If $u$ and $v$ are complex-valued and satisfy the Cauchy-Riemann equations, does it follow that $u$ and $v$...
$begingroup$
I am trying to follow an example in my literature and I am pretty lost. It says that if $u$ and $v$ are complex-valued functions that are $C^1$-smooth in some open set, and $$frac{partial u}{partial x} = frac{partial v}{partial y}, hspace{5mm}frac{partial u}{partial y} = -frac{partial v}{partial x}, $$ then we can show that $u$ and $v$ are $C^{infty}$-smooth. Then they go on to show it, by stating that it follows from the assumptions that $f=u+iv$ and $g = bar{u} + ibar{v}$ are analytic, and hence that $$u = frac{f + bar{g}}{2} $$ and $$v=frac{f-bar{g}}{2i} $$ are analytic, and so $u$ and $v$ must be $C^{infty}$-smooth, since analytic functions always are.
The issues I have with this is:
1) Why would it even follow that $f=u+iv$ is analytic? I mean, if $u$ and $v$ were real-valued, then I know that it does follow. But $u$ and $v$ aren't said to be real-valued in this case.
2) I also don't see how we can draw from the assumption the conclusion that $g =bar{u} + ibar{v}$ would be analytic. I suppose that if I understood (1), then maybe I would understand (2) as well.
3) And lastly, I also don't get how we can assume that for example $frac{f-bar{g}}{2}$ is analytic. I mean, $g$ being analytic doesn't imply that $bar{g}$ is, does it?
I wish I had any of my own work to show, but at this point I am simply looking through the literature, trying to find something that explains these conclusions that are drawn in the example and I can't find any... So I'm hoping for help here.
EDIT
I think I have figured out why $f=u+iv$ and $g=bar{u} + ibar{v}$ are analytic. So I suppose that what I need help with right now is to understand why that implies that for example $$u = frac{f + bar{g}}{2} $$ is $C^{infty}$-smooth.
complex-analysis
asked Dec 7 '18 at 0:53
j.eeej.eee
766
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add a comment |
$begingroup$
I am trying to follow an example in my literature and I am pretty lost. It says that if $u$ and $v$ are complex-valued functions that are $C^1$-smooth in some open set, and $$frac{partial u}{partial x} = frac{partial v}{partial y}, hspace{5mm}frac{partial u}{partial y} = -frac{partial v}{partial x}, $$ then we can show that $u$ and $v$ are $C^{infty}$-smooth. Then they go on to show it, by stating that it follows from the assumptions that $f=u+iv$ and $g = bar{u} + ibar{v}$ are analytic, and hence that $$u = frac{f + bar{g}}{2} $$ and $$v=frac{f-bar{g}}{2i} $$ are analytic, and so $u$ and $v$ must be $C^{infty}$-smooth, since analytic functions always are.
The issues I have with this is:
1) Why would it even follow that $f=u+iv$ is analytic? I mean, if $u$ and $v$ were real-valued, then I know that it does follow. But $u$ and $v$ aren't said to be real-valued in this case.
2) I also don't see how we can draw from the assumption the conclusion that $g =bar{u} + ibar{v}$ would be analytic. I suppose that if I understood (1), then maybe I would understand (2) as well.
3) And lastly, I also don't get how we can assume that for example $frac{f-bar{g}}{2}$ is analytic. I mean, $g$ being analytic doesn't imply that $bar{g}$ is, does it?
I wish I had any of my own work to show, but at this point I am simply looking through the literature, trying to find something that explains these conclusions that are drawn in the example and I can't find any... So I'm hoping for help here.
EDIT
I think I have figured out why $f=u+iv$ and $g=bar{u} + ibar{v}$ are analytic. So I suppose that what I need help with right now is to understand why that implies that for example $$u = frac{f + bar{g}}{2} $$ is $C^{infty}$-smooth.
