Mixing
A variation of the Casorati-Weierstrass Theorem
Clash Royale CLAN TAG#URR8PPP
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I'm trying to prove the following claim:
Suppose $f$ is analytic in a deleted neighborhood $D$ of $z_0$ except for poles at all points of a sequence
$z_n longrightarrow z_0$. (Note that $z_0$ is not an isolated singularity.) Show that $f(D)$ is dense in the complex
plane.
Here's my attempt.
Let $epsilon > 0$ such that $D = < epsilon $. Assume that $f(D)$ isn't dense in $mathbbC$. Then there exists a $w in mathbbC$ and $delta > 0$ such that,
beginalign
|f(z) - w| > delta Longleftrightarrow 0 < frac1f(z) - w < frac1delta equiv g(z) < frac1delta
endalign
Now, for $epsilon/2 > 0$, we have that there exists a $N in mathbbN$ such that,
beginalign
z_n in B(z_0, epsilon/2), quad n geq N.
endalign
Now let $epsilon' < textmin epsilon/2, $. We then have that $B(z_1, epsilon')$ is contained in $D$ such that $z_1$ is an essential singularity of f(z), since the open ball $B(z_1, epsilon'$) contains no other points from the sequence.
Consider the annular region $D' = z - z_1$. From the first equation, we have that,
beginalign
g(z) < frac1delta, quad z in D'.
endalign
Since $g(z)$ is bounded on $D'$, we have that $g(z)$ has (at most) an isolated singularity in $D'$. Therefore, we have that there is a function, $h(z)$, such that,
beginalign
g(z) = frac1f(z) - w = h(z), quad z in D',
endalign
and that $h(z_1)$ is analytic. Rearranging the previous equation, we have that,
beginalign
f(z) = w + frac1h(z) = frach(z)w + 1h(z) equiv = fracA(z)B(z), quad z in D'.
endalign
Since $f$ has a pole at $z_1$, we have that $h(z_1) = 0$, implying that $|g(z_1)| = |h(z_1)| = 0$, contradicting that $|g(z)$ is never zero on $D' subset D$.
complex-analysis
asked Jul 14 at 18:42
user82261
356
add a comment |Â
up vote
0
down vote
favorite
I'm trying to prove the following claim:
Suppose $f$ is analytic in a deleted neighborhood $D$ of $z_0$ except for poles at all points of a sequence
$z_n longrightarrow z_0$. (Note that $z_0$ is not an isolated singularity.) Show that $f(D)$ is dense in the complex
plane.
Here's my attempt.
Let $epsilon > 0$ such that $D = < epsilon $. Assume that $f(D)$ isn't dense in $mathbbC$. Then there exists a $w in mathbbC$ and $delta > 0$ such that,
beginalign
|f(z) - w| > delta Longleftrightarrow 0 < frac1f(z) - w < frac1delta equiv g(z) < frac1delta
endalign
Now, for $epsilon/2 > 0$, we have that there exists a $N in mathbbN$ such that,
beginalign
z_n in B(z_0, epsilon/2), quad n geq N.
endalign
Now let $epsilon' < textmin epsilon/2, $. We then have that $B(z_1, epsilon')$ is contained in $D$ such that $z_1$ is an essential singularity of f(z), since the open ball $B(z_1, epsilon'$) contains no other points from the sequence.
Consider the annular region $D' = z - z_1$. From the first equation, we have that,
beginalign
g(z) < frac1delta, quad z in D'.
endalign
Since $g(z)$ is bounded on $D'$, we have that $g(z)$ has (at most) an isolated singularity in $D'$. Therefore, we have that there is a function, $h(z)$, such that,
beginalign
g(z) = frac1f(z) - w = h(z), quad z in D',
endalign
and that $h(z_1)$ is analytic. Rearranging the previous equation, we have that,
beginalign
f(z) = w + frac1h(z) = frach(z)w + 1h(z) equiv = fracA(z)B(z), quad z in D'.
endalign
Since $f$ has a pole at $z_1$, we have that $h(z_1) = 0$, implying that $|g(z_1)| = |h(z_1)| = 0$, contradicting that $|g(z)$ is never zero on $D' subset D$.
complex-analysis
asked Jul 14 at 18:42
user82261
356
This statement is covered by Casorati-Weierstrass's theorem, and its proofs. Were you given a more restrictive version in class? Wow! Wikipedia also has a restricted statement (assuming holomorphic). Ignore the statement, just look at the proof.
