[Reference Request]: Elementary proof of: Analytic function on $mathbb{C}$ implies infinite radius of...
$begingroup$
In trying to find a self-contained argument for the non-emptiness of the spectrum in a Banach algebra, I found this question, where the accepted answer referenced this other question (see in particular both answers by
Alexandre Eremenko).
The idea, if I'm not mistaken, is that from the spectrum being empty, one deduces that the resolvent function, which admits a development as a power series in a neighbourhood of each point, must be holomorphic everywhere, and from this (using what I think may be called “Cauchy's Theorem”), the fact that the development around zero has an infinite radius of convergence.
Applying Liouville's theorem then shows that the resolvent function must be constant (since bounded), which is a contradiction.
The author of the accepted answer (for the first question) alludes to an elementary argument for my question:
The weak form of Cauchy's theorem that I mentioned uses only elementary manipulation with absolutely convergent series, no integral is involved
Question: What is this elementary argument being referred to?
And as a remark:
If the goal is to prove non-emptiness of the spectrum without using any outside “named” theorem of complex analysis, the missing piece is necessary if I'm not mistaken: The bare hand proof of Liouville given in the second link assumes infinite radius of convergence for the expansion at zero, but to show the latter (infinite radius) usually requires Cauchy's integral formula, and breaks the elementary part of the argument.
Comments/corrections welcome!
complex-analysis
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
$endgroup$
add a comment |
$begingroup$
In trying to find a self-contained argument for the non-emptiness of the spectrum in a Banach algebra, I found this question, where the accepted answer referenced this other question (see in particular both answers by
Alexandre Eremenko).
The idea, if I'm not mistaken, is that from the spectrum being empty, one deduces that the resolvent function, which admits a development as a power series in a neighbourhood of each point, must be holomorphic everywhere, and from this (using what I think may be called “Cauchy's Theorem”), the fact that the development around zero has an infinite radius of convergence.
Applying Liouville's theorem then shows that the resolvent function must be constant (since bounded), which is a contradiction.
The author of the accepted answer (for the first question) alludes to an elementary argument for my question:
The weak form of Cauchy's theorem that I mentioned uses only elementary manipulation with absolutely convergent series, no integral is involved
Question: What is this elementary argument being referred to?
And as a remark:
If the goal is to prove non-emptiness of the spectrum without using any outside “named” theorem of complex analysis, the missing piece is necessary if I'm not mistaken: The bare hand proof of Liouville given in the second link assumes infinite radius of convergence for the expansion at zero, but to show the latter (infinite radius) usually requires Cauchy's integral formula, and breaks the elementary part of the argument.
Comments/corrections welcome!
complex-analysis
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
$endgroup$
$begingroup$
Cauchy integral formula for analytic functions is really not hard : start from $int_gamma frac{f(z)}{z-a}dz$, massage the closed contour $gamma$ to obtain $sum_j int_{gamma_j} frac{f(z)}{z-a}dz$ where the $gamma_j$ are closed contours contained in regions where $frac{f(z)}{z-a}$ is a power series (Laurent series for the $gamma_j$ containing $a$), invert $sum$ and $int$, integrate each term.
$endgroup$
– reuns
Dec 6 '18 at 13:28
$begingroup$
I understand Cauchy is not so hard, but the point of my question was mainly to fill in the hole in the argument exposed in the links. Because with the usual Cauchy argument, it seems that the simplicity of the argument given sort of collapses because you end up still using named theorems. Well, maybe I'm misguided!
$endgroup$
– ouimerci
Dec 7 '18 at 7:34
$begingroup$
So if I understand right, you are asking what an elementary proof looks like for the following: if $f$ is analytic in a disk of radius $r$ centered at $z_0,$ then the power series at $z_0$ converges in the disk? I recall seeing such a proof in the past, but can't recall where. I'll see if I can find it sometime this week.
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 13:05
$begingroup$
Right, maybe I should actually have written that more explicitely in the question.
$endgroup$
– ouimerci
Dec 10 '18 at 13:47
add a comment |
$begingroup$
In trying to find a self-contained argument for the non-emptiness of the spectrum in a Banach algebra, I found this question, where the accepted answer referenced this other question (see in particular both answers by
Alexandre Eremenko).
The idea, if I'm not mistaken, is that from the spectrum being empty, one deduces that the resolvent function, which admits a development as a power series in a neighbourhood of each point, must be holomorphic everywhere, and from this (using what I think may be called “Cauchy's Theorem”), the fact that the development around zero has an infinite radius of convergence.
