Holomorphic function $F:Bbb Hto Bbb C$ having the bounded sequence ${ir_n}$ as zeros.












2












$begingroup$


Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.





  1. Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
    $F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
    a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.

  2. If $sum r_n<infty$, it is possible to construct a bounded function
    on the upper half-plane with zeros precisely at the points $ir_n$.




There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?






complex-analysis







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.





    1. Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
      $F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
      a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.

    2. If $sum r_n<infty$, it is possible to construct a bounded function
      on the upper half-plane with zeros precisely at the points $ir_n$.




    There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?






    complex-analysis







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.





      1. Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
        $F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
        a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.

      2. If $sum r_n<infty$, it is possible to construct a bounded function
        on the upper half-plane with zeros precisely at the points $ir_n$.




      There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?






      complex-analysis







      share|cite|improve this question











      $endgroup$




      Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.





      1. Suppose that $F:mathbb{H}tomathbb{C}$ is holomorphic and bounded. Also, suppose
        $F(z)$ vanishes when $z=ir_n$, $n=1,2,3,ldots,$ where ${r_n}$ is
        a bounded sequence of positive numbers. Prove that if $sum r_n=infty$ then $F=0$.

      2. If $sum r_n<infty$, it is possible to construct a bounded function
        on the upper half-plane with zeros precisely at the points $ir_n$.




      There is something weird about the first part. If the sequence ${ir_n}$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $F$ accumulate in $Bbb H$ and $F$ is zero. Am I missing something here?






      complex-analysis




      complex-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 20:18







      UserA

















      asked Dec 16 '18 at 20:08









      UserAUserA

      505216




      505216






















          2 Answers
          2






          active

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          1












          $begingroup$

          If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
          sum_n Im (z_n)<infty.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:36










          • $begingroup$
            That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
            $endgroup$
            – Song
            Dec 16 '18 at 20:38












          • $begingroup$
            Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:40












          • $begingroup$
            I guess that will work :)
            $endgroup$
            – Song
            Dec 16 '18 at 20:43










          • $begingroup$
            what about the second part?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:46



















          1












          $begingroup$

          The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            1












            $begingroup$

            If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
            sum_n Im (z_n)<infty.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:36










            • $begingroup$
              That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
              $endgroup$
              – Song
              Dec 16 '18 at 20:38












            • $begingroup$
              Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:40












            • $begingroup$
              I guess that will work :)
              $endgroup$
              – Song
              Dec 16 '18 at 20:43










            • $begingroup$
              what about the second part?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:46
















            1












            $begingroup$

            If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
            sum_n Im (z_n)<infty.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:36










            • $begingroup$
              That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
              $endgroup$
              – Song
              Dec 16 '18 at 20:38












            • $begingroup$
              Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:40












            • $begingroup$
              I guess that will work :)
              $endgroup$
              – Song
              Dec 16 '18 at 20:43










            • $begingroup$
              what about the second part?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:46














            1












            1








            1





            $begingroup$

            If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
            sum_n Im (z_n)<infty.$$






            share|cite|improve this answer











            $endgroup$



            If ${ir_n}$ accumulates at a point other than $0$, then $F=0$ is trivial as you said. But the problem is requiring you to show that if ${ir_n}$ accumulates at $0$ and the convergence $r_nto 0$ is "slow" enough to make $sum_n r_n =infty$, then $F$ must be $0$. Put differently, if $Fneq 0$ is bounded on the upper half plane and ${z_n}$ are zeros of $F$, then it must be that $Im(z_n)to 0 $ fast enough to make $$
            sum_n Im (z_n)<infty.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 16 '18 at 20:30

























            answered Dec 16 '18 at 20:22









            SongSong

            8,924627




            8,924627












            • $begingroup$
              what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:36










            • $begingroup$
              That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
              $endgroup$
              – Song
              Dec 16 '18 at 20:38












            • $begingroup$
              Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:40












            • $begingroup$
              I guess that will work :)
              $endgroup$
              – Song
              Dec 16 '18 at 20:43










            • $begingroup$
              what about the second part?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:46


















            • $begingroup$
              what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:36










            • $begingroup$
              That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
              $endgroup$
              – Song
              Dec 16 '18 at 20:38












            • $begingroup$
              Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:40












            • $begingroup$
              I guess that will work :)
              $endgroup$
              – Song
              Dec 16 '18 at 20:43










            • $begingroup$
              what about the second part?
              $endgroup$
              – UserA
              Dec 16 '18 at 20:46
















            $begingroup$
            what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:36




            $begingroup$
            what do you suggest in this case? Does the Jensen formula help? Should we work with the unit disk instead of the upper half plane?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:36












            $begingroup$
            That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
            $endgroup$
            – Song
            Dec 16 '18 at 20:38






            $begingroup$
            That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = frac{aw+b}{cw+d}$ or something like this.
            $endgroup$
            – Song
            Dec 16 '18 at 20:38














            $begingroup$
            Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:40






            $begingroup$
            Use the map $T:Bbb Dto Bbb H$ defined by $zmapsto ifrac{1-z}{1+z}$?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:40














            $begingroup$
            I guess that will work :)
            $endgroup$
            – Song
            Dec 16 '18 at 20:43




            $begingroup$
            I guess that will work :)
            $endgroup$
            – Song
            Dec 16 '18 at 20:43












            $begingroup$
            what about the second part?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:46




            $begingroup$
            what about the second part?
            $endgroup$
            – UserA
            Dec 16 '18 at 20:46











            1












            $begingroup$

            The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.






                share|cite|improve this answer









                $endgroup$



                The sequence $(ir_n)$ accumulates at a point of $mathbb{C}$, but not necessarily at a point of $mathbb{H}$. Indeed, if $r_nto 0$ then they accumulate only at $0$, which is not in $mathbb{H}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 16 '18 at 20:16









                Eric WofseyEric Wofsey

                182k12209337




                182k12209337






























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