– user574380
Jul 14 at 19:22
@scentofthetrees I'm self-studying from a book. The theorem as stated in the textbook is: "If f has an essential singularity at $z_0$ and if $D = z in mathbbC : 0 < $ , then $f(R)$ is dense in the complex plane.
– user82261
Jul 14 at 19:26
@scentofthetrees I thought the proof of the CW theorem has to be amended in this case, since, in the context of the question, $f(z)$ doesn't have an isolated singularity at $z_0$ due to the presence of infinitely many poles in $D$. Therefore, I think I had to further zoom into a small enough neighborhood in $D$ in which $f$ may be considered to have an isolated singularity.
– user82261
Jul 14 at 19:28
It doesn't have to be isolated, only a point at which on a punctured neighborhood the function is meromorphic. See the proof in the link.
– user574380
Jul 14 at 19:31
Oh. I have read a more restricted version of the proof as of yet now then, I guess. I'll see the link.
– user82261
Jul 14 at 19:32
add a comment |Â
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0
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up vote
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I'm trying to prove the following claim:
Suppose $f$ is analytic in a deleted neighborhood $D$ of $z_0$ except for poles at all points of a sequence
$z_n longrightarrow z_0$. (Note that $z_0$ is not an isolated singularity.) Show that $f(D)$ is dense in the complex
plane.
Here's my attempt.
Let $epsilon > 0$ such that $D = < epsilon $. Assume that $f(D)$ isn't dense in $mathbbC$. Then there exists a $w in mathbbC$ and $delta > 0$ such that,
beginalign
|f(z) - w| > delta Longleftrightarrow 0 < frac1f(z) - w < frac1delta equiv g(z) < frac1delta
endalign
Now, for $epsilon/2 > 0$, we have that there exists a $N in mathbbN$ such that,
beginalign
z_n in B(z_0, epsilon/2), quad n geq N.
endalign
Now let $epsilon' < textmin epsilon/2, $. We then have that $B(z_1, epsilon')$ is contained in $D$ such that $z_1$ is an essential singularity of f(z), since the open ball $B(z_1, epsilon'$) contains no other points from the sequence.
Consider the annular region $D' = z - z_1$. From the first equation, we have that,
beginalign
g(z) < frac1delta, quad z in D'.
endalign
Since $g(z)$ is bounded on $D'$, we have that $g(z)$ has (at most) an isolated singularity in $D'$. Therefore, we have that there is a function, $h(z)$, such that,
beginalign
g(z) = frac1f(z) - w = h(z), quad z in D',
endalign
and that $h(z_1)$ is analytic. Rearranging the previous equation, we have that,
beginalign
f(z) = w + frac1h(z) = frach(z)w + 1h(z) equiv = fracA(z)B(z), quad z in D'.
endalign
Since $f$ has a pole at $z_1$, we have that $h(z_1) = 0$, implying that $|g(z_1)| = |h(z_1)| = 0$, contradicting that $|g(z)$ is never zero on $D' subset D$.
complex-analysis
asked Jul 14 at 18:42
user82261
356
I'm trying to prove the following claim:
Suppose $f$ is analytic in a deleted neighborhood $D$ of $z_0$ except for poles at all points of a sequence
$z_n longrightarrow z_0$. (Note that $z_0$ is not an isolated singularity.) Show that $f(D)$ is dense in the complex
plane.
Here's my attempt.
