Show convergence of inner product of an operator and a weak convergent sequence
$begingroup$
Let $(u_n)_{ninmathbb{N}}subseteq (W^{1,p}_0(Omega))$, $u_n rightharpoonup u in (W^{1,p}_0(Omega))$ and $B$ be an operator on $(W^{1,p}_0(Omega))times(W^{1,p}_0(Omega))$.
B is defined by $$langle B(w,u), vrangle := int_Omega ||nabla u - h(w)||^{p-2}(nabla u - h(w))nabla v dx,$$ where $h:mathbb{R}tomathbb{R}^n$ is continuous and bounded.
I want to show that $langle B(u_n, v), u_n -urangle to 0$ where $vin (W^{1,p}_0(Omega))$.
Is it right, that I just have to show that $B(u_n, v)in(W^{1,p}_0(Omega))^*$ $forall ninmathbb{N}$?
Then the statement follows from the weak continuity of $(u_n)_{ninmathbb{N}}$ (?)
functional-analysis ordinary-differential-equations convergence sobolev-spaces weak-convergence
asked Jan 5 at 13:33
user3766553user3766553
526
$endgroup$
|
show 8 more comments
$begingroup$
Let $(u_n)_{ninmathbb{N}}subseteq (W^{1,p}_0(Omega))$, $u_n rightharpoonup u in (W^{1,p}_0(Omega))$ and $B$ be an operator on $(W^{1,p}_0(Omega))times(W^{1,p}_0(Omega))$.
B is defined by $$langle B(w,u), vrangle := int_Omega ||nabla u - h(w)||^{p-2}(nabla u - h(w))nabla v dx,$$ where $h:mathbb{R}tomathbb{R}^n$ is continuous and bounded.
I want to show that $langle B(u_n, v), u_n -urangle to 0$ where $vin (W^{1,p}_0(Omega))$.
Is it right, that I just have to show that $B(u_n, v)in(W^{1,p}_0(Omega))^*$ $forall ninmathbb{N}$?
Then the statement follows from the weak continuity of $(u_n)_{ninmathbb{N}}$ (?)
functional-analysis ordinary-differential-equations convergence sobolev-spaces weak-convergence
asked Jan 5 at 13:33
user3766553user3766553
526
$endgroup$
$begingroup$
Is it intended that you first write $B(w,u)$ and then $B(u_n,v)$? Or did you swap the two variables by any chance?
$endgroup$
– Lorenzo Quarisa
Jan 5 at 15:53
$begingroup$
This is the way it is written in my exercise. So I think it is intended to be like this @LorenzoQuarisa
$endgroup$
– user3766553
Jan 5 at 16:55
$begingroup$
$Omega$ is bounded, right? Because otherwise for $h(x)=1$ the integral defining $left langle B(w,u),vrightrangle$ might fail to converge
$endgroup$
– Lorenzo Quarisa
Jan 5 at 20:39
$begingroup$
Actually there is no information about $Omega$ in my exercise. But is my idea right? When $B(u_n, v)in (W^{1,p}_0(Omega ))^*$ $forall nin mathbb{N}$ it holds that $langle B(u_n, v), u_n-urangle to 0$?
$endgroup$
– user3766553
Jan 6 at 15:54
$begingroup$
No, it does not. You need some information on the map $u_nmapsto B(u_n,v)$. A priori, it could be that this dependance makes the evaluation $left langle B(u_n,v),u_n-urightrangle$ blow-up even if $left langle F,u_n-urightrangle$ for all $Fin W_0^{1,p}(Omega)^*$. I think you need to work on the specific expression of $B(w,v)$ here. I tried myself and was not able to solve it, probably because I don't know how to use the condition that $u,vin W_0^{1,p}$ rather than just $W^{1,p}$. Any ideas on that?
$endgroup$
– Lorenzo Quarisa
Jan 6 at 16:00
|
show 8 more comments
$begingroup$
Let $(u_n)_{ninmathbb{N}}subseteq (W^{1,p}_0(Omega))$, $u_n rightharpoonup u in (W^{1,p}_0(Omega))$ and $B$ be an operator on $(W^{1,p}_0(Omega))times(W^{1,p}_0(Omega))$.
B is defined by $$langle B(w,u), vrangle := int_Omega ||nabla u - h(w)||^{p-2}(nabla u - h(w))nabla v dx,$$ where $h:mathbb{R}tomathbb{R}^n$ is continuous and bounded.
