Mixing
Questions About the Proof of Cauchy–Pompeiu Integral Formula.
$begingroup$
I am studying function theory in several complex variables and the book I am using is "Tasty Bits of Several Complex Variables" by Jiří Lebl: https://www.jirka.org/scv/scv.pdf.
At the moment I am reading chapter 4, where he introduces the $bar{partial}$-problem. However, he begins by proving a generalized form of the Cauchy Integral formula and I have some questions about the proof.
Theorem:
Let $Usubsetmathbb{C}$ be a bounded domain with piecewise $C^1$-smooth boundary $partial U$ oriented positively, and let $f:bar{U}tomathbb{C}$ be a $ C^1$-smooth function. Then for $ zin U$:
$ f(z)=frac{1}{2pi i}int_{partial U}frac{f(zeta)}{zeta-z}dzeta+frac{1}{2pi i}int_Udfrac{frac{partial f}{partialbar{z}}(zeta)}{zeta-z}dzetawedge dbar{zeta}$.
"Proof":
This is how he does it.
He begins by fixing $zin U$ and a small disc $Delta_r(z)$ such that $Delta_r(z) subsetsubset U$. He then applies Stokes Theorem, which gives us
$$int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big )=int_{UbackslashDelta_r(z)}dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta.quadquadquad (*)$$
The above equation begs us to let $r$ tend to $0$. If we manage to get anything out of this, we are done.
Notice that $dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}$ is integrable over all of $U$ (this follows from an exercise right before the proof). Therefore
$$lim_{rto 0}int_{UbackslashDelta_r(z)} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=-int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dzeta wedge dbar{zeta},$$
where we used the asymmetry of the operator $wedge$ in the last equality.
As a last step. Jiří argues that, by continuity of $f$, we find
$$lim_{rto 0}frac{1}{2pi i}int_{partialDelta_r(z)} dfrac{f(zeta)}{zeta-z}dbar{zeta}wedge dzeta = lim_{rto 0}frac{1}{2pi}int_{0}^{2pi} f(z+re^{itheta})dtheta=f(z).quadquadquad (**)$$
Questions:
(1) My first question is about the Stokes Theorem.
It's quite a long time since I last used Stokes theorem and back then it was in a basic calculus class. I have looked around and I cannot manage to find something which seems to give me
$$ int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big ).$$
I suspect there is a differential form version of Stokes Theorem I should use?
I found this on wikipedia:
Stokes' theorem says that the integral of a differential form ω over
the boundary of some orientable manifold Ω is equal to the integral of
its exterior derivative dω over the whole of Ω, i.e.,
$$int_{partialOmega}omega=int_{Omega}domega.$$
Is it the above I am using in some way, if so, could you please tell me how this is being used? Also, as a bonus question, I do not see why we choose a small disc $Delta_r(z)subsetsubset U$. But perhaps this will be answered at the same time as the explanation of Stokes Theorem.
(2) He argues that the second equality in $(*)$ follows since holomorphic derivatives in $zeta$ will have a $dzeta$ and when we wedge with $dzeta$ we just get zero.
I, honestly, do not understand what I just wrote. Can someone explain to me what this means or give another argument why the second equality is true?
(3) The computations in $(**)$ seems to be fundamental ones. For instance, it seems to me that we are using the Cauchy integral formula(?) And doing a variable substitution - polar coordinates(?) However, I am not sure about this. I would be really happy if you could tell me how the first equality in $(**)$ holds and where the continuity of $f$ is being used.
Just as a last thought. Perhaps the continuity is being used when we compute the limit to obtain the last equality?
As you see, I find it quite difficult to understand the proof, so I would be really happy if you could help me by answering (1), (2) and/or (3). Thanks for taking your time!
complex-analysis holomorphic-functions exterior-algebra cauchy-integral-formula several-complex-variables
asked Jan 3 at 23:48
JoeJoe
4214
$endgroup$
add a comment |
$begingroup$
I am studying function theory in several complex variables and the book I am using is "Tasty Bits of Several Complex Variables" by Jiří Lebl: https://www.jirka.org/scv/scv.pdf.
At the moment I am reading chapter 4, where he introduces the $bar{partial}$-problem. However, he begins by proving a generalized form of the Cauchy Integral formula and I have some questions about the proof.
