A condition implying that a holomorphic function is the identity map
Let $Omega$ be an open, connected and bounded subset of $mathbb{C}$, and $varphi : Omega rightarrow Omega$ a holomorphic function.
If there exists a $z_{0} in Omega$, such that
$$
varphi(z_0) = z_{0}quadtext{and}quadvarphi'(z_0) = 1,
$$
then prove that $varphi$ is the identity function,
i.e. $varphi(z)=z$, for all $zinOmega$.
Any hints?
complex-analysis analysis fixed-point-theorems analyticity holomorphic-functions
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
asked Feb 4 '14 at 10:34
WLOG
7,21932258
add a comment |
Let $Omega$ be an open, connected and bounded subset of $mathbb{C}$, and $varphi : Omega rightarrow Omega$ a holomorphic function.
If there exists a $z_{0} in Omega$, such that
$$
varphi(z_0) = z_{0}quadtext{and}quadvarphi'(z_0) = 1,
$$
then prove that $varphi$ is the identity function,
i.e. $varphi(z)=z$, for all $zinOmega$.
Any hints?
complex-analysis analysis fixed-point-theorems analyticity holomorphic-functions
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
asked Feb 4 '14 at 10:34
WLOG
7,21932258
Hint: First try the simply connected case and Schartz lemma.
– Moishe Cohen
Feb 4 '14 at 14:20
Apparently $Omega$ is connected as well.
– Yiorgos S. Smyrlis
Feb 4 '14 at 16:03
add a comment |
Let $Omega$ be an open, connected and bounded subset of $mathbb{C}$, and $varphi : Omega rightarrow Omega$ a holomorphic function.
If there exists a $z_{0} in Omega$, such that
$$
varphi(z_0) = z_{0}quadtext{and}quadvarphi'(z_0) = 1,
$$
then prove that $varphi$ is the identity function,
i.e. $varphi(z)=z$, for all $zinOmega$.
Any hints?
complex-analysis analysis fixed-point-theorems analyticity holomorphic-functions
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
asked Feb 4 '14 at 10:34
WLOG
7,21932258
Let $Omega$ be an open, connected and bounded subset of $mathbb{C}$, and $varphi : Omega rightarrow Omega$ a holomorphic function.
If there exists a $z_{0} in Omega$, such that
$$
varphi(z_0) = z_{0}quadtext{and}quadvarphi'(z_0) = 1,
$$
then prove that $varphi$ is the identity function,
i.e. $varphi(z)=z$, for all $zinOmega$.
Any hints?
complex-analysis analysis fixed-point-theorems analyticity holomorphic-functions
complex-analysis analysis fixed-point-theorems analyticity holomorphic-functions
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
asked Feb 4 '14 at 10:34
WLOG
7,21932258
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
asked Feb 4 '14 at 10:34
WLOG
7,21932258
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
edited Dec 9 '18 at 19:32
Yiorgos S. Smyrlis
62.5k1383162
62.5k1383162
asked Feb 4 '14 at 10:34
WLOG
7,21932258
asked Feb 4 '14 at 10:34
WLOG
7,21932258
asked Feb 4 '14 at 10:34
WLOG
7,21932258
7,21932258
Hint: First try the simply connected case and Schartz lemma.
– Moishe Cohen
Feb 4 '14 at 14:20
Apparently $Omega$ is connected as well.
– Yiorgos S. Smyrlis
Feb 4 '14 at 16:03
add a comment |
Hint: First try the simply connected case and Schartz lemma.
– Moishe Cohen
Feb 4 '14 at 14:20
Apparently $Omega$ is connected as well.
– Yiorgos S. Smyrlis
Feb 4 '14 at 16:03
Hint: First try the simply connected case and Schartz lemma.
– Moishe Cohen
Feb 4 '14 at 14:20
Hint: First try the simply connected case and Schartz lemma.
