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Extension of real or imaginary part of entire function to complex valued functions
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In this forum post on MSE we were trying to find the imaginary part of an entire function given it's real parte. One of the answers called for this method $$f(z) = 2uleft(zover 2,-izover 2right)-u(0,0) tag1$$ which I've never seen before. This caught a lot of attention for the fact that, based on it's definition $$u(x,y):mathbbR^2rightarrowmathbbR$$ and by using equation $(1)$ we are implying that we can extend the domain of $u(x,y)$ from $mathbbR^2$ to $mathbbC^2$, mainly $$u(z,w):mathbbC^2rightarrowmathbbC$$ As you can see in the forum post, this started some back and forth between some users, me as well, to try and find some mathematical rigour to equation $(1)$. Because of this I'm now here asking for that so desired mathematical proof: when and why is it possible to extend real-valued function to complex-valued functions? Moreover: when is it possible to apply equation $(1)$ to find an entire function from it's real part?
My approach
I copy here my attempt of an explanation and a sort of proof that I've given even in the comment of the MSE post cited:
The real part of an entire function can be expressed as
$$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex-iy) = f(x+iy)+overlinef(x-iy)tag2$$
by choosing $$z=x+iy;;w=x-iy Rightarrow x=1over 2(z+w);;y=1over2i(z-w)$$ then $$2uleft(1over 2(z+w), 1over2i(z-w)right) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$
So we can choose $w$ arbitrarily and set it to $w = z_0$ which for the problem given was $z_0=0+i0$, then $$f(z)=2uleft(1over 2(z+z_0), 1over2i(z-z_0)right)-overlinef(z_0) Rightarrow f(z)=2uleft(1over 2z, 1over2izright)-u(0,0)$$
This same proof can be used to find that $$f(z) = 2ivleft(zover 2,zover2iright)+v(0,0)$$
The problem comes again when substituting in equation $(2)$ for the fact that $z,winmathbbC$ and my answer was that the right hand side has sense for every $x$ and $y$,real or complex, such that $x+iy$ and $x-iy$ are in the domain of $f$ which clearly are being $f$ an entire function, so analytic an all $mathbbC$.
Final thoughts
So in the end, I would like some comments by you! If it's possible, I'd love to have a clear and mathematical rigorous explanation.
complex-analysis complex-numbers entire-functions
asked Jul 17 at 11:06
Davide Morgante
1,875220
add a comment |Â
up vote
2
down vote
favorite
In this forum post on MSE we were trying to find the imaginary part of an entire function given it's real parte. One of the answers called for this method $$f(z) = 2uleft(zover 2,-izover 2right)-u(0,0) tag1$$ which I've never seen before. This caught a lot of attention for the fact that, based on it's definition $$u(x,y):mathbbR^2rightarrowmathbbR$$ and by using equation $(1)$ we are implying that we can extend the domain of $u(x,y)$ from $mathbbR^2$ to $mathbbC^2$, mainly $$u(z,w):mathbbC^2rightarrowmathbbC$$ As you can see in the forum post, this started some back and forth between some users, me as well, to try and find some mathematical rigour to equation $(1)$. Because of this I'm now here asking for that so desired mathematical proof: when and why is it possible to extend real-valued function to complex-valued functions? Moreover: when is it possible to apply equation $(1)$ to find an entire function from it's real part?
My approach
I copy here my attempt of an explanation and a sort of proof that I've given even in the comment of the MSE post cited:
The real part of an entire function can be expressed as
$$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex-iy) = f(x+iy)+overlinef(x-iy)tag2$$
by choosing $$z=x+iy;;w=x-iy Rightarrow x=1over 2(z+w);;y=1over2i(z-w)$$ then $$2uleft(1over 2(z+w), 1over2i(z-w)right) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$
So we can choose $w$ arbitrarily and set it to $w = z_0$ which for the problem given was $z_0=0+i0$, then $$f(z)=2uleft(1over 2(z+z_0), 1over2i(z-z_0)right)-overlinef(z_0) Rightarrow f(z)=2uleft(1over 2z, 1over2izright)-u(0,0)$$
This same proof can be used to find that $$f(z) = 2ivleft(zover 2,zover2iright)+v(0,0)$$
The problem comes again when substituting in equation $(2)$ for the fact that $z,winmathbbC$ and my answer was that the right hand side has sense for every $x$ and $y$,real or complex, such that $x+iy$ and $x-iy$ are in the domain of $f$ which clearly are being $f$ an entire function, so analytic an all $mathbbC$.
