Must every holomorphic function $f: mathbb{D} to mathbb{D}$ have a fixed point?
(a) Prove that if $f: mathbb{D} to mathbb{D}$ is analytic and has two distinct fixed points, then $f$ is identity.
(b) Must every holomorphic function $f: mathbb{D} to mathbb{D}$ have a fixed point?
Notation. $mathbb{D}$ is the open unit disk.
My attempt.
(a) Let $displaystyle psi_{z_{1}} = frac{z_{1} - z}{1 - bar{z_{1}}z}$ and $z_{1},z_{2}$ the fixed points of $f$, and define $g: mathbb{D} to mathbb{D}$ by $g(z) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(z)$. Since $psi_{z_{1}}$ is holomorphic and an automorphism of $mathbb{D}$ (such that $psi_{z_{1}}^{2} = id$), $g$ maps $mathbb{D}$ into itself. Note that
$$g(0) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(0) = (psi_{z_{1}} circ f)(psi_{z_{1}}(0)) = psi_{z_{1}}(f(z_{1})) = 0.$$
Since $psi_{z_{1}}$ is bijective, there is $alpha$ such that $psi_{z_{1}}(alpha) = z_{2}$, moreover, $alpha = psi_{z_{1}}^{2}(alpha) = psi_{z_{1}}(z_{2})$. Then,
$$g(alpha) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(alpha) = (psi_{z_{1}} circ f)(psi_{z_{1}}(alpha)) = psi_{z_{1}}(f(z_{2})) = alpha.$$
Also, if $alpha = 0$, $z_{1} = z_{2}$, a contradiction. By Schwarz lemma, $g(z) = cz$ where $c = e^{itheta}$. But, since $g(alpha) = alpha$, $calpha = alpha$ and so, $c=1$.
(b) I know that, with tha same ideia that I use in (a), if the answer is yes, I can always apply the Schwarz lemma in every holomorphic function from $mathbb{D}$ to $mathbb{D}$. So, I think the answer is no, but I could not find a counterexample. I'm trying to get an example where Schwarz lemma fails.
complex-analysis conformal-geometry
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(a) Prove that if $f: mathbb{D} to mathbb{D}$ is analytic and has two distinct fixed points, then $f$ is identity.
(b) Must every holomorphic function $f: mathbb{D} to mathbb{D}$ have a fixed point?
Notation. $mathbb{D}$ is the open unit disk.
My attempt.
(a) Let $displaystyle psi_{z_{1}} = frac{z_{1} - z}{1 - bar{z_{1}}z}$ and $z_{1},z_{2}$ the fixed points of $f$, and define $g: mathbb{D} to mathbb{D}$ by $g(z) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(z)$. Since $psi_{z_{1}}$ is holomorphic and an automorphism of $mathbb{D}$ (such that $psi_{z_{1}}^{2} = id$), $g$ maps $mathbb{D}$ into itself. Note that
$$g(0) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(0) = (psi_{z_{1}} circ f)(psi_{z_{1}}(0)) = psi_{z_{1}}(f(z_{1})) = 0.$$
Since $psi_{z_{1}}$ is bijective, there is $alpha$ such that $psi_{z_{1}}(alpha) = z_{2}$, moreover, $alpha = psi_{z_{1}}^{2}(alpha) = psi_{z_{1}}(z_{2})$. Then,
$$g(alpha) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(alpha) = (psi_{z_{1}} circ f)(psi_{z_{1}}(alpha)) = psi_{z_{1}}(f(z_{2})) = alpha.$$
Also, if $alpha = 0$, $z_{1} = z_{2}$, a contradiction. By Schwarz lemma, $g(z) = cz$ where $c = e^{itheta}$. But, since $g(alpha) = alpha$, $calpha = alpha$ and so, $c=1$.
(b) I know that, with tha same ideia that I use in (a), if the answer is yes, I can always apply the Schwarz lemma in every holomorphic function from $mathbb{D}$ to $mathbb{D}$. So, I think the answer is no, but I could not find a counterexample. I'm trying to get an example where Schwarz lemma fails.
complex-analysis conformal-geometry
add a comment |
(a) Prove that if $f: mathbb{D} to mathbb{D}$ is analytic and has two distinct fixed points, then $f$ is identity.
(b) Must every holomorphic function $f: mathbb{D} to mathbb{D}$ have a fixed point?
Notation. $mathbb{D}$ is the open unit disk.
My attempt.