– Moishe Cohen
Feb 4 '14 at 14:20
Apparently $Omega$ is connected as well.
– Yiorgos S. Smyrlis
Feb 4 '14 at 16:03
Apparently $Omega$ is connected as well.
– Yiorgos S. Smyrlis
Feb 4 '14 at 16:03
add a comment |
1 Answer
1
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oldest
votes
As $Omega$ is bounded, then so is $varphi$, i.e., $|varphi(z)|le M$, for some $M>0$.
Without loss of generality assume that $z_0=0$. If $varphi$ is not the identity map, then near $z=0$
$$
varphi(z)=z+a_kz^k+{mathcal O} (z^{k+1}),
$$
where $a_k$ is the first nonzero coefficient of its expansion. Clearly, if $varphi^{ncirc}=underbrace{varphicirccdotscircvarphi}_{text{$n$ times}}$, then $big|varphi^{ncirc}(z)big|le M$, and
$$
varphi^{ncirc}(z)=z+na_kz^k+{mathcal O} (z^{k+1}).
$$
This means that
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=n k!a_k
$$
As $0inOmega$, there is an $r>0$, such that $bar D_rsubsetOmega$, where $bar D_r$ is the closed disc of radius $r$ and centered at the origin. According to Cauchy integral formula
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=frac{k!}{2pi i}int_{|z|=r}frac{varphi^{ncirc}(z),dz}{z^{k+1}},
$$
and hence
$$
n,|a_k| le frac{M}{r^k},
$$
for every $ninmathbb N$, which is a contradiction, as the left hand side tends to infinity, when $ntoinfty$ and the right hand side is constant.
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
Why we need composition?
– 1256
Dec 9 '18 at 18:22
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
As $Omega$ is bounded, then so is $varphi$, i.e., $|varphi(z)|le M$, for some $M>0$.
Without loss of generality assume that $z_0=0$. If $varphi$ is not the identity map, then near $z=0$
$$
varphi(z)=z+a_kz^k+{mathcal O} (z^{k+1}),
$$
where $a_k$ is the first nonzero coefficient of its expansion. Clearly, if $varphi^{ncirc}=underbrace{varphicirccdotscircvarphi}_{text{$n$ times}}$, then $big|varphi^{ncirc}(z)big|le M$, and
$$
varphi^{ncirc}(z)=z+na_kz^k+{mathcal O} (z^{k+1}).
$$
This means that
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=n k!a_k
$$
As $0inOmega$, there is an $r>0$, such that $bar D_rsubsetOmega$, where $bar D_r$ is the closed disc of radius $r$ and centered at the origin. According to Cauchy integral formula
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=frac{k!}{2pi i}int_{|z|=r}frac{varphi^{ncirc}(z),dz}{z^{k+1}},
$$
and hence
$$
n,|a_k| le frac{M}{r^k},
$$
for every $ninmathbb N$, which is a contradiction, as the left hand side tends to infinity, when $ntoinfty$ and the right hand side is constant.
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
Why we need composition?
– 1256
Dec 9 '18 at 18:22
add a comment |
As $Omega$ is bounded, then so is $varphi$, i.e., $|varphi(z)|le M$, for some $M>0$.
Without loss of generality assume that $z_0=0$. If $varphi$ is not the identity map, then near $z=0$
$$
varphi(z)=z+a_kz^k+{mathcal O} (z^{k+1}),
$$
where $a_k$ is the first nonzero coefficient of its expansion. Clearly, if $varphi^{ncirc}=underbrace{varphicirccdotscircvarphi}_{text{$n$ times}}$, then $big|varphi^{ncirc}(z)big|le M$, and
$$
varphi^{ncirc}(z)=z+na_kz^k+{mathcal O} (z^{k+1}).
$$
This means that
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=n k!a_k
$$
As $0inOmega$, there is an $r>0$, such that $bar D_rsubsetOmega$, where $bar D_r$ is the closed disc of radius $r$ and centered at the origin. According to Cauchy integral formula
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=frac{k!}{2pi i}int_{|z|=r}frac{varphi^{ncirc}(z),dz}{z^{k+1}},
$$
and hence
$$
n,|a_k| le frac{M}{r^k},
$$
for every $ninmathbb N$, which is a contradiction, as the left hand side tends to infinity, when $ntoinfty$ and the right hand side is constant.
