All boundary points of the unit ball in $\mathbb C^n$ are local peak points 
Claim: All boundary points of the unit ball in $\mathbb C^n$ are local peak points, i.e. $\forall p\in \partial B\ \exists N$ ball centered in $p$, $h\in \mathcal O(N)$ (holomorphic functions in $\mathbb C$) such that $h( p)=1, |h(z)|<1$ for all $z\in \overline{B\cap N}\setminus\{p\}$.

It should be true for all convex sets: there should exist a function $f\in\mathcal O(N)$ s.t. $f( p)=0, Re\ f(z)>0$ for all $z\in \overline{B\cap N}\setminus\{p\}$ for some $N$ centred in $p$. Then pick $$h(z)=\frac{1}{1+f}$$
Can this work?
The only point I'm not really sure is the existence of such $f$. Am I right about it?
 A: No, not every boundary point of a convex domain is a peak point. Consider the polydisk $\mathbb{D}^n \subset \mathbb{C}^n$. If $f$ is holomorphic on the closure of $\mathbb{D}^n$, the maximum principle implies that the maximum of $|f|$ on any "face" like $\{\zeta\}\times \mathbb{D}^{n-1}$, $|\zeta|=1$, is attained on the relative boundary of that face, i.e, on $\{\zeta\}\times \partial (\mathbb{D}^{n-1})$. Repeating this process, one concludes that the maximum of $|f|$ is attained somewhere on the torus $\mathbb{T}^n$. By a limiting argument, the assumption on $f$ can be relaxed to "holomorphic on \mathbb{D}^n$ and continuous on its closure". 
The above shows that a point of the boundary of $\mathbb{D}^n$ that does not lie on $\mathbb{T}^n$ is not a peak point. For a concrete example, $(1,0)\in \partial \mathbb{D}^2$ is not a peak point. 
In technical terms, the Shilov boundary of a polydisk $\mathbb D^n$ is the torus $\mathbb T^n$ rather than the topological boundary $\partial \mathbb D^n$. Every peak point lies on the Shilov boundary. (For a discussion of the converse, see Bergman-Shilov Boundary and Peak Points.)

There are other notions of convexity, adapted to several complex variables, which help with  peak points more. I won't go into them here; see Hörmander's book Notions of Convexity.
