Find point spectrum and spectrum of integral operator
$begingroup$
Let $A:L^2(0,pi) to L^2(0,pi)$ be defined by $(Af)(x)=displaystyleint_{0}^pi sin(x-y)f(y)dy$. Find the point spectrum and spectrum of $A$.
I am not sure how to go about this. I thought to start with, I should find the point spectrum (i.e. the set of eigenvalues). So not knowing any techniques for integral operators, I would just set $displaystyleint_{0}^pi sin(x-y)f(y)dy=lambda f(x)$ and try to solve for $f$. Unfortunately I have no idea how to, or even if it is possible, to solve for $f$ here. I am wondering if there is a trick specific to this integrand or a more general technique regarding integral operators that I should know about. I have a result that tells me $A$ is compact but it's not clear how this could be used. Any hints would be appreciated.
functional-analysis spectral-theory integral-operators
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
$endgroup$
|
show 1 more comment
$begingroup$
Let $A:L^2(0,pi) to L^2(0,pi)$ be defined by $(Af)(x)=displaystyleint_{0}^pi sin(x-y)f(y)dy$. Find the point spectrum and spectrum of $A$.
I am not sure how to go about this. I thought to start with, I should find the point spectrum (i.e. the set of eigenvalues). So not knowing any techniques for integral operators, I would just set $displaystyleint_{0}^pi sin(x-y)f(y)dy=lambda f(x)$ and try to solve for $f$. Unfortunately I have no idea how to, or even if it is possible, to solve for $f$ here. I am wondering if there is a trick specific to this integrand or a more general technique regarding integral operators that I should know about. I have a result that tells me $A$ is compact but it's not clear how this could be used. Any hints would be appreciated.
functional-analysis spectral-theory integral-operators
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
$endgroup$
$begingroup$
Note that $Af(x) = (sinstar f)(x)$, where $star$ denotes convolution.
$endgroup$
– Math1000
Dec 24 '18 at 22:04
$begingroup$
See here: math.stackexchange.com/questions/1117098/…
$endgroup$
– Math1000
Dec 24 '18 at 22:05
$begingroup$
We have never actually seen or used Fourier transforms so far, so it is strange that we would be set such an exercise. I suppose there is no more elementary way of finding the spectrum, or by using the spectral theory for compact operators?
$endgroup$
– AlephNull
Dec 24 '18 at 22:09
$begingroup$
$A$ is a compact operator, so it is not invertible. Hence $0$ is in the spectrum. For a compact operator non-zero points in the spectrum are all eigen values. Hence it is enough to find eigen values which has been done by Jacky Chong.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 23:24
2
$begingroup$
@Math1000 If you expand $sin (x-y)$ as $sin,x cos ,y-cos, x sin, y$ you will see that the range of $A$ is contained in the span of sin and cos. Any finite rank operator is compact.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 0:17
|
show 1 more comment
$begingroup$
Let $A:L^2(0,pi) to L^2(0,pi)$ be defined by $(Af)(x)=displaystyleint_{0}^pi sin(x-y)f(y)dy$. Find the point spectrum and spectrum of $A$.
I am not sure how to go about this. I thought to start with, I should find the point spectrum (i.e. the set of eigenvalues). So not knowing any techniques for integral operators, I would just set $displaystyleint_{0}^pi sin(x-y)f(y)dy=lambda f(x)$ and try to solve for $f$. Unfortunately I have no idea how to, or even if it is possible, to solve for $f$ here. I am wondering if there is a trick specific to this integrand or a more general technique regarding integral operators that I should know about. I have a result that tells me $A$ is compact but it's not clear how this could be used. Any hints would be appreciated.
functional-analysis spectral-theory integral-operators
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
$endgroup$
Let $A:L^2(0,pi) to L^2(0,pi)$ be defined by $(Af)(x)=displaystyleint_{0}^pi sin(x-y)f(y)dy$. Find the point spectrum and spectrum of $A$.
