Mixing
Existence of a bounded function on the disk with specified values
$begingroup$
I am trying to answer the question:
“Does there exist a bounded holomorphic function on the unit disk such that for all $nin mathbb{N}$,
$f(1-frac{1}{n}) = frac{(-1)^n}{n}$?”
My solution goes roughly likes this: using Jensen’s formula we find that any holomorphic function with zeroes at $1-frac{1}{n}$ cannot be bounded, so if there was such a function as the problem desired, it would be the only one, lest the difference between two such functions violate Poisson Jensen. But using some algebra tricks, given any one such function we mess it around to get another. This is a contradiction.
I just want to ask if there is a more straightforward way. Any other ideas are much appreciated.
complex-analysis holomorphic-functions
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
$endgroup$
add a comment |
$begingroup$
I am trying to answer the question:
“Does there exist a bounded holomorphic function on the unit disk such that for all $nin mathbb{N}$,
$f(1-frac{1}{n}) = frac{(-1)^n}{n}$?”
My solution goes roughly likes this: using Jensen’s formula we find that any holomorphic function with zeroes at $1-frac{1}{n}$ cannot be bounded, so if there was such a function as the problem desired, it would be the only one, lest the difference between two such functions violate Poisson Jensen. But using some algebra tricks, given any one such function we mess it around to get another. This is a contradiction.
I just want to ask if there is a more straightforward way. Any other ideas are much appreciated.
complex-analysis holomorphic-functions
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
$endgroup$
add a comment |
$begingroup$
I am trying to answer the question:
“Does there exist a bounded holomorphic function on the unit disk such that for all $nin mathbb{N}$,
$f(1-frac{1}{n}) = frac{(-1)^n}{n}$?”
My solution goes roughly likes this: using Jensen’s formula we find that any holomorphic function with zeroes at $1-frac{1}{n}$ cannot be bounded, so if there was such a function as the problem desired, it would be the only one, lest the difference between two such functions violate Poisson Jensen. But using some algebra tricks, given any one such function we mess it around to get another. This is a contradiction.
I just want to ask if there is a more straightforward way. Any other ideas are much appreciated.
complex-analysis holomorphic-functions
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
$endgroup$
I am trying to answer the question:
“Does there exist a bounded holomorphic function on the unit disk such that for all $nin mathbb{N}$,
$f(1-frac{1}{n}) = frac{(-1)^n}{n}$?”
My solution goes roughly likes this: using Jensen’s formula we find that any holomorphic function with zeroes at $1-frac{1}{n}$ cannot be bounded, so if there was such a function as the problem desired, it would be the only one, lest the difference between two such functions violate Poisson Jensen. But using some algebra tricks, given any one such function we mess it around to get another. This is a contradiction.
I just want to ask if there is a more straightforward way. Any other ideas are much appreciated.
complex-analysis holomorphic-functions
complex-analysis holomorphic-functions
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
edited Jan 7 at 1:30
Van Latimer
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
asked Jan 6 at 15:12
Van LatimerVan Latimer
321110
321110
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Here's another way. If we let $g(z) = [f(z)]^2 -(1-z)^2$, then $g$ is bounded on the unit disk and $g(x_n)=0$ for all $x_n = 1-frac{1}{n}$. By Blaschke condition, we must have $g equiv 0$. But this implies $h(z)= frac{f(z)}{1-z}$ is a continuous functions taking values in a disconnected set ${-1,1}$. It says that $h$ must be a constant. This leads to a contradiction to that $h(x_n)=(-1)^n$.
answered Jan 6 at 15:26
SongSong
10.2k627
$endgroup$
$begingroup$
I don’t think this is quite right, but it’s given me an idea which I’ll add below.
$endgroup$
– Van Latimer
Jan 7 at 1:15
$begingroup$
Yes, I made a quite silly mistake ... But the method can be applied in the right direction as you suggested.
$endgroup$
– Song
Jan 7 at 3:44
$begingroup$
Right. Thank you so much for the help!
