Understanding Proposition $3.14$ in Ullrich's Complex Made Simple
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In Ullrich's Complex Made Simple, Proposition $3.14$ part $ii$ states
If $f in H(D'(z,r))$ (holomorphic in the punctured disk) has an essential singularity at $z$ if and only if $f(D'(z,rho))$ is dense in $mathbb C$ for all $rho in (0,r)$.
I understand the if part which refers as Casorati–Weierstrass theorem. But in the proof to only if part I am having trouble to fill a gap. The proof proceeds by contradiction.
If $f$ has a pole or removable singularity at $z$ then $f(w)$ has (finite or infinite) limit as $ w to z$. [This part is OK. The limit would be $0$ if $z$ is removable singularity and infinite if $z$ is a pole.] Then it claims the range $f(D'(z, rho))$ would be obviously not dense. I am having trouble to understand the 'obvious part'.
complex-analysis singularity
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
$endgroup$
add a comment |
$begingroup$
In Ullrich's Complex Made Simple, Proposition $3.14$ part $ii$ states
If $f in H(D'(z,r))$ (holomorphic in the punctured disk) has an essential singularity at $z$ if and only if $f(D'(z,rho))$ is dense in $mathbb C$ for all $rho in (0,r)$.
I understand the if part which refers as Casorati–Weierstrass theorem. But in the proof to only if part I am having trouble to fill a gap. The proof proceeds by contradiction.
If $f$ has a pole or removable singularity at $z$ then $f(w)$ has (finite or infinite) limit as $ w to z$. [This part is OK. The limit would be $0$ if $z$ is removable singularity and infinite if $z$ is a pole.] Then it claims the range $f(D'(z, rho))$ would be obviously not dense. I am having trouble to understand the 'obvious part'.
complex-analysis singularity
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
$endgroup$
add a comment |
$begingroup$
In Ullrich's Complex Made Simple, Proposition $3.14$ part $ii$ states
If $f in H(D'(z,r))$ (holomorphic in the punctured disk) has an essential singularity at $z$ if and only if $f(D'(z,rho))$ is dense in $mathbb C$ for all $rho in (0,r)$.
I understand the if part which refers as Casorati–Weierstrass theorem. But in the proof to only if part I am having trouble to fill a gap. The proof proceeds by contradiction.
If $f$ has a pole or removable singularity at $z$ then $f(w)$ has (finite or infinite) limit as $ w to z$. [This part is OK. The limit would be $0$ if $z$ is removable singularity and infinite if $z$ is a pole.] Then it claims the range $f(D'(z, rho))$ would be obviously not dense. I am having trouble to understand the 'obvious part'.
complex-analysis singularity
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
$endgroup$
In Ullrich's Complex Made Simple, Proposition $3.14$ part $ii$ states
If $f in H(D'(z,r))$ (holomorphic in the punctured disk) has an essential singularity at $z$ if and only if $f(D'(z,rho))$ is dense in $mathbb C$ for all $rho in (0,r)$.
I understand the if part which refers as Casorati–Weierstrass theorem. But in the proof to only if part I am having trouble to fill a gap. The proof proceeds by contradiction.
If $f$ has a pole or removable singularity at $z$ then $f(w)$ has (finite or infinite) limit as $ w to z$. [This part is OK. The limit would be $0$ if $z$ is removable singularity and infinite if $z$ is a pole.] Then it claims the range $f(D'(z, rho))$ would be obviously not dense. I am having trouble to understand the 'obvious part'.
complex-analysis singularity
complex-analysis singularity
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
edited Dec 18 '18 at 19:55
José Carlos Santos
171k23132240
171k23132240
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
asked Dec 18 '18 at 19:33
user1101010user1101010
9011830
9011830
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Suppose that $lim_{wto z}f(w)=l(inmathbb{C})$. Take $varepsilon>0$. There is the a $delta>0$ such that $lvert w-zrvert<deltawedge wneq zimpliesbigllvert f(w)-lbigrrvert<1$. That is,$$fbigl(D'(z,delta)bigr)subset D(l,1).$$But then$$overline{fbigl(D'(z,delta)bigr)}subsetoverline{D(l,1)}varsubsetneqmathbb C.$$Can we deal with case in which $f$ has a pole on $z$ now?