Theorem:
Let $Usubsetmathbb{C}$ be a bounded domain with piecewise $C^1$-smooth boundary $partial U$ oriented positively, and let $f:bar{U}tomathbb{C}$ be a $ C^1$-smooth function. Then for $ zin U$:
$ f(z)=frac{1}{2pi i}int_{partial U}frac{f(zeta)}{zeta-z}dzeta+frac{1}{2pi i}int_Udfrac{frac{partial f}{partialbar{z}}(zeta)}{zeta-z}dzetawedge dbar{zeta}$.
"Proof":
This is how he does it.
He begins by fixing $zin U$ and a small disc $Delta_r(z)$ such that $Delta_r(z) subsetsubset U$. He then applies Stokes Theorem, which gives us
$$int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big )=int_{UbackslashDelta_r(z)}dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta.quadquadquad (*)$$
The above equation begs us to let $r$ tend to $0$. If we manage to get anything out of this, we are done.
Notice that $dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}$ is integrable over all of $U$ (this follows from an exercise right before the proof). Therefore
$$lim_{rto 0}int_{UbackslashDelta_r(z)} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=-int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dzeta wedge dbar{zeta},$$
where we used the asymmetry of the operator $wedge$ in the last equality.
As a last step. Jiří argues that, by continuity of $f$, we find
$$lim_{rto 0}frac{1}{2pi i}int_{partialDelta_r(z)} dfrac{f(zeta)}{zeta-z}dbar{zeta}wedge dzeta = lim_{rto 0}frac{1}{2pi}int_{0}^{2pi} f(z+re^{itheta})dtheta=f(z).quadquadquad (**)$$
Questions:
(1) My first question is about the Stokes Theorem.
It's quite a long time since I last used Stokes theorem and back then it was in a basic calculus class. I have looked around and I cannot manage to find something which seems to give me
$$ int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big ).$$
I suspect there is a differential form version of Stokes Theorem I should use?
I found this on wikipedia:
Stokes' theorem says that the integral of a differential form ω over
the boundary of some orientable manifold Ω is equal to the integral of
its exterior derivative dω over the whole of Ω, i.e.,
$$int_{partialOmega}omega=int_{Omega}domega.$$
Is it the above I am using in some way, if so, could you please tell me how this is being used? Also, as a bonus question, I do not see why we choose a small disc $Delta_r(z)subsetsubset U$. But perhaps this will be answered at the same time as the explanation of Stokes Theorem.
(2) He argues that the second equality in $(*)$ follows since holomorphic derivatives in $zeta$ will have a $dzeta$ and when we wedge with $dzeta$ we just get zero.
I, honestly, do not understand what I just wrote. Can someone explain to me what this means or give another argument why the second equality is true?
(3) The computations in $(**)$ seems to be fundamental ones. For instance, it seems to me that we are using the Cauchy integral formula(?) And doing a variable substitution - polar coordinates(?) However, I am not sure about this. I would be really happy if you could tell me how the first equality in $(**)$ holds and where the continuity of $f$ is being used.
Just as a last thought. Perhaps the continuity is being used when we compute the limit to obtain the last equality?
As you see, I find it quite difficult to understand the proof, so I would be really happy if you could help me by answering (1), (2) and/or (3). Thanks for taking your time!
complex-analysis holomorphic-functions exterior-algebra cauchy-integral-formula several-complex-variables
asked Jan 3 at 23:48
JoeJoe
4214
$endgroup$
add a comment |
$begingroup$
I am studying function theory in several complex variables and the book I am using is "Tasty Bits of Several Complex Variables" by Jiří Lebl: https://www.jirka.org/scv/scv.pdf.
At the moment I am reading chapter 4, where he introduces the $bar{partial}$-problem. However, he begins by proving a generalized form of the Cauchy Integral formula and I have some questions about the proof.
Theorem:
Let $Usubsetmathbb{C}$ be a bounded domain with piecewise $C^1$-smooth boundary $partial U$ oriented positively, and let $f:bar{U}tomathbb{C}$ be a $ C^1$-smooth function. Then for $ zin U$:
$ f(z)=frac{1}{2pi i}int_{partial U}frac{f(zeta)}{zeta-z}dzeta+frac{1}{2pi i}int_Udfrac{frac{partial f}{partialbar{z}}(zeta)}{zeta-z}dzetawedge dbar{zeta}$.