Final thoughts
So in the end, I would like some comments by you! If it's possible, I'd love to have a clear and mathematical rigorous explanation.
complex-analysis complex-numbers entire-functions
asked Jul 17 at 11:06
Davide Morgante
1,875220
1
That's all. There is nothing mysterious about it. If $f$ exists, then $overlinef(overlinez)$ is also analytic. Therefore $tildeu(z_1,z_2)=frac12(f(z_1+iz_2))+overlinef(overlinez_1-iz_2)$ is analytic. The restriction of this to the real plane $z_1=overlinez_1,z_2=overlinez_2$ coincides with $u=Re(f)$. Therefore, if $u$ is real analytic, and $hatu$ is its complexification, then $tildeu=hatu$ and $hatu(z/2,z/(2i))=frac12(f(z)+overlinef(0))$. What is the complexification? Just take the power series of $u$ and look at it as a series over the complex.
– user574889
Jul 17 at 12:13
You have a typo in $(2)$, $overlinef(x+iy) = overlinefbigl(overlinex-iybigr)$, not as you wrote $overlinefbigl(overlinex+iybigr)$.
– Daniel Fischer♦
Jul 17 at 12:38
A little below that you have "$dotsc = f(z) + overlinef(w) = f(z) + overlinefbigl(overlinewbigr)$". There's no reason why $overlinef(w) = overlinefbigl(overlinewbigr)$ should hold. But maybe it's just a typo and should have been $overlinefbigl(overlinewbigr)$.
– Daniel Fischer♦
Jul 17 at 12:42
@DanielFischer thank's for all the corrections, they where just typos!
– Davide Morgante
Jul 17 at 14:06
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In this forum post on MSE we were trying to find the imaginary part of an entire function given it's real parte. One of the answers called for this method $$f(z) = 2uleft(zover 2,-izover 2right)-u(0,0) tag1$$ which I've never seen before. This caught a lot of attention for the fact that, based on it's definition $$u(x,y):mathbbR^2rightarrowmathbbR$$ and by using equation $(1)$ we are implying that we can extend the domain of $u(x,y)$ from $mathbbR^2$ to $mathbbC^2$, mainly $$u(z,w):mathbbC^2rightarrowmathbbC$$ As you can see in the forum post, this started some back and forth between some users, me as well, to try and find some mathematical rigour to equation $(1)$. Because of this I'm now here asking for that so desired mathematical proof: when and why is it possible to extend real-valued function to complex-valued functions? Moreover: when is it possible to apply equation $(1)$ to find an entire function from it's real part?
My approach
I copy here my attempt of an explanation and a sort of proof that I've given even in the comment of the MSE post cited:
The real part of an entire function can be expressed as
$$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex-iy) = f(x+iy)+overlinef(x-iy)tag2$$
by choosing $$z=x+iy;;w=x-iy Rightarrow x=1over 2(z+w);;y=1over2i(z-w)$$ then $$2uleft(1over 2(z+w), 1over2i(z-w)right) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$
So we can choose $w$ arbitrarily and set it to $w = z_0$ which for the problem given was $z_0=0+i0$, then $$f(z)=2uleft(1over 2(z+z_0), 1over2i(z-z_0)right)-overlinef(z_0) Rightarrow f(z)=2uleft(1over 2z, 1over2izright)-u(0,0)$$
This same proof can be used to find that $$f(z) = 2ivleft(zover 2,zover2iright)+v(0,0)$$
The problem comes again when substituting in equation $(2)$ for the fact that $z,winmathbbC$ and my answer was that the right hand side has sense for every $x$ and $y$,real or complex, such that $x+iy$ and $x-iy$ are in the domain of $f$ which clearly are being $f$ an entire function, so analytic an all $mathbbC$.
Final thoughts
So in the end, I would like some comments by you! If it's possible, I'd love to have a clear and mathematical rigorous explanation.
complex-analysis complex-numbers entire-functions
asked Jul 17 at 11:06
Davide Morgante
1,875220
In this forum post on MSE we were trying to find the imaginary part of an entire function given it's real parte. One of the answers called for this method $$f(z) = 2uleft(zover 2,-izover 2right)-u(0,0) tag1$$ which I've never seen before. This caught a lot of attention for the fact that, based on it's definition $$u(x,y):mathbbR^2rightarrowmathbbR$$ and by using equation $(1)$ we are implying that we can extend the domain of $u(x,y)$ from $mathbbR^2$ to $mathbbC^2$, mainly $$u(z,w):mathbbC^2rightarrowmathbbC$$ As you can see in the forum post, this started some back and forth between some users, me as well, to try and find some mathematical rigour to equation $(1)$. Because of this I'm now here asking for that so desired mathematical proof: when and why is it possible to extend real-valued function to complex-valued functions? Moreover: when is it possible to apply equation $(1)$ to find an entire function from it's real part?
