Mixing
Is there a “monotonicity” property for analytic continuation?
$begingroup$
If I have two complex functions defined by power series $A(z) = sum a_n z^n $, $B(z) = sum b_n z^n$ with $|a_n| ge |b_n|$ for all $n$, and I know that $A$ converges in some set $U_1$ and defines a function of $z$ with an analytic extension to set $U_2$, then of course $B$ converges in $U_1$ also, but does this imply that $B$ has an analytic extension to $U_2$?
It would be nice if this were true, but I am a bit concerned because it's not hard to imagine a situation in which $B(z)$ that slightly shifts the region where $A(z)$ fails to have an analytic continuation. Can anyone help me think of a concrete counter example, or give an idea towards a proof?
complex-analysis power-series analytic-functions analytic-continuation
asked Jan 15 at 18:02
DylanDylan
33
$endgroup$
add a comment |
$begingroup$
If I have two complex functions defined by power series $A(z) = sum a_n z^n $, $B(z) = sum b_n z^n$ with $|a_n| ge |b_n|$ for all $n$, and I know that $A$ converges in some set $U_1$ and defines a function of $z$ with an analytic extension to set $U_2$, then of course $B$ converges in $U_1$ also, but does this imply that $B$ has an analytic extension to $U_2$?
It would be nice if this were true, but I am a bit concerned because it's not hard to imagine a situation in which $B(z)$ that slightly shifts the region where $A(z)$ fails to have an analytic continuation. Can anyone help me think of a concrete counter example, or give an idea towards a proof?
complex-analysis power-series analytic-functions analytic-continuation
asked Jan 15 at 18:02
DylanDylan
33
$endgroup$
2
$begingroup$
As stated, what you say about convergence is false, never mind analytic continuation. (Consider $b_n=-n!$, $a_n=1/n!$.) Surely you meant $a_nge b_nge 0$, or maybe just $|a_n|ge |b_n|$. With that change it seems like a reasonable question - my guess is the answer is "of course not", but I don't have a counterexample handy.
$endgroup$
– David C. Ullrich
Jan 15 at 18:15
$begingroup$
Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence.
$endgroup$
– David C. Ullrich
Jan 15 at 18:40
$begingroup$
Good catch, I did indeed mean to have absolute values; edited that. My guess is also "of course not" but I am having trouble thinking of an example.
$endgroup$
– Dylan
Jan 15 at 18:57
add a comment |
$begingroup$
If I have two complex functions defined by power series $A(z) = sum a_n z^n $, $B(z) = sum b_n z^n$ with $|a_n| ge |b_n|$ for all $n$, and I know that $A$ converges in some set $U_1$ and defines a function of $z$ with an analytic extension to set $U_2$, then of course $B$ converges in $U_1$ also, but does this imply that $B$ has an analytic extension to $U_2$?
It would be nice if this were true, but I am a bit concerned because it's not hard to imagine a situation in which $B(z)$ that slightly shifts the region where $A(z)$ fails to have an analytic continuation. Can anyone help me think of a concrete counter example, or give an idea towards a proof?
complex-analysis power-series analytic-functions analytic-continuation
asked Jan 15 at 18:02
DylanDylan
33
$endgroup$
If I have two complex functions defined by power series $A(z) = sum a_n z^n $, $B(z) = sum b_n z^n$ with $|a_n| ge |b_n|$ for all $n$, and I know that $A$ converges in some set $U_1$ and defines a function of $z$ with an analytic extension to set $U_2$, then of course $B$ converges in $U_1$ also, but does this imply that $B$ has an analytic extension to $U_2$?
It would be nice if this were true, but I am a bit concerned because it's not hard to imagine a situation in which $B(z)$ that slightly shifts the region where $A(z)$ fails to have an analytic continuation. Can anyone help me think of a concrete counter example, or give an idea towards a proof?
complex-analysis power-series analytic-functions analytic-continuation
complex-analysis power-series analytic-functions analytic-continuation
asked Jan 15 at 18:02
DylanDylan
33
asked Jan 15 at 18:02
DylanDylan
33
edited Jan 15 at 18:56
Dylan
asked Jan 15 at 18:02
DylanDylan
33
asked Jan 15 at 18:02
DylanDylan
33
asked Jan 15 at 18:02
DylanDylan
33
33
2
$begingroup$
As stated, what you say about convergence is false, never mind analytic continuation. (Consider $b_n=-n!$, $a_n=1/n!$.) Surely you meant $a_nge b_nge 0$, or maybe just $|a_n|ge |b_n|$. With that change it seems like a reasonable question - my guess is the answer is "of course not", but I don't have a counterexample handy.
$endgroup$
– David C. Ullrich
Jan 15 at 18:15
$begingroup$
Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence.
