Mixing
Bounded holomorphic function in unit disk
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Does there exist a bounded holomorphic function in the unit disk $D$ such that
$$fleft(1-frac1nright) = frac(-1)^nn$$
for $n = 1,2,3,dots$ ?
Sorry it's not homework so I've no idea how should I begin. I only noticed that $1-1/n$ has a limit point out of $D$. Also $f'(1)$ does not exist.
complex-analysis holomorphic-functions
asked Aug 6 at 5:48
MonkeyKing
2,2291029
add a comment |Â
up vote
1
down vote
favorite
Does there exist a bounded holomorphic function in the unit disk $D$ such that
$$fleft(1-frac1nright) = frac(-1)^nn$$
for $n = 1,2,3,dots$ ?
Sorry it's not homework so I've no idea how should I begin. I only noticed that $1-1/n$ has a limit point out of $D$. Also $f'(1)$ does not exist.
complex-analysis holomorphic-functions
asked Aug 6 at 5:48
MonkeyKing
2,2291029
2
The obvious candidate for such $f$ is $ f(z)=(1-z)cosfracpi1-z$, but that is not bounded
– Hagen von Eitzen
Aug 6 at 6:08
I'm still wondering about the proof from fred, because if $f$ is bounded and analytic then why should the value at $1$ on the boundary not exist? If it would not I would even say the function is not holomorph.
– Diger
Aug 6 at 6:41
Diger, f(z)=1/(1-z) is holomorphic on the open unit disk and is not bounded there.
– zokomoko
Aug 6 at 6:58
@HagenvonEitzen Is guessing enough? Seems identity theorem doesn't apply here.
– MonkeyKing
Aug 6 at 17:21
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Does there exist a bounded holomorphic function in the unit disk $D$ such that
$$fleft(1-frac1nright) = frac(-1)^nn$$
for $n = 1,2,3,dots$ ?
Sorry it's not homework so I've no idea how should I begin. I only noticed that $1-1/n$ has a limit point out of $D$. Also $f'(1)$ does not exist.
complex-analysis holomorphic-functions
asked Aug 6 at 5:48
MonkeyKing
2,2291029
Does there exist a bounded holomorphic function in the unit disk $D$ such that
$$fleft(1-frac1nright) = frac(-1)^nn$$
for $n = 1,2,3,dots$ ?
Sorry it's not homework so I've no idea how should I begin. I only noticed that $1-1/n$ has a limit point out of $D$. Also $f'(1)$ does not exist.
complex-analysis holomorphic-functions
asked Aug 6 at 5:48
MonkeyKing
2,2291029
asked Aug 6 at 5:48
MonkeyKing
2,2291029
asked Aug 6 at 5:48
MonkeyKing
2,2291029
asked Aug 6 at 5:48
MonkeyKing
2,2291029
2,2291029
2
The obvious candidate for such $f$ is $ f(z)=(1-z)cosfracpi1-z$, but that is not bounded
– Hagen von Eitzen
Aug 6 at 6:08
I'm still wondering about the proof from fred, because if $f$ is bounded and analytic then why should the value at $1$ on the boundary not exist? If it would not I would even say the function is not holomorph.
– Diger
Aug 6 at 6:41
Diger, f(z)=1/(1-z) is holomorphic on the open unit disk and is not bounded there.
– zokomoko
Aug 6 at 6:58
@HagenvonEitzen Is guessing enough? Seems identity theorem doesn't apply here.
– MonkeyKing
Aug 6 at 17:21
add a comment |Â
2
The obvious candidate for such $f$ is $ f(z)=(1-z)cosfracpi1-z$, but that is not bounded
– Hagen von Eitzen
Aug 6 at 6:08
I'm still wondering about the proof from fred, because if $f$ is bounded and analytic then why should the value at $1$ on the boundary not exist? If it would not I would even say the function is not holomorph.
– Diger
Aug 6 at 6:41
Diger, f(z)=1/(1-z) is holomorphic on the open unit disk and is not bounded there.
– zokomoko
Aug 6 at 6:58
@HagenvonEitzen Is guessing enough? Seems identity theorem doesn't apply here.
– MonkeyKing
Aug 6 at 17:21
2
2
The obvious candidate for such $f$ is $ f(z)=(1-z)cosfracpi1-z$, but that is not bounded
– Hagen von Eitzen
Aug 6 at 6:08
The obvious candidate for such $f$ is $ f(z)=(1-z)cosfracpi1-z$, but that is not bounded
– Hagen von Eitzen
Aug 6 at 6:08
I'm still wondering about the proof from fred, because if $f$ is bounded and analytic then why should the value at $1$ on the boundary not exist? If it would not I would even say the function is not holomorph.
