Let $f$ be holomorphic on an open connected subset $Omegasubseteq mathbb{C}$.












0












$begingroup$


Let $f$ be holomorphic on an open connected subset $Omegasubseteq mathbb{C}$. Assume $fnotequiv 0$ on $Omega$. If $mathcal{Z}={zinOmega:f^{(n)}(z)=0 hbox{ for all }n=0,1,2,...}$ then prove that $mathcal{Z}=emptyset$ without using identity principle.



Clearly if $mathcal{Z}neqemptyset$, then it implies that $f$ is identically zero. But I'm having trouble showing this in a formal manner.






complex-analysis







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$endgroup$












  • $begingroup$
    you did not define the set phi
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:34










  • $begingroup$
    Nevermind, it is the emptyset.
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:35










  • $begingroup$
    $mathcal{Z}$ is clearly closed in $Omega$ because it is the intersection of all of the closed sets given by $left(f^{(n)}right)^{-1}(0)$. That $mathcal{Z}$ is open follows by considering the radius of convergence of power series expansion of $f$ at each point of $mathcal{Z}$. Hence it is an open and closed subset of a connected set, and since it isn’t the entire set, it must be the empty set.
    $endgroup$
    – Adam Higgins
    Dec 29 '18 at 20:49


















0












$begingroup$


Let $f$ be holomorphic on an open connected subset $Omegasubseteq mathbb{C}$. Assume $fnotequiv 0$ on $Omega$. If $mathcal{Z}={zinOmega:f^{(n)}(z)=0 hbox{ for all }n=0,1,2,...}$ then prove that $mathcal{Z}=emptyset$ without using identity principle.



Clearly if $mathcal{Z}neqemptyset$, then it implies that $f$ is identically zero. But I'm having trouble showing this in a formal manner.






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    you did not define the set phi
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:34










  • $begingroup$
    Nevermind, it is the emptyset.
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:35










  • $begingroup$
    $mathcal{Z}$ is clearly closed in $Omega$ because it is the intersection of all of the closed sets given by $left(f^{(n)}right)^{-1}(0)$. That $mathcal{Z}$ is open follows by considering the radius of convergence of power series expansion of $f$ at each point of $mathcal{Z}$. Hence it is an open and closed subset of a connected set, and since it isn’t the entire set, it must be the empty set.
    $endgroup$
    – Adam Higgins
    Dec 29 '18 at 20:49
















0












0








0


1



$begingroup$


Let $f$ be holomorphic on an open connected subset $Omegasubseteq mathbb{C}$. Assume $fnotequiv 0$ on $Omega$. If $mathcal{Z}={zinOmega:f^{(n)}(z)=0 hbox{ for all }n=0,1,2,...}$ then prove that $mathcal{Z}=emptyset$ without using identity principle.



Clearly if $mathcal{Z}neqemptyset$, then it implies that $f$ is identically zero. But I'm having trouble showing this in a formal manner.






complex-analysis







share|cite|improve this question











$endgroup$




Let $f$ be holomorphic on an open connected subset $Omegasubseteq mathbb{C}$. Assume $fnotequiv 0$ on $Omega$. If $mathcal{Z}={zinOmega:f^{(n)}(z)=0 hbox{ for all }n=0,1,2,...}$ then prove that $mathcal{Z}=emptyset$ without using identity principle.



Clearly if $mathcal{Z}neqemptyset$, then it implies that $f$ is identically zero. But I'm having trouble showing this in a formal manner.






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 29 '18 at 20:44









mathcounterexamples.net

27k22158




27k22158










asked Dec 29 '18 at 20:31









Ya GYa G

536211




536211












  • $begingroup$
    you did not define the set phi
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:34










  • $begingroup$
    Nevermind, it is the emptyset.
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:35










  • $begingroup$
    $mathcal{Z}$ is clearly closed in $Omega$ because it is the intersection of all of the closed sets given by $left(f^{(n)}right)^{-1}(0)$. That $mathcal{Z}$ is open follows by considering the radius of convergence of power series expansion of $f$ at each point of $mathcal{Z}$. Hence it is an open and closed subset of a connected set, and since it isn’t the entire set, it must be the empty set.
    $endgroup$
    – Adam Higgins
    Dec 29 '18 at 20:49




















  • $begingroup$
    you did not define the set phi
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:34










  • $begingroup$
    Nevermind, it is the emptyset.
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:35