My approach
I copy here my attempt of an explanation and a sort of proof that I've given even in the comment of the MSE post cited:
The real part of an entire function can be expressed as
$$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex-iy) = f(x+iy)+overlinef(x-iy)tag2$$
by choosing $$z=x+iy;;w=x-iy Rightarrow x=1over 2(z+w);;y=1over2i(z-w)$$ then $$2uleft(1over 2(z+w), 1over2i(z-w)right) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$
So we can choose $w$ arbitrarily and set it to $w = z_0$ which for the problem given was $z_0=0+i0$, then $$f(z)=2uleft(1over 2(z+z_0), 1over2i(z-z_0)right)-overlinef(z_0) Rightarrow f(z)=2uleft(1over 2z, 1over2izright)-u(0,0)$$
This same proof can be used to find that $$f(z) = 2ivleft(zover 2,zover2iright)+v(0,0)$$
The problem comes again when substituting in equation $(2)$ for the fact that $z,winmathbbC$ and my answer was that the right hand side has sense for every $x$ and $y$,real or complex, such that $x+iy$ and $x-iy$ are in the domain of $f$ which clearly are being $f$ an entire function, so analytic an all $mathbbC$.
Final thoughts
So in the end, I would like some comments by you! If it's possible, I'd love to have a clear and mathematical rigorous explanation.
complex-analysis complex-numbers entire-functions
asked Jul 17 at 11:06
Davide Morgante
1,875220
edited Jul 17 at 14:08
asked Jul 17 at 11:06
Davide Morgante
1,875220
asked Jul 17 at 11:06
Davide Morgante
1,875220
asked Jul 17 at 11:06
Davide Morgante
1,875220
1,875220
1
That's all. There is nothing mysterious about it. If $f$ exists, then $overlinef(overlinez)$ is also analytic. Therefore $tildeu(z_1,z_2)=frac12(f(z_1+iz_2))+overlinef(overlinez_1-iz_2)$ is analytic. The restriction of this to the real plane $z_1=overlinez_1,z_2=overlinez_2$ coincides with $u=Re(f)$. Therefore, if $u$ is real analytic, and $hatu$ is its complexification, then $tildeu=hatu$ and $hatu(z/2,z/(2i))=frac12(f(z)+overlinef(0))$. What is the complexification? Just take the power series of $u$ and look at it as a series over the complex.
– user574889
Jul 17 at 12:13
You have a typo in $(2)$, $overlinef(x+iy) = overlinefbigl(overlinex-iybigr)$, not as you wrote $overlinefbigl(overlinex+iybigr)$.
– Daniel Fischer♦
Jul 17 at 12:38
A little below that you have "$dotsc = f(z) + overlinef(w) = f(z) + overlinefbigl(overlinewbigr)$". There's no reason why $overlinef(w) = overlinefbigl(overlinewbigr)$ should hold. But maybe it's just a typo and should have been $overlinefbigl(overlinewbigr)$.
– Daniel Fischer♦
Jul 17 at 12:42
@DanielFischer thank's for all the corrections, they where just typos!
– Davide Morgante
Jul 17 at 14:06
add a comment |Â
1
That's all. There is nothing mysterious about it. If $f$ exists, then $overlinef(overlinez)$ is also analytic. Therefore $tildeu(z_1,z_2)=frac12(f(z_1+iz_2))+overlinef(overlinez_1-iz_2)$ is analytic. The restriction of this to the real plane $z_1=overlinez_1,z_2=overlinez_2$ coincides with $u=Re(f)$. Therefore, if $u$ is real analytic, and $hatu$ is its complexification, then $tildeu=hatu$ and $hatu(z/2,z/(2i))=frac12(f(z)+overlinef(0))$. What is the complexification? Just take the power series of $u$ and look at it as a series over the complex.
– user574889
Jul 17 at 12:13
You have a typo in $(2)$, $overlinef(x+iy) = overlinefbigl(overlinex-iybigr)$, not as you wrote $overlinefbigl(overlinex+iybigr)$.