I want to show that $langle B(u_n, v), u_n -urangle to 0$ where $vin (W^{1,p}_0(Omega))$.
Is it right, that I just have to show that $B(u_n, v)in(W^{1,p}_0(Omega))^*$ $forall ninmathbb{N}$?
Then the statement follows from the weak continuity of $(u_n)_{ninmathbb{N}}$ (?)
functional-analysis ordinary-differential-equations convergence sobolev-spaces weak-convergence
asked Jan 5 at 13:33
user3766553user3766553
526
$endgroup$
Let $(u_n)_{ninmathbb{N}}subseteq (W^{1,p}_0(Omega))$, $u_n rightharpoonup u in (W^{1,p}_0(Omega))$ and $B$ be an operator on $(W^{1,p}_0(Omega))times(W^{1,p}_0(Omega))$.
B is defined by $$langle B(w,u), vrangle := int_Omega ||nabla u - h(w)||^{p-2}(nabla u - h(w))nabla v dx,$$ where $h:mathbb{R}tomathbb{R}^n$ is continuous and bounded.
I want to show that $langle B(u_n, v), u_n -urangle to 0$ where $vin (W^{1,p}_0(Omega))$.
Is it right, that I just have to show that $B(u_n, v)in(W^{1,p}_0(Omega))^*$ $forall ninmathbb{N}$?
Then the statement follows from the weak continuity of $(u_n)_{ninmathbb{N}}$ (?)
functional-analysis ordinary-differential-equations convergence sobolev-spaces weak-convergence
functional-analysis ordinary-differential-equations convergence sobolev-spaces weak-convergence
asked Jan 5 at 13:33
user3766553user3766553
526
asked Jan 5 at 13:33
user3766553user3766553
526
asked Jan 5 at 13:33
user3766553user3766553
526
asked Jan 5 at 13:33
user3766553user3766553
526
asked Jan 5 at 13:33
user3766553user3766553
526
526
$begingroup$
Is it intended that you first write $B(w,u)$ and then $B(u_n,v)$? Or did you swap the two variables by any chance?
$endgroup$
– Lorenzo Quarisa
Jan 5 at 15:53
$begingroup$
This is the way it is written in my exercise. So I think it is intended to be like this @LorenzoQuarisa
$endgroup$
– user3766553
Jan 5 at 16:55
$begingroup$
$Omega$ is bounded, right? Because otherwise for $h(x)=1$ the integral defining $left langle B(w,u),vrightrangle$ might fail to converge
$endgroup$
– Lorenzo Quarisa
Jan 5 at 20:39
$begingroup$
Actually there is no information about $Omega$ in my exercise. But is my idea right? When $B(u_n, v)in (W^{1,p}_0(Omega ))^*$ $forall nin mathbb{N}$ it holds that $langle B(u_n, v), u_n-urangle to 0$?
$endgroup$
– user3766553
Jan 6 at 15:54
$begingroup$
No, it does not. You need some information on the map $u_nmapsto B(u_n,v)$. A priori, it could be that this dependance makes the evaluation $left langle B(u_n,v),u_n-urightrangle$ blow-up even if $left langle F,u_n-urightrangle$ for all $Fin W_0^{1,p}(Omega)^*$. I think you need to work on the specific expression of $B(w,v)$ here. I tried myself and was not able to solve it, probably because I don't know how to use the condition that $u,vin W_0^{1,p}$ rather than just $W^{1,p}$. Any ideas on that?
$endgroup$
– Lorenzo Quarisa
Jan 6 at 16:00
|
show 8 more comments
$begingroup$
Is it intended that you first write $B(w,u)$ and then $B(u_n,v)$? Or did you swap the two variables by any chance?
$endgroup$
– Lorenzo Quarisa
Jan 5 at 15:53
$begingroup$
This is the way it is written in my exercise. So I think it is intended to be like this @LorenzoQuarisa
$endgroup$
– user3766553
Jan 5 at 16:55
$begingroup$
$Omega$ is bounded, right? Because otherwise for $h(x)=1$ the integral defining $left langle B(w,u),vrightrangle$ might fail to converge
$endgroup$
– Lorenzo Quarisa
Jan 5 at 20:39
$begingroup$
Actually there is no information about $Omega$ in my exercise. But is my idea right? When $B(u_n, v)in (W^{1,p}_0(Omega ))^*$ $forall nin mathbb{N}$ it holds that $langle B(u_n, v), u_n-urangle to 0$?
