A function holomorphic on $mathbb{D}(0,2)$ and bounded on a unit circle is a polynomial












3















Suppose function $f(z)$ is holomorphic on $mathbb{D}(0,2)$ and $N>0$
is an integer such that: $$ |f^{(N)}(0)| = N! sup{|f(z)|: |z|=1} $$
show that $f(z) = cz^N$, $c in mathbb{C}$.




I have shown that since $f(z)$ is holomorphic in $mathbb{D}(0,2)$, then it has a power series expansion around zero.



$$ f(z) = sum_{n=0}^{infty} a_n z^n $$



Calculating the $N$-th derivative I got:



$$ |f^{(N)}(0)| = N! a_N$$



from which I conclude that $f(z)$ is bounded by $a_N$ on the unit circle. Therefore by maximum principle for holomorphic function we may also conclude that it is bounded in the unit disc. But I still can't figure out how to show that $f(z) = cz^N$.






complex-analysis power-series holomorphic-functions







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    3















    Suppose function $f(z)$ is holomorphic on $mathbb{D}(0,2)$ and $N>0$
    is an integer such that: $$ |f^{(N)}(0)| = N! sup{|f(z)|: |z|=1} $$
    show that $f(z) = cz^N$, $c in mathbb{C}$.




    I have shown that since $f(z)$ is holomorphic in $mathbb{D}(0,2)$, then it has a power series expansion around zero.



    $$ f(z) = sum_{n=0}^{infty} a_n z^n $$



    Calculating the $N$-th derivative I got:



    $$ |f^{(N)}(0)| = N! a_N$$



    from which I conclude that $f(z)$ is bounded by $a_N$ on the unit circle. Therefore by maximum principle for holomorphic function we may also conclude that it is bounded in the unit disc. But I still can't figure out how to show that $f(z) = cz^N$.






    complex-analysis power-series holomorphic-functions







    share|cite|improve this question









    New contributor




    s.kovalska is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      3












      3








      3








      Suppose function $f(z)$ is holomorphic on $mathbb{D}(0,2)$ and $N>0$
      is an integer such that: $$ |f^{(N)}(0)| = N! sup{|f(z)|: |z|=1} $$
      show that $f(z) = cz^N$, $c in mathbb{C}$.




      I have shown that since $f(z)$ is holomorphic in $mathbb{D}(0,2)$, then it has a power series expansion around zero.



      $$ f(z) = sum_{n=0}^{infty} a_n z^n $$



      Calculating the $N$-th derivative I got:



      $$ |f^{(N)}(0)| = N! a_N$$



      from which I conclude that $f(z)$ is bounded by $a_N$ on the unit circle. Therefore by maximum principle for holomorphic function we may also conclude that it is bounded in the unit disc. But I still can't figure out how to show that $f(z) = cz^N$.






      complex-analysis power-series holomorphic-functions







      share|cite|improve this question









      New contributor




      s.kovalska is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.












      Suppose function $f(z)$ is holomorphic on $mathbb{D}(0,2)$ and $N>0$
      is an integer such that: $$ |f^{(N)}(0)| = N! sup{|f(z)|: |z|=1} $$
      show that $f(z) = cz^N$, $c in mathbb{C}$.




      I have shown that since $f(z)$ is holomorphic in $mathbb{D}(0,2)$, then it has a power series expansion around zero.



      $$ f(z) = sum_{n=0}^{infty} a_n z^n $$



      Calculating the $N$-th derivative I got:



      $$ |f^{(N)}(0)| = N! a_N$$



      from which I conclude that $f(z)$ is bounded by $a_N$ on the unit circle. Therefore by maximum principle for holomorphic function we may also conclude that it is bounded in the unit disc. But I still can't figure out how to show that $f(z) = cz^N$.






      complex-analysis power-series holomorphic-functions




      complex-analysis power-series holomorphic-functions






      share|cite|improve this question









      New contributor




      s.kovalska is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      share|cite|improve this question








      edited 2 days ago







      s.kovalska













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      asked 2 days ago









      s.kovalskas.kovalska

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          2 Answers
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          3














