Mixing
Space of univalent mappings $f: mathbb{D} to mathbb{C}$ has no nesting
$begingroup$
Let $S$ be the space of univalent (i.e. injective) mappings from the disk $mathbb{D}$ to the plane $mathbb{C}$ normalized so that $f(0) =0$ and $f'(0)=1$. So
$$f(z) = z+a_2z^2+a_3z^3+cdots.$$
I'd like to show that if $f,g in S$ and $f(mathbb{D}) = g(mathbb{D})$ then $f=g$.
Attempt: I'm aware the Schwarz lemma implies this. Let $U = f(mathbb{D})= g(mathbb{D})$ and let $F: U to mathbb{D}$ be a Riemann conformal mapping with $F(0)=0$. Then the Schwarz lemma applies to $Fcirc f$ and $Fcirc g$. But other than seeing that $F'(0)leq 1$, I do not see what this gets me. Any help would be appreciated.
complex-analysis riemann-surfaces conformal-geometry
asked Jan 2 at 20:49
DwaggDwagg
307111
$endgroup$
add a comment |
$begingroup$
Let $S$ be the space of univalent (i.e. injective) mappings from the disk $mathbb{D}$ to the plane $mathbb{C}$ normalized so that $f(0) =0$ and $f'(0)=1$. So
$$f(z) = z+a_2z^2+a_3z^3+cdots.$$
I'd like to show that if $f,g in S$ and $f(mathbb{D}) = g(mathbb{D})$ then $f=g$.
Attempt: I'm aware the Schwarz lemma implies this. Let $U = f(mathbb{D})= g(mathbb{D})$ and let $F: U to mathbb{D}$ be a Riemann conformal mapping with $F(0)=0$. Then the Schwarz lemma applies to $Fcirc f$ and $Fcirc g$. But other than seeing that $F'(0)leq 1$, I do not see what this gets me. Any help would be appreciated.
complex-analysis riemann-surfaces conformal-geometry
asked Jan 2 at 20:49
DwaggDwagg
307111
$endgroup$
add a comment |
$begingroup$
Let $S$ be the space of univalent (i.e. injective) mappings from the disk $mathbb{D}$ to the plane $mathbb{C}$ normalized so that $f(0) =0$ and $f'(0)=1$. So
$$f(z) = z+a_2z^2+a_3z^3+cdots.$$
I'd like to show that if $f,g in S$ and $f(mathbb{D}) = g(mathbb{D})$ then $f=g$.
Attempt: I'm aware the Schwarz lemma implies this. Let $U = f(mathbb{D})= g(mathbb{D})$ and let $F: U to mathbb{D}$ be a Riemann conformal mapping with $F(0)=0$. Then the Schwarz lemma applies to $Fcirc f$ and $Fcirc g$. But other than seeing that $F'(0)leq 1$, I do not see what this gets me. Any help would be appreciated.
complex-analysis riemann-surfaces conformal-geometry
asked Jan 2 at 20:49
DwaggDwagg
307111
$endgroup$
Let $S$ be the space of univalent (i.e. injective) mappings from the disk $mathbb{D}$ to the plane $mathbb{C}$ normalized so that $f(0) =0$ and $f'(0)=1$. So
$$f(z) = z+a_2z^2+a_3z^3+cdots.$$
I'd like to show that if $f,g in S$ and $f(mathbb{D}) = g(mathbb{D})$ then $f=g$.
Attempt: I'm aware the Schwarz lemma implies this. Let $U = f(mathbb{D})= g(mathbb{D})$ and let $F: U to mathbb{D}$ be a Riemann conformal mapping with $F(0)=0$. Then the Schwarz lemma applies to $Fcirc f$ and $Fcirc g$. But other than seeing that $F'(0)leq 1$, I do not see what this gets me. Any help would be appreciated.
complex-analysis riemann-surfaces conformal-geometry
complex-analysis riemann-surfaces conformal-geometry
asked Jan 2 at 20:49
DwaggDwagg
307111
asked Jan 2 at 20:49
DwaggDwagg
307111
asked Jan 2 at 20:49
DwaggDwagg
307111
asked Jan 2 at 20:49
DwaggDwagg
307111
asked Jan 2 at 20:49
DwaggDwagg
307111
307111
add a comment |
add a comment |
1 Answer
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$begingroup$
If $f(mathbb{D}) = g(mathbb{D})$ then $h = g^{-1} circ f$ is a holomorphic function from $mathbb{D}$ into $mathbb{D}$ with $h(0) = 0$ and $h'(0) = 1$. Now use the Schwarz Lemma to conclude that $h$ is the identity.
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
If $f(mathbb{D}) = g(mathbb{D})$ then $h = g^{-1} circ f$ is a holomorphic function from $mathbb{D}$ into $mathbb{D}$ with $h(0) = 0$ and $h'(0) = 1$. Now use the Schwarz Lemma to conclude that $h$ is the identity.
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
$endgroup$
add a comment |
$begingroup$
If $f(mathbb{D}) = g(mathbb{D})$ then $h = g^{-1} circ f$ is a holomorphic function from $mathbb{D}$ into $mathbb{D}$ with $h(0) = 0$ and $h'(0) = 1$. Now use the Schwarz Lemma to conclude that $h$ is the identity.
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
$endgroup$
add a comment |
$begingroup$
If $f(mathbb{D}) = g(mathbb{D})$ then $h = g^{-1} circ f$ is a holomorphic function from $mathbb{D}$ into $mathbb{D}$ with $h(0) = 0$ and $h'(0) = 1$. Now use the Schwarz Lemma to conclude that $h$ is the identity.
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
$endgroup$
If $f(mathbb{D}) = g(mathbb{D})$ then $h = g^{-1} circ f$ is a holomorphic function from $mathbb{D}$ into $mathbb{D}$ with $h(0) = 0$ and $h'(0) = 1$. Now use the Schwarz Lemma to conclude that $h$ is the identity.
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
answered Jan 2 at 20:55
Martin RMartin R
27.6k33255
27.6k33255
add a comment |
add a comment |
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