 1-}f(rw)\) (a.e. in \({\mathbb {T}}\)) satisfies that

$$\begin{aligned} \{w\in {\mathbb {T}}:\, |f^*(w)|=1\} \end{aligned}$$

has positive Lebesgue measure on \({\mathbb {T}}\).

In this paper, in Sect. 2, we explore several variable versions of Fisher’s result. We also examine, in Sects. 3 and 4, similar questions for the polydisc algebra \(H^\infty ({\mathbb {D}}^N)\) and for the space of Dirichlet series \({\mathcal {D}}^\infty ({\mathbb {C}}_+).\)

2 The Case of the Euclidean Ball

Recall that the Euclidean open unit ball in \({\mathbb {C}}^N\) is:

$$\begin{aligned} B_N:=\left\{ z=(z_1,\ldots , z_N)\in {\mathbb {C}}^N:\, \Vert z\Vert _N:=\root 2 \of {|z_1|^2+\cdots + |z_N|^2}<1\right\} . \end{aligned}$$

The unit sphere in \({\mathbb {C}}^N\) is:

$$\begin{aligned} S_N:=\left\{ z=(z_1,\ldots , z_N)\in {\mathbb {C}}^N:\, \Vert z\Vert _N:=\root 2 \of {|z_1|^2+\cdots + |z_N|^2}=1\right\} . \end{aligned}$$

(Observe that this is not completely standard notation since the usual notation for the N-dimensional real sphere in \({\mathbb {R}}^N\) is \(S_{N-1}\).)

By \(\sigma _N\) we denote the unique rotation-invariant positive Borel measure on \(S_N\) for which

$$\begin{aligned} \sigma _N(S_N)=1. \end{aligned}$$

In other words, \(\sigma _N\) is the Haar measure of the N-dimensional sphere.

In [15, p.84], the space \(H^\infty (B_N)\), is defined as

$$\begin{aligned} H^\infty (B_N):=\left\{ f\in H(B_N):\, \Vert f\Vert _\infty :=\sup _{z\in B_N}|f(z)|<\infty \right\} . \end{aligned}$$

The ball algebra is the Banach subalgebra of \(H^\infty (B_N)\) given by

$$\begin{aligned} A(B_N):=\{f:{\overline{B}}_N\rightarrow {\mathbb {C}}:\, f \text { is continuous on }{\overline{B}}_N \text { and holomorphic on } B_N\}. \end{aligned}$$

Finally, by \(A(S_N)=A(B_N)\cap C(S_N)\), we understand the restrictions of the elements of \(A(B_N)\) to the sphere \(S_N\), i.e.

$$\begin{aligned} A(S_N):=\{f_{|S_N}:\ f \in A(B_N)\}. \end{aligned}$$

By the maximum modulus theorem, the mapping \(\pi : A(B_N)\rightarrow A(S_N)\) defined by \(\pi (f):=f_{|S_N}\) is an isometry.

Hardy spaces have a dual definition. The Hardy space \(H^\infty (S_N)\) is the weak-star closure of \(A(S_N)\) in \(L^\infty (S_N,\sigma _N)\). i.e.

$$\begin{aligned} H^\infty (S_N):=\overline{A(S_N)}^{w(L_\infty (S_N),L_1(S_N))}. \end{aligned}$$

As the polynomials are dense in \(A(B_N)\) we have that \(\text {span}\{z^\beta :\beta \in {\mathbb {N}}_0^N\}\) is a \(\Vert .\Vert _\infty \) dense subspace of \(A(B_N)\). Hence, \(\text {span}\{w^\beta :\beta \in {\mathbb {N}}_0^N\}\) is \(\Vert .\Vert _\infty \) dense in \(A(S_N)\). Thus

$$\begin{aligned} H^\infty (S_N)=\overline{\text {span}\{w^\beta :\beta \in {\mathbb {N}}_0^N\}}^{w(L_\infty (S_N),L_1(S_N))}. \end{aligned}$$

At this point, we show that \(H^\infty (S_N)\) and \(H^\infty (B_N)\) are isometrically isomorphic. We need some notation and results that can be found, for example, in the books [15] and [16]. The invariant Poisson kernel of \(B_N\) is the kernel function \(P_N:B_N\times S_N\rightarrow [0,+\infty [\)

$$\begin{aligned} P_N(z,w):=\frac{(1-|z|^2)^N}{|1-<z,w>|^{2N}}. \end{aligned}$$

The Poisson integral P(g) of a function g in \(L^1(S_N, \sigma _N)\) is defined, for \(z\in B_N\), by

$$\begin{aligned} P_N(g)(z):=\int _{S_N}P(z,w)g(w)d\sigma _N(w). \end{aligned}$$

We have that \(P_N:H^\infty (S_N)\longrightarrow H^\infty (B_N)\) is a linear isometry onto.

To prove that this mapping is onto, the concept of Korányi, or K-limit, of a holomorphic function on \(B_N\) is needed. For \(\alpha >1\) and \(w\in S_N\) we set

$$\begin{aligned} D_\alpha (w):=\left\{ z\in {\mathbb {C}}^N:\, |w-z|<\frac{\alpha }{2}(1-|z|^2)\right\} . \end{aligned}$$

Clearly \(D_\alpha (w) \subset B_N\). We say that a function \(F:B_N\rightarrow {\mathbb {C}}\) has K-limit \(\lambda \in {\mathbb {C}}\) at \(w\in S_N\) if the following is true: For every \(\alpha >1\) and for every sequence \((z_j)\) in \(D_\alpha (w)\) that converges to a point \(w \in S_N\), we have that \(F(z_j)\) converges to \(\lambda \) and write

$$\begin{aligned} (K-\lim F)(w)=\lambda . \end{aligned}$$

The following result (see e.g. [15, Section 5.4.]) is important and very useful for our paper.

Theorem 3

If f is a function in \(H^\infty (B_N)\) then f has finite K-limits \(f^*\) \(\sigma _N\)-almost everywhere on \(S_N\). Moreover, \(f^*\in H^\infty (S_N)\), \(\Vert f^*\Vert _\infty =\Vert f\Vert _\infty \) and

$$\begin{aligned} P_N(f^*)=f. \end{aligned}$$

In other words, the mapping \(f\rightarrow f^*\) is a linear isometry from \(H^\infty (B_N)\) onto \(H^\infty (S_N)\).

We also need the following well known fact, a proof of which is given for the sake of completeness.

