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2

1

$begingroup$

Schwarz Kernel Theorem states that:

If $A: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ is a linear continuous operator, then there exists a unique kernel $Kin mathcal{D}'(mathbb{R}^ntimes mathbb{R}^n)$ such that $langle Au, vrangle_{mathcal{D}', mathcal{D}} = langle K, votimes urangle_{mathcal{D}',mathcal{D}}$ for all $u, vin C_0^infty(mathbb{R}^n)$.

My question is:

Suppose we have a sequence of linear continuous operators $A_n: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ with kernels $K_n$. And $A: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ is also continuous and linear with kernel $K$. If $A_nu to Au$ in $mathcal{D}'$ for every $u in C_0^infty$, can we conclude that $K_n to K$ in $mathcal{D}'(mathbb{R}^ntimes mathbb{R}^n)$?

By definition of kernels in the theorem, one can check that $K_n to K$ when restricted on $C_0^infty(mathbb{R}^n)otimes C_0^infty(mathbb{R}^n)$. So can we use a certain density argument to prove it?

distribution-theory

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asked Mar 20 at 0:34

user655213user655213

161

$endgroup$

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  $begingroup$
  What topology are you using on the space $mathcal{D}'(mathbb{R}^n)$?
  $endgroup$
  – Abdelmalek Abdesselam
  Mar 20 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AbdelmalekAbdesselam the usual topology for the distribution space (I don’t how to describe this topology in a precise way)
  $endgroup$
  – user655213
  Mar 21 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The reason I ask is there is no "usual topology" for this space. The two most common ones are the weak-$ast$ and the strong topology. However, it is my opinion that only the strong should be used especially in matters related to the nuclear theorem.
  $endgroup$
  – Abdelmalek Abdesselam
  Mar 21 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The distribution space is with the weak * topology. (And the test function space is with the topology induced by Fréchet spaces $mathcal{D}_K$, and is complete) I use the definition in Rudin’s Functional Analysis.
  $endgroup$
  – user655213
  Mar 26 at 12:18
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Schwarz Kernel Theorem states that:

If $A: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ is a linear continuous operator, then there exists a unique kernel $Kin mathcal{D}'(mathbb{R}^ntimes mathbb{R}^n)$ such that $langle Au, vrangle_{mathcal{D}', mathcal{D}} = langle K, votimes urangle_{mathcal{D}',mathcal{D}}$ for all $u, vin C_0^infty(mathbb{R}^n)$.

My question is:

Suppose we have a sequence of linear continuous operators $A_n: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ with kernels $K_n$. And $A: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ is also continuous and linear with kernel $K$. If $A_nu to Au$ in $mathcal{D}'$ for every $u in C_0^infty$, can we conclude that $K_n to K$ in $mathcal{D}'(mathbb{R}^ntimes mathbb{R}^n)$?

By definition of kernels in the theorem, one can check that $K_n to K$ when restricted on $C_0^infty(mathbb{R}^n)otimes C_0^infty(mathbb{R}^n)$. So can we use a certain density argument to prove it?

distribution-theory

share|cite|improve this question

asked Mar 20 at 0:34

user655213user655213

161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What topology are you using on the space $mathcal{D}'(mathbb{R}^n)$?
  $endgroup$
  – Abdelmalek Abdesselam
  Mar 20 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AbdelmalekAbdesselam the usual topology for the distribution space (I don’t how to describe this topology in a precise way)
  $endgroup$
  – user655213
  Mar 21 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The reason I ask is there is no "usual topology" for this space. The two most common ones are the weak-$ast$ and the strong topology. However, it is my opinion that only the strong should be used especially in matters related to the nuclear theorem.
  $endgroup$
  – Abdelmalek Abdesselam
  Mar 21 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The distribution space is with the weak * topology. (And the test function space is with the topology induced by Fréchet spaces $mathcal{D}_K$, and is complete) I use the definition in Rudin’s Functional Analysis.
  $endgroup$
  – user655213
  Mar 26 at 12:18
  

  
  
  
  

add a comment | 

$begingroup$

Schwarz Kernel Theorem states that:

If $A: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ is a linear continuous operator, then there exists a unique kernel $Kin mathcal{D}'(mathbb{R}^ntimes mathbb{R}^n)$ such that $langle Au, vrangle_{mathcal{D}', mathcal{D}} = langle K, votimes urangle_{mathcal{D}',mathcal{D}}$ for all $u, vin C_0^infty(mathbb{R}^n)$.

My question is:

Suppose we have a sequence of linear continuous operators $A_n: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ with kernels $K_n$. And $A: C_0^infty(mathbb{R}^n) to mathcal{D}'(mathbb{R}^n)$ is also continuous and linear with kernel $K$. If $A_nu to Au$ in $mathcal{D}'$ for every $u in C_0^infty$, can we conclude that $K_n to K$ in $mathcal{D}'(mathbb{R}^ntimes mathbb{R}^n)$?

By definition of kernels in the theorem, one can check that $K_n to K$ when restricted on $C_0^infty(mathbb{R}^n)otimes C_0^infty(mathbb{R}^n)$. So can we use a certain density argument to prove it?

distribution-theory

share|cite|improve this question

asked Mar 20 at 0:34

user655213user655213

161

$endgroup$

share|cite|improve this question

asked Mar 20 at 0:34

user655213user655213

161

share|cite|improve this question

asked Mar 20 at 0:34

user655213user655213

161

asked Mar 20 at 0:34

user655213user655213

161

asked Mar 20 at 0:34

user655213user655213

161