  • $begingroup$
    $mathcal{Z}$ is clearly closed in $Omega$ because it is the intersection of all of the closed sets given by $left(f^{(n)}right)^{-1}(0)$. That $mathcal{Z}$ is open follows by considering the radius of convergence of power series expansion of $f$ at each point of $mathcal{Z}$. Hence it is an open and closed subset of a connected set, and since it isn’t the entire set, it must be the empty set.
    $endgroup$
    – Adam Higgins
    Dec 29 '18 at 20:49


















$begingroup$
you did not define the set phi
$endgroup$
– M. Van
Dec 29 '18 at 20:34




$begingroup$
you did not define the set phi
$endgroup$
– M. Van
Dec 29 '18 at 20:34












$begingroup$
Nevermind, it is the emptyset.
$endgroup$
– M. Van
Dec 29 '18 at 20:35




$begingroup$
Nevermind, it is the emptyset.
$endgroup$
– M. Van
Dec 29 '18 at 20:35












$begingroup$
$mathcal{Z}$ is clearly closed in $Omega$ because it is the intersection of all of the closed sets given by $left(f^{(n)}right)^{-1}(0)$. That $mathcal{Z}$ is open follows by considering the radius of convergence of power series expansion of $f$ at each point of $mathcal{Z}$. Hence it is an open and closed subset of a connected set, and since it isn’t the entire set, it must be the empty set.
$endgroup$
– Adam Higgins
Dec 29 '18 at 20:49






$begingroup$
$mathcal{Z}$ is clearly closed in $Omega$ because it is the intersection of all of the closed sets given by $left(f^{(n)}right)^{-1}(0)$. That $mathcal{Z}$ is open follows by considering the radius of convergence of power series expansion of $f$ at each point of $mathcal{Z}$. Hence it is an open and closed subset of a connected set, and since it isn’t the entire set, it must be the empty set.
$endgroup$
– Adam Higgins
Dec 29 '18 at 20:49












2 Answers
2






active

oldest

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2












$begingroup$

Hint: By using power series expansions prove that $Z$ is open in $mathbb{C}$ and use a continuity argument to show $Omega-Z$ is open in $mathbb{C}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Presumably you only need that $Z$ and $Omega - Z$ are both open in $Omega$?
    $endgroup$
    – Adam Higgins
    Dec 29 '18 at 20:46










  • $begingroup$
    Yes you do character fill.
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:47



















0












$begingroup$

Suppose toward a contradiction, that $Zneq emptyset$ and let $z_0in Z$. Then, there is a disk $D(z_0,r)subset Omega$ such that $f(z)=sum a_n(z-z_0)^n$ for all $zin D$. Thus, $f=0$ in $D$ and so $Z$ is open. On the other hand, if $zin overline Z$, then there is a sequence $(z_k)subseteq Z$ such that $z_kto z$. As $f^{(n)}$ are continuous, we have $f^{(n)}(z_k)to f^{(n)}(z)$ so in fact $zin Z$, which implies that $Z$ is closed. But then, $Z$ and $Omegasetminus Z$ form a disconnection of $Omega$. So $Omegasetminus Z=emptyset$ and $f=0$ on $Omega$, which is a contradiction.






share|cite|improve this answer









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    2 Answers
    2






    active

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    2 Answers
    2






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    active

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    2












    $begingroup$

    Hint: By using power series expansions prove that $Z$ is open in $mathbb{C}$ and use a continuity argument to show $Omega-Z$ is open in $mathbb{C}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Presumably you only need that $Z$ and $Omega - Z$ are both open in $Omega$?
      $endgroup$
      – Adam Higgins
      Dec 29 '18 at 20:46










    • $begingroup$
      Yes you do character fill.
      $endgroup$
      – M. Van
      Dec 29 '18 at 20:47
















    2












    $begingroup$

    Hint: By using power series expansions prove that $Z$ is open in $mathbb{C}$ and use a continuity argument to show $Omega-Z$ is open in $mathbb{C}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Presumably you only need that $Z$ and $Omega - Z$ are both open in $Omega$?
      $endgroup$
      – Adam Higgins
      Dec 29 '18 at 20:46










    • $begingroup$
      Yes you do character fill.
      $endgroup$
      – M. Van
      Dec 29 '18 at 20:47














    2












    2








    2





    $begingroup$

    Hint: By using power series expansions prove that $Z$ is open in $mathbb{C}$ and use a continuity argument to show $Omega-Z$ is open in $mathbb{C}$.






