Holomorphic function $F:Bbb Hto Bbb C$ having the bounded sequence ${ir_n}$ as zeros.
$begingroup$
Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.
- Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
$F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.
- If $sum r_n<infty$, it is possible to construct a bounded function
on the upper half-plane with zeros precisely at the points $ir_n$.
There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?
complex-analysis
$endgroup$
add a comment |
$begingroup$
Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.
- Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
$F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.
- If $sum r_n<infty$, it is possible to construct a bounded function
on the upper half-plane with zeros precisely at the points $ir_n$.
There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?
complex-analysis
$endgroup$
add a comment |
$begingroup$
Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.
- Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
$F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.
- If $sum r_n<infty$, it is possible to construct a bounded function
on the upper half-plane with zeros precisely at the points $ir_n$.
There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?
complex-analysis
$endgroup$
Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.
- Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
$F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.
- If $sum r_n<infty$, it is possible to construct a bounded function
on the upper half-plane with zeros precisely at the points $ir_n$.
There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?
complex-analysis
complex-analysis
edited Dec 16 '18 at 20:18
UserA
asked Dec 16 '18 at 20:08
UserAUserA
542216
542216
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
sum_n Im (z_n)<infty.$$
$endgroup$
$begingroup$
what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
$endgroup$
– UserA
Dec 16 '18 at 20:36
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That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
$endgroup$
– Song
Dec 16 '18 at 20:38
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Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
$endgroup$
– UserA
Dec 16 '18 at 20:40
$begingroup$
I guess that will work :)
$endgroup$
– Song
Dec 16 '18 at 20:43
$begingroup$
what about the second part?
$endgroup$
– UserA
Dec 16 '18 at 20:46
|
show 1 more comment
$begingroup$
The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
sum_n Im (z_n)<infty.$$
$endgroup$
$begingroup$
what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
$endgroup$
– UserA
Dec 16 '18 at 20:36
$begingroup$
That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
$endgroup$
– Song
Dec 16 '18 at 20:38
$begingroup$
Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
$endgroup$
– UserA
Dec 16 '18 at 20:40
$begingroup$
I guess that will work :)
$endgroup$
– Song
Dec 16 '18 at 20:43
$begingroup$
what about the second part?
$endgroup$
– UserA
Dec 16 '18 at 20:46
|
show 1 more comment
$begingroup$
If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
sum_n Im (z_n)<infty.$$
$endgroup$
$begingroup$
what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
$endgroup$
– UserA
Dec 16 '18 at 20:36
$begingroup$
That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
$endgroup$
– Song
Dec 16 '18 at 20:38
$begingroup$
Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
$endgroup$
– UserA
Dec 16 '18 at 20:40
$begingroup$
I guess that will work :)
$endgroup$
– Song
Dec 16 '18 at 20:43
$begingroup$
what about the second part?
$endgroup$
– UserA
Dec 16 '18 at 20:46
|
show 1 more comment
$begingroup$
If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
sum_n Im (z_n)<infty.$$
$endgroup$
If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
sum_n Im (z_n)<infty.$$
edited Dec 16 '18 at 20:30
answered Dec 16 '18 at 20:22
SongSong
14.4k1635
14.4k1635
$begingroup$
what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
$endgroup$
– UserA
Dec 16 '18 at 20:36
$begingroup$
That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
$endgroup$
– Song
Dec 16 '18 at 20:38
$begingroup$
Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
$endgroup$
– UserA
Dec 16 '18 at 20:40
$begingroup$
I guess that will work :)
$endgroup$
– Song
Dec 16 '18 at 20:43
$begingroup$
what about the second part?
$endgroup$
– UserA
Dec 16 '18 at 20:46
|
show 1 more comment
$begingroup$
what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
$endgroup$
– UserA
Dec 16 '18 at 20:36
$begingroup$
That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
$endgroup$
– Song
Dec 16 '18 at 20:38
$begingroup$
Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
$endgroup$
– UserA
Dec 16 '18 at 20:40
$begingroup$
I guess that will work :)
$endgroup$
– Song
Dec 16 '18 at 20:43
$begingroup$
what about the second part?
$endgroup$
– UserA
Dec 16 '18 at 20:46
$begingroup$
what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
$endgroup$
– UserA
Dec 16 '18 at 20:36
$begingroup$
what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
$endgroup$
– UserA
Dec 16 '18 at 20:36
$begingroup$
That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
$endgroup$
– Song
Dec 16 '18 at 20:38
$begingroup$
That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
$endgroup$
– Song
Dec 16 '18 at 20:38
$begingroup$
Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
$endgroup$
– UserA
Dec 16 '18 at 20:40
$begingroup$
Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
$endgroup$
– UserA
Dec 16 '18 at 20:40
$begingroup$
I guess that will work :)
$endgroup$
– Song
Dec 16 '18 at 20:43
$begingroup$
I guess that will work :)
$endgroup$
– Song
Dec 16 '18 at 20:43
$begingroup$
what about the second part?
$endgroup$
– UserA
Dec 16 '18 at 20:46
$begingroup$
what about the second part?
$endgroup$
– UserA
Dec 16 '18 at 20:46
|
show 1 more comment
$begingroup$
The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.
$endgroup$
add a comment |
$begingroup$
The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.
$endgroup$
add a comment |
$begingroup$
The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.
$endgroup$
The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.
answered Dec 16 '18 at 20:16
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
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