Is taking the positive part of a measure a continuous operation?

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Here is question I tried to answer for some time - it seems to be straightforward, but I have trouble figuring it out.

Let $Omega$ be a compact domain in $mathbbR^n$. For any signed Borel measure $mu$ on $Omega$ let $mu_+$ denote its positive part (obtained by Hahn-Jordan decomposition). My question is:

Is taking the positive part a continuous operation, i.e. does $mu^nto mu$ in the space signed Borel measures imply $mu^n_+to mu_+$ in the same space?

Of course, the answer would be positive if
$$|w^n_+-w_+|_mathfrakMleq |w_n-w|_mathfrakM$$
where $|cdot|_mathfrakM$ denotes the variation norm, but I could not see this.

fa.functional-analysis measure-theory

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asked 1 hour ago

Dirk

6,88443264

add a comment | 

up vote
2
down vote

favorite

1

Here is question I tried to answer for some time - it seems to be straightforward, but I have trouble figuring it out.

Let $Omega$ be a compact domain in $mathbbR^n$. For any signed Borel measure $mu$ on $Omega$ let $mu_+$ denote its positive part (obtained by Hahn-Jordan decomposition). My question is:

Is taking the positive part a continuous operation, i.e. does $mu^nto mu$ in the space signed Borel measures imply $mu^n_+to mu_+$ in the same space?

Of course, the answer would be positive if
$$|w^n_+-w_+|_mathfrakMleq |w_n-w|_mathfrakM$$
where $|cdot|_mathfrakM$ denotes the variation norm, but I could not see this.

fa.functional-analysis measure-theory

share|cite|improve this question

asked 1 hour ago

Dirk

6,88443264

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Here is question I tried to answer for some time - it seems to be straightforward, but I have trouble figuring it out.

Let $Omega$ be a compact domain in $mathbbR^n$. For any signed Borel measure $mu$ on $Omega$ let $mu_+$ denote its positive part (obtained by Hahn-Jordan decomposition). My question is:

Is taking the positive part a continuous operation, i.e. does $mu^nto mu$ in the space signed Borel measures imply $mu^n_+to mu_+$ in the same space?

Of course, the answer would be positive if
$$|w^n_+-w_+|_mathfrakMleq |w_n-w|_mathfrakM$$
where $|cdot|_mathfrakM$ denotes the variation norm, but I could not see this.

fa.functional-analysis measure-theory

share|cite|improve this question

asked 1 hour ago

Dirk

6,88443264

Here is question I tried to answer for some time - it seems to be straightforward, but I have trouble figuring it out.

Let $Omega$ be a compact domain in $mathbbR^n$. For any signed Borel measure $mu$ on $Omega$ let $mu_+$ denote its positive part (obtained by Hahn-Jordan decomposition). My question is:

Is taking the positive part a continuous operation, i.e. does $mu^nto mu$ in the space signed Borel measures imply $mu^n_+to mu_+$ in the same space?

Of course, the answer would be positive if
$$|w^n_+-w_+|_mathfrakMleq |w_n-w|_mathfrakM$$
where $|cdot|_mathfrakM$ denotes the variation norm, but I could not see this.

fa.functional-analysis measure-theory

fa.functional-analysis measure-theory

share|cite|improve this question

asked 1 hour ago

Dirk

6,88443264

share|cite|improve this question

asked 1 hour ago

Dirk

6,88443264

share|cite|improve this question

share|cite|improve this question

asked 1 hour ago

Dirk

6,88443264

asked 1 hour ago

Dirk

6,88443264

asked 1 hour ago

Dirk

6,88443264

6,88443264

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
5
down vote



accepted

If $mu =fcdot lambda$ for a positive measure $lambda$ (i.e., $mu(A)=int_A fdlambda$), isn't then $mu_+= f_+ dlambda$ (where $f_+$ is the positive part $maxf,0$ of $f$) and $|mu|=int|f|dlambda$? Then $|mu_+-nu_+| le |mu-nu|$ just follows from Radon-Nikodym (applied to $lambda=|mu|+|nu|$) and $|f_+-g_+|le |f-g|$.

