Prove that there exists complex numbers $a$ and $c$ such that $f(z)=ae^{cz}$.












1












$begingroup$


Let $f$ be an entire function such that $|f'(z)|leq|f(z)|$ for all $zin mathbb{C}$. Prove that there exist complex numbers $a$ and $c$ such that $f(z)=ae^{cz}$ for all $zinmathbb{C}$.



My try: take $g(z)$ such that
$$
f(z)=e^{g(z)}.
$$
Then by chain rule
$$
f'(z)=e^{g(z)}g'(z).
$$
Now we have
$$
|g'(z)||f(z)|=|g'(z)||e^{g(z)}|=|f'(z)|leq |f(z)|
$$
which implies
$$
|g'(z)|leq 1.
$$
Thus $g(z)=cz$, and then $f(z)=e^{cz}$. However, how to make $g(z)$ is entire function? and where is the constant $a$?






complex-analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $f$ be an entire function such that $|f'(z)|leq|f(z)|$ for all $zin mathbb{C}$. Prove that there exist complex numbers $a$ and $c$ such that $f(z)=ae^{cz}$ for all $zinmathbb{C}$.



    My try: take $g(z)$ such that
    $$
    f(z)=e^{g(z)}.
    $$
    Then by chain rule
    $$
    f'(z)=e^{g(z)}g'(z).
    $$
    Now we have
    $$
    |g'(z)||f(z)|=|g'(z)||e^{g(z)}|=|f'(z)|leq |f(z)|
    $$
    which implies
    $$
    |g'(z)|leq 1.
    $$
    Thus $g(z)=cz$, and then $f(z)=e^{cz}$. However, how to make $g(z)$ is entire function? and where is the constant $a$?






    complex-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $f$ be an entire function such that $|f'(z)|leq|f(z)|$ for all $zin mathbb{C}$. Prove that there exist complex numbers $a$ and $c$ such that $f(z)=ae^{cz}$ for all $zinmathbb{C}$.



      My try: take $g(z)$ such that
      $$
      f(z)=e^{g(z)}.
      $$
      Then by chain rule
      $$
      f'(z)=e^{g(z)}g'(z).
      $$
      Now we have
      $$
      |g'(z)||f(z)|=|g'(z)||e^{g(z)}|=|f'(z)|leq |f(z)|
      $$
      which implies
      $$
      |g'(z)|leq 1.
      $$
      Thus $g(z)=cz$, and then $f(z)=e^{cz}$. However, how to make $g(z)$ is entire function? and where is the constant $a$?






      complex-analysis







      share|cite|improve this question









      $endgroup$




      Let $f$ be an entire function such that $|f'(z)|leq|f(z)|$ for all $zin mathbb{C}$. Prove that there exist complex numbers $a$ and $c$ such that $f(z)=ae^{cz}$ for all $zinmathbb{C}$.



      My try: take $g(z)$ such that
      $$
      f(z)=e^{g(z)}.
      $$
      Then by chain rule
      $$
      f'(z)=e^{g(z)}g'(z).
      $$
      Now we have
      $$
      |g'(z)||f(z)|=|g'(z)||e^{g(z)}|=|f'(z)|leq |f(z)|
      $$
      which implies
      $$
      |g'(z)|leq 1.
      $$
      Thus $g(z)=cz$, and then $f(z)=e^{cz}$. However, how to make $g(z)$ is entire function? and where is the constant $a$?






      complex-analysis




      complex-analysis






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 6 '18 at 22:51









      whereamIwhereamI

      313114




      313114






















          2 Answers
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          $begingroup$

          You have to justify writing $f$ as $e^{g}$. Here is a correct solution: If $f equiv 0$ we can take $a=0$ and $c=1$. Otherwise the zeros of $f$ form a set ${z_n}$ with no limit points. On $mathbb C setminus {z_n}$ we have $|frac {f'(z)} {f(z)}| leq 1$. This implies that $frac {f'(z)} {f(z)}$ has a removable singularity at each $z_n$. Hence it extends to a bounded entire function. By Louiville's Theorem we have $frac {f'(z)} {f(z)}=A$ for some constant $A$. Now observe that $(e^{-Az}f(z))'=0$ so $e^{-Az}f(z)=B$ for some constant $B$. Hence $f(z)=Be^{Az}$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Note that $f$ has not zeros. In fact, if $a$ is a zero of $f$ of order $ninmathbb N$, then $a$ is a zero of $f'$ os order $n-1$. But then, near $a$, the inequality $bigllvert f'(z)bigrrvertleqslantbigllvert f(z)bigrrvert$ means$$bigllvert na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+cdotsbigrrvertleqslantbigllvert a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+cdotsbigrrvert.$$Now, dividing both sides by $bigllvert(z-a)^nbigrrvert$, you get that$$bigllvert na_n(z-a)^{-1}+(n+1)a_{n+1}+cdotsbigrrvertleqslantbigllvert a_n+a_{n+1}(z-a)+cdotsbigrrvert,$$which is impossible, because the LHS tends to $+infty$ when $zto a$, whereas the RHS tends to $lvert a_nrvert$.



