I know that in general for $u,v\in PSH$ (plurisubharmonic) $\min\{u,v\}$ is not a $PSH$ function. Are there any known results under which conditions on $u$,$v$ a function $\min\{u,v\}$ is $PSH$? I know about the Kiselman's minimum principle, and some work of Poletsky on the disc envelope...

The problem that I have is the following: Suppose $D\subset\mathbb{C}^n$ is a Stein domain and $f$, $g$ are continuous on $\overline{D}$ and holomorphic on $D$.

Is it true that for $\Phi(z)=\min\{ ||f(z)||, ||g(z)||\}$, we have $\Phi|_D\lneq\sup_D\Phi$? (i.e. is global maximum attained only on $\partial D$) (under what conditions)

If you have any good reference I will be more than happy to look at it. Thank you

  • $\begingroup$ It suffices that min {u,v} = u everywhere. More seriously (it is easier to visualize the real analogue of convex functions), if $u$ and $v$ are distinct linear functions on the real line and $x\in\mathbf R$ is such that $u(x)=v(x)$, then $\min\{u,v\}$ is concave and is not convex. $\endgroup$
    – ACL
    Sep 22, 2014 at 16:26
  • $\begingroup$ Yes thank you I am familiar with that. I've just seen that I've asked the same question twice. The edited version can be seen above. Sorry for inconvenience. $\endgroup$ Sep 22, 2014 at 18:00
  • $\begingroup$ In the real case, take $D=\mathopen]-1;1\mathclose[$, $u(x)=x$ and $v(x)=-x$. One has $\Phi(x)=-|x|$, $\Phi|_{\partial D}=-1$ hence $\Phi(x)>\sup_{\partial D}\Phi$ for every $x\in D$. $\endgroup$
    – ACL
    Sep 23, 2014 at 12:43

1 Answer 1


Here is a counterexample to your statement for $n=1$. Let $D=\{ x+iy:|x|<1,|y|<\pi/2\}$. Functions $u_1(x,y)=e^x\cos y$ and $u_2(x,y)=e^{-x}\cos y$ are harmonic. Their minimum is $e^{-|x|}\cos y$ which is equal to $1$ at $0$, and the maximum on $\partial D$ is $e^{-1}<1$. It remains to notice that $\exp(u_1)=|\exp\exp z|$ and $\exp(u_2)=|\exp\exp(-z)|,$ so for $f(z)=\exp\exp(z)$ and $g(z)=\exp\exp(-z)$ your statement does not hold.


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