Why needed Open disk in domain?
$begingroup$
Why is in above theorem, it is assumed that $D$ an open disk? Is it to make sure that we can surely apply mean value Theorem, ie, walking parallel to x or y axis, we are not moving out of domain? Can a convex open set do that job?
Also, this may sound silly, but why open disk? Is it so that derivative is defined without fuss, that is, without considering boundary of that disk??
complex-analysis derivatives holomorphic-functions
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
asked Jan 8 at 13:30
SilentSilent
2,83632152
$endgroup$
add a comment |
$begingroup$
Why is in above theorem, it is assumed that $D$ an open disk? Is it to make sure that we can surely apply mean value Theorem, ie, walking parallel to x or y axis, we are not moving out of domain? Can a convex open set do that job?
Also, this may sound silly, but why open disk? Is it so that derivative is defined without fuss, that is, without considering boundary of that disk??
complex-analysis derivatives holomorphic-functions
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
asked Jan 8 at 13:30
SilentSilent
2,83632152
$endgroup$
1
$begingroup$
If it's not open the derivative may not exist, e.g. singletons. It's just a simple way of avoiding such issues without really losing any knowledge.
$endgroup$
– Git Gud
Jan 8 at 13:33
$begingroup$
Also if $D$ is any open set (not necessarily a disc) it seems one could apply the theorem to any open disc contained in $D$ and get that conclusion, I think [haven't gone through details...]
$endgroup$
– coffeemath
Jan 8 at 13:37
add a comment |
$begingroup$
Why is in above theorem, it is assumed that $D$ an open disk? Is it to make sure that we can surely apply mean value Theorem, ie, walking parallel to x or y axis, we are not moving out of domain? Can a convex open set do that job?
Also, this may sound silly, but why open disk? Is it so that derivative is defined without fuss, that is, without considering boundary of that disk??
complex-analysis derivatives holomorphic-functions
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
asked Jan 8 at 13:30
SilentSilent
2,83632152
$endgroup$
Why is in above theorem, it is assumed that $D$ an open disk? Is it to make sure that we can surely apply mean value Theorem, ie, walking parallel to x or y axis, we are not moving out of domain? Can a convex open set do that job?
Also, this may sound silly, but why open disk? Is it so that derivative is defined without fuss, that is, without considering boundary of that disk??
complex-analysis derivatives holomorphic-functions
complex-analysis derivatives holomorphic-functions
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
asked Jan 8 at 13:30
SilentSilent
2,83632152
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
asked Jan 8 at 13:30
SilentSilent
2,83632152
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
edited Jan 8 at 13:54
José Carlos Santos
163k22130233
163k22130233
asked Jan 8 at 13:30
SilentSilent
2,83632152
asked Jan 8 at 13:30
SilentSilent
2,83632152
asked Jan 8 at 13:30
SilentSilent
2,83632152
2,83632152
1
$begingroup$
If it's not open the derivative may not exist, e.g. singletons. It's just a simple way of avoiding such issues without really losing any knowledge.
$endgroup$
– Git Gud
Jan 8 at 13:33
$begingroup$
Also if $D$ is any open set (not necessarily a disc) it seems one could apply the theorem to any open disc contained in $D$ and get that conclusion, I think [haven't gone through details...]
$endgroup$
– coffeemath
Jan 8 at 13:37
add a comment |
1
$begingroup$
If it's not open the derivative may not exist, e.g. singletons. It's just a simple way of avoiding such issues without really losing any knowledge.
$endgroup$
– Git Gud
Jan 8 at 13:33
$begingroup$
Also if $D$ is any open set (not necessarily a disc) it seems one could apply the theorem to any open disc contained in $D$ and get that conclusion, I think [haven't gone through details...]
$endgroup$
– coffeemath
Jan 8 at 13:37
1
1
$begingroup$
If it's not open the derivative may not exist, e.g. singletons. It's just a simple way of avoiding such issues without really losing any knowledge.