"Proof":
This is how he does it.
He begins by fixing $zin U$ and a small disc $Delta_r(z)$ such that $Delta_r(z) subsetsubset U$. He then applies Stokes Theorem, which gives us
$$int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big )=int_{UbackslashDelta_r(z)}dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta.quadquadquad (*)$$
The above equation begs us to let $r$ tend to $0$. If we manage to get anything out of this, we are done.
Notice that $dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}$ is integrable over all of $U$ (this follows from an exercise right before the proof). Therefore
$$lim_{rto 0}int_{UbackslashDelta_r(z)} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=-int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dzeta wedge dbar{zeta},$$
where we used the asymmetry of the operator $wedge$ in the last equality.
As a last step. Jiří argues that, by continuity of $f$, we find
$$lim_{rto 0}frac{1}{2pi i}int_{partialDelta_r(z)} dfrac{f(zeta)}{zeta-z}dbar{zeta}wedge dzeta = lim_{rto 0}frac{1}{2pi}int_{0}^{2pi} f(z+re^{itheta})dtheta=f(z).quadquadquad (**)$$
Questions:
(1) My first question is about the Stokes Theorem.
It's quite a long time since I last used Stokes theorem and back then it was in a basic calculus class. I have looked around and I cannot manage to find something which seems to give me
$$ int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big ).$$
I suspect there is a differential form version of Stokes Theorem I should use?
I found this on wikipedia:
Stokes' theorem says that the integral of a differential form ω over
the boundary of some orientable manifold Ω is equal to the integral of
its exterior derivative dω over the whole of Ω, i.e.,
$$int_{partialOmega}omega=int_{Omega}domega.$$
Is it the above I am using in some way, if so, could you please tell me how this is being used? Also, as a bonus question, I do not see why we choose a small disc $Delta_r(z)subsetsubset U$. But perhaps this will be answered at the same time as the explanation of Stokes Theorem.
(2) He argues that the second equality in $(*)$ follows since holomorphic derivatives in $zeta$ will have a $dzeta$ and when we wedge with $dzeta$ we just get zero.
I, honestly, do not understand what I just wrote. Can someone explain to me what this means or give another argument why the second equality is true?
(3) The computations in $(**)$ seems to be fundamental ones. For instance, it seems to me that we are using the Cauchy integral formula(?) And doing a variable substitution - polar coordinates(?) However, I am not sure about this. I would be really happy if you could tell me how the first equality in $(**)$ holds and where the continuity of $f$ is being used.
Just as a last thought. Perhaps the continuity is being used when we compute the limit to obtain the last equality?
As you see, I find it quite difficult to understand the proof, so I would be really happy if you could help me by answering (1), (2) and/or (3). Thanks for taking your time!
complex-analysis holomorphic-functions exterior-algebra cauchy-integral-formula several-complex-variables
asked Jan 3 at 23:48
JoeJoe
4214
$endgroup$
I am studying function theory in several complex variables and the book I am using is "Tasty Bits of Several Complex Variables" by Jiří Lebl: https://www.jirka.org/scv/scv.pdf.
At the moment I am reading chapter 4, where he introduces the $bar{partial}$-problem. However, he begins by proving a generalized form of the Cauchy Integral formula and I have some questions about the proof.
Theorem:
Let $Usubsetmathbb{C}$ be a bounded domain with piecewise $C^1$-smooth boundary $partial U$ oriented positively, and let $f:bar{U}tomathbb{C}$ be a $ C^1$-smooth function. Then for $ zin U$:
$ f(z)=frac{1}{2pi i}int_{partial U}frac{f(zeta)}{zeta-z}dzeta+frac{1}{2pi i}int_Udfrac{frac{partial f}{partialbar{z}}(zeta)}{zeta-z}dzetawedge dbar{zeta}$.
"Proof":
This is how he does it.
