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Definition of closed Ideal in the space of finite measures.

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Let $M(mathbbR)$ be the space of finite signed measure on $(mathbbR)$. What is the norm defined on this space? From theorem 1.3.5 in Rudin Fourier analysis on groups. Interscience Tracts in Pure and Applies Mathematics, It is proved that $L^1(mathbbR)$ is a closed ideal in the space $M(mathbbR)$. What is the definition of closed ideal means here?

general-topology measure-theory definition

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edited Mar 21 at 5:38

Ajsal Shereef

asked Mar 21 at 5:29

Ajsal ShereefAjsal Shereef

11

$endgroup$

add a comment | 

0

$begingroup$

Let $M(mathbbR)$ be the space of finite signed measure on $(mathbbR)$. What is the norm defined on this space? From theorem 1.3.5 in Rudin Fourier analysis on groups. Interscience Tracts in Pure and Applies Mathematics, It is proved that $L^1(mathbbR)$ is a closed ideal in the space $M(mathbbR)$. What is the definition of closed ideal means here?

general-topology measure-theory definition

share|cite|improve this question

edited Mar 21 at 5:38

Ajsal Shereef

asked Mar 21 at 5:29

Ajsal ShereefAjsal Shereef

11

$endgroup$

add a comment | 

$begingroup$

Let $M(mathbbR)$ be the space of finite signed measure on $(mathbbR)$. What is the norm defined on this space? From theorem 1.3.5 in Rudin Fourier analysis on groups. Interscience Tracts in Pure and Applies Mathematics, It is proved that $L^1(mathbbR)$ is a closed ideal in the space $M(mathbbR)$. What is the definition of closed ideal means here?

general-topology measure-theory definition

share|cite|improve this question

edited Mar 21 at 5:38

Ajsal Shereef

asked Mar 21 at 5:29

Ajsal ShereefAjsal Shereef

11

$endgroup$

share|cite|improve this question

edited Mar 21 at 5:38

Ajsal Shereef

asked Mar 21 at 5:29

Ajsal ShereefAjsal Shereef

11

share|cite|improve this question

edited Mar 21 at 5:38

Ajsal Shereef

asked Mar 21 at 5:29

Ajsal ShereefAjsal Shereef

11

asked Mar 21 at 5:29

Ajsal ShereefAjsal Shereef

11

asked Mar 21 at 5:29

Ajsal ShereefAjsal Shereef

11

1 Answer
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$begingroup$

A fairly standard norm in this case would be the total variation; in particular, for $nu in M(mathbbR)$, we define
$$| nu| = lvert nu rvert (mathbbR).$$
If I'm not mistaken, a closed ideal is simply an ideal (very much like those from ring theory) which is closed in the topology induced by the norm on $M(mathbbR)$.

Edit: To flesh out a few more details, first note that we can embed $L^1(mathbbR) hookrightarrow M(mathbbR)$ via the identification $f mapsto nu$, where $dnu = fdm$ (or, equivalently, $nu(E) = int_E f dm$), where $m$ denotes Lebesgue measure. As Rudin notes in his text, multiplication of measures is defined by convolution. In particular, for $lambda, mu in M(mathbbR)$, we define
$$(lambda * mu)(E) = (lambda times mu)(E_2),$$
where, in our case, $E_2 = (x,y) in mathbbR^2 : x+y in E $. We are almost ready to show that $L^1$ is an ideal of $M$. We just need to think about how to show that $mu * nu in L^1$ for $mu in M$ and $nu in L^1$. Recall that the Radon-Nikodym theorem implies that if $mu in M$ is absolutely continuous with respect to $m$, then $dmu = g dm$ for some $g in L^1(mathbbR)$.

We can now show that $L^1$ is an ideal of $M$ by taking arbitrary $nu "in" L^1$ (ie, $dnu = fdm$ for $f in L^1$) and showing that $mu * nu ll m$ for all $mu in M$ (see Rudin for the proof). This, of course, implies (again by Radon-Nikodym) that $d(mu * nu) = gdm$ for some $L^1$ function $g$. Therefore, $mu * nu in L^1$ and so $L^1$ is an ideal of $M$. That $L^1$ is complete implies that $L^1$ is closed in $M$ and therefore $L^1$ is in fact a closed ideal of $M$.

share|cite|improve this answer

edited Mar 21 at 20:38

answered Mar 21 at 6:38

Gary MoonGary Moon

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0

$begingroup$

A fairly standard norm in this case would be the total variation; in particular, for $nu in M(mathbbR)$, we define
$$| nu| = lvert nu rvert (mathbbR).$$
If I'm not mistaken, a closed ideal is simply an ideal (very much like those from ring theory) which is closed in the topology induced by the norm on $M(mathbbR)$.

Edit: To flesh out a few more details, first note that we can embed $L^1(mathbbR) hookrightarrow M(mathbbR)$ via the identification $f mapsto nu$, where $dnu = fdm$ (or, equivalently, $nu(E) = int_E f dm$), where $m$ denotes Lebesgue measure. As Rudin notes in his text, multiplication of measures is defined by convolution. In particular, for $lambda, mu in M(mathbbR)$, we define
$$(lambda * mu)(E) = (lambda times mu)(E_2),$$
where, in our case, $E_2 = (x,y) in mathbbR^2 : x+y in E $. We are almost ready to show that $L^1$ is an ideal of $M$. We just need to think about how to show that $mu * nu in L^1$ for $mu in M$ and $nu in L^1$. Recall that the Radon-Nikodym theorem implies that if $mu in M$ is absolutely continuous with respect to $m$, then $dmu = g dm$ for some $g in L^1(mathbbR)$.