    share|cite|improve this answer











    $endgroup$



    Hint: By using power series expansions prove that $Z$ is open in $mathbb{C}$ and use a continuity argument to show $Omega-Z$ is open in $mathbb{C}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 31 '18 at 7:09

























    answered Dec 29 '18 at 20:41









    M. VanM. Van

    2,680311




    2,680311












    • $begingroup$
      Presumably you only need that $Z$ and $Omega - Z$ are both open in $Omega$?
      $endgroup$
      – Adam Higgins
      Dec 29 '18 at 20:46










    • $begingroup$
      Yes you do character fill.
      $endgroup$
      – M. Van
      Dec 29 '18 at 20:47


















    • $begingroup$
      Presumably you only need that $Z$ and $Omega - Z$ are both open in $Omega$?
      $endgroup$
      – Adam Higgins
      Dec 29 '18 at 20:46










    • $begingroup$
      Yes you do character fill.
      $endgroup$
      – M. Van
      Dec 29 '18 at 20:47
















    $begingroup$
    Presumably you only need that $Z$ and $Omega - Z$ are both open in $Omega$?
    $endgroup$
    – Adam Higgins
    Dec 29 '18 at 20:46




    $begingroup$
    Presumably you only need that $Z$ and $Omega - Z$ are both open in $Omega$?
    $endgroup$
    – Adam Higgins
    Dec 29 '18 at 20:46












    $begingroup$
    Yes you do character fill.
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:47




    $begingroup$
    Yes you do character fill.
    $endgroup$
    – M. Van
    Dec 29 '18 at 20:47











    0












    $begingroup$

    Suppose toward a contradiction, that $Zneq emptyset$ and let $z_0in Z$. Then, there is a disk $D(z_0,r)subset Omega$ such that $f(z)=sum a_n(z-z_0)^n$ for all $zin D$. Thus, $f=0$ in $D$ and so $Z$ is open. On the other hand, if $zin overline Z$, then there is a sequence $(z_k)subseteq Z$ such that $z_kto z$. As $f^{(n)}$ are continuous, we have $f^{(n)}(z_k)to f^{(n)}(z)$ so in fact $zin Z$, which implies that $Z$ is closed. But then, $Z$ and $Omegasetminus Z$ form a disconnection of $Omega$. So $Omegasetminus Z=emptyset$ and $f=0$ on $Omega$, which is a contradiction.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Suppose toward a contradiction, that $Zneq emptyset$ and let $z_0in Z$. Then, there is a disk $D(z_0,r)subset Omega$ such that $f(z)=sum a_n(z-z_0)^n$ for all $zin D$. Thus, $f=0$ in $D$ and so $Z$ is open. On the other hand, if $zin overline Z$, then there is a sequence $(z_k)subseteq Z$ such that $z_kto z$. As $f^{(n)}$ are continuous, we have $f^{(n)}(z_k)to f^{(n)}(z)$ so in fact $zin Z$, which implies that $Z$ is closed. But then, $Z$ and $Omegasetminus Z$ form a disconnection of $Omega$. So $Omegasetminus Z=emptyset$ and $f=0$ on $Omega$, which is a contradiction.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Suppose toward a contradiction, that $Zneq emptyset$ and let $z_0in Z$. Then, there is a disk $D(z_0,r)subset Omega$ such that $f(z)=sum a_n(z-z_0)^n$ for all $zin D$. Thus, $f=0$ in $D$ and so $Z$ is open. On the other hand, if $zin overline Z$, then there is a sequence $(z_k)subseteq Z$ such that $z_kto z$. As $f^{(n)}$ are continuous, we have $f^{(n)}(z_k)to f^{(n)}(z)$ so in fact $zin Z$, which implies that $Z$ is closed. But then, $Z$ and $Omegasetminus Z$ form a disconnection of $Omega$. So $Omegasetminus Z=emptyset$ and $f=0$ on $Omega$, which is a contradiction.






        share|cite|improve this answer









        $endgroup$



        Suppose toward a contradiction, that $Zneq emptyset$ and let $z_0in Z$. Then, there is a disk $D(z_0,r)subset Omega$ such that $f(z)=sum a_n(z-z_0)^n$ for all $zin D$. Thus, $f=0$ in $D$ and so $Z$ is open. On the other hand, if $zin overline Z$, then there is a sequence $(z_k)subseteq Z$ such that $z_kto z$. As $f^{(n)}$ are continuous, we have $f^{(n)}(z_k)to f^{(n)}(z)$ so in fact $zin Z$, which implies that $Z$ is closed. But then, $Z$ and $Omegasetminus Z$ form a disconnection of $Omega$. So $Omegasetminus Z=emptyset$ and $f=0$ on $Omega$, which is a contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 20:53









        MatematletaMatematleta

        11.5k2920




        11.5k2920






























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