He begins by fixing $zin U$ and a small disc $Delta_r(z)$ such that $Delta_r(z) subsetsubset U$. He then applies Stokes Theorem, which gives us
$$int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big )=int_{UbackslashDelta_r(z)}dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta.quadquadquad (*)$$
The above equation begs us to let $r$ tend to $0$. If we manage to get anything out of this, we are done.
Notice that $dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}$ is integrable over all of $U$ (this follows from an exercise right before the proof). Therefore
$$lim_{rto 0}int_{UbackslashDelta_r(z)} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dbar{zeta}wedge dzeta=-int_{U} dfrac{frac{partial f}{partialzeta}(zeta)}{zeta-z}dzeta wedge dbar{zeta},$$
where we used the asymmetry of the operator $wedge$ in the last equality.
As a last step. Jiří argues that, by continuity of $f$, we find
$$lim_{rto 0}frac{1}{2pi i}int_{partialDelta_r(z)} dfrac{f(zeta)}{zeta-z}dbar{zeta}wedge dzeta = lim_{rto 0}frac{1}{2pi}int_{0}^{2pi} f(z+re^{itheta})dtheta=f(z).quadquadquad (**)$$
Questions:
(1) My first question is about the Stokes Theorem.
It's quite a long time since I last used Stokes theorem and back then it was in a basic calculus class. I have looked around and I cannot manage to find something which seems to give me
$$ int_{partial U}frac{f(zeta)}{zeta-z}dzeta-int_{partial Delta_r(z)}frac{f(zeta)}{zeta-z}dzeta=int_{UbackslashDelta_r(z)}dBig ( frac{f(zeta)}{zeta-z}dzeta Big ).$$
I suspect there is a differential form version of Stokes Theorem I should use?
I found this on wikipedia:
Stokes' theorem says that the integral of a differential form ω over
the boundary of some orientable manifold Ω is equal to the integral of
its exterior derivative dω over the whole of Ω, i.e.,
$$int_{partialOmega}omega=int_{Omega}domega.$$
Is it the above I am using in some way, if so, could you please tell me how this is being used? Also, as a bonus question, I do not see why we choose a small disc $Delta_r(z)subsetsubset U$. But perhaps this will be answered at the same time as the explanation of Stokes Theorem.
(2) He argues that the second equality in $(*)$ follows since holomorphic derivatives in $zeta$ will have a $dzeta$ and when we wedge with $dzeta$ we just get zero.
I, honestly, do not understand what I just wrote. Can someone explain to me what this means or give another argument why the second equality is true?
(3) The computations in $(**)$ seems to be fundamental ones. For instance, it seems to me that we are using the Cauchy integral formula(?) And doing a variable substitution - polar coordinates(?) However, I am not sure about this. I would be really happy if you could tell me how the first equality in $(**)$ holds and where the continuity of $f$ is being used.
Just as a last thought. Perhaps the continuity is being used when we compute the limit to obtain the last equality?
As you see, I find it quite difficult to understand the proof, so I would be really happy if you could help me by answering (1), (2) and/or (3). Thanks for taking your time!
complex-analysis holomorphic-functions exterior-algebra cauchy-integral-formula several-complex-variables
complex-analysis holomorphic-functions exterior-algebra cauchy-integral-formula several-complex-variables
asked Jan 3 at 23:48
JoeJoe
4214
asked Jan 3 at 23:48
JoeJoe
4214
asked Jan 3 at 23:48
JoeJoe
4214
asked Jan 3 at 23:48
JoeJoe
4214
asked Jan 3 at 23:48
JoeJoe
4214
4214
add a comment |
add a comment |
1 Answer
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$begingroup$
While in the first part of the book (until chapter 4), one gets by without differential forms, in this part they become quite useful. In fact, Chapter 5 goes full on into differential form mayhem even on the first page :), but there it is totally unavoidable. I am currently writing up some appendixes for the book and one of them will be a very short (without proofs) review of differential forms and Stokes theorem. Though that's not done yet. Best is to look through the chapter on differential forms in a book like baby Rudin.
(1) Yes, it is the differential form version of the Stokes theorem. Really in this setting it is the Green's theorem (which is Stoke's theorem in two variables, or perhaps, classical Stokes in the $xy$-plane). As to why we take out a small disc, that is because the function that we would want to integrate over $U$ has a singularity in $U$, so Stokes does not apply. So we take out a small disc around the singularity. Then the theorem does apply since everything is nicely $C^1$ on the $U setminus Delta_r$.
