If a sequence of distinct points in a bounded connected open $Omega$ doesn't converge in $overlineOmega$, why...
$begingroup$
Let $gammasubsetBbb C$ be a closed, simple (if $gamma:[a,b]mapstoBbb C$, $gamma$ is injective on $(a,b)$, so the curve doesn't intersect except for $gamma(a)=gamma(b)$) and piece-wise regular curve and $z_1,z_2,...$ is an infinite set of distinct points all inside the domain (call it $Omega$) defined by $gamma$ ($partialOmega=gamma$) such that $z_n$ does not converge towards any limit in $overlineOmega$.
The analytic continuation principle requires a set a distinct points that converge to a point in the open $Omega$ such that a certain holomorphic function $f(z_n)=0~forall ninBbb N$ and $z_nnelim z_n~forall n$.
And in an exercise about the formula ${1over 2pi i}int_{gamma}frac{f'}{f}={text{the number of zeros of } f$ inside $Omega$ counted with their multiplicity (order)$}$, the following explentation confused me a little bit:
Let $z_1,...,z_m$ be in $Omega$ the zeros of $f$ of order $k_1,...,k_m ge 1$ respectively. Then $m$ is finite because otherwise, by the analytic continuation principle, $f$ would be identically zero, which is not the case because $f$ is not constant.
general-topology complex-analysis analytic-continuation
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
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add a comment |
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Let $gammasubsetBbb C$ be a closed, simple (if $gamma:[a,b]mapstoBbb C$, $gamma$ is injective on $(a,b)$, so the curve doesn't intersect except for $gamma(a)=gamma(b)$) and piece-wise regular curve and $z_1,z_2,...$ is an infinite set of distinct points all inside the domain (call it $Omega$) defined by $gamma$ ($partialOmega=gamma$) such that $z_n$ does not converge towards any limit in $overlineOmega$.
The analytic continuation principle requires a set a distinct points that converge to a point in the open $Omega$ such that a certain holomorphic function $f(z_n)=0~forall ninBbb N$ and $z_nnelim z_n~forall n$.
And in an exercise about the formula ${1over 2pi i}int_{gamma}frac{f'}{f}={text{the number of zeros of } f$ inside $Omega$ counted with their multiplicity (order)$}$, the following explentation confused me a little bit:
Let $z_1,...,z_m$ be in $Omega$ the zeros of $f$ of order $k_1,...,k_m ge 1$ respectively. Then $m$ is finite because otherwise, by the analytic continuation principle, $f$ would be identically zero, which is not the case because $f$ is not constant.
general-topology complex-analysis analytic-continuation
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
$endgroup$
$begingroup$
$Omega$ is not countable.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 10:35
$begingroup$
You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit.
$endgroup$
– James
Jan 9 at 13:27
$begingroup$
Is that a theorem? What is it called?
$endgroup$
– John Cataldo
Jan 9 at 13:28
$begingroup$
My advice would be to prove those things assuming $f$ is analytic. If the $ a_n$ are distinct $a_n to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $forall n, f(a_n) =0 implies forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $implies$ analytic.
$endgroup$
– reuns
Jan 9 at 16:38
add a comment |
$begingroup$
Let $gammasubsetBbb C$ be a closed, simple (if $gamma:[a,b]mapstoBbb C$, $gamma$ is injective on $(a,b)$, so the curve doesn't intersect except for $gamma(a)=gamma(b)$) and piece-wise regular curve and $z_1,z_2,...$ is an infinite set of distinct points all inside the domain (call it $Omega$) defined by $gamma$ ($partialOmega=gamma$) such that $z_n$ does not converge towards any limit in $overlineOmega$.
The analytic continuation principle requires a set a distinct points that converge to a point in the open $Omega$ such that a certain holomorphic function $f(z_n)=0~forall ninBbb N$ and $z_nnelim z_n~forall n$.