share|cite|improve this answer

answered 1 hour ago

Jochen Wengenroth

7,0572147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
  – Dirk
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here you tacitly assume that both $|mu|$ and $|nu|$ are absolutely continuous with respect to the same positive measure $lambda$ so you can invoke the densities $f$ and $g$.
  – Liviu Nicolaescu
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think he just takes $lambda =|mu| + |nu|$.
  – Dirk
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK. I get now. Nice
  – Liviu Nicolaescu
  20 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

The Borel measures on $Omega$ can be identified with continuous linear functionals $newcommandbRmathbbR$ $mu:C(Omega)tobR$. Assume that $Omega$ is compact so $C(Omega)$ is Banach space. Denote by $Vert-Vert$ the sup norm on $C(Omega)$ and by $newcommandeMmathscrM$ $eM(Omega)$ the dual of $C(Omega)$ equipped with the dual norm
$$
Vert muVert_*:=sup_Vert fVertleq 1|mu(f)|.
$$
Set
$$
C(Omega)_+:=big fin C(Omega):;; f(x)geq 0,;;forall xinOmega)big.
$$Let $muin eM(Omega)$ and $fin C(Omega)$. Then, for any $fin C(Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis)
$$
mu_+(f)=sup_0leq gleq f mu(g).
$$
Let $mu,nuin eM(Omega)$, and $fin C(Omega)_+$. Then for any $0leq gleq f$ we have

$$
mu(g)-nu(g)leq Vertmu-nuVert_*Vert gVert leq Vertmu-nuVert_*Vert fVert.
$$
Hence, for any $0leq gleq f$
$$
mu(g)leq Vertmu-nuVert_*Vert fVert+nu(g),
$$
and, symmetrically,
$$
nu(g)leq Vertmu-nuVert_*Vert fVert+mu(g).
$$
Taking the sup on both sides of the above inequalities we deduce
$$
mu_+(f)leq Vertmu-nuVert_*Vert fVert+nu_+(f)implies mu_+(f)- nu_+(f)leq Vertmu-nuVert_*Vert fVert,
$$
$$
nu_+(f)leq Vertmu-nuVert_*Vert fVert+mu_+(f)implies nu_+(f)- mu_+(f)leq Vertmu-nuVert_*Vert fVert.
$$
Hence
$$
bigvert (mu_+-nu_+)fbigvertleq Vertmu-nuVert_*Vert fVert.
$$
This implies
$$
Vert mu_+-nu_+Vert_*leq Vertmu-nuVert_*.
$$

share|cite|improve this answer

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

If $mu =fcdot lambda$ for a positive measure $lambda$ (i.e., $mu(A)=int_A fdlambda$), isn't then $mu_+= f_+ dlambda$ (where $f_+$ is the positive part $maxf,0$ of $f$) and $|mu|=int|f|dlambda$? Then $|mu_+-nu_+| le |mu-nu|$ just follows from Radon-Nikodym (applied to $lambda=|mu|+|nu|$) and $|f_+-g_+|le |f-g|$.

share|cite|improve this answer

answered 1 hour ago

Jochen Wengenroth

7,0572147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
  – Dirk
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here you tacitly assume that both $|mu|$ and $|nu|$ are absolutely continuous with respect to the same positive measure $lambda$ so you can invoke the densities $f$ and $g$.
  – Liviu Nicolaescu
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think he just takes $lambda =|mu| + |nu|$.
  – Dirk
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK. I get now. Nice
  – Liviu Nicolaescu
  20 mins ago
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

If $mu =fcdot lambda$ for a positive measure $lambda$ (i.e., $mu(A)=int_A fdlambda$), isn't then $mu_+= f_+ dlambda$ (where $f_+$ is the positive part $maxf,0$ of $f$) and $|mu|=int|f|dlambda$? Then $|mu_+-nu_+| le |mu-nu|$ just follows from Radon-Nikodym (applied to $lambda=|mu|+|nu|$) and $|f_+-g_+|le |f-g|$.

