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I am interested in a vector-valued form of the disc "algebra" (which in this setting is not in general an algebra, hence the scare quotes). Let $E$ be a Banach space, and let $A(\mathbb D,E)$ be the space of bounded continuous functions $f:\overline{\mathbb D}\rightarrow E$ which are holomorphic on $\mathbb D$.

Has this space been systematically studied?

I am particularity interested in different continuity assumptions.

If $f$ is only weakly-continouous, i.e. $\mu\circ f\in A(\mathbb D)$ for each $\mu\in E^*$, is there a counter-example showing that $f$ need not be norm continuous?

If $E = F^*$ is a dual space, we can ask the same question with $\mu\in F$, i.e. consider weak$^*$-continuity.

There are of course different notions of Banach-space valued Holomorphic functions (weakly or weak$^*$-holomorphic) but it is well-known that these all agree and imply norm analytic. Thus it is continuity at the boundary which is of interest. Probably it is just my lack of knowledge about the classical disc algebra which is a problem, and so I'm hoping a few references might help me?

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  • $\begingroup$ My first instinct for Q1 is that there might be relevant stuff in the work of people like Richard Aron, but I haven't gone looking for specific mention of the space you're after $\endgroup$ – Yemon Choi Jul 30 '18 at 11:54
  • $\begingroup$ As for Q2: what are the starting assumptions on $F$ exactly? Do you mean that $f$ is weakly continuous $\overline{\mathbb D}\to E$ and holomorphic on the interior, and the question is whether this gets us norm continuity at the boundary? $\endgroup$ – Yemon Choi Jul 30 '18 at 12:01
  • $\begingroup$ @YemonChoi Yes, exactly that. $\endgroup$ – Matthew Daws Jul 30 '18 at 16:07
  • $\begingroup$ Vague thought/comment: I did wonder if something about your question(s) was related to the analytic Radon-Nikodym property but I haven't had time to check the definition and see if it applies. $\endgroup$ – Yemon Choi Aug 8 '18 at 2:52
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    $\begingroup$ Not sure if you are still interested, but I think the space that you define above is $A({\mathbb D}) \check{\otimes} E$ by a Fejer-kernel kind of approximation argument (plus the identification $C({\mathbb T},E) = C({\mathbb T}) \check{\otimes} E$) -- but I haven't sat down to check the details $\endgroup$ – Yemon Choi Sep 9 '18 at 1:09
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Edit 2018-08-08: Partial positive result added (Theorem 2).

First, here is a counterexample for the case of weak${}^*$-continuity:

Counterexample 1. Let $E = \ell^\infty := \ell^\infty(\mathbb{N}_0)$ and let $f: \overline{\mathbb{D}} \to E$ be given by \begin{align*} z \mapsto f(z) := (z^n)_{n \in \mathbb{N_0}}. \end{align*} For every $\mu \in E_* = \ell^1$ the function \begin{align*} \mu \circ f: z \mapsto \sum_{n=0}^\infty \mu_n z^n \end{align*} is in $A(\mathbb{D})$, i.e. $f$ is weakly${}^*$-continuous on $\overline{\mathbb{D}}$ and weakly${}^*$-holomorphic on $\mathbb{D}$. Hence, $f$ is actually holomorphic on $\mathbb{D}$ (since $f$ is bounded and $E_*$ is norming in $E^*$ or, more generally, since $f$ is bounded and $E_* \subseteq E^*$ separates $E$; see e.g. [Arendt, Nikolski: Vector-valued holomorphic functions revisisted (2000), Theorems 1.3 and 3.1]).

However, $f$ is not norm continuous at $z=1$ as we have $\|f(1) - f(z)\|_\infty \ge 1$ for every $z \in \mathbb{D}$.

Now, let us prove a positive (no pun intended) result.

Theorem 2. Let $E$ be a complex Banach lattice and let $f: \overline{\mathbb{D}} \to E$ be a function which is holomorphic on $\mathbb{D}$ and weakly continuous on $\overline{\mathbb{D}}$. If $f^{(k)}(0) \ge 0$ for each $k \in \mathbb{N}_0$, then $f$ is continuous on $\overline{\mathbb{D}}$.

Proof. We set $a_k := f^{(k)}(0)/k!$ for each $k \in \mathbb{N}_0$. For every $0 \le \mu \in E^*$ we have \begin{align*} \sum_{k=0}^\infty \langle \mu, a_k\rangle = \lim_{r \uparrow 1} \sum_{k=0}^\infty r^k \langle \mu, a_k\rangle = \lim_{r \uparrow 1} \langle \mu, \sum_{k=0}^\infty a_k r^k \rangle = \lim_{r \uparrow 1} \langle \mu, f(r)\rangle = \langle \mu, f(1) \rangle, \end{align*} where the first equality follows from the monotone convergence theorem. Hence, the increasing sequence $(\sum_{k=0}^n a_k)_{n \in \mathbb{N}_0}$ converges weakly to $f(1)$, and thus the sequence is even norm convergent to $f(1)$ by the Banach lattice version of Dini's theorem (see e.g. [Schaefer: Banach Lattices and Positive Operators (1974), Corollary to Theorem II.5.9]).