– Daniel Fischer♦
Jul 17 at 12:38
A little below that you have "$dotsc = f(z) + overlinef(w) = f(z) + overlinefbigl(overlinewbigr)$". There's no reason why $overlinef(w) = overlinefbigl(overlinewbigr)$ should hold. But maybe it's just a typo and should have been $overlinefbigl(overlinewbigr)$.
– Daniel Fischer♦
Jul 17 at 12:42
@DanielFischer thank's for all the corrections, they where just typos!
– Davide Morgante
Jul 17 at 14:06
1
1
That's all. There is nothing mysterious about it. If $f$ exists, then $overlinef(overlinez)$ is also analytic. Therefore $tildeu(z_1,z_2)=frac12(f(z_1+iz_2))+overlinef(overlinez_1-iz_2)$ is analytic. The restriction of this to the real plane $z_1=overlinez_1,z_2=overlinez_2$ coincides with $u=Re(f)$. Therefore, if $u$ is real analytic, and $hatu$ is its complexification, then $tildeu=hatu$ and $hatu(z/2,z/(2i))=frac12(f(z)+overlinef(0))$. What is the complexification? Just take the power series of $u$ and look at it as a series over the complex.
– user574889
Jul 17 at 12:13
That's all. There is nothing mysterious about it. If $f$ exists, then $overlinef(overlinez)$ is also analytic. Therefore $tildeu(z_1,z_2)=frac12(f(z_1+iz_2))+overlinef(overlinez_1-iz_2)$ is analytic. The restriction of this to the real plane $z_1=overlinez_1,z_2=overlinez_2$ coincides with $u=Re(f)$. Therefore, if $u$ is real analytic, and $hatu$ is its complexification, then $tildeu=hatu$ and $hatu(z/2,z/(2i))=frac12(f(z)+overlinef(0))$. What is the complexification? Just take the power series of $u$ and look at it as a series over the complex.
– user574889
Jul 17 at 12:13
You have a typo in $(2)$, $overlinef(x+iy) = overlinefbigl(overlinex-iybigr)$, not as you wrote $overlinefbigl(overlinex+iybigr)$.
– Daniel Fischer♦
Jul 17 at 12:38
You have a typo in $(2)$, $overlinef(x+iy) = overlinefbigl(overlinex-iybigr)$, not as you wrote $overlinefbigl(overlinex+iybigr)$.
– Daniel Fischer♦
Jul 17 at 12:38
A little below that you have "$dotsc = f(z) + overlinef(w) = f(z) + overlinefbigl(overlinewbigr)$". There's no reason why $overlinef(w) = overlinefbigl(overlinewbigr)$ should hold. But maybe it's just a typo and should have been $overlinefbigl(overlinewbigr)$.
– Daniel Fischer♦
Jul 17 at 12:42
A little below that you have "$dotsc = f(z) + overlinef(w) = f(z) + overlinefbigl(overlinewbigr)$". There's no reason why $overlinef(w) = overlinefbigl(overlinewbigr)$ should hold. But maybe it's just a typo and should have been $overlinefbigl(overlinewbigr)$.
– Daniel Fischer♦
Jul 17 at 12:42
@DanielFischer thank's for all the corrections, they where just typos!
– Davide Morgante
Jul 17 at 14:06
@DanielFischer thank's for all the corrections, they where just typos!
– Davide Morgante
Jul 17 at 14:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
when and why is it possible to extend real-valued function to complex-valued functions?
is the wrong question, it's too general. We can't work with just any extension, we need certain regularity properties. We want a holomorphic extension. Technically, for $f$ defined by $(1)$ to be holomorphic it is not neccessary that $u$ is holomorphic, $z mapsto u(z/2,-iz/2)$ can be holomorphic even if $(z,w) mapsto u(z,w)$ isn't. But we can always work with a holomorphic extension.
The local version of the extension question becomes
Let $U subset mathbbR^2$ be open, and $u colon U to mathbbR$. When does there exist an open $Omega subset mathbbC^2$ with $U subsetOmega cap mathbbR^2$ and a holomorphic $tildeu colon Omega to mathbbC$ such that $tildeurvert_U = u$?