Applying Liouville's theorem then shows that the resolvent function must be constant (since bounded), which is a contradiction.
The author of the accepted answer (for the first question) alludes to an elementary argument for my question:
The weak form of Cauchy's theorem that I mentioned uses only elementary manipulation with absolutely convergent series, no integral is involved
Question: What is this elementary argument being referred to?
And as a remark:
If the goal is to prove non-emptiness of the spectrum without using any outside “named” theorem of complex analysis, the missing piece is necessary if I'm not mistaken: The bare hand proof of Liouville given in the second link assumes infinite radius of convergence for the expansion at zero, but to show the latter (infinite radius) usually requires Cauchy's integral formula, and breaks the elementary part of the argument.
Comments/corrections welcome!
complex-analysis
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
$endgroup$
In trying to find a self-contained argument for the non-emptiness of the spectrum in a Banach algebra, I found this question, where the accepted answer referenced this other question (see in particular both answers by
Alexandre Eremenko).
The idea, if I'm not mistaken, is that from the spectrum being empty, one deduces that the resolvent function, which admits a development as a power series in a neighbourhood of each point, must be holomorphic everywhere, and from this (using what I think may be called “Cauchy's Theorem”), the fact that the development around zero has an infinite radius of convergence.
Applying Liouville's theorem then shows that the resolvent function must be constant (since bounded), which is a contradiction.
The author of the accepted answer (for the first question) alludes to an elementary argument for my question:
The weak form of Cauchy's theorem that I mentioned uses only elementary manipulation with absolutely convergent series, no integral is involved
Question: What is this elementary argument being referred to?
And as a remark:
If the goal is to prove non-emptiness of the spectrum without using any outside “named” theorem of complex analysis, the missing piece is necessary if I'm not mistaken: The bare hand proof of Liouville given in the second link assumes infinite radius of convergence for the expansion at zero, but to show the latter (infinite radius) usually requires Cauchy's integral formula, and breaks the elementary part of the argument.
Comments/corrections welcome!
complex-analysis
complex-analysis
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
edited Dec 6 '18 at 7:22
ouimerci
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
asked Dec 6 '18 at 6:24
ouimerciouimerci
62
62
$begingroup$
Cauchy integral formula for analytic functions is really not hard : start from $int_gamma frac{f(z)}{z-a}dz$, massage the closed contour $gamma$ to obtain $sum_j int_{gamma_j} frac{f(z)}{z-a}dz$ where the $gamma_j$ are closed contours contained in regions where $frac{f(z)}{z-a}$ is a power series (Laurent series for the $gamma_j$ containing $a$), invert $sum$ and $int$, integrate each term.
$endgroup$
– reuns
Dec 6 '18 at 13:28
$begingroup$
I understand Cauchy is not so hard, but the point of my question was mainly to fill in the hole in the argument exposed in the links. Because with the usual Cauchy argument, it seems that the simplicity of the argument given sort of collapses because you end up still using named theorems. Well, maybe I'm misguided!
$endgroup$
– ouimerci
Dec 7 '18 at 7:34
$begingroup$
So if I understand right, you are asking what an elementary proof looks like for the following: if $f$ is analytic in a disk of radius $r$ centered at $z_0,$ then the power series at $z_0$ converges in the disk? I recall seeing such a proof in the past, but can't recall where. I'll see if I can find it sometime this week.
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 13:05
$begingroup$
Right, maybe I should actually have written that more explicitely in the question.
$endgroup$
– ouimerci
Dec 10 '18 at 13:47
add a comment |
$begingroup$
Cauchy integral formula for analytic functions is really not hard : start from $int_gamma frac{f(z)}{z-a}dz$, massage the closed contour $gamma$ to obtain $sum_j int_{gamma_j} frac{f(z)}{z-a}dz$ where the $gamma_j$ are closed contours contained in regions where $frac{f(z)}{z-a}$ is a power series (Laurent series for the $gamma_j$ containing $a$), invert $sum$ and $int$, integrate each term.
$endgroup$
– reuns
Dec 6 '18 at 13:28
$begingroup$
I understand Cauchy is not so hard, but the point of my question was mainly to fill in the hole in the argument exposed in the links. Because with the usual Cauchy argument, it seems that the simplicity of the argument given sort of collapses because you end up still using named theorems. Well, maybe I'm misguided!