Lemma 4

Let X be a Banach space and let Y be a weak-star closed subspace of \(X^*\). The subspace

$$\begin{aligned} Y_{\perp }:=\{x\in X:\, y^*(x)=0, \text { for all } y^*\in Y\},\end{aligned}$$

satisfies

$$\begin{aligned} Y_{\perp }^\perp :=\{x^*\in X^*:\, x^*(x)=0, \text { for all } x \in Y_\perp \}=Y, \end{aligned}$$

and Y is isometrically isomorphic to \((X/Y_{\perp })^*\).

Proof

Clearly, by the definition, \(Y\subset Y_{\perp }^\perp \). Assume that the reverse inclusion is not true. Hence there exists \(x_0^*\in Y_{\perp }^\perp {\setminus } Y\).

Since Y is \(w(X^*,X)\) closed and convex we can find \(\varphi :X^*\rightarrow {\mathbb {C}},\) \(w(X^*,X)\)-continuous, such that

$$\begin{aligned}\varphi (x_0^*)=1\text { and } \varphi (y^*)=0, \end{aligned}$$

for all \(y^*\in Y\). Since \(\varphi \) is weak-star continuous, there exists \(x_0\in X\) such that

$$\begin{aligned} \varphi (x^*)=x^*(x_0), \end{aligned}$$

for all \(x^*\in X^*\). Thus, \(x_0^*(x_0)=1\) and \(y^*(x_0)=0\) for all \(y^*\in Y\). Hence \(x_0\) belongs \(Y_{\perp }\). But, \(x_0^*\in Y_{\perp }^\perp \), which, by definition implies

$$\begin{aligned} x_0^*(x_0)=0. \end{aligned}$$

This is a contradiction.

Finally, we have \((X/Y_{\perp })^*{\mathop {=}\limits ^{1}} Y_{\perp }^\perp =Y\), as follows from [10, Theorem 1.10.17] for example. \(\square \)

Now we define

$$\begin{aligned} H^1_0(S_N)=\left\{ g\in L_1(S_N)\,:\, \int _{S_N}g(w)f(w)d\sigma _N(w)=0 \text { for all } f\in A(S_N)\right\} . \end{aligned}$$

Since

$$\begin{aligned} H^\infty (S_N):=\overline{A(S_N)}^{w(L_\infty (S_N),L_1(S_N))}=\overline{\text {span}\{w^\beta :\beta \in {\mathbb {N}}_0^N\}}^{w(L_\infty (S_N),L_1(S_N))}, \end{aligned}$$

the subspace \(H^\infty (S_N)\subset L_\infty (S_N)\) is \({w(L_\infty (S_N),L_1(S_N))}\)-closed in \(L_\infty (S_N)\) and

$$\begin{aligned} \begin{aligned} H_0^1(S_N)=&\left\{ g\in L_1(S_N):\, \int _{S_N}g(w)f(w)d\sigma _N(w)=0,\ \text {for all } f\in H^\infty (S_N)\right\} \\=&\ \left\{ g\in L_1(S_N):\, {\hat{g}}(-\beta ):=\int _{S_N}g(w)w^{\beta }d\sigma _N(w)=0,\ \text {for all } \beta \in {\mathbb {N}}_0^N\right\} . \end{aligned} \end{aligned}$$

In the notation of Lemma 4, with \(X=L_1(S_N)\), \(X^*=L_\infty (S_N)\) and \(Y=H^\infty (S_N)\) (which is weak-star closed in \(X^*\)), we have

$$\begin{aligned}{} & {} Y_\perp =H^\infty (S_N)_\perp =H^1_0(S_N), \\{} & {} Y_\perp ^\perp =H^1_0(S_N)^\perp =H^\infty (S_N). \end{aligned}$$

Lemma 4 implies the isometric isomorphism

$$\begin{aligned} H^\infty (S_N){\mathop {=}\limits ^{1}} \big (L_1(S_N)/ H_0^1(S_N)\big )^*. \end{aligned}$$

Next we show that \(G^\infty (B_N)\) and \(L^1(S_N)/ H_0^1(S_N)\) are isometrically isomorphic. Thus, these two natural preduals of \(H^\infty (B_N)\) coincide, and so the extension of Ando’s result on the uniqueness of the predual of \(H^\infty ({\mathbb {D}})\) to several variables is still open.

Theorem 5

For every \(N\in {\mathbb {N}}\) we have that

$$\begin{aligned} L^1(S_N)/ H_0^1(S_N)= G^\infty (B_N) \end{aligned}$$

isometrically.

Proof

First we prove that \(L^1(S_N)/ H_0^1(S_N)\subset G^\infty (B_N)\).

Let \([\varphi ]\in L^1(S_N)/ H_0^1(S_N)\) and \(g\in H^\infty (S_N)\). The duality is given by

$$\begin{aligned} <[\varphi ],g>=\int _{S_N}\varphi (w)g(w)d\sigma _N(w)=\int _{S_N}(\varphi (w)+\eta (w))g(w)d\sigma _N(w), \end{aligned}$$

for every \(\varphi \in L_1(S_N)\) and every \(\eta \in H_0^1(S_N)\).

We identify \(L^1(S_N)/ H_0^1(S_N)\) as a subspace of the dual of \(H^\infty (S_N)\) in the following natural way. Define \(T_{[\varphi ]}: H^\infty (B_N)\longrightarrow {\mathbb {C}}\) by

$$\begin{aligned} T_{[\varphi ]}(f):= <[\varphi ],f^*>=\int _{S_N}\varphi (w)f^*(w)d\sigma _N(w). \end{aligned}$$

We check that \(T_{[\varphi ]}\) belongs to \(G^\infty (B_N)\) for every equivalence class \([\varphi ]\in L^1(S_N)/ H_0^1(S_N)\).

Clearly

$$\begin{aligned} |T_{[\varphi ]}(f)|\le \int _{S_N}|\varphi (w)\Vert f^*\Vert _\infty d\sigma _N(w)=\Vert \varphi \Vert _1\Vert f\Vert _\infty . \end{aligned}$$

Hence, \(T_{[\varphi ]}\) belongs to \(H^\infty (B_N)^*\). This fact and the equality \(\Vert T_{[\varphi ]}\Vert =\Vert [\varphi ]\Vert \) are consequences of the isometric isomorphism \(H^\infty (S_N){\mathop {=}\limits ^{1}} \big (L_1(S_N)/ H_0^1(S_N)\big )^*\) and Theorem 3.

Let us check that \(T_{[\varphi ]}\) is \(\tau _0\)-continuous when restricted to the closed unit ball \(U_{ H^\infty (B_N)}\) of \(H^\infty (B_N)\).