The answer is "if and only if $u$ is real-analytic". That's clearly necessary, since the Taylor expansion of $tildeu$ about a point in $U$ gives a power series expansion of $u$ about that point by restricting to real arguments. And it is sufficient because the Taylor series of $u$ about a point $p in U$ converges on some neighbourhood $W_p$ of $p$ in $mathbbC^2$ and thus provides a holomorphic extension of $urvert_W_p cap mathbbR^2$ to $W_p$. If we take appropriate $W_p$, say balls or polydisks with centre $p$, then all these extensions fit together and provide a holomorphic extension $tildeu$ with domain
$$bigcup_p in U W_p,.$$
A global version of the extension question is
Let $u colon mathbbR^2 to mathbbR$. When does there exist a holomorphic $tildeu colon mathbbC^2 to mathbbC$ with $tildeurvert_mathbbR^2 = u$?
The answer to this question is easy too. If such a $tildeu$ exists, its Taylor series (about $0$, say) converges on all of $mathbbC^2$, and thus $u$ has a globally convergent power series expansion. Conversely if $u$ has a power series expansion that converges on all of $mathbbR^2$, this series converges on all of $mathbbC^2$ and thus provides an entire $tildeu$.
Now, if $u colon mathbbR^2 to mathbbR$ is real-analytic, it need not be the case that $u$ has a globally convergent series expansion. Take for example $u(x,y) = frac11+x^2+y^2$. Its extension is meromorphic with the nonempty pole set $ (z,w) in mathbbC^2 : z^2 + w^2 = -1$.
But, the $u$ we want to extend is not an arbitrary real-analytic function. It is supposed to be the real part of a holomorphic function, and hence it must be harmonic. [Which yields sanity check 1 for such problems: Is the given function harmonic? If it isn't, it cannot be the real (or imaginary) part of a holomorphic function.] And an entire harmonic function has a globally convergent power series representation. (Take the Poisson integral over a sphere of radius $R$ and expand the Poisson kernel into a power series. This shows that the Taylor series about $0$ converges at least on the ball of radius $R$.)
Thus when we're looking for an entire function with a prescribed real part - and the given function is harmonic - the method always works.
The method can fail if we want to find a holomorphic $f colon Omega to mathbbC$ with prescribed real part $u$ when $Omega subsetneq mathbbC$, since $(z/2, -iz/2)$ need not belong to the domain of the holomorphic extension $tildeu$ for $zin Omega$ in that case.
Ahlfors gives a short discussion of the method in section 1.2 of chapter 2 in his Complex Analysis.
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
This was everything that I could ask for. Thank you very much for the answer!
– Davide Morgante
Jul 17 at 15:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
when and why is it possible to extend real-valued function to complex-valued functions?
is the wrong question, it's too general. We can't work with just any extension, we need certain regularity properties. We want a holomorphic extension. Technically, for $f$ defined by $(1)$ to be holomorphic it is not neccessary that $u$ is holomorphic, $z mapsto u(z/2,-iz/2)$ can be holomorphic even if $(z,w) mapsto u(z,w)$ isn't. But we can always work with a holomorphic extension.
The local version of the extension question becomes
Let $U subset mathbbR^2$ be open, and $u colon U to mathbbR$. When does there exist an open $Omega subset mathbbC^2$ with $U subsetOmega cap mathbbR^2$ and a holomorphic $tildeu colon Omega to mathbbC$ such that $tildeurvert_U = u$?
The answer is "if and only if $u$ is real-analytic". That's clearly necessary, since the Taylor expansion of $tildeu$ about a point in $U$ gives a power series expansion of $u$ about that point by restricting to real arguments. And it is sufficient because the Taylor series of $u$ about a point $p in U$ converges on some neighbourhood $W_p$ of $p$ in $mathbbC^2$ and thus provides a holomorphic extension of $urvert_W_p cap mathbbR^2$ to $W_p$. If we take appropriate $W_p$, say balls or polydisks with centre $p$, then all these extensions fit together and provide a holomorphic extension $tildeu$ with domain
$$bigcup_p in U W_p,.$$
A global version of the extension question is
Let $u colon mathbbR^2 to mathbbR$. When does there exist a holomorphic $tildeu colon mathbbC^2 to mathbbC$ with $tildeurvert_mathbbR^2 = u$?
The answer to this question is easy too. If such a $tildeu$ exists, its Taylor series (about $0$, say) converges on all of $mathbbC^2$, and thus $u$ has a globally convergent power series expansion. Conversely if $u$ has a power series expansion that converges on all of $mathbbR^2$, this series converges on all of $mathbbC^2$ and thus provides an entire $tildeu$.
