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Existence of functionals on $L^0$

3

$begingroup$

Studying a paper about risk measures by F. Delbaen, I bumped into this statement:

Let $(Omega,mathcalF,mathbbP)$ be a probability space: if $mathbbP$ is atomless, then there exists no functional $rho:L^0tomathbbR$ such that:

• $rho(X+a)=rho(X)-a quad forall a in mathbbR,$


• $rho(X+Y)le rho(X)+rho(Y),$


• $rho(lambda X)=lambdarho(X),$


• $Xge 0 implies rho(X)le 0,$

for every $Xin L^0.$

Here we denote by $L^0$ the linear space of all random variables on $Omega$ with the metric of the convergence in probability.

Then the author assesses that this is a consequence of the analytic Hahn-Banach theorem and of the fact that a continuous functional on $L^0$ must be necessarily null if $mathbbP$ is atomless.

Now, I'm full of doubts: first of all I didn't know the statement about the linear functionals on the $L^0$ space: could you give me some reference where to read about it? I tried to google something but didn't find anything.

Secondly I didn't really undestand how to use in a clever way the Hahn-Banach theorem: this risk functionals were previously introduced on the space $L^infty,$ where it is easy to check that they are continuous (wrt to $|cdot|_infty$), so I thought it was natural to use $L^infty$ as subspace where to use Hahn-Banach, but I don't know which linear functional on $L^infty$ I shoul use to be sure that it will be continuous on $L^0$ when extended.

Any help would be a lot appreciated. Thanks to everybody.

probability functional-analysis measure-theory lp-spaces risk-assessment

share|cite|improve this question

edited Mar 19 at 15:03

Riccardo Ceccon

asked Mar 19 at 11:41

Riccardo CecconRiccardo Ceccon

1,080321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the topological structure on $L^0$?
  $endgroup$
  – rubikscube09
  Mar 19 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The metric is given by $d(X,Y)=mathbbE[|X-Y|wedge 1],$ that induces the convergence in probability.
  $endgroup$
  – Riccardo Ceccon
  Mar 19 at 14:56
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Studying a paper about risk measures by F. Delbaen, I bumped into this statement:

Let $(Omega,mathcalF,mathbbP)$ be a probability space: if $mathbbP$ is atomless, then there exists no functional $rho:L^0tomathbbR$ such that:

• $rho(X+a)=rho(X)-a quad forall a in mathbbR,$


• $rho(X+Y)le rho(X)+rho(Y),$


• $rho(lambda X)=lambdarho(X),$


• $Xge 0 implies rho(X)le 0,$

for every $Xin L^0.$

Here we denote by $L^0$ the linear space of all random variables on $Omega$ with the metric of the convergence in probability.

Then the author assesses that this is a consequence of the analytic Hahn-Banach theorem and of the fact that a continuous functional on $L^0$ must be necessarily null if $mathbbP$ is atomless.

Now, I'm full of doubts: first of all I didn't know the statement about the linear functionals on the $L^0$ space: could you give me some reference where to read about it? I tried to google something but didn't find anything.

Secondly I didn't really undestand how to use in a clever way the Hahn-Banach theorem: this risk functionals were previously introduced on the space $L^infty,$ where it is easy to check that they are continuous (wrt to $|cdot|_infty$), so I thought it was natural to use $L^infty$ as subspace where to use Hahn-Banach, but I don't know which linear functional on $L^infty$ I shoul use to be sure that it will be continuous on $L^0$ when extended.

Any help would be a lot appreciated. Thanks to everybody.

probability functional-analysis measure-theory lp-spaces risk-assessment

share|cite|improve this question

edited Mar 19 at 15:03

Riccardo Ceccon

asked Mar 19 at 11:41

Riccardo CecconRiccardo Ceccon

1,080321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the topological structure on $L^0$?
  $endgroup$
  – rubikscube09
  Mar 19 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The metric is given by $d(X,Y)=mathbbE[|X-Y|wedge 1],$ that induces the convergence in probability.
  $endgroup$
  – Riccardo Ceccon
  Mar 19 at 14:56
  
  

  
  
  
  

add a comment | 

$begingroup$

Studying a paper about risk measures by F. Delbaen, I bumped into this statement:

Let $(Omega,mathcalF,mathbbP)$ be a probability space: if $mathbbP$ is atomless, then there exists no functional $rho:L^0tomathbbR$ such that:

• $rho(X+a)=rho(X)-a quad forall a in mathbbR,$


• $rho(X+Y)le rho(X)+rho(Y),$


• $rho(lambda X)=lambdarho(X),$


• $Xge 0 implies rho(X)le 0,$

for every $Xin L^0.$

Here we denote by $L^0$ the linear space of all random variables on $Omega$ with the metric of the convergence in probability.

Then the author assesses that this is a consequence of the analytic Hahn-Banach theorem and of the fact that a continuous functional on $L^0$ must be necessarily null if $mathbbP$ is atomless.

Now, I'm full of doubts: first of all I didn't know the statement about the linear functionals on the $L^0$ space: could you give me some reference where to read about it? I tried to google something but didn't find anything.

Secondly I didn't really undestand how to use in a clever way the Hahn-Banach theorem: this risk functionals were previously introduced on the space $L^infty,$ where it is easy to check that they are continuous (wrt to $|cdot|_infty$), so I thought it was natural to use $L^infty$ as subspace where to use Hahn-Banach, but I don't know which linear functional on $L^infty$ I shoul use to be sure that it will be continuous on $L^0$ when extended.

Any help would be a lot appreciated. Thanks to everybody.

probability functional-analysis measure-theory lp-spaces risk-assessment

share|cite|improve this question

edited Mar 19 at 15:03

Riccardo Ceccon

asked Mar 19 at 11:41

Riccardo CecconRiccardo Ceccon

1,080321

$endgroup$

share|cite|improve this question

edited Mar 19 at 15:03

Riccardo Ceccon

asked Mar 19 at 11:41

Riccardo CecconRiccardo Ceccon

1,080321

share|cite|improve this question

edited Mar 19 at 15:03

Riccardo Ceccon

asked Mar 19 at 11:41

Riccardo CecconRiccardo Ceccon

1,080321

asked Mar 19 at 11:41

Riccardo CecconRiccardo Ceccon

1,080321

asked Mar 19 at 11:41

Riccardo CecconRiccardo Ceccon

1,080321