Given $f(z)$ is entire and a bound on f at each $ z in mathbb{C}$, prove that the integral is 0
$begingroup$
The Problem:
It's given that $f(z)$ is an entire function on $mathbb{C}$ and that there exist $M > 0$ and $A > 0$ such that $|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$ for all $x,y in mathbb{R}$. The task is to prove that for all $xi in R$ with $|xi|>M$, $int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx=0$.
My thoughts so far:
So my initial thought was that I have to cleverly choose an entire function and a closed path in $mathbb{C}$ to evaluate the chosen function over. Given that the line integral around closed paths for an entire function is $0$, I can theoretically show that the contribution of the line integral along the real line is $0$ by showing that the contribution from every other part of the line integral is $0$. I've been playing with different paths and functions, but haven't been successful yet.
The other thing I noticed is that if
$|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$
then that implies that
$e^{-2 pi i xi (x+iy)}|f(x+iy)|lefrac{Ae^{2pi(M|y|-ixi(x+iy))}}{1+x^2}$
which certainly seems important since the left hand side is the integrand that I want if I let $y$ be $0$ (which it is on the real axis). It doesn't seem that this bound on the function is very useful though, since the bound doesn't go to $0$ except when $x$ is large and $y$ is around $0$, and so it doesn't seem to tell me anything useful about the function closer to the origin.
In all of this, I haven't really appealed to the requirement that $|xi|$ be larger than $M$, which I'm sure is important.
A push in the right direction would be appreciated.
complex-analysis complex-integration
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
asked Dec 15 '18 at 3:01
JoeJoe
447
$endgroup$
add a comment |
$begingroup$
The Problem:
It's given that $f(z)$ is an entire function on $mathbb{C}$ and that there exist $M > 0$ and $A > 0$ such that $|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$ for all $x,y in mathbb{R}$. The task is to prove that for all $xi in R$ with $|xi|>M$, $int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx=0$.
My thoughts so far:
So my initial thought was that I have to cleverly choose an entire function and a closed path in $mathbb{C}$ to evaluate the chosen function over. Given that the line integral around closed paths for an entire function is $0$, I can theoretically show that the contribution of the line integral along the real line is $0$ by showing that the contribution from every other part of the line integral is $0$. I've been playing with different paths and functions, but haven't been successful yet.
The other thing I noticed is that if
$|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$
then that implies that
$e^{-2 pi i xi (x+iy)}|f(x+iy)|lefrac{Ae^{2pi(M|y|-ixi(x+iy))}}{1+x^2}$
which certainly seems important since the left hand side is the integrand that I want if I let $y$ be $0$ (which it is on the real axis). It doesn't seem that this bound on the function is very useful though, since the bound doesn't go to $0$ except when $x$ is large and $y$ is around $0$, and so it doesn't seem to tell me anything useful about the function closer to the origin.
In all of this, I haven't really appealed to the requirement that $|xi|$ be larger than $M$, which I'm sure is important.
A push in the right direction would be appreciated.
complex-analysis complex-integration
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
asked Dec 15 '18 at 3:01
JoeJoe
447
$endgroup$
$begingroup$
"then that implies that " You can't multiply inequalities by a complex number. Do you perhaps mean the norm of the LHS is less than the norm of the RHS?
$endgroup$
– Winther
Dec 15 '18 at 3:12
add a comment |
$begingroup$
The Problem:
It's given that $f(z)$ is an entire function on $mathbb{C}$ and that there exist $M > 0$ and $A > 0$ such that $|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$ for all $x,y in mathbb{R}$. The task is to prove that for all $xi in R$ with $|xi|>M$, $int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx=0$.
My thoughts so far:
So my initial thought was that I have to cleverly choose an entire function and a closed path in $mathbb{C}$ to evaluate the chosen function over. Given that the line integral around closed paths for an entire function is $0$, I can theoretically show that the contribution of the line integral along the real line is $0$ by showing that the contribution from every other part of the line integral is $0$. I've been playing with different paths and functions, but haven't been successful yet.
The other thing I noticed is that if
$|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$
then that implies that
$e^{-2 pi i xi (x+iy)}|f(x+iy)|lefrac{Ae^{2pi(M|y|-ixi(x+iy))}}{1+x^2}$
which certainly seems important since the left hand side is the integrand that I want if I let $y$ be $0$ (which it is on the real axis). It doesn't seem that this bound on the function is very useful though, since the bound doesn't go to $0$ except when $x$ is large and $y$ is around $0$, and so it doesn't seem to tell me anything useful about the function closer to the origin.
