Given two biholomorphic maps such that $f(z_0)=g(z_0)=0$, prove there exists $c$ such that $f(z)=cg(z)$
$begingroup$
Given two biholomorphic maps $f:Omegarightarrowmathbb{D}$ and $g:Omegarightarrowmathbb{D}$ such that $f(z_0)=g(z_0)=0$, prove that there exists $cinmathbb{C}$ with $|c|=1$ such that $f(z)=cg(z)$
If $f$ or $g$ is identically zero, it is trivial as $0=c0$, so assume they are both not identically zero. Assume WLOG,$|f|leq|g|$. Then, $f(z)=(z-z_0)^mk(z)$ and $g(z)=(z-z_0)^nh(z)$ where $k(z_0)$ and $h(z_0)$ are both not zero. Then, for $zneq z_0$, $$left|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq1$$ and there exists some constant $k$ such that $left|frac{k(z)}{h(z)}right|geq frac{1}{k}$ so we have $$frac{|z-z_0|^{m-n}}{k}leqleft|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq 1Rightarrow|z-z_0|^{m-n}leq k$$
How do I proceed further to show that there is a constant $c$? or am I totally wrong?
complex-analysis
asked Dec 19 '18 at 17:14
Ya GYa G
514210
$endgroup$
|
show 7 more comments
$begingroup$
Given two biholomorphic maps $f:Omegarightarrowmathbb{D}$ and $g:Omegarightarrowmathbb{D}$ such that $f(z_0)=g(z_0)=0$, prove that there exists $cinmathbb{C}$ with $|c|=1$ such that $f(z)=cg(z)$
If $f$ or $g$ is identically zero, it is trivial as $0=c0$, so assume they are both not identically zero. Assume WLOG,$|f|leq|g|$. Then, $f(z)=(z-z_0)^mk(z)$ and $g(z)=(z-z_0)^nh(z)$ where $k(z_0)$ and $h(z_0)$ are both not zero. Then, for $zneq z_0$, $$left|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq1$$ and there exists some constant $k$ such that $left|frac{k(z)}{h(z)}right|geq frac{1}{k}$ so we have $$frac{|z-z_0|^{m-n}}{k}leqleft|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq 1Rightarrow|z-z_0|^{m-n}leq k$$
How do I proceed further to show that there is a constant $c$? or am I totally wrong?
complex-analysis
asked Dec 19 '18 at 17:14
Ya GYa G
514210
$endgroup$
$begingroup$
Why do you think that you can assume that $lvert frvertleqslantlvert grvert$ without loss of generality?
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:25
$begingroup$
@JoséCarlosSantos because if they are equal to each other, it's very trivial and always true for some $c$ where $|c|=1$. Maybe I'm not interpreting the problem correctly.
$endgroup$
– Ya G
Dec 19 '18 at 17:43
$begingroup$
Indeed, if $lvert frvert=lvert grvert$, then the problem is trivial. But is is perfectly possible that for some $z$ you have $bigllvert f(z)bigrrvert<bigllvert g(z)bigrrvert$, where as for some $w$ you have $bigllvert f(w)bigrrvert>bigllvert g(w)bigrrvert$.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:47
$begingroup$
To be clear, are you saying that for all $zinmathbb{C}$ there is a unique $cinmathbb{C}$ such that $f(z)=cg(z)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:55
$begingroup$
@R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $zinmathbb{C}$
$endgroup$
– Ya G
Dec 19 '18 at 17:57
|
show 7 more comments
$begingroup$
Given two biholomorphic maps $f:Omegarightarrowmathbb{D}$ and $g:Omegarightarrowmathbb{D}$ such that $f(z_0)=g(z_0)=0$, prove that there exists $cinmathbb{C}$ with $|c|=1$ such that $f(z)=cg(z)$
If $f$ or $g$ is identically zero, it is trivial as $0=c0$, so assume they are both not identically zero. Assume WLOG,$|f|leq|g|$. Then, $f(z)=(z-z_0)^mk(z)$ and $g(z)=(z-z_0)^nh(z)$ where $k(z_0)$ and $h(z_0)$ are both not zero. Then, for $zneq z_0$, $$left|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq1$$ and there exists some constant $k$ such that $left|frac{k(z)}{h(z)}right|geq frac{1}{k}$ so we have $$frac{|z-z_0|^{m-n}}{k}leqleft|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq 1Rightarrow|z-z_0|^{m-n}leq k$$
How do I proceed further to show that there is a constant $c$? or am I totally wrong?
