Power series is locally normally convergent in its convergence radius $B(0,R)$
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, 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
56.3k42054
asked Dec 30 '18 at 16:38
dan
489413
add a comment |
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, 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
56.3k42054
asked Dec 30 '18 at 16:38
dan
489413
@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...
– dan
Dec 30 '18 at 17:36
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
– Martin R
Dec 30 '18 at 17:41
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
– dan
Dec 30 '18 at 17:43
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
– LutzL
Dec 31 '18 at 12:37
add a comment |
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, 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
56.3k42054
asked Dec 30 '18 at 16:38
dan
489413
$sum_0^infty a_nz^n$ , $zin mathbb{C}$, 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
56.3k42054
asked Dec 30 '18 at 16:38
dan
489413
edited Dec 31 '18 at 12:35
LutzL
56.3k42054
asked Dec 30 '18 at 16:38
dan
489413
edited Dec 31 '18 at 12:35
LutzL
56.3k42054
edited Dec 31 '18 at 12:35
LutzL
56.3k42054
edited Dec 31 '18 at 12:35
LutzL
56.3k42054
56.3k42054
asked Dec 30 '18 at 16:38
dan
489413
asked Dec 30 '18 at 16:38
dan
489413
asked Dec 30 '18 at 16:38
dan
489413
489413
@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...
– dan
Dec 30 '18 at 17:36
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
– Martin R
Dec 30 '18 at 17:41
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
– dan
Dec 30 '18 at 17:43
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
– LutzL
Dec 31 '18 at 12:37
add a comment |
@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...
– dan
Dec 30 '18 at 17:36
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
– Martin R
Dec 30 '18 at 17:41
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
– dan
Dec 30 '18 at 17:43
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
– LutzL
Dec 31 '18 at 12:37
@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...
– dan
Dec 30 '18 at 17:36
@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...
– dan
Dec 30 '18 at 17:36
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
– Martin R
Dec 30 '18 at 17:41
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
– Martin R
Dec 30 '18 at 17:41
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
– dan
Dec 30 '18 at 17:43
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
– dan
Dec 30 '18 at 17:43
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
– LutzL
Dec 31 '18 at 12:37
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
– LutzL
Dec 31 '18 at 12:37
add a comment |
1 Answer
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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 R
27.2k33254
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
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 R
27.2k33254
add a comment |
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 R
27.2k33254
add a comment |
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 R
27.2k33254
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 R
27.2k33254
answered Dec 30 '18 at 17:58
Martin R
27.2k33254
answered Dec 30 '18 at 17:58
Martin R
27.2k33254
answered Dec 30 '18 at 17:58
Martin R
27.2k33254
27.2k33254
add a comment |
add a comment |
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@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...
– dan
Dec 30 '18 at 17:36
Actually what I said is nonsense. You need to show that $sum sup_U |a_n z^n| $ converges, not $sum sup_U |f|$.
– Martin R
Dec 30 '18 at 17:41
@MartinR ,Right! I'm sorry I miss typed, just corrected it.
– dan
Dec 30 '18 at 17:43
Or in short, for any $z_0in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment.
– LutzL
Dec 31 '18 at 12:37