Moreover, the net $(f(r))_{r \in [0,1)}$ is increasing and weakly convergent to $f(1)$, so it is in fact norm convergent to $f(1)$, again by Dini's theorem.

The convergence of the sequence $(\sum_{k=0}^n a_k)_{n \in \mathbb{N}_0}$ implies that, for every complex number $\lambda$ of modulus $1$, the sequence $(\sum_{k=0}^n \lambda^k a_k )_{n \in \mathbb{N}_0}$ is a Cauchy-sequence (and thus convergent) for we have \begin{align*} \lvert \sum_{k=n_1+1}^{n_2} \lambda^k a_k \rvert \le \sum_{k=n_1+1}^{n_2} a_k \end{align*} whenever $0 \le n_1 \le n_2$.

Now we show that the set $f(\mathbb{D})$ is relatively compact in $E$. To this end, let $\varepsilon > 0$ and choose $r_0 \in (0,1)$ such that $\lVert f(1) - f(r)\rVert < \varepsilon$ whenever $r \ge r_0$. We show that every vector in $f(\mathbb{D})$ is closer than $2\varepsilon$ to the compact set $f(r_0 \overline{\mathbb{D}})$

Let $\lambda$ be a complex number of modulus $1$ and let $r \in [r_0,1)$. Then we have \begin{align*} \lvert \sum_{k=0}^\infty \lambda^k a_k - f(r\lambda) \rvert \le \sum_{k=0}^\infty (1 - r^k)a_k = f(1) - f(r) \end{align*} and thus \begin{align*} \lVert \sum_{k=0}^\infty \lambda^k a_k - f(r\lambda) \rVert \le \lVert f(1) - f(r) \rVert < \varepsilon. \end{align*} This implies that $\lVert f(r\lambda ) - f(r_0 \lambda) \rVert < 2\varepsilon$ for every $r \in [r_0,1)$ and every complex number $\lambda$ of modulus $1$.

Thus, every vector in $f(\mathbb{D})$ is indeed closer than $2\varepsilon$ to a vector in the compact set $f(r_0 \overline{\mathbb{D}})$, as claimed. This proves that $f(\mathbb{D})$ is totally bounded and thus relatively compact.

Now, let $z_0 \in \partial \mathbb{D}$. First consider a sequence $(z_n)$ in $\mathbb{D}$ which converges to $z_0$. Then every subsequence of $(f(z_n))$ has a norm convergent subsequence, and the limit of this latter subsequence equals $f(z_0)$ since $(f(z_n))$ converges weakly to $f(z_0)$ by assumption.

Hence, we conclude that $(f(z_n))$ is actually norm convergent to $f(z_0)$.

A simple approximation argument now shows that we also have $\lim_{n \to \infty} f(z_n) = f(z_0)$ for every sequence in $\overline{\mathbb{D}}$ which converges to $z_0$. The theorem is proved.

Remark 3. Theorem 2 shows in particular that the function $f$ from Counterexample 1 cannot be weakly continuous on $\overline{\mathbb{D}}$.

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    $\begingroup$ Jochen, you've shown that $f$ is point-to-weakstar continuous, but I thought Matt was aking about point-to-weak continuous? $\endgroup$ – Yemon Choi Aug 7 '18 at 22:42
  • $\begingroup$ This is nice! But yes I am mostly interested in point-weak continuity, as Yemon says. $\endgroup$ – Matthew Daws Aug 8 '18 at 0:52
  • $\begingroup$ @YemonChoi: Thanks, you're right of course. Corrected. I must have been somewhat delirious when I wrote the first version of the answer... :-( $\endgroup$ – Jochen Glueck Aug 8 '18 at 4:49
  • $\begingroup$ @MatthewDaws: Thanks for your response. I've added a sufficient condition for weak continuity to imply norm continuity (Theorem 2). $\endgroup$ – Jochen Glueck Aug 8 '18 at 19:27
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I believe the following provides a counter-example for "weak implies norm continuity" in the case $E=c_0$.