          By Cauchy's formula,
          $$ f^{(N)} (0) = frac{N!}{2pi i} oint_{|z| = 1} frac{f(z) dz}{z^{N+1}} = frac{N!}{2pi } int_{0}^{2pi} e^{-iNtheta}f(e^{itheta}) dtheta.$$



          Thus
          $$ |f^{(N)}(0)| leq N!sup_{theta in [0, 2pi)} |f(e^{itheta})|,$$



          with equality if and only if



          $$ e^{-iNtheta} f(e^{itheta}) = c$$



          for some constant $c in mathbb C$.



          As the equality does hold by assumption, we have $f(z) = cz^N$ when $|z| = 1$.






          share|cite|improve this answer





















          • By the way, you wrote $f^{(N)}(0) = N! sup_{theta in [0, 2pi)} |f(e^{itheta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...)
            – Kenny Wong
            2 days ago












          • Yes, sure. That was a typo,it should have been a modulus. Thank you!
            – s.kovalska
            2 days ago



















          2














          For every $0 leq theta leq 1$, $f(e^{2ipitheta})e^{-2iNpitheta}$ has an absolute value not greater than $a_N$, but the integral over $[0;1]$ is exactly $a_N$.



          So it is constant equal to $a_N$, thus $f(z)=a_Nz^N$ on the unit circle, thus the equality holds on the whole disc.






          share|cite|improve this answer























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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            3














            By Cauchy's formula,
            $$ f^{(N)} (0) = frac{N!}{2pi i} oint_{|z| = 1} frac{f(z) dz}{z^{N+1}} = frac{N!}{2pi } int_{0}^{2pi} e^{-iNtheta}f(e^{itheta}) dtheta.$$



            Thus
            $$ |f^{(N)}(0)| leq N!sup_{theta in [0, 2pi)} |f(e^{itheta})|,$$



            with equality if and only if



            $$ e^{-iNtheta} f(e^{itheta}) = c$$



            for some constant $c in mathbb C$.



            As the equality does hold by assumption, we have $f(z) = cz^N$ when $|z| = 1$.






            share|cite|improve this answer





















            • By the way, you wrote $f^{(N)}(0) = N! sup_{theta in [0, 2pi)} |f(e^{itheta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...)
              – Kenny Wong
              2 days ago












            • Yes, sure. That was a typo,it should have been a modulus. Thank you!
              – s.kovalska
              2 days ago
















            3














            By Cauchy's formula,
            $$ f^{(N)} (0) = frac{N!}{2pi i} oint_{|z| = 1} frac{f(z) dz}{z^{N+1}} = frac{N!}{2pi } int_{0}^{2pi} e^{-iNtheta}f(e^{itheta}) dtheta.$$



            Thus
            $$ |f^{(N)}(0)| leq N!sup_{theta in [0, 2pi)} |f(e^{itheta})|,$$



            with equality if and only if



            $$ e^{-iNtheta} f(e^{itheta}) = c$$



            for some constant $c in mathbb C$.



            As the equality does hold by assumption, we have $f(z) = cz^N$ when $|z| = 1$.






            share|cite|improve this answer





















            • By the way, you wrote $f^{(N)}(0) = N! sup_{theta in [0, 2pi)} |f(e^{itheta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...)
              – Kenny Wong
              2 days ago












            • Yes, sure. That was a typo,it should have been a modulus. Thank you!
              – s.kovalska
              2 days ago














            3












            3








            3






            By Cauchy's formula,
            $$ f^{(N)} (0) = frac{N!}{2pi i} oint_{|z| = 1} frac{f(z) dz}{z^{N+1}} = frac{N!}{2pi } int_{0}^{2pi} e^{-iNtheta}f(e^{itheta}) dtheta.$$



            Thus
            $$ |f^{(N)}(0)| leq N!sup_{theta in [0, 2pi)} |f(e^{itheta})|,$$



            with equality if and only if



            $$ e^{-iNtheta} f(e^{itheta}) = c$$



            for some constant $c in mathbb C$.



