Prove a bounded harmonic function $u$ on $0<|z|<1$ has removable singularity at the origin ...
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Prove a bounded harmonic function $u$ on $0
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$begingroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
$endgroup$
add a comment |
$begingroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
$endgroup$
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
add a comment |
$begingroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
$endgroup$
This is Exercise no.1, page 166 in Complex Analysis by Ahlfors.
Prove that a bounded harmonic function $u$ on $0<|z|<1,$ has removable singularity at the origin?
I need a solution without using Poisson integral.
By the mean value property and boundedness of $u$, we get
$int_{|z|=r}udtheta $ is a constant, say $beta$.
What remains is to show that $u(z)rightarrow beta$ as $zrightarrow 0.$
I am struggling at this point. How to proceed? Thanks.
complex-analysis
complex-analysis
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
edited Mar 17 at 7:16
Andrews
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
edited Mar 17 at 7:16
Andrews
1,2812422
edited Mar 17 at 7:16
Andrews
1,2812422
edited Mar 17 at 7:16
Andrews
1,2812422
1,2812422
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
asked Mar 17 at 6:57
KIRAN KUMARKIRAN KUMAR
1
1
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
add a comment |
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29
add a comment |
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$begingroup$
I don't see how you got the integrals of $u$ constant.
$endgroup$
– zhw.
Mar 17 at 23:51
$begingroup$
This integral will be a linear function of $log(r)$. By boundedness of $u$, the log part becomes absent.
$endgroup$
– KIRAN KUMAR
Mar 18 at 6:29