# Restricted Perron-Bremermann envelopes

Consider an upper semicontinuous function $$\phi: \Omega \to (-\infty, \infty]$$, in the sense that $$\phi = \phi^*$$, where $$\phi^*$$ denotes the upper semicontinuous regularization $$\phi^*(z) = \varlimsup_{\zeta \to z} \phi(\zeta)$$ and $$\Omega$$ is some domain in $$\mathbb{C}^n$$, let's say strictly pseudoconvex if it helps.

Forming the usual Perron-Bremermann envelope of $$\phi$$, i.e. $$U(z) = \sup\{ u(z) : u \in \mathcal{PSH}(\Omega), u \le \phi \},$$ it is clear (assuming that $$U$$ is locally bounded) that $$U^* \le \phi^* = \phi$$, so $$U^*$$ is a member in the defining family for $$U$$, and thus $$U \ge U^*$$ and consequently, $$U = U^*$$. In particular, $$U$$ is already upper semi-continuous, and thus plurisubharmonic without having to take the upper semicontinuous regularization.

Now, let us consider the following variation, where we take the envelope only using upper bounded functions: $$V(z) = \sup\{ u(z) : u \in \mathcal{PSH}(\Omega), u \le \phi, \sup_\Omega u < +\infty \}.$$ Note that in general, $$V < U$$ (take for example $$\phi$$ as the Poisson kernel with a pole at $$z=1$$ on the unit disc: then $$U = \phi$$, but $$V = 0$$).

Question: Is it true that $$V$$ is upper semicontinuous and thus plurisubharmonic? The problem being, of course, that $$V$$ is in general not upper bounded, so it doesn't follow immediately that $$V^*$$ is a member of the defining family for $$V$$.

• Upper semicontinuous functions take their values in $[-\infty,\infty)$, not $(-\infty,\infty]$... Commented Apr 4, 2021 at 13:11
• @user111 That's the most common convention, for sure, but I specifically want to allow things like $\phi = -\log|f|$ where $f$ is holomorphic. (Allowing $\phi$ to take the value $-\infty$ is ok.)
– mrf
Commented Apr 5, 2021 at 21:24

It is not true, even for strictly pseudoconvex domains. Consider for $$1>a>0$$ the function $$u_a(z,w) = (-\log(1-|w|^2))^a+ \max\{-1, \sum_{k=1}^{\infty} \frac{1}{2^k}\log|z-\frac{1}{2^k}|\}$$ defined on the unit ball $$B$$ in $$\mathbb{C}^2$$, and denote its restriction to $$\partial B$$ by $$\phi$$. Then $$\phi: \partial B \rightarrow [-1, \infty]$$ with $$\phi^* = \phi$$, and there exists a harmonic extension $$h_\phi: B \rightarrow [-1, \infty]$$ since $$u_a$$ is dominated by the plurisuperharmonic function $$v_b(z,w) = (-\log(|z|^2))^b,$$ with $$1>b > a$$. With $$h_\phi$$ as dominating function, we have $$V\leq u_a$$ by the comparison principle (this requires a little bit of work), but $$u_a - \epsilon v_b \leq V$$ for all $$\epsilon >0$$. Hence $$V=u_a$$ outside $$\{z=0\}$$, and in particular, $$V$$ is not continuous on $$\Omega$$.
This constitutes a counterexample since on strictly pseudoconvex domains with $$h_\phi$$ lower semicontinuous on $$\bar \Omega$$, $$V$$ is always lower semicontinuous, so upper semicontinuity would imply that $$V$$ above is continuous. To see that $$V$$ is lower semicontinuous, note that we for any element $$u$$ in the defining family may associate an upper semicontinuous function $$u^*\mid_{\partial \Omega}$$ on the boundary. By Katětov–Tong insertion, there exists a continuous function on the boundary between $$u^*\mid_{\partial \Omega}$$ and $$\phi$$, which one may extend to a maximal, continuous plurisubharmonic function $$\tilde u$$ such that $$u \leq \tilde u \leq h_\phi$$. This shows that $$V$$ is lower continuous, as it may be written as an envelope of a family of continuous functions.