            So, since $f$ has no zeros and $mathbb C$ is simply connected, there is indeed an entire function $g$ such that $f=e^g$. You proved that $g'$ is bounded. Therefore, by Liouville's therem, it is constant. So there are constants $k$ and $c$ such that $(forall zinmathbb{C}):g(z)=cz+k$and therefore$$(forall zinmathbb{C}):f(z)=e^{g(z)}=e^{cz+k}=e^ke^{cz}.$$






            share|cite|improve this answer









            $endgroup$













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              3












              $begingroup$

              You have to justify writing $f$ as $e^{g}$. Here is a correct solution: If $f equiv 0$ we can take $a=0$ and $c=1$. Otherwise the zeros of $f$ form a set ${z_n}$ with no limit points. On $mathbb C setminus {z_n}$ we have $|frac {f'(z)} {f(z)}| leq 1$. This implies that $frac {f'(z)} {f(z)}$ has a removable singularity at each $z_n$. Hence it extends to a bounded entire function. By Louiville's Theorem we have $frac {f'(z)} {f(z)}=A$ for some constant $A$. Now observe that $(e^{-Az}f(z))'=0$ so $e^{-Az}f(z)=B$ for some constant $B$. Hence $f(z)=Be^{Az}$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                You have to justify writing $f$ as $e^{g}$. Here is a correct solution: If $f equiv 0$ we can take $a=0$ and $c=1$. Otherwise the zeros of $f$ form a set ${z_n}$ with no limit points. On $mathbb C setminus {z_n}$ we have $|frac {f'(z)} {f(z)}| leq 1$. This implies that $frac {f'(z)} {f(z)}$ has a removable singularity at each $z_n$. Hence it extends to a bounded entire function. By Louiville's Theorem we have $frac {f'(z)} {f(z)}=A$ for some constant $A$. Now observe that $(e^{-Az}f(z))'=0$ so $e^{-Az}f(z)=B$ for some constant $B$. Hence $f(z)=Be^{Az}$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You have to justify writing $f$ as $e^{g}$. Here is a correct solution: If $f equiv 0$ we can take $a=0$ and $c=1$. Otherwise the zeros of $f$ form a set ${z_n}$ with no limit points. On $mathbb C setminus {z_n}$ we have $|frac {f'(z)} {f(z)}| leq 1$. This implies that $frac {f'(z)} {f(z)}$ has a removable singularity at each $z_n$. Hence it extends to a bounded entire function. By Louiville's Theorem we have $frac {f'(z)} {f(z)}=A$ for some constant $A$. Now observe that $(e^{-Az}f(z))'=0$ so $e^{-Az}f(z)=B$ for some constant $B$. Hence $f(z)=Be^{Az}$.






                  share|cite|improve this answer









                  $endgroup$



                  You have to justify writing $f$ as $e^{g}$. Here is a correct solution: If $f equiv 0$ we can take $a=0$ and $c=1$. Otherwise the zeros of $f$ form a set ${z_n}$ with no limit points. On $mathbb C setminus {z_n}$ we have $|frac {f'(z)} {f(z)}| leq 1$. This implies that $frac {f'(z)} {f(z)}$ has a removable singularity at each $z_n$. Hence it extends to a bounded entire function. By Louiville's Theorem we have $frac {f'(z)} {f(z)}=A$ for some constant $A$. Now observe that $(e^{-Az}f(z))'=0$ so $e^{-Az}f(z)=B$ for some constant $B$. Hence $f(z)=Be^{Az}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 23:23









                  Kavi Rama MurthyKavi Rama Murthy

                  53.2k32055




                  53.2k32055























                      2












                      $begingroup$

                      Note that $f$ has not zeros. In fact, if $a$ is a zero of $f$ of order $ninmathbb N$, then $a$ is a zero of $f'$ os order $n-1$. But then, near $a$, the inequality $bigllvert f'(z)bigrrvertleqslantbigllvert f(z)bigrrvert$ means$$bigllvert na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+cdotsbigrrvertleqslantbigllvert a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+cdotsbigrrvert.$$Now, dividing both sides by $bigllvert(z-a)^nbigrrvert$, you get that$$bigllvert na_n(z-a)^{-1}+(n+1)a_{n+1}+cdotsbigrrvertleqslantbigllvert a_n+a_{n+1}(z-a)+cdotsbigrrvert,$$which is impossible, because the LHS tends to $+infty$ when $zto a$, whereas the RHS tends to $lvert a_nrvert$.