– Diger
Aug 6 at 6:41
I'm still wondering about the proof from fred, because if $f$ is bounded and analytic then why should the value at $1$ on the boundary not exist? If it would not I would even say the function is not holomorph.
– Diger
Aug 6 at 6:41
Diger, f(z)=1/(1-z) is holomorphic on the open unit disk and is not bounded there.
– zokomoko
Aug 6 at 6:58
Diger, f(z)=1/(1-z) is holomorphic on the open unit disk and is not bounded there.
– zokomoko
Aug 6 at 6:58
@HagenvonEitzen Is guessing enough? Seems identity theorem doesn't apply here.
– MonkeyKing
Aug 6 at 17:21
@HagenvonEitzen Is guessing enough? Seems identity theorem doesn't apply here.
– MonkeyKing
Aug 6 at 17:21
add a comment |Â
1 Answer
1
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up vote
4
down vote
accepted
It is a well known consequence of Jensen's inequality that if $g$ is a non-zero bounded analytic function in the unit disk and its zeros are $a_1,a_2,...$ the $sum [1-|a_n|] <infty$. Let $g(z)=f(z)-(1-z)$. Then its zeros include $1-frac 1 2n$ which do not satisfy the summability condition. Hence $f(z)equiv (1-z)$. But this does not satisfy the hypothesis for odd $n$ so there is no bounded analytic function with the given property.
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
2
The section on Blaschke Products in Rudin's RCA has a proof of the result I have used in my answer.
– Kavi Rama Murthy
Aug 6 at 7:28
1
Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set.
– Jyrki Lahtonen
Aug 6 at 7:52
@JyrkiLahtonen I didn't give you the theorem number because I have an Indain edition of the book and theorem numbers don't match with international editions.
– Kavi Rama Murthy
Aug 6 at 7:55
That's fine. Mentioning Blaschke products is sufficient. After all, that section is not too long. Also, theorem numbers may vary from one edition to another. Mine is the 3rd.
– Jyrki Lahtonen
Aug 6 at 7:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
It is a well known consequence of Jensen's inequality that if $g$ is a non-zero bounded analytic function in the unit disk and its zeros are $a_1,a_2,...$ the $sum [1-|a_n|] <infty$. Let $g(z)=f(z)-(1-z)$. Then its zeros include $1-frac 1 2n$ which do not satisfy the summability condition. Hence $f(z)equiv (1-z)$. But this does not satisfy the hypothesis for odd $n$ so there is no bounded analytic function with the given property.
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
2
The section on Blaschke Products in Rudin's RCA has a proof of the result I have used in my answer.
– Kavi Rama Murthy
Aug 6 at 7:28
1
Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set.
– Jyrki Lahtonen
Aug 6 at 7:52
@JyrkiLahtonen I didn't give you the theorem number because I have an Indain edition of the book and theorem numbers don't match with international editions.
– Kavi Rama Murthy
Aug 6 at 7:55
That's fine. Mentioning Blaschke products is sufficient. After all, that section is not too long. Also, theorem numbers may vary from one edition to another. Mine is the 3rd.
– Jyrki Lahtonen
Aug 6 at 7:58
add a comment |Â
up vote
4
down vote
accepted
It is a well known consequence of Jensen's inequality that if $g$ is a non-zero bounded analytic function in the unit disk and its zeros are $a_1,a_2,...$ the $sum [1-|a_n|] <infty$. Let $g(z)=f(z)-(1-z)$. Then its zeros include $1-frac 1 2n$ which do not satisfy the summability condition. Hence $f(z)equiv (1-z)$. But this does not satisfy the hypothesis for odd $n$ so there is no bounded analytic function with the given property.
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
2
The section on Blaschke Products in Rudin's RCA has a proof of the result I have used in my answer.
– Kavi Rama Murthy
Aug 6 at 7:28
1
Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set.
– Jyrki Lahtonen
Aug 6 at 7:52
@JyrkiLahtonen I didn't give you the theorem number because I have an Indain edition of the book and theorem numbers don't match with international editions.
– Kavi Rama Murthy
Aug 6 at 7:55
That's fine. Mentioning Blaschke products is sufficient. After all, that section is not too long. Also, theorem numbers may vary from one edition to another. Mine is the 3rd.