share|cite|improve this answer

answered 1 hour ago

Jochen Wengenroth

7,0572147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
  – Dirk
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here you tacitly assume that both $|mu|$ and $|nu|$ are absolutely continuous with respect to the same positive measure $lambda$ so you can invoke the densities $f$ and $g$.
  – Liviu Nicolaescu
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think he just takes $lambda =|mu| + |nu|$.
  – Dirk
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK. I get now. Nice
  – Liviu Nicolaescu
  20 mins ago
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

If $mu =fcdot lambda$ for a positive measure $lambda$ (i.e., $mu(A)=int_A fdlambda$), isn't then $mu_+= f_+ dlambda$ (where $f_+$ is the positive part $maxf,0$ of $f$) and $|mu|=int|f|dlambda$? Then $|mu_+-nu_+| le |mu-nu|$ just follows from Radon-Nikodym (applied to $lambda=|mu|+|nu|$) and $|f_+-g_+|le |f-g|$.

share|cite|improve this answer

answered 1 hour ago

Jochen Wengenroth

7,0572147

If $mu =fcdot lambda$ for a positive measure $lambda$ (i.e., $mu(A)=int_A fdlambda$), isn't then $mu_+= f_+ dlambda$ (where $f_+$ is the positive part $maxf,0$ of $f$) and $|mu|=int|f|dlambda$? Then $|mu_+-nu_+| le |mu-nu|$ just follows from Radon-Nikodym (applied to $lambda=|mu|+|nu|$) and $|f_+-g_+|le |f-g|$.

share|cite|improve this answer

answered 1 hour ago

Jochen Wengenroth

7,0572147

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Jochen Wengenroth

7,0572147

answered 1 hour ago

Jochen Wengenroth

7,0572147

answered 1 hour ago

Jochen Wengenroth

7,0572147

7,0572147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
  – Dirk
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here you tacitly assume that both $|mu|$ and $|nu|$ are absolutely continuous with respect to the same positive measure $lambda$ so you can invoke the densities $f$ and $g$.
  – Liviu Nicolaescu
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think he just takes $lambda =|mu| + |nu|$.
  – Dirk
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK. I get now. Nice
  – Liviu Nicolaescu
  20 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
  – Dirk
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Here you tacitly assume that both $|mu|$ and $|nu|$ are absolutely continuous with respect to the same positive measure $lambda$ so you can invoke the densities $f$ and $g$.
  – Liviu Nicolaescu
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think he just takes $lambda =|mu| + |nu|$.
  – Dirk
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK. I get now. Nice
  – Liviu Nicolaescu
  20 mins ago
  

  
  
  
  

Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
– Dirk
1 hour ago

Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
– Dirk
1 hour ago

Here you tacitly assume that both $|mu|$ and $|nu|$ are absolutely continuous with respect to the same positive measure $lambda$ so you can invoke the densities $f$ and $g$.
– Liviu Nicolaescu
1 hour ago

Here you tacitly assume that both $|mu|$ and $|nu|$ are absolutely continuous with respect to the same positive measure $lambda$ so you can invoke the densities $f$ and $g$.
– Liviu Nicolaescu
1 hour ago

I think he just takes $lambda =|mu| + |nu|$.
– Dirk
47 mins ago

I think he just takes $lambda =|mu| + |nu|$.
– Dirk
47 mins ago

OK. I get now. Nice
– Liviu Nicolaescu
20 mins ago

OK. I get now. Nice
– Liviu Nicolaescu
20 mins ago

add a comment | 

up vote
0
down vote

The Borel measures on $Omega$ can be identified with continuous linear functionals $newcommandbRmathbbR$ $mu:C(Omega)tobR$. Assume that $Omega$ is compact so $C(Omega)$ is Banach space. Denote by $Vert-Vert$ the sup norm on $C(Omega)$ and by $newcommandeMmathscrM$ $eM(Omega)$ the dual of $C(Omega)$ equipped with the dual norm
$$
Vert muVert_*:=sup_Vert fVertleq 1|mu(f)|.
$$
Set
$$
C(Omega)_+:=big fin C(Omega):;; f(x)geq 0,;;forall xinOmega)big.
$$Let $muin eM(Omega)$ and $fin C(Omega)$. Then, for any $fin C(Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis)
$$
mu_+(f)=sup_0leq gleq f mu(g).
$$
Let $mu,nuin eM(Omega)$, and $fin C(Omega)_+$. Then for any $0leq gleq f$ we have

