Verification of Alternate Proof for Identity Theorem in Conplex Analysis. The Next CEO of...
Increase performance creating Mandelbrot set in python
Why do professional authors make "consistency" mistakes? And how to avoid them?
How can I open an app using Terminal?
Anatomically Correct Mesopelagic Aves
Should I tutor a student who I know has cheated on their homework?
How did people program for Consoles with multiple CPUs?
The King's new dress
Return the Closest Prime Number
Why is there a PLL in CPU?
Grabbing quick drinks
Is a stroke of luck acceptable after a series of unfavorable events?
If the heap is initialized for security, then why is the stack uninitialized?
How do I go from 300 unfinished/half written blog posts, to published posts?
How do I construct this japanese bowl?
Why here is plural "We went to the movies last night."
Why didn't Khan get resurrected in the Genesis Explosion?
How easy is it to start Magic from scratch?
How should I support this large drywall patch?
Go Pregnant or Go Home
What is the point of a new vote on May's deal when the indicative votes suggest she will not win?
Science fiction (dystopian) short story set after WWIII
Is it my responsibility to learn a new technology in my own time my employer wants to implement?
Can I equip Skullclamp on a creature I am sacrificing?
Opposite of a diet
Verification of Alternate Proof for Identity Theorem in Conplex Analysis.
The Next CEO of Stack OverflowOpen mapping principle complex?Corollary of identity theorem without connectedness assumptionTheorem related to the zeros of a function (Stein's Complex Analysis textbook)Can a non-constant analytic function have infinitely many zeros on a closed disk?problem on identity theorem for analytic functionsIf $f^2$ is analytic in $Omega$ then so is $f$-Problem in proof UnderstandingExercise for the identity theoremOpen mapping theorem using wikipedia proof$f(z)=0$ for all $z in mathbb{R}$Connectedness of the domain in Identity Theorem
$begingroup$
Can't we prove Identity Theorem like this.
IDENTITY THEOREM.
Consider a function which is analytic on an open connected domain $D$, if the set of zeros of $f$ has a limit point in $D$ then $f=0$ on $D$.
My Proof Idea.
If it is not $0$ identically then since the zeros of $f$ has a limit point in $D$ say $z_0$.
By continuity of $f$ we have $f(z_0)=0$ and since $z_0$ is a limit point of set of zeros it is not isolated.
Thus $z_0$ being a zero of the non-constant analytic function $f$ is not isolated.
This contradicts the fact that An non-constant analytic function has isolated zeros.
Hence our assumption is wrong and f is a constant function with value being zero.
complex-analysis analytic-functions
asked Mar 16 at 7:07
SunShineSunShine
1727
$endgroup$
add a comment |
$begingroup$
Can't we prove Identity Theorem like this.
IDENTITY THEOREM.
Consider a function which is analytic on an open connected domain $D$, if the set of zeros of $f$ has a limit point in $D$ then $f=0$ on $D$.
My Proof Idea.
If it is not $0$ identically then since the zeros of $f$ has a limit point in $D$ say $z_0$.
By continuity of $f$ we have $f(z_0)=0$ and since $z_0$ is a limit point of set of zeros it is not isolated.
Thus $z_0$ being a zero of the non-constant analytic function $f$ is not isolated.
This contradicts the fact that An non-constant analytic function has isolated zeros.
Hence our assumption is wrong and f is a constant function with value being zero.
complex-analysis analytic-functions
asked Mar 16 at 7:07
SunShineSunShine
1727
$endgroup$
1
$begingroup$
Where have you used connectedness of $D$?
$endgroup$
– Lord Shark the Unknown
Mar 16 at 7:10
$begingroup$
@LordSharktheUnknown . So that may work even when D is not connected? By the way is there any fault in reasoning?
$endgroup$
– SunShine
Mar 16 at 7:15
$begingroup$
It's a bit hard to check this since your proof isn't clearly worded. Please try to be more detailed and careful in your wording. That being said, connectedness is a technical condition here, since if our set is the disjoint union of the sets $A$ and $B$ and $f$ is identically zero on $A$ and identically one on $B$ then $f$ is in fact analytic on the domain $A cup B. $
$endgroup$
– Brevan Ellefsen
Mar 16 at 9:15
add a comment |
$begingroup$
Can't we prove Identity Theorem like this.
IDENTITY THEOREM.
Consider a function which is analytic on an open connected domain $D$, if the set of zeros of $f$ has a limit point in $D$ then $f=0$ on $D$.
My Proof Idea.
If it is not $0$ identically then since the zeros of $f$ has a limit point in $D$ say $z_0$.
By continuity of $f$ we have $f(z_0)=0$ and since $z_0$ is a limit point of set of zeros it is not isolated.
Thus $z_0$ being a zero of the non-constant analytic function $f$ is not isolated.
This contradicts the fact that An non-constant analytic function has isolated zeros.
Hence our assumption is wrong and f is a constant function with value being zero.
complex-analysis analytic-functions
asked Mar 16 at 7:07
SunShineSunShine
1727
$endgroup$
Can't we prove Identity Theorem like this.
IDENTITY THEOREM.
