Power series and local minimum?












1














Let consider that $f(z)=sumlimits_{nge 0}a_nz^n$ in $mathcal{D}(O,R)$ with $0<Rle +infty$. If $mid fmid$ has a local minimum in $0$ then $fequiv 0$.



Is it linked to the maximum modulus principle ?



Thanks in advance !






complex-analysis power-series holomorphic-functions analytic-functions







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  • 2




    What does it mean for a complex-valued function to have a minimum? Do you mean $vert f vert$ has a minimum at 0? Cheers!
    – Robert Lewis
    58 mins ago






  • 1




    @RobertLewis Indeed my mistake !
    – Maman
    57 mins ago
















1














Let consider that $f(z)=sumlimits_{nge 0}a_nz^n$ in $mathcal{D}(O,R)$ with $0<Rle +infty$. If $mid fmid$ has a local minimum in $0$ then $fequiv 0$.



Is it linked to the maximum modulus principle ?



Thanks in advance !






complex-analysis power-series holomorphic-functions analytic-functions







share|cite|improve this question




















  • 2




    What does it mean for a complex-valued function to have a minimum? Do you mean $vert f vert$ has a minimum at 0? Cheers!
    – Robert Lewis
    58 mins ago






  • 1




    @RobertLewis Indeed my mistake !
    – Maman
    57 mins ago














1












1








1


1





Let consider that $f(z)=sumlimits_{nge 0}a_nz^n$ in $mathcal{D}(O,R)$ with $0<Rle +infty$. If $mid fmid$ has a local minimum in $0$ then $fequiv 0$.



Is it linked to the maximum modulus principle ?



Thanks in advance !






complex-analysis power-series holomorphic-functions analytic-functions







share|cite|improve this question















Let consider that $f(z)=sumlimits_{nge 0}a_nz^n$ in $mathcal{D}(O,R)$ with $0<Rle +infty$. If $mid fmid$ has a local minimum in $0$ then $fequiv 0$.



Is it linked to the maximum modulus principle ?



Thanks in advance !






complex-analysis power-series holomorphic-functions analytic-functions




complex-analysis power-series holomorphic-functions analytic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 56 mins ago

























asked 1 hour ago









Maman

1,144722




1,144722








  • 2




    What does it mean for a complex-valued function to have a minimum? Do you mean $vert f vert$ has a minimum at 0? Cheers!
    – Robert Lewis
    58 mins ago






  • 1




    @RobertLewis Indeed my mistake !
    – Maman
    57 mins ago














  • 2




    What does it mean for a complex-valued function to have a minimum? Do you mean $vert f vert$ has a minimum at 0? Cheers!
    – Robert Lewis
    58 mins ago






  • 1




    @RobertLewis Indeed my mistake !
    – Maman
    57 mins ago








2




2




What does it mean for a complex-valued function to have a minimum? Do you mean $vert f vert$ has a minimum at 0? Cheers!
– Robert Lewis
58 mins ago




What does it mean for a complex-valued function to have a minimum? Do you mean $vert f vert$ has a minimum at 0? Cheers!
– Robert Lewis
58 mins ago




1




1




@RobertLewis Indeed my mistake !
– Maman
57 mins ago




@RobertLewis Indeed my mistake !
– Maman
57 mins ago










2 Answers
2






active

oldest

votes


















4














The identity map $f : z mapsto z$ is such that $vert fvert$ has a local minimum at $0$. However $f$ is not the always vanishing map.



Hence the statement of the initial question is wrong.






share|cite|improve this answer





























    1














    The assertion appears to be false.



    Consider



    $f(z) = z^n, 1 le n in Bbb N; tag 1$



    writing



    $z = re^{itheta}, tag 2$



    we have



    $z^n = r^n e^{i n theta}, tag 3$



    whence



    $vert f(z) vert = vert z^n vert = r^n; tag 4$



    clearly then,



    $vert f(0) vert = 0 tag 5$



    so $0$ is an absolute minimum of $vert f(z) vert$; however,



    $f(z) ne 0, ; text{for} ; z ne 0. tag 6$






    share|cite|improve this answer

















    • 1




      Thank you maybe a hypothesis is missing...
      – Maman
      40 mins ago










    • That very well may be; but what?
      – Robert Lewis
      36 mins ago











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    2 Answers
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    2 Answers
    2






    active

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    active

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    active

    oldest

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    4














    The identity map $f : z mapsto z$ is such that $vert fvert$ has a local minimum at $0$. However $f$ is not the always vanishing map.



