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Calculate supreme of $((T+T^*)f,f)$
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Let $0<theta <pi $, and continuous linear operator $T:L^2(0,1)to L^2(0,1)$ $Tf(x):=int _0^x e^itheta f(t)dt$
Then, what is $sup_int_0 ^1 (T+T^*)f(t) barf(t)dt$ ?
My idea :$T+T^*$ is symmetry, so $int (T+T^*)f(t)barf(t)dt$ is real value.
I calculated $int_0 ^1 (T+T^*)f(t) barf(t)dt=int int costhetaf(t)barf(x)+isintheta barf(x)(chi _[0,x](t)-chi_[0,t](x))dtdx$ by Fubini's theorem, but I don't know how to find sup.
I guess supremum is $sqrt2costheta$
functional-analysis operator-theory hilbert-spaces lebesgue-integral
asked Aug 1 at 12:38
Yan Wen Lie
867
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up vote
2
down vote
favorite
Let $0<theta <pi $, and continuous linear operator $T:L^2(0,1)to L^2(0,1)$ $Tf(x):=int _0^x e^itheta f(t)dt$
Then, what is $sup_int_0 ^1 (T+T^*)f(t) barf(t)dt$ ?
My idea :$T+T^*$ is symmetry, so $int (T+T^*)f(t)barf(t)dt$ is real value.
I calculated $int_0 ^1 (T+T^*)f(t) barf(t)dt=int int costhetaf(t)barf(x)+isintheta barf(x)(chi _[0,x](t)-chi_[0,t](x))dtdx$ by Fubini's theorem, but I don't know how to find sup.
I guess supremum is $sqrt2costheta$
functional-analysis operator-theory hilbert-spaces lebesgue-integral
asked Aug 1 at 12:38
Yan Wen Lie
867
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $0<theta <pi $, and continuous linear operator $T:L^2(0,1)to L^2(0,1)$ $Tf(x):=int _0^x e^itheta f(t)dt$
Then, what is $sup_int_0 ^1 (T+T^*)f(t) barf(t)dt$ ?
My idea :$T+T^*$ is symmetry, so $int (T+T^*)f(t)barf(t)dt$ is real value.
I calculated $int_0 ^1 (T+T^*)f(t) barf(t)dt=int int costhetaf(t)barf(x)+isintheta barf(x)(chi _[0,x](t)-chi_[0,t](x))dtdx$ by Fubini's theorem, but I don't know how to find sup.
I guess supremum is $sqrt2costheta$
functional-analysis operator-theory hilbert-spaces lebesgue-integral
asked Aug 1 at 12:38
Yan Wen Lie
867
Let $0<theta <pi $, and continuous linear operator $T:L^2(0,1)to L^2(0,1)$ $Tf(x):=int _0^x e^itheta f(t)dt$
Then, what is $sup_int_0 ^1 (T+T^*)f(t) barf(t)dt$ ?
My idea :$T+T^*$ is symmetry, so $int (T+T^*)f(t)barf(t)dt$ is real value.
I calculated $int_0 ^1 (T+T^*)f(t) barf(t)dt=int int costhetaf(t)barf(x)+isintheta barf(x)(chi _[0,x](t)-chi_[0,t](x))dtdx$ by Fubini's theorem, but I don't know how to find sup.
I guess supremum is $sqrt2costheta$
functional-analysis operator-theory hilbert-spaces lebesgue-integral
asked Aug 1 at 12:38
Yan Wen Lie
867
asked Aug 1 at 12:38
Yan Wen Lie
867
asked Aug 1 at 12:38
Yan Wen Lie
867
asked Aug 1 at 12:38
Yan Wen Lie
867
867
add a comment |Â
add a comment |Â
2 Answers
2
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0
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The adjoint of $Tf = e^ithetaint_0^xf(t)dt$ is $T^*f = e^-ithetaint_x^1f(t)dt$. Both $T$ and $T^*$ are compact operators and, therefore $T+T^*$ is selfadjoint and compact. Therefore,
$$
sup_=1langle (T+T^*)f,frangle
$$
is the largest eigenvalue of $T+T^*$. The eigenvalue problem for $T+T^*$ is
$$ e^ithetaint_0^xf(t)dt+e^-ithetaint_x^1f(t)dt=lambda f
$$
Any such $f$ must satisfy $lambda f(0)=e^-ithetaint_0^1f(t)dt$ and $lambda f(1)=e^ithetaint_0^1f(t)dt$, or
$$
e^ithetaf(0)-e^-ithetaf(1)=0.
$$
$$
f(1)=e^2ithetaf(0).
$$
And, $f$ must satisfy the ODE
$$
e^ithetaf(x)-e^-ithetaf(x)=lambda f'(x)
$$
$$
2isintheta f(x)=lambda f'(x)
$$
$$
fracf'(x)f(x)=frac2isinthetalambda
$$
$$
f(x) = Ce^2isin(theta) x/lambda
$$
The endpoint condition is satisfied for $Cne 0$ iff
$$
e^2isin(theta)/lambda=e^2itheta
$$
$$
sin(theta)/lambda=theta+pi n
$$
$$
lambda = fracsin(theta)theta+pi n.
$$
So the supremum is the largest value of $lambda$ given above for $n=0,pm 1,pm 2,cdots$.
