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?
complex-analysis
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 15 '18 at 9:13
UserA
asked Dec 15 '18 at 8:22
UserAUserA
551216
551216
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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 $r<0$.
$endgroup$
add a comment |
$begingroup$
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 $r<0$.
$endgroup$
add a comment |
$begingroup$
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 $r<0$.
$endgroup$
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 $r<0$.
edited Dec 15 '18 at 11:50
answered Dec 15 '18 at 10:23
tonychow0929tonychow0929
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