Prove a bounded harmonic function $u$ on $0<|z|<1$ has removable singularity at the origin ...
How did people program for Consoles with multiple CPUs?
Why do airplanes bank sharply to the right after air-to-air refueling?
Won the lottery - how do I keep the money?
Are police here, aren't itthey?
Does soap repel water?
Does Germany produce more waste than the US?
What happened in Rome, when the western empire "fell"?
Is it convenient to ask the journal's editor for two additional days to complete a review?
Should I tutor a student who I know has cheated on their homework?
Is it ever safe to open a suspicious HTML file (e.g. email attachment)?
I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin
Easy to read palindrome checker
Is it okay to majorly distort historical facts while writing a fiction story?
A small doubt about the dominated convergence theorem
Can we say or write : "No, it'sn't"?
How to edit “Name” property in GCI output?
Proper way to express "He disappeared them"
TikZ: How to reverse arrow direction without switching start/end point?
Solving system of ODEs with extra parameter
How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?
What did we know about the Kessel run before the prequels?
How to get from Geneva Airport to Metabief, Doubs, France by public transport?
Does increasing your ability score affect your main stat?
How to write a definition with variants?
Prove a bounded harmonic function $u$ on $0
The Next CEO of Stack OverflowRemovable singularity at 0 if the image of the punctured unit disc has finite areaIf $F$ is entire with removable singularity at $infty$, then $F$ is constant?Mean value property for a harmonic function$u(x,y)$ harmonic and bounded in punctured disc; show $0$ is a removable singularityIf $u$ is harmonic and bounded in $0 < |z| < rho$, show that the origin is a removable singularityIntegrating $int_0^infty frac{1-cos x }{x^2}dx$ via contour integral.Harmonic except at a pointProving that an Isolated Singularity is removable.Harmonic function with vanishing partial derivativeRepresenting a function as a Poisson Integral.
$begingroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
$endgroup$
add a comment |
$begingroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
$endgroup$
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
add a comment |
$begingroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
$endgroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
edited Mar 17 at 7:16
Andrews
1,2812422
edited Mar 17 at 7:16
Andrews
1,2812422
edited Mar 17 at 7:16
Andrews
1,2812422
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
1
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
add a comment |
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
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%2f3151213%2fprove-a-bounded-harmonic-function-u-on-0z1-has-removable-singularity-at%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%2f3151213%2fprove-a-bounded-harmonic-function-u-on-0z1-has-removable-singularity-at%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
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29