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
Why we need composition?
– 1256
Dec 9 '18 at 18:22
add a comment |
As $Omega$ is bounded, then so is $varphi$, i.e., $|varphi(z)|le M$, for some $M>0$.
Without loss of generality assume that $z_0=0$. If $varphi$ is not the identity map, then near $z=0$
$$
varphi(z)=z+a_kz^k+{mathcal O} (z^{k+1}),
$$
where $a_k$ is the first nonzero coefficient of its expansion. Clearly, if $varphi^{ncirc}=underbrace{varphicirccdotscircvarphi}_{text{$n$ times}}$, then $big|varphi^{ncirc}(z)big|le M$, and
$$
varphi^{ncirc}(z)=z+na_kz^k+{mathcal O} (z^{k+1}).
$$
This means that
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=n k!a_k
$$
As $0inOmega$, there is an $r>0$, such that $bar D_rsubsetOmega$, where $bar D_r$ is the closed disc of radius $r$ and centered at the origin. According to Cauchy integral formula
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=frac{k!}{2pi i}int_{|z|=r}frac{varphi^{ncirc}(z),dz}{z^{k+1}},
$$
and hence
$$
n,|a_k| le frac{M}{r^k},
$$
for every $ninmathbb N$, which is a contradiction, as the left hand side tends to infinity, when $ntoinfty$ and the right hand side is constant.
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
As $Omega$ is bounded, then so is $varphi$, i.e., $|varphi(z)|le M$, for some $M>0$.
Without loss of generality assume that $z_0=0$. If $varphi$ is not the identity map, then near $z=0$
$$
varphi(z)=z+a_kz^k+{mathcal O} (z^{k+1}),
$$
where $a_k$ is the first nonzero coefficient of its expansion. Clearly, if $varphi^{ncirc}=underbrace{varphicirccdotscircvarphi}_{text{$n$ times}}$, then $big|varphi^{ncirc}(z)big|le M$, and
$$
varphi^{ncirc}(z)=z+na_kz^k+{mathcal O} (z^{k+1}).
$$
This means that
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=n k!a_k
$$
As $0inOmega$, there is an $r>0$, such that $bar D_rsubsetOmega$, where $bar D_r$ is the closed disc of radius $r$ and centered at the origin. According to Cauchy integral formula
$$
frac{d^k}{dz^k}big(varphi^{ncirc}(z)big)_{z=0}=frac{k!}{2pi i}int_{|z|=r}frac{varphi^{ncirc}(z),dz}{z^{k+1}},
$$
and hence
$$
n,|a_k| le frac{M}{r^k},
$$
for every $ninmathbb N$, which is a contradiction, as the left hand side tends to infinity, when $ntoinfty$ and the right hand side is constant.
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
edited Feb 4 '14 at 21:09
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
answered Feb 4 '14 at 16:28
Yiorgos S. Smyrlis
62.5k1383162
62.5k1383162
Why we need composition?
– 1256
Dec 9 '18 at 18:22
add a comment |
Why we need composition?
– 1256
Dec 9 '18 at 18:22
Why we need composition?
– 1256
Dec 9 '18 at 18:22
Why we need composition?
– 1256
Dec 9 '18 at 18:22
add a comment |
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Hint: First try the simply connected case and Schartz lemma.
– Moishe Cohen
Feb 4 '14 at 14:20
Apparently $Omega$ is connected as well.
– Yiorgos S. Smyrlis
Feb 4 '14 at 16:03