Power series is locally normally convergent in its convergence radius $B(0,R)$
$begingroup$
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.
What I did so far:
- Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
$$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$ - Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.
Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.
complex-analysis power-series
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
asked Dec 30 '18 at 16:38
dandan
623613
$endgroup$
add a comment |
$begingroup$
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.
What I did so far:
- Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
$$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$ - Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.
Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.
complex-analysis power-series
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
asked Dec 30 '18 at 16:38
dandan
623613
$endgroup$
$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36
$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41
$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43
$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37
add a comment |
$begingroup$
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.
What I did so far:
- Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
$$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$ - Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.
Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.
complex-analysis power-series
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
asked Dec 30 '18 at 16:38
dandan
623613
$endgroup$
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, a power series with $R:=sup { tge0 : a_n t^n$ is bounded $}$ as its convergence radius. I wish to prove that $sum_0^infty a_nz^n$ is locally normally convergent over $B(0,R)$.
What I did so far:
- Let $r_1 <r_2<R$ , then for all $zin B(0,r_1)$ :
$$sum_0^infty|a_nz^n| = sum _0^infty|a_n|frac{r_2^n}{r_2^n}|z|^n le sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n .$$ - Now, because $r_2<R$ , $|a_n|r_2^n <M$ , so: $$sum _0^infty|a_n|r_2^n(frac{r_1}{r_2})^n le Msum _0^infty(frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $r_1<r_2$.
So any power series for $zin B(0,r_1)$ is mutually bounded by $Msum_0^infty(frac{r_1}{r_2})^n$.
Yet I don't succeed to make the next step, and show that any $zin B(0,r_1)$ has a neighborhood $U_z$ such that $sum_0^infty sup_{U_z} |a_n z^n|$ converges.
complex-analysis power-series
complex-analysis power-series
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
asked Dec 30 '18 at 16:38
dandan
623613
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
asked Dec 30 '18 at 16:38
dandan
623613
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
edited Dec 31 '18 at 12:35
LutzL
60.7k42057
60.7k42057
asked Dec 30 '18 at 16:38
dandan
623613
asked Dec 30 '18 at 16:38
dandan
623613
asked Dec 30 '18 at 16:38
dandan
623613
623613
$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36
$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41
$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43
$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37
add a comment |
$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36
$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41
$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43
$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37
$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36
$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36
$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41
$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41
$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43
$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43
$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37
$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have done all the necessary work, you just need to put it together:
For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
$U=B(0, r_1)$ is a neighbourhood of $z_0$, and
$ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
$$
sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
$$
for some $M > 0$.
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
$endgroup$
add a comment |
Your Answer
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%2f3056998%2fpower-series-is-locally-normally-convergent-in-its-convergence-radius-b0-r%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
$begingroup$
You have done all the necessary work, you just need to put it together:
For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
$U=B(0, r_1)$ is a neighbourhood of $z_0$, and
$ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
$$
sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
$$
for some $M > 0$.
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
$endgroup$
add a comment |
$begingroup$
You have done all the necessary work, you just need to put it together:
For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
$U=B(0, r_1)$ is a neighbourhood of $z_0$, and
$ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
$$
sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
$$
for some $M > 0$.
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
$endgroup$
add a comment |
$begingroup$
You have done all the necessary work, you just need to put it together:
For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
$U=B(0, r_1)$ is a neighbourhood of $z_0$, and
$ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
$$
sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
$$
for some $M > 0$.
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
$endgroup$
You have done all the necessary work, you just need to put it together:
For a given $z_0 in B(0, R)$ choose $r_1, r_2$ with $|z_0| < r_1 < r_2 < R$. Then
$U=B(0, r_1)$ is a neighbourhood of $z_0$, and
$ sum_{n=0}^infty sup_U |a_n z^n| $ is convergent because
$$
sup_U |a_n z^n| le M left( frac{r_1}{r_2} right)^n
$$
for some $M > 0$.
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
answered Dec 30 '18 at 17:58
Martin RMartin R
31k33561
31k33561
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%2f3056998%2fpower-series-is-locally-normally-convergent-in-its-convergence-radius-b0-r%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
$begingroup$
@MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it...
$endgroup$
– dan
Dec 30 '18 at 17:36
$begingroup$
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
$endgroup$
– Martin R
Dec 30 '18 at 17:41
$begingroup$
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
$endgroup$
– dan
Dec 30 '18 at 17:43
$begingroup$
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
$endgroup$
– LutzL
Dec 31 '18 at 12:37