Given $f$ is complete and $lim_{z rightarrow infty} frac{f(z)}{z} = 0$, prove that $f$ is constant [closed]
0
Does anyone know how to solve the following question without zeros poles and Macloren series?
Given $f$ is complete and $lim_{z rightarrow infty} frac{f(z)}{z} = 0$, > prove that $f$ is constant
Thanks!
complex-analysis complex-integration
asked Dec 28 '18 at 7:36
user135172
43429
closed as off-topic by RRL, Holo, KReiser, metamorphy, José Carlos Santos Dec 29 '18 at 9:11
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- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
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0
Does anyone know how to solve the following question without zeros poles and Macloren series?
Given $f$ is complete and $lim_{z rightarrow infty} frac{f(z)}{z} = 0$, > prove that $f$ is constant
Thanks!
complex-analysis complex-integration
asked Dec 28 '18 at 7:36
user135172
43429
closed as off-topic by RRL, Holo, KReiser, metamorphy, José Carlos Santos Dec 29 '18 at 9:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
0
0
Does anyone know how to solve the following question without zeros poles and Macloren series?
Given $f$ is complete and $lim_{z rightarrow infty} frac{f(z)}{z} = 0$, > prove that $f$ is constant
Thanks!
complex-analysis complex-integration
asked Dec 28 '18 at 7:36
user135172
43429
Does anyone know how to solve the following question without zeros poles and Macloren series?
Given $f$ is complete and $lim_{z rightarrow infty} frac{f(z)}{z} = 0$, > prove that $f$ is constant
Thanks!
complex-analysis complex-integration
complex-analysis complex-integration
asked Dec 28 '18 at 7:36
user135172
43429
asked Dec 28 '18 at 7:36
user135172
43429
asked Dec 28 '18 at 7:36
user135172
43429
asked Dec 28 '18 at 7:36
user135172
43429
asked Dec 28 '18 at 7:36
user135172
43429
43429
closed as off-topic by RRL, Holo, KReiser, metamorphy, José Carlos Santos Dec 29 '18 at 9:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Holo, KReiser, metamorphy, José Carlos Santos Dec 29 '18 at 9:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, KReiser, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
4
Hints: let $g(z)=frac {f(z)-f(0)} z$ for $z neq 0$ and $g(0)=f'(0)$. Show that $g$ is an entire function and that it is bounded. Apply Louiville's Theorem. You will get $f(z)=az+b$ but the hypothesis can hold for this function only when $a=0$.
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
Proof with negtivity? Suppose $f'(z)$ is not bounded then $frac {f(z) - f(0)}{z}$ is not bouned? how do I justify that?
– user135172
Dec 28 '18 at 7:52
$g(z) to 0$ as $|z| to infty$. Since it is a continuous function it follows that it is bounded.
– Kavi Rama Murthy
Dec 28 '18 at 7:54
BTW you may be translating from some other language. 'Complete function' is not a term used in English. 'Entire function' is the correct name.
– Kavi Rama Murthy
Dec 28 '18 at 7:57
Is there a theorem that states the following? g(z)→0 as |z|→∞. Since it is a continuous function it follows that it is bounded. It is indeed very intuitive but How to prove? Suppose not true then for every $M in mathbb R$ there is $z in mathbb C$ such that $|z| > M $ and then what? In $R$ I would prove it by weierstrass on a closed segment and small values on th complement which is a neighborhood of $infty$ but hoe to prove it in $C$? Thanks
– user135172
Dec 28 '18 at 8:04
1
Any continuous function is bounded on compact sets. We can choose $R$ such that $|g(z)| <1$ for $|z| >R$. Since $g$ is bounded for $|z| leq R$ we see that $g$ is bounded on the whole complex plane.
– Kavi Rama Murthy
Dec 28 '18 at 8:06
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
Hints: let $g(z)=frac {f(z)-f(0)} z$ for $z neq 0$ and $g(0)=f'(0)$. Show that $g$ is an entire function and that it is bounded. Apply Louiville's Theorem. You will get $f(z)=az+b$ but the hypothesis can hold for this function only when $a=0$.
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
Proof with negtivity? Suppose $f'(z)$ is not bounded then $frac {f(z) - f(0)}{z}$ is not bouned? how do I justify that?
– user135172
Dec 28 '18 at 7:52
$g(z) to 0$ as $|z| to infty$. Since it is a continuous function it follows that it is bounded.