Consider a function which is analytic on an open connected domain $D$, if the set of zeros of $f$ has a limit point in $D$ then $f=0$ on $D$.
My Proof Idea.
If it is not $0$ identically then since the zeros of $f$ has a limit point in $D$ say $z_0$.
By continuity of $f$ we have $f(z_0)=0$ and since $z_0$ is a limit point of set of zeros it is not isolated.
Thus $z_0$ being a zero of the non-constant analytic function $f$ is not isolated.
This contradicts the fact that An non-constant analytic function has isolated zeros.
Hence our assumption is wrong and f is a constant function with value being zero.
complex-analysis analytic-functions
complex-analysis analytic-functions
asked Mar 16 at 7:07
SunShineSunShine
1727
asked Mar 16 at 7:07
SunShineSunShine
1727
asked Mar 16 at 7:07
SunShineSunShine
1727
asked Mar 16 at 7:07
SunShineSunShine
1727
asked Mar 16 at 7:07
SunShineSunShine
1727
1727
1
$begingroup$
Where have you used connectedness of $D$?
$endgroup$
– Lord Shark the Unknown
Mar 16 at 7:10
$begingroup$
@LordSharktheUnknown . So that may work even when D is not connected? By the way is there any fault in reasoning?
$endgroup$
– SunShine
Mar 16 at 7:15
$begingroup$
It's a bit hard to check this since your proof isn't clearly worded. Please try to be more detailed and careful in your wording. That being said, connectedness is a technical condition here, since if our set is the disjoint union of the sets $A$ and $B$ and $f$ is identically zero on $A$ and identically one on $B$ then $f$ is in fact analytic on the domain $A cup B. $
$endgroup$
– Brevan Ellefsen
Mar 16 at 9:15
add a comment |
1
$begingroup$
Where have you used connectedness of $D$?
$endgroup$
– Lord Shark the Unknown
Mar 16 at 7:10
$begingroup$
@LordSharktheUnknown . So that may work even when D is not connected? By the way is there any fault in reasoning?
$endgroup$
– SunShine
Mar 16 at 7:15
$begingroup$
It's a bit hard to check this since your proof isn't clearly worded. Please try to be more detailed and careful in your wording. That being said, connectedness is a technical condition here, since if our set is the disjoint union of the sets $A$ and $B$ and $f$ is identically zero on $A$ and identically one on $B$ then $f$ is in fact analytic on the domain $A cup B. $
$endgroup$
– Brevan Ellefsen
Mar 16 at 9:15
1
1
$begingroup$
Where have you used connectedness of $D$?
$endgroup$
– Lord Shark the Unknown
Mar 16 at 7:10
$begingroup$
Where have you used connectedness of $D$?
$endgroup$
– Lord Shark the Unknown
Mar 16 at 7:10
$begingroup$
@LordSharktheUnknown . So that may work even when D is not connected? By the way is there any fault in reasoning?
$endgroup$
– SunShine
Mar 16 at 7:15
$begingroup$
@LordSharktheUnknown . So that may work even when D is not connected? By the way is there any fault in reasoning?
$endgroup$
– SunShine
Mar 16 at 7:15
$begingroup$
It's a bit hard to check this since your proof isn't clearly worded. Please try to be more detailed and careful in your wording. That being said, connectedness is a technical condition here, since if our set is the disjoint union of the sets $A$ and $B$ and $f$ is identically zero on $A$ and identically one on $B$ then $f$ is in fact analytic on the domain $A cup B. $
$endgroup$
– Brevan Ellefsen
Mar 16 at 9:15
$begingroup$
It's a bit hard to check this since your proof isn't clearly worded. Please try to be more detailed and careful in your wording. That being said, connectedness is a technical condition here, since if our set is the disjoint union of the sets $A$ and $B$ and $f$ is identically zero on $A$ and identically one on $B$ then $f$ is in fact analytic on the domain $A cup B. $
$endgroup$
– Brevan Ellefsen
Mar 16 at 9:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As far as I can tell, the error in your proof is in the last line, when you use the fact that a non-constant function has isolated zeros. This is definitely true if your domain is $mathbb{C}, $ which is probably the version you are thinking of, but is false in general. For example, let $A$ and $B$ be disjoint and define a function $f$ by $f equiv 0$ on $A$ and $f equiv 1$ on $B.$ Then $f$ is analytic on $A cup B$ and has non-isolated zeros but is not constant.
The theorem you are looking for is this:
given a function $fcolon A to B$, if $f$ has an accumulation point of zeros on some domain (i.e. open and connected) $C subseteq A$ then $f$ is constant on $C$.
This phenomenon is more general throughout math, and is related to topology. For example, via purely topological methods we can show that if a function is locally constant on a connected component then it is globally constant on the whole connected component, but we cannot conclude the function is constant elsewhere.
Note: you should also use the assumption that $D$ is open, since if our limit point is on the boundary the Identity Theorem can fail. Note this is part of the theorem you should be using above.