Let $epsilon > 0$ such that $D = < epsilon $. Assume that $f(D)$ isn't dense in $mathbbC$. Then there exists a $w in mathbbC$ and $delta > 0$ such that,
beginalign
|f(z) - w| > delta Longleftrightarrow 0 < frac1f(z) - w < frac1delta equiv g(z) < frac1delta
endalign
Now, for $epsilon/2 > 0$, we have that there exists a $N in mathbbN$ such that,
beginalign
z_n in B(z_0, epsilon/2), quad n geq N.
endalign
Now let $epsilon' < textmin epsilon/2, $. We then have that $B(z_1, epsilon')$ is contained in $D$ such that $z_1$ is an essential singularity of f(z), since the open ball $B(z_1, epsilon'$) contains no other points from the sequence.
Consider the annular region $D' = z - z_1$. From the first equation, we have that,
beginalign
g(z) < frac1delta, quad z in D'.
endalign
Since $g(z)$ is bounded on $D'$, we have that $g(z)$ has (at most) an isolated singularity in $D'$. Therefore, we have that there is a function, $h(z)$, such that,
beginalign
g(z) = frac1f(z) - w = h(z), quad z in D',
endalign
and that $h(z_1)$ is analytic. Rearranging the previous equation, we have that,
beginalign
f(z) = w + frac1h(z) = frach(z)w + 1h(z) equiv = fracA(z)B(z), quad z in D'.
endalign
Since $f$ has a pole at $z_1$, we have that $h(z_1) = 0$, implying that $|g(z_1)| = |h(z_1)| = 0$, contradicting that $|g(z)$ is never zero on $D' subset D$.
complex-analysis
asked Jul 14 at 18:42
user82261
356
asked Jul 14 at 18:42
user82261
356
asked Jul 14 at 18:42
user82261
356
asked Jul 14 at 18:42
user82261
356
356
This statement is covered by Casorati-Weierstrass's theorem, and its proofs. Were you given a more restrictive version in class? Wow! Wikipedia also has a restricted statement (assuming holomorphic). Ignore the statement, just look at the proof.
– user574380
Jul 14 at 19:22
@scentofthetrees I'm self-studying from a book. The theorem as stated in the textbook is: "If f has an essential singularity at $z_0$ and if $D = z in mathbbC : 0 < $ , then $f(R)$ is dense in the complex plane.
– user82261
Jul 14 at 19:26
@scentofthetrees I thought the proof of the CW theorem has to be amended in this case, since, in the context of the question, $f(z)$ doesn't have an isolated singularity at $z_0$ due to the presence of infinitely many poles in $D$. Therefore, I think I had to further zoom into a small enough neighborhood in $D$ in which $f$ may be considered to have an isolated singularity.
– user82261
Jul 14 at 19:28
It doesn't have to be isolated, only a point at which on a punctured neighborhood the function is meromorphic. See the proof in the link.
– user574380
Jul 14 at 19:31
Oh. I have read a more restricted version of the proof as of yet now then, I guess. I'll see the link.
– user82261
Jul 14 at 19:32
add a comment |Â
This statement is covered by Casorati-Weierstrass's theorem, and its proofs. Were you given a more restrictive version in class? Wow! Wikipedia also has a restricted statement (assuming holomorphic). Ignore the statement, just look at the proof.
– user574380
Jul 14 at 19:22
@scentofthetrees I'm self-studying from a book. The theorem as stated in the textbook is: "If f has an essential singularity at $z_0$ and if $D = z in mathbbC : 0 < $ , then $f(R)$ is dense in the complex plane.
– user82261
Jul 14 at 19:26
@scentofthetrees I thought the proof of the CW theorem has to be amended in this case, since, in the context of the question, $f(z)$ doesn't have an isolated singularity at $z_0$ due to the presence of infinitely many poles in $D$. Therefore, I think I had to further zoom into a small enough neighborhood in $D$ in which $f$ may be considered to have an isolated singularity.