By Theorem 3, we know that if \(f\in H^\infty (B_N)\) and \(f^*\in H^\infty (S_N)\) is its K-limit that exists a.e. in \(S_N\), then

$$\begin{aligned} f(z)=\int _{S_N}P_N(z,w)f^*(w)d\sigma _N(w) \end{aligned}$$

for all \(z\in B_N.\) Conversely, if \(h\in H^\infty (S_N)\), then \(P_N(h)\in H^\infty (B_N)\) and we have

$$\begin{aligned} P_N(h)^*(w)=h(w) \end{aligned}$$

a.e. on \(S_N\).

For each \(z\in B_N\) the mapping \(P_N(z,.):S_N\rightarrow ]0,+\infty [\) is continuous on \(S_N\). Hence \(P_N(z,.)\in L^1(S_N)\).

Given \((f_n)\cup \{f\}\subset U_{ H^\infty (B_N)}\) such that \((f_n)\) converges to f with respect to the compact-open topology on \(B_N\), we have \((f^*_n)\cup \{f^*\}\subset U_{ H^\infty (S_N)}\). But \(U_{ H^\infty (S_N)}\) is a weak-star closed subset of \(U_{ L^\infty (S_N)}\) which, in turn, is a \(w(L^\infty (S_N),L^1(S_N))\)-compact set. Since \(L^1(S_N)\) is separable, it follows that \(U_{ H^\infty (S_N)}\) is a metrizable compact set with the weak-star topology. Consider now any subsequence \((f^*_{n_k})\) that is \(w(L^\infty (S_N),L^1(S_N))\)-convergent to some \(h\in U_{ H^\infty (S_N)}\). We will have

$$\begin{aligned} \begin{aligned} P_N(h)(z)&=\int _{S_N}P_N(z,w)h(w)d\sigma _N(w) \\= & {} lt;P_N(z,.),h> =\lim _{k\rightarrow \infty }<P_N(z,.),f_{n_k}> \\&=\lim _{k\rightarrow \infty }\int _{S_N}P_N(z,w)f_{n_k}^*(w)d\sigma _N(w)\\&=\lim _{k\rightarrow \infty }f_{n_k}(z)=f(z), \end{aligned} \end{aligned}$$

for all \(z\in B_N\). Hence,

$$\begin{aligned} h(w)=P_N(h)^*(w)=f^*(w) \end{aligned}$$

a.e in \(S_N\). We have just proved that the only weak-star adherent point of \((f_n^*)\) is \(f^*\). Thus \((f^*_n)\) weak-star converges to \(f^*\). In particular

$$\begin{aligned} \begin{aligned} T_{[\varphi ]}(f)&=\int _{S_N}f^*(w)\varphi (w)d\sigma _N(w) \\= & {} lt;[\varphi ],f^*>=\lim _{n\rightarrow \infty }<[\varphi ],f_n^*> \\&\lim _{n\rightarrow \infty }T_{[\varphi ]} (f_n), \end{aligned} \end{aligned}$$

and \(T_{[\varphi ]}\) is continuous with the compact-open topology when restricted to the closed unit ball of \(H^\infty (B_N);\) i.e. \(T_{[\varphi ]}\in G^\infty (B_N)\).

For the other inclusion observe that

$$\begin{aligned} \delta _z(f)=P_N(f^*)(z)=\int _{S_N}P_N(z,w)f^*(w)d\sigma _N(w)=T_{[P_N(z,.)]}(f), \end{aligned}$$

for every \(z\in B_N\) and every \(f\in H^\infty (B_N)\). Thus

$$\begin{aligned} \text {span}\{\delta _z:\, z\in B_N\}\subset L^1(S_N)/ H_0^1(S_N). \end{aligned}$$

The conclusion follows from Lemma 1. \(\square \)

Theorem 5 permits us to get a sufficient condition for a function on \(H^\infty (B_N)\) to attain the norm.

Proposition 6

If f is an element of \(H^\infty (B_N)\) of norm one such that the set

$$\begin{aligned} E:= \{w\in S_N:\, |f^*(w)|=1\}, \end{aligned}$$

has positive \(\sigma _N\) measure in \(S_N\), then f attains its norm as an element of the dual of \(L^1(S_N)/ H_0^1(S_N) = G^\infty (B_N)\).

Proof

Define \(\varphi :S_N\longrightarrow {\mathbb {C}}\) by

$$\begin{aligned} \varphi (w)= {\left\{ \begin{array}{ll} \frac{|f^*(w)|}{f^*(w)}\frac{1}{\sigma _N(E)}, &{} \text{ if } w\in E \\ 0, &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

We have that \(\varphi \) is a bounded measurable function on \(S_N\). Thus \(\varphi \in L^1(S_N)\) and

$$\begin{aligned} \int _{S_N}{|\varphi (w)|}d\sigma _N(w)=\frac{1}{\sigma _N(E)}\int _{E}d\sigma _N(w)=1. \end{aligned}$$

Define \(T_{[\varphi ]}: H^\infty (B_N)\longrightarrow {\mathbb {C}}\) by

$$\begin{aligned} T_{[\varphi ]}(g):= <[\varphi ],g^*>=\int _{S_N}\varphi (w)g^*(w)d\sigma _N(w). \end{aligned}$$

By Theorem 5, \(T_{[\varphi ]} \in L^1(S_N)/ H_0^1(S_N)= G^\infty (B_N)\) and

$$\begin{aligned} |T_{[\varphi ]}(g)|\le \Vert g^*\Vert _\infty \Vert \varphi \Vert _1=\Vert g\Vert _\infty \Vert \varphi \Vert _1=\Vert g\Vert _\infty , \end{aligned}$$

for every \(g\in H^\infty (B_N)\). Hence

$$\begin{aligned} \Vert T_{[\varphi ]}\Vert \le 1. \end{aligned}$$

But

$$\begin{aligned} T_{[\varphi ]}(f)=\int _{S_N}\varphi (w)f^*(w)d\sigma _N(w)=\frac{1}{\sigma _N(E)}\int _{E}|f^*(w)|d\sigma _N(w)=1=||f||. \end{aligned}$$

and f in the dual of \(G^\infty (B_N)\) attains its norm at \(T_{[\varphi ]}\). \(\square \)

A partial converse to the above proposition is the following.

Proposition 7

If f is an element of \(H^\infty (B_N)\) of norm one such that there exists \(\varphi \in L^1(S_N)\) with \(\Vert \varphi \Vert _1=1\) and \(T_{[\varphi ]}(f)=1\), then

$$\begin{aligned} \sigma _N(\{w\in S_N:\, |f^*(w)|=1\})>0. \end{aligned}$$

Proof

We denote \(E=\{w\in S_N:\, |f^*(w)|=1\}\).

