Finding spectrum of an integral operator
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Hello I'm wondering about how to deal with a situation like this:
I have $mathcal{H} = L^2(0,pi /2)$ and
begin{equation}
Tf(x) = int_0^{pi/2} f(y)cos(y)sin(x),dy
end{equation}
First I have to show that $T$ belongs to the space of finite rank operators , but this is trivial cause I can identify Im$(T)$ as span of $h(x) := sin(x)|_{(0,pi/2)}$ (right?). Then I have to determine its spectrum. So I know that
begin{equation}
lambda in sigma_p (T) Leftrightarrowexists fin L^2(0,pi /2), f neq 0
end{equation}
such that $Tf = lambda f$. In this case I have
begin{equation}
sin(x)int_0^{pi/2} f(y)cos(y),dy = lambda f(x)
end{equation}
Should I now derive both sides of equation, assuming $f in C^1 (0,pi/2)$?
functional-analysis spectral-theory
edited Nov 27 at 13:43
MisterRiemann
5,7031624
asked Nov 27 at 13:34
James Arten
579
|
show 3 more comments
up vote
0
down vote
favorite
Hello I'm wondering about how to deal with a situation like this:
I have $mathcal{H} = L^2(0,pi /2)$ and
begin{equation}
Tf(x) = int_0^{pi/2} f(y)cos(y)sin(x),dy
end{equation}
First I have to show that $T$ belongs to the space of finite rank operators , but this is trivial cause I can identify Im$(T)$ as span of $h(x) := sin(x)|_{(0,pi/2)}$ (right?). Then I have to determine its spectrum. So I know that
begin{equation}
lambda in sigma_p (T) Leftrightarrowexists fin L^2(0,pi /2), f neq 0
end{equation}
such that $Tf = lambda f$. In this case I have
begin{equation}
sin(x)int_0^{pi/2} f(y)cos(y),dy = lambda f(x)
end{equation}
Should I now derive both sides of equation, assuming $f in C^1 (0,pi/2)$?
functional-analysis spectral-theory
edited Nov 27 at 13:43
MisterRiemann
5,7031624
asked Nov 27 at 13:34
James Arten
579
1
You don't need to assume $f in C^1$. For eignevectors you know that already because if $f$ is an eigenvector then $f in operatorname{Im}(T) = operatorname{span}(sin) subset C^infty$.
– Rhys Steele
Nov 27 at 14:10
right.. but how should I isolate $f$?
– James Arten
Nov 27 at 14:14
Well try plugging $f = alpha sin$ into your equation since you know that's what $f$ must be if it is an eigenvector. What happens? (in fact you don't even need to know that $f in C^1$ for this)
– Rhys Steele
Nov 27 at 14:19
well I get $lambda = frac{int_0^{pi /2} f(y) cos(y),dy}{alpha}$.. is that right?
– James Arten
Nov 27 at 14:25
1
Yes. What you've done is assumed that $lambda$ is an eigenvalue with eigenvector $f$ and shown that $lambda = 1/2$. Taking $f = sin$ shows that $lambda = 1/2$ is in fact an eigenvalue.
– Rhys Steele
Nov 27 at 16:34
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Hello I'm wondering about how to deal with a situation like this:
I have $mathcal{H} = L^2(0,pi /2)$ and
begin{equation}
Tf(x) = int_0^{pi/2} f(y)cos(y)sin(x),dy
end{equation}
First I have to show that $T$ belongs to the space of finite rank operators , but this is trivial cause I can identify Im$(T)$ as span of $h(x) := sin(x)|_{(0,pi/2)}$ (right?). Then I have to determine its spectrum. So I know that
begin{equation}
lambda in sigma_p (T) Leftrightarrowexists fin L^2(0,pi /2), f neq 0
end{equation}
such that $Tf = lambda f$. In this case I have
begin{equation}
sin(x)int_0^{pi/2} f(y)cos(y),dy = lambda f(x)
end{equation}
Should I now derive both sides of equation, assuming $f in C^1 (0,pi/2)$?
functional-analysis spectral-theory
edited Nov 27 at 13:43
MisterRiemann
5,7031624
asked Nov 27 at 13:34
James Arten
579
Hello I'm wondering about how to deal with a situation like this:
I have $mathcal{H} = L^2(0,pi /2)$ and
begin{equation}
Tf(x) = int_0^{pi/2} f(y)cos(y)sin(x),dy
end{equation}
First I have to show that $T$ belongs to the space of finite rank operators , but this is trivial cause I can identify Im$(T)$ as span of $h(x) := sin(x)|_{(0,pi/2)}$ (right?). Then I have to determine its spectrum. So I know that
begin{equation}
lambda in sigma_p (T) Leftrightarrowexists fin L^2(0,pi /2), f neq 0
end{equation}
such that $Tf = lambda f$. In this case I have
begin{equation}
sin(x)int_0^{pi/2} f(y)cos(y),dy = lambda f(x)
end{equation}
Should I now derive both sides of equation, assuming $f in C^1 (0,pi/2)$?
functional-analysis spectral-theory
functional-analysis spectral-theory
edited Nov 27 at 13:43
MisterRiemann
5,7031624
asked Nov 27 at 13:34
James Arten
579
edited Nov 27 at 13:43
MisterRiemann
5,7031624
asked Nov 27 at 13:34
James Arten
579
edited Nov 27 at 13:43
MisterRiemann
5,7031624
edited Nov 27 at 13:43
MisterRiemann
5,7031624
edited Nov 27 at 13:43
MisterRiemann
5,7031624
5,7031624
asked Nov 27 at 13:34
James Arten
579
asked Nov 27 at 13:34
James Arten
579
asked Nov 27 at 13:34
James Arten
579
579
1
You don't need to assume $f in C^1$. For eignevectors you know that already because if $f$ is an eigenvector then $f in operatorname{Im}(T) = operatorname{span}(sin) subset C^infty$.