Now, if $u colon mathbbR^2 to mathbbR$ is real-analytic, it need not be the case that $u$ has a globally convergent series expansion. Take for example $u(x,y) = frac11+x^2+y^2$. Its extension is meromorphic with the nonempty pole set $ (z,w) in mathbbC^2 : z^2 + w^2 = -1$.
But, the $u$ we want to extend is not an arbitrary real-analytic function. It is supposed to be the real part of a holomorphic function, and hence it must be harmonic. [Which yields sanity check 1 for such problems: Is the given function harmonic? If it isn't, it cannot be the real (or imaginary) part of a holomorphic function.] And an entire harmonic function has a globally convergent power series representation. (Take the Poisson integral over a sphere of radius $R$ and expand the Poisson kernel into a power series. This shows that the Taylor series about $0$ converges at least on the ball of radius $R$.)
Thus when we're looking for an entire function with a prescribed real part - and the given function is harmonic - the method always works.
The method can fail if we want to find a holomorphic $f colon Omega to mathbbC$ with prescribed real part $u$ when $Omega subsetneq mathbbC$, since $(z/2, -iz/2)$ need not belong to the domain of the holomorphic extension $tildeu$ for $zin Omega$ in that case.
Ahlfors gives a short discussion of the method in section 1.2 of chapter 2 in his Complex Analysis.
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
This was everything that I could ask for. Thank you very much for the answer!
– Davide Morgante
Jul 17 at 15:06
add a comment |Â
up vote
3
down vote
accepted
when and why is it possible to extend real-valued function to complex-valued functions?
is the wrong question, it's too general. We can't work with just any extension, we need certain regularity properties. We want a holomorphic extension. Technically, for $f$ defined by $(1)$ to be holomorphic it is not neccessary that $u$ is holomorphic, $z mapsto u(z/2,-iz/2)$ can be holomorphic even if $(z,w) mapsto u(z,w)$ isn't. But we can always work with a holomorphic extension.
The local version of the extension question becomes
Let $U subset mathbbR^2$ be open, and $u colon U to mathbbR$. When does there exist an open $Omega subset mathbbC^2$ with $U subsetOmega cap mathbbR^2$ and a holomorphic $tildeu colon Omega to mathbbC$ such that $tildeurvert_U = u$?
The answer is "if and only if $u$ is real-analytic". That's clearly necessary, since the Taylor expansion of $tildeu$ about a point in $U$ gives a power series expansion of $u$ about that point by restricting to real arguments. And it is sufficient because the Taylor series of $u$ about a point $p in U$ converges on some neighbourhood $W_p$ of $p$ in $mathbbC^2$ and thus provides a holomorphic extension of $urvert_W_p cap mathbbR^2$ to $W_p$. If we take appropriate $W_p$, say balls or polydisks with centre $p$, then all these extensions fit together and provide a holomorphic extension $tildeu$ with domain
$$bigcup_p in U W_p,.$$
A global version of the extension question is
Let $u colon mathbbR^2 to mathbbR$. When does there exist a holomorphic $tildeu colon mathbbC^2 to mathbbC$ with $tildeurvert_mathbbR^2 = u$?
The answer to this question is easy too. If such a $tildeu$ exists, its Taylor series (about $0$, say) converges on all of $mathbbC^2$, and thus $u$ has a globally convergent power series expansion. Conversely if $u$ has a power series expansion that converges on all of $mathbbR^2$, this series converges on all of $mathbbC^2$ and thus provides an entire $tildeu$.
Now, if $u colon mathbbR^2 to mathbbR$ is real-analytic, it need not be the case that $u$ has a globally convergent series expansion. Take for example $u(x,y) = frac11+x^2+y^2$. Its extension is meromorphic with the nonempty pole set $ (z,w) in mathbbC^2 : z^2 + w^2 = -1$.
But, the $u$ we want to extend is not an arbitrary real-analytic function. It is supposed to be the real part of a holomorphic function, and hence it must be harmonic. [Which yields sanity check 1 for such problems: Is the given function harmonic? If it isn't, it cannot be the real (or imaginary) part of a holomorphic function.] And an entire harmonic function has a globally convergent power series representation. (Take the Poisson integral over a sphere of radius $R$ and expand the Poisson kernel into a power series. This shows that the Taylor series about $0$ converges at least on the ball of radius $R$.)
Thus when we're looking for an entire function with a prescribed real part - and the given function is harmonic - the method always works.
The method can fail if we want to find a holomorphic $f colon Omega to mathbbC$ with prescribed real part $u$ when $Omega subsetneq mathbbC$, since $(z/2, -iz/2)$ need not belong to the domain of the holomorphic extension $tildeu$ for $zin Omega$ in that case.
