Proving a certain holomorphic function is polinomic












0












$begingroup$


Suppose we have an holomorphic function $f$ on the open unit disk $D(0,1)$ s.t.:



$$forall rin (0,1) exists nin mathbb{N}| max_{C(0,r)}|f|=r^n$$



prove that $f$ is polinomic. Honestly I don't know where to start, every approach I take leads me to triviality or useless formulas.






complex-analysis holomorphic-functions cauchy-integral-formula maximum-principle







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Suppose we have an holomorphic function $f$ on the open unit disk $D(0,1)$ s.t.:



    $$forall rin (0,1) exists nin mathbb{N}| max_{C(0,r)}|f|=r^n$$



    prove that $f$ is polinomic. Honestly I don't know where to start, every approach I take leads me to triviality or useless formulas.






    complex-analysis holomorphic-functions cauchy-integral-formula maximum-principle







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose we have an holomorphic function $f$ on the open unit disk $D(0,1)$ s.t.:



      $$forall rin (0,1) exists nin mathbb{N}| max_{C(0,r)}|f|=r^n$$



      prove that $f$ is polinomic. Honestly I don't know where to start, every approach I take leads me to triviality or useless formulas.






      complex-analysis holomorphic-functions cauchy-integral-formula maximum-principle







      share|cite|improve this question









      $endgroup$




      Suppose we have an holomorphic function $f$ on the open unit disk $D(0,1)$ s.t.:



      $$forall rin (0,1) exists nin mathbb{N}| max_{C(0,r)}|f|=r^n$$



      prove that $f$ is polinomic. Honestly I don't know where to start, every approach I take leads me to triviality or useless formulas.






      complex-analysis holomorphic-functions cauchy-integral-formula maximum-principle




      complex-analysis holomorphic-functions cauchy-integral-formula maximum-principle






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 10:29









      Renato FaraoneRenato Faraone

      2,33111627




      2,33111627






















          1 Answer
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          $begingroup$

          First note that $f$ has a zero of $n$-th order at $z=0$. Let $g(z)=frac{f(z)}{z^n}$ and observe that $g$ attains maximum modulus in the interior.



          $textbf{EDIT:}$ I've misread the problem. We cannot draw directly from the statement the fact that there is an independent $n$ such that $max_{|z|=r} |f(z)|=r^n$ holds. What we have is $n(r)geq 1$ such that $$ max_{|z|=r}|f(z)|=r^{n(r)}.$$To see there is independent $n$, note that
          $$
          phi(r) = max_{|z|=r}|f(z)|
          $$           is continuous in $rin(0,1)$ by the continuity of $f$. Thus, a positive integer-valued function
          $$
          n(r) = frac{phi(r)}{log r}
          $$           is also continuous, establishing $n(r) equiv n$. If you feel this argument tricky, you can solve the problem inductively, by noting that if there are $r_1<r_2$ such that $n(r_1)=n(r_2)=m$ attains minimum of $n(cdot)$, then we can argue by maximum modulus principle that $frac{f(z)}{z^m}$ is a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I can't see why it has a zero at the origin
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:18






          • 1




            $begingroup$
            Just take $rto 0$ and use continuity. Divide $f$ by $z$ and do the same thing $n$ times.
            $endgroup$
            – Song
            Dec 10 '18 at 11:21










          • $begingroup$
            Why does $g$ attains its maximum modulus in the interior?
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:33






          • 1




            $begingroup$
            It is because $max_{|z|=r} |g(z)|=1$ for $0<r<1$. I'll explain it in more detail. So there is $w$ with $|w|=frac{1}{3}$ such that $|g(w)|=1=max_{|z|=frac{1}{2}}|g(z)|$. Now, apply Maximum modulus principle.
            $endgroup$
            – Song
            Dec 10 '18 at 11:39










          • $begingroup$
            @Song The hypothesis says for each $r$ there exist $n$ such that $cdots$; so $n$ varies with $r$. Aren't you assuming that $n$ is independent of $r$?.
            $endgroup$
            – Kavi Rama Murthy
            Dec 11 '18 at 7:46













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          2












          $begingroup$

          First note that $f$ has a zero of $n$-th order at $z=0$. Let $g(z)=frac{f(z)}{z^n}$ and observe that $g$ attains maximum modulus in the interior.



          $textbf{EDIT:}$ I've misread the problem. We cannot draw directly from the statement the fact that there is an independent $n$ such that $max_{|z|=r} |f(z)|=r^n$ holds. What we have is $n(r)geq 1$ such that $$ max_{|z|=r}|f(z)|=r^{n(r)}.$$To see there is independent $n$, note that
          $$
          phi(r) = max_{|z|=r}|f(z)|
          $$           is continuous in $rin(0,1)$ by the continuity of $f$. Thus, a positive integer-valued function
          $$
          n(r) = frac{phi(r)}{log r}
          $$           is also continuous, establishing $n(r) equiv n$. If you feel this argument tricky, you can solve the problem inductively, by noting that if there are $r_1<r_2$ such that $n(r_1)=n(r_2)=m$ attains minimum of $n(cdot)$, then we can argue by maximum modulus principle that $frac{f(z)}{z^m}$ is a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I can't see why it has a zero at the origin
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:18