complex-analysis
asked Dec 19 '18 at 17:14
Ya GYa G
514210
$endgroup$
Given two biholomorphic maps $f:Omegarightarrowmathbb{D}$ and $g:Omegarightarrowmathbb{D}$ such that $f(z_0)=g(z_0)=0$, prove that there exists $cinmathbb{C}$ with $|c|=1$ such that $f(z)=cg(z)$
If $f$ or $g$ is identically zero, it is trivial as $0=c0$, so assume they are both not identically zero. Assume WLOG,$|f|leq|g|$. Then, $f(z)=(z-z_0)^mk(z)$ and $g(z)=(z-z_0)^nh(z)$ where $k(z_0)$ and $h(z_0)$ are both not zero. Then, for $zneq z_0$, $$left|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq1$$ and there exists some constant $k$ such that $left|frac{k(z)}{h(z)}right|geq frac{1}{k}$ so we have $$frac{|z-z_0|^{m-n}}{k}leqleft|frac{(z-z_0)^{m-n}k(z)}{h(z)}right|leq 1Rightarrow|z-z_0|^{m-n}leq k$$
How do I proceed further to show that there is a constant $c$? or am I totally wrong?
complex-analysis
complex-analysis
asked Dec 19 '18 at 17:14
Ya GYa G
514210
asked Dec 19 '18 at 17:14
Ya GYa G
514210
asked Dec 19 '18 at 17:14
Ya GYa G
514210
asked Dec 19 '18 at 17:14
Ya GYa G
514210
asked Dec 19 '18 at 17:14
Ya GYa G
514210
514210
$begingroup$
Why do you think that you can assume that $lvert frvertleqslantlvert grvert$ without loss of generality?
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:25
$begingroup$
@JoséCarlosSantos because if they are equal to each other, it's very trivial and always true for some $c$ where $|c|=1$. Maybe I'm not interpreting the problem correctly.
$endgroup$
– Ya G
Dec 19 '18 at 17:43
$begingroup$
Indeed, if $lvert frvert=lvert grvert$, then the problem is trivial. But is is perfectly possible that for some $z$ you have $bigllvert f(z)bigrrvert<bigllvert g(z)bigrrvert$, where as for some $w$ you have $bigllvert f(w)bigrrvert>bigllvert g(w)bigrrvert$.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:47
$begingroup$
To be clear, are you saying that for all $zinmathbb{C}$ there is a unique $cinmathbb{C}$ such that $f(z)=cg(z)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:55
$begingroup$
@R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $zinmathbb{C}$
$endgroup$
– Ya G
Dec 19 '18 at 17:57
|
show 7 more comments
$begingroup$
Why do you think that you can assume that $lvert frvertleqslantlvert grvert$ without loss of generality?
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:25
$begingroup$
@JoséCarlosSantos because if they are equal to each other, it's very trivial and always true for some $c$ where $|c|=1$. Maybe I'm not interpreting the problem correctly.
$endgroup$
– Ya G
Dec 19 '18 at 17:43
$begingroup$
Indeed, if $lvert frvert=lvert grvert$, then the problem is trivial. But is is perfectly possible that for some $z$ you have $bigllvert f(z)bigrrvert<bigllvert g(z)bigrrvert$, where as for some $w$ you have $bigllvert f(w)bigrrvert>bigllvert g(w)bigrrvert$.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:47
$begingroup$
To be clear, are you saying that for all $zinmathbb{C}$ there is a unique $cinmathbb{C}$ such that $f(z)=cg(z)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:55
$begingroup$
@R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $zinmathbb{C}$
$endgroup$
– Ya G
Dec 19 '18 at 17:57
$begingroup$
Why do you think that you can assume that $lvert frvertleqslantlvert grvert$ without loss of generality?
