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Maximal Natural Domain of An Analytic Function on Complex Plane
I want to ask a question of the maximal natural domain of an analytic function. We know that on $mathbb{C}^n$, a domain $U$ is a domain of holomorphy if and only if $U$ is a pseudoconvex domain, and in particular in complex plane, we have that every open set is pseudoconvex.
Here I want to ask for the converse for $mathbb{C}$:
Q: For an analytic function $f$, is the maximal set $W$ for which $f$ can be extended necessarily open?
And moreover, does this generalize in $mathbb{C}^n$?
Notice that for some particular cases, this does not raise a question. For instance, if $f$ is defined on some closed set $W$ with $C^1$ boundary(say, a Jordan curve), then an application of Caratheodory theorem and Schwartz reflection could be enough.
It might be a stupid question, but it seems that I could not find a counterexample or a proof yet.
complex-analysis analysis several-complex-variables analytic-continuation
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
add a comment |
I want to ask a question of the maximal natural domain of an analytic function. We know that on $mathbb{C}^n$, a domain $U$ is a domain of holomorphy if and only if $U$ is a pseudoconvex domain, and in particular in complex plane, we have that every open set is pseudoconvex.
Here I want to ask for the converse for $mathbb{C}$:
Q: For an analytic function $f$, is the maximal set $W$ for which $f$ can be extended necessarily open?
And moreover, does this generalize in $mathbb{C}^n$?
Notice that for some particular cases, this does not raise a question. For instance, if $f$ is defined on some closed set $W$ with $C^1$ boundary(say, a Jordan curve), then an application of Caratheodory theorem and Schwartz reflection could be enough.
It might be a stupid question, but it seems that I could not find a counterexample or a proof yet.
complex-analysis analysis several-complex-variables analytic-continuation
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
1
What is your definition of analytic on non-open sets?
– mrf
Nov 21 '18 at 6:13
@mrf I guess for this case, I am considering $pdv{f}{z}$ is still bounded when $z_j to z$ for $z_j in W$, where $W$ is a domain for $f$.
– The Hong Kong Journalist
Nov 21 '18 at 6:18
@TheHongKongJournalist there need not be a maximal set $W$ over which $f$ can be extended. take the example of the complex log; it can be extended on any domain consisting of the plane with a half-line starting at $0$ removed, but it cannot be defined over $mathbb{C}^*$
– Glougloubarbaki
Nov 26 '18 at 12:12
add a comment |
I want to ask a question of the maximal natural domain of an analytic function. We know that on $mathbb{C}^n$, a domain $U$ is a domain of holomorphy if and only if $U$ is a pseudoconvex domain, and in particular in complex plane, we have that every open set is pseudoconvex.
Here I want to ask for the converse for $mathbb{C}$:
Q: For an analytic function $f$, is the maximal set $W$ for which $f$ can be extended necessarily open?
And moreover, does this generalize in $mathbb{C}^n$?
Notice that for some particular cases, this does not raise a question. For instance, if $f$ is defined on some closed set $W$ with $C^1$ boundary(say, a Jordan curve), then an application of Caratheodory theorem and Schwartz reflection could be enough.
It might be a stupid question, but it seems that I could not find a counterexample or a proof yet.
complex-analysis analysis several-complex-variables analytic-continuation
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
I want to ask a question of the maximal natural domain of an analytic function. We know that on $mathbb{C}^n$, a domain $U$ is a domain of holomorphy if and only if $U$ is a pseudoconvex domain, and in particular in complex plane, we have that every open set is pseudoconvex.
Here I want to ask for the converse for $mathbb{C}$:
Q: For an analytic function $f$, is the maximal set $W$ for which $f$ can be extended necessarily open?
And moreover, does this generalize in $mathbb{C}^n$?
Notice that for some particular cases, this does not raise a question. For instance, if $f$ is defined on some closed set $W$ with $C^1$ boundary(say, a Jordan curve), then an application of Caratheodory theorem and Schwartz reflection could be enough.
It might be a stupid question, but it seems that I could not find a counterexample or a proof yet.
complex-analysis analysis several-complex-variables analytic-continuation
complex-analysis analysis several-complex-variables analytic-continuation
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
asked Nov 21 '18 at 6:05
The Hong Kong Journalist
11
11
1
What is your definition of analytic on non-open sets?
– mrf
Nov 21 '18 at 6:13
@mrf I guess for this case, I am considering $pdv{f}{z}$ is still bounded when $z_j to z$ for $z_j in W$, where $W$ is a domain for $f$.
– The Hong Kong Journalist
Nov 21 '18 at 6:18
@TheHongKongJournalist there need not be a maximal set $W$ over which $f$ can be extended. take the example of the complex log; it can be extended on any domain consisting of the plane with a half-line starting at $0$ removed, but it cannot be defined over $mathbb{C}^*$
– Glougloubarbaki
Nov 26 '18 at 12:12
add a comment |
1
What is your definition of analytic on non-open sets?
– mrf
Nov 21 '18 at 6:13
@mrf I guess for this case, I am considering $pdv{f}{z}$ is still bounded when $z_j to z$ for $z_j in W$, where $W$ is a domain for $f$.
– The Hong Kong Journalist
Nov 21 '18 at 6:18
@TheHongKongJournalist there need not be a maximal set $W$ over which $f$ can be extended. take the example of the complex log; it can be extended on any domain consisting of the plane with a half-line starting at $0$ removed, but it cannot be defined over $mathbb{C}^*$
– Glougloubarbaki
Nov 26 '18 at 12:12
1
1
What is your definition of analytic on non-open sets?
– mrf
Nov 21 '18 at 6:13
What is your definition of analytic on non-open sets?
– mrf
Nov 21 '18 at 6:13
@mrf I guess for this case, I am considering $pdv{f}{z}$ is still bounded when $z_j to z$ for $z_j in W$, where $W$ is a domain for $f$.
– The Hong Kong Journalist
Nov 21 '18 at 6:18
@mrf I guess for this case, I am considering $pdv{f}{z}$ is still bounded when $z_j to z$ for $z_j in W$, where $W$ is a domain for $f$.
– The Hong Kong Journalist
Nov 21 '18 at 6:18
@TheHongKongJournalist there need not be a maximal set $W$ over which $f$ can be extended. take the example of the complex log; it can be extended on any domain consisting of the plane with a half-line starting at $0$ removed, but it cannot be defined over $mathbb{C}^*$
– Glougloubarbaki
Nov 26 '18 at 12:12
@TheHongKongJournalist there need not be a maximal set $W$ over which $f$ can be extended. take the example of the complex log; it can be extended on any domain consisting of the plane with a half-line starting at $0$ removed, but it cannot be defined over $mathbb{C}^*$
– Glougloubarbaki
Nov 26 '18 at 12:12
add a comment |
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1
What is your definition of analytic on non-open sets?
– mrf
Nov 21 '18 at 6:13
@mrf I guess for this case, I am considering $pdv{f}{z}$ is still bounded when $z_j to z$ for $z_j in W$, where $W$ is a domain for $f$.
– The Hong Kong Journalist
Nov 21 '18 at 6:18
@TheHongKongJournalist there need not be a maximal set $W$ over which $f$ can be extended. take the example of the complex log; it can be extended on any domain consisting of the plane with a half-line starting at $0$ removed, but it cannot be defined over $mathbb{C}^*$
– Glougloubarbaki
Nov 26 '18 at 12:12