In all of this, I haven't really appealed to the requirement that $|xi|$ be larger than $M$, which I'm sure is important.
A push in the right direction would be appreciated.
complex-analysis complex-integration
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
asked Dec 15 '18 at 3:01
JoeJoe
447
$endgroup$
The Problem:
It's given that $f(z)$ is an entire function on $mathbb{C}$ and that there exist $M > 0$ and $A > 0$ such that $|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$ for all $x,y in mathbb{R}$. The task is to prove that for all $xi in R$ with $|xi|>M$, $int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx=0$.
My thoughts so far:
So my initial thought was that I have to cleverly choose an entire function and a closed path in $mathbb{C}$ to evaluate the chosen function over. Given that the line integral around closed paths for an entire function is $0$, I can theoretically show that the contribution of the line integral along the real line is $0$ by showing that the contribution from every other part of the line integral is $0$. I've been playing with different paths and functions, but haven't been successful yet.
The other thing I noticed is that if
$|f(x+iy)|lefrac{Ae^{2pi M|y|}}{1+x^2}$
then that implies that
$e^{-2 pi i xi (x+iy)}|f(x+iy)|lefrac{Ae^{2pi(M|y|-ixi(x+iy))}}{1+x^2}$
which certainly seems important since the left hand side is the integrand that I want if I let $y$ be $0$ (which it is on the real axis). It doesn't seem that this bound on the function is very useful though, since the bound doesn't go to $0$ except when $x$ is large and $y$ is around $0$, and so it doesn't seem to tell me anything useful about the function closer to the origin.
In all of this, I haven't really appealed to the requirement that $|xi|$ be larger than $M$, which I'm sure is important.
A push in the right direction would be appreciated.
complex-analysis complex-integration
complex-analysis complex-integration
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
asked Dec 15 '18 at 3:01
JoeJoe
447
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
asked Dec 15 '18 at 3:01
JoeJoe
447
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
edited Dec 15 '18 at 3:09
Kemono Chen
2,9181739
2,9181739
asked Dec 15 '18 at 3:01
JoeJoe
447
asked Dec 15 '18 at 3:01
JoeJoe
447
asked Dec 15 '18 at 3:01
JoeJoe
447
447
$begingroup$
"then that implies that " You can't multiply inequalities by a complex number. Do you perhaps mean the norm of the LHS is less than the norm of the RHS?
$endgroup$
– Winther
Dec 15 '18 at 3:12
add a comment |
$begingroup$
"then that implies that " You can't multiply inequalities by a complex number. Do you perhaps mean the norm of the LHS is less than the norm of the RHS?
$endgroup$
– Winther
Dec 15 '18 at 3:12
$begingroup$
"then that implies that " You can't multiply inequalities by a complex number. Do you perhaps mean the norm of the LHS is less than the norm of the RHS?
$endgroup$
– Winther
Dec 15 '18 at 3:12
$begingroup$
"then that implies that " You can't multiply inequalities by a complex number. Do you perhaps mean the norm of the LHS is less than the norm of the RHS?
$endgroup$
– Winther
Dec 15 '18 at 3:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$textbf{Hint:}$ Move the domain of integral to $mathbb{R}+iy$ from $mathbb{R}$.
$textbf{EDIT:}$ If you are struggling, I'll give you a detailed hint. Let $xi$ be fixed with $|xi|>M$ and let
$$
g(z) = f(z)e^{-2pi ixi z}.
$$ Consider a rectangle
$$
R_N = [-N,N] times [0, y].
$$ Since $g$ is entire, we have
$$int_{partial R_N}g(z)dz=0quadcdots(*).
$$ Arrange $(*)$ to get
$$begin{eqnarray}
int_{-N}^{N} f(x)e^{−2pi ixi x}dx &=&int_{-N}^{N} f(x+iy)e^{−2pi ixi (x+iy)}\
&&- iint_0^y f(N+it)e^{−2pi ixi (N+it)}dt\&& + iint_0^y f(-N+it)e^{−2pi ixi (-N+it)}dt .
end{eqnarray}$$ Take $Nto infty$. We end up with
$$
int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx =e^{2pixi y}int_{-infty}^{infty} f(x+iy)e^{−2pi ixi x }dx.
$$ Assume $xi>M$. Take $yto -infty$ to get the desired conclusion. If $xi<-M$, it can be dealt with similarly by taking $yto infty$. (cf. Paley-Wiener theorem)
answered Dec 15 '18 at 3:15
SongSong
8,096524
$endgroup$
$begingroup$
To be clear: Are you suggesting that I integrate along a line parallel to the real axis?