(a) Let $displaystyle psi_{z_{1}} = frac{z_{1} - z}{1 - bar{z_{1}}z}$ and $z_{1},z_{2}$ the fixed points of $f$, and define $g: mathbb{D} to mathbb{D}$ by $g(z) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(z)$. Since $psi_{z_{1}}$ is holomorphic and an automorphism of $mathbb{D}$ (such that $psi_{z_{1}}^{2} = id$), $g$ maps $mathbb{D}$ into itself. Note that
$$g(0) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(0) = (psi_{z_{1}} circ f)(psi_{z_{1}}(0)) = psi_{z_{1}}(f(z_{1})) = 0.$$
Since $psi_{z_{1}}$ is bijective, there is $alpha$ such that $psi_{z_{1}}(alpha) = z_{2}$, moreover, $alpha = psi_{z_{1}}^{2}(alpha) = psi_{z_{1}}(z_{2})$. Then,
$$g(alpha) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(alpha) = (psi_{z_{1}} circ f)(psi_{z_{1}}(alpha)) = psi_{z_{1}}(f(z_{2})) = alpha.$$
Also, if $alpha = 0$, $z_{1} = z_{2}$, a contradiction. By Schwarz lemma, $g(z) = cz$ where $c = e^{itheta}$. But, since $g(alpha) = alpha$, $calpha = alpha$ and so, $c=1$.
(b) I know that, with tha same ideia that I use in (a), if the answer is yes, I can always apply the Schwarz lemma in every holomorphic function from $mathbb{D}$ to $mathbb{D}$. So, I think the answer is no, but I could not find a counterexample. I'm trying to get an example where Schwarz lemma fails.
complex-analysis conformal-geometry
(a) Prove that if $f: mathbb{D} to mathbb{D}$ is analytic and has two distinct fixed points, then $f$ is identity.
(b) Must every holomorphic function $f: mathbb{D} to mathbb{D}$ have a fixed point?
Notation. $mathbb{D}$ is the open unit disk.
My attempt.
(a) Let $displaystyle psi_{z_{1}} = frac{z_{1} - z}{1 - bar{z_{1}}z}$ and $z_{1},z_{2}$ the fixed points of $f$, and define $g: mathbb{D} to mathbb{D}$ by $g(z) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(z)$. Since $psi_{z_{1}}$ is holomorphic and an automorphism of $mathbb{D}$ (such that $psi_{z_{1}}^{2} = id$), $g$ maps $mathbb{D}$ into itself. Note that
$$g(0) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(0) = (psi_{z_{1}} circ f)(psi_{z_{1}}(0)) = psi_{z_{1}}(f(z_{1})) = 0.$$
Since $psi_{z_{1}}$ is bijective, there is $alpha$ such that $psi_{z_{1}}(alpha) = z_{2}$, moreover, $alpha = psi_{z_{1}}^{2}(alpha) = psi_{z_{1}}(z_{2})$. Then,
$$g(alpha) = (psi_{z_{1}} circ f circ psi_{z_{1}}^{-1})(alpha) = (psi_{z_{1}} circ f)(psi_{z_{1}}(alpha)) = psi_{z_{1}}(f(z_{2})) = alpha.$$
Also, if $alpha = 0$, $z_{1} = z_{2}$, a contradiction. By Schwarz lemma, $g(z) = cz$ where $c = e^{itheta}$. But, since $g(alpha) = alpha$, $calpha = alpha$ and so, $c=1$.
(b) I know that, with tha same ideia that I use in (a), if the answer is yes, I can always apply the Schwarz lemma in every holomorphic function from $mathbb{D}$ to $mathbb{D}$. So, I think the answer is no, but I could not find a counterexample. I'm trying to get an example where Schwarz lemma fails.
complex-analysis conformal-geometry
complex-analysis conformal-geometry
edited Nov 22 at 21:26
asked Nov 22 at 21:22
Lucas Corrêa
1,327321
1,327321
add a comment |
add a comment |
2 Answers
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The unit disc $Bbb D$ is conformally equivalent to the upper half-plane $Bbb H$.
That has a holomorphic map $Bbb Hto Bbb H$ without fixed points, for instance
$zmapsto z+1$. So if $h:Bbb DtoBbb H$ is a holomorphic map, with a holomorphic inverse, then you can take $f(z)=h^{-1}(h(z)+1)$. I'll leave it to you to find
such an $h$.
$h(z) = ifrac{1-z}{1+z}$ is a holomorphic map from $mathbb{D}$ to $mathbb{H}$ (since $mathrm{Im}(h(z)) > 0$) with inverse $g(z) = frac{i-z}{i+z}$, right?
– Lucas Corrêa
Nov 22 at 21:34
1
@LucasCorrêa That looks very plausible.
– Lord Shark the Unknown
Nov 22 at 21:43
add a comment |
Another Counterexample :
$f(Z)=frac{z+1}{2}$
add a comment |
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2 Answers
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2 Answers
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The unit disc $Bbb D$ is conformally equivalent to the upper half-plane $Bbb H$.