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
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-1
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1) The set of which you are computing supremum of, is a subset of the real numbers.
2) Therefore, the supremum is the maximum of the closure.
3) Every point in the spectrum of $T+T^*$ is in that subset.
4) Since $T+T^*$ is normal, then its spectral radius is $|T+T^*|$.
5) $((T+T^*)f,f)leq |T+T^*|$ for $|f|=1$.
It follows that the supremum that you need is $|T+T^*|$.
answered Aug 1 at 12:48
whatsgoingon
152
1
I think it is trivial. And it is difficult to calculate $||T+T^*||$
– B.T.O
Aug 1 at 13:01
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The adjoint of $Tf = e^ithetaint_0^xf(t)dt$ is $T^*f = e^-ithetaint_x^1f(t)dt$. Both $T$ and $T^*$ are compact operators and, therefore $T+T^*$ is selfadjoint and compact. Therefore,
$$
sup_=1langle (T+T^*)f,frangle
$$
is the largest eigenvalue of $T+T^*$. The eigenvalue problem for $T+T^*$ is
$$ e^ithetaint_0^xf(t)dt+e^-ithetaint_x^1f(t)dt=lambda f
$$
Any such $f$ must satisfy $lambda f(0)=e^-ithetaint_0^1f(t)dt$ and $lambda f(1)=e^ithetaint_0^1f(t)dt$, or
$$
e^ithetaf(0)-e^-ithetaf(1)=0.
$$
$$
f(1)=e^2ithetaf(0).
$$
And, $f$ must satisfy the ODE
$$
e^ithetaf(x)-e^-ithetaf(x)=lambda f'(x)
$$
$$
2isintheta f(x)=lambda f'(x)
$$
$$
fracf'(x)f(x)=frac2isinthetalambda
$$
$$
f(x) = Ce^2isin(theta) x/lambda
$$
The endpoint condition is satisfied for $Cne 0$ iff
$$
e^2isin(theta)/lambda=e^2itheta
$$
$$
sin(theta)/lambda=theta+pi n
$$
$$
lambda = fracsin(theta)theta+pi n.
$$
So the supremum is the largest value of $lambda$ given above for $n=0,pm 1,pm 2,cdots$.
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
add a comment |Â
up vote
0
down vote
accepted
The adjoint of $Tf = e^ithetaint_0^xf(t)dt$ is $T^*f = e^-ithetaint_x^1f(t)dt$. Both $T$ and $T^*$ are compact operators and, therefore $T+T^*$ is selfadjoint and compact. Therefore,
$$
sup_=1langle (T+T^*)f,frangle
$$
is the largest eigenvalue of $T+T^*$. The eigenvalue problem for $T+T^*$ is
$$ e^ithetaint_0^xf(t)dt+e^-ithetaint_x^1f(t)dt=lambda f
$$
Any such $f$ must satisfy $lambda f(0)=e^-ithetaint_0^1f(t)dt$ and $lambda f(1)=e^ithetaint_0^1f(t)dt$, or
$$
e^ithetaf(0)-e^-ithetaf(1)=0.
$$
$$
f(1)=e^2ithetaf(0).
$$
And, $f$ must satisfy the ODE
$$
e^ithetaf(x)-e^-ithetaf(x)=lambda f'(x)
$$
$$
2isintheta f(x)=lambda f'(x)
$$
$$
fracf'(x)f(x)=frac2isinthetalambda
$$
$$
f(x) = Ce^2isin(theta) x/lambda
$$
The endpoint condition is satisfied for $Cne 0$ iff
$$
e^2isin(theta)/lambda=e^2itheta
$$
$$
sin(theta)/lambda=theta+pi n
$$
$$
lambda = fracsin(theta)theta+pi n.
$$
So the supremum is the largest value of $lambda$ given above for $n=0,pm 1,pm 2,cdots$.
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The adjoint of $Tf = e^ithetaint_0^xf(t)dt$ is $T^*f = e^-ithetaint_x^1f(t)dt$. Both $T$ and $T^*$ are compact operators and, therefore $T+T^*$ is selfadjoint and compact. Therefore,
$$
sup_=1langle (T+T^*)f,frangle
$$
is the largest eigenvalue of $T+T^*$. The eigenvalue problem for $T+T^*$ is
$$ e^ithetaint_0^xf(t)dt+e^-ithetaint_x^1f(t)dt=lambda f
$$
Any such $f$ must satisfy $lambda f(0)=e^-ithetaint_0^1f(t)dt$ and $lambda f(1)=e^ithetaint_0^1f(t)dt$, or
$$
e^ithetaf(0)-e^-ithetaf(1)=0.
$$
$$
f(1)=e^2ithetaf(0).
$$
And, $f$ must satisfy the ODE
$$
e^ithetaf(x)-e^-ithetaf(x)=lambda f'(x)
$$
$$
2isintheta f(x)=lambda f'(x)
$$
$$
fracf'(x)f(x)=frac2isinthetalambda
$$
$$
f(x) = Ce^2isin(theta) x/lambda
$$
The endpoint condition is satisfied for $Cne 0$ iff
$$
e^2isin(theta)/lambda=e^2itheta
$$
$$
sin(theta)/lambda=theta+pi n
$$
$$
lambda = fracsin(theta)theta+pi n.
$$
So the supremum is the largest value of $lambda$ given above for $n=0,pm 1,pm 2,cdots$.