– Rhys Steele
Nov 27 at 14:10
right.. but how should I isolate $f$?
– James Arten
Nov 27 at 14:14
Well try plugging $f = alpha sin$ into your equation since you know that's what $f$ must be if it is an eigenvector. What happens? (in fact you don't even need to know that $f in C^1$ for this)
– Rhys Steele
Nov 27 at 14:19
well I get $lambda = frac{int_0^{pi /2} f(y) cos(y),dy}{alpha}$.. is that right?
– James Arten
Nov 27 at 14:25
1
Yes. What you've done is assumed that $lambda$ is an eigenvalue with eigenvector $f$ and shown that $lambda = 1/2$. Taking $f = sin$ shows that $lambda = 1/2$ is in fact an eigenvalue.
– Rhys Steele
Nov 27 at 16:34
|
show 3 more comments
1
You don't need to assume $f in C^1$. For eignevectors you know that already because if $f$ is an eigenvector then $f in operatorname{Im}(T) = operatorname{span}(sin) subset C^infty$.
– Rhys Steele
Nov 27 at 14:10
right.. but how should I isolate $f$?
– James Arten
Nov 27 at 14:14
Well try plugging $f = alpha sin$ into your equation since you know that's what $f$ must be if it is an eigenvector. What happens? (in fact you don't even need to know that $f in C^1$ for this)
– Rhys Steele
Nov 27 at 14:19
well I get $lambda = frac{int_0^{pi /2} f(y) cos(y),dy}{alpha}$.. is that right?
– James Arten
Nov 27 at 14:25
1
Yes. What you've done is assumed that $lambda$ is an eigenvalue with eigenvector $f$ and shown that $lambda = 1/2$. Taking $f = sin$ shows that $lambda = 1/2$ is in fact an eigenvalue.
– Rhys Steele
Nov 27 at 16:34
1
1
You don't need to assume $f in C^1$. For eignevectors you know that already because if $f$ is an eigenvector then $f in operatorname{Im}(T) = operatorname{span}(sin) subset C^infty$.
– Rhys Steele
Nov 27 at 14:10
You don't need to assume $f in C^1$. For eignevectors you know that already because if $f$ is an eigenvector then $f in operatorname{Im}(T) = operatorname{span}(sin) subset C^infty$.
– Rhys Steele
Nov 27 at 14:10
right.. but how should I isolate $f$?
– James Arten
Nov 27 at 14:14
right.. but how should I isolate $f$?
– James Arten
Nov 27 at 14:14
Well try plugging $f = alpha sin$ into your equation since you know that's what $f$ must be if it is an eigenvector. What happens? (in fact you don't even need to know that $f in C^1$ for this)
– Rhys Steele
Nov 27 at 14:19
Well try plugging $f = alpha sin$ into your equation since you know that's what $f$ must be if it is an eigenvector. What happens? (in fact you don't even need to know that $f in C^1$ for this)
– Rhys Steele
Nov 27 at 14:19
well I get $lambda = frac{int_0^{pi /2} f(y) cos(y),dy}{alpha}$.. is that right?
– James Arten
Nov 27 at 14:25
well I get $lambda = frac{int_0^{pi /2} f(y) cos(y),dy}{alpha}$.. is that right?
– James Arten
Nov 27 at 14:25
1
1
Yes. What you've done is assumed that $lambda$ is an eigenvalue with eigenvector $f$ and shown that $lambda = 1/2$. Taking $f = sin$ shows that $lambda = 1/2$ is in fact an eigenvalue.
– Rhys Steele
Nov 27 at 16:34
Yes. What you've done is assumed that $lambda$ is an eigenvalue with eigenvector $f$ and shown that $lambda = 1/2$. Taking $f = sin$ shows that $lambda = 1/2$ is in fact an eigenvalue.
– Rhys Steele
Nov 27 at 16:34
|
show 3 more comments
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1
You don't need to assume $f in C^1$. For eignevectors you know that already because if $f$ is an eigenvector then $f in operatorname{Im}(T) = operatorname{span}(sin) subset C^infty$.
– Rhys Steele
Nov 27 at 14:10
right.. but how should I isolate $f$?
– James Arten
Nov 27 at 14:14
Well try plugging $f = alpha sin$ into your equation since you know that's what $f$ must be if it is an eigenvector. What happens? (in fact you don't even need to know that $f in C^1$ for this)
– Rhys Steele
Nov 27 at 14:19
well I get $lambda = frac{int_0^{pi /2} f(y) cos(y),dy}{alpha}$.. is that right?
– James Arten
Nov 27 at 14:25
1
Yes. What you've done is assumed that $lambda$ is an eigenvalue with eigenvector $f$ and shown that $lambda = 1/2$. Taking $f = sin$ shows that $lambda = 1/2$ is in fact an eigenvalue.
– Rhys Steele
Nov 27 at 16:34