$endgroup$
– Joe
Dec 15 '18 at 3:33
$begingroup$
Yes. I will add some details when I have time.
$endgroup$
– Song
Dec 15 '18 at 3:54
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$textbf{Hint:}$ Move the domain of integral to $mathbb{R}+iy$ from $mathbb{R}$.
$textbf{EDIT:}$ If you are struggling, I'll give you a detailed hint. Let $xi$ be fixed with $|xi|>M$ and let
$$
g(z) = f(z)e^{-2pi ixi z}.
$$ Consider a rectangle
$$
R_N = [-N,N] times [0, y].
$$ Since $g$ is entire, we have
$$int_{partial R_N}g(z)dz=0quadcdots(*).
$$ Arrange $(*)$ to get
$$begin{eqnarray}
int_{-N}^{N} f(x)e^{−2pi ixi x}dx &=&int_{-N}^{N} f(x+iy)e^{−2pi ixi (x+iy)}\
&&- iint_0^y f(N+it)e^{−2pi ixi (N+it)}dt\&& + iint_0^y f(-N+it)e^{−2pi ixi (-N+it)}dt .
end{eqnarray}$$ Take $Nto infty$. We end up with
$$
int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx =e^{2pixi y}int_{-infty}^{infty} f(x+iy)e^{−2pi ixi x }dx.
$$ Assume $xi>M$. Take $yto -infty$ to get the desired conclusion. If $xi<-M$, it can be dealt with similarly by taking $yto infty$. (cf. Paley-Wiener theorem)
answered Dec 15 '18 at 3:15
SongSong
8,096524
$endgroup$
$begingroup$
To be clear: Are you suggesting that I integrate along a line parallel to the real axis?
$endgroup$
– Joe
Dec 15 '18 at 3:33
$begingroup$
Yes. I will add some details when I have time.
$endgroup$
– Song
Dec 15 '18 at 3:54
add a comment |
$begingroup$
$textbf{Hint:}$ Move the domain of integral to $mathbb{R}+iy$ from $mathbb{R}$.
$textbf{EDIT:}$ If you are struggling, I'll give you a detailed hint. Let $xi$ be fixed with $|xi|>M$ and let
$$
g(z) = f(z)e^{-2pi ixi z}.
$$ Consider a rectangle
$$
R_N = [-N,N] times [0, y].
$$ Since $g$ is entire, we have
$$int_{partial R_N}g(z)dz=0quadcdots(*).
$$ Arrange $(*)$ to get
$$begin{eqnarray}
int_{-N}^{N} f(x)e^{−2pi ixi x}dx &=&int_{-N}^{N} f(x+iy)e^{−2pi ixi (x+iy)}\
&&- iint_0^y f(N+it)e^{−2pi ixi (N+it)}dt\&& + iint_0^y f(-N+it)e^{−2pi ixi (-N+it)}dt .
end{eqnarray}$$ Take $Nto infty$. We end up with
$$
int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx =e^{2pixi y}int_{-infty}^{infty} f(x+iy)e^{−2pi ixi x }dx.
$$ Assume $xi>M$. Take $yto -infty$ to get the desired conclusion. If $xi<-M$, it can be dealt with similarly by taking $yto infty$. (cf. Paley-Wiener theorem)
answered Dec 15 '18 at 3:15
SongSong
8,096524
$endgroup$
$begingroup$
To be clear: Are you suggesting that I integrate along a line parallel to the real axis?
$endgroup$
– Joe
Dec 15 '18 at 3:33
$begingroup$
Yes. I will add some details when I have time.
$endgroup$
– Song
Dec 15 '18 at 3:54
add a comment |
$begingroup$
$textbf{Hint:}$ Move the domain of integral to $mathbb{R}+iy$ from $mathbb{R}$.