Assume that \(\sigma _N(E) = 0.\)

Let

$$\begin{aligned} K_n=\left\{ w\in S_N:\, |f^*(w)|<\frac{n-1}{n} \right\} . \end{aligned}$$

Clearly \(S_N\setminus E=\cup _{n=1}^\infty K_n.\)

We have that \(T_{[\varphi ]}\in G^\infty (B_N)\) and is of norm one since

$$\begin{aligned} 1=T_{[\varphi ]}(f)\le \Vert [\varphi ]\Vert \Vert f\Vert _\infty =\Vert [\varphi ]\Vert \le \Vert \varphi \Vert _1=1. \end{aligned}$$

For each n, we get

$$\begin{aligned} \int _{S_N\setminus K_n}|\varphi (w)|d\sigma _N(w)&+ \int _{ K_n}|\varphi (w)|d\sigma _N(w)= 1=\int _{S_N}f(w)\varphi (w)d\sigma _N(w) \\&= \int _{S_N\setminus K_n}f(w)\varphi (w)d\sigma _N(w)+ \int _{K_n}f(w)\varphi (w)d\sigma _N(w) \\&\le \int _{S_N\setminus K_n}|f(w)\varphi (w)|d\sigma _N(w)+ \int _{ K_n}|f(w)\varphi (w)|d\sigma _N(w)\\&\le \int _{S_N\setminus K_n}|\varphi (w)|d\sigma _N(w)+ \frac{n-1}{n} \int _{K_n}|\varphi (w)|d\sigma _N(w). \end{aligned}$$

Thus, \(\int _{K_n}|\varphi (w)|d\sigma _N(w)=0\). Since n is arbitrary, we get

$$\begin{aligned} \int _{S_N\setminus E}|\varphi (w)|d\sigma _N(w)=0. \end{aligned}$$

But, by hypothesis \(\sigma _N(E)=0\) and finally we arrive at the contradiction

$$\begin{aligned} 1=\int _{S_N}|\varphi (w)|d\sigma _N(w)=0. \end{aligned}$$

\(\square \)

A subset E of \(S_N\) is called a peak set if there exists \(f \in A(B_N)\) such that \(f(z)=1\) for every \(z\in E\) and \(|f(z)|<1\) for every \(z\in {\overline{B}}_N\setminus E\). Every peak set is a null set.

A result by Fatou states that every compact subset of \({\mathbb {T}}\) of Lebesgue measure zero is a peak set of \(A({\mathbb {D}})\), a fact which is instrumental in the proof of Fisher’s Theorem  2. On the other hand, there are null sets on \(S_N\) (respectively in \({\mathbb {T}}^N\)), which are not peak sets [15, 10.1.1 and 11.2.5] (respectively [14, Theorem 6.3.4, p. 149-150]). We do not know if the converse of Proposition 6 is true or not. But, if we restrict ourselves to functions in \(A(B_N)\) that attain their norm, we get the following characterization in terms of peak sets.

Theorem 8

Let f be an element of \(A(B_N)\) of norm one. The function f attains its norm as an element of \(H^\infty (B_N)\) if and only if the set

$$\begin{aligned} E(f)=\{w\in S_N:\, |f(w)|=1\} \end{aligned}$$

is not a peak set.

Before presenting the proof we need some notation and a lemma.

We recall that a complex Borel measure \(\mu \) on \(S_N\) is a Henkin measure (See [15, 9.1.5, p. 186]) if

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{S_N}f_n(w)d\mu (w)=0, \end{aligned}$$

for every sequence \((f_n)\) contained in the closed unit ball \(U_{A(B_N)}\) of \(A(B_N)\) that converges uniformly to 0 on the compact subsets of \(B_N\), that is, converges to 0 in the \(\tau _0\) topology in \(B_N\). (By the Montel theorem, a sequence \((f_n)\) contained in \(U_{A(B_N)}\) converges to 0 in \(\tau _0\) if and only if converges to 0 pointwise on \(B_N\)).

Lemma 9

  1. (1)

    For every Henkin measure \(\mu \) there is \(T\in G^\infty (B_N)\) such that

    $$\begin{aligned} T(f)=\int _{S_N}f(w)d\mu (w) \end{aligned}$$

    for each \(f\in A(B_N),\) and \(\Vert \mu \Vert \ge \Vert T\Vert \).

  2. (2)

    If \(T\in G^\infty (B_N)\), then there is a Henkin measure \(\mu \) on \(S_N\) such that

    $$\begin{aligned} T(f)=\int _{S_N}f(w)d\mu (w) \end{aligned}$$

    for each \(f\in A(B_N),\) and \(\Vert \mu \Vert = \Vert T\Vert \).

Proof

  1. (1)

    Define \(T_1:A(B_N)\longrightarrow {\mathbb {C}}\) by

    $$\begin{aligned} T_1(g):=\int _{S_N}g(w)d\mu (w). \end{aligned}$$

    Clearly, \(T_1\) is a continuous linear form on \(A(B_N)\) which is \(\tau _0\)-continuous on \(U_{A(B_N)}\) and

    $$\begin{aligned} \Vert T_1\Vert \le \Vert \mu \Vert . \end{aligned}$$

    Given \(f\in H^\infty (B_N)\), the function \(f_r(z):=f(rz)\), \(0\le r<1\), belongs to \(A(B_N).\) In addition, \((f_r)\) converges to f uniformly on the compact subsets of \(B_N\) and

    $$\begin{aligned} \Vert f_r\Vert \le \Vert f\Vert ,\ \ \ \Vert f\Vert =\sup _r\Vert f_r\Vert . \end{aligned}$$
    (1)

    By [15, 11.3.1], since \(\mu \) is a Henkin measure, the limit

    $$\begin{aligned} \lim _{r\rightarrow 1-}\int _{S_N}f_r(w)d\mu (w)=\lim _{r\rightarrow 1-}T_1(f_r)\in {\mathbb {C}}, \end{aligned}$$

    exists for every \(f\in H^\infty (B_N)\).

We define \(T:H^\infty (B_N)\longrightarrow {\mathbb {C}}\), by

$$\begin{aligned} T(f):=\lim _{r\rightarrow 1-}T_1(f_r). \end{aligned}$$

T is linear and \(T\in (H^\infty (B_N))^*\), since

$$\begin{aligned} |T(f)|\le \sup _r|T_1(f_r)|\le \Vert T_1\Vert \Vert f\Vert , \end{aligned}$$

for every \(f\in H^\infty (B_N)\). Moreover, \(\Vert T\Vert =\Vert T_1\Vert \).

