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Let $U\subset\mathbb{C}^n$ be a domain of holomorphy.

The following proposition is true?

There is a holomorphic function $f\in H(U)$ such that for all $a\in\partial U$, $\lim_{z\to a}f(z)=\infty$ $(z\in U)$.

Any hint would be appreciated.

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  • $\begingroup$ You want a single $f\in H(U)$ to blow up at each boundary point? $\endgroup$
    – zhw.
    Jul 13 '15 at 3:01
  • $\begingroup$ Yes, this is possible? $\endgroup$
    – felipeuni
    Jul 13 '15 at 3:57
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This can't even be done in $\mathbb {C}^1.$ Take the open unit disc $\mathbb {D}.$ Then there is no $f\in H(\mathbb {D})$ such that the radial limit of $f$ is $\infty$ at every point of the boundary, or even at every point of an open arc on the boundary. This is a consequence of Baire's theorem: If there were such an $f,$ then by Baire the radial limit of $\infty$ would have to occur on some open arc at a uniform rate. Then nearby that arc, $1/f$ would be holomorphic and extend continuously to the arc, with boundary values $0$ on the arc. That can't happen, contradiction.

Now you could ask for a holomorphic $f$ that is unbounded in $U\cap B(a,r)$ for each $a \in \partial U$ and each $r>0.$ This can be done in one complex variable. Not sure about several variables.

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  • $\begingroup$ But $f(z)=\sum_{n=1}^{\infty}z^{2^n}$ has radial limit $\infty$ at every point of $\partial \mathbb{D}$. $\endgroup$
    – felipeuni
    Jul 16 '15 at 16:44
  • $\begingroup$ And why do you think that? $\endgroup$
    – zhw.
    Jul 16 '15 at 17:21
  • $\begingroup$ Correction $f(z)=\sum_{n=1}^{\infty}z^{2^n}$ has radial limit $\infty$ in a dense set $E$, $\overline{E}=\partial\mathbb{D}$. $\endgroup$
    – felipeuni
    Jul 18 '15 at 17:56

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