Mixing
Definition clarification of analytic continuation of holomorphic function
$begingroup$
I was given the definition of anlytic continuation as in Stein's book- given two regions $Omegasubset Omega'$
for analytic functions $F:Omega'to mathbb{C}$ is an analytic continuation of $f:Omega to mathbb{C}$ if $F$ agrees with $f$ on $Omega$. I know that $sin:mathbb{C}to mathbb{C}$ is an analytic continuation of $sin:mathbb{R}to mathbb{R}$ because they agree on $mathbb{R}$, but this is not a region in $mathbb{C}$, so how does that add up with the defintion?
complex-analysis definition holomorphic-functions analytic-continuation
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
asked Jan 18 at 10:44
Simon GreenSimon Green
845
$endgroup$
add a comment |
$begingroup$
I was given the definition of anlytic continuation as in Stein's book- given two regions $Omegasubset Omega'$
for analytic functions $F:Omega'to mathbb{C}$ is an analytic continuation of $f:Omega to mathbb{C}$ if $F$ agrees with $f$ on $Omega$. I know that $sin:mathbb{C}to mathbb{C}$ is an analytic continuation of $sin:mathbb{R}to mathbb{R}$ because they agree on $mathbb{R}$, but this is not a region in $mathbb{C}$, so how does that add up with the defintion?
complex-analysis definition holomorphic-functions analytic-continuation
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
asked Jan 18 at 10:44
Simon GreenSimon Green
845
$endgroup$
add a comment |
$begingroup$
I was given the definition of anlytic continuation as in Stein's book- given two regions $Omegasubset Omega'$
for analytic functions $F:Omega'to mathbb{C}$ is an analytic continuation of $f:Omega to mathbb{C}$ if $F$ agrees with $f$ on $Omega$. I know that $sin:mathbb{C}to mathbb{C}$ is an analytic continuation of $sin:mathbb{R}to mathbb{R}$ because they agree on $mathbb{R}$, but this is not a region in $mathbb{C}$, so how does that add up with the defintion?
complex-analysis definition holomorphic-functions analytic-continuation
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
asked Jan 18 at 10:44
Simon GreenSimon Green
845
$endgroup$
I was given the definition of anlytic continuation as in Stein's book- given two regions $Omegasubset Omega'$
for analytic functions $F:Omega'to mathbb{C}$ is an analytic continuation of $f:Omega to mathbb{C}$ if $F$ agrees with $f$ on $Omega$. I know that $sin:mathbb{C}to mathbb{C}$ is an analytic continuation of $sin:mathbb{R}to mathbb{R}$ because they agree on $mathbb{R}$, but this is not a region in $mathbb{C}$, so how does that add up with the defintion?
complex-analysis definition holomorphic-functions analytic-continuation
complex-analysis definition holomorphic-functions analytic-continuation
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
asked Jan 18 at 10:44
Simon GreenSimon Green
845
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
asked Jan 18 at 10:44
Simon GreenSimon Green
845
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
edited Jan 18 at 11:21
José Carlos Santos
164k22131234
164k22131234
asked Jan 18 at 10:44
Simon GreenSimon Green
845
asked Jan 18 at 10:44
Simon GreenSimon Green
845
asked Jan 18 at 10:44
Simon GreenSimon Green
845
845
add a comment |
add a comment |
1 Answer
1
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$begingroup$
There is some ambiguity here (not in what you wrote, which is perfectly clear, but in Stein's textbook). Sometimes, while writing about analytic continuation, Stein indeed assumes that $Omega$ and $Omega'$ are regions. But when finally Stein states a theorem about analytic continuation, the statement is:
Theorem: Suppose $f$ is a holomorphic function in a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$. Then $f$ is identically $0$.
As you can see, when it is stated like this, the theorem on analytic continuation applies indeed to the situation that you described.
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I think it answered my question; We know that $sin(z)$ agrees with $sin(x)$ on $mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer!
$endgroup$
– Simon Green
Jan 18 at 11:10
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 18 at 11:20
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
There is some ambiguity here (not in what you wrote, which is perfectly clear, but in Stein's textbook). Sometimes, while writing about analytic continuation, Stein indeed assumes that $Omega$ and $Omega'$ are regions. But when finally Stein states a theorem about analytic continuation, the statement is:
Theorem: Suppose $f$ is a holomorphic function in a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$. Then $f$ is identically $0$.
As you can see, when it is stated like this, the theorem on analytic continuation applies indeed to the situation that you described.
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I think it answered my question; We know that $sin(z)$ agrees with $sin(x)$ on $mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer!
$endgroup$
– Simon Green
Jan 18 at 11:10
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 18 at 11:20
add a comment |
$begingroup$
There is some ambiguity here (not in what you wrote, which is perfectly clear, but in Stein's textbook). Sometimes, while writing about analytic continuation, Stein indeed assumes that $Omega$ and $Omega'$ are regions. But when finally Stein states a theorem about analytic continuation, the statement is:
Theorem: Suppose $f$ is a holomorphic function in a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$. Then $f$ is identically $0$.
As you can see, when it is stated like this, the theorem on analytic continuation applies indeed to the situation that you described.
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I think it answered my question; We know that $sin(z)$ agrees with $sin(x)$ on $mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer!
$endgroup$
– Simon Green
Jan 18 at 11:10
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 18 at 11:20
add a comment |
$begingroup$
There is some ambiguity here (not in what you wrote, which is perfectly clear, but in Stein's textbook). Sometimes, while writing about analytic continuation, Stein indeed assumes that $Omega$ and $Omega'$ are regions. But when finally Stein states a theorem about analytic continuation, the statement is:
Theorem: Suppose $f$ is a holomorphic function in a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$. Then $f$ is identically $0$.
As you can see, when it is stated like this, the theorem on analytic continuation applies indeed to the situation that you described.
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
There is some ambiguity here (not in what you wrote, which is perfectly clear, but in Stein's textbook). Sometimes, while writing about analytic continuation, Stein indeed assumes that $Omega$ and $Omega'$ are regions. But when finally Stein states a theorem about analytic continuation, the statement is:
Theorem: Suppose $f$ is a holomorphic function in a region $Omega$ that vanishes on a sequence of distinct points with a limit point in $Omega$. Then $f$ is identically $0$.
As you can see, when it is stated like this, the theorem on analytic continuation applies indeed to the situation that you described.
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 10:56
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
I think it answered my question; We know that $sin(z)$ agrees with $sin(x)$ on $mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer!
$endgroup$
– Simon Green
Jan 18 at 11:10
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 18 at 11:20
add a comment |
$begingroup$
I think it answered my question; We know that $sin(z)$ agrees with $sin(x)$ on $mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer!
$endgroup$
– Simon Green
Jan 18 at 11:10
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 18 at 11:20
$begingroup$
I think it answered my question; We know that $sin(z)$ agrees with $sin(x)$ on $mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer!
$endgroup$
– Simon Green
Jan 18 at 11:10
$begingroup$
I think it answered my question; We know that $sin(z)$ agrees with $sin(x)$ on $mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer!
$endgroup$
– Simon Green
Jan 18 at 11:10
1
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 18 at 11:20
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 18 at 11:20
add a comment |
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