For which subsets does the harmonic to analytic connection hold?












1












$begingroup$


I'm a bit confused on the choice of sets that authors choose and why. For example :



"Any harmonic function $u$ on an open subset $Omega$ of $R^2$ is locally the real part of a holomorphic function."



Also



"Given $h$ positive, $(a,b) in R^2$. Let $U$ be harmonic on $I=(a-h,a+h) times (b-h, b+h)$. Then there exists $f$ analytic on $sigma(I)$ such that $U=Re F$, $F$ being the associate of $f$"



I'm not sure what restrictions are necessary? It seems like it would be true for any open set of any size (connected or otherwise). Since the harmonic and analytic definitions are local in nature, I'm not sure why it is called a "local" converse, or what that really means in this context? Also, I'm not sure why the second author chose the set he did? I feel like I am missing something.










share|cite|improve this question









$endgroup$












  • $begingroup$
    what is $sigma$ and what does "associate" mean?
    $endgroup$
    – zhw.
    Dec 2 '18 at 21:01










  • $begingroup$
    Did you look at $g(z)= frac{1}{2ipi}int_gamma frac{f(u)}{z-u}du$ with $f$ harmonic inside $gamma$
    $endgroup$
    – reuns
    Dec 2 '18 at 23:38










  • $begingroup$
    @zhw $sigma$ is the canonical map between $R times R$ and the complex plane. $F$ is $f(sigma)$, the equivalent function defined on $R times R$
    $endgroup$
    – Darren
    Dec 3 '18 at 1:39
















1












$begingroup$


I'm a bit confused on the choice of sets that authors choose and why. For example :



"Any harmonic function $u$ on an open subset $Omega$ of $R^2$ is locally the real part of a holomorphic function."



Also



"Given $h$ positive, $(a,b) in R^2$. Let $U$ be harmonic on $I=(a-h,a+h) times (b-h, b+h)$. Then there exists $f$ analytic on $sigma(I)$ such that $U=Re F$, $F$ being the associate of $f$"



I'm not sure what restrictions are necessary? It seems like it would be true for any open set of any size (connected or otherwise). Since the harmonic and analytic definitions are local in nature, I'm not sure why it is called a "local" converse, or what that really means in this context? Also, I'm not sure why the second author chose the set he did? I feel like I am missing something.










share|cite|improve this question









$endgroup$












  • $begingroup$
    what is $sigma$ and what does "associate" mean?
    $endgroup$
    – zhw.
    Dec 2 '18 at 21:01










  • $begingroup$
    Did you look at $g(z)= frac{1}{2ipi}int_gamma frac{f(u)}{z-u}du$ with $f$ harmonic inside $gamma$
    $endgroup$
    – reuns
    Dec 2 '18 at 23:38










  • $begingroup$
    @zhw $sigma$ is the canonical map between $R times R$ and the complex plane. $F$ is $f(sigma)$, the equivalent function defined on $R times R$
    $endgroup$
    – Darren
    Dec 3 '18 at 1:39














1












1








1


1



$begingroup$


I'm a bit confused on the choice of sets that authors choose and why. For example :



"Any harmonic function $u$ on an open subset $Omega$ of $R^2$ is locally the real part of a holomorphic function."



Also



"Given $h$ positive, $(a,b) in R^2$. Let $U$ be harmonic on $I=(a-h,a+h) times (b-h, b+h)$. Then there exists $f$ analytic on $sigma(I)$ such that $U=Re F$, $F$ being the associate of $f$"



I'm not sure what restrictions are necessary? It seems like it would be true for any open set of any size (connected or otherwise). Since the harmonic and analytic definitions are local in nature, I'm not sure why it is called a "local" converse, or what that really means in this context? Also, I'm not sure why the second author chose the set he did? I feel like I am missing something.










share|cite|improve this question









$endgroup$




I'm a bit confused on the choice of sets that authors choose and why. For example :



"Any harmonic function $u$ on an open subset $Omega$ of $R^2$ is locally the real part of a holomorphic function."



