Application of Reflection Principle for Holomorphic functionsUnderstanding the Schwarz reflection principleSchwarz reflection principle and bounded derivativesThe continuity assumption in Schwarz's reflection principleUnderstanding the Schwarz Reflection Principle,Schwarz Reflection Principle on a unit diskUsing Schwarz Reflection principle and Identity principle to show that a mapping vanishes.Holomorphic extension using reflection principleUnderstanding this wording of the Schwarz Reflection PrincipleGeneralization of Schwarz Reflection Principle for Harmonic FunctionsHarmonic functions and Maximum Modulus Principle. (Real harmonic function implies holomorphic function?)
In 'Revenger,' what does 'cove' come from?
Is there a hemisphere-neutral way of specifying a season?
All in one piece, we mend holes in your socks
Is it acceptable for a professor to tell male students to not think that they are smarter than female students?
How badly should I try to prevent a user from XSSing themselves?
What does the expression "A Mann!" means
Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?
Im going to France and my passport expires June 19th
Expand and Contract
Would Slavery Reparations be considered Bills of Attainder and hence Illegal?
How to tell a function to use the default argument values?
Arrow those variables!
Can a virus destroy the BIOS of a modern computer?
What do you call someone who asks many questions?
Is "remove commented out code" correct English?
How do conventional missiles fly?
Size of subfigure fitting its content (tikzpicture)
Why can't we play rap on piano?
Why no variance term in Bayesian logistic regression?
Is it inappropriate for a student to attend their mentor's dissertation defense?
How much of data wrangling is a data scientist's job?
What exploit Are these user agents trying to use?
Detention in 1997
How dangerous is XSS?
Application of Reflection Principle for Holomorphic functions
Understanding the Schwarz reflection principleSchwarz reflection principle and bounded derivativesThe continuity assumption in Schwarz's reflection principleUnderstanding the Schwarz Reflection Principle,Schwarz Reflection Principle on a unit diskUsing Schwarz Reflection principle and Identity principle to show that a mapping vanishes.Holomorphic extension using reflection principleUnderstanding this wording of the Schwarz Reflection PrincipleGeneralization of Schwarz Reflection Principle for Harmonic FunctionsHarmonic functions and Maximum Modulus Principle. (Real harmonic function implies holomorphic function?)
$begingroup$
Let $f$ be holomorphic on $D'(0,1)=0<$ and $f$ is continuous and real valued on $=1.$ Show $f$ can be extended to $mathbbC-0$ such that $f(z)= overlinef(1/overlinez), forall z neq 0.$
My attempt: $f=u+iv=u$ on $=1.$ By Max/Min Principle for Harmonic functions, $v equiv 0$ in $D^prime(0,1) $ and so $f equiv $ constant in $D'(0,1).$ By Identity Theorem, $f equiv$ constant in $mathbbC ?$ I suspect I have wrongly applied the Max/Min Principle because $f$ is not necessarily continuous on boundary of $D^prime(0,1) ?$
I'm trying to make use the following theorem:
Let $Omega$ be a region which is symmetric wrt to real axis and define $Omega^+, Omega^-$ and $L$ as intersection of $Omega$ with upper half plane, lower half plane and real axis respectively. If $f$ is continuous on $Omega^+ cup L$ which is analytic on $Omega^+$ and real on $L,$ then $f$ can be uniquely extended to holomorphic $F$ on all of $Omega$ such that $F(z)=f(z), z in Omega^+ cup L; F(z)= overlinef(overlinez), z in Omega^-.$
Could anyone advise please? Thank you.
complex-analysis
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
$endgroup$
|
show 2 more comments
$begingroup$
Let $f$ be holomorphic on $D'(0,1)=0<$ and $f$ is continuous and real valued on $=1.$ Show $f$ can be extended to $mathbbC-0$ such that $f(z)= overlinef(1/overlinez), forall z neq 0.$
My attempt: $f=u+iv=u$ on $=1.$ By Max/Min Principle for Harmonic functions, $v equiv 0$ in $D^prime(0,1) $ and so $f equiv $ constant in $D'(0,1).$ By Identity Theorem, $f equiv$ constant in $mathbbC ?$ I suspect I have wrongly applied the Max/Min Principle because $f$ is not necessarily continuous on boundary of $D^prime(0,1) ?$
I'm trying to make use the following theorem:
Let $Omega$ be a region which is symmetric wrt to real axis and define $Omega^+, Omega^-$ and $L$ as intersection of $Omega$ with upper half plane, lower half plane and real axis respectively. If $f$ is continuous on $Omega^+ cup L$ which is analytic on $Omega^+$ and real on $L,$ then $f$ can be uniquely extended to holomorphic $F$ on all of $Omega$ such that $F(z)=f(z), z in Omega^+ cup L; F(z)= overlinef(overlinez), z in Omega^-.$
Could anyone advise please? Thank you.
complex-analysis
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
$endgroup$
1
$begingroup$
$u$ and $v$ are not defined in $0$, so you can't apply the maximum principle to deduce that $vequiv 0$ from its vanishing on $lvert zrvert = 1$. It can be unbounded at $0$.
