Mixing
Does Radon-Nikodym imply Riesz Representation Theorem?
$begingroup$
In Axler's Linear Algebra Done Right we have the theorem
6.42: (Riesz Representation Theorem) Suppose $V$ is a finite dimensional inner product space and $phi$ is a linear functional on $V$. Then there is a unique vector $u in V$ such that $$phi(v) = langle v, urangle$$ for every $v in V$
I'm currently learning measure theory and have came across Radon-Nikodym
(Radon-Nikodym) Consider a measurable space $(X,mathcal{M})$ on which two $sigma$-finite signed measures $mu,nu$ are defined such that $nu << mu$ ($nu$ is absolutely continuous with respect to $mu$) then there is a $mu$-integrable function $f: X to mathbb{R}$ such that $$nu(E) = int_E f dmu$$ for every $E in mathcal{M}$ and any other function $g$ satisfying this is equal to $f$ almost everywhere with respect to $mu$.
These two theorems seems very similar. Is it possible to go from Radon-Nikodym and get Riesz Representation? The integral in Radon-Nikodym "acts" like the inner product in Riesz Representation, the function $f$ "acts" like the vector $u$ in Riesz, and the signed measure $nu$ acts like the linear functional $phi$.
I am inclined to think that somehow we can recover Riesz from Radon-Nikodym. To start, we would need to somehow get a $sigma$-algebra, $mathcal{M}$ on $V$ so that $(V, mathcal{M})$ is a measurable space. This has to be a very particular $sigma$-algebra so that somehow the integral can be reduced to the inner product on $V$. We would also need to show that the linear functional is absolutely continuous with respect to the inner product.
So is it possible to recover Riesz from Radon-Nikodym? If so, how? If not, what's the issue?
real-analysis linear-algebra analysis measure-theory riesz-representation-theorem
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
$endgroup$
add a comment |
$begingroup$
In Axler's Linear Algebra Done Right we have the theorem
6.42: (Riesz Representation Theorem) Suppose $V$ is a finite dimensional inner product space and $phi$ is a linear functional on $V$. Then there is a unique vector $u in V$ such that $$phi(v) = langle v, urangle$$ for every $v in V$
I'm currently learning measure theory and have came across Radon-Nikodym
(Radon-Nikodym) Consider a measurable space $(X,mathcal{M})$ on which two $sigma$-finite signed measures $mu,nu$ are defined such that $nu << mu$ ($nu$ is absolutely continuous with respect to $mu$) then there is a $mu$-integrable function $f: X to mathbb{R}$ such that $$nu(E) = int_E f dmu$$ for every $E in mathcal{M}$ and any other function $g$ satisfying this is equal to $f$ almost everywhere with respect to $mu$.
These two theorems seems very similar. Is it possible to go from Radon-Nikodym and get Riesz Representation? The integral in Radon-Nikodym "acts" like the inner product in Riesz Representation, the function $f$ "acts" like the vector $u$ in Riesz, and the signed measure $nu$ acts like the linear functional $phi$.
I am inclined to think that somehow we can recover Riesz from Radon-Nikodym. To start, we would need to somehow get a $sigma$-algebra, $mathcal{M}$ on $V$ so that $(V, mathcal{M})$ is a measurable space. This has to be a very particular $sigma$-algebra so that somehow the integral can be reduced to the inner product on $V$. We would also need to show that the linear functional is absolutely continuous with respect to the inner product.
So is it possible to recover Riesz from Radon-Nikodym? If so, how? If not, what's the issue?
real-analysis linear-algebra analysis measure-theory riesz-representation-theorem
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
$endgroup$
$begingroup$
if $phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $lambda(E)=phi(chi_E)$, if $phi$ is positive we can (generalize after) show that $lambda$ is a measure and if the set $E$ has measure $mu(E)=0$ than $lambda(E)=0$ because $phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $phi(f)=int fg dmu$ for some $gin L^q$
$endgroup$
– Robson
Nov 18 '18 at 22:20
add a comment |
$begingroup$
In Axler's Linear Algebra Done Right we have the theorem
6.42: (Riesz Representation Theorem) Suppose $V$ is a finite dimensional inner product space and $phi$ is a linear functional on $V$. Then there is a unique vector $u in V$ such that $$phi(v) = langle v, urangle$$ for every $v in V$
I'm currently learning measure theory and have came across Radon-Nikodym
(Radon-Nikodym) Consider a measurable space $(X,mathcal{M})$ on which two $sigma$-finite signed measures $mu,nu$ are defined such that $nu << mu$ ($nu$ is absolutely continuous with respect to $mu$) then there is a $mu$-integrable function $f: X to mathbb{R}$ such that $$nu(E) = int_E f dmu$$ for every $E in mathcal{M}$ and any other function $g$ satisfying this is equal to $f$ almost everywhere with respect to $mu$.
