Upper envelope of plurisubharmonic functions
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Suppose that ${u_alpha}_{alpha in A}$ is a family of plurisubharmonic functions (psh function) on $Omega subsetsubset mathbb{C}^n$. Then, let $u(z) = sup_{alpha in A} u_{alpha}(z)$ be the upper envelope.
I found in lots of literature that
$$u^ast(z) := lim_{epsilon rightarrow 0} sup_{B_{epsilon}(z)} u(z)$$ (usc regularization)
is still a psh fucntion.
Is there any counterexample that the upper envelope $u$ for psh functions $u_alpha$ before doing usc regulatization is not psh, I wonder?
real-analysis complex-analysis reference-request several-complex-variables potential-theory
asked Nov 26 at 6:03
Pan
606
add a comment |
up vote
1
down vote
favorite
Suppose that ${u_alpha}_{alpha in A}$ is a family of plurisubharmonic functions (psh function) on $Omega subsetsubset mathbb{C}^n$. Then, let $u(z) = sup_{alpha in A} u_{alpha}(z)$ be the upper envelope.
I found in lots of literature that
$$u^ast(z) := lim_{epsilon rightarrow 0} sup_{B_{epsilon}(z)} u(z)$$ (usc regularization)
is still a psh fucntion.
Is there any counterexample that the upper envelope $u$ for psh functions $u_alpha$ before doing usc regulatization is not psh, I wonder?
real-analysis complex-analysis reference-request several-complex-variables potential-theory
asked Nov 26 at 6:03
Pan
606
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that ${u_alpha}_{alpha in A}$ is a family of plurisubharmonic functions (psh function) on $Omega subsetsubset mathbb{C}^n$. Then, let $u(z) = sup_{alpha in A} u_{alpha}(z)$ be the upper envelope.
I found in lots of literature that
$$u^ast(z) := lim_{epsilon rightarrow 0} sup_{B_{epsilon}(z)} u(z)$$ (usc regularization)
is still a psh fucntion.
Is there any counterexample that the upper envelope $u$ for psh functions $u_alpha$ before doing usc regulatization is not psh, I wonder?
real-analysis complex-analysis reference-request several-complex-variables potential-theory
asked Nov 26 at 6:03
Pan
606
Suppose that ${u_alpha}_{alpha in A}$ is a family of plurisubharmonic functions (psh function) on $Omega subsetsubset mathbb{C}^n$. Then, let $u(z) = sup_{alpha in A} u_{alpha}(z)$ be the upper envelope.
I found in lots of literature that
$$u^ast(z) := lim_{epsilon rightarrow 0} sup_{B_{epsilon}(z)} u(z)$$ (usc regularization)
is still a psh fucntion.
Is there any counterexample that the upper envelope $u$ for psh functions $u_alpha$ before doing usc regulatization is not psh, I wonder?
real-analysis complex-analysis reference-request several-complex-variables potential-theory
real-analysis complex-analysis reference-request several-complex-variables potential-theory
asked Nov 26 at 6:03
Pan
606
asked Nov 26 at 6:03
Pan
606
edited Nov 26 at 7:27
asked Nov 26 at 6:03
Pan
606
asked Nov 26 at 6:03
Pan
606
asked Nov 26 at 6:03
Pan
606
606
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
Here is an example that works already for $n=1$. Let $Omega$ be the unit disc and let
$$
u_n(z) = frac1n log|z|.
$$
Then
$$
u(z) = sup_n u_n(z) = begin{cases} 0, & z neq 0 \ -infty, & z = 0, end{cases}
$$
which is not upper semicontinuous, and therefore not (pluri-)subharmonic.
It may also be helpful to know that $u$ is "almost" psh already before the usc regularization; assuming that the family ${ u_alpha }$ is locally upper bounded, then $u = u^*$ outside a pluripolar set, so in particular outside a set of Lebesgue measure $0$.
answered Nov 26 at 10:22
mrf
37.3k54685
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here is an example that works already for $n=1$. Let $Omega$ be the unit disc and let
$$
u_n(z) = frac1n log|z|.
$$
Then
$$
u(z) = sup_n u_n(z) = begin{cases} 0, & z neq 0 \ -infty, & z = 0, end{cases}
$$
which is not upper semicontinuous, and therefore not (pluri-)subharmonic.
It may also be helpful to know that $u$ is "almost" psh already before the usc regularization; assuming that the family ${ u_alpha }$ is locally upper bounded, then $u = u^*$ outside a pluripolar set, so in particular outside a set of Lebesgue measure $0$.
answered Nov 26 at 10:22
mrf
37.3k54685
add a comment |
up vote
1
down vote
accepted
Here is an example that works already for $n=1$. Let $Omega$ be the unit disc and let
$$
u_n(z) = frac1n log|z|.
$$
Then
$$
u(z) = sup_n u_n(z) = begin{cases} 0, & z neq 0 \ -infty, & z = 0, end{cases}
$$
which is not upper semicontinuous, and therefore not (pluri-)subharmonic.
It may also be helpful to know that $u$ is "almost" psh already before the usc regularization; assuming that the family ${ u_alpha }$ is locally upper bounded, then $u = u^*$ outside a pluripolar set, so in particular outside a set of Lebesgue measure $0$.
answered Nov 26 at 10:22
mrf
37.3k54685
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here is an example that works already for $n=1$. Let $Omega$ be the unit disc and let
$$
u_n(z) = frac1n log|z|.
$$
Then
$$
u(z) = sup_n u_n(z) = begin{cases} 0, & z neq 0 \ -infty, & z = 0, end{cases}
$$
which is not upper semicontinuous, and therefore not (pluri-)subharmonic.
It may also be helpful to know that $u$ is "almost" psh already before the usc regularization; assuming that the family ${ u_alpha }$ is locally upper bounded, then $u = u^*$ outside a pluripolar set, so in particular outside a set of Lebesgue measure $0$.
answered Nov 26 at 10:22
mrf
37.3k54685
Here is an example that works already for $n=1$. Let $Omega$ be the unit disc and let
$$
u_n(z) = frac1n log|z|.
$$
Then
$$
u(z) = sup_n u_n(z) = begin{cases} 0, & z neq 0 \ -infty, & z = 0, end{cases}
$$
which is not upper semicontinuous, and therefore not (pluri-)subharmonic.
It may also be helpful to know that $u$ is "almost" psh already before the usc regularization; assuming that the family ${ u_alpha }$ is locally upper bounded, then $u = u^*$ outside a pluripolar set, so in particular outside a set of Lebesgue measure $0$.
answered Nov 26 at 10:22
mrf
37.3k54685
edited Nov 26 at 10:30
answered Nov 26 at 10:22
mrf
37.3k54685
answered Nov 26 at 10:22
mrf
37.3k54685
answered Nov 26 at 10:22
mrf
37.3k54685
37.3k54685
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