I am not sure how to go about this. I thought to start with, I should find the point spectrum (i.e. the set of eigenvalues). So not knowing any techniques for integral operators, I would just set $displaystyleint_{0}^pi sin(x-y)f(y)dy=lambda f(x)$ and try to solve for $f$. Unfortunately I have no idea how to, or even if it is possible, to solve for $f$ here. I am wondering if there is a trick specific to this integrand or a more general technique regarding integral operators that I should know about. I have a result that tells me $A$ is compact but it's not clear how this could be used. Any hints would be appreciated.
functional-analysis spectral-theory integral-operators
functional-analysis spectral-theory integral-operators
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
asked Dec 24 '18 at 20:40
AlephNullAlephNull
559110
559110
$begingroup$
Note that $Af(x) = (sinstar f)(x)$, where $star$ denotes convolution.
$endgroup$
– Math1000
Dec 24 '18 at 22:04
$begingroup$
See here: math.stackexchange.com/questions/1117098/…
$endgroup$
– Math1000
Dec 24 '18 at 22:05
$begingroup$
We have never actually seen or used Fourier transforms so far, so it is strange that we would be set such an exercise. I suppose there is no more elementary way of finding the spectrum, or by using the spectral theory for compact operators?
$endgroup$
– AlephNull
Dec 24 '18 at 22:09
$begingroup$
$A$ is a compact operator, so it is not invertible. Hence $0$ is in the spectrum. For a compact operator non-zero points in the spectrum are all eigen values. Hence it is enough to find eigen values which has been done by Jacky Chong.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 23:24
2
$begingroup$
@Math1000 If you expand $sin (x-y)$ as $sin,x cos ,y-cos, x sin, y$ you will see that the range of $A$ is contained in the span of sin and cos. Any finite rank operator is compact.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 0:17
|
show 1 more comment
$begingroup$
Note that $Af(x) = (sinstar f)(x)$, where $star$ denotes convolution.
$endgroup$
– Math1000
Dec 24 '18 at 22:04
$begingroup$
See here: math.stackexchange.com/questions/1117098/…
$endgroup$
– Math1000
Dec 24 '18 at 22:05
$begingroup$
We have never actually seen or used Fourier transforms so far, so it is strange that we would be set such an exercise. I suppose there is no more elementary way of finding the spectrum, or by using the spectral theory for compact operators?
$endgroup$
– AlephNull
Dec 24 '18 at 22:09
$begingroup$
$A$ is a compact operator, so it is not invertible. Hence $0$ is in the spectrum. For a compact operator non-zero points in the spectrum are all eigen values. Hence it is enough to find eigen values which has been done by Jacky Chong.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 23:24
2
$begingroup$
@Math1000 If you expand $sin (x-y)$ as $sin,x cos ,y-cos, x sin, y$ you will see that the range of $A$ is contained in the span of sin and cos. Any finite rank operator is compact.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 0:17
$begingroup$
Note that $Af(x) = (sinstar f)(x)$, where $star$ denotes convolution.
$endgroup$
– Math1000
Dec 24 '18 at 22:04
$begingroup$
Note that $Af(x) = (sinstar f)(x)$, where $star$ denotes convolution.
$endgroup$
– Math1000
Dec 24 '18 at 22:04
$begingroup$
See here: math.stackexchange.com/questions/1117098/…
$endgroup$
– Math1000
Dec 24 '18 at 22:05
$begingroup$
See here: math.stackexchange.com/questions/1117098/…
$endgroup$
– Math1000
Dec 24 '18 at 22:05
$begingroup$
We have never actually seen or used Fourier transforms so far, so it is strange that we would be set such an exercise. I suppose there is no more elementary way of finding the spectrum, or by using the spectral theory for compact operators?
$endgroup$
– AlephNull
Dec 24 '18 at 22:09
$begingroup$
We have never actually seen or used Fourier transforms so far, so it is strange that we would be set such an exercise. I suppose there is no more elementary way of finding the spectrum, or by using the spectral theory for compact operators?
