Proving that all entire & injective functions take the form $f = ax + b$?
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I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?
Also, more specific questions about the different cases:
Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?
Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?
complex-analysis
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
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add a comment |
$begingroup$
I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?
Also, more specific questions about the different cases:
Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?
Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?
complex-analysis
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
$endgroup$
1
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Can you give the reference to the picture you took? like the source of the problem?
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– BAYMAX
Jul 12 '18 at 11:39
add a comment |
$begingroup$
I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?
Also, more specific questions about the different cases:
Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?
Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?
complex-analysis
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
$endgroup$
I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?
Also, more specific questions about the different cases:
Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?
Essential singularity case: Why exactly is $f({|z| > r } cap f({|z|<r}) neq emptyset$? $f({|z| > r })$ is dense, but how does $f({|z|<r}$ being open guarantee that their union is non-empty?
complex-analysis
complex-analysis
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
asked Oct 9 '14 at 0:21
Ryan YuRyan Yu
7681018
7681018
1
$begingroup$
Can you give the reference to the picture you took? like the source of the problem?
$endgroup$
– BAYMAX
Jul 12 '18 at 11:39
add a comment |
1
$begingroup$
Can you give the reference to the picture you took? like the source of the problem?
$endgroup$
– BAYMAX
Jul 12 '18 at 11:39
1
1
$begingroup$
Can you give the reference to the picture you took? like the source of the problem?
$endgroup$
– BAYMAX
Jul 12 '18 at 11:39
$begingroup$
Can you give the reference to the picture you took? like the source of the problem?
$endgroup$
– BAYMAX
Jul 12 '18 at 11:39
add a comment |
1 Answer
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The first part of what you say its right, thats what the author is trying to do.
For the other 2 specific questions.
1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.
2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.
answered Oct 9 '14 at 0:42
AramAram
1,4451022
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Sorry, why is the closed circle a compact set?
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– Ryan Yu
Oct 9 '14 at 0:52
1
$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
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– JHance
Oct 9 '14 at 1:19
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The first part of what you say its right, thats what the author is trying to do.
For the other 2 specific questions.
1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.
2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.
answered Oct 9 '14 at 0:42
AramAram
1,4451022
$endgroup$
$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52
1
$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19
add a comment |
$begingroup$
The first part of what you say its right, thats what the author is trying to do.
For the other 2 specific questions.
1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.
2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.
answered Oct 9 '14 at 0:42
AramAram
1,4451022
$endgroup$
$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52
1
$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19
add a comment |
$begingroup$
The first part of what you say its right, thats what the author is trying to do.
For the other 2 specific questions.
1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.
2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.
answered Oct 9 '14 at 0:42
AramAram
1,4451022
$endgroup$
The first part of what you say its right, thats what the author is trying to do.
For the other 2 specific questions.
1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.
2.Because the set $f({|z|<r})$ is open it means that there is some ball inside $f({|z|<r})$ that only take points from $f({|z|<r})$, now because Cassorati-Weierstrass assures you that $f({|z| > r }$ is dense in all $mathbb{C}$, it must have some point inside that ball, thus the intersection, $f({|z| > r } cap f({|z|<r})$ is non-empty.
answered Oct 9 '14 at 0:42
AramAram
1,4451022
answered Oct 9 '14 at 0:42
AramAram
1,4451022
answered Oct 9 '14 at 0:42
AramAram
1,4451022
answered Oct 9 '14 at 0:42
AramAram
1,4451022
1,4451022
$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52
1
$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19
add a comment |
$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52
1
$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19
$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52
$begingroup$
Sorry, why is the closed circle a compact set?
$endgroup$
– Ryan Yu
Oct 9 '14 at 0:52
1
1
$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19
$begingroup$
The Heine-Borel theorem tells us that in $mathbb{R}^n$, and thus in $mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
$endgroup$
– JHance
Oct 9 '14 at 1:19
add a comment |
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$begingroup$
Can you give the reference to the picture you took? like the source of the problem?
$endgroup$
– BAYMAX
Jul 12 '18 at 11:39