0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Do you know Blaschke products? $\endgroup$ Aug 18 '15 at 18:50
  • $\begingroup$ Yes, some reference? $\endgroup$
    – felipeuni
    Aug 18 '15 at 18:53
  • $\begingroup$ If you have Rudin's Real and Complex Analysis, looking at theorem 15.21 would be a spoiler. $\endgroup$ Aug 18 '15 at 18:55
1
$\begingroup$

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 $w_n.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.