Deduce from Hadamard's Theorem that if $F$ is entire and has non-integral order of growth then it has...












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Deduce from Hadamard's Factorization Theorem that if $F$ is entire and has finite non-integral order of growth then it has infinitely many zeros. This is exercise 14 of chapter 5 in Stein and Shakarchi's Complex Analysis.



Hadamard's theorem states that if $F$ is entire and has growth order $rho_0$, and $k=lfloor rho_0 rfloor$, $a_1,a_2,ldots$ are the zeros of $F$ then
$$ F(z)=e^{P(z)}z^mprod_{n=1}^infty E_k(z/a_n),$$
where $P$ is a polynomial of degree at most $k$ and $m$ is the order of the zero of $f$ at $z=0$.



How can I use the theorem? Assume that $F$ has finitely many zeros and get a contradiction perhaps?






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    Deduce from Hadamard's Factorization Theorem that if $F$ is entire and has finite non-integral order of growth then it has infinitely many zeros. This is exercise 14 of chapter 5 in Stein and Shakarchi's Complex Analysis.



    Hadamard's theorem states that if $F$ is entire and has growth order $rho_0$, and $k=lfloor rho_0 rfloor$, $a_1,a_2,ldots$ are the zeros of $F$ then
    $$ F(z)=e^{P(z)}z^mprod_{n=1}^infty E_k(z/a_n),$$
    where $P$ is a polynomial of degree at most $k$ and $m$ is the order of the zero of $f$ at $z=0$.



    How can I use the theorem? Assume that $F$ has finitely many zeros and get a contradiction perhaps?






    complex-analysis







    share|cite|improve this question











    $endgroup$















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      0





      $begingroup$


      Deduce from Hadamard's Factorization Theorem that if $F$ is entire and has finite non-integral order of growth then it has infinitely many zeros. This is exercise 14 of chapter 5 in Stein and Shakarchi's Complex Analysis.



      Hadamard's theorem states that if $F$ is entire and has growth order $rho_0$, and $k=lfloor rho_0 rfloor$, $a_1,a_2,ldots$ are the zeros of $F$ then
      $$ F(z)=e^{P(z)}z^mprod_{n=1}^infty E_k(z/a_n),$$
      where $P$ is a polynomial of degree at most $k$ and $m$ is the order of the zero of $f$ at $z=0$.



      How can I use the theorem? Assume that $F$ has finitely many zeros and get a contradiction perhaps?






      complex-analysis







      share|cite|improve this question











      $endgroup$




      Deduce from Hadamard's Factorization Theorem that if $F$ is entire and has finite non-integral order of growth then it has infinitely many zeros. This is exercise 14 of chapter 5 in Stein and Shakarchi's Complex Analysis.



      Hadamard's theorem states that if $F$ is entire and has growth order $rho_0$, and $k=lfloor rho_0 rfloor$, $a_1,a_2,ldots$ are the zeros of $F$ then
      $$ F(z)=e^{P(z)}z^mprod_{n=1}^infty E_k(z/a_n),$$
      where $P$ is a polynomial of degree at most $k$ and $m$ is the order of the zero of $f$ at $z=0$.



      How can I use the theorem? Assume that $F$ has finitely many zeros and get a contradiction perhaps?






      complex-analysis




      complex-analysis






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      share|cite|improve this question













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      edited Dec 15 '18 at 9:13







      UserA

















      asked Dec 15 '18 at 8:22









      UserAUserA

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          EDIT: The old answer is wrong. $f=f_1f_2$ where $f_1(z)=e^z$ and $f_2(z)=e^{-z}$ is a counter-example.



