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I came across this theorem:

Every convex domain $D$ in $\mathbb C^n$ is a weak domain of holomorphy.

But I couldn't understand some points on the following proof:

Let $p\in \partial D$. Since $D$ is convex, one can find an $\mathbb R$-linear function $l=l_p:\mathbb C^n\to \mathbb R$ such that the hyperplane $\{z:l(z)=l(p)\}$ separates $D$ and $p$, i.e., we may assume $l(z)<l(p)$ for $z\in D$. We can write $l(z)=\sum_{j=1}^n \alpha_jz_j+\sum_{j=1}^n \beta_j\bar z_j$, with $\alpha,\beta\in \mathbb C$, and since $l$ is real valued, one must have $\beta_j=\bar\alpha_j$. Hence $l(z)=Re \ h(z)$, where $h(z)=2\sum_{j=1}^n \alpha_jz_j$ is complex linear. it follows that $f_p=[h-h(p)]^{-1}$ is holomorphic on $D$ and completely singular at $p$.

Now my questions, why convexity implies this linear map? And why $f_p=[h-h(p)]^{-1}$ can't be extended to $p$? I mean why it is not like in Hartogs domain?

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The existence of $l$ comes from the supporting hyperplane theorem, https://en.wikipedia.org/wiki/Supporting_hyperplane, and a real hyperplane is defined by a real linear map, call it $l$.

Now as to why $f_p$ cannot be extended: WLOG after translation and rotation we can assume that $h(z) = z_1$ and $p=0$. If $f_p$ would extend, the extension would be holomorphic in a neighbourhood $U$ of the origin, and would equal $f_p$ on the intersection $U \cap D$. There is no function defined in any neighbourhood of the origin that can equal $\frac{1}{z_1}$ arbitrarily close to the origin (that is at points of $U \cap D$, as the modulus of that expression must go to infinity as $z_1$ tends to zero no matter how you approach $p=0$.

You should notice that in the Hartogs domain there is no way to fit a zero set of a holomorphic function (what in our case is $h$) inside the "hole" that is to be filled.

There is a more general concept here. If you can find a zero set of a holomorphic function defined on a larger domain which touches your boundary, then that's a point through with you cannot extend. Although do note that the converse is not true, you might have such a point but no such $h$.

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  • $\begingroup$ but i dont understand why it is necessary to find $l$? why it is not enough to consider the function $h(z)=\frac 1z_1$ directly? $\endgroup$
    – Ronald
    Aug 27 '16 at 10:23
  • $\begingroup$ But how do you know $\frac{1}{z_1}$ will work. You have to find a complex hyperplane that is outside your domain. Convexity gives you the $l$, not the $h$. Only after you know the $l$ to you know what transmission and rotation to do. The zero set of h, is a complex hypersurface inside the supporting hyperplane. $\endgroup$
    – Jiri Lebl
    Aug 28 '16 at 4:24

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