It is not a bad exercise to use the classical Green's theorem formulation and see if you can make the proof work. That is, use Green's theorem and then see if you can end up with the second integral in (**) using the classical formulation. It is slightly more tedious to write down that way, but it is not that terrible.
(2) For differential one-forms, $dx wedge dy = - dy wedge dx$ (let's say $x$ and $y$ are two coordinates). This means that $dx wedge dx = -dx wedge dx = 0$. Same thing works for the complex forms $dz$ and $dbar{z}$ instead of $x$ and $y$ (if $z=x+iy$, then $dz = dx + i ,dy$, and $dbar{z} = dx - i,dy$). Then the equality just follows from the definition of the $d$ operator on differential forms:
$d(a, dz + b, dbar{z}) =
frac{partial a}{partial z} dz wedge dz +
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z} +
frac{partial b}{partial bar{z}} dbar{z} wedge dbar{z}
=
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z}
=
left(frac{partial b}{partial z}-frac{partial a}{partial bar{z}} right) dbar{z} wedge dz$
(3) In the first guy on the left in (**) that is just $dzeta$ (no wedge there) that's evaluated over the boundary of the disc. This is just the normal path integral from one variable complex analysis. In fact, this is precisely the same argument from the standard proof of Cauchy formula that is usually given (you reduce to a disc centered at a point). The idea is to now actually compute the integral, so parametrize by saying that $zeta = z+re^{it}$, then $zeta-z = re^{it}$ and $dzeta = rie^{it}dt$. The denominator gets canceled, the $i$ also gets canceled, and you get the integral in the middle. Now this is where you use continuity, the integral is just the average of the value of $f$ over a tiny circle around $z$, so as $r to 0$, the limit must be the value of $f$ at $z$.
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
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While in the first part of the book (until chapter 4), one gets by without differential forms, in this part they become quite useful. In fact, Chapter 5 goes full on into differential form mayhem even on the first page :), but there it is totally unavoidable. I am currently writing up some appendixes for the book and one of them will be a very short (without proofs) review of differential forms and Stokes theorem. Though that's not done yet. Best is to look through the chapter on differential forms in a book like baby Rudin.
(1) Yes, it is the differential form version of the Stokes theorem. Really in this setting it is the Green's theorem (which is Stoke's theorem in two variables, or perhaps, classical Stokes in the $xy$-plane). As to why we take out a small disc, that is because the function that we would want to integrate over $U$ has a singularity in $U$, so Stokes does not apply. So we take out a small disc around the singularity. Then the theorem does apply since everything is nicely $C^1$ on the $U setminus Delta_r$.
It is not a bad exercise to use the classical Green's theorem formulation and see if you can make the proof work. That is, use Green's theorem and then see if you can end up with the second integral in (**) using the classical formulation. It is slightly more tedious to write down that way, but it is not that terrible.
(2) For differential one-forms, $dx wedge dy = - dy wedge dx$ (let's say $x$ and $y$ are two coordinates). This means that $dx wedge dx = -dx wedge dx = 0$. Same thing works for the complex forms $dz$ and $dbar{z}$ instead of $x$ and $y$ (if $z=x+iy$, then $dz = dx + i ,dy$, and $dbar{z} = dx - i,dy$). Then the equality just follows from the definition of the $d$ operator on differential forms:
$d(a, dz + b, dbar{z}) =
frac{partial a}{partial z} dz wedge dz +
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z} +
frac{partial b}{partial bar{z}} dbar{z} wedge dbar{z}
=
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z}
=
left(frac{partial b}{partial z}-frac{partial a}{partial bar{z}} right) dbar{z} wedge dz$
(3) In the first guy on the left in (**) that is just $dzeta$ (no wedge there) that's evaluated over the boundary of the disc. This is just the normal path integral from one variable complex analysis. In fact, this is precisely the same argument from the standard proof of Cauchy formula that is usually given (you reduce to a disc centered at a point). The idea is to now actually compute the integral, so parametrize by saying that $zeta = z+re^{it}$, then $zeta-z = re^{it}$ and $dzeta = rie^{it}dt$. The denominator gets canceled, the $i$ also gets canceled, and you get the integral in the middle. Now this is where you use continuity, the integral is just the average of the value of $f$ over a tiny circle around $z$, so as $r to 0$, the limit must be the value of $f$ at $z$.