$endgroup$
– AlephNull
Dec 24 '18 at 22:09
$begingroup$
$A$ is a compact operator, so it is not invertible. Hence $0$ is in the spectrum. For a compact operator non-zero points in the spectrum are all eigen values. Hence it is enough to find eigen values which has been done by Jacky Chong.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 23:24
$begingroup$
$A$ is a compact operator, so it is not invertible. Hence $0$ is in the spectrum. For a compact operator non-zero points in the spectrum are all eigen values. Hence it is enough to find eigen values which has been done by Jacky Chong.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 23:24
2
2
$begingroup$
@Math1000 If you expand $sin (x-y)$ as $sin,x cos ,y-cos, x sin, y$ you will see that the range of $A$ is contained in the span of sin and cos. Any finite rank operator is compact.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 0:17
$begingroup$
@Math1000 If you expand $sin (x-y)$ as $sin,x cos ,y-cos, x sin, y$ you will see that the range of $A$ is contained in the span of sin and cos. Any finite rank operator is compact.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 0:17
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Observe
begin{align}
Tf(x)=int^pi_0 sin(x-y) f(y) dy = left(int^pi_0 cos(y)f(y) dyright) sin(x)-left(int^pi_0 sin(y)f(y) dyright)cos(x)
end{align}
which means $mathcal{R}(T) = operatorname{span}{cos(x), sin(x)}$ and $dimmathcal{R}(T)=2$. Next, consider $f(x) = Acos(x)+ Bsin(x)$, then we see that
begin{align}
Tf(x) = frac{pi A}{2}sin(x)-frac{pi B}{2}cos(x) = lambda (Acos(x)+ Bsin(x))
end{align}
iff $lambda A = -frac{pi B}{2}$ and $lambda B= frac{pi A}{2}$. Solving for $A$ and $B$ yields
begin{align}
lambda^2 +frac{pi^2}{4} =0 implies lambda = pm frac{pi}{2}i
end{align}
which means $f_1(x) = cos(x)-isin(x)$ and $f_2(x) = cos(x)+isin(x)$ are both eigenfunctions of $T$.
Remark: Note that if $L^2(0, pi)$ is viewed as a real vector space, then $T$ has no real eigenvalue other than $0$.
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
$endgroup$
$begingroup$
Great answer, might be worth mentioning that the first equality is due to $sin(x-y)=sin(x)cos(y) - sin(y)cos(x)$ though. At first I thought you invoked integration by parts which was confusing.
$endgroup$
– Math1000
Dec 24 '18 at 23:58
$begingroup$
Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum.
$endgroup$
– AlephNull
Dec 25 '18 at 10:38
$begingroup$
@AlephNull Note the operator is compact.
$endgroup$
– Jacky Chong
Dec 25 '18 at 19:07
$begingroup$
Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately.
$endgroup$
– AlephNull
Dec 25 '18 at 19:15
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
votes
$begingroup$
Observe
begin{align}
Tf(x)=int^pi_0 sin(x-y) f(y) dy = left(int^pi_0 cos(y)f(y) dyright) sin(x)-left(int^pi_0 sin(y)f(y) dyright)cos(x)
end{align}
which means $mathcal{R}(T) = operatorname{span}{cos(x), sin(x)}$ and $dimmathcal{R}(T)=2$. Next, consider $f(x) = Acos(x)+ Bsin(x)$, then we see that
begin{align}
Tf(x) = frac{pi A}{2}sin(x)-frac{pi B}{2}cos(x) = lambda (Acos(x)+ Bsin(x))
end{align}
iff $lambda A = -frac{pi B}{2}$ and $lambda B= frac{pi A}{2}$. Solving for $A$ and $B$ yields
begin{align}
lambda^2 +frac{pi^2}{4} =0 implies lambda = pm frac{pi}{2}i
end{align}
which means $f_1(x) = cos(x)-isin(x)$ and $f_2(x) = cos(x)+isin(x)$ are both eigenfunctions of $T$.
Remark: Note that if $L^2(0, pi)$ is viewed as a real vector space, then $T$ has no real eigenvalue other than $0$.
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
$endgroup$
$begingroup$
Great answer, might be worth mentioning that the first equality is due to $sin(x-y)=sin(x)cos(y) - sin(y)cos(x)$ though. At first I thought you invoked integration by parts which was confusing.
$endgroup$
– Math1000
Dec 24 '18 at 23:58
$begingroup$
Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum.
$endgroup$
– AlephNull
Dec 25 '18 at 10:38
$begingroup$
@AlephNull Note the operator is compact.
$endgroup$
– Jacky Chong
Dec 25 '18 at 19:07
$begingroup$
Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately.
