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
1,1881316
$endgroup$
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
1,1881316
$endgroup$
$begingroup$
$Omega$ 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
1,1881316
$endgroup$
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
1,1881316
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1881316
edited Jan 9 at 13:23
John Cataldo
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1881316
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1881316
asked Jan 9 at 10:31
John CataldoJohn Cataldo
1,1881316
1,1881316
$begingroup$
$Omega$ 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 |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067302%2fif-a-sequence-of-distinct-points-in-a-bounded-connected-open-omega-doesnt-co%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067302%2fif-a-sequence-of-distinct-points-in-a-bounded-connected-open-omega-doesnt-co%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$Omega$ 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