$$
mu(g)-nu(g)leq Vertmu-nuVert_*Vert gVert leq Vertmu-nuVert_*Vert fVert.
$$
Hence, for any $0leq gleq f$
$$
mu(g)leq Vertmu-nuVert_*Vert fVert+nu(g),
$$
and, symmetrically,
$$
nu(g)leq Vertmu-nuVert_*Vert fVert+mu(g).
$$
Taking the sup on both sides of the above inequalities we deduce
$$
mu_+(f)leq Vertmu-nuVert_*Vert fVert+nu_+(f)implies mu_+(f)- nu_+(f)leq Vertmu-nuVert_*Vert fVert,
$$
$$
nu_+(f)leq Vertmu-nuVert_*Vert fVert+mu_+(f)implies nu_+(f)- mu_+(f)leq Vertmu-nuVert_*Vert fVert.
$$
Hence
$$
bigvert (mu_+-nu_+)fbigvertleq Vertmu-nuVert_*Vert fVert.
$$
This implies
$$
Vert mu_+-nu_+Vert_*leq Vertmu-nuVert_*.
$$

share|cite|improve this answer

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

add a comment | 

up vote
0
down vote

The Borel measures on $Omega$ can be identified with continuous linear functionals $newcommandbRmathbbR$ $mu:C(Omega)tobR$. Assume that $Omega$ is compact so $C(Omega)$ is Banach space. Denote by $Vert-Vert$ the sup norm on $C(Omega)$ and by $newcommandeMmathscrM$ $eM(Omega)$ the dual of $C(Omega)$ equipped with the dual norm
$$
Vert muVert_*:=sup_Vert fVertleq 1|mu(f)|.
$$
Set
$$
C(Omega)_+:=big fin C(Omega):;; f(x)geq 0,;;forall xinOmega)big.
$$Let $muin eM(Omega)$ and $fin C(Omega)$. Then, for any $fin C(Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis)
$$
mu_+(f)=sup_0leq gleq f mu(g).
$$
Let $mu,nuin eM(Omega)$, and $fin C(Omega)_+$. Then for any $0leq gleq f$ we have

$$
mu(g)-nu(g)leq Vertmu-nuVert_*Vert gVert leq Vertmu-nuVert_*Vert fVert.
$$
Hence, for any $0leq gleq f$
$$
mu(g)leq Vertmu-nuVert_*Vert fVert+nu(g),
$$
and, symmetrically,
$$
nu(g)leq Vertmu-nuVert_*Vert fVert+mu(g).
$$
Taking the sup on both sides of the above inequalities we deduce
$$
mu_+(f)leq Vertmu-nuVert_*Vert fVert+nu_+(f)implies mu_+(f)- nu_+(f)leq Vertmu-nuVert_*Vert fVert,
$$
$$
nu_+(f)leq Vertmu-nuVert_*Vert fVert+mu_+(f)implies nu_+(f)- mu_+(f)leq Vertmu-nuVert_*Vert fVert.
$$
Hence
$$
bigvert (mu_+-nu_+)fbigvertleq Vertmu-nuVert_*Vert fVert.
$$
This implies
$$
Vert mu_+-nu_+Vert_*leq Vertmu-nuVert_*.
$$