          • 1




            $begingroup$
            Just take $rto 0$ and use continuity. Divide $f$ by $z$ and do the same thing $n$ times.
            $endgroup$
            – Song
            Dec 10 '18 at 11:21










          • $begingroup$
            Why does $g$ attains its maximum modulus in the interior?
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:33






          • 1




            $begingroup$
            It is because $max_{|z|=r} |g(z)|=1$ for $0<r<1$. I'll explain it in more detail. So there is $w$ with $|w|=frac{1}{3}$ such that $|g(w)|=1=max_{|z|=frac{1}{2}}|g(z)|$. Now, apply Maximum modulus principle.
            $endgroup$
            – Song
            Dec 10 '18 at 11:39










          • $begingroup$
            @Song The hypothesis says for each $r$ there exist $n$ such that $cdots$; so $n$ varies with $r$. Aren't you assuming that $n$ is independent of $r$?.
            $endgroup$
            – Kavi Rama Murthy
            Dec 11 '18 at 7:46


















          2












          $begingroup$

          First note that $f$ has a zero of $n$-th order at $z=0$. Let $g(z)=frac{f(z)}{z^n}$ and observe that $g$ attains maximum modulus in the interior.



          $textbf{EDIT:}$ I've misread the problem. We cannot draw directly from the statement the fact that there is an independent $n$ such that $max_{|z|=r} |f(z)|=r^n$ holds. What we have is $n(r)geq 1$ such that $$ max_{|z|=r}|f(z)|=r^{n(r)}.$$To see there is independent $n$, note that
          $$
          phi(r) = max_{|z|=r}|f(z)|
          $$           is continuous in $rin(0,1)$ by the continuity of $f$. Thus, a positive integer-valued function
          $$
          n(r) = frac{phi(r)}{log r}
          $$           is also continuous, establishing $n(r) equiv n$. If you feel this argument tricky, you can solve the problem inductively, by noting that if there are $r_1<r_2$ such that $n(r_1)=n(r_2)=m$ attains minimum of $n(cdot)$, then we can argue by maximum modulus principle that $frac{f(z)}{z^m}$ is a constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I can't see why it has a zero at the origin
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:18






          • 1




            $begingroup$
            Just take $rto 0$ and use continuity. Divide $f$ by $z$ and do the same thing $n$ times.
            $endgroup$
            – Song
            Dec 10 '18 at 11:21










          • $begingroup$
            Why does $g$ attains its maximum modulus in the interior?
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:33






          • 1




            $begingroup$
            It is because $max_{|z|=r} |g(z)|=1$ for $0<r<1$. I'll explain it in more detail. So there is $w$ with $|w|=frac{1}{3}$ such that $|g(w)|=1=max_{|z|=frac{1}{2}}|g(z)|$. Now, apply Maximum modulus principle.
            $endgroup$
            – Song
            Dec 10 '18 at 11:39










          • $begingroup$
            @Song The hypothesis says for each $r$ there exist $n$ such that $cdots$; so $n$ varies with $r$. Aren't you assuming that $n$ is independent of $r$?.
            $endgroup$
            – Kavi Rama Murthy
            Dec 11 '18 at 7:46
















          2












          2








          2





          $begingroup$

          First note that $f$ has a zero of $n$-th order at $z=0$. Let $g(z)=frac{f(z)}{z^n}$ and observe that $g$ attains maximum modulus in the interior.



          $textbf{EDIT:}$ I've misread the problem. We cannot draw directly from the statement the fact that there is an independent $n$ such that $max_{|z|=r} |f(z)|=r^n$ holds. What we have is $n(r)geq 1$ such that $$ max_{|z|=r}|f(z)|=r^{n(r)}.$$To see there is independent $n$, note that
          $$
          phi(r) = max_{|z|=r}|f(z)|
          $$           is continuous in $rin(0,1)$ by the continuity of $f$. Thus, a positive integer-valued function
          $$
          n(r) = frac{phi(r)}{log r}
          $$           is also continuous, establishing $n(r) equiv n$. If you feel this argument tricky, you can solve the problem inductively, by noting that if there are $r_1<r_2$ such that $n(r_1)=n(r_2)=m$ attains minimum of $n(cdot)$, then we can argue by maximum modulus principle that $frac{f(z)}{z^m}$ is a constant.






          share|cite|improve this answer











          $endgroup$



          First note that $f$ has a zero of $n$-th order at $z=0$. Let $g(z)=frac{f(z)}{z^n}$ and observe that $g$ attains maximum modulus in the interior.