$endgroup$
– David C. Ullrich
Jan 15 at 18:40
$begingroup$
Good catch, I did indeed mean to have absolute values; edited that. My guess is also "of course not" but I am having trouble thinking of an example.
$endgroup$
– Dylan
Jan 15 at 18:57
add a comment |
2
$begingroup$
As stated, what you say about convergence is false, never mind analytic continuation. (Consider $b_n=-n!$, $a_n=1/n!$.) Surely you meant $a_nge b_nge 0$, or maybe just $|a_n|ge |b_n|$. With that change it seems like a reasonable question - my guess is the answer is "of course not", but I don't have a counterexample handy.
$endgroup$
– David C. Ullrich
Jan 15 at 18:15
$begingroup$
Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence.
$endgroup$
– David C. Ullrich
Jan 15 at 18:40
$begingroup$
Good catch, I did indeed mean to have absolute values; edited that. My guess is also "of course not" but I am having trouble thinking of an example.
$endgroup$
– Dylan
Jan 15 at 18:57
2
2
$begingroup$
As stated, what you say about convergence is false, never mind analytic continuation. (Consider $b_n=-n!$, $a_n=1/n!$.) Surely you meant $a_nge b_nge 0$, or maybe just $|a_n|ge |b_n|$. With that change it seems like a reasonable question - my guess is the answer is "of course not", but I don't have a counterexample handy.
$endgroup$
– David C. Ullrich
Jan 15 at 18:15
$begingroup$
As stated, what you say about convergence is false, never mind analytic continuation. (Consider $b_n=-n!$, $a_n=1/n!$.) Surely you meant $a_nge b_nge 0$, or maybe just $|a_n|ge |b_n|$. With that change it seems like a reasonable question - my guess is the answer is "of course not", but I don't have a counterexample handy.
$endgroup$
– David C. Ullrich
Jan 15 at 18:15
$begingroup$
Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence.
$endgroup$
– David C. Ullrich
Jan 15 at 18:40
$begingroup$
Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence.
$endgroup$
– David C. Ullrich
Jan 15 at 18:40
$begingroup$
Good catch, I did indeed mean to have absolute values; edited that. My guess is also "of course not" but I am having trouble thinking of an example.
$endgroup$
– Dylan
Jan 15 at 18:57
$begingroup$
Good catch, I did indeed mean to have absolute values; edited that. My guess is also "of course not" but I am having trouble thinking of an example.
$endgroup$
– Dylan
Jan 15 at 18:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We need to modify the hypotheses to get a reasonable question; if $b_n<0$ and $a_n>0$ then $b_nle a_n$ but there's obviously no inclusion that follows even for the region of convergence.
Two reasonable questions might be the above assuming $a_nge b_nge0$ or assuming just $|a_n|ge |b_n|$. The answer is no for the second version: If $A(z)=1/(z-1)$ and $B(z)=1/(z+1)$ then $|b_n|=|a_n|$ but there is no inclusion between the two regions of continuation.
Ah, a more interesting counterexample with $a_nge b_nge0$: $$A(z)=sum z^n,$$ $$B(z)=sum z^{2^n}.$$Then $A$ extends to a function holomorphic in $Bbb Csetminus{1}$, while it's well known that $B$ does not extend past any point of the unit circle.
(Heh: Replacing $B$ by $(A+B)/3$ gives a counterexample with $a_n>b_n>0$, the strongest version of the hypothesis I can think of. So it's really no, with no way to fix it.)
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
thank you for the examples! They are both very good.
$endgroup$
– Dylan
Jan 15 at 18:58
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%2f3074729%2fis-there-a-monotonicity-property-for-analytic-continuation%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
$begingroup$
We need to modify the hypotheses to get a reasonable question; if $b_n<0$ and $a_n>0$ then $b_nle a_n$ but there's obviously no inclusion that follows even for the region of convergence.
Two reasonable questions might be the above assuming $a_nge b_nge0$ or assuming just $|a_n|ge |b_n|$. The answer is no for the second version: If $A(z)=1/(z-1)$ and $B(z)=1/(z+1)$ then $|b_n|=|a_n|$ but there is no inclusion between the two regions of continuation.
Ah, a more interesting counterexample with $a_nge b_nge0$: $$A(z)=sum z^n,$$ $$B(z)=sum z^{2^n}.$$Then $A$ extends to a function holomorphic in $Bbb Csetminus{1}$, while it's well known that $B$ does not extend past any point of the unit circle.
(Heh: Replacing $B$ by $(A+B)/3$ gives a counterexample with $a_n>b_n>0$, the strongest version of the hypothesis I can think of. So it's really no, with no way to fix it.)
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
thank you for the examples! They are both very good.