          Suppose $f$ has only finitely many zeroes in $mathbb{C}$, say, ${a_1,...,a_N}$. Let $g(z)=prod_{n=1}^N (z-a_n) ,forall z in mathbb{C}.$ Then $frac{f}{g}$ has no zeroes in $mathbb{C}.$



          Now apply Hadamard factorizaton: $frac{f(z)}{g(z)}=e^{P(z)} ,forall z in mathbb{C}$, where $P(z)$ is a polynomial. Hence $f(z)=g(z)e^{P(z)} forall z in mathbb{C}.$



          Since $g$ is a polynomial of degree $N$, the order of growth of $f$ equals $deg(P)$. By the infimum property it cannot be larger, for, if $f=f_1f_2$ where $|f_1(z)| le A_1e^{B_1|z|^{rho_1}} forall z in mathbb{C}$ and $|f_2(z)| le A_2e^{B_2|z|^{rho_2}} forall z in mathbb{C}$, we must have $|f(z)|=|f_1(z)f_2(z)| le A_1A_2e^{(B_1+B_2)|z|^{max{rho_1,rho_2}}} forall z in mathbb{C}$. If it were smaller, say, $rho_0$, taking $z to infty, z in mathbb{R}$, you get a contradiction, since $|g(x)| le Ae^{{B|x|^{rho_0}-P(x)}} le A'e^{B'|x|^r}$ for sufficiently large $x$, for some $r<0$.






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            $begingroup$

            EDIT: The old answer is wrong. $f=f_1f_2$ where $f_1(z)=e^z$ and $f_2(z)=e^{-z}$ is a counter-example.



            Suppose $f$ has only finitely many zeroes in $mathbb{C}$, say, ${a_1,...,a_N}$. Let $g(z)=prod_{n=1}^N (z-a_n) ,forall z in mathbb{C}.$ Then $frac{f}{g}$ has no zeroes in $mathbb{C}.$



            Now apply Hadamard factorizaton: $frac{f(z)}{g(z)}=e^{P(z)} ,forall z in mathbb{C}$, where $P(z)$ is a polynomial. Hence $f(z)=g(z)e^{P(z)} forall z in mathbb{C}.$



            Since $g$ is a polynomial of degree $N$, the order of growth of $f$ equals $deg(P)$. By the infimum property it cannot be larger, for, if $f=f_1f_2$ where $|f_1(z)| le A_1e^{B_1|z|^{rho_1}} forall z in mathbb{C}$ and $|f_2(z)| le A_2e^{B_2|z|^{rho_2}} forall z in mathbb{C}$, we must have $|f(z)|=|f_1(z)f_2(z)| le A_1A_2e^{(B_1+B_2)|z|^{max{rho_1,rho_2}}} forall z in mathbb{C}$. If it were smaller, say, $rho_0$, taking $z to infty, z in mathbb{R}$, you get a contradiction, since $|g(x)| le Ae^{{B|x|^{rho_0}-P(x)}} le A'e^{B'|x|^r}$ for sufficiently large $x$, for some $r<0$.






            share|cite|improve this answer











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              1












              $begingroup$

              EDIT: The old answer is wrong. $f=f_1f_2$ where $f_1(z)=e^z$ and $f_2(z)=e^{-z}$ is a counter-example.



              Suppose $f$ has only finitely many zeroes in $mathbb{C}$, say, ${a_1,...,a_N}$. Let $g(z)=prod_{n=1}^N (z-a_n) ,forall z in mathbb{C}.$ Then $frac{f}{g}$ has no zeroes in $mathbb{C}.$



              Now apply Hadamard factorizaton: $frac{f(z)}{g(z)}=e^{P(z)} ,forall z in mathbb{C}$, where $P(z)$ is a polynomial. Hence $f(z)=g(z)e^{P(z)} forall z in mathbb{C}.$



