Mixing
Cousin I problem in $mathbb{C}$ and Mittag-Leffler theorem
$begingroup$
In the wikipedia page about Cousin's problems (https://en.wikipedia.org/wiki/Cousin_problems) is stated that "the case of one variable is the Mittag-Leffler theorem"; I'm a bit in trouble with this statement.
I'm able to show that, assuming that Cousin I problem can be solved in every (connected) open subset of $mathbb{C}$, the Mittag-Leffler theorem follows, but I can't prove the other implication; it seems to me that, proving that the Cousin I problem in one variable can always be solved on an open domain, you don't use Mittag-Leffler theorem at all, but only the existence of partitions of unity and the fact that every open domain $Omega subset mathbb{C}$ is a holomorphy domain (that is, there is a holomorphic function on $Omega$ that cannot be extended outside $Omega$). For a proof of this fact see for example Krantz "Function theory of several complex variables".
complex-analysis holomorphic-functions sheaf-cohomology meromorphic-functions
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
$endgroup$
add a comment |
$begingroup$
In the wikipedia page about Cousin's problems (https://en.wikipedia.org/wiki/Cousin_problems) is stated that "the case of one variable is the Mittag-Leffler theorem"; I'm a bit in trouble with this statement.
I'm able to show that, assuming that Cousin I problem can be solved in every (connected) open subset of $mathbb{C}$, the Mittag-Leffler theorem follows, but I can't prove the other implication; it seems to me that, proving that the Cousin I problem in one variable can always be solved on an open domain, you don't use Mittag-Leffler theorem at all, but only the existence of partitions of unity and the fact that every open domain $Omega subset mathbb{C}$ is a holomorphy domain (that is, there is a holomorphic function on $Omega$ that cannot be extended outside $Omega$). For a proof of this fact see for example Krantz "Function theory of several complex variables".
complex-analysis holomorphic-functions sheaf-cohomology meromorphic-functions
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
$endgroup$
$begingroup$
Mittag-Leffler is not immediate : if $|a_k| to infty$ but $sum_k |a_k|^{-1} = infty$ how do you construct a meromorphic function such that $f(z)-frac{1}{z-a_k}$ is analytic at $a_k$ for every $k$ ? It is a problem of "analytic regularization" of $sum_k frac{1}{z-a_k}$. What do you mean with partitions of unity ?
$endgroup$
– reuns
Feb 3 at 12:30
add a comment |
$begingroup$
In the wikipedia page about Cousin's problems (https://en.wikipedia.org/wiki/Cousin_problems) is stated that "the case of one variable is the Mittag-Leffler theorem"; I'm a bit in trouble with this statement.
I'm able to show that, assuming that Cousin I problem can be solved in every (connected) open subset of $mathbb{C}$, the Mittag-Leffler theorem follows, but I can't prove the other implication; it seems to me that, proving that the Cousin I problem in one variable can always be solved on an open domain, you don't use Mittag-Leffler theorem at all, but only the existence of partitions of unity and the fact that every open domain $Omega subset mathbb{C}$ is a holomorphy domain (that is, there is a holomorphic function on $Omega$ that cannot be extended outside $Omega$). For a proof of this fact see for example Krantz "Function theory of several complex variables".
complex-analysis holomorphic-functions sheaf-cohomology meromorphic-functions
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
$endgroup$
In the wikipedia page about Cousin's problems (https://en.wikipedia.org/wiki/Cousin_problems) is stated that "the case of one variable is the Mittag-Leffler theorem"; I'm a bit in trouble with this statement.
I'm able to show that, assuming that Cousin I problem can be solved in every (connected) open subset of $mathbb{C}$, the Mittag-Leffler theorem follows, but I can't prove the other implication; it seems to me that, proving that the Cousin I problem in one variable can always be solved on an open domain, you don't use Mittag-Leffler theorem at all, but only the existence of partitions of unity and the fact that every open domain $Omega subset mathbb{C}$ is a holomorphy domain (that is, there is a holomorphic function on $Omega$ that cannot be extended outside $Omega$). For a proof of this fact see for example Krantz "Function theory of several complex variables".
complex-analysis holomorphic-functions sheaf-cohomology meromorphic-functions
complex-analysis holomorphic-functions sheaf-cohomology meromorphic-functions
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
asked Feb 3 at 12:09
Giuseppe BargagnatiGiuseppe Bargagnati
1,251514
1,251514
$begingroup$
Mittag-Leffler is not immediate : if $|a_k| to infty$ but $sum_k |a_k|^{-1} = infty$ how do you construct a meromorphic function such that $f(z)-frac{1}{z-a_k}$ is analytic at $a_k$ for every $k$ ? It is a problem of "analytic regularization" of $sum_k frac{1}{z-a_k}$. What do you mean with partitions of unity ?
