I need to show that the each boundary point of a convex domain $\Omega$ is regular where a point $x_0$ on the boundary is said to be regular if there exist a barrier function at $x_0$. And by a barrier function at $x_0$ we mean that a superharmonic function $u$ such that $u>0$ in $\bar{\Omega}\setminus{\{x_0\}}$ and $u(x_0)=0$.
Any type of help will be appreciated. Thanks in advance.
Generally, each boundary point $z_0$ of a convex set in $\Bbb R^n$ has a supporting hyperplane containing $z_0$. For a convex set $\Omega \subset \Bbb C$ this means that $\Omega$ is contained in some half-plane $H$ with $z_0 \in \partial H$.
So by translating/rotating $\Omega$ we can assume that it is contained in the right half-plane and $z_0 = 0$ is the boundary point in question.
Now define $u(z) = \operatorname{Re}(\sqrt z)$ where $\sqrt z$ is the principal branch of the square root which maps the right half-plane onto the sector $\{ -\pi/4 < \arg z < \pi/4 \}$. Then $u$ is harmonic in $\Omega$ with $\lim_{z \to 0} u(z) = 0$ and $\lim_{z \to \zeta} u(z) = u(\zeta) > 0$ for all boundary points $\zeta \ne 0$.