Why $V_{K_\epsilon}^*|_{K_\epsilon}\equiv 0$?

Question

Here are few corollaries I was reading in Pluripotential theory by M. Klimek. Here $V_K$ is the extremal function.

1. If $\{K_j\}$ are compact sets with $K_{j+1}\subset K_j$ and $K:=\cap_j K_j$, then $\lim_{j\rightarrow \infty} V_{K_j}=V_{K}$;
2. For $K$ compact, $V_{K}$ is lower semicontinuous;
3. For $K$ compact, if $V^*_{K}|_K \equiv 0$ then $V_{K}$ is continuous on $\mathbb{C}^d;$
4. For $K$ compact, $V_{K_\epsilon}$ is continuous and $\lim_{\epsilon \rightarrow 0} V_{K_\epsilon}=V_{K}$, where $K_\epsilon:=\{z: dist(z,K)\leq \epsilon\}$.

Klimek states proofs of all except the 4th one. Since $K_\epsilon$ is compact, we have $\lim_{\epsilon \rightarrow 0} V_{K_\epsilon}=V_{K}$ by 1st one. However, to prove that $V_{K_\epsilon}$ is continuous, we need to show that $V^*_{K_\epsilon}|_{K_\epsilon}\equiv 0$ so we can use 2nd one. But I am having trouble to show it. Could someone please help me to figure it out?

Answer 1

$\textbf{Propositision.}$ If $K$ is compact, then $V_{K}$ is lower semicontinuous in $\mathbb{C}^{d}$.

Proof. Fix $u \in L(K)$ and $\varepsilon>0$. By the compactness of $K$ one can find $j=j(\varepsilon)>0$ so small that $u_{j}=u*w_{j}\le\varepsilon$ on $K$. Hence $u_{j}-\varepsilon\in L(K)$ and so $u_{j}-\varepsilon\le V_{K}$ in $\mathbb{C}^{n}$. It follows that $V_{K}$ is an upper envelope of continuous functions $u*w_{j}-\varepsilon$, where $u \in L(K),\ \ \varepsilon>0,\ \ j=j(\varepsilon, u)>0$. Therefore $V_{K}$ is lower semicontinuous.

Answer 2

Define $u_{j}(z):=0-\frac{1}{j}.$ Note that $u_{j}(z)\in L(K_{\varepsilon})$. Consequently, $V^{*}(z,K_{\varepsilon})|_{K_{\varepsilon}}=\{\sup\{u(z): u(z)\in L(K_{\varepsilon})\},\ z\in K_{\varepsilon}\}^{*}=\{\sup\limits_{j}\{u_{j}(z),\ z\in K_{\varepsilon}\}^{*}\equiv 0$.