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 $V$.

  • $\begingroup$ Upper semicontinuous functions take their values in $[-\infty,\infty)$, not $(-\infty,\infty]$... $\endgroup$
    – user111
    Commented Apr 4, 2021 at 13:11
  • $\begingroup$ @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.) $\endgroup$
    – mrf
    Commented Apr 5, 2021 at 21:24

1 Answer 1


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.


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