              Since $g$ is a polynomial of degree $N$, the order of growth of $f$ equals $deg(P)$. By the infimum property it cannot be larger, for, if $f=f_1f_2$ where $|f_1(z)| le A_1e^{B_1|z|^{rho_1}} forall z in mathbb{C}$ and $|f_2(z)| le A_2e^{B_2|z|^{rho_2}} forall z in mathbb{C}$, we must have $|f(z)|=|f_1(z)f_2(z)| le A_1A_2e^{(B_1+B_2)|z|^{max{rho_1,rho_2}}} forall z in mathbb{C}$. If it were smaller, say, $rho_0$, taking $z to infty, z in mathbb{R}$, you get a contradiction, since $|g(x)| le Ae^{{B|x|^{rho_0}-P(x)}} le A'e^{B'|x|^r}$ for sufficiently large $x$, for some $r<0$.






              share|cite|improve this answer











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                1





                $begingroup$

                EDIT: The old answer is wrong. $f=f_1f_2$ where $f_1(z)=e^z$ and $f_2(z)=e^{-z}$ is a counter-example.



                Suppose $f$ has only finitely many zeroes in $mathbb{C}$, say, ${a_1,...,a_N}$. Let $g(z)=prod_{n=1}^N (z-a_n) ,forall z in mathbb{C}.$ Then $frac{f}{g}$ has no zeroes in $mathbb{C}.$



                Now apply Hadamard factorizaton: $frac{f(z)}{g(z)}=e^{P(z)} ,forall z in mathbb{C}$, where $P(z)$ is a polynomial. Hence $f(z)=g(z)e^{P(z)} forall z in mathbb{C}.$



                Since $g$ is a polynomial of degree $N$, the order of growth of $f$ equals $deg(P)$. By the infimum property it cannot be larger, for, if $f=f_1f_2$ where $|f_1(z)| le A_1e^{B_1|z|^{rho_1}} forall z in mathbb{C}$ and $|f_2(z)| le A_2e^{B_2|z|^{rho_2}} forall z in mathbb{C}$, we must have $|f(z)|=|f_1(z)f_2(z)| le A_1A_2e^{(B_1+B_2)|z|^{max{rho_1,rho_2}}} forall z in mathbb{C}$. If it were smaller, say, $rho_0$, taking $z to infty, z in mathbb{R}$, you get a contradiction, since $|g(x)| le Ae^{{B|x|^{rho_0}-P(x)}} le A'e^{B'|x|^r}$ for sufficiently large $x$, for some $r<0$.






                share|cite|improve this answer











                $endgroup$



                EDIT: The old answer is wrong. $f=f_1f_2$ where $f_1(z)=e^z$ and $f_2(z)=e^{-z}$ is a counter-example.



                Suppose $f$ has only finitely many zeroes in $mathbb{C}$, say, ${a_1,...,a_N}$. Let $g(z)=prod_{n=1}^N (z-a_n) ,forall z in mathbb{C}.$ Then $frac{f}{g}$ has no zeroes in $mathbb{C}.$



                Now apply Hadamard factorizaton: $frac{f(z)}{g(z)}=e^{P(z)} ,forall z in mathbb{C}$, where $P(z)$ is a polynomial. Hence $f(z)=g(z)e^{P(z)} forall z in mathbb{C}.$



                Since $g$ is a polynomial of degree $N$, the order of growth of $f$ equals $deg(P)$. By the infimum property it cannot be larger, for, if $f=f_1f_2$ where $|f_1(z)| le A_1e^{B_1|z|^{rho_1}} forall z in mathbb{C}$ and $|f_2(z)| le A_2e^{B_2|z|^{rho_2}} forall z in mathbb{C}$, we must have $|f(z)|=|f_1(z)f_2(z)| le A_1A_2e^{(B_1+B_2)|z|^{max{rho_1,rho_2}}} forall z in mathbb{C}$. If it were smaller, say, $rho_0$, taking $z to infty, z in mathbb{R}$, you get a contradiction, since $|g(x)| le Ae^{{B|x|^{rho_0}-P(x)}} le A'e^{B'|x|^r}$ for sufficiently large $x$, for some $r<0$.







                share|cite|improve this answer














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                edited Dec 15 '18 at 11:50

























                answered Dec 15 '18 at 10:23









                tonychow0929tonychow0929

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