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:25
$begingroup$
Why do you think that you can assume that $lvert frvertleqslantlvert grvert$ without loss of generality?
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:25
$begingroup$
@JoséCarlosSantos because if they are equal to each other, it's very trivial and always true for some $c$ where $|c|=1$. Maybe I'm not interpreting the problem correctly.
$endgroup$
– Ya G
Dec 19 '18 at 17:43
$begingroup$
@JoséCarlosSantos because if they are equal to each other, it's very trivial and always true for some $c$ where $|c|=1$. Maybe I'm not interpreting the problem correctly.
$endgroup$
– Ya G
Dec 19 '18 at 17:43
$begingroup$
Indeed, if $lvert frvert=lvert grvert$, then the problem is trivial. But is is perfectly possible that for some $z$ you have $bigllvert f(z)bigrrvert<bigllvert g(z)bigrrvert$, where as for some $w$ you have $bigllvert f(w)bigrrvert>bigllvert g(w)bigrrvert$.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:47
$begingroup$
Indeed, if $lvert frvert=lvert grvert$, then the problem is trivial. But is is perfectly possible that for some $z$ you have $bigllvert f(z)bigrrvert<bigllvert g(z)bigrrvert$, where as for some $w$ you have $bigllvert f(w)bigrrvert>bigllvert g(w)bigrrvert$.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:47
$begingroup$
To be clear, are you saying that for all $zinmathbb{C}$ there is a unique $cinmathbb{C}$ such that $f(z)=cg(z)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:55
$begingroup$
To be clear, are you saying that for all $zinmathbb{C}$ there is a unique $cinmathbb{C}$ such that $f(z)=cg(z)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:55
$begingroup$
@R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $zinmathbb{C}$
$endgroup$
– Ya G
Dec 19 '18 at 17:57
$begingroup$
@R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $zinmathbb{C}$
$endgroup$
– Ya G
Dec 19 '18 at 17:57
|
show 7 more comments
1 Answer
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$begingroup$
Consider $f circ g^{-1}$ this is an automorphisms of $mathbb{D}$ since $f$ and $g$ are biholomorphic. In particular, we know all automorphisms of the unit disk are given by Blaschke factors (http://mathworld.wolfram.com/BlaschkeFactor.html). Therefore, one has begin{equation} f circ g^{-1} = e^{itheta} frac{z - alpha}{1-overline{alpha}z} end{equation} In particular as $f circ g^{-1} (0) = f(z_0) = 0$, one sees $alpha = 0$ (this can also be seen by the Blaschke Factor inter swaps $0$ and $alpha$). So in particular, begin{equation} f circ g^{-1} = z e^{i theta} end{equation} So it follows from composing more that begin{equation} f = cg end{equation} for some $c$ with magnitude $1$.
answered Dec 21 '18 at 18:52
Story123Story123
22017
$endgroup$
add a comment |
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$begingroup$
Consider $f circ g^{-1}$ this is an automorphisms of $mathbb{D}$ since $f$ and $g$ are biholomorphic. In particular, we know all automorphisms of the unit disk are given by Blaschke factors (http://mathworld.wolfram.com/BlaschkeFactor.html). Therefore, one has begin{equation} f circ g^{-1} = e^{itheta} frac{z - alpha}{1-overline{alpha}z} end{equation} In particular as $f circ g^{-1} (0) = f(z_0) = 0$, one sees $alpha = 0$ (this can also be seen by the Blaschke Factor inter swaps $0$ and $alpha$). So in particular, begin{equation} f circ g^{-1} = z e^{i theta} end{equation} So it follows from composing more that begin{equation} f = cg end{equation} for some $c$ with magnitude $1$.
