Why does not the Montel Theorem apply in this case?












-1














enter image description here



I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?






complex-analysis







share|cite|improve this question


















  • 2




    Probably because $sin(nx)$ is not holomorphic.
    – Lukas Kofler
    Nov 29 at 12:55
















-1














enter image description here



I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?






complex-analysis







share|cite|improve this question


















  • 2




    Probably because $sin(nx)$ is not holomorphic.
    – Lukas Kofler
    Nov 29 at 12:55














-1












-1








-1







enter image description here



I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?






complex-analysis







share|cite|improve this question













enter image description here



I have proved that the family is not equicontinuous, but it is not understanding why the theorem 3.3 can not be applied. Can someone explain me?






complex-analysis




complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 at 12:53









Ricardo Freire

392110




392110








  • 2




    Probably because $sin(nx)$ is not holomorphic.
    – Lukas Kofler
    Nov 29 at 12:55














  • 2




    Probably because $sin(nx)$ is not holomorphic.
    – Lukas Kofler
    Nov 29 at 12:55








2




2




Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55




Probably because $sin(nx)$ is not holomorphic.
– Lukas Kofler
Nov 29 at 12:55










1 Answer
1






active

oldest

votes


















1














This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$






share|cite|improve this answer





















  • The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
    – Ricardo Freire
    Nov 29 at 13:37






  • 1




    In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
    – David C. Ullrich
    Nov 29 at 15:08













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018582%2fwhy-does-not-the-montel-theorem-apply-in-this-case%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









1














This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$






share|cite|improve this answer





















  • The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
    – Ricardo Freire
    Nov 29 at 13:37






  • 1




    In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
    – David C. Ullrich
    Nov 29 at 15:08


















1














This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$






share|cite|improve this answer





















  • The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
    – Ricardo Freire
    Nov 29 at 13:37






  • 1




    In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
    – David C. Ullrich
    Nov 29 at 15:08
















1












1








1






This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$






share|cite|improve this answer












This is because if we take any open set $A$ in the complex plane such that $(0,1)subset A$ then $A$ must have an element $z=a+bi$ with $bneq 0.$
Therefore $|sin nz|=|(2i)^{-1} (e^{nia -nb} -e^{-nia +nb})|to infty $ as $nto infty.$
So the family is not uniformly bounded in any region $A$ in complex plane, such that $(0,1)subset A.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 at 13:11









MotylaNogaTomkaMazura

6,564917




6,564917












  • The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
    – Ricardo Freire
    Nov 29 at 13:37






  • 1




    In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
    – David C. Ullrich
    Nov 29 at 15:08




















  • The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
    – Ricardo Freire
    Nov 29 at 13:37






  • 1




    In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
    – David C. Ullrich
    Nov 29 at 15:08


















The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37




The book itself says that the family is uniformly limited, do not you think it can be something else? Being uniformly limited does not refer to the open containing the set, but to the compacts that are contained in the set.
– Ricardo Freire
Nov 29 at 13:37




1




1




In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08






In the theorem $Omega$ is an open subset of the plane. If $Omega$ is an open subset of the plane and $(0,1)subsetOmega$ then there exists a compact $Ksubset Omega$ such that $\sin(nz)$ is not uniformly bounded on $K$. That's clear from the above, since singletons are compact.
– David C. Ullrich
Nov 29 at 15:08




















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018582%2fwhy-does-not-the-montel-theorem-apply-in-this-case%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
Read more

Fiat S.p.A.

Image
Cet article court présente un sujet plus développé dans : Fiat et Fiat Chrysler Automobiles. Fiat S.p.A. Fiat S.p.A. ( Fabbrica Italiana Automobili Torino , acronyme FIAT) est un groupe, notamment connu comme société-mère du constructeur automobile italien, Fiat. Portail des entreprises Portail de Turin Portail de l’automobile This page is only for reference, If you need detailed information, please check here
Read more

Type 'String' is not a subtype of type 'int' of 'index'

Image
up vote 0 down vote favorite My App can authenticate successfully when debugging against the Android emulator, but if I try to authenticate using debugging against the physical device (with same OS version), an error message appears after more than one minute waiting: Exception has occurred. _TypeError (type 'String' is not a subtype of type 'int' of 'index') The error message points to the following code: if (responseJson[AuthUtils.authTokenKey] != null) { AuthUtils.insertDetails( userNameController.text, passwordController.text, responseJson); ... } else { ... }; And in the DEBUG CONSOLE I get the following: I/flutter ( 9531): Auth: SocketException: OS Error: Connection timed out, errno = 110, address = example.com, port = 3897
Read more