Ahlfors gives a short discussion of the method in section 1.2 of chapter 2 in his Complex Analysis.
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
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Jul 17 at 15:06
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when and why is it possible to extend real-valued function to complex-valued functions?
is the wrong question, it's too general. We can't work with just any extension, we need certain regularity properties. We want a holomorphic extension. Technically, for $f$ defined by $(1)$ to be holomorphic it is not neccessary that $u$ is holomorphic, $z mapsto u(z/2,-iz/2)$ can be holomorphic even if $(z,w) mapsto u(z,w)$ isn't. But we can always work with a holomorphic extension.
The local version of the extension question becomes
Let $U subset mathbbR^2$ be open, and $u colon U to mathbbR$. When does there exist an open $Omega subset mathbbC^2$ with $U subsetOmega cap mathbbR^2$ and a holomorphic $tildeu colon Omega to mathbbC$ such that $tildeurvert_U = u$?
The answer is "if and only if $u$ is real-analytic". That's clearly necessary, since the Taylor expansion of $tildeu$ about a point in $U$ gives a power series expansion of $u$ about that point by restricting to real arguments. And it is sufficient because the Taylor series of $u$ about a point $p in U$ converges on some neighbourhood $W_p$ of $p$ in $mathbbC^2$ and thus provides a holomorphic extension of $urvert_W_p cap mathbbR^2$ to $W_p$. If we take appropriate $W_p$, say balls or polydisks with centre $p$, then all these extensions fit together and provide a holomorphic extension $tildeu$ with domain
$$bigcup_p in U W_p,.$$
A global version of the extension question is
Let $u colon mathbbR^2 to mathbbR$. When does there exist a holomorphic $tildeu colon mathbbC^2 to mathbbC$ with $tildeurvert_mathbbR^2 = u$?
The answer to this question is easy too. If such a $tildeu$ exists, its Taylor series (about $0$, say) converges on all of $mathbbC^2$, and thus $u$ has a globally convergent power series expansion. Conversely if $u$ has a power series expansion that converges on all of $mathbbR^2$, this series converges on all of $mathbbC^2$ and thus provides an entire $tildeu$.
Now, if $u colon mathbbR^2 to mathbbR$ is real-analytic, it need not be the case that $u$ has a globally convergent series expansion. Take for example $u(x,y) = frac11+x^2+y^2$. Its extension is meromorphic with the nonempty pole set $ (z,w) in mathbbC^2 : z^2 + w^2 = -1$.
But, the $u$ we want to extend is not an arbitrary real-analytic function. It is supposed to be the real part of a holomorphic function, and hence it must be harmonic. [Which yields sanity check 1 for such problems: Is the given function harmonic? If it isn't, it cannot be the real (or imaginary) part of a holomorphic function.] And an entire harmonic function has a globally convergent power series representation. (Take the Poisson integral over a sphere of radius $R$ and expand the Poisson kernel into a power series. This shows that the Taylor series about $0$ converges at least on the ball of radius $R$.)
Thus when we're looking for an entire function with a prescribed real part - and the given function is harmonic - the method always works.
The method can fail if we want to find a holomorphic $f colon Omega to mathbbC$ with prescribed real part $u$ when $Omega subsetneq mathbbC$, since $(z/2, -iz/2)$ need not belong to the domain of the holomorphic extension $tildeu$ for $zin Omega$ in that case.
Ahlfors gives a short discussion of the method in section 1.2 of chapter 2 in his Complex Analysis.
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
when and why is it possible to extend real-valued function to complex-valued functions?
is the wrong question, it's too general. We can't work with just any extension, we need certain regularity properties. We want a holomorphic extension. Technically, for $f$ defined by $(1)$ to be holomorphic it is not neccessary that $u$ is holomorphic, $z mapsto u(z/2,-iz/2)$ can be holomorphic even if $(z,w) mapsto u(z,w)$ isn't. But we can always work with a holomorphic extension.
The local version of the extension question becomes
Let $U subset mathbbR^2$ be open, and $u colon U to mathbbR$. When does there exist an open $Omega subset mathbbC^2$ with $U subsetOmega cap mathbbR^2$ and a holomorphic $tildeu colon Omega to mathbbC$ such that $tildeurvert_U = u$?