These two theorems seems very similar. Is it possible to go from Radon-Nikodym and get Riesz Representation? The integral in Radon-Nikodym "acts" like the inner product in Riesz Representation, the function $f$ "acts" like the vector $u$ in Riesz, and the signed measure $nu$ acts like the linear functional $phi$.
I am inclined to think that somehow we can recover Riesz from Radon-Nikodym. To start, we would need to somehow get a $sigma$-algebra, $mathcal{M}$ on $V$ so that $(V, mathcal{M})$ is a measurable space. This has to be a very particular $sigma$-algebra so that somehow the integral can be reduced to the inner product on $V$. We would also need to show that the linear functional is absolutely continuous with respect to the inner product.
So is it possible to recover Riesz from Radon-Nikodym? If so, how? If not, what's the issue?
real-analysis linear-algebra analysis measure-theory riesz-representation-theorem
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
$endgroup$
In Axler's Linear Algebra Done Right we have the theorem
6.42: (Riesz Representation Theorem) Suppose $V$ is a finite dimensional inner product space and $phi$ is a linear functional on $V$. Then there is a unique vector $u in V$ such that $$phi(v) = langle v, urangle$$ for every $v in V$
I'm currently learning measure theory and have came across Radon-Nikodym
(Radon-Nikodym) Consider a measurable space $(X,mathcal{M})$ on which two $sigma$-finite signed measures $mu,nu$ are defined such that $nu << mu$ ($nu$ is absolutely continuous with respect to $mu$) then there is a $mu$-integrable function $f: X to mathbb{R}$ such that $$nu(E) = int_E f dmu$$ for every $E in mathcal{M}$ and any other function $g$ satisfying this is equal to $f$ almost everywhere with respect to $mu$.
These two theorems seems very similar. Is it possible to go from Radon-Nikodym and get Riesz Representation? The integral in Radon-Nikodym "acts" like the inner product in Riesz Representation, the function $f$ "acts" like the vector $u$ in Riesz, and the signed measure $nu$ acts like the linear functional $phi$.
I am inclined to think that somehow we can recover Riesz from Radon-Nikodym. To start, we would need to somehow get a $sigma$-algebra, $mathcal{M}$ on $V$ so that $(V, mathcal{M})$ is a measurable space. This has to be a very particular $sigma$-algebra so that somehow the integral can be reduced to the inner product on $V$. We would also need to show that the linear functional is absolutely continuous with respect to the inner product.
So is it possible to recover Riesz from Radon-Nikodym? If so, how? If not, what's the issue?
real-analysis linear-algebra analysis measure-theory riesz-representation-theorem
real-analysis linear-algebra analysis measure-theory riesz-representation-theorem
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
asked Nov 18 '18 at 17:54
carsandpulsarscarsandpulsars
37817
37817
$begingroup$
if $phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $lambda(E)=phi(chi_E)$, if $phi$ is positive we can (generalize after) show that $lambda$ is a measure and if the set $E$ has measure $mu(E)=0$ than $lambda(E)=0$ because $phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $phi(f)=int fg dmu$ for some $gin L^q$
$endgroup$
– Robson
Nov 18 '18 at 22:20
add a comment |
$begingroup$
if $phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $lambda(E)=phi(chi_E)$, if $phi$ is positive we can (generalize after) show that $lambda$ is a measure and if the set $E$ has measure $mu(E)=0$ than $lambda(E)=0$ because $phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $phi(f)=int fg dmu$ for some $gin L^q$
$endgroup$
– Robson
Nov 18 '18 at 22:20
$begingroup$
if $phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $lambda(E)=phi(chi_E)$, if $phi$ is positive we can (generalize after) show that $lambda$ is a measure and if the set $E$ has measure $mu(E)=0$ than $lambda(E)=0$ because $phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $phi(f)=int fg dmu$ for some $gin L^q$
$endgroup$
– Robson
Nov 18 '18 at 22:20
$begingroup$
if $phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $lambda(E)=phi(chi_E)$, if $phi$ is positive we can (generalize after) show that $lambda$ is a measure and if the set $E$ has measure $mu(E)=0$ than $lambda(E)=0$ because $phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $phi(f)=int fg dmu$ for some $gin L^q$
$endgroup$
– Robson
Nov 18 '18 at 22:20
add a comment |
1 Answer
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$begingroup$
I am not sure about your specific claim, but certainly there is a connection in the "opposite" direction. That is, a general version of the Riesz Representation Theorem is used to prove the Radon-Nikodym Theorem. See "von Neumann's proof" of the Radon-Nikodym Theorem
(for instance, Section 8.1, in particular Theorem 8.1.3, of Daniel Stroock's "Essentials of Integration Theory for Analysis," DOI:10.1007/978-1-4614-1135-2, or a neat summary at https://blameitontheanalyst.wordpress.com/2010/06/23/von-neumanns-proof-of-radon-nikodym/).