$endgroup$
– Van Latimer
Jan 7 at 18:13
add a comment |
$begingroup$
Playing off of @Song’s answer, $f^2$ agrees with $(1-z)^2$ at all the points $1-frac{1}{n}$, so that since both are bounded, Jensen applied to their difference says they agree everywhere. But then the holomorphic function $frac{f}{1-z}$ has constant modulus and is therefore constant, so that $fequiv z mapsto a(1-z)$ for $a$ equal to $pm 1$, independently of $z$ of course, which is a contradiction.
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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votes
$begingroup$
Here's another way. If we let $g(z) = [f(z)]^2 -(1-z)^2$, then $g$ is bounded on the unit disk and $g(x_n)=0$ for all $x_n = 1-frac{1}{n}$. By Blaschke condition, we must have $g equiv 0$. But this implies $h(z)= frac{f(z)}{1-z}$ is a continuous functions taking values in a disconnected set ${-1,1}$. It says that $h$ must be a constant. This leads to a contradiction to that $h(x_n)=(-1)^n$.
answered Jan 6 at 15:26
SongSong
10.2k627
$endgroup$
$begingroup$
I don’t think this is quite right, but it’s given me an idea which I’ll add below.
$endgroup$
– Van Latimer
Jan 7 at 1:15
$begingroup$
Yes, I made a quite silly mistake ... But the method can be applied in the right direction as you suggested.
$endgroup$
– Song
Jan 7 at 3:44
$begingroup$
Right. Thank you so much for the help!
$endgroup$
– Van Latimer
Jan 7 at 18:13
add a comment |
$begingroup$
Here's another way. If we let $g(z) = [f(z)]^2 -(1-z)^2$, then $g$ is bounded on the unit disk and $g(x_n)=0$ for all $x_n = 1-frac{1}{n}$. By Blaschke condition, we must have $g equiv 0$. But this implies $h(z)= frac{f(z)}{1-z}$ is a continuous functions taking values in a disconnected set ${-1,1}$. It says that $h$ must be a constant. This leads to a contradiction to that $h(x_n)=(-1)^n$.
answered Jan 6 at 15:26
SongSong
10.2k627
$endgroup$
$begingroup$
I don’t think this is quite right, but it’s given me an idea which I’ll add below.
$endgroup$
– Van Latimer
Jan 7 at 1:15
$begingroup$
Yes, I made a quite silly mistake ... But the method can be applied in the right direction as you suggested.
$endgroup$
– Song
Jan 7 at 3:44
$begingroup$
Right. Thank you so much for the help!
$endgroup$
– Van Latimer
Jan 7 at 18:13
add a comment |
$begingroup$
Here's another way. If we let $g(z) = [f(z)]^2 -(1-z)^2$, then $g$ is bounded on the unit disk and $g(x_n)=0$ for all $x_n = 1-frac{1}{n}$. By Blaschke condition, we must have $g equiv 0$. But this implies $h(z)= frac{f(z)}{1-z}$ is a continuous functions taking values in a disconnected set ${-1,1}$. It says that $h$ must be a constant. This leads to a contradiction to that $h(x_n)=(-1)^n$.
answered Jan 6 at 15:26
SongSong
10.2k627
$endgroup$
Here's another way. If we let $g(z) = [f(z)]^2 -(1-z)^2$, then $g$ is bounded on the unit disk and $g(x_n)=0$ for all $x_n = 1-frac{1}{n}$. By Blaschke condition, we must have $g equiv 0$. But this implies $h(z)= frac{f(z)}{1-z}$ is a continuous functions taking values in a disconnected set ${-1,1}$. It says that $h$ must be a constant. This leads to a contradiction to that $h(x_n)=(-1)^n$.
answered Jan 6 at 15:26
SongSong
10.2k627
edited Jan 7 at 3:41
answered Jan 6 at 15:26
SongSong
10.2k627
answered Jan 6 at 15:26
SongSong
10.2k627
answered Jan 6 at 15:26
SongSong
10.2k627
10.2k627
$begingroup$
I don’t think this is quite right, but it’s given me an idea which I’ll add below.