$endgroup$
– reuns
Feb 3 at 12:30
add a comment |
$begingroup$
Mittag-Leffler is not immediate : if $|a_k| to infty$ but $sum_k |a_k|^{-1} = infty$ how do you construct a meromorphic function such that $f(z)-frac{1}{z-a_k}$ is analytic at $a_k$ for every $k$ ? It is a problem of "analytic regularization" of $sum_k frac{1}{z-a_k}$. What do you mean with partitions of unity ?
$endgroup$
– reuns
Feb 3 at 12:30
$begingroup$
Mittag-Leffler is not immediate : if $|a_k| to infty$ but $sum_k |a_k|^{-1} = infty$ how do you construct a meromorphic function such that $f(z)-frac{1}{z-a_k}$ is analytic at $a_k$ for every $k$ ? It is a problem of "analytic regularization" of $sum_k frac{1}{z-a_k}$. What do you mean with partitions of unity ?
$endgroup$
– reuns
Feb 3 at 12:30
$begingroup$
Mittag-Leffler is not immediate : if $|a_k| to infty$ but $sum_k |a_k|^{-1} = infty$ how do you construct a meromorphic function such that $f(z)-frac{1}{z-a_k}$ is analytic at $a_k$ for every $k$ ? It is a problem of "analytic regularization" of $sum_k frac{1}{z-a_k}$. What do you mean with partitions of unity ?
$endgroup$
– reuns
Feb 3 at 12:30
add a comment |
1 Answer
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Assume the Mittag-Leffler theorem, and let $(f_i)$ be some Cousin data, $f_i$ meromorphic on $U_isubset mathbb{C}$; $f_i-f_j$ holomorphic on $U_icap U_j$; and $displaystylebigcup_{iin I}U_i = mathbb{C}$.
Consider $E={ain mathbb{C} mid exists iin I, ain U_i land f_i$ is not holomorphic at $a}$. Then $E$ is discrete and closed.
Indeed take $ain E$, $i$ as in the definition of $E$. $a$ is isolated in the set ${zin U_imid f_i$ is not holomorphic at $z}$ because $f_i$ is meromorphic.
So let $U$ be an open set containing $a$, $Usubset U_i$ such that $f_i$ is holomorphic on $Usetminus{a}$.
Now assiume $zin Ecap U$. Then for some $j, zin U_j$ and $f_j$ is not holomorphic at $z$. Let $V= Ucap U_j$, then both $f_i$ and $f_j$ are defined on $V$ and $f_i-f_j$ is holomorphic, so $f_i$ isn't holomorphic at $z$; thus since $zin U$, it follows that $z=a$.
Thus $Ecap U = {a}$, so $E$ is discrete.
Moreover, if $znotin E$, then take some $i$ such that $zin U_i$, and take an open set $Usubset U_i$ on which $f_i$ is holomorphic. Then clearly any $f_j$ is holomorphic on $Ucap U_j$, so that $Usubset mathbb{C}setminus E$ : $E$ is closed.
Finally, for $ain E$ consider $i$ such that $a in U_i$; and let $p_a(z)$ be the principal part of $f_i$ at $a$. By the hypothesis on $f_i-f_j$, it doesn't depend on the chosen $i$, as long as $ain U_i$.
We can now apply Mittag-Leffler's theorem to get $f$ meromorphic on $mathbb{C}$ such that $f-p_a$ has no singularity at $a$, for all $ain E$; and such that the set of poles of $f$ is included in $E$.
Then $f$ is a solution for the Cousin problem, because $f-f_i = (f-p_a)-(f_i-p_a)$, which is holomorphic at $a$ as a difference of holomorphic functions; so $f-f_i$ is holomorphic at every point of $Ecap U_i$, and since both are holomorphic on $U_isetminus E$, $f-f_i$ is holomorphic on $U_i$.
answered Feb 3 at 13:39
MaxMax
16.3k11144
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add a comment |
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$begingroup$
Assume the Mittag-Leffler theorem, and let $(f_i)$ be some Cousin data, $f_i$ meromorphic on $U_isubset mathbb{C}$; $f_i-f_j$ holomorphic on $U_icap U_j$; and $displaystylebigcup_{iin I}U_i = mathbb{C}$.
Consider $E={ain mathbb{C} mid exists iin I, ain U_i land f_i$ is not holomorphic at $a}$. Then $E$ is discrete and closed.
Indeed take $ain E$, $i$ as in the definition of $E$. $a$ is isolated in the set ${zin U_imid f_i$ is not holomorphic at $z}$ because $f_i$ is meromorphic.
So let $U$ be an open set containing $a$, $Usubset U_i$ such that $f_i$ is holomorphic on $Usetminus{a}$.
