# $\Delta:$ $H^{\infty}(\Delta)$- domain of holomorphy

How to prove that $\Delta$ is an $H^{\infty}(\Delta)$- domain of holomorphy (i.e. there is a bounded holomorphic function $f\in H(\Delta)$ such that $\partial \Delta$ is the natural boundary).

$\Delta$: unit disc in $\mathbb{C}$.

Any help would be appreciated.

• Do you know Blaschke products? Aug 18 '15 at 18:50
• Yes, some reference? Aug 18 '15 at 18:53
• If you have Rudin's Real and Complex Analysis, looking at theorem 15.21 would be a spoiler. Aug 18 '15 at 18:55

Take a dense subset $\{w_1,w_2, \dots \}$ of the unit circle and let $\mu$ be the sum of point masses on the circle that puts mass $1/2^n$ at $w_n.$ Then let $u= P[\mu],$ where $P$ denotes the Poisson integral. At each $w_n,$ there is $c_n > 0$ such that $u(rw_n) \ge c_n/(1-r)$ as $r \to 1^-.$ (That is quite simple to verify.)
Now let $v$ be any harmonic conjugate of $u$ in the disc, and set $f = \exp [-(u+iv)].$ Then $f\in H^\infty.$ This $f$ cannot extend holomorphically to any larger open set containing a point on the circle, because such an open set will contain some $w_n,$ and $|f(rw_n)| \le e^{-c_n/(1-r)},$ which $\to 0$ much too fast for a function homomorphic in a neighborhood of $w_n.$