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
$endgroup$
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%2f3150155%2fverification-of-alternate-proof-for-identity-theorem-in-conplex-analysis%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
$begingroup$
As far as I can tell, the error in your proof is in the last line, when you use the fact that a non-constant function has isolated zeros. This is definitely true if your domain is $mathbb{C}, $ which is probably the version you are thinking of, but is false in general. For example, let $A$ and $B$ be disjoint and define a function $f$ by $f equiv 0$ on $A$ and $f equiv 1$ on $B.$ Then $f$ is analytic on $A cup B$ and has non-isolated zeros but is not constant.
The theorem you are looking for is this:
given a function $fcolon A to B$, if $f$ has an accumulation point of zeros on some domain (i.e. open and connected) $C subseteq A$ then $f$ is constant on $C$.
This phenomenon is more general throughout math, and is related to topology. For example, via purely topological methods we can show that if a function is locally constant on a connected component then it is globally constant on the whole connected component, but we cannot conclude the function is constant elsewhere.
Note: you should also use the assumption that $D$ is open, since if our limit point is on the boundary the Identity Theorem can fail. Note this is part of the theorem you should be using above.
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
$endgroup$
add a comment |
$begingroup$
As far as I can tell, the error in your proof is in the last line, when you use the fact that a non-constant function has isolated zeros. This is definitely true if your domain is $mathbb{C}, $ which is probably the version you are thinking of, but is false in general. For example, let $A$ and $B$ be disjoint and define a function $f$ by $f equiv 0$ on $A$ and $f equiv 1$ on $B.$ Then $f$ is analytic on $A cup B$ and has non-isolated zeros but is not constant.
The theorem you are looking for is this:
given a function $fcolon A to B$, if $f$ has an accumulation point of zeros on some domain (i.e. open and connected) $C subseteq A$ then $f$ is constant on $C$.
This phenomenon is more general throughout math, and is related to topology. For example, via purely topological methods we can show that if a function is locally constant on a connected component then it is globally constant on the whole connected component, but we cannot conclude the function is constant elsewhere.
Note: you should also use the assumption that $D$ is open, since if our limit point is on the boundary the Identity Theorem can fail. Note this is part of the theorem you should be using above.
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
$endgroup$
add a comment |
$begingroup$
As far as I can tell, the error in your proof is in the last line, when you use the fact that a non-constant function has isolated zeros. This is definitely true if your domain is $mathbb{C}, $ which is probably the version you are thinking of, but is false in general. For example, let $A$ and $B$ be disjoint and define a function $f$ by $f equiv 0$ on $A$ and $f equiv 1$ on $B.$ Then $f$ is analytic on $A cup B$ and has non-isolated zeros but is not constant.
The theorem you are looking for is this:
given a function $fcolon A to B$, if $f$ has an accumulation point of zeros on some domain (i.e. open and connected) $C subseteq A$ then $f$ is constant on $C$.
This phenomenon is more general throughout math, and is related to topology. For example, via purely topological methods we can show that if a function is locally constant on a connected component then it is globally constant on the whole connected component, but we cannot conclude the function is constant elsewhere.
Note: you should also use the assumption that $D$ is open, since if our limit point is on the boundary the Identity Theorem can fail. Note this is part of the theorem you should be using above.
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
$endgroup$
As far as I can tell, the error in your proof is in the last line, when you use the fact that a non-constant function has isolated zeros. This is definitely true if your domain is $mathbb{C}, $ which is probably the version you are thinking of, but is false in general. For example, let $A$ and $B$ be disjoint and define a function $f$ by $f equiv 0$ on $A$ and $f equiv 1$ on $B.$ Then $f$ is analytic on $A cup B$ and has non-isolated zeros but is not constant.
The theorem you are looking for is this:
given a function $fcolon A to B$, if $f$ has an accumulation point of zeros on some domain (i.e. open and connected) $C subseteq A$ then $f$ is constant on $C$.
This phenomenon is more general throughout math, and is related to topology. For example, via purely topological methods we can show that if a function is locally constant on a connected component then it is globally constant on the whole connected component, but we cannot conclude the function is constant elsewhere.
Note: you should also use the assumption that $D$ is open, since if our limit point is on the boundary the Identity Theorem can fail. Note this is part of the theorem you should be using above.
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
edited Mar 16 at 9:39
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
answered Mar 16 at 9:25
Brevan EllefsenBrevan Ellefsen
12k31650
12k31650
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%2f3150155%2fverification-of-alternate-proof-for-identity-theorem-in-conplex-analysis%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
$begingroup$
Where have you used connectedness of $D$?
$endgroup$
– Lord Shark the Unknown
Mar 16 at 7:10
$begingroup$
@LordSharktheUnknown . So that may work even when D is not connected? By the way is there any fault in reasoning?
$endgroup$
– SunShine
Mar 16 at 7:15
$begingroup$
It's a bit hard to check this since your proof isn't clearly worded. Please try to be more detailed and careful in your wording. That being said, connectedness is a technical condition here, since if our set is the disjoint union of the sets $A$ and $B$ and $f$ is identically zero on $A$ and identically one on $B$ then $f$ is in fact analytic on the domain $A cup B. $
$endgroup$
– Brevan Ellefsen
Mar 16 at 9:15