Show that if $f(z)$ is entire and $leftvert f(z) rightvert > M$ for all $z in mathbb{C}$, then $f(z)$ is...
$begingroup$
I am trying to prove that if $f(z)$ is entire and $leftvert f(z) rightvert > M$ for all $z in mathbb{C}$, then $f(z)$ is constant. Is my proof correct?
Since $f(z)$ is entire, we know that $frac{1}{f(z)}$ is entire and satisfies $leftvert frac{1}{f(z)} rightvert < 1/M$. So, by Liouville's Theorem, $frac{1}{f(z)}$, and hence $f(z)$, is constant.
complex-analysis proof-verification
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
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add a comment |
$begingroup$
I am trying to prove that if $f(z)$ is entire and $leftvert f(z) rightvert > M$ for all $z in mathbb{C}$, then $f(z)$ is constant. Is my proof correct?
Since $f(z)$ is entire, we know that $frac{1}{f(z)}$ is entire and satisfies $leftvert frac{1}{f(z)} rightvert < 1/M$. So, by Liouville's Theorem, $frac{1}{f(z)}$, and hence $f(z)$, is constant.
complex-analysis proof-verification
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
$endgroup$
$begingroup$
I think nonconstant entire functions can only miss 1 point?
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– mathworker21
Dec 2 '18 at 15:34
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Although that is true @mathworker21 the proof given is much more elementary and only uses Liouville.
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– Paul K
Dec 2 '18 at 15:35
3
$begingroup$
What is your question?
$endgroup$
– Did
Dec 2 '18 at 15:36
add a comment |
$begingroup$
I am trying to prove that if $f(z)$ is entire and $leftvert f(z) rightvert > M$ for all $z in mathbb{C}$, then $f(z)$ is constant. Is my proof correct?
Since $f(z)$ is entire, we know that $frac{1}{f(z)}$ is entire and satisfies $leftvert frac{1}{f(z)} rightvert < 1/M$. So, by Liouville's Theorem, $frac{1}{f(z)}$, and hence $f(z)$, is constant.
complex-analysis proof-verification
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
$endgroup$
I am trying to prove that if $f(z)$ is entire and $leftvert f(z) rightvert > M$ for all $z in mathbb{C}$, then $f(z)$ is constant. Is my proof correct?
Since $f(z)$ is entire, we know that $frac{1}{f(z)}$ is entire and satisfies $leftvert frac{1}{f(z)} rightvert < 1/M$. So, by Liouville's Theorem, $frac{1}{f(z)}$, and hence $f(z)$, is constant.
complex-analysis proof-verification
complex-analysis proof-verification
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
edited Dec 2 '18 at 17:38
Yunus Syed
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
asked Dec 2 '18 at 15:33
Yunus SyedYunus Syed
1,217218
1,217218
$begingroup$
I think nonconstant entire functions can only miss 1 point?
$endgroup$
– mathworker21
Dec 2 '18 at 15:34
$begingroup$
Although that is true @mathworker21 the proof given is much more elementary and only uses Liouville.
$endgroup$
– Paul K
Dec 2 '18 at 15:35
3
$begingroup$
What is your question?
$endgroup$
– Did
Dec 2 '18 at 15:36
add a comment |
$begingroup$
I think nonconstant entire functions can only miss 1 point?
$endgroup$
– mathworker21
Dec 2 '18 at 15:34
$begingroup$
Although that is true @mathworker21 the proof given is much more elementary and only uses Liouville.
$endgroup$
– Paul K
Dec 2 '18 at 15:35
3
$begingroup$
What is your question?
$endgroup$
– Did
Dec 2 '18 at 15:36
$begingroup$
I think nonconstant entire functions can only miss 1 point?
$endgroup$
– mathworker21
Dec 2 '18 at 15:34
$begingroup$
I think nonconstant entire functions can only miss 1 point?
$endgroup$
– mathworker21
Dec 2 '18 at 15:34
$begingroup$
Although that is true @mathworker21 the proof given is much more elementary and only uses Liouville.
$endgroup$
– Paul K
Dec 2 '18 at 15:35
$begingroup$
Although that is true @mathworker21 the proof given is much more elementary and only uses Liouville.
$endgroup$
– Paul K
Dec 2 '18 at 15:35
3
3
$begingroup$
What is your question?
$endgroup$
– Did
Dec 2 '18 at 15:36
$begingroup$
What is your question?
$endgroup$
– Did
Dec 2 '18 at 15:36
add a comment |
1 Answer
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$begingroup$
Is my proof correct?
The correct start of the proof should be:
Since $f$ is entire and nowhere zero, we know that $frac{1}{f}$ is entire ...
Perhaps you meant that without writing it explicitly. You should
also distinguish between $f$ (the function) and $f(z)$ (a complex number, the function evaluated at $z$).
Apart from that, the proof is correct.
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Is my proof correct?
The correct start of the proof should be:
Since $f$ is entire and nowhere zero, we know that $frac{1}{f}$ is entire ...
Perhaps you meant that without writing it explicitly. You should
also distinguish between $f$ (the function) and $f(z)$ (a complex number, the function evaluated at $z$).
Apart from that, the proof is correct.
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
$begingroup$
Is my proof correct?
The correct start of the proof should be:
Since $f$ is entire and nowhere zero, we know that $frac{1}{f}$ is entire ...
Perhaps you meant that without writing it explicitly. You should
also distinguish between $f$ (the function) and $f(z)$ (a complex number, the function evaluated at $z$).
Apart from that, the proof is correct.
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
$endgroup$
add a comment |
$begingroup$
Is my proof correct?
The correct start of the proof should be:
Since $f$ is entire and nowhere zero, we know that $frac{1}{f}$ is entire ...
Perhaps you meant that without writing it explicitly. You should
also distinguish between $f$ (the function) and $f(z)$ (a complex number, the function evaluated at $z$).
Apart from that, the proof is correct.
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
$endgroup$
Is my proof correct?
The correct start of the proof should be:
Since $f$ is entire and nowhere zero, we know that $frac{1}{f}$ is entire ...
Perhaps you meant that without writing it explicitly. You should
also distinguish between $f$ (the function) and $f(z)$ (a complex number, the function evaluated at $z$).
Apart from that, the proof is correct.
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
answered Dec 2 '18 at 17:49
Martin RMartin R
27.9k33255
27.9k33255
add a comment |
add a comment |
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$begingroup$
I think nonconstant entire functions can only miss 1 point?
$endgroup$
– mathworker21
Dec 2 '18 at 15:34
$begingroup$
Although that is true @mathworker21 the proof given is much more elementary and only uses Liouville.
$endgroup$
– Paul K
Dec 2 '18 at 15:35
3
$begingroup$
What is your question?
$endgroup$
– Did
Dec 2 '18 at 15:36