answered Dec 21 '18 at 18:52
Story123Story123
22017
$endgroup$
add a comment |
$begingroup$
Consider $f circ g^{-1}$ this is an automorphisms of $mathbb{D}$ since $f$ and $g$ are biholomorphic. In particular, we know all automorphisms of the unit disk are given by Blaschke factors (http://mathworld.wolfram.com/BlaschkeFactor.html). Therefore, one has begin{equation} f circ g^{-1} = e^{itheta} frac{z - alpha}{1-overline{alpha}z} end{equation} In particular as $f circ g^{-1} (0) = f(z_0) = 0$, one sees $alpha = 0$ (this can also be seen by the Blaschke Factor inter swaps $0$ and $alpha$). So in particular, begin{equation} f circ g^{-1} = z e^{i theta} end{equation} So it follows from composing more that begin{equation} f = cg end{equation} for some $c$ with magnitude $1$.
answered Dec 21 '18 at 18:52
Story123Story123
22017
$endgroup$
add a comment |
$begingroup$
Consider $f circ g^{-1}$ this is an automorphisms of $mathbb{D}$ since $f$ and $g$ are biholomorphic. In particular, we know all automorphisms of the unit disk are given by Blaschke factors (http://mathworld.wolfram.com/BlaschkeFactor.html). Therefore, one has begin{equation} f circ g^{-1} = e^{itheta} frac{z - alpha}{1-overline{alpha}z} end{equation} In particular as $f circ g^{-1} (0) = f(z_0) = 0$, one sees $alpha = 0$ (this can also be seen by the Blaschke Factor inter swaps $0$ and $alpha$). So in particular, begin{equation} f circ g^{-1} = z e^{i theta} end{equation} So it follows from composing more that begin{equation} f = cg end{equation} for some $c$ with magnitude $1$.
answered Dec 21 '18 at 18:52
Story123Story123
22017
$endgroup$
Consider $f circ g^{-1}$ this is an automorphisms of $mathbb{D}$ since $f$ and $g$ are biholomorphic. In particular, we know all automorphisms of the unit disk are given by Blaschke factors (http://mathworld.wolfram.com/BlaschkeFactor.html). Therefore, one has begin{equation} f circ g^{-1} = e^{itheta} frac{z - alpha}{1-overline{alpha}z} end{equation} In particular as $f circ g^{-1} (0) = f(z_0) = 0$, one sees $alpha = 0$ (this can also be seen by the Blaschke Factor inter swaps $0$ and $alpha$). So in particular, begin{equation} f circ g^{-1} = z e^{i theta} end{equation} So it follows from composing more that begin{equation} f = cg end{equation} for some $c$ with magnitude $1$.
answered Dec 21 '18 at 18:52
Story123Story123
22017
answered Dec 21 '18 at 18:52
Story123Story123
22017
answered Dec 21 '18 at 18:52
Story123Story123
22017
answered Dec 21 '18 at 18:52
Story123Story123
22017
22017
add a comment |
add a comment |
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$begingroup$
Why do you think that you can assume that $lvert frvertleqslantlvert grvert$ without loss of generality?
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:25
$begingroup$
@JoséCarlosSantos because if they are equal to each other, it's very trivial and always true for some $c$ where $|c|=1$. Maybe I'm not interpreting the problem correctly.
$endgroup$
– Ya G
Dec 19 '18 at 17:43
$begingroup$
Indeed, if $lvert frvert=lvert grvert$, then the problem is trivial. But is is perfectly possible that for some $z$ you have $bigllvert f(z)bigrrvert<bigllvert g(z)bigrrvert$, where as for some $w$ you have $bigllvert f(w)bigrrvert>bigllvert g(w)bigrrvert$.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 17:47
$begingroup$
To be clear, are you saying that for all $zinmathbb{C}$ there is a unique $cinmathbb{C}$ such that $f(z)=cg(z)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:55
$begingroup$
@R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $zinmathbb{C}$
$endgroup$
– Ya G
Dec 19 '18 at 17:57