$endgroup$
– Van Latimer
Jan 7 at 1:15
$begingroup$
Yes, I made a quite silly mistake ... But the method can be applied in the right direction as you suggested.
$endgroup$
– Song
Jan 7 at 3:44
$begingroup$
Right. Thank you so much for the help!
$endgroup$
– Van Latimer
Jan 7 at 18:13
add a comment |
$begingroup$
I don’t think this is quite right, but it’s given me an idea which I’ll add below.
$endgroup$
– Van Latimer
Jan 7 at 1:15
$begingroup$
Yes, I made a quite silly mistake ... But the method can be applied in the right direction as you suggested.
$endgroup$
– Song
Jan 7 at 3:44
$begingroup$
Right. Thank you so much for the help!
$endgroup$
– Van Latimer
Jan 7 at 18:13
$begingroup$
I don’t think this is quite right, but it’s given me an idea which I’ll add below.
$endgroup$
– Van Latimer
Jan 7 at 1:15
$begingroup$
I don’t think this is quite right, but it’s given me an idea which I’ll add below.
$endgroup$
– Van Latimer
Jan 7 at 1:15
$begingroup$
Yes, I made a quite silly mistake ... But the method can be applied in the right direction as you suggested.
$endgroup$
– Song
Jan 7 at 3:44
$begingroup$
Yes, I made a quite silly mistake ... But the method can be applied in the right direction as you suggested.
$endgroup$
– Song
Jan 7 at 3:44
$begingroup$
Right. Thank you so much for the help!
$endgroup$
– Van Latimer
Jan 7 at 18:13
$begingroup$
Right. Thank you so much for the help!
$endgroup$
– Van Latimer
Jan 7 at 18:13
add a comment |
$begingroup$
Playing off of @Song’s answer, $f^2$ agrees with $(1-z)^2$ at all the points $1-frac{1}{n}$, so that since both are bounded, Jensen applied to their difference says they agree everywhere. But then the holomorphic function $frac{f}{1-z}$ has constant modulus and is therefore constant, so that $fequiv z mapsto a(1-z)$ for $a$ equal to $pm 1$, independently of $z$ of course, which is a contradiction.
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
$endgroup$
add a comment |
$begingroup$
Playing off of @Song’s answer, $f^2$ agrees with $(1-z)^2$ at all the points $1-frac{1}{n}$, so that since both are bounded, Jensen applied to their difference says they agree everywhere. But then the holomorphic function $frac{f}{1-z}$ has constant modulus and is therefore constant, so that $fequiv z mapsto a(1-z)$ for $a$ equal to $pm 1$, independently of $z$ of course, which is a contradiction.
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
$endgroup$
add a comment |
$begingroup$
Playing off of @Song’s answer, $f^2$ agrees with $(1-z)^2$ at all the points $1-frac{1}{n}$, so that since both are bounded, Jensen applied to their difference says they agree everywhere. But then the holomorphic function $frac{f}{1-z}$ has constant modulus and is therefore constant, so that $fequiv z mapsto a(1-z)$ for $a$ equal to $pm 1$, independently of $z$ of course, which is a contradiction.
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
$endgroup$
Playing off of @Song’s answer, $f^2$ agrees with $(1-z)^2$ at all the points $1-frac{1}{n}$, so that since both are bounded, Jensen applied to their difference says they agree everywhere. But then the holomorphic function $frac{f}{1-z}$ has constant modulus and is therefore constant, so that $fequiv z mapsto a(1-z)$ for $a$ equal to $pm 1$, independently of $z$ of course, which is a contradiction.
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
edited Jan 7 at 1:30
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
answered Jan 7 at 1:20
Van LatimerVan Latimer
321110
321110
add a comment |
add a comment |
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