            As the equality does hold by assumption, we have $f(z) = cz^N$ when $|z| = 1$.






            share|cite|improve this answer












            By Cauchy's formula,
            $$ f^{(N)} (0) = frac{N!}{2pi i} oint_{|z| = 1} frac{f(z) dz}{z^{N+1}} = frac{N!}{2pi } int_{0}^{2pi} e^{-iNtheta}f(e^{itheta}) dtheta.$$



            Thus
            $$ |f^{(N)}(0)| leq N!sup_{theta in [0, 2pi)} |f(e^{itheta})|,$$



            with equality if and only if



            $$ e^{-iNtheta} f(e^{itheta}) = c$$



            for some constant $c in mathbb C$.



            As the equality does hold by assumption, we have $f(z) = cz^N$ when $|z| = 1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Kenny WongKenny Wong

            18.3k21438




            18.3k21438












            • By the way, you wrote $f^{(N)}(0) = N! sup_{theta in [0, 2pi)} |f(e^{itheta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...)
              – Kenny Wong
              2 days ago












            • Yes, sure. That was a typo,it should have been a modulus. Thank you!
              – s.kovalska
              2 days ago


















            • By the way, you wrote $f^{(N)}(0) = N! sup_{theta in [0, 2pi)} |f(e^{itheta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...)
              – Kenny Wong
              2 days ago












            • Yes, sure. That was a typo,it should have been a modulus. Thank you!
              – s.kovalska
              2 days ago
















            By the way, you wrote $f^{(N)}(0) = N! sup_{theta in [0, 2pi)} |f(e^{itheta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...)
            – Kenny Wong
            2 days ago






            By the way, you wrote $f^{(N)}(0) = N! sup_{theta in [0, 2pi)} |f(e^{itheta})|$ (no modulus sign on the $f^{(N)}(0)$). This implies that $f^{(N)}(0)$ is real and positive, so $c$ is real and positive too. (Not sure if that was a typo though...)
            – Kenny Wong
            2 days ago














            Yes, sure. That was a typo,it should have been a modulus. Thank you!
            – s.kovalska
            2 days ago




            Yes, sure. That was a typo,it should have been a modulus. Thank you!
            – s.kovalska
            2 days ago











            2














            For every $0 leq theta leq 1$, $f(e^{2ipitheta})e^{-2iNpitheta}$ has an absolute value not greater than $a_N$, but the integral over $[0;1]$ is exactly $a_N$.



            So it is constant equal to $a_N$, thus $f(z)=a_Nz^N$ on the unit circle, thus the equality holds on the whole disc.






            share|cite|improve this answer




























              2














              For every $0 leq theta leq 1$, $f(e^{2ipitheta})e^{-2iNpitheta}$ has an absolute value not greater than $a_N$, but the integral over $[0;1]$ is exactly $a_N$.



              So it is constant equal to $a_N$, thus $f(z)=a_Nz^N$ on the unit circle, thus the equality holds on the whole disc.






              share|cite|improve this answer


























                2












                2








                2






                For every $0 leq theta leq 1$, $f(e^{2ipitheta})e^{-2iNpitheta}$ has an absolute value not greater than $a_N$, but the integral over $[0;1]$ is exactly $a_N$.



                So it is constant equal to $a_N$, thus $f(z)=a_Nz^N$ on the unit circle, thus the equality holds on the whole disc.






                share|cite|improve this answer














                For every $0 leq theta leq 1$, $f(e^{2ipitheta})e^{-2iNpitheta}$ has an absolute value not greater than $a_N$, but the integral over $[0;1]$ is exactly $a_N$.



                So it is constant equal to $a_N$, thus $f(z)=a_Nz^N$ on the unit circle, thus the equality holds on the whole disc.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered 2 days ago









                MindlackMindlack

                2,05217




                2,05217






















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