We claim that the restriction of \(T_1\) to \(U_{A(B_N)}\) is \(\tau _0\)-uniformly continuous. Indeed, given \(\varepsilon >0\) there are a compact subset K of \(B_N\) and \(\delta >0\) such that \(|T_1(g)|<\varepsilon \) if \(g\in U_{A(B_N)}\) and \(\sup _{z\in K}|g(z)|<\delta \). Hence, if \(g,h\in U_{A(B_N)}\) and \(\sup _{z\in K}|g(z)-h(z)|<\delta \), then

$$\begin{aligned} |T_1(g)-T_1(h))|=|2T_1(\frac{g-h}{2})|<2\varepsilon . \end{aligned}$$

Since \(U_{A(B_N)}\) is \(\tau _0\)-dense in \(U_{H^\infty (B_N)},\) there exists a unique \(\widetilde{T_1}: U_{H^\infty (B_N)}\longrightarrow {\mathbb {C}}\) that is \(\tau _0\)-continuous and such that

$$\begin{aligned} \widetilde{T_1}(g)=T_1(g), \end{aligned}$$

for all \(g\in U_{A(B_N)}\). Given \(f\in U_{H^\infty (B_N)}\), then \((f_r)\subset U_{A(B_N)}\), and \((f_r)\) converges to f in \(\tau _0\) as \(r\rightarrow 1-\). Thus

$$\begin{aligned} \widetilde{T_1}(f_r)\rightarrow \widetilde{T_1}(f), \end{aligned}$$

in \({\mathbb {C}}\). But \(\widetilde{T_1}(f_r)=T_1(f_r)\) for each \(r\in [0,1[\) and \(T_1(f_r)\) converges to T(f) by definition. This implies \(\widetilde{T_1}(f)=T(f)\) for each \(f\in U_{H^\infty (B_N)}\). We have obtained that the restriction of T to \(U_{H^\infty (B_N)}\) is \(\tau _0\) continuous. This, by definition, implies that T belongs to \(G^\infty (B_N)\).

  1. (2)

    If \(T\in G^\infty (B_N)\), the restriction of T to \(U_{H^\infty (B_N)}\) is \(\tau _0\) continuous. If we define \(T_1:A(B_N)\longrightarrow {\mathbb {C}}\), by \(T_1(g):=T(g)\), then \(T_1\) is continuous for the sup-norm, \(\Vert T_1\Vert \le \Vert T\Vert \) and \(T_1\) is \(\tau _0\)-continuous on \(U_{A(B_N)}\). By (1), we have

    $$\begin{aligned} |T(f)|=\lim _{r\rightarrow 1-}|T(f_r)|=\lim _{r\rightarrow 1-}|T_1(f_r)|\le \Vert T_1\Vert \sup _{r<1}\Vert f_r\Vert =\Vert T_1\Vert \Vert f\Vert , \end{aligned}$$

    for every \(f\in H^\infty (B_N)\). Thus, \(\Vert T_1\Vert = \Vert T\Vert \). We can consider \(T_1:A(S_N)\longrightarrow {\mathbb {C}}\). By the Riesz theorem, there is a complex Borel measure \(\mu \) on \(S_N\) such that

    $$\begin{aligned} T_1(h):=\int _{S_n}h(w)d\mu (w), \end{aligned}$$

    for every \(h\in A(B_N)\) with \(\Vert T_1\Vert =|\mu |(S_N)=\Vert \mu \Vert .\)

The properties of \(T_1\) imply that \(\mu \) is a Henkin measure. \(\square \)

Now we give the proof of Theorem 8:

Proof

Assume that \(f\in A(B_N)\), \(\Vert f\Vert =1\) and that E(f) is not a peak set. By [15, 10.1.1] there exists a Borel measure \(\rho \), such that \(\rho (E(f))\ne 0\) such that

$$\begin{aligned} h(0)=\int _{S_N}h(w)d\rho (w), \end{aligned}$$
(2)

for every \(h\in A(B_N)\).

Define

$$\begin{aligned} g(z)= {\left\{ \begin{array}{ll} 0 &{} \text {if } z\in S_N\setminus E(f) \\ \frac{\overline{f(z)}}{|\rho |(E(f))} &{} \text {if } z\in E(f)\ \ \ . \end{array}\right. } \end{aligned}$$

Since g is bounded and measurable, \(g\in L^1(|\rho |)\). Hence, the measure \(g|\rho |\) defined by \(g|\rho |(M)=\int _{M}gd|\rho |\) for Borel measurable sets M is absolutely continuous with respect to \(\rho \). The measure \(\rho \) is Henkin (this fact is a direct consequence of (2) and the definition of Henkin measure as given in [15, p. 187]), and so \(g|\rho |\) is also a Henkin measure by [15, 9.3.1]. We set \(T_1:A(B_N)\longrightarrow {\mathbb {C}}\).

$$\begin{aligned} T_1(h)=\int _{S_N}h(w)g(w)d|\rho |(w). \end{aligned}$$

We have

$$\begin{aligned} T_1(f)=\int _{S_N}f(w)g(w)d|\rho |(w)=1, \end{aligned}$$

and

$$\begin{aligned} |T_1(h)|\le \int _{S_N}|h(w)||g(w)|d|\rho |(w)\le \frac{|\rho |(E(f))}{|\rho |(E(f))}=1, \end{aligned}$$

for every \(h\in A(B_N)\) with \(\Vert h\Vert \le 1\), and we have \(g|\rho |\) a Henkin measure such that

$$\begin{aligned} T_1(f)=1 \ \ \ \ \text {and } \ \ \Vert T_1\Vert =1. \end{aligned}$$

By Lemma 9.(1), there is \(T\in G^\infty (B_N)\) with \(\Vert T\Vert =\Vert T_1\Vert =1\) such that

$$\begin{aligned} T(h)=T_1(h), \end{aligned}$$

for every \(h\in A(B_N)\).

In particular, \(T(f)=1\) and f attains its norm on \(G^\infty (B_N)\).

Suppose now that \(f\in A(B_N)\), \(\Vert f\Vert =1\), satisfies that E(f) is a peak set and that there is \(T\in G^\infty (B_N)\), \(\Vert T\Vert =1\) with \(T(f)=1\). To get a contradiction we are going to give an argument that follows closely the one given in Proposition 7:

By Lemma 9.(2) there exists \(\mu \) a Henkin measure such that

$$\begin{aligned} T(h)=\int _{S_N}h(w)d\mu (w), \end{aligned}$$

for every \(h\in A(B_N)\) and

$$\begin{aligned} \Vert T\Vert =\Vert T_{\big |A(B_N)}\Vert =|\mu |(S_N)=1. \end{aligned}$$

By [15, 9.3.1] \(|\mu |\) is also a Henkin measure. Hence, by [15, Lemma 11.3.3] (see also [15, Lemma 11.3.1]), \(|\mu |(E(f))=0\). Let

$$\begin{aligned} K_n=\left\{ w\in S_N:\, |f(w)|<\frac{n-1}{n} \right\} . \end{aligned}$$

Clearly \(S_N\setminus E(f)=\cup _{n=1}^\infty K_n.\)

We have that, for each n,

$$\begin{aligned} |\mu |(S_N\setminus K_n)+|\mu |(K_n)&=|\mu |(S_N)=1=T(f)=\int _{S_N}|f(w)|d|\mu |(w)\\&=\int _{S_N\setminus K_n}|f(w)|d|\mu |(w)+ \int _{ K_n}|f(w)|d|\mu |(w) \\&\le \int _{S_N\setminus K_n}d|\mu |(w)+ \frac{n-1}{n} \int _{K_n}d|\mu |(w)\\&=|\mu |(S_N\setminus K_n)+\frac{n-1}{n} |\mu |(K_n). \end{aligned}$$

This implies \(|\mu |(K_n)=0\) for each n and \(|\mu |(S_N{\setminus } E(f))=0\).