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
$endgroup$
add a comment |
$begingroup$
While in the first part of the book (until chapter 4), one gets by without differential forms, in this part they become quite useful. In fact, Chapter 5 goes full on into differential form mayhem even on the first page :), but there it is totally unavoidable. I am currently writing up some appendixes for the book and one of them will be a very short (without proofs) review of differential forms and Stokes theorem. Though that's not done yet. Best is to look through the chapter on differential forms in a book like baby Rudin.
(1) Yes, it is the differential form version of the Stokes theorem. Really in this setting it is the Green's theorem (which is Stoke's theorem in two variables, or perhaps, classical Stokes in the $xy$-plane). As to why we take out a small disc, that is because the function that we would want to integrate over $U$ has a singularity in $U$, so Stokes does not apply. So we take out a small disc around the singularity. Then the theorem does apply since everything is nicely $C^1$ on the $U setminus Delta_r$.
It is not a bad exercise to use the classical Green's theorem formulation and see if you can make the proof work. That is, use Green's theorem and then see if you can end up with the second integral in (**) using the classical formulation. It is slightly more tedious to write down that way, but it is not that terrible.
(2) For differential one-forms, $dx wedge dy = - dy wedge dx$ (let's say $x$ and $y$ are two coordinates). This means that $dx wedge dx = -dx wedge dx = 0$. Same thing works for the complex forms $dz$ and $dbar{z}$ instead of $x$ and $y$ (if $z=x+iy$, then $dz = dx + i ,dy$, and $dbar{z} = dx - i,dy$). Then the equality just follows from the definition of the $d$ operator on differential forms:
$d(a, dz + b, dbar{z}) =
frac{partial a}{partial z} dz wedge dz +
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z} +
frac{partial b}{partial bar{z}} dbar{z} wedge dbar{z}
=
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z}
=
left(frac{partial b}{partial z}-frac{partial a}{partial bar{z}} right) dbar{z} wedge dz$
(3) In the first guy on the left in (**) that is just $dzeta$ (no wedge there) that's evaluated over the boundary of the disc. This is just the normal path integral from one variable complex analysis. In fact, this is precisely the same argument from the standard proof of Cauchy formula that is usually given (you reduce to a disc centered at a point). The idea is to now actually compute the integral, so parametrize by saying that $zeta = z+re^{it}$, then $zeta-z = re^{it}$ and $dzeta = rie^{it}dt$. The denominator gets canceled, the $i$ also gets canceled, and you get the integral in the middle. Now this is where you use continuity, the integral is just the average of the value of $f$ over a tiny circle around $z$, so as $r to 0$, the limit must be the value of $f$ at $z$.
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
$endgroup$
add a comment |
$begingroup$
While in the first part of the book (until chapter 4), one gets by without differential forms, in this part they become quite useful. In fact, Chapter 5 goes full on into differential form mayhem even on the first page :), but there it is totally unavoidable. I am currently writing up some appendixes for the book and one of them will be a very short (without proofs) review of differential forms and Stokes theorem. Though that's not done yet. Best is to look through the chapter on differential forms in a book like baby Rudin.
(1) Yes, it is the differential form version of the Stokes theorem. Really in this setting it is the Green's theorem (which is Stoke's theorem in two variables, or perhaps, classical Stokes in the $xy$-plane). As to why we take out a small disc, that is because the function that we would want to integrate over $U$ has a singularity in $U$, so Stokes does not apply. So we take out a small disc around the singularity. Then the theorem does apply since everything is nicely $C^1$ on the $U setminus Delta_r$.
It is not a bad exercise to use the classical Green's theorem formulation and see if you can make the proof work. That is, use Green's theorem and then see if you can end up with the second integral in (**) using the classical formulation. It is slightly more tedious to write down that way, but it is not that terrible.
