Bochner's tube theorem

Remarks

Uniqueness holds because $T_{\operatorname{ch}(B)}$ is connected and contains the nonempty open set $T_B$. Classical proofs push the function across the convex hull with Cauchy integrals over families of analytic discs, in the spirit of the continuity principle. Another route uses that a tube which is a domain of holomorphy has a convex base. On such a tube, $-\log$ of the distance to the boundary is plurisubharmonic. It does not depend on $y$, so it is a convex function of $x$.

Example: let $B = \set{x \in \R^2 : 1 < \abs{x} < 2}$, an annulus. Its convex hull is the disc $\abs{x} < 2$, so every holomorphic function on $T_B$ extends to the tube over the disc.

Consequently, a tube $T_B$ with connected open base is a domain of holomorphy exactly when $B$ is convex. For $n = 1$ the theorem is empty, since connected open subsets of $\R$ are intervals and already convex.

Going deeper

For a student

Tube domains appear in Fourier and Laplace analysis: the Laplace transform $z \mapsto \int e^{z_1 t_1 + \cdots + z_n t_n} \, d\mu(t)$ of a measure $\mu$ on $\R^n$ is holomorphic on a tube whose base is convex. The theorem explains why convex bases are the natural ones. It is a basic example of the computation of an envelope of holomorphy: the envelope of $T_B$ is $T_{\operatorname{ch}(B)}$.