If we know that the unit ball $\mathbb B_n\subset \mathbb C^n$ is a domain of holomorphy. Can we use this fact to conclude the fact that very convex domain in $\mathbb C^n$ is a domain of holomorphy?
Convex domains are domains of holomorphy
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Ball is not the right domain here. A convex set is not really like a ball. For easy proofs here you want domains that contain your domain, and a convex domain need not be contained in any ball.
One way to do it though would be to think of the halfspace (place where a real affine function is negative) as a ball of infinite radius. Then once you know that half space is a domain of holomorphy, which it is, the convex set is an intersection of half spaces, it is therefore a domain of holomorphy.
Although underneath it is really the standard proof of the fact that a convex set is a domain of holomorphy. Really it all boils down to showing that a half space is a domain of holomorphy. It is the same proof dressed up in different clothing.
I don't know a simple way you can you can use the fact that the ball is convex and a domain of holomorphy to show all convex sets are domains of holomorphy - I think you need a little more machinery first. We can use the following fundamental equivalence to show convex sets are domains of holomorphy, though. First, we need to define the holomorphic hull. Let $U \subset \mathbb{C}^n$ be a domain, and let $K$ be some relatively compact subset of $U$. Define the holomorphic hull of $U$ to be the set
$$\hat K_{U} :=\{z \in U: |f(z)| \leq \sup_{z \in K} |f|\,\, \mbox{for all holomorphic functions $f$ on U}\}.$$
A good reference for the proof of this fact can be found in chapter 2 of Hormander's "Several Complex Variables."
Using this characterization, you should check that the holomorphic hull $K$ is contained inside of the convex hull of $K$. It follows immediately that convex domains will be domains of holomorphy.