The answer is "if and only if $u$ is real-analytic". That's clearly necessary, since the Taylor expansion of $tildeu$ about a point in $U$ gives a power series expansion of $u$ about that point by restricting to real arguments. And it is sufficient because the Taylor series of $u$ about a point $p in U$ converges on some neighbourhood $W_p$ of $p$ in $mathbbC^2$ and thus provides a holomorphic extension of $urvert_W_p cap mathbbR^2$ to $W_p$. If we take appropriate $W_p$, say balls or polydisks with centre $p$, then all these extensions fit together and provide a holomorphic extension $tildeu$ with domain
$$bigcup_p in U W_p,.$$
A global version of the extension question is
Let $u colon mathbbR^2 to mathbbR$. When does there exist a holomorphic $tildeu colon mathbbC^2 to mathbbC$ with $tildeurvert_mathbbR^2 = u$?
The answer to this question is easy too. If such a $tildeu$ exists, its Taylor series (about $0$, say) converges on all of $mathbbC^2$, and thus $u$ has a globally convergent power series expansion. Conversely if $u$ has a power series expansion that converges on all of $mathbbR^2$, this series converges on all of $mathbbC^2$ and thus provides an entire $tildeu$.
Now, if $u colon mathbbR^2 to mathbbR$ is real-analytic, it need not be the case that $u$ has a globally convergent series expansion. Take for example $u(x,y) = frac11+x^2+y^2$. Its extension is meromorphic with the nonempty pole set $ (z,w) in mathbbC^2 : z^2 + w^2 = -1$.
But, the $u$ we want to extend is not an arbitrary real-analytic function. It is supposed to be the real part of a holomorphic function, and hence it must be harmonic. [Which yields sanity check 1 for such problems: Is the given function harmonic? If it isn't, it cannot be the real (or imaginary) part of a holomorphic function.] And an entire harmonic function has a globally convergent power series representation. (Take the Poisson integral over a sphere of radius $R$ and expand the Poisson kernel into a power series. This shows that the Taylor series about $0$ converges at least on the ball of radius $R$.)
Thus when we're looking for an entire function with a prescribed real part - and the given function is harmonic - the method always works.
The method can fail if we want to find a holomorphic $f colon Omega to mathbbC$ with prescribed real part $u$ when $Omega subsetneq mathbbC$, since $(z/2, -iz/2)$ need not belong to the domain of the holomorphic extension $tildeu$ for $zin Omega$ in that case.
Ahlfors gives a short discussion of the method in section 1.2 of chapter 2 in his Complex Analysis.
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
answered Jul 17 at 14:53
Daniel Fischer♦
171k16154274
171k16154274
This was everything that I could ask for. Thank you very much for the answer!
– Davide Morgante
Jul 17 at 15:06
add a comment |Â
This was everything that I could ask for. Thank you very much for the answer!
– Davide Morgante
Jul 17 at 15:06
This was everything that I could ask for. Thank you very much for the answer!
– Davide Morgante
Jul 17 at 15:06
This was everything that I could ask for. Thank you very much for the answer!
– Davide Morgante
Jul 17 at 15:06
add a comment |Â
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1
That's all. There is nothing mysterious about it. If $f$ exists, then $overlinef(overlinez)$ is also analytic. Therefore $tildeu(z_1,z_2)=frac12(f(z_1+iz_2))+overlinef(overlinez_1-iz_2)$ is analytic. The restriction of this to the real plane $z_1=overlinez_1,z_2=overlinez_2$ coincides with $u=Re(f)$. Therefore, if $u$ is real analytic, and $hatu$ is its complexification, then $tildeu=hatu$ and $hatu(z/2,z/(2i))=frac12(f(z)+overlinef(0))$. What is the complexification? Just take the power series of $u$ and look at it as a series over the complex.
– user574889
Jul 17 at 12:13
You have a typo in $(2)$, $overlinef(x+iy) = overlinefbigl(overlinex-iybigr)$, not as you wrote $overlinefbigl(overlinex+iybigr)$.
– Daniel Fischer♦
Jul 17 at 12:38
A little below that you have "$dotsc = f(z) + overlinef(w) = f(z) + overlinefbigl(overlinewbigr)$". There's no reason why $overlinef(w) = overlinefbigl(overlinewbigr)$ should hold. But maybe it's just a typo and should have been $overlinefbigl(overlinewbigr)$.
– Daniel Fischer♦
Jul 17 at 12:42
@DanielFischer thank's for all the corrections, they where just typos!
– Davide Morgante
Jul 17 at 14:06