complex-analysis
asked Dec 7 '18 at 0:53
j.eeej.eee
766
$endgroup$
$begingroup$
It's not really clear what you have available, but I'll try to add some thoughts. First, there is a difference between real- and complex-analytic that's important here, and neither are equivalent to (nor implied by) $C^{infty}$. Since you're talking about the real and imaginary parts, real-analyticity is probably the right thing to look at. The proof that solutions to the Cauchy-Riemann equations are analytic is non-trivial and typically goes through the Cauchy integral formula or a variant on Weyl's lemma.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:59
$begingroup$
(I've never heard of Weyl's lemma before.) In my literature, there hasn't really been a distinction made between real-analytic and complex-analytic. Which now makes me feel really frustrated. :p But there is a theorem stating that if a function $f$ is analytic on some domain, then it is $C^{infty}$-smooth there... And also a theorem stating that if $u,v$ are real-valued and satisfy the Cauchy-Riemann equations, then $f=u+iv$ is analytic. Are you saying that this is true even if $u$ and $v$ are not necessarily real? @T.Bongers
$endgroup$
– j.eee
Dec 7 '18 at 1:10
$begingroup$
The real part of a complex-analytic function is generally not complex-analytic, so there is an important distinction to make. Perhaps this question will help.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:12
$begingroup$
One should be careful. If $f$ is analytic, then $overline{f(bar z)} $ is analytic. The real and imaginary parts of a complex analytic function aren't in general complex analytic, so you can't expect them to be a sum of complex analytic functions. We do have that the real and imaginary parts are Harmonic though, which minimally implies $C^infty$
$endgroup$
– Brevan Ellefsen
Dec 7 '18 at 16:51
add a comment |
$begingroup$
I am trying to follow an example in my literature and I am pretty lost. It says that if $u$ and $v$ are complex-valued functions that are $C^1$-smooth in some open set, and $$frac{partial u}{partial x} = frac{partial v}{partial y}, hspace{5mm}frac{partial u}{partial y} = -frac{partial v}{partial x}, $$ then we can show that $u$ and $v$ are $C^{infty}$-smooth. Then they go on to show it, by stating that it follows from the assumptions that $f=u+iv$ and $g = bar{u} + ibar{v}$ are analytic, and hence that $$u = frac{f + bar{g}}{2} $$ and $$v=frac{f-bar{g}}{2i} $$ are analytic, and so $u$ and $v$ must be $C^{infty}$-smooth, since analytic functions always are.
The issues I have with this is:
1) Why would it even follow that $f=u+iv$ is analytic? I mean, if $u$ and $v$ were real-valued, then I know that it does follow. But $u$ and $v$ aren't said to be real-valued in this case.
2) I also don't see how we can draw from the assumption the conclusion that $g =bar{u} + ibar{v}$ would be analytic. I suppose that if I understood (1), then maybe I would understand (2) as well.
3) And lastly, I also don't get how we can assume that for example $frac{f-bar{g}}{2}$ is analytic. I mean, $g$ being analytic doesn't imply that $bar{g}$ is, does it?
I wish I had any of my own work to show, but at this point I am simply looking through the literature, trying to find something that explains these conclusions that are drawn in the example and I can't find any... So I'm hoping for help here.
EDIT
I think I have figured out why $f=u+iv$ and $g=bar{u} + ibar{v}$ are analytic. So I suppose that what I need help with right now is to understand why that implies that for example $$u = frac{f + bar{g}}{2} $$ is $C^{infty}$-smooth.
complex-analysis
asked Dec 7 '18 at 0:53
j.eeej.eee
766
$endgroup$
I am trying to follow an example in my literature and I am pretty lost. It says that if $u$ and $v$ are complex-valued functions that are $C^1$-smooth in some open set, and $$frac{partial u}{partial x} = frac{partial v}{partial y}, hspace{5mm}frac{partial u}{partial y} = -frac{partial v}{partial x}, $$ then we can show that $u$ and $v$ are $C^{infty}$-smooth. Then they go on to show it, by stating that it follows from the assumptions that $f=u+iv$ and $g = bar{u} + ibar{v}$ are analytic, and hence that $$u = frac{f + bar{g}}{2} $$ and $$v=frac{f-bar{g}}{2i} $$ are analytic, and so $u$ and $v$ must be $C^{infty}$-smooth, since analytic functions always are.
The issues I have with this is:
1) Why would it even follow that $f=u+iv$ is analytic? I mean, if $u$ and $v$ were real-valued, then I know that it does follow. But $u$ and $v$ aren't said to be real-valued in this case.
2) I also don't see how we can draw from the assumption the conclusion that $g =bar{u} + ibar{v}$ would be analytic. I suppose that if I understood (1), then maybe I would understand (2) as well.
3) And lastly, I also don't get how we can assume that for example $frac{f-bar{g}}{2}$ is analytic. I mean, $g$ being analytic doesn't imply that $bar{g}$ is, does it?