Also



"Given $h$ positive, $(a,b) in R^2$. Let $U$ be harmonic on $I=(a-h,a+h) times (b-h, b+h)$. Then there exists $f$ analytic on $sigma(I)$ such that $U=Re F$, $F$ being the associate of $f$"



I'm not sure what restrictions are necessary? It seems like it would be true for any open set of any size (connected or otherwise). Since the harmonic and analytic definitions are local in nature, I'm not sure why it is called a "local" converse, or what that really means in this context? Also, I'm not sure why the second author chose the set he did? I feel like I am missing something.







complex-analysis harmonic-functions






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 19:24









DarrenDarren

345110




345110












  • $begingroup$
    what is $sigma$ and what does "associate" mean?
    $endgroup$
    – zhw.
    Dec 2 '18 at 21:01










  • $begingroup$
    Did you look at $g(z)= frac{1}{2ipi}int_gamma frac{f(u)}{z-u}du$ with $f$ harmonic inside $gamma$
    $endgroup$
    – reuns
    Dec 2 '18 at 23:38










  • $begingroup$
    @zhw $sigma$ is the canonical map between $R times R$ and the complex plane. $F$ is $f(sigma)$, the equivalent function defined on $R times R$
    $endgroup$
    – Darren
    Dec 3 '18 at 1:39


















  • $begingroup$
    what is $sigma$ and what does "associate" mean?
    $endgroup$
    – zhw.
    Dec 2 '18 at 21:01










  • $begingroup$
    Did you look at $g(z)= frac{1}{2ipi}int_gamma frac{f(u)}{z-u}du$ with $f$ harmonic inside $gamma$
    $endgroup$
    – reuns
    Dec 2 '18 at 23:38










  • $begingroup$
    @zhw $sigma$ is the canonical map between $R times R$ and the complex plane. $F$ is $f(sigma)$, the equivalent function defined on $R times R$
    $endgroup$
    – Darren
    Dec 3 '18 at 1:39
















$begingroup$
what is $sigma$ and what does "associate" mean?
$endgroup$
– zhw.
Dec 2 '18 at 21:01




$begingroup$
what is $sigma$ and what does "associate" mean?
$endgroup$
– zhw.
Dec 2 '18 at 21:01












$begingroup$
Did you look at $g(z)= frac{1}{2ipi}int_gamma frac{f(u)}{z-u}du$ with $f$ harmonic inside $gamma$
$endgroup$
– reuns
Dec 2 '18 at 23:38




$begingroup$
Did you look at $g(z)= frac{1}{2ipi}int_gamma frac{f(u)}{z-u}du$ with $f$ harmonic inside $gamma$
$endgroup$
– reuns
Dec 2 '18 at 23:38












$begingroup$
@zhw $sigma$ is the canonical map between $R times R$ and the complex plane. $F$ is $f(sigma)$, the equivalent function defined on $R times R$
$endgroup$
– Darren
Dec 3 '18 at 1:39




$begingroup$
@zhw $sigma$ is the canonical map between $R times R$ and the complex plane. $F$ is $f(sigma)$, the equivalent function defined on $R times R$
$endgroup$
– Darren
Dec 3 '18 at 1:39










1 Answer
1






active

oldest

votes


















1












$begingroup$

I just post my comment as an answer because it has grown too long.



If you are given any holomorphic function, its real and imaginary part are harmonic (rewrite the Laplacian in complex coordinates using Wirtinger's equations).



Conversely, if you are given any harmonic function and you want to construct a holomorphic function with the given harmonic function as real or imaginary part, there is an integral involved and this imposes the condition of simple connectedness on the defining domain of the harmonic function.



I'm not entirely sure wether this is the end of the story. The connection between holomorphic and harmonic functions plays an important role in what is called Levi's problem. You find a complete and nice overview of this topic in Grauert-Fritzsche: From holomorphic functions to complex manifolds, Chapter II. This is almost 60 pages because the topic is, as many things in complex analysis, in parts very technical.



EDIT: According to your comment, here are some more details on where we need a simply connected domain.



Let $Gsubset mathbb C$ be any domain. By a domain I mean a connected, non-empty open subset.



Definition. Let $u,v:Gto mathbb R$ be harmonic functions. If the function $f=u+iv$ is holomorphic on $G$ we call $(u,v)$ an adjoint harmonic pair.



We need a simple connected domain in the proof of the following theorem.