Therefore, \(1=|\mu |(S_N)=|\mu |(S_N\setminus E(f))+|\mu |(E(f))=0\), a contradiction. \(\square \)

3 The Case of the Polydisc

For a fixed \(N\in {\mathbb {N}}\), the N-dimensional Poisson kernel [14, p. 17] \(P_N:{\mathbb {D}}^N \times {\mathbb {T}}^N \rightarrow (0,\infty )\) is defined as

$$\begin{aligned} P_N(z,w):=\prod _{j=1}^{N}P_1(z_j,w_j)=\prod _{j=1}^{N}\frac{1-|z_j|^2}{|1-z_j{\overline{w}}_j|^2}. \end{aligned}$$

It is well known ( [14, Theorem 3.3.3, p.45]) that if \(f\in H^\infty ({\mathbb {D}}^N)\) then the limit

$$\begin{aligned} f^*(w)=\lim _{r\rightarrow 1-}f(rw) \end{aligned}$$

exists almost everywhere in \({\mathbb {T}}^N\), and

$$\begin{aligned} f(z)=\int _{{\mathbb {T}}^N}P_N(z,w)f^*(w)dm_N(w) \end{aligned}$$
(3)

for all \(z \in {\mathbb {D}}^N.\) As a consequence there exists an isometric isomorphism

$$\begin{aligned}{} & {} H^\infty ({\mathbb {D}}^N) \longrightarrow H^\infty ({\mathbb {T}}^N) \\{} & {} f\longrightarrow f^* \end{aligned}$$

where \(H^\infty ({\mathbb {T}}^N):=\overline{A({\mathbb {T}}^N)}^{w(L_\infty ({\mathbb {T}}^N),L_1({\mathbb {T}}^N))}\),

$$\begin{aligned} A({\mathbb {D}}^N)=\{f:{\overline{{\mathbb {D}}}}^N\rightarrow {\mathbb {C}}:\, f \text { is continuous on }{\overline{{\mathbb {D}}}}^N \text { and holomorphic on } {\mathbb {D}}^N\} \end{aligned}$$

and

$$\begin{aligned} A({\mathbb {T}}^N):=\{f_{|{\mathbb {T}}^N}:\ f \in A({\mathbb {D}}^N)\}. \end{aligned}$$

By the maximum modulus theorem \(A({\mathbb {D}}^N)\) and \(A({\mathbb {T}}^N)\) are isometrically isomorphic. By Fejer’s theory for the polydisc we have

$$\begin{aligned} H^\infty ({\mathbb {T}}^N){}:= & {} \Bigg \{ g\in L^\infty ({\mathbb {T}}):\; {\hat{g}}(\alpha ) \\= & {} \int _{{\mathbb {T}}^N}w^{-\alpha }g(w)dm_N(w)=0, \text{ for } \text{ all } \alpha \in {\mathbb {Z}}^N\setminus {\mathbb {N}}_0^N\Bigg \} . \end{aligned}$$

On the other hand, by applying Lemma 4,

$$\begin{aligned} H^\infty ({\mathbb {T}}^N){\mathop {=}\limits ^{1}}(L^1({\mathbb {T}}^N)/ H_0^1({\mathbb {T}}^N)\big )^*, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} H_0^1({\mathbb {T}}^N)=&\left\{ f\in L_1({\mathbb {T}}^N):\, {\hat{f}}(-\beta )=\int _{{\mathbb {T}}}f(w)w^{\beta }dm_N(w)=0,\ \text {for all } \beta \in {\mathbb {N}}_0^N\right\} \\=&\ \overline{\text {span}\{w^{-\alpha }:\, \text { for all } \alpha \in {\mathbb {Z}}^N\setminus {\mathbb {N}}_0^N\}}. \end{aligned} \end{aligned}$$

Very similar arguments to the ones given for the N-dimensional Euclidean ball can be given for the N-polydisc to obtain the following results.

Theorem 10

For every \(N\in {\mathbb {N}}\) we have

$$\begin{aligned} L^1({\mathbb {T}}^N)/ H_0^1({\mathbb {T}}^N){\mathop {=}\limits ^{1}} G^\infty ({\mathbb {D}}^N). \end{aligned}$$

Proposition 11

Let f be an element of \(H^\infty ({\mathbb {D}}^N)\) of norm one such that the set

$$\begin{aligned} E:= \{w\in {\mathbb {T}}^N:\, |f^*(w)|=1\}, \end{aligned}$$

has positive Lebesgue measure (in \({\mathbb {T}}^N\)). Then f attains its norm as an element of the dual of \(L^1({\mathbb {T}}^N)/ H_0^1({\mathbb {T}}^N)\).

Proposition 12

If f is an element of \(H^\infty ({\mathbb {D}}^N)\) of norm one such that there exists \(\varphi \in L^1({\mathbb {T}}^N)\) with \(\Vert \varphi \Vert _1=1\) and \(T_{[\varphi ]}(f)=1\), then the set

$$\begin{aligned} \{w\in {\mathbb {T}}^N:\, |f^*(w)|=1\}, \end{aligned}$$

has positive Lebesgue measure in the polytorus \({\mathbb {T}}^N.\)

