Show that $\widetilde{u} \in PSH(\Omega)$

Question

EDITED

I'm reading the book:

(Oxford science publications._ London Mathematical Society monographs, new ser., no. 6) Maciej Klimek -Pluripotential theory -OUP (1992).

• I don't understand proof for Theorem 2.7.1! Can anyone explain it?

• Example:

1/ Why $\boxed{(\sup_{\epsilon >0}u_\epsilon)^*=\widetilde{u}}$?

2/ Why $(\sup_{\epsilon >0}u_\epsilon)=u \in \Omega \setminus F$.

3/ Why can we suppose that $v<0$ in $\Omega$.

Any help will be appreciated. Thanks!

Answer 1

First note that $$\sup_{\varepsilon > 0} u_\varepsilon = \begin{cases} u, & \text{on $\Omega \setminus F$} \\ -\infty, & \text{on $F$}\end{cases} $$

Since $u$ is subharmonic (and hence upper semicontinuous), the upper semicontinuous regularization, $(\sup_{\varepsilon > 0} u_\varepsilon)^* = u$ on (the open set) $\Omega \setminus F$. For points in $F$, it's just the definition of the usc regularization $(\cdot)^*$.