Why does this complex analytic function statement hold?
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I don’t really have any experience in complex analysis so I’m hoping someone can explain why the following statement is true (it is used to prove a lemma necessary to prove Itos representation theorem)
“Since $G()=0$ on $mathbb{R}^n$ and $G$ is analytic, it follows $G()=0$ on $mathbb{C}^n$.”
Why is this the case? If it is the result of a well known theorem, could someone give the actual theorem?
As I understand it, a complex analytic function is simply a function that can be written as a power series at any point. I don’t understand how it being zero on the reals means it must be zero for all complex numbers.
complex-analysis functional-analysis
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
$endgroup$
add a comment |
$begingroup$
I don’t really have any experience in complex analysis so I’m hoping someone can explain why the following statement is true (it is used to prove a lemma necessary to prove Itos representation theorem)
“Since $G()=0$ on $mathbb{R}^n$ and $G$ is analytic, it follows $G()=0$ on $mathbb{C}^n$.”
Why is this the case? If it is the result of a well known theorem, could someone give the actual theorem?
As I understand it, a complex analytic function is simply a function that can be written as a power series at any point. I don’t understand how it being zero on the reals means it must be zero for all complex numbers.
complex-analysis functional-analysis
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
$endgroup$
$begingroup$
On the one dimensional case, it is a simple conclusion from the Cauchy-Reimann equations, perhaps there is a generalization of the concept to n dimensions that still allows the same conclusion.
$endgroup$
– Sar
Dec 18 '18 at 6:39
add a comment |
$begingroup$
I don’t really have any experience in complex analysis so I’m hoping someone can explain why the following statement is true (it is used to prove a lemma necessary to prove Itos representation theorem)
“Since $G()=0$ on $mathbb{R}^n$ and $G$ is analytic, it follows $G()=0$ on $mathbb{C}^n$.”
Why is this the case? If it is the result of a well known theorem, could someone give the actual theorem?
As I understand it, a complex analytic function is simply a function that can be written as a power series at any point. I don’t understand how it being zero on the reals means it must be zero for all complex numbers.
complex-analysis functional-analysis
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
$endgroup$
I don’t really have any experience in complex analysis so I’m hoping someone can explain why the following statement is true (it is used to prove a lemma necessary to prove Itos representation theorem)
“Since $G()=0$ on $mathbb{R}^n$ and $G$ is analytic, it follows $G()=0$ on $mathbb{C}^n$.”
Why is this the case? If it is the result of a well known theorem, could someone give the actual theorem?
As I understand it, a complex analytic function is simply a function that can be written as a power series at any point. I don’t understand how it being zero on the reals means it must be zero for all complex numbers.
complex-analysis functional-analysis
complex-analysis functional-analysis
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
asked Dec 18 '18 at 6:24
XiaomiXiaomi
1,071115
1,071115
$begingroup$
On the one dimensional case, it is a simple conclusion from the Cauchy-Reimann equations, perhaps there is a generalization of the concept to n dimensions that still allows the same conclusion.
$endgroup$
– Sar
Dec 18 '18 at 6:39
add a comment |
$begingroup$
On the one dimensional case, it is a simple conclusion from the Cauchy-Reimann equations, perhaps there is a generalization of the concept to n dimensions that still allows the same conclusion.
$endgroup$
– Sar
Dec 18 '18 at 6:39
$begingroup$
On the one dimensional case, it is a simple conclusion from the Cauchy-Reimann equations, perhaps there is a generalization of the concept to n dimensions that still allows the same conclusion.
$endgroup$
– Sar
Dec 18 '18 at 6:39
$begingroup$
On the one dimensional case, it is a simple conclusion from the Cauchy-Reimann equations, perhaps there is a generalization of the concept to n dimensions that still allows the same conclusion.
$endgroup$
– Sar
Dec 18 '18 at 6:39
add a comment |
1 Answer
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If $x_2,cdots,x_n$ are real then $f(z_1,x_2,cdots,x_n)$ is an entire function of the first variable which vanishes on the real line, so we get $f(z_1,x_2,cdots,x_n)=0$ for all $z_1 in mathbb C$. Now consider $f(z_1,z_2,cdots,x_n)$ and apply the same argument to conclude that $f(z_1,z_2,x_3,cdots,x_n)=0$ for all complex $z_2$, and so on.
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
The basic idea is to vary only one variable at a time, keeping $n-1$ variables fixed. Please go through my proof for $n=2$ and you will see what is going on.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 9:12
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
If $x_2,cdots,x_n$ are real then $f(z_1,x_2,cdots,x_n)$ is an entire function of the first variable which vanishes on the real line, so we get $f(z_1,x_2,cdots,x_n)=0$ for all $z_1 in mathbb C$. Now consider $f(z_1,z_2,cdots,x_n)$ and apply the same argument to conclude that $f(z_1,z_2,x_3,cdots,x_n)=0$ for all complex $z_2$, and so on.
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
The basic idea is to vary only one variable at a time, keeping $n-1$ variables fixed. Please go through my proof for $n=2$ and you will see what is going on.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 9:12
add a comment |
$begingroup$
If $x_2,cdots,x_n$ are real then $f(z_1,x_2,cdots,x_n)$ is an entire function of the first variable which vanishes on the real line, so we get $f(z_1,x_2,cdots,x_n)=0$ for all $z_1 in mathbb C$. Now consider $f(z_1,z_2,cdots,x_n)$ and apply the same argument to conclude that $f(z_1,z_2,x_3,cdots,x_n)=0$ for all complex $z_2$, and so on.
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
The basic idea is to vary only one variable at a time, keeping $n-1$ variables fixed. Please go through my proof for $n=2$ and you will see what is going on.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 9:12
add a comment |
$begingroup$
If $x_2,cdots,x_n$ are real then $f(z_1,x_2,cdots,x_n)$ is an entire function of the first variable which vanishes on the real line, so we get $f(z_1,x_2,cdots,x_n)=0$ for all $z_1 in mathbb C$. Now consider $f(z_1,z_2,cdots,x_n)$ and apply the same argument to conclude that $f(z_1,z_2,x_3,cdots,x_n)=0$ for all complex $z_2$, and so on.
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
If $x_2,cdots,x_n$ are real then $f(z_1,x_2,cdots,x_n)$ is an entire function of the first variable which vanishes on the real line, so we get $f(z_1,x_2,cdots,x_n)=0$ for all $z_1 in mathbb C$. Now consider $f(z_1,z_2,cdots,x_n)$ and apply the same argument to conclude that $f(z_1,z_2,x_3,cdots,x_n)=0$ for all complex $z_2$, and so on.
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 18 '18 at 6:38
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
69.5k53170
$begingroup$
The basic idea is to vary only one variable at a time, keeping $n-1$ variables fixed. Please go through my proof for $n=2$ and you will see what is going on.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 9:12
add a comment |
$begingroup$
The basic idea is to vary only one variable at a time, keeping $n-1$ variables fixed. Please go through my proof for $n=2$ and you will see what is going on.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 9:12
$begingroup$
The basic idea is to vary only one variable at a time, keeping $n-1$ variables fixed. Please go through my proof for $n=2$ and you will see what is going on.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 9:12
$begingroup$
The basic idea is to vary only one variable at a time, keeping $n-1$ variables fixed. Please go through my proof for $n=2$ and you will see what is going on.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 9:12
add a comment |
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$begingroup$
On the one dimensional case, it is a simple conclusion from the Cauchy-Reimann equations, perhaps there is a generalization of the concept to n dimensions that still allows the same conclusion.
$endgroup$
– Sar
Dec 18 '18 at 6:39