$endgroup$
– AlephNull
Dec 25 '18 at 19:15
add a comment |
$begingroup$
Observe
begin{align}
Tf(x)=int^pi_0 sin(x-y) f(y) dy = left(int^pi_0 cos(y)f(y) dyright) sin(x)-left(int^pi_0 sin(y)f(y) dyright)cos(x)
end{align}
which means $mathcal{R}(T) = operatorname{span}{cos(x), sin(x)}$ and $dimmathcal{R}(T)=2$. Next, consider $f(x) = Acos(x)+ Bsin(x)$, then we see that
begin{align}
Tf(x) = frac{pi A}{2}sin(x)-frac{pi B}{2}cos(x) = lambda (Acos(x)+ Bsin(x))
end{align}
iff $lambda A = -frac{pi B}{2}$ and $lambda B= frac{pi A}{2}$. Solving for $A$ and $B$ yields
begin{align}
lambda^2 +frac{pi^2}{4} =0 implies lambda = pm frac{pi}{2}i
end{align}
which means $f_1(x) = cos(x)-isin(x)$ and $f_2(x) = cos(x)+isin(x)$ are both eigenfunctions of $T$.
Remark: Note that if $L^2(0, pi)$ is viewed as a real vector space, then $T$ has no real eigenvalue other than $0$.
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
$endgroup$
$begingroup$
Great answer, might be worth mentioning that the first equality is due to $sin(x-y)=sin(x)cos(y) - sin(y)cos(x)$ though. At first I thought you invoked integration by parts which was confusing.
$endgroup$
– Math1000
Dec 24 '18 at 23:58
$begingroup$
Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum.
$endgroup$
– AlephNull
Dec 25 '18 at 10:38
$begingroup$
@AlephNull Note the operator is compact.
$endgroup$
– Jacky Chong
Dec 25 '18 at 19:07
$begingroup$
Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately.
$endgroup$
– AlephNull
Dec 25 '18 at 19:15
add a comment |
$begingroup$
Observe
begin{align}
Tf(x)=int^pi_0 sin(x-y) f(y) dy = left(int^pi_0 cos(y)f(y) dyright) sin(x)-left(int^pi_0 sin(y)f(y) dyright)cos(x)
end{align}
which means $mathcal{R}(T) = operatorname{span}{cos(x), sin(x)}$ and $dimmathcal{R}(T)=2$. Next, consider $f(x) = Acos(x)+ Bsin(x)$, then we see that
begin{align}
Tf(x) = frac{pi A}{2}sin(x)-frac{pi B}{2}cos(x) = lambda (Acos(x)+ Bsin(x))
end{align}
iff $lambda A = -frac{pi B}{2}$ and $lambda B= frac{pi A}{2}$. Solving for $A$ and $B$ yields
begin{align}
lambda^2 +frac{pi^2}{4} =0 implies lambda = pm frac{pi}{2}i
end{align}
which means $f_1(x) = cos(x)-isin(x)$ and $f_2(x) = cos(x)+isin(x)$ are both eigenfunctions of $T$.
Remark: Note that if $L^2(0, pi)$ is viewed as a real vector space, then $T$ has no real eigenvalue other than $0$.
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
$endgroup$
Observe
begin{align}
Tf(x)=int^pi_0 sin(x-y) f(y) dy = left(int^pi_0 cos(y)f(y) dyright) sin(x)-left(int^pi_0 sin(y)f(y) dyright)cos(x)
end{align}
which means $mathcal{R}(T) = operatorname{span}{cos(x), sin(x)}$ and $dimmathcal{R}(T)=2$. Next, consider $f(x) = Acos(x)+ Bsin(x)$, then we see that
begin{align}
Tf(x) = frac{pi A}{2}sin(x)-frac{pi B}{2}cos(x) = lambda (Acos(x)+ Bsin(x))
end{align}
iff $lambda A = -frac{pi B}{2}$ and $lambda B= frac{pi A}{2}$. Solving for $A$ and $B$ yields
begin{align}
lambda^2 +frac{pi^2}{4} =0 implies lambda = pm frac{pi}{2}i
end{align}
which means $f_1(x) = cos(x)-isin(x)$ and $f_2(x) = cos(x)+isin(x)$ are both eigenfunctions of $T$.
Remark: Note that if $L^2(0, pi)$ is viewed as a real vector space, then $T$ has no real eigenvalue other than $0$.