                      So, since $f$ has no zeros and $mathbb C$ is simply connected, there is indeed an entire function $g$ such that $f=e^g$. You proved that $g'$ is bounded. Therefore, by Liouville's therem, it is constant. So there are constants $k$ and $c$ such that $(forall zinmathbb{C}):g(z)=cz+k$and therefore$$(forall zinmathbb{C}):f(z)=e^{g(z)}=e^{cz+k}=e^ke^{cz}.$$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Note that $f$ has not zeros. In fact, if $a$ is a zero of $f$ of order $ninmathbb N$, then $a$ is a zero of $f'$ os order $n-1$. But then, near $a$, the inequality $bigllvert f'(z)bigrrvertleqslantbigllvert f(z)bigrrvert$ means$$bigllvert na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+cdotsbigrrvertleqslantbigllvert a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+cdotsbigrrvert.$$Now, dividing both sides by $bigllvert(z-a)^nbigrrvert$, you get that$$bigllvert na_n(z-a)^{-1}+(n+1)a_{n+1}+cdotsbigrrvertleqslantbigllvert a_n+a_{n+1}(z-a)+cdotsbigrrvert,$$which is impossible, because the LHS tends to $+infty$ when $zto a$, whereas the RHS tends to $lvert a_nrvert$.



                        So, since $f$ has no zeros and $mathbb C$ is simply connected, there is indeed an entire function $g$ such that $f=e^g$. You proved that $g'$ is bounded. Therefore, by Liouville's therem, it is constant. So there are constants $k$ and $c$ such that $(forall zinmathbb{C}):g(z)=cz+k$and therefore$$(forall zinmathbb{C}):f(z)=e^{g(z)}=e^{cz+k}=e^ke^{cz}.$$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Note that $f$ has not zeros. In fact, if $a$ is a zero of $f$ of order $ninmathbb N$, then $a$ is a zero of $f'$ os order $n-1$. But then, near $a$, the inequality $bigllvert f'(z)bigrrvertleqslantbigllvert f(z)bigrrvert$ means$$bigllvert na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+cdotsbigrrvertleqslantbigllvert a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+cdotsbigrrvert.$$Now, dividing both sides by $bigllvert(z-a)^nbigrrvert$, you get that$$bigllvert na_n(z-a)^{-1}+(n+1)a_{n+1}+cdotsbigrrvertleqslantbigllvert a_n+a_{n+1}(z-a)+cdotsbigrrvert,$$which is impossible, because the LHS tends to $+infty$ when $zto a$, whereas the RHS tends to $lvert a_nrvert$.



                          So, since $f$ has no zeros and $mathbb C$ is simply connected, there is indeed an entire function $g$ such that $f=e^g$. You proved that $g'$ is bounded. Therefore, by Liouville's therem, it is constant. So there are constants $k$ and $c$ such that $(forall zinmathbb{C}):g(z)=cz+k$and therefore$$(forall zinmathbb{C}):f(z)=e^{g(z)}=e^{cz+k}=e^ke^{cz}.$$






                          share|cite|improve this answer









                          $endgroup$



                          Note that $f$ has not zeros. In fact, if $a$ is a zero of $f$ of order $ninmathbb N$, then $a$ is a zero of $f'$ os order $n-1$. But then, near $a$, the inequality $bigllvert f'(z)bigrrvertleqslantbigllvert f(z)bigrrvert$ means$$bigllvert na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+cdotsbigrrvertleqslantbigllvert a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+cdotsbigrrvert.$$Now, dividing both sides by $bigllvert(z-a)^nbigrrvert$, you get that$$bigllvert na_n(z-a)^{-1}+(n+1)a_{n+1}+cdotsbigrrvertleqslantbigllvert a_n+a_{n+1}(z-a)+cdotsbigrrvert,$$which is impossible, because the LHS tends to $+infty$ when $zto a$, whereas the RHS tends to $lvert a_nrvert$.



                          So, since $f$ has no zeros and $mathbb C$ is simply connected, there is indeed an entire function $g$ such that $f=e^g$. You proved that $g'$ is bounded. Therefore, by Liouville's therem, it is constant. So there are constants $k$ and $c$ such that $(forall zinmathbb{C}):g(z)=cz+k$and therefore$$(forall zinmathbb{C}):f(z)=e^{g(z)}=e^{cz+k}=e^ke^{cz}.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 6 '18 at 23:19









                          José Carlos SantosJosé Carlos Santos

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