$endgroup$
– ouimerci
Dec 7 '18 at 7:34
$begingroup$
So if I understand right, you are asking what an elementary proof looks like for the following: if $f$ is analytic in a disk of radius $r$ centered at $z_0,$ then the power series at $z_0$ converges in the disk? I recall seeing such a proof in the past, but can't recall where. I'll see if I can find it sometime this week.
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 13:05
$begingroup$
Right, maybe I should actually have written that more explicitely in the question.
$endgroup$
– ouimerci
Dec 10 '18 at 13:47
$begingroup$
Cauchy integral formula for analytic functions is really not hard : start from $int_gamma frac{f(z)}{z-a}dz$, massage the closed contour $gamma$ to obtain $sum_j int_{gamma_j} frac{f(z)}{z-a}dz$ where the $gamma_j$ are closed contours contained in regions where $frac{f(z)}{z-a}$ is a power series (Laurent series for the $gamma_j$ containing $a$), invert $sum$ and $int$, integrate each term.
$endgroup$
– reuns
Dec 6 '18 at 13:28
$begingroup$
Cauchy integral formula for analytic functions is really not hard : start from $int_gamma frac{f(z)}{z-a}dz$, massage the closed contour $gamma$ to obtain $sum_j int_{gamma_j} frac{f(z)}{z-a}dz$ where the $gamma_j$ are closed contours contained in regions where $frac{f(z)}{z-a}$ is a power series (Laurent series for the $gamma_j$ containing $a$), invert $sum$ and $int$, integrate each term.
$endgroup$
– reuns
Dec 6 '18 at 13:28
$begingroup$
I understand Cauchy is not so hard, but the point of my question was mainly to fill in the hole in the argument exposed in the links. Because with the usual Cauchy argument, it seems that the simplicity of the argument given sort of collapses because you end up still using named theorems. Well, maybe I'm misguided!
$endgroup$
– ouimerci
Dec 7 '18 at 7:34
$begingroup$
I understand Cauchy is not so hard, but the point of my question was mainly to fill in the hole in the argument exposed in the links. Because with the usual Cauchy argument, it seems that the simplicity of the argument given sort of collapses because you end up still using named theorems. Well, maybe I'm misguided!
$endgroup$
– ouimerci
Dec 7 '18 at 7:34
$begingroup$
So if I understand right, you are asking what an elementary proof looks like for the following: if $f$ is analytic in a disk of radius $r$ centered at $z_0,$ then the power series at $z_0$ converges in the disk? I recall seeing such a proof in the past, but can't recall where. I'll see if I can find it sometime this week.
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 13:05
$begingroup$
So if I understand right, you are asking what an elementary proof looks like for the following: if $f$ is analytic in a disk of radius $r$ centered at $z_0,$ then the power series at $z_0$ converges in the disk? I recall seeing such a proof in the past, but can't recall where. I'll see if I can find it sometime this week.
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 13:05
$begingroup$
Right, maybe I should actually have written that more explicitely in the question.
$endgroup$
– ouimerci
Dec 10 '18 at 13:47
$begingroup$
Right, maybe I should actually have written that more explicitely in the question.
$endgroup$
– ouimerci
Dec 10 '18 at 13:47
add a comment |
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$begingroup$
Cauchy integral formula for analytic functions is really not hard : start from $int_gamma frac{f(z)}{z-a}dz$, massage the closed contour $gamma$ to obtain $sum_j int_{gamma_j} frac{f(z)}{z-a}dz$ where the $gamma_j$ are closed contours contained in regions where $frac{f(z)}{z-a}$ is a power series (Laurent series for the $gamma_j$ containing $a$), invert $sum$ and $int$, integrate each term.
$endgroup$
– reuns
Dec 6 '18 at 13:28
$begingroup$
I understand Cauchy is not so hard, but the point of my question was mainly to fill in the hole in the argument exposed in the links. Because with the usual Cauchy argument, it seems that the simplicity of the argument given sort of collapses because you end up still using named theorems. Well, maybe I'm misguided!
$endgroup$
– ouimerci
Dec 7 '18 at 7:34
$begingroup$
So if I understand right, you are asking what an elementary proof looks like for the following: if $f$ is analytic in a disk of radius $r$ centered at $z_0,$ then the power series at $z_0$ converges in the disk? I recall seeing such a proof in the past, but can't recall where. I'll see if I can find it sometime this week.
$endgroup$
– Brevan Ellefsen
Dec 10 '18 at 13:05
$begingroup$
Right, maybe I should actually have written that more explicitely in the question.
$endgroup$
– ouimerci
Dec 10 '18 at 13:47