– Jyrki Lahtonen
Aug 6 at 7:58
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
It is a well known consequence of Jensen's inequality that if $g$ is a non-zero bounded analytic function in the unit disk and its zeros are $a_1,a_2,...$ the $sum [1-|a_n|] <infty$. Let $g(z)=f(z)-(1-z)$. Then its zeros include $1-frac 1 2n$ which do not satisfy the summability condition. Hence $f(z)equiv (1-z)$. But this does not satisfy the hypothesis for odd $n$ so there is no bounded analytic function with the given property.
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
It is a well known consequence of Jensen's inequality that if $g$ is a non-zero bounded analytic function in the unit disk and its zeros are $a_1,a_2,...$ the $sum [1-|a_n|] <infty$. Let $g(z)=f(z)-(1-z)$. Then its zeros include $1-frac 1 2n$ which do not satisfy the summability condition. Hence $f(z)equiv (1-z)$. But this does not satisfy the hypothesis for odd $n$ so there is no bounded analytic function with the given property.
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
answered Aug 6 at 7:22
Kavi Rama Murthy
21k2830
21k2830
2
The section on Blaschke Products in Rudin's RCA has a proof of the result I have used in my answer.
– Kavi Rama Murthy
Aug 6 at 7:28
1
Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set.
– Jyrki Lahtonen
Aug 6 at 7:52
@JyrkiLahtonen I didn't give you the theorem number because I have an Indain edition of the book and theorem numbers don't match with international editions.
– Kavi Rama Murthy
Aug 6 at 7:55
That's fine. Mentioning Blaschke products is sufficient. After all, that section is not too long. Also, theorem numbers may vary from one edition to another. Mine is the 3rd.
– Jyrki Lahtonen
Aug 6 at 7:58
add a comment |Â
2
The section on Blaschke Products in Rudin's RCA has a proof of the result I have used in my answer.
– Kavi Rama Murthy
Aug 6 at 7:28
1
Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set.
– Jyrki Lahtonen
Aug 6 at 7:52
@JyrkiLahtonen I didn't give you the theorem number because I have an Indain edition of the book and theorem numbers don't match with international editions.
– Kavi Rama Murthy
Aug 6 at 7:55
That's fine. Mentioning Blaschke products is sufficient. After all, that section is not too long. Also, theorem numbers may vary from one edition to another. Mine is the 3rd.
– Jyrki Lahtonen
Aug 6 at 7:58
2
2
The section on Blaschke Products in Rudin's RCA has a proof of the result I have used in my answer.
– Kavi Rama Murthy
Aug 6 at 7:28
The section on Blaschke Products in Rudin's RCA has a proof of the result I have used in my answer.
– Kavi Rama Murthy
Aug 6 at 7:28
1
1
Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set.
– Jyrki Lahtonen
Aug 6 at 7:52
Thanks for the reference. It's theorem 15.23 (in my edition). Took me a while to dig up what $U$ means in that setting. I had a guess, and Rudin does provide a list of symbols :-). I was a bit surprised that $U$ is reserved for the open unit disk because in topology books it is more often a generic open set.
– Jyrki Lahtonen
Aug 6 at 7:52
@JyrkiLahtonen I didn't give you the theorem number because I have an Indain edition of the book and theorem numbers don't match with international editions.
– Kavi Rama Murthy
Aug 6 at 7:55
@JyrkiLahtonen I didn't give you the theorem number because I have an Indain edition of the book and theorem numbers don't match with international editions.
– Kavi Rama Murthy
Aug 6 at 7:55
That's fine. Mentioning Blaschke products is sufficient. After all, that section is not too long. Also, theorem numbers may vary from one edition to another. Mine is the 3rd.
– Jyrki Lahtonen
Aug 6 at 7:58
That's fine. Mentioning Blaschke products is sufficient. After all, that section is not too long. Also, theorem numbers may vary from one edition to another. Mine is the 3rd.
– Jyrki Lahtonen
Aug 6 at 7:58
add a comment |Â
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2
The obvious candidate for such $f$ is $ f(z)=(1-z)cosfracpi1-z$, but that is not bounded
– Hagen von Eitzen
Aug 6 at 6:08
I'm still wondering about the proof from fred, because if $f$ is bounded and analytic then why should the value at $1$ on the boundary not exist? If it would not I would even say the function is not holomorph.
– Diger
Aug 6 at 6:41
Diger, f(z)=1/(1-z) is holomorphic on the open unit disk and is not bounded there.
– zokomoko
Aug 6 at 6:58
@HagenvonEitzen Is guessing enough? Seems identity theorem doesn't apply here.
– MonkeyKing
Aug 6 at 17:21