$endgroup$
– Git Gud
Jan 8 at 13:33
$begingroup$
If it's not open the derivative may not exist, e.g. singletons. It's just a simple way of avoiding such issues without really losing any knowledge.
$endgroup$
– Git Gud
Jan 8 at 13:33
$begingroup$
Also if $D$ is any open set (not necessarily a disc) it seems one could apply the theorem to any open disc contained in $D$ and get that conclusion, I think [haven't gone through details...]
$endgroup$
– coffeemath
Jan 8 at 13:37
$begingroup$
Also if $D$ is any open set (not necessarily a disc) it seems one could apply the theorem to any open disc contained in $D$ and get that conclusion, I think [haven't gone through details...]
$endgroup$
– coffeemath
Jan 8 at 13:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The domain can in fact be an arbitrary open connected subset of $mathbb C$ and the statement will still hold. Of course, the proof will have to be changed, because then we will not be able to go from an arbitrary point to another arbitrary point of the domain applying twice the mean value theorem (in other words: your guess was good).
The domain doesn't have to be open, but in order to have a sensible notion of derivative, at least the domain cannot have isolated points.
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Thank you very much, sir.
$endgroup$
– Silent
Jan 8 at 14:44
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%2f3066169%2fwhy-needed-open-disk-in-domain%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
$begingroup$
The domain can in fact be an arbitrary open connected subset of $mathbb C$ and the statement will still hold. Of course, the proof will have to be changed, because then we will not be able to go from an arbitrary point to another arbitrary point of the domain applying twice the mean value theorem (in other words: your guess was good).
The domain doesn't have to be open, but in order to have a sensible notion of derivative, at least the domain cannot have isolated points.
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Thank you very much, sir.
$endgroup$
– Silent
Jan 8 at 14:44
add a comment |
$begingroup$
The domain can in fact be an arbitrary open connected subset of $mathbb C$ and the statement will still hold. Of course, the proof will have to be changed, because then we will not be able to go from an arbitrary point to another arbitrary point of the domain applying twice the mean value theorem (in other words: your guess was good).
The domain doesn't have to be open, but in order to have a sensible notion of derivative, at least the domain cannot have isolated points.
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Thank you very much, sir.
$endgroup$
– Silent
Jan 8 at 14:44
add a comment |
$begingroup$
The domain can in fact be an arbitrary open connected subset of $mathbb C$ and the statement will still hold. Of course, the proof will have to be changed, because then we will not be able to go from an arbitrary point to another arbitrary point of the domain applying twice the mean value theorem (in other words: your guess was good).
The domain doesn't have to be open, but in order to have a sensible notion of derivative, at least the domain cannot have isolated points.
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
The domain can in fact be an arbitrary open connected subset of $mathbb C$ and the statement will still hold. Of course, the proof will have to be changed, because then we will not be able to go from an arbitrary point to another arbitrary point of the domain applying twice the mean value theorem (in other words: your guess was good).
The domain doesn't have to be open, but in order to have a sensible notion of derivative, at least the domain cannot have isolated points.
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 8 at 13:37
José Carlos SantosJosé Carlos Santos
163k22130233
163k22130233
$begingroup$
Thank you very much, sir.
$endgroup$
– Silent
Jan 8 at 14:44
add a comment |
$begingroup$
Thank you very much, sir.
$endgroup$
– Silent
Jan 8 at 14:44
$begingroup$
Thank you very much, sir.
$endgroup$
– Silent
Jan 8 at 14:44
$begingroup$
Thank you very much, sir.
$endgroup$
– Silent
Jan 8 at 14:44
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%2f3066169%2fwhy-needed-open-disk-in-domain%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
$begingroup$
If it's not open the derivative may not exist, e.g. singletons. It's just a simple way of avoiding such issues without really losing any knowledge.
$endgroup$
– Git Gud
Jan 8 at 13:33
$begingroup$
Also if $D$ is any open set (not necessarily a disc) it seems one could apply the theorem to any open disc contained in $D$ and get that conclusion, I think [haven't gone through details...]
$endgroup$
– coffeemath
Jan 8 at 13:37