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
The adjoint of $Tf = e^ithetaint_0^xf(t)dt$ is $T^*f = e^-ithetaint_x^1f(t)dt$. Both $T$ and $T^*$ are compact operators and, therefore $T+T^*$ is selfadjoint and compact. Therefore,
$$
sup_=1langle (T+T^*)f,frangle
$$
is the largest eigenvalue of $T+T^*$. The eigenvalue problem for $T+T^*$ is
$$ e^ithetaint_0^xf(t)dt+e^-ithetaint_x^1f(t)dt=lambda f
$$
Any such $f$ must satisfy $lambda f(0)=e^-ithetaint_0^1f(t)dt$ and $lambda f(1)=e^ithetaint_0^1f(t)dt$, or
$$
e^ithetaf(0)-e^-ithetaf(1)=0.
$$
$$
f(1)=e^2ithetaf(0).
$$
And, $f$ must satisfy the ODE
$$
e^ithetaf(x)-e^-ithetaf(x)=lambda f'(x)
$$
$$
2isintheta f(x)=lambda f'(x)
$$
$$
fracf'(x)f(x)=frac2isinthetalambda
$$
$$
f(x) = Ce^2isin(theta) x/lambda
$$
The endpoint condition is satisfied for $Cne 0$ iff
$$
e^2isin(theta)/lambda=e^2itheta
$$
$$
sin(theta)/lambda=theta+pi n
$$
$$
lambda = fracsin(theta)theta+pi n.
$$
So the supremum is the largest value of $lambda$ given above for $n=0,pm 1,pm 2,cdots$.
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
answered Aug 4 at 7:16
DisintegratingByParts
55.5k42273
55.5k42273
add a comment |Â
add a comment |Â
up vote
-1
down vote
1) The set of which you are computing supremum of, is a subset of the real numbers.
2) Therefore, the supremum is the maximum of the closure.
3) Every point in the spectrum of $T+T^*$ is in that subset.
4) Since $T+T^*$ is normal, then its spectral radius is $|T+T^*|$.
5) $((T+T^*)f,f)leq |T+T^*|$ for $|f|=1$.
It follows that the supremum that you need is $|T+T^*|$.
answered Aug 1 at 12:48
whatsgoingon
152
1
I think it is trivial. And it is difficult to calculate $||T+T^*||$
– B.T.O
Aug 1 at 13:01
add a comment |Â
up vote
-1
down vote
1) The set of which you are computing supremum of, is a subset of the real numbers.
2) Therefore, the supremum is the maximum of the closure.
3) Every point in the spectrum of $T+T^*$ is in that subset.
4) Since $T+T^*$ is normal, then its spectral radius is $|T+T^*|$.
5) $((T+T^*)f,f)leq |T+T^*|$ for $|f|=1$.
It follows that the supremum that you need is $|T+T^*|$.
answered Aug 1 at 12:48
whatsgoingon
152
1
I think it is trivial. And it is difficult to calculate $||T+T^*||$
– B.T.O
Aug 1 at 13:01
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
1) The set of which you are computing supremum of, is a subset of the real numbers.
2) Therefore, the supremum is the maximum of the closure.
3) Every point in the spectrum of $T+T^*$ is in that subset.
4) Since $T+T^*$ is normal, then its spectral radius is $|T+T^*|$.
5) $((T+T^*)f,f)leq |T+T^*|$ for $|f|=1$.
It follows that the supremum that you need is $|T+T^*|$.
answered Aug 1 at 12:48
whatsgoingon
152
1) The set of which you are computing supremum of, is a subset of the real numbers.
2) Therefore, the supremum is the maximum of the closure.
3) Every point in the spectrum of $T+T^*$ is in that subset.
4) Since $T+T^*$ is normal, then its spectral radius is $|T+T^*|$.
5) $((T+T^*)f,f)leq |T+T^*|$ for $|f|=1$.
It follows that the supremum that you need is $|T+T^*|$.
answered Aug 1 at 12:48
whatsgoingon
152
answered Aug 1 at 12:48
whatsgoingon
152
answered Aug 1 at 12:48
whatsgoingon
152
answered Aug 1 at 12:48
whatsgoingon
152
152
1
I think it is trivial. And it is difficult to calculate $||T+T^*||$
– B.T.O
Aug 1 at 13:01
add a comment |Â
1
I think it is trivial. And it is difficult to calculate $||T+T^*||$
– B.T.O
Aug 1 at 13:01
1
1
I think it is trivial. And it is difficult to calculate $||T+T^*||$
– B.T.O
Aug 1 at 13:01
I think it is trivial. And it is difficult to calculate $||T+T^*||$
– B.T.O
Aug 1 at 13:01
add a comment |Â
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