We can now show that $L^1$ is an ideal of $M$ by taking arbitrary $nu "in" L^1$ (ie, $dnu = fdm$ for $f in L^1$) and showing that $mu * nu ll m$ for all $mu in M$ (see Rudin for the proof). This, of course, implies (again by Radon-Nikodym) that $d(mu * nu) = gdm$ for some $L^1$ function $g$. Therefore, $mu * nu in L^1$ and so $L^1$ is an ideal of $M$. That $L^1$ is complete implies that $L^1$ is closed in $M$ and therefore $L^1$ is in fact a closed ideal of $M$.

share|cite|improve this answer

edited Mar 21 at 20:38

answered Mar 21 at 6:38

Gary MoonGary Moon

92127

$endgroup$

add a comment | 

0

$begingroup$

A fairly standard norm in this case would be the total variation; in particular, for $nu in M(mathbbR)$, we define
$$| nu| = lvert nu rvert (mathbbR).$$
If I'm not mistaken, a closed ideal is simply an ideal (very much like those from ring theory) which is closed in the topology induced by the norm on $M(mathbbR)$.

Edit: To flesh out a few more details, first note that we can embed $L^1(mathbbR) hookrightarrow M(mathbbR)$ via the identification $f mapsto nu$, where $dnu = fdm$ (or, equivalently, $nu(E) = int_E f dm$), where $m$ denotes Lebesgue measure. As Rudin notes in his text, multiplication of measures is defined by convolution. In particular, for $lambda, mu in M(mathbbR)$, we define
$$(lambda * mu)(E) = (lambda times mu)(E_2),$$
where, in our case, $E_2 = (x,y) in mathbbR^2 : x+y in E $. We are almost ready to show that $L^1$ is an ideal of $M$. We just need to think about how to show that $mu * nu in L^1$ for $mu in M$ and $nu in L^1$. Recall that the Radon-Nikodym theorem implies that if $mu in M$ is absolutely continuous with respect to $m$, then $dmu = g dm$ for some $g in L^1(mathbbR)$.

We can now show that $L^1$ is an ideal of $M$ by taking arbitrary $nu "in" L^1$ (ie, $dnu = fdm$ for $f in L^1$) and showing that $mu * nu ll m$ for all $mu in M$ (see Rudin for the proof). This, of course, implies (again by Radon-Nikodym) that $d(mu * nu) = gdm$ for some $L^1$ function $g$. Therefore, $mu * nu in L^1$ and so $L^1$ is an ideal of $M$. That $L^1$ is complete implies that $L^1$ is closed in $M$ and therefore $L^1$ is in fact a closed ideal of $M$.

share|cite|improve this answer

edited Mar 21 at 20:38

answered Mar 21 at 6:38

Gary MoonGary Moon

92127

$endgroup$

add a comment | 

$begingroup$

A fairly standard norm in this case would be the total variation; in particular, for $nu in M(mathbbR)$, we define
$$| nu| = lvert nu rvert (mathbbR).$$
If I'm not mistaken, a closed ideal is simply an ideal (very much like those from ring theory) which is closed in the topology induced by the norm on $M(mathbbR)$.

Edit: To flesh out a few more details, first note that we can embed $L^1(mathbbR) hookrightarrow M(mathbbR)$ via the identification $f mapsto nu$, where $dnu = fdm$ (or, equivalently, $nu(E) = int_E f dm$), where $m$ denotes Lebesgue measure. As Rudin notes in his text, multiplication of measures is defined by convolution. In particular, for $lambda, mu in M(mathbbR)$, we define
$$(lambda * mu)(E) = (lambda times mu)(E_2),$$
where, in our case, $E_2 = (x,y) in mathbbR^2 : x+y in E $. We are almost ready to show that $L^1$ is an ideal of $M$. We just need to think about how to show that $mu * nu in L^1$ for $mu in M$ and $nu in L^1$. Recall that the Radon-Nikodym theorem implies that if $mu in M$ is absolutely continuous with respect to $m$, then $dmu = g dm$ for some $g in L^1(mathbbR)$.

We can now show that $L^1$ is an ideal of $M$ by taking arbitrary $nu "in" L^1$ (ie, $dnu = fdm$ for $f in L^1$) and showing that $mu * nu ll m$ for all $mu in M$ (see Rudin for the proof). This, of course, implies (again by Radon-Nikodym) that $d(mu * nu) = gdm$ for some $L^1$ function $g$. Therefore, $mu * nu in L^1$ and so $L^1$ is an ideal of $M$. That $L^1$ is complete implies that $L^1$ is closed in $M$ and therefore $L^1$ is in fact a closed ideal of $M$.

share|cite|improve this answer

edited Mar 21 at 20:38

answered Mar 21 at 6:38

Gary MoonGary Moon

92127

$endgroup$

share|cite|improve this answer

edited Mar 21 at 20:38

answered Mar 21 at 6:38

Gary MoonGary Moon

92127

answered Mar 21 at 6:38

Gary MoonGary Moon

92127

answered Mar 21 at 6:38

Gary MoonGary Moon

92127