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:35
1
$begingroup$
Let $Tz = fracz-iz+i$. What do you know about $fcirc T$?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:36
$begingroup$
$f circ T : mathbbH-i$ is a holomorphic map?
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:43
$begingroup$
And, what about its behaviour at the real axis?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:44
$begingroup$
$T$ maps real axis to boundary of the unit disk and $f$ maps those boundary points into $mathbbR.$
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:49
|
show 2 more comments
$begingroup$
Let $f$ be holomorphic on $D'(0,1)=0<$ and $f$ is continuous and real valued on $=1.$ Show $f$ can be extended to $mathbbC-0$ such that $f(z)= overlinef(1/overlinez), forall z neq 0.$
My attempt: $f=u+iv=u$ on $=1.$ By Max/Min Principle for Harmonic functions, $v equiv 0$ in $D^prime(0,1) $ and so $f equiv $ constant in $D'(0,1).$ By Identity Theorem, $f equiv$ constant in $mathbbC ?$ I suspect I have wrongly applied the Max/Min Principle because $f$ is not necessarily continuous on boundary of $D^prime(0,1) ?$
I'm trying to make use the following theorem:
Let $Omega$ be a region which is symmetric wrt to real axis and define $Omega^+, Omega^-$ and $L$ as intersection of $Omega$ with upper half plane, lower half plane and real axis respectively. If $f$ is continuous on $Omega^+ cup L$ which is analytic on $Omega^+$ and real on $L,$ then $f$ can be uniquely extended to holomorphic $F$ on all of $Omega$ such that $F(z)=f(z), z in Omega^+ cup L; F(z)= overlinef(overlinez), z in Omega^-.$
Could anyone advise please? Thank you.
complex-analysis
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
$endgroup$
Let $f$ be holomorphic on $D'(0,1)=0<$ and $f$ is continuous and real valued on $=1.$ Show $f$ can be extended to $mathbbC-0$ such that $f(z)= overlinef(1/overlinez), forall z neq 0.$
My attempt: $f=u+iv=u$ on $=1.$ By Max/Min Principle for Harmonic functions, $v equiv 0$ in $D^prime(0,1) $ and so $f equiv $ constant in $D'(0,1).$ By Identity Theorem, $f equiv$ constant in $mathbbC ?$ I suspect I have wrongly applied the Max/Min Principle because $f$ is not necessarily continuous on boundary of $D^prime(0,1) ?$
I'm trying to make use the following theorem:
Let $Omega$ be a region which is symmetric wrt to real axis and define $Omega^+, Omega^-$ and $L$ as intersection of $Omega$ with upper half plane, lower half plane and real axis respectively. If $f$ is continuous on $Omega^+ cup L$ which is analytic on $Omega^+$ and real on $L,$ then $f$ can be uniquely extended to holomorphic $F$ on all of $Omega$ such that $F(z)=f(z), z in Omega^+ cup L; F(z)= overlinef(overlinez), z in Omega^-.$
Could anyone advise please? Thank you.
complex-analysis
complex-analysis
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
edited Nov 4 '14 at 20:31
Alexy Vincenzo
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
asked Nov 4 '14 at 20:26
Alexy VincenzoAlexy Vincenzo
2,1903926
2,1903926
1
$begingroup$
$u$ and $v$ are not defined in $0$, so you can't apply the maximum principle to deduce that $vequiv 0$ from its vanishing on $lvert zrvert = 1$. It can be unbounded at $0$.
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:35
1
$begingroup$
Let $Tz = fracz-iz+i$. What do you know about $fcirc T$?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:36
$begingroup$
$f circ T : mathbbH-i$ is a holomorphic map?