Now assiume $zin Ecap U$. Then for some $j, zin U_j$ and $f_j$ is not holomorphic at $z$. Let $V= Ucap U_j$, then both $f_i$ and $f_j$ are defined on $V$ and $f_i-f_j$ is holomorphic, so $f_i$ isn't holomorphic at $z$; thus since $zin U$, it follows that $z=a$.
Thus $Ecap U = {a}$, so $E$ is discrete.
Moreover, if $znotin E$, then take some $i$ such that $zin U_i$, and take an open set $Usubset U_i$ on which $f_i$ is holomorphic. Then clearly any $f_j$ is holomorphic on $Ucap U_j$, so that $Usubset mathbb{C}setminus E$ : $E$ is closed.
Finally, for $ain E$ consider $i$ such that $a in U_i$; and let $p_a(z)$ be the principal part of $f_i$ at $a$. By the hypothesis on $f_i-f_j$, it doesn't depend on the chosen $i$, as long as $ain U_i$.
We can now apply Mittag-Leffler's theorem to get $f$ meromorphic on $mathbb{C}$ such that $f-p_a$ has no singularity at $a$, for all $ain E$; and such that the set of poles of $f$ is included in $E$.
Then $f$ is a solution for the Cousin problem, because $f-f_i = (f-p_a)-(f_i-p_a)$, which is holomorphic at $a$ as a difference of holomorphic functions; so $f-f_i$ is holomorphic at every point of $Ecap U_i$, and since both are holomorphic on $U_isetminus E$, $f-f_i$ is holomorphic on $U_i$.
answered Feb 3 at 13:39
MaxMax
16.3k11144
$endgroup$
add a comment |
$begingroup$
Assume the Mittag-Leffler theorem, and let $(f_i)$ be some Cousin data, $f_i$ meromorphic on $U_isubset mathbb{C}$; $f_i-f_j$ holomorphic on $U_icap U_j$; and $displaystylebigcup_{iin I}U_i = mathbb{C}$.
Consider $E={ain mathbb{C} mid exists iin I, ain U_i land f_i$ is not holomorphic at $a}$. Then $E$ is discrete and closed.
Indeed take $ain E$, $i$ as in the definition of $E$. $a$ is isolated in the set ${zin U_imid f_i$ is not holomorphic at $z}$ because $f_i$ is meromorphic.
So let $U$ be an open set containing $a$, $Usubset U_i$ such that $f_i$ is holomorphic on $Usetminus{a}$.
Now assiume $zin Ecap U$. Then for some $j, zin U_j$ and $f_j$ is not holomorphic at $z$. Let $V= Ucap U_j$, then both $f_i$ and $f_j$ are defined on $V$ and $f_i-f_j$ is holomorphic, so $f_i$ isn't holomorphic at $z$; thus since $zin U$, it follows that $z=a$.
Thus $Ecap U = {a}$, so $E$ is discrete.
Moreover, if $znotin E$, then take some $i$ such that $zin U_i$, and take an open set $Usubset U_i$ on which $f_i$ is holomorphic. Then clearly any $f_j$ is holomorphic on $Ucap U_j$, so that $Usubset mathbb{C}setminus E$ : $E$ is closed.
Finally, for $ain E$ consider $i$ such that $a in U_i$; and let $p_a(z)$ be the principal part of $f_i$ at $a$. By the hypothesis on $f_i-f_j$, it doesn't depend on the chosen $i$, as long as $ain U_i$.
We can now apply Mittag-Leffler's theorem to get $f$ meromorphic on $mathbb{C}$ such that $f-p_a$ has no singularity at $a$, for all $ain E$; and such that the set of poles of $f$ is included in $E$.
Then $f$ is a solution for the Cousin problem, because $f-f_i = (f-p_a)-(f_i-p_a)$, which is holomorphic at $a$ as a difference of holomorphic functions; so $f-f_i$ is holomorphic at every point of $Ecap U_i$, and since both are holomorphic on $U_isetminus E$, $f-f_i$ is holomorphic on $U_i$.
answered Feb 3 at 13:39
MaxMax
16.3k11144
$endgroup$
add a comment |
$begingroup$
Assume the Mittag-Leffler theorem, and let $(f_i)$ be some Cousin data, $f_i$ meromorphic on $U_isubset mathbb{C}$; $f_i-f_j$ holomorphic on $U_icap U_j$; and $displaystylebigcup_{iin I}U_i = mathbb{C}$.
Consider $E={ain mathbb{C} mid exists iin I, ain U_i land f_i$ is not holomorphic at $a}$. Then $E$ is discrete and closed.
Indeed take $ain E$, $i$ as in the definition of $E$. $a$ is isolated in the set ${zin U_imid f_i$ is not holomorphic at $z}$ because $f_i$ is meromorphic.
So let $U$ be an open set containing $a$, $Usubset U_i$ such that $f_i$ is holomorphic on $Usetminus{a}$.