    Hence the statement of the initial question is wrong.






    share|cite|improve this answer


























      4














      The identity map $f : z mapsto z$ is such that $vert fvert$ has a local minimum at $0$. However $f$ is not the always vanishing map.



      Hence the statement of the initial question is wrong.






      share|cite|improve this answer
























        4












        4








        4






        The identity map $f : z mapsto z$ is such that $vert fvert$ has a local minimum at $0$. However $f$ is not the always vanishing map.



        Hence the statement of the initial question is wrong.






        share|cite|improve this answer












        The identity map $f : z mapsto z$ is such that $vert fvert$ has a local minimum at $0$. However $f$ is not the always vanishing map.



        Hence the statement of the initial question is wrong.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 53 mins ago









        mathcounterexamples.net

        24k21753




        24k21753























            1














            The assertion appears to be false.



            Consider



            $f(z) = z^n, 1 le n in Bbb N; tag 1$



            writing



            $z = re^{itheta}, tag 2$



            we have



            $z^n = r^n e^{i n theta}, tag 3$



            whence



            $vert f(z) vert = vert z^n vert = r^n; tag 4$



            clearly then,



            $vert f(0) vert = 0 tag 5$



            so $0$ is an absolute minimum of $vert f(z) vert$; however,



            $f(z) ne 0, ; text{for} ; z ne 0. tag 6$






            share|cite|improve this answer

















            • 1




              Thank you maybe a hypothesis is missing...
              – Maman
              40 mins ago










            • That very well may be; but what?
              – Robert Lewis
              36 mins ago
















            1














            The assertion appears to be false.



            Consider



            $f(z) = z^n, 1 le n in Bbb N; tag 1$



            writing



            $z = re^{itheta}, tag 2$



            we have



            $z^n = r^n e^{i n theta}, tag 3$



            whence



            $vert f(z) vert = vert z^n vert = r^n; tag 4$



            clearly then,



            $vert f(0) vert = 0 tag 5$



            so $0$ is an absolute minimum of $vert f(z) vert$; however,



            $f(z) ne 0, ; text{for} ; z ne 0. tag 6$






            share|cite|improve this answer

















            • 1




              Thank you maybe a hypothesis is missing...
              – Maman
              40 mins ago










            • That very well may be; but what?
              – Robert Lewis
              36 mins ago














            1












            1








            1






            The assertion appears to be false.



            Consider



            $f(z) = z^n, 1 le n in Bbb N; tag 1$



            writing



            $z = re^{itheta}, tag 2$



            we have



            $z^n = r^n e^{i n theta}, tag 3$



            whence



            $vert f(z) vert = vert z^n vert = r^n; tag 4$



            clearly then,



            $vert f(0) vert = 0 tag 5$



            so $0$ is an absolute minimum of $vert f(z) vert$; however,



            $f(z) ne 0, ; text{for} ; z ne 0. tag 6$






            share|cite|improve this answer












            The assertion appears to be false.



            Consider



            $f(z) = z^n, 1 le n in Bbb N; tag 1$



            writing



            $z = re^{itheta}, tag 2$



            we have



            $z^n = r^n e^{i n theta}, tag 3$



            whence



            $vert f(z) vert = vert z^n vert = r^n; tag 4$



            clearly then,



            $vert f(0) vert = 0 tag 5$



            so $0$ is an absolute minimum of $vert f(z) vert$; however,



            $f(z) ne 0, ; text{for} ; z ne 0. tag 6$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 46 mins ago









            Robert Lewis

            43.5k22863




            43.5k22863








            • 1




              Thank you maybe a hypothesis is missing...
              – Maman
              40 mins ago










            • That very well may be; but what?
              – Robert Lewis
              36 mins ago














            • 1




              Thank you maybe a hypothesis is missing...
              – Maman
              40 mins ago










            • That very well may be; but what?
              – Robert Lewis
              36 mins ago








            1




            1




            Thank you maybe a hypothesis is missing...
            – Maman
            40 mins ago




            Thank you maybe a hypothesis is missing...
            – Maman
            40 mins ago












            That very well may be; but what?
            – Robert Lewis
            36 mins ago




            That very well may be; but what?
            – Robert Lewis
            36 mins ago


















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