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
answered Dec 24 '18 at 23:17
Jacky ChongJacky Chong
19.3k21129
19.3k21129
$begingroup$
Great answer, might be worth mentioning that the first equality is due to $sin(x-y)=sin(x)cos(y) - sin(y)cos(x)$ though. At first I thought you invoked integration by parts which was confusing.
$endgroup$
– Math1000
Dec 24 '18 at 23:58
$begingroup$
Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum.
$endgroup$
– AlephNull
Dec 25 '18 at 10:38
$begingroup$
@AlephNull Note the operator is compact.
$endgroup$
– Jacky Chong
Dec 25 '18 at 19:07
$begingroup$
Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately.
$endgroup$
– AlephNull
Dec 25 '18 at 19:15
add a comment |
$begingroup$
Great answer, might be worth mentioning that the first equality is due to $sin(x-y)=sin(x)cos(y) - sin(y)cos(x)$ though. At first I thought you invoked integration by parts which was confusing.
$endgroup$
– Math1000
Dec 24 '18 at 23:58
$begingroup$
Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum.
$endgroup$
– AlephNull
Dec 25 '18 at 10:38
$begingroup$
@AlephNull Note the operator is compact.
$endgroup$
– Jacky Chong
Dec 25 '18 at 19:07
$begingroup$
Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately.
$endgroup$
– AlephNull
Dec 25 '18 at 19:15
$begingroup$
Great answer, might be worth mentioning that the first equality is due to $sin(x-y)=sin(x)cos(y) - sin(y)cos(x)$ though. At first I thought you invoked integration by parts which was confusing.
$endgroup$
– Math1000
Dec 24 '18 at 23:58
$begingroup$
Great answer, might be worth mentioning that the first equality is due to $sin(x-y)=sin(x)cos(y) - sin(y)cos(x)$ though. At first I thought you invoked integration by parts which was confusing.
$endgroup$
– Math1000
Dec 24 '18 at 23:58
$begingroup$
Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum.
$endgroup$
– AlephNull
Dec 25 '18 at 10:38
$begingroup$
Somehow I forgot that I could expand the sine. Very nice solution for the point spectrum. However I don't see how to get the full spectrum from here. The operator isn't self-adjoint so I can't use that the spectrum is the closure of the point spectrum.
$endgroup$
– AlephNull
Dec 25 '18 at 10:38
$begingroup$
@AlephNull Note the operator is compact.
$endgroup$
– Jacky Chong
Dec 25 '18 at 19:07
$begingroup$
@AlephNull Note the operator is compact.
$endgroup$
– Jacky Chong
Dec 25 '18 at 19:07
$begingroup$
Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately.
$endgroup$
– AlephNull
Dec 25 '18 at 19:15
$begingroup$
Yeah I did some googling and found results such as '$0$ is in the spectrum of a compact operator on an infinite-dimensional space' and 'for compact operators, every nonzero element of the spectrum is an eigenvalue'. Strangely I have not come across such results before. Not even sure why this type of operator is compact. But I suppose I should look into these things separately.
$endgroup$
– AlephNull
Dec 25 '18 at 19:15
add a comment |
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Note that $Af(x) = (sinstar f)(x)$, where $star$ denotes convolution.
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– Math1000
Dec 24 '18 at 22:04
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See here: math.stackexchange.com/questions/1117098/…
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– Math1000
Dec 24 '18 at 22:05
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We have never actually seen or used Fourier transforms so far, so it is strange that we would be set such an exercise. I suppose there is no more elementary way of finding the spectrum, or by using the spectral theory for compact operators?
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– AlephNull
Dec 24 '18 at 22:09
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$A$ is a compact operator, so it is not invertible. Hence $0$ is in the spectrum. For a compact operator non-zero points in the spectrum are all eigen values. Hence it is enough to find eigen values which has been done by Jacky Chong.
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– Kavi Rama Murthy
Dec 24 '18 at 23:24
2
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@Math1000 If you expand $sin (x-y)$ as $sin,x cos ,y-cos, x sin, y$ you will see that the range of $A$ is contained in the span of sin and cos. Any finite rank operator is compact.
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– Kavi Rama Murthy
Dec 25 '18 at 0:17