Theorem. If $G$ is simply connected and $u:Gto mathbb R$ harmonic, then there is a harmonic function $v:Gto mathbb R$ such that $(u,v)$ is an adjoint harmonic pair. In other words, every harmonic function on a simply connected domain is the real part of some holomorphic function.



Proof. Notice that in complex coordinates, $$Delta=4partial overline partial = 4overline partial partial.$$
Set $g:=partial u$. It follows from $Delta=4partial overline partial = 4overline partial partial$ that $g$ is holomorphic on $G$. Since $G$ is simply connected there exists a primitive function $f$ of $g$. Using the Cauchy Riemann equations and Wirtinger calculus we see that
begin{align}
g=partial f&=frac{1}{2}left(partial_x-ipartial_yright)(Re(f)+iIm(f))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)+i(partial_xIm(f)-ipartial_yIm(f)))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)-ipartial_yRe(f)+partial_xRe(f)))\
&=partial Re(f),
end{align}

where $Re,Im$ denote the real, resp. imaginary part.
Moreover
begin{equation*}
partial (u-Re(f))=g-g=0,
end{equation*}

hence $u-Re(f)=cin mathbb C$. As $f$ is only unique up to a constant we can, if necessary, change $f$ to $f-c$ and the claim follows.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the answer. Can you explain why the domain has to be connected?
    $endgroup$
    – Darren
    Dec 5 '18 at 3:45










  • $begingroup$
    @Darren I expanded my answer, does this help you?
    $endgroup$
    – James
    Dec 5 '18 at 9:02










  • $begingroup$
    Thanks for taking the time to write an answer. After more reading, I think I was getting stuck because of the following: The definition of a derivative is a local one, yet the concept of an anti-derivative has topological considerations. For example, the function $1/z$ does not possess an anti-derivative, yet presumably, each point on the domain $C - {0}$ can be surrounded by an open set where the anti-derivative exists? So the fact that the anti-derivative doesn't exist is because the domain is not simply connected.
    $endgroup$
    – Darren
    Dec 9 '18 at 15:40













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

I just post my comment as an answer because it has grown too long.



If you are given any holomorphic function, its real and imaginary part are harmonic (rewrite the Laplacian in complex coordinates using Wirtinger's equations).



Conversely, if you are given any harmonic function and you want to construct a holomorphic function with the given harmonic function as real or imaginary part, there is an integral involved and this imposes the condition of simple connectedness on the defining domain of the harmonic function.



I'm not entirely sure wether this is the end of the story. The connection between holomorphic and harmonic functions plays an important role in what is called Levi's problem. You find a complete and nice overview of this topic in Grauert-Fritzsche: From holomorphic functions to complex manifolds, Chapter II. This is almost 60 pages because the topic is, as many things in complex analysis, in parts very technical.



EDIT: According to your comment, here are some more details on where we need a simply connected domain.



Let $Gsubset mathbb C$ be any domain. By a domain I mean a connected, non-empty open subset.



Definition. Let $u,v:Gto mathbb R$ be harmonic functions. If the function $f=u+iv$ is holomorphic on $G$ we call $(u,v)$ an adjoint harmonic pair.



We need a simple connected domain in the proof of the following theorem.



Theorem. If $G$ is simply connected and $u:Gto mathbb R$ harmonic, then there is a harmonic function $v:Gto mathbb R$ such that $(u,v)$ is an adjoint harmonic pair. In other words, every harmonic function on a simply connected domain is the real part of some holomorphic function.



Proof. Notice that in complex coordinates, $$Delta=4partial overline partial = 4overline partial partial.$$
Set $g:=partial u$. It follows from $Delta=4partial overline partial = 4overline partial partial$ that $g$ is holomorphic on $G$. Since $G$ is simply connected there exists a primitive function $f$ of $g$. Using the Cauchy Riemann equations and Wirtinger calculus we see that
begin{align}
g=partial f&=frac{1}{2}left(partial_x-ipartial_yright)(Re(f)+iIm(f))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)+i(partial_xIm(f)-ipartial_yIm(f)))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)-ipartial_yRe(f)+partial_xRe(f)))\
&=partial Re(f),
end{align}

where $Re,Im$ denote the real, resp. imaginary part.
Moreover
begin{equation*}
partial (u-Re(f))=g-g=0,
end{equation*}

hence $u-Re(f)=cin mathbb C$. As $f$ is only unique up to a constant we can, if necessary, change $f$ to $f-c$ and the claim follows.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the answer. Can you explain why the domain has to be connected?
    $endgroup$
    – Darren
    Dec 5 '18 at 3:45