– user82261
Jul 14 at 19:28
It doesn't have to be isolated, only a point at which on a punctured neighborhood the function is meromorphic. See the proof in the link.
– user574380
Jul 14 at 19:31
Oh. I have read a more restricted version of the proof as of yet now then, I guess. I'll see the link.
– user82261
Jul 14 at 19:32
This statement is covered by Casorati-Weierstrass's theorem, and its proofs. Were you given a more restrictive version in class? Wow! Wikipedia also has a restricted statement (assuming holomorphic). Ignore the statement, just look at the proof.
– user574380
Jul 14 at 19:22
This statement is covered by Casorati-Weierstrass's theorem, and its proofs. Were you given a more restrictive version in class? Wow! Wikipedia also has a restricted statement (assuming holomorphic). Ignore the statement, just look at the proof.
– user574380
Jul 14 at 19:22
@scentofthetrees I'm self-studying from a book. The theorem as stated in the textbook is: "If f has an essential singularity at $z_0$ and if $D = z in mathbbC : 0 < $ , then $f(R)$ is dense in the complex plane.
– user82261
Jul 14 at 19:26
@scentofthetrees I'm self-studying from a book. The theorem as stated in the textbook is: "If f has an essential singularity at $z_0$ and if $D = z in mathbbC : 0 < $ , then $f(R)$ is dense in the complex plane.
– user82261
Jul 14 at 19:26
@scentofthetrees I thought the proof of the CW theorem has to be amended in this case, since, in the context of the question, $f(z)$ doesn't have an isolated singularity at $z_0$ due to the presence of infinitely many poles in $D$. Therefore, I think I had to further zoom into a small enough neighborhood in $D$ in which $f$ may be considered to have an isolated singularity.
– user82261
Jul 14 at 19:28
@scentofthetrees I thought the proof of the CW theorem has to be amended in this case, since, in the context of the question, $f(z)$ doesn't have an isolated singularity at $z_0$ due to the presence of infinitely many poles in $D$. Therefore, I think I had to further zoom into a small enough neighborhood in $D$ in which $f$ may be considered to have an isolated singularity.
– user82261
Jul 14 at 19:28
It doesn't have to be isolated, only a point at which on a punctured neighborhood the function is meromorphic. See the proof in the link.
– user574380
Jul 14 at 19:31
It doesn't have to be isolated, only a point at which on a punctured neighborhood the function is meromorphic. See the proof in the link.
– user574380
Jul 14 at 19:31
Oh. I have read a more restricted version of the proof as of yet now then, I guess. I'll see the link.
– user82261
Jul 14 at 19:32
Oh. I have read a more restricted version of the proof as of yet now then, I guess. I'll see the link.
– user82261
Jul 14 at 19:32
add a comment |Â
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This statement is covered by Casorati-Weierstrass's theorem, and its proofs. Were you given a more restrictive version in class? Wow! Wikipedia also has a restricted statement (assuming holomorphic). Ignore the statement, just look at the proof.
– user574380
Jul 14 at 19:22
@scentofthetrees I'm self-studying from a book. The theorem as stated in the textbook is: "If f has an essential singularity at $z_0$ and if $D = z in mathbbC : 0 < $ , then $f(R)$ is dense in the complex plane.
– user82261
Jul 14 at 19:26
@scentofthetrees I thought the proof of the CW theorem has to be amended in this case, since, in the context of the question, $f(z)$ doesn't have an isolated singularity at $z_0$ due to the presence of infinitely many poles in $D$. Therefore, I think I had to further zoom into a small enough neighborhood in $D$ in which $f$ may be considered to have an isolated singularity.
– user82261
Jul 14 at 19:28
It doesn't have to be isolated, only a point at which on a punctured neighborhood the function is meromorphic. See the proof in the link.
– user574380
Jul 14 at 19:31
Oh. I have read a more restricted version of the proof as of yet now then, I guess. I'll see the link.
– user82261
Jul 14 at 19:32