Given any $F:\overline{\mathbb D} \rightarrow c_0$ let $F_n:\overline{\mathbb D} \rightarrow \mathbb C$ be the $n$th coordinate. If $F$ is weakly continuous on $\overline{\mathbb D}$ and analytic on $\mathbb D$, then each $F_n$ is a member of $A(\mathbb D)$ and $\|F_n\|_\infty\leq K$ for some $K$ independent of $n$. Conversely, if these two conditions hold on the $F_n$ then for any $a=(a_n)\in\ell^1 = c_0^*$ we have that $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle}\ip{a}{F(z)} = \sum_n a_n F_n(z) $$ and so $z\mapsto \ip{a}{F(z)}$ is in $A(\mathbb D)$ as it is the absolutely convergent sum of members of $A(\mathbb D)$. By properties of functions in Hardy spaces (see e.g. page 50 of these notes) we know that $$ F_n(\xi) = \int_{\mathbb T} \frac{z F_n(z)}{z-\xi} \ dz \qquad (\xi\in\mathbb D). $$

We now seek to construct $F$ by constructing a suitable sequence $(F_n)$. That $F(z)\in c_0$ for all $z$ implies that $F_n(z)\rightarrow 0$ pointwise. Under the assumption that $\|F_n\|_\infty\leq K$ for all $n$, for fixed $\xi\in\mathbb D$ we can apply the dominated convergence theorem to the integral above to see that $F_n(z)\rightarrow 0$ pointwise for $z\in\mathbb T$ implies that $F_n(\xi)\rightarrow 0$.

We shall use Outer functions (page 35 in the notes linked above). Choose a positive function $h$ on $\mathbb T$ with $\log h$ being integrable; then we can form an outer function $F$ with $|F|$ having nontangential limit $h$ at the boundary, almost everywhere. It is not clear to me that just because $h$ is continuous it will follow that $F$ is in $A(\mathbb D)$ instead of just a Hardy space; so we work a bit harder.

Firstly, choose $t\mapsto h_n(e^{it})$ to be the piecewise linear function with $h_n(1) = h_n(e^{i2\pi/n})=1/n$ and $h_n(e^{i\pi/n})=1$. Let $F_n$ be the Outer function with $|F_n|$ having nontangential limit $h_n$ on the boundary, almost everywhere. We can find $r_n<1$ such that if we define $G_n(z) = F_n(r_nz)$ for $z\in\overline{\mathbb D}$ then there will be $a_n, b_n\in\mathbb T$ with $|a_n-1|<\frac1n, |b_n-e^{i\pi/n}|<\frac1n, ||G_n(a_n)|-\frac1n|<\frac1n$ and $||G_n(b_n)|-1|<\frac1n$. Certainly $G_n\in A(\mathbb D)$ with $\|G_n\|\leq 1$.

Finally, form $G\in A(\mathbb D, c_0)$ using the sequence $(G_n)$, and observe that $$ \| G(b_n) - G(a_n) \|_{\infty} \geq | G_n(b_n) - G_n(a_n) | \geq 1 - 3/n $$ so as $a_n \rightarrow 1$ and $b_n\rightarrow 1$, we conclude that $G$ cannot be norm continuous.

Alternative construction (from a hint from Yemon Choi): Simply directly define $$ F_n(z) = \exp\big(k_n(e^{-i\pi/n}z-1)\big). $$ where $(k_n)$ is a rapidly increasing sequence. Then obviously $F_n$ is in the disc algebra. We have that $$ |F_n(e^{it})| = \exp\big( k_n(\cos(t-\pi/n)-1) \big). $$ For $t\not=0$ this obviously converges to $0$; we chose $(k_n)$ increasing fast enough so that $\exp( k_n(\cos(-\pi/n)-1) )\rightarrow 0$. Thus $F_n(z)\rightarrow 0$ pointwise on $\mathbb T$, and so on $\mathbb D$, as before. The remainder of the argument also runs the same way, as $F_n(1)\rightarrow 0$ while $F_n(e^{i\pi/n}) = 1$.

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  • $\begingroup$ I'll leave this here, but I am currently not 100% sure that outer functions, with continuous boundary values, are in $A(\mathbb D)$. $\endgroup$ – Matthew Daws Aug 9 '18 at 12:52
  • $\begingroup$ Hmm, for outer functions I think you can prescribe the modulus of $F$ on the boundary a.e. but not its argument. Otherwise you'd be saying something a bit odd about harmonic extensions to the disc of prescribed functions on the boundary $\endgroup$ – Yemon Choi Aug 9 '18 at 13:19
  • $\begingroup$ @YemonChoi Yes indeed, my mistake. This would not change the argument, but it does cast further doubt on my fears about whether Outer functions, whose radial limits are in, absolute value, continuous functions, are in $A(\mathbb D)$. $\endgroup$ – Matthew Daws Aug 9 '18 at 13:39
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    $\begingroup$ I think you could adapt your idea to cook up $F_n$ by hand which are definitely in the disc algebra and which converge pointwise to zero on the closed disc. You just need to arrange that $e^{\pi i n}$ is a peak point for $F_n$, so I'd take the usual kinds of function which peak at 1, say $\exp n(z-1)$, and then rotate a bit. Does this work? $\endgroup$ – Yemon Choi Aug 9 '18 at 13:45
  • $\begingroup$ @MatthewDaws: I'm somewhat confused by the "Cauchy type" integral formula for $F_n$ that you quoted from the linked notes. If, say, $F_n$ is constantly $1$ and $\xi= 0$, then $F_n(\xi) = 1$ but the integral vanishes. $\endgroup$ – Jochen Glueck Aug 9 '18 at 16:51

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