  • $begingroup$
    @Darren I expanded my answer, does this help you?
    $endgroup$
    – James
    Dec 5 '18 at 9:02










  • $begingroup$
    Thanks for taking the time to write an answer. After more reading, I think I was getting stuck because of the following: The definition of a derivative is a local one, yet the concept of an anti-derivative has topological considerations. For example, the function $1/z$ does not possess an anti-derivative, yet presumably, each point on the domain $C - {0}$ can be surrounded by an open set where the anti-derivative exists? So the fact that the anti-derivative doesn't exist is because the domain is not simply connected.
    $endgroup$
    – Darren
    Dec 9 '18 at 15:40


















1












$begingroup$

I just post my comment as an answer because it has grown too long.



If you are given any holomorphic function, its real and imaginary part are harmonic (rewrite the Laplacian in complex coordinates using Wirtinger's equations).



Conversely, if you are given any harmonic function and you want to construct a holomorphic function with the given harmonic function as real or imaginary part, there is an integral involved and this imposes the condition of simple connectedness on the defining domain of the harmonic function.



I'm not entirely sure wether this is the end of the story. The connection between holomorphic and harmonic functions plays an important role in what is called Levi's problem. You find a complete and nice overview of this topic in Grauert-Fritzsche: From holomorphic functions to complex manifolds, Chapter II. This is almost 60 pages because the topic is, as many things in complex analysis, in parts very technical.



EDIT: According to your comment, here are some more details on where we need a simply connected domain.



Let $Gsubset mathbb C$ be any domain. By a domain I mean a connected, non-empty open subset.



Definition. Let $u,v:Gto mathbb R$ be harmonic functions. If the function $f=u+iv$ is holomorphic on $G$ we call $(u,v)$ an adjoint harmonic pair.



We need a simple connected domain in the proof of the following theorem.



Theorem. If $G$ is simply connected and $u:Gto mathbb R$ harmonic, then there is a harmonic function $v:Gto mathbb R$ such that $(u,v)$ is an adjoint harmonic pair. In other words, every harmonic function on a simply connected domain is the real part of some holomorphic function.



Proof. Notice that in complex coordinates, $$Delta=4partial overline partial = 4overline partial partial.$$
Set $g:=partial u$. It follows from $Delta=4partial overline partial = 4overline partial partial$ that $g$ is holomorphic on $G$. Since $G$ is simply connected there exists a primitive function $f$ of $g$. Using the Cauchy Riemann equations and Wirtinger calculus we see that
begin{align}
g=partial f&=frac{1}{2}left(partial_x-ipartial_yright)(Re(f)+iIm(f))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)+i(partial_xIm(f)-ipartial_yIm(f)))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)-ipartial_yRe(f)+partial_xRe(f)))\
&=partial Re(f),
end{align}

where $Re,Im$ denote the real, resp. imaginary part.
Moreover
begin{equation*}
partial (u-Re(f))=g-g=0,
end{equation*}

hence $u-Re(f)=cin mathbb C$. As $f$ is only unique up to a constant we can, if necessary, change $f$ to $f-c$ and the claim follows.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the answer. Can you explain why the domain has to be connected?
    $endgroup$
    – Darren
    Dec 5 '18 at 3:45










  • $begingroup$
    @Darren I expanded my answer, does this help you?
    $endgroup$
    – James
    Dec 5 '18 at 9:02










  • $begingroup$
    Thanks for taking the time to write an answer. After more reading, I think I was getting stuck because of the following: The definition of a derivative is a local one, yet the concept of an anti-derivative has topological considerations. For example, the function $1/z$ does not possess an anti-derivative, yet presumably, each point on the domain $C - {0}$ can be surrounded by an open set where the anti-derivative exists? So the fact that the anti-derivative doesn't exist is because the domain is not simply connected.
    $endgroup$
    – Darren
    Dec 9 '18 at 15:40
















1












1








1





$begingroup$

I just post my comment as an answer because it has grown too long.