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
$endgroup$
$begingroup$
I think this is a good answer though.
$endgroup$
– user370967
Jan 30 at 15:52
$begingroup$
Although not the exact thing I was looking for, this is a very good answer. Thank you.
$endgroup$
– carsandpulsars
Feb 4 at 4:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I am not sure about your specific claim, but certainly there is a connection in the "opposite" direction. That is, a general version of the Riesz Representation Theorem is used to prove the Radon-Nikodym Theorem. See "von Neumann's proof" of the Radon-Nikodym Theorem
(for instance, Section 8.1, in particular Theorem 8.1.3, of Daniel Stroock's "Essentials of Integration Theory for Analysis," DOI:10.1007/978-1-4614-1135-2, or a neat summary at https://blameitontheanalyst.wordpress.com/2010/06/23/von-neumanns-proof-of-radon-nikodym/).
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
$endgroup$
$begingroup$
I think this is a good answer though.
$endgroup$
– user370967
Jan 30 at 15:52
$begingroup$
Although not the exact thing I was looking for, this is a very good answer. Thank you.
$endgroup$
– carsandpulsars
Feb 4 at 4:10
add a comment |
$begingroup$
I am not sure about your specific claim, but certainly there is a connection in the "opposite" direction. That is, a general version of the Riesz Representation Theorem is used to prove the Radon-Nikodym Theorem. See "von Neumann's proof" of the Radon-Nikodym Theorem
(for instance, Section 8.1, in particular Theorem 8.1.3, of Daniel Stroock's "Essentials of Integration Theory for Analysis," DOI:10.1007/978-1-4614-1135-2, or a neat summary at https://blameitontheanalyst.wordpress.com/2010/06/23/von-neumanns-proof-of-radon-nikodym/).
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
$endgroup$
$begingroup$
I think this is a good answer though.
$endgroup$
– user370967
Jan 30 at 15:52
$begingroup$
Although not the exact thing I was looking for, this is a very good answer. Thank you.
$endgroup$
– carsandpulsars
Feb 4 at 4:10
add a comment |
$begingroup$
I am not sure about your specific claim, but certainly there is a connection in the "opposite" direction. That is, a general version of the Riesz Representation Theorem is used to prove the Radon-Nikodym Theorem. See "von Neumann's proof" of the Radon-Nikodym Theorem
(for instance, Section 8.1, in particular Theorem 8.1.3, of Daniel Stroock's "Essentials of Integration Theory for Analysis," DOI:10.1007/978-1-4614-1135-2, or a neat summary at https://blameitontheanalyst.wordpress.com/2010/06/23/von-neumanns-proof-of-radon-nikodym/).
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
$endgroup$
I am not sure about your specific claim, but certainly there is a connection in the "opposite" direction. That is, a general version of the Riesz Representation Theorem is used to prove the Radon-Nikodym Theorem. See "von Neumann's proof" of the Radon-Nikodym Theorem
(for instance, Section 8.1, in particular Theorem 8.1.3, of Daniel Stroock's "Essentials of Integration Theory for Analysis," DOI:10.1007/978-1-4614-1135-2, or a neat summary at https://blameitontheanalyst.wordpress.com/2010/06/23/von-neumanns-proof-of-radon-nikodym/).
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
edited Jan 30 at 19:45
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
answered Jan 30 at 14:16
UserNamesAreHardUserNamesAreHard
365
365
$begingroup$
I think this is a good answer though.
$endgroup$
– user370967
Jan 30 at 15:52
$begingroup$
Although not the exact thing I was looking for, this is a very good answer. Thank you.
$endgroup$
– carsandpulsars
Feb 4 at 4:10
add a comment |
$begingroup$
I think this is a good answer though.
$endgroup$
– user370967
Jan 30 at 15:52
$begingroup$
Although not the exact thing I was looking for, this is a very good answer. Thank you.
$endgroup$
– carsandpulsars
Feb 4 at 4:10
$begingroup$
I think this is a good answer though.
$endgroup$
– user370967
Jan 30 at 15:52
$begingroup$
I think this is a good answer though.
$endgroup$
– user370967
Jan 30 at 15:52
$begingroup$
Although not the exact thing I was looking for, this is a very good answer. Thank you.
$endgroup$
– carsandpulsars
Feb 4 at 4:10
$begingroup$
Although not the exact thing I was looking for, this is a very good answer. Thank you.
$endgroup$
– carsandpulsars
Feb 4 at 4:10
add a comment |
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$begingroup$
if $phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $lambda(E)=phi(chi_E)$, if $phi$ is positive we can (generalize after) show that $lambda$ is a measure and if the set $E$ has measure $mu(E)=0$ than $lambda(E)=0$ because $phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $phi(f)=int fg dmu$ for some $gin L^q$
$endgroup$
– Robson
Nov 18 '18 at 22:20