That has a holomorphic map $Bbb Hto Bbb H$ without fixed points, for instance
$zmapsto z+1$. So if $h:Bbb DtoBbb H$ is a holomorphic map, with a holomorphic inverse, then you can take $f(z)=h^{-1}(h(z)+1)$. I'll leave it to you to find
such an $h$.
$h(z) = ifrac{1-z}{1+z}$ is a holomorphic map from $mathbb{D}$ to $mathbb{H}$ (since $mathrm{Im}(h(z)) > 0$) with inverse $g(z) = frac{i-z}{i+z}$, right?
– Lucas Corrêa
Nov 22 at 21:34
1
@LucasCorrêa That looks very plausible.
– Lord Shark the Unknown
Nov 22 at 21:43
add a comment |
The unit disc $Bbb D$ is conformally equivalent to the upper half-plane $Bbb H$.
That has a holomorphic map $Bbb Hto Bbb H$ without fixed points, for instance
$zmapsto z+1$. So if $h:Bbb DtoBbb H$ is a holomorphic map, with a holomorphic inverse, then you can take $f(z)=h^{-1}(h(z)+1)$. I'll leave it to you to find
such an $h$.
$h(z) = ifrac{1-z}{1+z}$ is a holomorphic map from $mathbb{D}$ to $mathbb{H}$ (since $mathrm{Im}(h(z)) > 0$) with inverse $g(z) = frac{i-z}{i+z}$, right?
– Lucas Corrêa
Nov 22 at 21:34
1
@LucasCorrêa That looks very plausible.
– Lord Shark the Unknown
Nov 22 at 21:43
add a comment |
The unit disc $Bbb D$ is conformally equivalent to the upper half-plane $Bbb H$.
That has a holomorphic map $Bbb Hto Bbb H$ without fixed points, for instance
$zmapsto z+1$. So if $h:Bbb DtoBbb H$ is a holomorphic map, with a holomorphic inverse, then you can take $f(z)=h^{-1}(h(z)+1)$. I'll leave it to you to find
such an $h$.
The unit disc $Bbb D$ is conformally equivalent to the upper half-plane $Bbb H$.
That has a holomorphic map $Bbb Hto Bbb H$ without fixed points, for instance
$zmapsto z+1$. So if $h:Bbb DtoBbb H$ is a holomorphic map, with a holomorphic inverse, then you can take $f(z)=h^{-1}(h(z)+1)$. I'll leave it to you to find
such an $h$.
answered Nov 22 at 21:25
Lord Shark the Unknown
100k958131
100k958131
$h(z) = ifrac{1-z}{1+z}$ is a holomorphic map from $mathbb{D}$ to $mathbb{H}$ (since $mathrm{Im}(h(z)) > 0$) with inverse $g(z) = frac{i-z}{i+z}$, right?
– Lucas Corrêa
Nov 22 at 21:34
1
@LucasCorrêa That looks very plausible.
– Lord Shark the Unknown
Nov 22 at 21:43
add a comment |
$h(z) = ifrac{1-z}{1+z}$ is a holomorphic map from $mathbb{D}$ to $mathbb{H}$ (since $mathrm{Im}(h(z)) > 0$) with inverse $g(z) = frac{i-z}{i+z}$, right?
– Lucas Corrêa
Nov 22 at 21:34
1
@LucasCorrêa That looks very plausible.
– Lord Shark the Unknown
Nov 22 at 21:43
$h(z) = ifrac{1-z}{1+z}$ is a holomorphic map from $mathbb{D}$ to $mathbb{H}$ (since $mathrm{Im}(h(z)) > 0$) with inverse $g(z) = frac{i-z}{i+z}$, right?
– Lucas Corrêa
Nov 22 at 21:34
$h(z) = ifrac{1-z}{1+z}$ is a holomorphic map from $mathbb{D}$ to $mathbb{H}$ (since $mathrm{Im}(h(z)) > 0$) with inverse $g(z) = frac{i-z}{i+z}$, right?
– Lucas Corrêa
Nov 22 at 21:34
1
1
@LucasCorrêa That looks very plausible.
– Lord Shark the Unknown
Nov 22 at 21:43
@LucasCorrêa That looks very plausible.
– Lord Shark the Unknown
Nov 22 at 21:43
add a comment |
Another Counterexample :
$f(Z)=frac{z+1}{2}$
add a comment |
Another Counterexample :
$f(Z)=frac{z+1}{2}$
add a comment |
Another Counterexample :
$f(Z)=frac{z+1}{2}$
Another Counterexample :
$f(Z)=frac{z+1}{2}$
answered Nov 23 at 8:09
Shubham
1,5921519
1,5921519
add a comment |
add a comment |
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