If you are given any holomorphic function, its real and imaginary part are harmonic (rewrite the Laplacian in complex coordinates using Wirtinger's equations).



Conversely, if you are given any harmonic function and you want to construct a holomorphic function with the given harmonic function as real or imaginary part, there is an integral involved and this imposes the condition of simple connectedness on the defining domain of the harmonic function.



I'm not entirely sure wether this is the end of the story. The connection between holomorphic and harmonic functions plays an important role in what is called Levi's problem. You find a complete and nice overview of this topic in Grauert-Fritzsche: From holomorphic functions to complex manifolds, Chapter II. This is almost 60 pages because the topic is, as many things in complex analysis, in parts very technical.



EDIT: According to your comment, here are some more details on where we need a simply connected domain.



Let $Gsubset mathbb C$ be any domain. By a domain I mean a connected, non-empty open subset.



Definition. Let $u,v:Gto mathbb R$ be harmonic functions. If the function $f=u+iv$ is holomorphic on $G$ we call $(u,v)$ an adjoint harmonic pair.



We need a simple connected domain in the proof of the following theorem.



Theorem. If $G$ is simply connected and $u:Gto mathbb R$ harmonic, then there is a harmonic function $v:Gto mathbb R$ such that $(u,v)$ is an adjoint harmonic pair. In other words, every harmonic function on a simply connected domain is the real part of some holomorphic function.



Proof. Notice that in complex coordinates, $$Delta=4partial overline partial = 4overline partial partial.$$
Set $g:=partial u$. It follows from $Delta=4partial overline partial = 4overline partial partial$ that $g$ is holomorphic on $G$. Since $G$ is simply connected there exists a primitive function $f$ of $g$. Using the Cauchy Riemann equations and Wirtinger calculus we see that
begin{align}
g=partial f&=frac{1}{2}left(partial_x-ipartial_yright)(Re(f)+iIm(f))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)+i(partial_xIm(f)-ipartial_yIm(f)))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)-ipartial_yRe(f)+partial_xRe(f)))\
&=partial Re(f),
end{align}

where $Re,Im$ denote the real, resp. imaginary part.
Moreover
begin{equation*}
partial (u-Re(f))=g-g=0,
end{equation*}

hence $u-Re(f)=cin mathbb C$. As $f$ is only unique up to a constant we can, if necessary, change $f$ to $f-c$ and the claim follows.






share|cite|improve this answer











$endgroup$



I just post my comment as an answer because it has grown too long.



If you are given any holomorphic function, its real and imaginary part are harmonic (rewrite the Laplacian in complex coordinates using Wirtinger's equations).



Conversely, if you are given any harmonic function and you want to construct a holomorphic function with the given harmonic function as real or imaginary part, there is an integral involved and this imposes the condition of simple connectedness on the defining domain of the harmonic function.



I'm not entirely sure wether this is the end of the story. The connection between holomorphic and harmonic functions plays an important role in what is called Levi's problem. You find a complete and nice overview of this topic in Grauert-Fritzsche: From holomorphic functions to complex manifolds, Chapter II. This is almost 60 pages because the topic is, as many things in complex analysis, in parts very technical.



EDIT: According to your comment, here are some more details on where we need a simply connected domain.



Let $Gsubset mathbb C$ be any domain. By a domain I mean a connected, non-empty open subset.



Definition. Let $u,v:Gto mathbb R$ be harmonic functions. If the function $f=u+iv$ is holomorphic on $G$ we call $(u,v)$ an adjoint harmonic pair.



We need a simple connected domain in the proof of the following theorem.



Theorem. If $G$ is simply connected and $u:Gto mathbb R$ harmonic, then there is a harmonic function $v:Gto mathbb R$ such that $(u,v)$ is an adjoint harmonic pair. In other words, every harmonic function on a simply connected domain is the real part of some holomorphic function.