And in an exercise about the formula ${1over 2pi i}int_{gamma}frac{f'}{f}={text{the number of zeros of } f$ inside $Omega$ counted with their multiplicity (order)$}$, the following explentation confused me a little bit:
Let $z_1,...,z_m$ be in $Omega$ the zeros of $f$ of order $k_1,...,k_m ge 1$ respectively. Then $m$ is finite because otherwise, by the analytic continuation principle, $f$ would be identically zero, which is not the case because $f$ is not constant.
general-topology complex-analysis analytic-continuation
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
$endgroup$
Let $gammasubsetBbb C$ be a closed, simple (if $gamma:[a,b]mapstoBbb C$, $gamma$ is injective on $(a,b)$, so the curve doesn't intersect except for $gamma(a)=gamma(b)$) and piece-wise regular curve and $z_1,z_2,...$ is an infinite set of distinct points all inside the domain (call it $Omega$) defined by $gamma$ ($partialOmega=gamma$) such that $z_n$ does not converge towards any limit in $overlineOmega$.
The analytic continuation principle requires a set a distinct points that converge to a point in the open $Omega$ such that a certain holomorphic function $f(z_n)=0~forall ninBbb N$ and $z_nnelim z_n~forall n$.
And in an exercise about the formula ${1over 2pi i}int_{gamma}frac{f'}{f}={text{the number of zeros of } f$ inside $Omega$ counted with their multiplicity (order)$}$, the following explentation confused me a little bit:
Let $z_1,...,z_m$ be in $Omega$ the zeros of $f$ of order $k_1,...,k_m ge 1$ respectively. Then $m$ is finite because otherwise, by the analytic continuation principle, $f$ would be identically zero, which is not the case because $f$ is not constant.
general-topology complex-analysis analytic-continuation
general-topology complex-analysis analytic-continuation
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
edited Jan 9 at 13:23
John Cataldo
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1271316
1,1271316
$begingroup$
$Omega$ is not countable.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 10:35
$begingroup$
You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit.
$endgroup$
– James
Jan 9 at 13:27
$begingroup$
Is that a theorem? What is it called?
$endgroup$
– John Cataldo
Jan 9 at 13:28
$begingroup$
My advice would be to prove those things assuming $f$ is analytic. If the $ a_n$ are distinct $a_n to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $forall n, f(a_n) =0 implies forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $implies$ analytic.
$endgroup$
– reuns
Jan 9 at 16:38
add a comment |
$begingroup$
$Omega$ is not countable.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 10:35
$begingroup$
You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit.
$endgroup$
– James
Jan 9 at 13:27
$begingroup$
Is that a theorem? What is it called?
$endgroup$
– John Cataldo
Jan 9 at 13:28
$begingroup$
My advice would be to prove those things assuming $f$ is analytic. If the $ a_n$ are distinct $a_n to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $forall n, f(a_n) =0 implies forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $implies$ analytic.
$endgroup$
– reuns
Jan 9 at 16:38
$begingroup$
$Omega$ is not countable.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 10:35
$begingroup$
$Omega$ is not countable.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 10:35
$begingroup$
You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit.
$endgroup$
– James
Jan 9 at 13:27
$begingroup$
You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit.
$endgroup$
– James
Jan 9 at 13:27
$begingroup$
Is that a theorem? What is it called?
$endgroup$
– John Cataldo
Jan 9 at 13:28
$begingroup$
Is that a theorem? What is it called?
$endgroup$
– John Cataldo
Jan 9 at 13:28
$begingroup$
My advice would be to prove those things assuming $f$ is analytic. If the $ a_n$ are distinct $a_n to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $forall n, f(a_n) =0 implies forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $implies$ analytic.
$endgroup$
– reuns
Jan 9 at 16:38
$begingroup$
My advice would be to prove those things assuming $f$ is analytic. If the $ a_n$ are distinct $a_n to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $forall n, f(a_n) =0 implies forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $implies$ analytic.
$endgroup$
– reuns
Jan 9 at 16:38
add a comment |
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$begingroup$
$Omega$ is not countable.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 10:35
$begingroup$
You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit.
$endgroup$
– James
Jan 9 at 13:27
$begingroup$
Is that a theorem? What is it called?
$endgroup$
– John Cataldo
Jan 9 at 13:28
$begingroup$
My advice would be to prove those things assuming $f$ is analytic. If the $ a_n$ are distinct $a_n to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $forall n, f(a_n) =0 implies forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $implies$ analytic.
$endgroup$
– reuns
Jan 9 at 16:38