Notation for singular measures
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I have a simple question: Is the notation $perp$ for singular measures related to orthogonality? For me, it would make sense if I can provide an inner product to the set of (complex) measures, but I have no idea if there is one (as far as I know there is no one)
Thanks in advance.
measure-theory
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
$endgroup$
add a comment |
$begingroup$
I have a simple question: Is the notation $perp$ for singular measures related to orthogonality? For me, it would make sense if I can provide an inner product to the set of (complex) measures, but I have no idea if there is one (as far as I know there is no one)
Thanks in advance.
measure-theory
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
$endgroup$
$begingroup$
Measures form a Banach algebra but not a Hilbert space.
$endgroup$
– Dunham
Nov 29 '18 at 2:08
1
$begingroup$
If you have an inner product for the space of complex measures corresponding to total variation norm then, by restriction (to absolutely continuous measures) you will get an inner product for $L^{1}$, but Parallelogram Identity fails in $L^{1}$.
$endgroup$
– Kavi Rama Murthy
Nov 29 '18 at 6:15
add a comment |
$begingroup$
I have a simple question: Is the notation $perp$ for singular measures related to orthogonality? For me, it would make sense if I can provide an inner product to the set of (complex) measures, but I have no idea if there is one (as far as I know there is no one)
Thanks in advance.
measure-theory
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
$endgroup$
I have a simple question: Is the notation $perp$ for singular measures related to orthogonality? For me, it would make sense if I can provide an inner product to the set of (complex) measures, but I have no idea if there is one (as far as I know there is no one)
Thanks in advance.
measure-theory
measure-theory
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
asked Nov 29 '18 at 1:57
Math GuyMath Guy
556
556
$begingroup$
Measures form a Banach algebra but not a Hilbert space.
$endgroup$
– Dunham
Nov 29 '18 at 2:08
1
$begingroup$
If you have an inner product for the space of complex measures corresponding to total variation norm then, by restriction (to absolutely continuous measures) you will get an inner product for $L^{1}$, but Parallelogram Identity fails in $L^{1}$.
$endgroup$
– Kavi Rama Murthy
Nov 29 '18 at 6:15
add a comment |
$begingroup$
Measures form a Banach algebra but not a Hilbert space.
$endgroup$
– Dunham
Nov 29 '18 at 2:08
1
$begingroup$
If you have an inner product for the space of complex measures corresponding to total variation norm then, by restriction (to absolutely continuous measures) you will get an inner product for $L^{1}$, but Parallelogram Identity fails in $L^{1}$.
$endgroup$
– Kavi Rama Murthy
Nov 29 '18 at 6:15
$begingroup$
Measures form a Banach algebra but not a Hilbert space.
$endgroup$
– Dunham
Nov 29 '18 at 2:08
$begingroup$
Measures form a Banach algebra but not a Hilbert space.
$endgroup$
– Dunham
Nov 29 '18 at 2:08
1
1
$begingroup$
If you have an inner product for the space of complex measures corresponding to total variation norm then, by restriction (to absolutely continuous measures) you will get an inner product for $L^{1}$, but Parallelogram Identity fails in $L^{1}$.
$endgroup$
– Kavi Rama Murthy
Nov 29 '18 at 6:15
$begingroup$
If you have an inner product for the space of complex measures corresponding to total variation norm then, by restriction (to absolutely continuous measures) you will get an inner product for $L^{1}$, but Parallelogram Identity fails in $L^{1}$.
$endgroup$
– Kavi Rama Murthy
Nov 29 '18 at 6:15
add a comment |
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$begingroup$
Measures form a Banach algebra but not a Hilbert space.
$endgroup$
– Dunham
Nov 29 '18 at 2:08
1
$begingroup$
If you have an inner product for the space of complex measures corresponding to total variation norm then, by restriction (to absolutely continuous measures) you will get an inner product for $L^{1}$, but Parallelogram Identity fails in $L^{1}$.
$endgroup$
– Kavi Rama Murthy
Nov 29 '18 at 6:15