Bergman-Shilov Boundary and Peak Points

Question

Let $\Omega$ be a domain in $\mathbb{C}^n.$ Consider the Banach algebra $A(\Omega):=\mathcal{C}({\overline{\Omega}})\cap\mathcal{O}(\Omega).$ Denote the Bergman-Shilov boundary of $A(\Omega)$ by $\partial_S(\Omega)$ . From the very definiton of peak points we know that every peak point belongs to $\partial_S(\Omega).$ Now the examples that I know the set of peak points coincides with the $\partial_S(\Omega).$ My question here is if there are domains for which the set of peak points are properly contained in $\partial_S(\Omega)$?

Answer 1

You ask difficult questions, and it's been a while since I've thought about peak points. I'm not sure if this is open or not, at least for smoothly bounded domains. Replacing $A(\Omega)$ with $A^1(\Omega) = C^1(\bar\Omega) \cap \mathcal{O}(\Omega)$ however, there are known examples.

Note that the Shilov boundary is closed by definition. Also, for a smoothly bounded pseudoconvex domain $\Omega$, it's well known that every strictly pseudoconvex point is a peak point for $A(\Omega)$ (in fact for $A^\infty(\Omega)$).

On the other hand, Kohn and Nirenberg (1973) constructed an example of a bounded pseduconvex domain $\Omega$ in $\mathbb{C}^2$ such that $\partial\Omega$ is strictly pseudoconvex at every point expect a point $P$ that doesn't admit a peak function that is holomorphic on a neighbourhood of $P$. This was later sharpened by several people to produce a smoothly bounded pseudoconvex domain with strictly pseudoconvex boundary except at one single point, admitting no peak function in $A^1(\Omega)$ at the bad point. Hence the Shilov boundary is the entire boundary, but the set of peak points for $A^1$ is a proper subset.

You should read Noell, Alan: "Peak points for pseudoconvex domains: a survey." J. Geom. Anal. 18 (2008), no. 4, 1058–1087 if you haven't already.

I'll think a bit about the non-smooth case and get back to you if I come up with something.

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