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:43
$begingroup$
And, what about its behaviour at the real axis?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:44
$begingroup$
$T$ maps real axis to boundary of the unit disk and $f$ maps those boundary points into $mathbbR.$
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:49
|
show 2 more comments
1
$begingroup$
$u$ and $v$ are not defined in $0$, so you can't apply the maximum principle to deduce that $vequiv 0$ from its vanishing on $lvert zrvert = 1$. It can be unbounded at $0$.
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:35
1
$begingroup$
Let $Tz = fracz-iz+i$. What do you know about $fcirc T$?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:36
$begingroup$
$f circ T : mathbbH-i$ is a holomorphic map?
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:43
$begingroup$
And, what about its behaviour at the real axis?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:44
$begingroup$
$T$ maps real axis to boundary of the unit disk and $f$ maps those boundary points into $mathbbR.$
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:49
1
1
$begingroup$
$u$ and $v$ are not defined in $0$, so you can't apply the maximum principle to deduce that $vequiv 0$ from its vanishing on $lvert zrvert = 1$. It can be unbounded at $0$.
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:35
$begingroup$
$u$ and $v$ are not defined in $0$, so you can't apply the maximum principle to deduce that $vequiv 0$ from its vanishing on $lvert zrvert = 1$. It can be unbounded at $0$.
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:35
1
1
$begingroup$
Let $Tz = fracz-iz+i$. What do you know about $fcirc T$?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:36
$begingroup$
Let $Tz = fracz-iz+i$. What do you know about $fcirc T$?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:36
$begingroup$
$f circ T : mathbbH-i$ is a holomorphic map?
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:43
$begingroup$
$f circ T : mathbbH-i$ is a holomorphic map?
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:43
$begingroup$
And, what about its behaviour at the real axis?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:44
$begingroup$
And, what about its behaviour at the real axis?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:44
$begingroup$
$T$ maps real axis to boundary of the unit disk and $f$ maps those boundary points into $mathbbR.$
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:49
$begingroup$
$T$ maps real axis to boundary of the unit disk and $f$ maps those boundary points into $mathbbR.$
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:49
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint. Let
$$
g(z)=fleft(fracz-i1-izright).
$$
Then $g$ takes real values on the real axis, and it is analytic in the upper half plane.
Use Schwarz Reflection for $g$, and then use the inverse transformation:
$$f(z)=gleft(fracz+i1+izright).$$
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
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%2f1006363%2fapplication-of-reflection-principle-for-holomorphic-functions%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint. Let
$$
g(z)=fleft(fracz-i1-izright).
$$
Then $g$ takes real values on the real axis, and it is analytic in the upper half plane.
Use Schwarz Reflection for $g$, and then use the inverse transformation:
$$f(z)=gleft(fracz+i1+izright).$$
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
add a comment |
$begingroup$
Hint. Let
$$
g(z)=fleft(fracz-i1-izright).
$$
Then $g$ takes real values on the real axis, and it is analytic in the upper half plane.
Use Schwarz Reflection for $g$, and then use the inverse transformation:
$$f(z)=gleft(fracz+i1+izright).$$
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
add a comment |
$begingroup$
Hint. Let
$$
g(z)=fleft(fracz-i1-izright).
$$
Then $g$ takes real values on the real axis, and it is analytic in the upper half plane.
Use Schwarz Reflection for $g$, and then use the inverse transformation:
$$f(z)=gleft(fracz+i1+izright).$$
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
Hint. Let
$$
g(z)=fleft(fracz-i1-izright).
$$
Then $g$ takes real values on the real axis, and it is analytic in the upper half plane.
Use Schwarz Reflection for $g$, and then use the inverse transformation:
$$f(z)=gleft(fracz+i1+izright).$$
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Nov 4 '14 at 20:41
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
63.7k1385165
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%2f1006363%2fapplication-of-reflection-principle-for-holomorphic-functions%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
1
$begingroup$
$u$ and $v$ are not defined in $0$, so you can't apply the maximum principle to deduce that $vequiv 0$ from its vanishing on $lvert zrvert = 1$. It can be unbounded at $0$.
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:35
1
$begingroup$
Let $Tz = fracz-iz+i$. What do you know about $fcirc T$?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:36
$begingroup$
$f circ T : mathbbH-i$ is a holomorphic map?
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:43
$begingroup$
And, what about its behaviour at the real axis?
$endgroup$
– Daniel Fischer
Nov 4 '14 at 20:44
$begingroup$
$T$ maps real axis to boundary of the unit disk and $f$ maps those boundary points into $mathbbR.$
$endgroup$
– Alexy Vincenzo
Nov 4 '14 at 20:49