I wish I had any of my own work to show, but at this point I am simply looking through the literature, trying to find something that explains these conclusions that are drawn in the example and I can't find any... So I'm hoping for help here.
EDIT
I think I have figured out why $f=u+iv$ and $g=bar{u} + ibar{v}$ are analytic. So I suppose that what I need help with right now is to understand why that implies that for example $$u = frac{f + bar{g}}{2} $$ is $C^{infty}$-smooth.
complex-analysis
complex-analysis
asked Dec 7 '18 at 0:53
j.eeej.eee
766
asked Dec 7 '18 at 0:53
j.eeej.eee
766
edited Dec 7 '18 at 2:23
j.eee
asked Dec 7 '18 at 0:53
j.eeej.eee
766
asked Dec 7 '18 at 0:53
j.eeej.eee
766
asked Dec 7 '18 at 0:53
j.eeej.eee
766
766
$begingroup$
It's not really clear what you have available, but I'll try to add some thoughts. First, there is a difference between real- and complex-analytic that's important here, and neither are equivalent to (nor implied by) $C^{infty}$. Since you're talking about the real and imaginary parts, real-analyticity is probably the right thing to look at. The proof that solutions to the Cauchy-Riemann equations are analytic is non-trivial and typically goes through the Cauchy integral formula or a variant on Weyl's lemma.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:59
$begingroup$
(I've never heard of Weyl's lemma before.) In my literature, there hasn't really been a distinction made between real-analytic and complex-analytic. Which now makes me feel really frustrated. :p But there is a theorem stating that if a function $f$ is analytic on some domain, then it is $C^{infty}$-smooth there... And also a theorem stating that if $u,v$ are real-valued and satisfy the Cauchy-Riemann equations, then $f=u+iv$ is analytic. Are you saying that this is true even if $u$ and $v$ are not necessarily real? @T.Bongers
$endgroup$
– j.eee
Dec 7 '18 at 1:10
$begingroup$
The real part of a complex-analytic function is generally not complex-analytic, so there is an important distinction to make. Perhaps this question will help.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:12
$begingroup$
One should be careful. If $f$ is analytic, then $overline{f(bar z)} $ is analytic. The real and imaginary parts of a complex analytic function aren't in general complex analytic, so you can't expect them to be a sum of complex analytic functions. We do have that the real and imaginary parts are Harmonic though, which minimally implies $C^infty$
$endgroup$
– Brevan Ellefsen
Dec 7 '18 at 16:51
add a comment |
$begingroup$
It's not really clear what you have available, but I'll try to add some thoughts. First, there is a difference between real- and complex-analytic that's important here, and neither are equivalent to (nor implied by) $C^{infty}$. Since you're talking about the real and imaginary parts, real-analyticity is probably the right thing to look at. The proof that solutions to the Cauchy-Riemann equations are analytic is non-trivial and typically goes through the Cauchy integral formula or a variant on Weyl's lemma.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:59
$begingroup$
(I've never heard of Weyl's lemma before.) In my literature, there hasn't really been a distinction made between real-analytic and complex-analytic. Which now makes me feel really frustrated. :p But there is a theorem stating that if a function $f$ is analytic on some domain, then it is $C^{infty}$-smooth there... And also a theorem stating that if $u,v$ are real-valued and satisfy the Cauchy-Riemann equations, then $f=u+iv$ is analytic. Are you saying that this is true even if $u$ and $v$ are not necessarily real? @T.Bongers
$endgroup$
– j.eee
Dec 7 '18 at 1:10
$begingroup$
The real part of a complex-analytic function is generally not complex-analytic, so there is an important distinction to make. Perhaps this question will help.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:12
$begingroup$
One should be careful. If $f$ is analytic, then $overline{f(bar z)} $ is analytic. The real and imaginary parts of a complex analytic function aren't in general complex analytic, so you can't expect them to be a sum of complex analytic functions. We do have that the real and imaginary parts are Harmonic though, which minimally implies $C^infty$
$endgroup$
– Brevan Ellefsen
Dec 7 '18 at 16:51
$begingroup$
It's not really clear what you have available, but I'll try to add some thoughts. First, there is a difference between real- and complex-analytic that's important here, and neither are equivalent to (nor implied by) $C^{infty}$. Since you're talking about the real and imaginary parts, real-analyticity is probably the right thing to look at. The proof that solutions to the Cauchy-Riemann equations are analytic is non-trivial and typically goes through the Cauchy integral formula or a variant on Weyl's lemma.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:59