$endgroup$
– Dylan
Jan 15 at 18:58
add a comment |
$begingroup$
We need to modify the hypotheses to get a reasonable question; if $b_n<0$ and $a_n>0$ then $b_nle a_n$ but there's obviously no inclusion that follows even for the region of convergence.
Two reasonable questions might be the above assuming $a_nge b_nge0$ or assuming just $|a_n|ge |b_n|$. The answer is no for the second version: If $A(z)=1/(z-1)$ and $B(z)=1/(z+1)$ then $|b_n|=|a_n|$ but there is no inclusion between the two regions of continuation.
Ah, a more interesting counterexample with $a_nge b_nge0$: $$A(z)=sum z^n,$$ $$B(z)=sum z^{2^n}.$$Then $A$ extends to a function holomorphic in $Bbb Csetminus{1}$, while it's well known that $B$ does not extend past any point of the unit circle.
(Heh: Replacing $B$ by $(A+B)/3$ gives a counterexample with $a_n>b_n>0$, the strongest version of the hypothesis I can think of. So it's really no, with no way to fix it.)
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
thank you for the examples! They are both very good.
$endgroup$
– Dylan
Jan 15 at 18:58
add a comment |
$begingroup$
We need to modify the hypotheses to get a reasonable question; if $b_n<0$ and $a_n>0$ then $b_nle a_n$ but there's obviously no inclusion that follows even for the region of convergence.
Two reasonable questions might be the above assuming $a_nge b_nge0$ or assuming just $|a_n|ge |b_n|$. The answer is no for the second version: If $A(z)=1/(z-1)$ and $B(z)=1/(z+1)$ then $|b_n|=|a_n|$ but there is no inclusion between the two regions of continuation.
Ah, a more interesting counterexample with $a_nge b_nge0$: $$A(z)=sum z^n,$$ $$B(z)=sum z^{2^n}.$$Then $A$ extends to a function holomorphic in $Bbb Csetminus{1}$, while it's well known that $B$ does not extend past any point of the unit circle.
(Heh: Replacing $B$ by $(A+B)/3$ gives a counterexample with $a_n>b_n>0$, the strongest version of the hypothesis I can think of. So it's really no, with no way to fix it.)
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
We need to modify the hypotheses to get a reasonable question; if $b_n<0$ and $a_n>0$ then $b_nle a_n$ but there's obviously no inclusion that follows even for the region of convergence.
Two reasonable questions might be the above assuming $a_nge b_nge0$ or assuming just $|a_n|ge |b_n|$. The answer is no for the second version: If $A(z)=1/(z-1)$ and $B(z)=1/(z+1)$ then $|b_n|=|a_n|$ but there is no inclusion between the two regions of continuation.
Ah, a more interesting counterexample with $a_nge b_nge0$: $$A(z)=sum z^n,$$ $$B(z)=sum z^{2^n}.$$Then $A$ extends to a function holomorphic in $Bbb Csetminus{1}$, while it's well known that $B$ does not extend past any point of the unit circle.
(Heh: Replacing $B$ by $(A+B)/3$ gives a counterexample with $a_n>b_n>0$, the strongest version of the hypothesis I can think of. So it's really no, with no way to fix it.)
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
edited Jan 15 at 19:03
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 15 at 18:35
David C. UllrichDavid C. Ullrich
60.9k43994
60.9k43994
$begingroup$
thank you for the examples! They are both very good.
$endgroup$
– Dylan
Jan 15 at 18:58
add a comment |
$begingroup$
thank you for the examples! They are both very good.
$endgroup$
– Dylan
Jan 15 at 18:58
$begingroup$
thank you for the examples! They are both very good.
$endgroup$
– Dylan
Jan 15 at 18:58
$begingroup$
thank you for the examples! They are both very good.
$endgroup$
– Dylan
Jan 15 at 18:58
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%2f3074729%2fis-there-a-monotonicity-property-for-analytic-continuation%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
2
$begingroup$
As stated, what you say about convergence is false, never mind analytic continuation. (Consider $b_n=-n!$, $a_n=1/n!$.) Surely you meant $a_nge b_nge 0$, or maybe just $|a_n|ge |b_n|$. With that change it seems like a reasonable question - my guess is the answer is "of course not", but I don't have a counterexample handy.
$endgroup$
– David C. Ullrich
Jan 15 at 18:15
$begingroup$
Better example: $b_n=-2^n$, $a_n=1$, so both series have positive radius of convergence.
$endgroup$
– David C. Ullrich
Jan 15 at 18:40
$begingroup$
Good catch, I did indeed mean to have absolute values; edited that. My guess is also "of course not" but I am having trouble thinking of an example.
$endgroup$
– Dylan
Jan 15 at 18:57