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.



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



Supremum over which sets makes $H^{\infty}$ non-separable?
Jochen Wengenroth suggested to look at Carleson's interpolation theorem, and it seems like it completely answers my question. Namely, the following is true.



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> 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.
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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.