          $textbf{EDIT:}$ I've misread the problem. We cannot draw directly from the statement the fact that there is an independent $n$ such that $max_{|z|=r} |f(z)|=r^n$ holds. What we have is $n(r)geq 1$ such that $$ max_{|z|=r}|f(z)|=r^{n(r)}.$$To see there is independent $n$, note that
          $$
          phi(r) = max_{|z|=r}|f(z)|
          $$           is continuous in $rin(0,1)$ by the continuity of $f$. Thus, a positive integer-valued function
          $$
          n(r) = frac{phi(r)}{log r}
          $$           is also continuous, establishing $n(r) equiv n$. If you feel this argument tricky, you can solve the problem inductively, by noting that if there are $r_1<r_2$ such that $n(r_1)=n(r_2)=m$ attains minimum of $n(cdot)$, then we can argue by maximum modulus principle that $frac{f(z)}{z^m}$ is a constant.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 9:03

























          answered Dec 10 '18 at 11:05









          SongSong

          11.3k628




          11.3k628












          • $begingroup$
            I can't see why it has a zero at the origin
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:18






          • 1




            $begingroup$
            Just take $rto 0$ and use continuity. Divide $f$ by $z$ and do the same thing $n$ times.
            $endgroup$
            – Song
            Dec 10 '18 at 11:21










          • $begingroup$
            Why does $g$ attains its maximum modulus in the interior?
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:33






          • 1




            $begingroup$
            It is because $max_{|z|=r} |g(z)|=1$ for $0<r<1$. I'll explain it in more detail. So there is $w$ with $|w|=frac{1}{3}$ such that $|g(w)|=1=max_{|z|=frac{1}{2}}|g(z)|$. Now, apply Maximum modulus principle.
            $endgroup$
            – Song
            Dec 10 '18 at 11:39










          • $begingroup$
            @Song The hypothesis says for each $r$ there exist $n$ such that $cdots$; so $n$ varies with $r$. Aren't you assuming that $n$ is independent of $r$?.
            $endgroup$
            – Kavi Rama Murthy
            Dec 11 '18 at 7:46




















          • $begingroup$
            I can't see why it has a zero at the origin
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:18






          • 1




            $begingroup$
            Just take $rto 0$ and use continuity. Divide $f$ by $z$ and do the same thing $n$ times.
            $endgroup$
            – Song
            Dec 10 '18 at 11:21










          • $begingroup$
            Why does $g$ attains its maximum modulus in the interior?
            $endgroup$
            – Renato Faraone
            Dec 10 '18 at 11:33






          • 1




            $begingroup$
            It is because $max_{|z|=r} |g(z)|=1$ for $0<r<1$. I'll explain it in more detail. So there is $w$ with $|w|=frac{1}{3}$ such that $|g(w)|=1=max_{|z|=frac{1}{2}}|g(z)|$. Now, apply Maximum modulus principle.
            $endgroup$
            – Song
            Dec 10 '18 at 11:39










          • $begingroup$
            @Song The hypothesis says for each $r$ there exist $n$ such that $cdots$; so $n$ varies with $r$. Aren't you assuming that $n$ is independent of $r$?.
            $endgroup$
            – Kavi Rama Murthy
            Dec 11 '18 at 7:46


















          $begingroup$
          I can't see why it has a zero at the origin
          $endgroup$
          – Renato Faraone
          Dec 10 '18 at 11:18




          $begingroup$
          I can't see why it has a zero at the origin
          $endgroup$
          – Renato Faraone
          Dec 10 '18 at 11:18




          1




          1




          $begingroup$
          Just take $rto 0$ and use continuity. Divide $f$ by $z$ and do the same thing $n$ times.
          $endgroup$
          – Song
          Dec 10 '18 at 11:21




          $begingroup$
          Just take $rto 0$ and use continuity. Divide $f$ by $z$ and do the same thing $n$ times.
          $endgroup$
          – Song
          Dec 10 '18 at 11:21












          $begingroup$
          Why does $g$ attains its maximum modulus in the interior?
          $endgroup$
          – Renato Faraone
          Dec 10 '18 at 11:33




          $begingroup$
          Why does $g$ attains its maximum modulus in the interior?
          $endgroup$
          – Renato Faraone
          Dec 10 '18 at 11:33




          1




          1




          $begingroup$
          It is because $max_{|z|=r} |g(z)|=1$ for $0<r<1$. I'll explain it in more detail. So there is $w$ with $|w|=frac{1}{3}$ such that $|g(w)|=1=max_{|z|=frac{1}{2}}|g(z)|$. Now, apply Maximum modulus principle.
          $endgroup$
          – Song
          Dec 10 '18 at 11:39




          $begingroup$
          It is because $max_{|z|=r} |g(z)|=1$ for $0<r<1$. I'll explain it in more detail. So there is $w$ with $|w|=frac{1}{3}$ such that $|g(w)|=1=max_{|z|=frac{1}{2}}|g(z)|$. Now, apply Maximum modulus principle.
          $endgroup$
          – Song
          Dec 10 '18 at 11:39












          $begingroup$
          @Song The hypothesis says for each $r$ there exist $n$ such that $cdots$; so $n$ varies with $r$. Aren't you assuming that $n$ is independent of $r$?.
          $endgroup$
          – Kavi Rama Murthy
          Dec 11 '18 at 7:46






          $begingroup$
          @Song The hypothesis says for each $r$ there exist $n$ such that $cdots$; so $n$ varies with $r$. Aren't you assuming that $n$ is independent of $r$?.
          $endgroup$
          – Kavi Rama Murthy
          Dec 11 '18 at 7:46




















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