Example 13

The following example, which is inspired by [3, Theorem 3.1], shows that a polydisc (for \(N>1\)) version of Theorem 8 does not hold. Let \(f: {\mathbb {D}}\times {\mathbb {D}}\rightarrow {\mathbb {C}}\) be the function \(f(z,w):=(1/2)(1+z)\), which belongs to \(A({\mathbb {D}}\times {\mathbb {D}})\). This function does not attain its norm on \(G^\infty ({\mathbb {D}}\times {\mathbb {D}})\). Indeed, if it did, the function \(g(z)=(1/2)(1+z)\), as an element of \(H^\infty ({\mathbb {D}})\), would attain its norm on \(G^\infty ({\mathbb {D}})\), because \(H^\infty ({\mathbb {D}})\) is canonically isometrically contained in \(H^\infty ({\mathbb {D}}\times {\mathbb {D}})\). But the function g does not attain its norm on \(G^\infty ({\mathbb {D}})\) by Fisher’s Theorem 2, because \(\{z \in {\mathbb {T}}\ | \ |g(z)| = 1 \} = \{ 1 \}\). On the other hand, \(E(f)= \{(z,w) \in {\mathbb {T}}\times {\mathbb {T}}\ | \ |f(z,w)|=1 \} = \{ 1 \} \times {\mathbb {T}}\), as it is easy to check. The set E(f) is not a peak set of \(A({\mathbb {D}}\times {\mathbb {D}})\). Otherwise, it would be a zero set; see [14, 6.1.2. Theorem, p.132]. But if a function \(h \in A({\mathbb {D}}\times {\mathbb {D}})\) vanishes on E(f), then \(h(1,w) \in A({\mathbb {D}})\) vanishes on \({\mathbb {T}}.\) The maximum principle then implies that h vanishes on \(\{ 1 \} \times {\mathbb {D}}\), and therefore, there is no function \(h \in A({\mathbb {D}}\times {\mathbb {D}})\) vanishing only in E(f). Observe that E(f) is a null set in \({\mathbb {T}}\times {\mathbb {T}}\) which is not a peak set.

4 The Case of the Space of Dirichelt Series \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\)

Let \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\) denote the Banach space of the Dirichlet series \(D(s)=\sum _{n=1}^{\infty }\frac{a_n}{n^s}\) convergent and bounded on the right half plane \({\mathbb {C}}_+\) endowed with the supremum norm. We refer the reader to [4] and [13] for detailed information about this space.

The space \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\) is a closed subspace of the Banach space \({H}^\infty ({\mathbb {C}}_+)\) of all bounded holomorphic functions in the right half plane \({\mathbb {C}}_+\) endowed with the supremum norm. Since, by the Montel theorem, the closed unit ball of \({H}^\infty ({\mathbb {C}}_+)\) is \(\tau _0\)-compact, we can apply the Dixmier-Ng theorem [12] to obtain that

$$\begin{aligned} G^\infty ({\mathbb {C}}_+)\!:=\!\{R\in {H}^\infty ({\mathbb {C}}_+)^*:\, \text{ the } \text{ restriction } \text{ of }\ R\ \text{ to } U_{{H}^\infty ({\mathbb {C}}_+)} \text{ is }\ \tau _0\ \text{ continuous }\}, \end{aligned}$$

is a predual of \({H}^\infty ({\mathbb {C}}_+)\).

It is well-known that the spaces \({H}^\infty ({\mathbb {C}}_+)\) and \({H}^\infty ({\mathbb {D}})\) are isometrically isomorphic. We are going to show that their preduals are also isometrically isomorphic.

Proposition 14

\({H}^\infty ({\mathbb {C}}_+)\) is isometrically isomorphic to \({H}^\infty ({\mathbb {D}})\), and \(G^\infty ({\mathbb {C}}_+)\) is isometrically isomorphic to \(G^\infty ({\mathbb {D}})\).

Proof

It is enough to consider the Cayley transformation \(\varphi :{\mathbb {C}}\setminus \{1\}\rightarrow {\mathbb {C}}\setminus \{-1\}\) defined by

$$\begin{aligned} \varphi (z)=\frac{1+z}{1-z} \end{aligned}$$

The Cayley transformation is a biholomorphic mapping with inverse

$$\begin{aligned} \varphi ^{-1}(s)=\frac{s-1}{s+1}. \end{aligned}$$

Actually it is also biholomorphic from \({\mathbb {D}}\) onto \({\mathbb {C}}_+\), and it is a homeomorphism from \({\mathbb {T}}\setminus \{1\}\) onto \(\{ti:\ t\in {\mathbb {R}}\}\). Clearly the composition operator \(T_\varphi : {H}^\infty ({\mathbb {C}}_+)\rightarrow {H}^\infty ({\mathbb {D}})\) defined by

$$\begin{aligned}T_\varphi (g)=g\circ \varphi , \end{aligned}$$

for \(g\in {H}^\infty ({\mathbb {C}}_+)\) is an isometry with inverse \((T_\varphi )^{-1}=T_{\varphi ^{-1}}\). Its adjoint \(T_\varphi ^*: {H}^\infty ({\mathbb {D}})^*\rightarrow {H}^\infty ({\mathbb {C}}_+)^*\) is also an isometric isomorphism with

$$\begin{aligned} (T_\varphi ^*)^{-1}=T_{\varphi ^{-1}}^*. \end{aligned}$$

It is enough to check that \(T_\varphi ^*(G^\infty ({\mathbb {D}}))=G^\infty ({\mathbb {C}}_+)\) to prove that \(G^\infty ({\mathbb {D}})\) and \(G^\infty ({\mathbb {C}}_+)\) are isometrically isomorphic.

Let \(R\in G^\infty ({\mathbb {D}}).\) We have

$$\begin{aligned} T_\varphi ^*(R)(g)=R(T_\varphi (g))=R(g\circ \varphi ) \end{aligned}$$

for all \(g \in {H}^\infty ({\mathbb {C}}_+)\). Let \(K\subset {\mathbb {D}}\) be a compact set. The set \(\varphi (K)\) is a compact subset of \({\mathbb {C}}_+\). Take \((g_n)\) and g in the closed unit ball of \({H}^\infty ({\mathbb {C}}_+)\) such that \((g_n)\) converges with respect to the compact open topology on \({\mathbb {C}}_+\) to g. Since \((g_n)\) converges to g uniformly on \(\varphi (K)\), we have \((g_n\circ \varphi )\) converges to \(g\circ \varphi \) uniformly on \(K=\varphi ^{-1}(\varphi (K))\), for every K. Thus, \((g_n\circ \varphi )\) converges to \(g\circ \varphi \) with respect to the compact open topology on \({\mathbb {D}}\). Hence, \((R(g_n\circ \varphi ))\) converges to \(R(g\circ \varphi )\) and we get

$$\begin{aligned} T_\varphi ^*(R)\in G^\infty ({\mathbb {C}}_+). \end{aligned}$$

Analogously we obtain \(T_{\varphi ^{-1}}^*(G^\infty ({\mathbb {C}}_+))\subset G^\infty ({\mathbb {D}})\), from which it follows that

$$\begin{aligned} G^\infty ({\mathbb {C}}_+)=T_\varphi ^*\circ T_{\varphi ^{-1}}^*(G^\infty ({\mathbb {C}}_+))\subset T_\varphi ^*( G^\infty ({\mathbb {D}})). \end{aligned}$$

\(\square \)

Remark 15

Recall that for any fixed \(\alpha >1\) and \(w\in {\mathbb {T}}\) the Stolz region is \(S(\alpha , w)=\{ z\in {\mathbb {D}}\,:\, |z-w|<\alpha (|1-|z|)\} \) ( [9, Definition 8.1.9. ]). Since w is an accumulation point of \(S(\alpha , w)\) it makes sense to speak about the limit at w of any function \(f:S(\alpha , w)\rightarrow {\mathbb {C}}\). Actually, in [9, Theorem 8.1.11], it is proved that if \(f\in H^\infty ({\mathbb {D}})\) the following equality holds on \({\mathbb {T}}\)

$$\begin{aligned} f^*(w)=\lim _{z\in S(\alpha , w)\rightarrow w }f(z), \end{aligned}$$

almost everywhere with respect to the Lebesgue measure.

