Suppose $f$ is a holomorphic function on a region $Omega$ that vanishes on a sequence of distinct points with...
$begingroup$
I am reviewing Stein's complex analysis , and there's one theorem I listed follows:
Suppose $ f$ is a holomorphic function on a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$ , then $f$ is identically $0$ . (Here region means both open and connect)
My question is : If we do not assume the limit point is in $Omega$ , the limit point might in the boundary of $Omega$ , can we get the similar conclusion ? Since the orginal proof using the power series of the limit point $z_0$ , I have no idea how to deal with this .
complex-analysis
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
$endgroup$
add a comment |
$begingroup$
I am reviewing Stein's complex analysis , and there's one theorem I listed follows:
Suppose $ f$ is a holomorphic function on a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$ , then $f$ is identically $0$ . (Here region means both open and connect)
My question is : If we do not assume the limit point is in $Omega$ , the limit point might in the boundary of $Omega$ , can we get the similar conclusion ? Since the orginal proof using the power series of the limit point $z_0$ , I have no idea how to deal with this .
complex-analysis
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
$endgroup$
add a comment |
$begingroup$
I am reviewing Stein's complex analysis , and there's one theorem I listed follows:
Suppose $ f$ is a holomorphic function on a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$ , then $f$ is identically $0$ . (Here region means both open and connect)
My question is : If we do not assume the limit point is in $Omega$ , the limit point might in the boundary of $Omega$ , can we get the similar conclusion ? Since the orginal proof using the power series of the limit point $z_0$ , I have no idea how to deal with this .
complex-analysis
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
$endgroup$
I am reviewing Stein's complex analysis , and there's one theorem I listed follows:
Suppose $ f$ is a holomorphic function on a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$ , then $f$ is identically $0$ . (Here region means both open and connect)
My question is : If we do not assume the limit point is in $Omega$ , the limit point might in the boundary of $Omega$ , can we get the similar conclusion ? Since the orginal proof using the power series of the limit point $z_0$ , I have no idea how to deal with this .
complex-analysis
complex-analysis
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
asked Dec 20 '18 at 17:15
J.GuoJ.Guo
4579
4579
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, if the limit point is outside the domain then we cannot conclude that the function is identically zero.
To be specific, the Blaschke product gives a very specific construction of functions with prescribed zeros within the unit disk (potentially infinitely many!) which are not identically zero. The zeros do need to satisfy some growth conditions, though - they have to be moving towards the boundary of the disk quickly enough.
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
$endgroup$
$begingroup$
I see it ! Thanks for the answer .
$endgroup$
– J.Guo
Dec 20 '18 at 17:20
add a comment |
$begingroup$
No, take $f(z)=sin(1/z)$ on the punctured plane. Then $f=0$ on $z_k=frac 1{kpi}$ and $z_kto 0$ but $f$ is not identically 0.
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
$endgroup$
add a comment |
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%2f3047768%2fsuppose-f-is-a-holomorphic-function-on-a-region-omega-that-vanishes-on-a-se%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, if the limit point is outside the domain then we cannot conclude that the function is identically zero.
To be specific, the Blaschke product gives a very specific construction of functions with prescribed zeros within the unit disk (potentially infinitely many!) which are not identically zero. The zeros do need to satisfy some growth conditions, though - they have to be moving towards the boundary of the disk quickly enough.
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
$endgroup$
$begingroup$
I see it ! Thanks for the answer .
$endgroup$
– J.Guo
Dec 20 '18 at 17:20
add a comment |
$begingroup$
No, if the limit point is outside the domain then we cannot conclude that the function is identically zero.
To be specific, the Blaschke product gives a very specific construction of functions with prescribed zeros within the unit disk (potentially infinitely many!) which are not identically zero. The zeros do need to satisfy some growth conditions, though - they have to be moving towards the boundary of the disk quickly enough.
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
$endgroup$
$begingroup$
I see it ! Thanks for the answer .
$endgroup$
– J.Guo
Dec 20 '18 at 17:20
add a comment |
$begingroup$
No, if the limit point is outside the domain then we cannot conclude that the function is identically zero.
To be specific, the Blaschke product gives a very specific construction of functions with prescribed zeros within the unit disk (potentially infinitely many!) which are not identically zero. The zeros do need to satisfy some growth conditions, though - they have to be moving towards the boundary of the disk quickly enough.
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
$endgroup$
No, if the limit point is outside the domain then we cannot conclude that the function is identically zero.
To be specific, the Blaschke product gives a very specific construction of functions with prescribed zeros within the unit disk (potentially infinitely many!) which are not identically zero. The zeros do need to satisfy some growth conditions, though - they have to be moving towards the boundary of the disk quickly enough.
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
answered Dec 20 '18 at 17:17
T. BongersT. Bongers
23.5k54762
23.5k54762
$begingroup$
I see it ! Thanks for the answer .
$endgroup$
– J.Guo
Dec 20 '18 at 17:20
add a comment |
$begingroup$
I see it ! Thanks for the answer .
$endgroup$
– J.Guo
Dec 20 '18 at 17:20
$begingroup$
I see it ! Thanks for the answer .
$endgroup$
– J.Guo
Dec 20 '18 at 17:20
$begingroup$
I see it ! Thanks for the answer .
$endgroup$
– J.Guo
Dec 20 '18 at 17:20
add a comment |
$begingroup$
No, take $f(z)=sin(1/z)$ on the punctured plane. Then $f=0$ on $z_k=frac 1{kpi}$ and $z_kto 0$ but $f$ is not identically 0.
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
$endgroup$
add a comment |
$begingroup$
No, take $f(z)=sin(1/z)$ on the punctured plane. Then $f=0$ on $z_k=frac 1{kpi}$ and $z_kto 0$ but $f$ is not identically 0.
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
$endgroup$
add a comment |
$begingroup$
No, take $f(z)=sin(1/z)$ on the punctured plane. Then $f=0$ on $z_k=frac 1{kpi}$ and $z_kto 0$ but $f$ is not identically 0.
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
$endgroup$
No, take $f(z)=sin(1/z)$ on the punctured plane. Then $f=0$ on $z_k=frac 1{kpi}$ and $z_kto 0$ but $f$ is not identically 0.
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
answered Dec 20 '18 at 17:20
Guacho PerezGuacho Perez
3,97911134
3,97911134
add a comment |
add a comment |
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%2f3047768%2fsuppose-f-is-a-holomorphic-function-on-a-region-omega-that-vanishes-on-a-se%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
