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It is known that the space $H^{\infty}$ of bounded holomorphic functions on the unit disk $D$ is non-separable with respect to the supremum norm $\|\cdot\|_{\infty}^{D}$. Let $E\subset D$ be connected and not a singleton. Then, $\|\cdot\|_{\infty}^{E}$ (the supremum norm over $E$) is in fact a norm.

What are geometric conditions on $E$ equivalent to $H^{\infty}$ being non-separable with respect to $\|\cdot\|_{\infty}^{E}$?

If $\overline{E}\subset D$, then polynomials are dense with respect to $\|\cdot\|_{\infty}^{E}$, and so in this case $H^{\infty}$ is separable with respect to $\|\cdot\|_{\infty}^{E}$. Hence, we need $\overline{E}$ to intersect the unit circle $C$.

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    $\begingroup$ Carleson's interpolation theorem gives a sufficient geometric condition. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 27, 2020 at 17:50
  • $\begingroup$ The proof I know to show the non separability of $H^\infty$ uses the functions $g_\xi (z)=\exp \left ((z+\xi)/(z-\xi) \right )$, $|\xi|=1$ . These are in $H^\infty$ and their mutual distance is bigger than 1; however this last is seen by taking radial (or non tangential limits). So we could consider the set $\Gamma$ of all points in $\partial D$ which are non tangential limits of points in $E$. If this set is uncountable, then $H^\infty$ is not separable under the norm $\|\cdot \|_\infty^E$. $\endgroup$
    – Giorgio Metafune
    Commented Jul 27, 2020 at 18:27
  • $\begingroup$ @GiorgioMetafune I thought the same thing, but i don't have a good sense of the non-tangential limits. Also, perhaps this condition is too strong, it's not clear to me if $[0,1)$ works for my purpose.. $\endgroup$
    – erz
    Commented Jul 27, 2020 at 18:33
  • $\begingroup$ You are right, the case of the interval is the first to be understood. $\endgroup$
    – Giorgio Metafune
    Commented Jul 27, 2020 at 19:01
  • $\begingroup$ @JochenWengenroth thank you for suggestion! It was enough to completely answer my question $\endgroup$
    – erz
    Commented Jul 28, 2020 at 8:29

1 Answer 1

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Jochen Wengenroth suggested to look at Carleson's interpolation theorem, and it seems like it completely answers my question. Namely, the following is true.

Let $E$ be a subset of $D$. Then $H^\infty$ is non-separable with respect to $\|\cdot\|_\infty^E$ if and only if $\overline{E}$ intersects the unit circle.

Necessity follows from the fact that if $\overline{E}\subset D$, the polynomials are dense in $H^\infty$ with respect to $\|\cdot\|_\infty^E$, and so $H^\infty$ is separable in this case.

To show sufficiency let us recall the following corollary from the Carleson's interpolation theorem (combine theorems 9.1 and 9.2 from the book Duren - Theory of Hp Spaces): Let $\{z_n\}_n$ be a sequence in $D$ such that there is $0<c<1$ such that $1-|z_{n+1}|\le c(1-|z_{n}|)$, for every $n$. Then $\{z_n\}_n$ is an interpolation sequence, i.e. the operator $T:H^\infty\to l^\infty$ defined by $Tf=\{f(z_n)\}_n$ is a surjection.

Now fix some $z_1\in E$ and $0<c<1$. If $z_1,...,z_n$ are already chosen then $F=\{z\in D, ~|z|\ge 1-c(1-|z_n|)\}$ intersects with $E$ as $\overline{E}$ intersects the unit circle. Choose $z_{n+1}\in F\cap E$.

The sequence $B=\{z_n\}_n$ constructed this way satisfies the condition above, and so the operator $T$ is surjective. Moreover, $\|\cdot\|_\infty^B\le\|\cdot\|_\infty^E$, and so $T$ is a continuous map from $(H^\infty, \|\cdot\|_\infty^E)$ onto a non-separable Banach space. Thus, $(H^\infty, \|\cdot\|_\infty^E)$ is non-separable.

Remark. Using this fact one can now show that $H^\infty(U)$ is non-separable for every bounded open connected set $U$. Indeed, $(H^\infty(V),\|\cdot\|_\infty^U)$ embeds isometrically into $H^\infty(U)$, where $V$ is the complement to the unbounded component of $\mathbb{C}\backslash U$, and the former is non-countable as we have shown.

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    $\begingroup$ I missed the connectedness assumption in your question. The Carleson interpolation theorem gives a sufficient condition also for disconnected sets. But then necessity seems to be difficult. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 28, 2020 at 9:33
  • $\begingroup$ I do not see the connectness assumption, too. If $E \subset D$ and its closure meets the boundary, you can always find a sequence $(z_n) \in E$ as above. Also, if $\bar E \subset D$, the density of polynomials follows without assuming connectness. $\endgroup$
    – Giorgio Metafune
    Commented Jul 28, 2020 at 10:37
  • $\begingroup$ @JochenWengenroth you are right. Thank you! $\endgroup$
    – erz
    Commented Jul 28, 2020 at 21:13
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    $\begingroup$ @GiorgioMetafune you are also right! This is what happens when one has an extra nice condition - gets blinded somehow $\endgroup$
    – erz
    Commented Jul 28, 2020 at 21:14

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