Now assiume $zin Ecap U$. Then for some $j, zin U_j$ and $f_j$ is not holomorphic at $z$. Let $V= Ucap U_j$, then both $f_i$ and $f_j$ are defined on $V$ and $f_i-f_j$ is holomorphic, so $f_i$ isn't holomorphic at $z$; thus since $zin U$, it follows that $z=a$.
Thus $Ecap U = {a}$, so $E$ is discrete.
Moreover, if $znotin E$, then take some $i$ such that $zin U_i$, and take an open set $Usubset U_i$ on which $f_i$ is holomorphic. Then clearly any $f_j$ is holomorphic on $Ucap U_j$, so that $Usubset mathbb{C}setminus E$ : $E$ is closed.
Finally, for $ain E$ consider $i$ such that $a in U_i$; and let $p_a(z)$ be the principal part of $f_i$ at $a$. By the hypothesis on $f_i-f_j$, it doesn't depend on the chosen $i$, as long as $ain U_i$.
We can now apply Mittag-Leffler's theorem to get $f$ meromorphic on $mathbb{C}$ such that $f-p_a$ has no singularity at $a$, for all $ain E$; and such that the set of poles of $f$ is included in $E$.
Then $f$ is a solution for the Cousin problem, because $f-f_i = (f-p_a)-(f_i-p_a)$, which is holomorphic at $a$ as a difference of holomorphic functions; so $f-f_i$ is holomorphic at every point of $Ecap U_i$, and since both are holomorphic on $U_isetminus E$, $f-f_i$ is holomorphic on $U_i$.
answered Feb 3 at 13:39
MaxMax
16.3k11144
$endgroup$
Assume the Mittag-Leffler theorem, and let $(f_i)$ be some Cousin data, $f_i$ meromorphic on $U_isubset mathbb{C}$; $f_i-f_j$ holomorphic on $U_icap U_j$; and $displaystylebigcup_{iin I}U_i = mathbb{C}$.
Consider $E={ain mathbb{C} mid exists iin I, ain U_i land f_i$ is not holomorphic at $a}$. Then $E$ is discrete and closed.
Indeed take $ain E$, $i$ as in the definition of $E$. $a$ is isolated in the set ${zin U_imid f_i$ is not holomorphic at $z}$ because $f_i$ is meromorphic.
So let $U$ be an open set containing $a$, $Usubset U_i$ such that $f_i$ is holomorphic on $Usetminus{a}$.
Now assiume $zin Ecap U$. Then for some $j, zin U_j$ and $f_j$ is not holomorphic at $z$. Let $V= Ucap U_j$, then both $f_i$ and $f_j$ are defined on $V$ and $f_i-f_j$ is holomorphic, so $f_i$ isn't holomorphic at $z$; thus since $zin U$, it follows that $z=a$.
Thus $Ecap U = {a}$, so $E$ is discrete.
Moreover, if $znotin E$, then take some $i$ such that $zin U_i$, and take an open set $Usubset U_i$ on which $f_i$ is holomorphic. Then clearly any $f_j$ is holomorphic on $Ucap U_j$, so that $Usubset mathbb{C}setminus E$ : $E$ is closed.
Finally, for $ain E$ consider $i$ such that $a in U_i$; and let $p_a(z)$ be the principal part of $f_i$ at $a$. By the hypothesis on $f_i-f_j$, it doesn't depend on the chosen $i$, as long as $ain U_i$.
We can now apply Mittag-Leffler's theorem to get $f$ meromorphic on $mathbb{C}$ such that $f-p_a$ has no singularity at $a$, for all $ain E$; and such that the set of poles of $f$ is included in $E$.
Then $f$ is a solution for the Cousin problem, because $f-f_i = (f-p_a)-(f_i-p_a)$, which is holomorphic at $a$ as a difference of holomorphic functions; so $f-f_i$ is holomorphic at every point of $Ecap U_i$, and since both are holomorphic on $U_isetminus E$, $f-f_i$ is holomorphic on $U_i$.
answered Feb 3 at 13:39
MaxMax
16.3k11144
answered Feb 3 at 13:39
MaxMax
16.3k11144
answered Feb 3 at 13:39
MaxMax
16.3k11144
answered Feb 3 at 13:39
MaxMax
16.3k11144
16.3k11144
add a comment |
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$begingroup$
Mittag-Leffler is not immediate : if $|a_k| to infty$ but $sum_k |a_k|^{-1} = infty$ how do you construct a meromorphic function such that $f(z)-frac{1}{z-a_k}$ is analytic at $a_k$ for every $k$ ? It is a problem of "analytic regularization" of $sum_k frac{1}{z-a_k}$. What do you mean with partitions of unity ?
$endgroup$
– reuns
Feb 3 at 12:30