(2) For differential one-forms, $dx wedge dy = - dy wedge dx$ (let's say $x$ and $y$ are two coordinates). This means that $dx wedge dx = -dx wedge dx = 0$. Same thing works for the complex forms $dz$ and $dbar{z}$ instead of $x$ and $y$ (if $z=x+iy$, then $dz = dx + i ,dy$, and $dbar{z} = dx - i,dy$). Then the equality just follows from the definition of the $d$ operator on differential forms:
$d(a, dz + b, dbar{z}) =
frac{partial a}{partial z} dz wedge dz +
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z} +
frac{partial b}{partial bar{z}} dbar{z} wedge dbar{z}
=
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z}
=
left(frac{partial b}{partial z}-frac{partial a}{partial bar{z}} right) dbar{z} wedge dz$
(3) In the first guy on the left in (**) that is just $dzeta$ (no wedge there) that's evaluated over the boundary of the disc. This is just the normal path integral from one variable complex analysis. In fact, this is precisely the same argument from the standard proof of Cauchy formula that is usually given (you reduce to a disc centered at a point). The idea is to now actually compute the integral, so parametrize by saying that $zeta = z+re^{it}$, then $zeta-z = re^{it}$ and $dzeta = rie^{it}dt$. The denominator gets canceled, the $i$ also gets canceled, and you get the integral in the middle. Now this is where you use continuity, the integral is just the average of the value of $f$ over a tiny circle around $z$, so as $r to 0$, the limit must be the value of $f$ at $z$.
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
$endgroup$
While in the first part of the book (until chapter 4), one gets by without differential forms, in this part they become quite useful. In fact, Chapter 5 goes full on into differential form mayhem even on the first page :), but there it is totally unavoidable. I am currently writing up some appendixes for the book and one of them will be a very short (without proofs) review of differential forms and Stokes theorem. Though that's not done yet. Best is to look through the chapter on differential forms in a book like baby Rudin.
(1) Yes, it is the differential form version of the Stokes theorem. Really in this setting it is the Green's theorem (which is Stoke's theorem in two variables, or perhaps, classical Stokes in the $xy$-plane). As to why we take out a small disc, that is because the function that we would want to integrate over $U$ has a singularity in $U$, so Stokes does not apply. So we take out a small disc around the singularity. Then the theorem does apply since everything is nicely $C^1$ on the $U setminus Delta_r$.
It is not a bad exercise to use the classical Green's theorem formulation and see if you can make the proof work. That is, use Green's theorem and then see if you can end up with the second integral in (**) using the classical formulation. It is slightly more tedious to write down that way, but it is not that terrible.
(2) For differential one-forms, $dx wedge dy = - dy wedge dx$ (let's say $x$ and $y$ are two coordinates). This means that $dx wedge dx = -dx wedge dx = 0$. Same thing works for the complex forms $dz$ and $dbar{z}$ instead of $x$ and $y$ (if $z=x+iy$, then $dz = dx + i ,dy$, and $dbar{z} = dx - i,dy$). Then the equality just follows from the definition of the $d$ operator on differential forms:
$d(a, dz + b, dbar{z}) =
frac{partial a}{partial z} dz wedge dz +
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z} +
frac{partial b}{partial bar{z}} dbar{z} wedge dbar{z}
=
frac{partial a}{partial bar{z}} dbar{z} wedge dz +
frac{partial b}{partial z} dz wedge dbar{z}
=
left(frac{partial b}{partial z}-frac{partial a}{partial bar{z}} right) dbar{z} wedge dz$
(3) In the first guy on the left in (**) that is just $dzeta$ (no wedge there) that's evaluated over the boundary of the disc. This is just the normal path integral from one variable complex analysis. In fact, this is precisely the same argument from the standard proof of Cauchy formula that is usually given (you reduce to a disc centered at a point). The idea is to now actually compute the integral, so parametrize by saying that $zeta = z+re^{it}$, then $zeta-z = re^{it}$ and $dzeta = rie^{it}dt$. The denominator gets canceled, the $i$ also gets canceled, and you get the integral in the middle. Now this is where you use continuity, the integral is just the average of the value of $f$ over a tiny circle around $z$, so as $r to 0$, the limit must be the value of $f$ at $z$.
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
answered Jan 6 at 19:26
Jiri LeblJiri Lebl
1,628412
1,628412
add a comment |
add a comment |
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