share|cite|improve this answer

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

add a comment | 

up vote
0
down vote

up vote
0
down vote

The Borel measures on $Omega$ can be identified with continuous linear functionals $newcommandbRmathbbR$ $mu:C(Omega)tobR$. Assume that $Omega$ is compact so $C(Omega)$ is Banach space. Denote by $Vert-Vert$ the sup norm on $C(Omega)$ and by $newcommandeMmathscrM$ $eM(Omega)$ the dual of $C(Omega)$ equipped with the dual norm
$$
Vert muVert_*:=sup_Vert fVertleq 1|mu(f)|.
$$
Set
$$
C(Omega)_+:=big fin C(Omega):;; f(x)geq 0,;;forall xinOmega)big.
$$Let $muin eM(Omega)$ and $fin C(Omega)$. Then, for any $fin C(Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis)
$$
mu_+(f)=sup_0leq gleq f mu(g).
$$
Let $mu,nuin eM(Omega)$, and $fin C(Omega)_+$. Then for any $0leq gleq f$ we have

$$
mu(g)-nu(g)leq Vertmu-nuVert_*Vert gVert leq Vertmu-nuVert_*Vert fVert.
$$
Hence, for any $0leq gleq f$
$$
mu(g)leq Vertmu-nuVert_*Vert fVert+nu(g),
$$
and, symmetrically,
$$
nu(g)leq Vertmu-nuVert_*Vert fVert+mu(g).
$$
Taking the sup on both sides of the above inequalities we deduce
$$
mu_+(f)leq Vertmu-nuVert_*Vert fVert+nu_+(f)implies mu_+(f)- nu_+(f)leq Vertmu-nuVert_*Vert fVert,
$$
$$
nu_+(f)leq Vertmu-nuVert_*Vert fVert+mu_+(f)implies nu_+(f)- mu_+(f)leq Vertmu-nuVert_*Vert fVert.
$$
Hence
$$
bigvert (mu_+-nu_+)fbigvertleq Vertmu-nuVert_*Vert fVert.
$$
This implies
$$
Vert mu_+-nu_+Vert_*leq Vertmu-nuVert_*.
$$

share|cite|improve this answer

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

The Borel measures on $Omega$ can be identified with continuous linear functionals $newcommandbRmathbbR$ $mu:C(Omega)tobR$. Assume that $Omega$ is compact so $C(Omega)$ is Banach space. Denote by $Vert-Vert$ the sup norm on $C(Omega)$ and by $newcommandeMmathscrM$ $eM(Omega)$ the dual of $C(Omega)$ equipped with the dual norm
$$
Vert muVert_*:=sup_Vert fVertleq 1|mu(f)|.
$$
Set
$$
C(Omega)_+:=big fin C(Omega):;; f(x)geq 0,;;forall xinOmega)big.
$$Let $muin eM(Omega)$ and $fin C(Omega)$. Then, for any $fin C(Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis)
$$
mu_+(f)=sup_0leq gleq f mu(g).
$$
Let $mu,nuin eM(Omega)$, and $fin C(Omega)_+$. Then for any $0leq gleq f$ we have

$$
mu(g)-nu(g)leq Vertmu-nuVert_*Vert gVert leq Vertmu-nuVert_*Vert fVert.
$$
Hence, for any $0leq gleq f$
$$
mu(g)leq Vertmu-nuVert_*Vert fVert+nu(g),
$$
and, symmetrically,
$$
nu(g)leq Vertmu-nuVert_*Vert fVert+mu(g).
$$
Taking the sup on both sides of the above inequalities we deduce
$$
mu_+(f)leq Vertmu-nuVert_*Vert fVert+nu_+(f)implies mu_+(f)- nu_+(f)leq Vertmu-nuVert_*Vert fVert,
$$
$$
nu_+(f)leq Vertmu-nuVert_*Vert fVert+mu_+(f)implies nu_+(f)- mu_+(f)leq Vertmu-nuVert_*Vert fVert.
$$
Hence
$$
bigvert (mu_+-nu_+)fbigvertleq Vertmu-nuVert_*Vert fVert.
$$
This implies
$$
Vert mu_+-nu_+Vert_*leq Vertmu-nuVert_*.
$$

share|cite|improve this answer

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

share|cite|improve this answer

share|cite|improve this answer

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

answered 30 mins ago

Liviu Nicolaescu

24.6k256107

24.6k256107

add a comment | 

add a comment | 

 

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