Proof. Notice that in complex coordinates, $$Delta=4partial overline partial = 4overline partial partial.$$
Set $g:=partial u$. It follows from $Delta=4partial overline partial = 4overline partial partial$ that $g$ is holomorphic on $G$. Since $G$ is simply connected there exists a primitive function $f$ of $g$. Using the Cauchy Riemann equations and Wirtinger calculus we see that
begin{align}
g=partial f&=frac{1}{2}left(partial_x-ipartial_yright)(Re(f)+iIm(f))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)+i(partial_xIm(f)-ipartial_yIm(f)))\
&=frac{1}{2}(partial_x Re(f)-ipartial_yRe(f)-ipartial_yRe(f)+partial_xRe(f)))\
&=partial Re(f),
end{align}

where $Re,Im$ denote the real, resp. imaginary part.
Moreover
begin{equation*}
partial (u-Re(f))=g-g=0,
end{equation*}

hence $u-Re(f)=cin mathbb C$. As $f$ is only unique up to a constant we can, if necessary, change $f$ to $f-c$ and the claim follows.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 9:01

























answered Dec 4 '18 at 9:48









JamesJames

798115




798115












  • $begingroup$
    Thank you for the answer. Can you explain why the domain has to be connected?
    $endgroup$
    – Darren
    Dec 5 '18 at 3:45










  • $begingroup$
    @Darren I expanded my answer, does this help you?
    $endgroup$
    – James
    Dec 5 '18 at 9:02










  • $begingroup$
    Thanks for taking the time to write an answer. After more reading, I think I was getting stuck because of the following: The definition of a derivative is a local one, yet the concept of an anti-derivative has topological considerations. For example, the function $1/z$ does not possess an anti-derivative, yet presumably, each point on the domain $C - {0}$ can be surrounded by an open set where the anti-derivative exists? So the fact that the anti-derivative doesn't exist is because the domain is not simply connected.
    $endgroup$
    – Darren
    Dec 9 '18 at 15:40




















  • $begingroup$
    Thank you for the answer. Can you explain why the domain has to be connected?
    $endgroup$
    – Darren
    Dec 5 '18 at 3:45










  • $begingroup$
    @Darren I expanded my answer, does this help you?
    $endgroup$
    – James
    Dec 5 '18 at 9:02










  • $begingroup$
    Thanks for taking the time to write an answer. After more reading, I think I was getting stuck because of the following: The definition of a derivative is a local one, yet the concept of an anti-derivative has topological considerations. For example, the function $1/z$ does not possess an anti-derivative, yet presumably, each point on the domain $C - {0}$ can be surrounded by an open set where the anti-derivative exists? So the fact that the anti-derivative doesn't exist is because the domain is not simply connected.
    $endgroup$
    – Darren
    Dec 9 '18 at 15:40


















$begingroup$
Thank you for the answer. Can you explain why the domain has to be connected?
$endgroup$
– Darren
Dec 5 '18 at 3:45




$begingroup$
Thank you for the answer. Can you explain why the domain has to be connected?
$endgroup$
– Darren
Dec 5 '18 at 3:45












$begingroup$
@Darren I expanded my answer, does this help you?
$endgroup$
– James
Dec 5 '18 at 9:02




$begingroup$
@Darren I expanded my answer, does this help you?
$endgroup$
– James
Dec 5 '18 at 9:02












$begingroup$
Thanks for taking the time to write an answer. After more reading, I think I was getting stuck because of the following: The definition of a derivative is a local one, yet the concept of an anti-derivative has topological considerations. For example, the function $1/z$ does not possess an anti-derivative, yet presumably, each point on the domain $C - {0}$ can be surrounded by an open set where the anti-derivative exists? So the fact that the anti-derivative doesn't exist is because the domain is not simply connected.
$endgroup$
– Darren
Dec 9 '18 at 15:40






$begingroup$
Thanks for taking the time to write an answer. After more reading, I think I was getting stuck because of the following: The definition of a derivative is a local one, yet the concept of an anti-derivative has topological considerations. For example, the function $1/z$ does not possess an anti-derivative, yet presumably, each point on the domain $C - {0}$ can be surrounded by an open set where the anti-derivative exists? So the fact that the anti-derivative doesn't exist is because the domain is not simply connected.
$endgroup$
– Darren
Dec 9 '18 at 15:40




















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