Must every holomorphic function $f: mathbb{D} to mathbb{D}$ have a fixed point?












3















(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







share|cite|improve this question





























    3















    (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







    share|cite|improve this question



























      3












      3








      3








      (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







      share|cite|improve this question
















      (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






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 22 at 21:26

























      asked Nov 22 at 21:22









      Lucas Corrêa

      1,327321




      1,327321






















          2 Answers
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          4














          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 $h$.






          share|cite|improve this answer





















          • $h(z) = ifrac{1-z}{1+z}$ 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



















          2














          Another Counterexample :



          $f(Z)=frac{z+1}{2}$






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

            oldest

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            4














            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 $h$.






            share|cite|improve this answer





















            • $h(z) = ifrac{1-z}{1+z}$ 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
















            4














            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 $h$.






            share|cite|improve this answer





















            • $h(z) = ifrac{1-z}{1+z}$ 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














            4












            4








            4






            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 $h$.






            share|cite|improve this answer












            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 $h$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • $h(z) = ifrac{1-z}{1+z}$ 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











            2














            Another Counterexample :



            $f(Z)=frac{z+1}{2}$






            share|cite|improve this answer


























              2














              Another Counterexample :



              $f(Z)=frac{z+1}{2}$






              share|cite|improve this answer
























                2












                2








                2






                Another Counterexample :



                $f(Z)=frac{z+1}{2}$






                share|cite|improve this answer












                Another Counterexample :



                $f(Z)=frac{z+1}{2}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 at 8:09









                Shubham

                1,5921519




                1,5921519






























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