In [4, p. 286 and 287] it is observed that if \(g \in {H}^\infty ({\mathbb {C}}_+)\), then there exists a Lebesgue null set \(A\subset {\mathbb {R}}\) such that the limit

$$\begin{aligned} \lim _{r\rightarrow 0+}g(r+it):=g^*(it) \end{aligned}$$

exists for every \(t\in {\mathbb {R}}\setminus A\) and actually that

$$\begin{aligned} g^*(it)=\lim _{z\in S(\alpha , \varphi ^{-1}(it))\rightarrow \varphi ^{-1}(it) }T_{\varphi }(g)(z). \end{aligned}$$

In other words, the “horizontal” limits of g exist a.e. and coincide with the Fatou radial limits of its associated function \(T_{\varphi }(g)\) belonging to \(H^\infty ({\mathbb {D}})\).

We can now get the following consequence of Ando’s Theorem [1] and Fisher’s Theorem 2.

Corollary 16

The space \(H^\infty ({\mathbb {C}}_+)\) has a unique predual. Moreover, \(g \in {H}^\infty ({\mathbb {C}}_+)\) with \(\Vert g\Vert _{{\mathbb {C}}_+}=1\) is norm attaining if and only if the set

$$\begin{aligned} E:=\{t\in {\mathbb {R}}:\, |g^*(it)|=1\} \end{aligned}$$

has positive (including \(+\infty \)) Lebesgue measure.

Proposition 17

\({\mathcal {D}}^\infty ({\mathbb {C}}_+)\) is a dual space.

Proof

By a result of F. Bayart (see e.g. [4, Theorem 3.11]), it is known that if \((D_n)\) is a bounded sequence in \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\) then there exists a subsequence \((D_{n_k})\) and a Dirichlet series \(D\in {\mathcal {D}}^\infty ({\mathbb {C}}_+)\) such that for every \(\sigma >0\) the sequence \((D_{n_k})\) converges to D uniformly on \({\mathbb {C}}_\sigma :=\{s\in {\mathbb {C}}\,;\, \text {Re}{s}\ge \sigma \)}. Thus, if we denote by \(\tau _+\) the topology of uniform convergence on these half planes \({\mathbb {C}}_\sigma \), Bayart’s result says that the closed unit ball of \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\) is a compact set. Now the Dixmier-Ng theorem [12] implies that

$$\begin{aligned} {\mathcal {G}}^\infty ({\mathbb {C}}_+):=\{R\in {\mathcal {D}}^\infty ({\mathbb {C}}_+)^*:\, \text{ the } \text{ restriction } \text{ of }\ R\ \text{ to } U_{{\mathcal {D}}^\infty ({\mathbb {C}}_+)} \text{ is }\ \tau _+\ \text{ continuous }\}, \end{aligned}$$
(4)

endowed with the topology induced by \({\mathcal {D}}^\infty ({\mathbb {C}}_+)^*\) is a predual of \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\).

\(\square \)

We can now get a positive result about norm attaining elements of \( {\mathcal {D}}^\infty ({\mathbb {C}}_+)\) with respect to that predual.

Proposition 18

Consider the space \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\) as the dual of \({\mathcal {G}}^\infty ({\mathbb {C}}_+).\) Given \(D \in {\mathcal {D}}^\infty ({\mathbb {C}}_+)\) of norm one, if the set

$$\begin{aligned} E:=\{t\in {\mathbb {R}}:\, |D^*(it)|=1\} \end{aligned}$$

has positive (including \(+\infty \)) Lebesgue measure, then D is norm attaining.

Proof

As \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\) is a closed subspace of \(H^\infty ({\mathbb {C}}_+)\), we can consider \(D\in H^\infty ({\mathbb {C}}_+)\). By Corollary 16, we know that there exists \(R\in G^\infty ({\mathbb {C}}_+)\) such that

$$\begin{aligned} \Vert R\Vert =1=R(D). \end{aligned}$$

Recall that \(R\in {H}^\infty ({\mathbb {C}}_+)^*\) and satisfies that the restriction of R to \(U_{{H}^\infty ({\mathbb {C}}_+)}\) is \(\tau _0\) continuous. We denote by S the restriction of R to \({\mathcal {D}}^\infty ({\mathbb {C}}_+)\). Since \(U_{{\mathcal {D}}^\infty ({\mathbb {C}}_+)}\subset U_{{H}^\infty ({\mathbb {C}}_+)}\) we have that S is \(\tau _0\) continuous when restricted to \(U_{{\mathcal {D}}^\infty ({\mathbb {C}}_+)}\). The theorem of Bayart [4, Theorem 3.11] implies that \(U_{{\mathcal {D}}^\infty ({\mathbb {C}}_+)}\) is a compact set with respect to \(\tau _+\). The compact open topology \(\tau _0\) on \({\mathbb {C}}_+\) is Hausdorff and weaker than \(\tau _+\) on that ball. Hence both topologies coincide on \(U_{ {\mathcal {D}}^\infty ({\mathbb {C}}_+)}\) and \(S\in {\mathcal {G}}^\infty ({\mathbb {C}}_+)\). Moreover

$$\begin{aligned} 1=\Vert R\Vert \ge \Vert S\Vert \ge |S(D)|=S(D)=R(D)=1, \end{aligned}$$

and D attains its norm. \(\square \)

It is natural to ask whether the converse of Proposition 18 holds. Actually, by the Hahn-Banach theorem one can extend R in \({\mathcal {G}}^\infty ({\mathbb {C}}_+)\) to an element T belonging to \(H^\infty ({\mathbb {C}}_+)^*\) with the same norm. But we don’t know if it is possible to choose an extension T in \({G}^\infty ({\mathbb {C}}_+)\).