Integration with Spectral Measure
$begingroup$
Now, suppose I have a compactly supported spectral measure $E$ defined on the Borel subsets of $mathbb{C}$. My question is this, suppose we have a sequence of functions ${f_n}$ non necessarily continuous, converging uniformly to $f$ continuous, is it true that $||int f_{n}dE-int fdE||to 0$? I know the result holds in strong operator topology.
functional-analysis operator-theory spectral-theory
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
$endgroup$
add a comment |
$begingroup$
Now, suppose I have a compactly supported spectral measure $E$ defined on the Borel subsets of $mathbb{C}$. My question is this, suppose we have a sequence of functions ${f_n}$ non necessarily continuous, converging uniformly to $f$ continuous, is it true that $||int f_{n}dE-int fdE||to 0$? I know the result holds in strong operator topology.
functional-analysis operator-theory spectral-theory
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
$endgroup$
add a comment |
$begingroup$
Now, suppose I have a compactly supported spectral measure $E$ defined on the Borel subsets of $mathbb{C}$. My question is this, suppose we have a sequence of functions ${f_n}$ non necessarily continuous, converging uniformly to $f$ continuous, is it true that $||int f_{n}dE-int fdE||to 0$? I know the result holds in strong operator topology.
functional-analysis operator-theory spectral-theory
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
$endgroup$
Now, suppose I have a compactly supported spectral measure $E$ defined on the Borel subsets of $mathbb{C}$. My question is this, suppose we have a sequence of functions ${f_n}$ non necessarily continuous, converging uniformly to $f$ continuous, is it true that $||int f_{n}dE-int fdE||to 0$? I know the result holds in strong operator topology.
functional-analysis operator-theory spectral-theory
functional-analysis operator-theory spectral-theory
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
edited Dec 9 '18 at 20:17
Martin Argerami
127k1182182
127k1182182
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
asked Dec 9 '18 at 19:06
Kesav KrishnanKesav Krishnan
201111
201111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes. Write $E_h(Delta)=langle E(Delta)h,hrangle$. Then
$$
leftlangle
left|left(int_{mathbb C} (f_n(t)-f(t)),dE(t)right)h,h
rightrangleright|
=left|int_{mathbb C}(f_n(t)-f(t)),dE_h right|
leqint_{mathbb C}|f_n(t)-f(t)|,dE_hleq|f_n-f|_infty.
$$
As you can do this for any $h$,
$$
left|int_{mathbb C} (f_n(t)-f(t)),dE(t)right|leq|f_n-f|_infty.
$$
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. Write $E_h(Delta)=langle E(Delta)h,hrangle$. Then
$$
leftlangle
left|left(int_{mathbb C} (f_n(t)-f(t)),dE(t)right)h,h
rightrangleright|
=left|int_{mathbb C}(f_n(t)-f(t)),dE_h right|
leqint_{mathbb C}|f_n(t)-f(t)|,dE_hleq|f_n-f|_infty.
$$
As you can do this for any $h$,
$$
left|int_{mathbb C} (f_n(t)-f(t)),dE(t)right|leq|f_n-f|_infty.
$$
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
add a comment |
$begingroup$
Yes. Write $E_h(Delta)=langle E(Delta)h,hrangle$. Then
$$
leftlangle
left|left(int_{mathbb C} (f_n(t)-f(t)),dE(t)right)h,h
rightrangleright|
=left|int_{mathbb C}(f_n(t)-f(t)),dE_h right|
leqint_{mathbb C}|f_n(t)-f(t)|,dE_hleq|f_n-f|_infty.
$$
As you can do this for any $h$,
$$
left|int_{mathbb C} (f_n(t)-f(t)),dE(t)right|leq|f_n-f|_infty.
$$
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
add a comment |
$begingroup$
Yes. Write $E_h(Delta)=langle E(Delta)h,hrangle$. Then
$$
leftlangle
left|left(int_{mathbb C} (f_n(t)-f(t)),dE(t)right)h,h
rightrangleright|
=left|int_{mathbb C}(f_n(t)-f(t)),dE_h right|
leqint_{mathbb C}|f_n(t)-f(t)|,dE_hleq|f_n-f|_infty.
$$
As you can do this for any $h$,
$$
left|int_{mathbb C} (f_n(t)-f(t)),dE(t)right|leq|f_n-f|_infty.
$$
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
Yes. Write $E_h(Delta)=langle E(Delta)h,hrangle$. Then
$$
leftlangle
left|left(int_{mathbb C} (f_n(t)-f(t)),dE(t)right)h,h
rightrangleright|
=left|int_{mathbb C}(f_n(t)-f(t)),dE_h right|
leqint_{mathbb C}|f_n(t)-f(t)|,dE_hleq|f_n-f|_infty.
$$
As you can do this for any $h$,
$$
left|int_{mathbb C} (f_n(t)-f(t)),dE(t)right|leq|f_n-f|_infty.
$$
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 9 '18 at 20:16
Martin ArgeramiMartin Argerami
127k1182182
127k1182182
add a comment |
add a comment |
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