Self adjoint operators and trace class property
$begingroup$
This is a variation of the problem questioned some time ago.
For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.
Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?
The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$
The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.
For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$
Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
functional-analysis operator-theory
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
$endgroup$
add a comment |
$begingroup$
This is a variation of the problem questioned some time ago.
For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.
Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?
The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$
The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.
For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$
Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
functional-analysis operator-theory
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
$endgroup$
1
$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31
$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35
$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53
$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37
$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10
add a comment |
$begingroup$
This is a variation of the problem questioned some time ago.
For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.
Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?
The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$
The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.
For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$
Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
functional-analysis operator-theory
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
$endgroup$
This is a variation of the problem questioned some time ago.
For a complex Hilbert space $H$ let $T: H rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum
$$sum_{i}{langle |T|e_{i}, e_{i} rangle} < infty$$
converges, where $|T| = (T T^{*})^{frac{1}{2}}$ is the absolute value of the operator.
Assume that
$$sum_{i}{langle Te_{i}, e_{i} rangle}$$
converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?
The progress on the problem is the following:
Given an arbitrary bounded operator $T: H rightarrow H$, one can use the following decomposition
$$T = big( frac{T + T^{*}}{2} big) ^{*} + i big( frac{T^{*} - T}{2i} big) ^{*}$$
The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.
For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to
$$(UT U^{-1})(f(x)) = g(x) f(x)$$
where
$$U: H rightarrow L^{2}(X, mu)$$
$$g in L^{infty}(X, mu)$$
Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
functional-analysis operator-theory
functional-analysis operator-theory
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
edited Dec 22 '18 at 16:33
hyperkahler
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
asked Dec 22 '18 at 15:57
hyperkahlerhyperkahler
1,471714
1,471714
1
$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31
$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35
$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53
$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37
$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10
add a comment |
1
$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31
$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35
$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53
$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37
$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10
1
1
$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31
$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31
$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35
$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35
$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53
$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53
$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37
$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37
$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10
$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10
add a comment |
1 Answer
1
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oldest
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$begingroup$
Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.
Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.
If $T$ is not trace-class, then
$$
operatorname{Tr}(T)=sum_j|lambda_j|=infty.
$$
If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
$$
sum_jlangle Te_j,e_jrangle
$$
cannot converge absolutely.
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
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$begingroup$
Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.
Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.
If $T$ is not trace-class, then
$$
operatorname{Tr}(T)=sum_j|lambda_j|=infty.
$$
If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
$$
sum_jlangle Te_j,e_jrangle
$$
cannot converge absolutely.
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.
Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.
If $T$ is not trace-class, then
$$
operatorname{Tr}(T)=sum_j|lambda_j|=infty.
$$
If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
$$
sum_jlangle Te_j,e_jrangle
$$
cannot converge absolutely.
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.
Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.
If $T$ is not trace-class, then
$$
operatorname{Tr}(T)=sum_j|lambda_j|=infty.
$$
If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
$$
sum_jlangle Te_j,e_jrangle
$$
cannot converge absolutely.
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $lambdainsigma(T)setminus{0}$ and $delta>0$ such that $lambda-delta>0$ and the spectral projection $E_T(lambda-delta,lambda+delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides ${f_j}$ such that $sum_n|langle Te_n,e_nrangle|=infty$.
Knowing that $T$ is compact, by the Spectral Theorem, we know that $$tag1T=sum_jlambda_jP_j,$$ where $lambda_jinmathbb Rsetminus{0}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.
If $T$ is not trace-class, then
$$
operatorname{Tr}(T)=sum_j|lambda_j|=infty.
$$
If we write $lambda_j^+$ for the positive eigenvalues and $lambda_j^-$ for the negative ones, at least one of $sum_jlambda_j^+$ and $sum_jlambda_j^-$ diverges. Then, with ${e_j}$ the orthonormal basis given by $(1)$ (i.e., $P_j=langlecdot,e_jrangle,e_j$), we have that
$$
sum_jlangle Te_j,e_jrangle
$$
cannot converge absolutely.
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
edited Jan 5 at 16:56
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 22 '18 at 19:20
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
add a comment |
add a comment |
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1
$begingroup$
As phrased right now, your are asking how to show that if $sumlangle|T|e_j,e_jrangle<infty$, then $T$ is trace-class; and that's exactly the definition of "trace-class".
$endgroup$
– Martin Argerami
Dec 22 '18 at 16:31
$begingroup$
@MartinArgerami I've made a mistake while typing, now fixed. We should ask that how to prove that the operator is a trace class, provided that $sum_{i}{langle T e_{i}, e_{i} rangle} < infty$
$endgroup$
– hyperkahler
Dec 22 '18 at 16:35
$begingroup$
It's not immediately obvious to me that the answer to my question answers yours.
$endgroup$
– Martin Argerami
Dec 25 '18 at 2:53
$begingroup$
@MartinArgerami Yes, you are right. This answer provides relevant information (math.stackexchange.com/questions/2036398/…)
$endgroup$
– hyperkahler
Jan 5 at 0:37
$begingroup$
hyperkahler: that answer is very wrong.
$endgroup$
– Martin Argerami
Jan 5 at 3:10