– Kavi Rama Murthy
Dec 28 '18 at 7:54
BTW you may be translating from some other language. 'Complete function' is not a term used in English. 'Entire function' is the correct name.
– Kavi Rama Murthy
Dec 28 '18 at 7:57
Is there a theorem that states the following? g(z)→0 as |z|→∞. Since it is a continuous function it follows that it is bounded. It is indeed very intuitive but How to prove? Suppose not true then for every $M in mathbb R$ there is $z in mathbb C$ such that $|z| > M $ and then what? In $R$ I would prove it by weierstrass on a closed segment and small values on th complement which is a neighborhood of $infty$ but hoe to prove it in $C$? Thanks
– user135172
Dec 28 '18 at 8:04
1
Any continuous function is bounded on compact sets. We can choose $R$ such that $|g(z)| <1$ for $|z| >R$. Since $g$ is bounded for $|z| leq R$ we see that $g$ is bounded on the whole complex plane.
– Kavi Rama Murthy
Dec 28 '18 at 8:06
|
show 1 more comment
4
Hints: let $g(z)=frac {f(z)-f(0)} z$ for $z neq 0$ and $g(0)=f'(0)$. Show that $g$ is an entire function and that it is bounded. Apply Louiville's Theorem. You will get $f(z)=az+b$ but the hypothesis can hold for this function only when $a=0$.
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
Proof with negtivity? Suppose $f'(z)$ is not bounded then $frac {f(z) - f(0)}{z}$ is not bouned? how do I justify that?
– user135172
Dec 28 '18 at 7:52
$g(z) to 0$ as $|z| to infty$. Since it is a continuous function it follows that it is bounded.
– Kavi Rama Murthy
Dec 28 '18 at 7:54
BTW you may be translating from some other language. 'Complete function' is not a term used in English. 'Entire function' is the correct name.
– Kavi Rama Murthy
Dec 28 '18 at 7:57
Is there a theorem that states the following? g(z)→0 as |z|→∞. Since it is a continuous function it follows that it is bounded. It is indeed very intuitive but How to prove? Suppose not true then for every $M in mathbb R$ there is $z in mathbb C$ such that $|z| > M $ and then what? In $R$ I would prove it by weierstrass on a closed segment and small values on th complement which is a neighborhood of $infty$ but hoe to prove it in $C$? Thanks
– user135172
Dec 28 '18 at 8:04
1
Any continuous function is bounded on compact sets. We can choose $R$ such that $|g(z)| <1$ for $|z| >R$. Since $g$ is bounded for $|z| leq R$ we see that $g$ is bounded on the whole complex plane.
– Kavi Rama Murthy
Dec 28 '18 at 8:06
|
show 1 more comment
4
4
4
Hints: let $g(z)=frac {f(z)-f(0)} z$ for $z neq 0$ and $g(0)=f'(0)$. Show that $g$ is an entire function and that it is bounded. Apply Louiville's Theorem. You will get $f(z)=az+b$ but the hypothesis can hold for this function only when $a=0$.
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
Hints: let $g(z)=frac {f(z)-f(0)} z$ for $z neq 0$ and $g(0)=f'(0)$. Show that $g$ is an entire function and that it is bounded. Apply Louiville's Theorem. You will get $f(z)=az+b$ but the hypothesis can hold for this function only when $a=0$.
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
answered Dec 28 '18 at 7:41
Kavi Rama Murthy
51.2k31855
51.2k31855
Proof with negtivity? Suppose $f'(z)$ is not bounded then $frac {f(z) - f(0)}{z}$ is not bouned? how do I justify that?
– user135172
Dec 28 '18 at 7:52
$g(z) to 0$ as $|z| to infty$. Since it is a continuous function it follows that it is bounded.
– Kavi Rama Murthy
Dec 28 '18 at 7:54
BTW you may be translating from some other language. 'Complete function' is not a term used in English. 'Entire function' is the correct name.
– Kavi Rama Murthy
Dec 28 '18 at 7:57
Is there a theorem that states the following? g(z)→0 as |z|→∞. Since it is a continuous function it follows that it is bounded. It is indeed very intuitive but How to prove? Suppose not true then for every $M in mathbb R$ there is $z in mathbb C$ such that $|z| > M $ and then what? In $R$ I would prove it by weierstrass on a closed segment and small values on th complement which is a neighborhood of $infty$ but hoe to prove it in $C$? Thanks
– user135172
Dec 28 '18 at 8:04
1
Any continuous function is bounded on compact sets. We can choose $R$ such that $|g(z)| <1$ for $|z| >R$. Since $g$ is bounded for $|z| leq R$ we see that $g$ is bounded on the whole complex plane.