$begingroup$
It's not really clear what you have available, but I'll try to add some thoughts. First, there is a difference between real- and complex-analytic that's important here, and neither are equivalent to (nor implied by) $C^{infty}$. Since you're talking about the real and imaginary parts, real-analyticity is probably the right thing to look at. The proof that solutions to the Cauchy-Riemann equations are analytic is non-trivial and typically goes through the Cauchy integral formula or a variant on Weyl's lemma.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:59
$begingroup$
(I've never heard of Weyl's lemma before.) In my literature, there hasn't really been a distinction made between real-analytic and complex-analytic. Which now makes me feel really frustrated. :p But there is a theorem stating that if a function $f$ is analytic on some domain, then it is $C^{infty}$-smooth there... And also a theorem stating that if $u,v$ are real-valued and satisfy the Cauchy-Riemann equations, then $f=u+iv$ is analytic. Are you saying that this is true even if $u$ and $v$ are not necessarily real? @T.Bongers
$endgroup$
– j.eee
Dec 7 '18 at 1:10
$begingroup$
(I've never heard of Weyl's lemma before.) In my literature, there hasn't really been a distinction made between real-analytic and complex-analytic. Which now makes me feel really frustrated. :p But there is a theorem stating that if a function $f$ is analytic on some domain, then it is $C^{infty}$-smooth there... And also a theorem stating that if $u,v$ are real-valued and satisfy the Cauchy-Riemann equations, then $f=u+iv$ is analytic. Are you saying that this is true even if $u$ and $v$ are not necessarily real? @T.Bongers
$endgroup$
– j.eee
Dec 7 '18 at 1:10
$begingroup$
The real part of a complex-analytic function is generally not complex-analytic, so there is an important distinction to make. Perhaps this question will help.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:12
$begingroup$
The real part of a complex-analytic function is generally not complex-analytic, so there is an important distinction to make. Perhaps this question will help.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:12
$begingroup$
One should be careful. If $f$ is analytic, then $overline{f(bar z)} $ is analytic. The real and imaginary parts of a complex analytic function aren't in general complex analytic, so you can't expect them to be a sum of complex analytic functions. We do have that the real and imaginary parts are Harmonic though, which minimally implies $C^infty$
$endgroup$
– Brevan Ellefsen
Dec 7 '18 at 16:51
$begingroup$
One should be careful. If $f$ is analytic, then $overline{f(bar z)} $ is analytic. The real and imaginary parts of a complex analytic function aren't in general complex analytic, so you can't expect them to be a sum of complex analytic functions. We do have that the real and imaginary parts are Harmonic though, which minimally implies $C^infty$
$endgroup$
– Brevan Ellefsen
Dec 7 '18 at 16:51
add a comment |
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$begingroup$
It's not really clear what you have available, but I'll try to add some thoughts. First, there is a difference between real- and complex-analytic that's important here, and neither are equivalent to (nor implied by) $C^{infty}$. Since you're talking about the real and imaginary parts, real-analyticity is probably the right thing to look at. The proof that solutions to the Cauchy-Riemann equations are analytic is non-trivial and typically goes through the Cauchy integral formula or a variant on Weyl's lemma.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:59
$begingroup$
(I've never heard of Weyl's lemma before.) In my literature, there hasn't really been a distinction made between real-analytic and complex-analytic. Which now makes me feel really frustrated. :p But there is a theorem stating that if a function $f$ is analytic on some domain, then it is $C^{infty}$-smooth there... And also a theorem stating that if $u,v$ are real-valued and satisfy the Cauchy-Riemann equations, then $f=u+iv$ is analytic. Are you saying that this is true even if $u$ and $v$ are not necessarily real? @T.Bongers
$endgroup$
– j.eee
Dec 7 '18 at 1:10
$begingroup$
The real part of a complex-analytic function is generally not complex-analytic, so there is an important distinction to make. Perhaps this question will help.
$endgroup$
– T. Bongers
Dec 7 '18 at 1:12
$begingroup$
One should be careful. If $f$ is analytic, then $overline{f(bar z)} $ is analytic. The real and imaginary parts of a complex analytic function aren't in general complex analytic, so you can't expect them to be a sum of complex analytic functions. We do have that the real and imaginary parts are Harmonic though, which minimally implies $C^infty$
$endgroup$
– Brevan Ellefsen
Dec 7 '18 at 16:51