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
If the limit is $L$ (finite), then $f(w)$ is within distance $1$ of $L$ if $w$ is sufficiently close to $z$. The disk of radius $1$ around $L$ is not dense in $mathbb C$.
Similarly, if the limit is $infty$, then $|f(w)| > 1$ if $w$ is sufficiently close to $z$.
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
Suppose that $lim_{wto z}f(w)=l(inmathbb{C})$. Take $varepsilon>0$. There is the a $delta>0$ such that $lvert w-zrvert<deltawedge wneq zimpliesbigllvert f(w)-lbigrrvert<1$. That is,$$fbigl(D'(z,delta)bigr)subset D(l,1).$$But then$$overline{fbigl(D'(z,delta)bigr)}subsetoverline{D(l,1)}varsubsetneqmathbb C.$$Can we deal with case in which $f$ has a pole on $z$ now?
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Suppose that $lim_{wto z}f(w)=l(inmathbb{C})$. Take $varepsilon>0$. There is the a $delta>0$ such that $lvert w-zrvert<deltawedge wneq zimpliesbigllvert f(w)-lbigrrvert<1$. That is,$$fbigl(D'(z,delta)bigr)subset D(l,1).$$But then$$overline{fbigl(D'(z,delta)bigr)}subsetoverline{D(l,1)}varsubsetneqmathbb C.$$Can we deal with case in which $f$ has a pole on $z$ now?
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Suppose that $lim_{wto z}f(w)=l(inmathbb{C})$. Take $varepsilon>0$. There is the a $delta>0$ such that $lvert w-zrvert<deltawedge wneq zimpliesbigllvert f(w)-lbigrrvert<1$. That is,$$fbigl(D'(z,delta)bigr)subset D(l,1).$$But then$$overline{fbigl(D'(z,delta)bigr)}subsetoverline{D(l,1)}varsubsetneqmathbb C.$$Can we deal with case in which $f$ has a pole on $z$ now?
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Suppose that $lim_{wto z}f(w)=l(inmathbb{C})$. Take $varepsilon>0$. There is the a $delta>0$ such that $lvert w-zrvert<deltawedge wneq zimpliesbigllvert f(w)-lbigrrvert<1$. That is,$$fbigl(D'(z,delta)bigr)subset D(l,1).$$But then$$overline{fbigl(D'(z,delta)bigr)}subsetoverline{D(l,1)}varsubsetneqmathbb C.$$Can we deal with case in which $f$ has a pole on $z$ now?
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
answered Dec 18 '18 at 19:50
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
If the limit is $L$ (finite), then $f(w)$ is within distance $1$ of $L$ if $w$ is sufficiently close to $z$. The disk of radius $1$ around $L$ is not dense in $mathbb C$.
Similarly, if the limit is $infty$, then $|f(w)| > 1$ if $w$ is sufficiently close to $z$.
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
If the limit is $L$ (finite), then $f(w)$ is within distance $1$ of $L$ if $w$ is sufficiently close to $z$. The disk of radius $1$ around $L$ is not dense in $mathbb C$.
Similarly, if the limit is $infty$, then $|f(w)| > 1$ if $w$ is sufficiently close to $z$.
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
$endgroup$
add a comment |
$begingroup$
If the limit is $L$ (finite), then $f(w)$ is within distance $1$ of $L$ if $w$ is sufficiently close to $z$. The disk of radius $1$ around $L$ is not dense in $mathbb C$.
Similarly, if the limit is $infty$, then $|f(w)| > 1$ if $w$ is sufficiently close to $z$.
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
$endgroup$
If the limit is $L$ (finite), then $f(w)$ is within distance $1$ of $L$ if $w$ is sufficiently close to $z$. The disk of radius $1$ around $L$ is not dense in $mathbb C$.
Similarly, if the limit is $infty$, then $|f(w)| > 1$ if $w$ is sufficiently close to $z$.
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
answered Dec 18 '18 at 19:49
Robert IsraelRobert Israel
330k23218473
330k23218473
add a comment |
add a comment |
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