– Kavi Rama Murthy
Dec 28 '18 at 8:06
|
show 1 more comment
Proof with negtivity? Suppose $f'(z)$ is not bounded then $frac {f(z) - f(0)}{z}$ is not bouned? how do I justify that?
– user135172
Dec 28 '18 at 7:52
$g(z) to 0$ as $|z| to infty$. Since it is a continuous function it follows that it is bounded.
– Kavi Rama Murthy
Dec 28 '18 at 7:54
BTW you may be translating from some other language. 'Complete function' is not a term used in English. 'Entire function' is the correct name.
– Kavi Rama Murthy
Dec 28 '18 at 7:57
Is there a theorem that states the following? g(z)→0 as |z|→∞. Since it is a continuous function it follows that it is bounded. It is indeed very intuitive but How to prove? Suppose not true then for every $M in mathbb R$ there is $z in mathbb C$ such that $|z| > M $ and then what? In $R$ I would prove it by weierstrass on a closed segment and small values on th complement which is a neighborhood of $infty$ but hoe to prove it in $C$? Thanks
– user135172
Dec 28 '18 at 8:04
1
Any continuous function is bounded on compact sets. We can choose $R$ such that $|g(z)| <1$ for $|z| >R$. Since $g$ is bounded for $|z| leq R$ we see that $g$ is bounded on the whole complex plane.
– Kavi Rama Murthy
Dec 28 '18 at 8:06
Proof with negtivity? Suppose $f'(z)$ is not bounded then $frac {f(z) - f(0)}{z}$ is not bouned? how do I justify that?
– user135172
Dec 28 '18 at 7:52
Proof with negtivity? Suppose $f'(z)$ is not bounded then $frac {f(z) - f(0)}{z}$ is not bouned? how do I justify that?
– user135172
Dec 28 '18 at 7:52
$g(z) to 0$ as $|z| to infty$. Since it is a continuous function it follows that it is bounded.
– Kavi Rama Murthy
Dec 28 '18 at 7:54
$g(z) to 0$ as $|z| to infty$. Since it is a continuous function it follows that it is bounded.
– Kavi Rama Murthy
Dec 28 '18 at 7:54
BTW you may be translating from some other language. 'Complete function' is not a term used in English. 'Entire function' is the correct name.
– Kavi Rama Murthy
Dec 28 '18 at 7:57
BTW you may be translating from some other language. 'Complete function' is not a term used in English. 'Entire function' is the correct name.
– Kavi Rama Murthy
Dec 28 '18 at 7:57
Is there a theorem that states the following? g(z)→0 as |z|→∞. Since it is a continuous function it follows that it is bounded. It is indeed very intuitive but How to prove? Suppose not true then for every $M in mathbb R$ there is $z in mathbb C$ such that $|z| > M $ and then what? In $R$ I would prove it by weierstrass on a closed segment and small values on th complement which is a neighborhood of $infty$ but hoe to prove it in $C$? Thanks
– user135172
Dec 28 '18 at 8:04
Is there a theorem that states the following? g(z)→0 as |z|→∞. Since it is a continuous function it follows that it is bounded. It is indeed very intuitive but How to prove? Suppose not true then for every $M in mathbb R$ there is $z in mathbb C$ such that $|z| > M $ and then what? In $R$ I would prove it by weierstrass on a closed segment and small values on th complement which is a neighborhood of $infty$ but hoe to prove it in $C$? Thanks
– user135172
Dec 28 '18 at 8:04
1
1
Any continuous function is bounded on compact sets. We can choose $R$ such that $|g(z)| <1$ for $|z| >R$. Since $g$ is bounded for $|z| leq R$ we see that $g$ is bounded on the whole complex plane.
– Kavi Rama Murthy
Dec 28 '18 at 8:06
Any continuous function is bounded on compact sets. We can choose $R$ such that $|g(z)| <1$ for $|z| >R$. Since $g$ is bounded for $|z| leq R$ we see that $g$ is bounded on the whole complex plane.
– Kavi Rama Murthy
Dec 28 '18 at 8:06
|
show 1 more comment
