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$intlimits_{Omega}{uvdx}

11

3

$begingroup$

Let $d=3$ and $Omegasubset mathbb R^d$ is a bounded Lipschitz domain and $u$ is a measurable function.
A sufficient condition for the integral $intlimits_{Omega}{uvdx}<infty,forall vin H_0^1(Omega)$ is that $uin L^{6/5}(Omega)$ which follows from Holder's inequality and the (continuous) embedding $H^1(Omega)hookrightarrow L^6(Omega)$.

Question: Is the opposite true, i.e is it true that
$$intlimits_{Omega}{uvdx}<infty,forall vin H_0^1(Omega)$$ implies $uin L^{6/5}(Omega)$ or at least $uin L^1(Omega)$ ?

My thoughts: It is easy to see that $uin L^1_{loc}(Omega)$ by taking $v$ to be smooth cut-off functions equal to $1$ in compact subsets of $Omega$ and $0$ in a neighborhood of the boundary $partial Omega$.

The motivation for this question is the "correct" weak formulation of a nonlinear problem - whether to formulate it as $(1)$ or as $(2)$:

$(1)$ Find $uin H_0^1(Omega)$ such that $f(u)in L^{6/5}(Omega)$ and
$$a(u,v)+intlimits_{Omega}{f(u)vdx}=0,forall vin H_0^1(Omega)$$

or

$(2)$ Find $uin H_0^1(Omega)$ such that $intlimits_{Omega}{f(u)vdx}<infty,forall vin H_0^1(Omega)$ and
$$a(u,v)+intlimits_{Omega}{f(u)vdx}=0,forall vin H_0^1(Omega)$$

where $a(.,.)$ is a bilinear form and $f(.)$ is in general a nonlinear function. If the answer to my question is affirmative then both formulations are equivalent.

Note that $(2)$ is less restrictive, because the set in which we search for a solution is bigger, so it might be easier to find such.

pde lebesgue-integral sobolev-spaces lp-spaces

share|cite|improve this question

edited Mar 23 at 12:57

Svetoslav

asked Jan 1 '16 at 15:24

SvetoslavSvetoslav

4,5872828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Tomás I reposted the question in MathOverflow here mathoverflow.net/questions/227892/… and I have an answer. I just can not understand how the important inequality in the comment is derived.
  $endgroup$
  – Svetoslav
  Jan 9 '16 at 19:28
  

  
  
  
  

add a comment | 

11

3

$begingroup$

Let $d=3$ and $Omegasubset mathbb R^d$ is a bounded Lipschitz domain and $u$ is a measurable function.
A sufficient condition for the integral $intlimits_{Omega}{uvdx}<infty,forall vin H_0^1(Omega)$ is that $uin L^{6/5}(Omega)$ which follows from Holder's inequality and the (continuous) embedding $H^1(Omega)hookrightarrow L^6(Omega)$.

Question: Is the opposite true, i.e is it true that
$$intlimits_{Omega}{uvdx}<infty,forall vin H_0^1(Omega)$$ implies $uin L^{6/5}(Omega)$ or at least $uin L^1(Omega)$ ?

My thoughts: It is easy to see that $uin L^1_{loc}(Omega)$ by taking $v$ to be smooth cut-off functions equal to $1$ in compact subsets of $Omega$ and $0$ in a neighborhood of the boundary $partial Omega$.

The motivation for this question is the "correct" weak formulation of a nonlinear problem - whether to formulate it as $(1)$ or as $(2)$:

$(1)$ Find $uin H_0^1(Omega)$ such that $f(u)in L^{6/5}(Omega)$ and
$$a(u,v)+intlimits_{Omega}{f(u)vdx}=0,forall vin H_0^1(Omega)$$

or

$(2)$ Find $uin H_0^1(Omega)$ such that $intlimits_{Omega}{f(u)vdx}<infty,forall vin H_0^1(Omega)$ and
$$a(u,v)+intlimits_{Omega}{f(u)vdx}=0,forall vin H_0^1(Omega)$$

where $a(.,.)$ is a bilinear form and $f(.)$ is in general a nonlinear function. If the answer to my question is affirmative then both formulations are equivalent.

Note that $(2)$ is less restrictive, because the set in which we search for a solution is bigger, so it might be easier to find such.

pde lebesgue-integral sobolev-spaces lp-spaces

share|cite|improve this question

edited Mar 23 at 12:57

Svetoslav

asked Jan 1 '16 at 15:24

SvetoslavSvetoslav

4,5872828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Tomás I reposted the question in MathOverflow here mathoverflow.net/questions/227892/… and I have an answer. I just can not understand how the important inequality in the comment is derived.
  $endgroup$
  – Svetoslav
  Jan 9 '16 at 19:28
  

  
  
  
  

add a comment | 

$begingroup$

Let $d=3$ and $Omegasubset mathbb R^d$ is a bounded Lipschitz domain and $u$ is a measurable function.
A sufficient condition for the integral $intlimits_{Omega}{uvdx}<infty,forall vin H_0^1(Omega)$ is that $uin L^{6/5}(Omega)$ which follows from Holder's inequality and the (continuous) embedding $H^1(Omega)hookrightarrow L^6(Omega)$.

Question: Is the opposite true, i.e is it true that
$$intlimits_{Omega}{uvdx}<infty,forall vin H_0^1(Omega)$$ implies $uin L^{6/5}(Omega)$ or at least $uin L^1(Omega)$ ?

My thoughts: It is easy to see that $uin L^1_{loc}(Omega)$ by taking $v$ to be smooth cut-off functions equal to $1$ in compact subsets of $Omega$ and $0$ in a neighborhood of the boundary $partial Omega$.

The motivation for this question is the "correct" weak formulation of a nonlinear problem - whether to formulate it as $(1)$ or as $(2)$:

$(1)$ Find $uin H_0^1(Omega)$ such that $f(u)in L^{6/5}(Omega)$ and
$$a(u,v)+intlimits_{Omega}{f(u)vdx}=0,forall vin H_0^1(Omega)$$

or

$(2)$ Find $uin H_0^1(Omega)$ such that $intlimits_{Omega}{f(u)vdx}<infty,forall vin H_0^1(Omega)$ and
$$a(u,v)+intlimits_{Omega}{f(u)vdx}=0,forall vin H_0^1(Omega)$$

where $a(.,.)$ is a bilinear form and $f(.)$ is in general a nonlinear function. If the answer to my question is affirmative then both formulations are equivalent.

Note that $(2)$ is less restrictive, because the set in which we search for a solution is bigger, so it might be easier to find such.

pde lebesgue-integral sobolev-spaces lp-spaces

share|cite|improve this question

edited Mar 23 at 12:57

Svetoslav

asked Jan 1 '16 at 15:24

SvetoslavSvetoslav

4,5872828

$endgroup$

share|cite|improve this question

edited Mar 23 at 12:57

Svetoslav

asked Jan 1 '16 at 15:24

SvetoslavSvetoslav

4,5872828

share|cite|improve this question

edited Mar 23 at 12:57

Svetoslav

asked Jan 1 '16 at 15:24

SvetoslavSvetoslav

4,5872828

asked Jan 1 '16 at 15:24

SvetoslavSvetoslav

4,5872828

asked Jan 1 '16 at 15:24

SvetoslavSvetoslav

4,5872828