The Euclidean ball is a domain of holomorphy

Question

How can one show that the Euclidean ball $B(0,R)$ with radius $R$ and center $0$ in $\mathbb C^n$ is a domain of holomorphy by finding a homomorphic function inside the ball which can't be extended over the boundary?

Answer 1

Hint: You do not need to find a single function which cannot be extended beyond every boundary point. The definition of domain of holomorphy only requires that for each point of the boundary, you can find a holomorphic function which does not extend beyond that point. It should be fairly easy to come up with a specific example for the Euclidean ball.

It is true that you can find a single function which works for all points on the boundary, but this is harder to prove.

Answer 2

Here's a way to do it with a ball (function not extending through any point of the boundary). At least an outline, which leads to a standard technique. Notice that $\frac{c}{Az - b}$ for a matrix $A$ and vector $b$ is really big near the line where $Az = b$. In particular notice the set where $\frac{|c|}{|Az-b|} = 1$. And now consider powers of this function. Construct a sequence $p_j \in B(0,R)$ going towards the boundary, with the limit set being the entire sphere. There should also exist a sequence of positive real numbers $r_j$, with $\|p_{j-1} \| < r_j < \|p_j\|$ (so $r_j \to R$). Next find a sequence of functions (using the idea above) $f_j$ holomorphic in $B(0,R)$, and such that $|f_j(z)| \leq 2^{-j}$ for $z \in B(0,r_j)$ and $\sum_{j=1}^k f_j(p_k) = k$. Then the function you want is $$ \phi(z) = \sum_{j=1}^\infty f_j(z) $$ Show that $\phi$ converges uniformly on compacta, and so is holomorphic, and further prove that $|\phi(p_j)| \to \infty$.

It is easy to generalize the above argument to any convex domain. Also, the argument leads to a definition of "holomorphic convexity", which can be proved (difficult) to be equivalent to pseudoconvexity.