Example of a Riemann mapping without continuous extension at the boundary.
Let $U$ be a simply connected open subset $U$ of $ mathbb{C}$ and $D={z in mathbb{C}: |z|<1 }$. I read a theorem which says that a Riemann mapping $f:U to D$ can be continuously extended to the boundary of $U$, that is, exists a continuous extension $tilde{f}:overline{U} to overline{D}$, if $partial U$ is a simple closed contour. But, is there a concrete example of a Riemann mapping without continuous extension at the boundary?.
Thank you for any help.
complex-analysis complex-numbers
asked Dec 3 '18 at 0:24
Minysh
13510
add a comment |
Let $U$ be a simply connected open subset $U$ of $ mathbb{C}$ and $D={z in mathbb{C}: |z|<1 }$. I read a theorem which says that a Riemann mapping $f:U to D$ can be continuously extended to the boundary of $U$, that is, exists a continuous extension $tilde{f}:overline{U} to overline{D}$, if $partial U$ is a simple closed contour. But, is there a concrete example of a Riemann mapping without continuous extension at the boundary?.
Thank you for any help.
complex-analysis complex-numbers
asked Dec 3 '18 at 0:24
Minysh
13510
1
$1-frac2{z+1}, Re(z) > 0$ ?
– reuns
Dec 3 '18 at 1:10
A simple example from a slit plane to a halfplane is $sqrt{z}$. With a fractional linear transformation, you can easily turn this into a conformal map onto the unit disk.
– Lukas Geyer
Dec 3 '18 at 5:52
add a comment |
Let $U$ be a simply connected open subset $U$ of $ mathbb{C}$ and $D={z in mathbb{C}: |z|<1 }$. I read a theorem which says that a Riemann mapping $f:U to D$ can be continuously extended to the boundary of $U$, that is, exists a continuous extension $tilde{f}:overline{U} to overline{D}$, if $partial U$ is a simple closed contour. But, is there a concrete example of a Riemann mapping without continuous extension at the boundary?.
Thank you for any help.
complex-analysis complex-numbers
asked Dec 3 '18 at 0:24
Minysh
13510
Let $U$ be a simply connected open subset $U$ of $ mathbb{C}$ and $D={z in mathbb{C}: |z|<1 }$. I read a theorem which says that a Riemann mapping $f:U to D$ can be continuously extended to the boundary of $U$, that is, exists a continuous extension $tilde{f}:overline{U} to overline{D}$, if $partial U$ is a simple closed contour. But, is there a concrete example of a Riemann mapping without continuous extension at the boundary?.
Thank you for any help.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Dec 3 '18 at 0:24
Minysh
13510
asked Dec 3 '18 at 0:24
Minysh
13510
edited Dec 3 '18 at 6:04
asked Dec 3 '18 at 0:24
Minysh
13510
asked Dec 3 '18 at 0:24
Minysh
13510
asked Dec 3 '18 at 0:24
Minysh
13510
13510
1
$1-frac2{z+1}, Re(z) > 0$ ?
– reuns
Dec 3 '18 at 1:10
A simple example from a slit plane to a halfplane is $sqrt{z}$. With a fractional linear transformation, you can easily turn this into a conformal map onto the unit disk.
– Lukas Geyer
Dec 3 '18 at 5:52
add a comment |
1
$1-frac2{z+1}, Re(z) > 0$ ?
– reuns
Dec 3 '18 at 1:10
A simple example from a slit plane to a halfplane is $sqrt{z}$. With a fractional linear transformation, you can easily turn this into a conformal map onto the unit disk.
– Lukas Geyer
Dec 3 '18 at 5:52
1
1
$1-frac2{z+1}, Re(z) > 0$ ?
– reuns
Dec 3 '18 at 1:10
$1-frac2{z+1}, Re(z) > 0$ ?
– reuns
Dec 3 '18 at 1:10
A simple example from a slit plane to a halfplane is $sqrt{z}$. With a fractional linear transformation, you can easily turn this into a conformal map onto the unit disk.
– Lukas Geyer
Dec 3 '18 at 5:52
A simple example from a slit plane to a halfplane is $sqrt{z}$. With a fractional linear transformation, you can easily turn this into a conformal map onto the unit disk.
– Lukas Geyer
Dec 3 '18 at 5:52
add a comment |
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1
$1-frac2{z+1}, Re(z) > 0$ ?
– reuns
Dec 3 '18 at 1:10
A simple example from a slit plane to a halfplane is $sqrt{z}$. With a fractional linear transformation, you can easily turn this into a conformal map onto the unit disk.
– Lukas Geyer
Dec 3 '18 at 5:52