$textbf{EDIT:}$ If you are struggling, I'll give you a detailed hint. Let $xi$ be fixed with $|xi|>M$ and let
$$
g(z) = f(z)e^{-2pi ixi z}.
$$ Consider a rectangle
$$
R_N = [-N,N] times [0, y].
$$ Since $g$ is entire, we have
$$int_{partial R_N}g(z)dz=0quadcdots(*).
$$ Arrange $(*)$ to get
$$begin{eqnarray}
int_{-N}^{N} f(x)e^{−2pi ixi x}dx &=&int_{-N}^{N} f(x+iy)e^{−2pi ixi (x+iy)}\
&&- iint_0^y f(N+it)e^{−2pi ixi (N+it)}dt\&& + iint_0^y f(-N+it)e^{−2pi ixi (-N+it)}dt .
end{eqnarray}$$ Take $Nto infty$. We end up with
$$
int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx =e^{2pixi y}int_{-infty}^{infty} f(x+iy)e^{−2pi ixi x }dx.
$$ Assume $xi>M$. Take $yto -infty$ to get the desired conclusion. If $xi<-M$, it can be dealt with similarly by taking $yto infty$. (cf. Paley-Wiener theorem)
answered Dec 15 '18 at 3:15
SongSong
8,096524
$endgroup$
$textbf{Hint:}$ Move the domain of integral to $mathbb{R}+iy$ from $mathbb{R}$.
$textbf{EDIT:}$ If you are struggling, I'll give you a detailed hint. Let $xi$ be fixed with $|xi|>M$ and let
$$
g(z) = f(z)e^{-2pi ixi z}.
$$ Consider a rectangle
$$
R_N = [-N,N] times [0, y].
$$ Since $g$ is entire, we have
$$int_{partial R_N}g(z)dz=0quadcdots(*).
$$ Arrange $(*)$ to get
$$begin{eqnarray}
int_{-N}^{N} f(x)e^{−2pi ixi x}dx &=&int_{-N}^{N} f(x+iy)e^{−2pi ixi (x+iy)}\
&&- iint_0^y f(N+it)e^{−2pi ixi (N+it)}dt\&& + iint_0^y f(-N+it)e^{−2pi ixi (-N+it)}dt .
end{eqnarray}$$ Take $Nto infty$. We end up with
$$
int_{-infty}^{infty} f(x)e^{−2pi ixi x}dx =e^{2pixi y}int_{-infty}^{infty} f(x+iy)e^{−2pi ixi x }dx.
$$ Assume $xi>M$. Take $yto -infty$ to get the desired conclusion. If $xi<-M$, it can be dealt with similarly by taking $yto infty$. (cf. Paley-Wiener theorem)
answered Dec 15 '18 at 3:15
SongSong
8,096524
edited Dec 15 '18 at 8:38
answered Dec 15 '18 at 3:15
SongSong
8,096524
answered Dec 15 '18 at 3:15
SongSong
8,096524
answered Dec 15 '18 at 3:15
SongSong
8,096524
8,096524
$begingroup$
To be clear: Are you suggesting that I integrate along a line parallel to the real axis?
$endgroup$
– Joe
Dec 15 '18 at 3:33
$begingroup$
Yes. I will add some details when I have time.
$endgroup$
– Song
Dec 15 '18 at 3:54
add a comment |
$begingroup$
To be clear: Are you suggesting that I integrate along a line parallel to the real axis?
$endgroup$
– Joe
Dec 15 '18 at 3:33
$begingroup$
Yes. I will add some details when I have time.
$endgroup$
– Song
Dec 15 '18 at 3:54
$begingroup$
To be clear: Are you suggesting that I integrate along a line parallel to the real axis?
$endgroup$
– Joe
Dec 15 '18 at 3:33
$begingroup$
To be clear: Are you suggesting that I integrate along a line parallel to the real axis?
$endgroup$
– Joe
Dec 15 '18 at 3:33
$begingroup$
Yes. I will add some details when I have time.
$endgroup$
– Song
Dec 15 '18 at 3:54
$begingroup$
Yes. I will add some details when I have time.
$endgroup$
– Song
Dec 15 '18 at 3:54
add a comment |
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$begingroup$
"then that implies that " You can't multiply inequalities by a complex number. Do you perhaps mean the norm of the LHS is less than the norm of the RHS?
$endgroup$
– Winther
Dec 15 '18 at 3:12