$endgroup$
– user3766553
Jan 6 at 15:54
$begingroup$
No, it does not. You need some information on the map $u_nmapsto B(u_n,v)$. A priori, it could be that this dependance makes the evaluation $left langle B(u_n,v),u_n-urightrangle$ blow-up even if $left langle F,u_n-urightrangle$ for all $Fin W_0^{1,p}(Omega)^*$. I think you need to work on the specific expression of $B(w,v)$ here. I tried myself and was not able to solve it, probably because I don't know how to use the condition that $u,vin W_0^{1,p}$ rather than just $W^{1,p}$. Any ideas on that?
$endgroup$
– Lorenzo Quarisa
Jan 6 at 16:00
$begingroup$
Is it intended that you first write $B(w,u)$ and then $B(u_n,v)$? Or did you swap the two variables by any chance?
$endgroup$
– Lorenzo Quarisa
Jan 5 at 15:53
$begingroup$
Is it intended that you first write $B(w,u)$ and then $B(u_n,v)$? Or did you swap the two variables by any chance?
$endgroup$
– Lorenzo Quarisa
Jan 5 at 15:53
$begingroup$
This is the way it is written in my exercise. So I think it is intended to be like this @LorenzoQuarisa
$endgroup$
– user3766553
Jan 5 at 16:55
$begingroup$
This is the way it is written in my exercise. So I think it is intended to be like this @LorenzoQuarisa
$endgroup$
– user3766553
Jan 5 at 16:55
$begingroup$
$Omega$ is bounded, right? Because otherwise for $h(x)=1$ the integral defining $left langle B(w,u),vrightrangle$ might fail to converge
$endgroup$
– Lorenzo Quarisa
Jan 5 at 20:39
$begingroup$
$Omega$ is bounded, right? Because otherwise for $h(x)=1$ the integral defining $left langle B(w,u),vrightrangle$ might fail to converge
$endgroup$
– Lorenzo Quarisa
Jan 5 at 20:39
$begingroup$
Actually there is no information about $Omega$ in my exercise. But is my idea right? When $B(u_n, v)in (W^{1,p}_0(Omega ))^*$ $forall nin mathbb{N}$ it holds that $langle B(u_n, v), u_n-urangle to 0$?
$endgroup$
– user3766553
Jan 6 at 15:54
$begingroup$
Actually there is no information about $Omega$ in my exercise. But is my idea right? When $B(u_n, v)in (W^{1,p}_0(Omega ))^*$ $forall nin mathbb{N}$ it holds that $langle B(u_n, v), u_n-urangle to 0$?
$endgroup$
– user3766553
Jan 6 at 15:54
$begingroup$
No, it does not. You need some information on the map $u_nmapsto B(u_n,v)$. A priori, it could be that this dependance makes the evaluation $left langle B(u_n,v),u_n-urightrangle$ blow-up even if $left langle F,u_n-urightrangle$ for all $Fin W_0^{1,p}(Omega)^*$. I think you need to work on the specific expression of $B(w,v)$ here. I tried myself and was not able to solve it, probably because I don't know how to use the condition that $u,vin W_0^{1,p}$ rather than just $W^{1,p}$. Any ideas on that?
$endgroup$
– Lorenzo Quarisa
Jan 6 at 16:00
$begingroup$
No, it does not. You need some information on the map $u_nmapsto B(u_n,v)$. A priori, it could be that this dependance makes the evaluation $left langle B(u_n,v),u_n-urightrangle$ blow-up even if $left langle F,u_n-urightrangle$ for all $Fin W_0^{1,p}(Omega)^*$. I think you need to work on the specific expression of $B(w,v)$ here. I tried myself and was not able to solve it, probably because I don't know how to use the condition that $u,vin W_0^{1,p}$ rather than just $W^{1,p}$. Any ideas on that?
$endgroup$
– Lorenzo Quarisa
Jan 6 at 16:00
|
show 8 more comments
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$begingroup$
Is it intended that you first write $B(w,u)$ and then $B(u_n,v)$? Or did you swap the two variables by any chance?
$endgroup$
– Lorenzo Quarisa
Jan 5 at 15:53
$begingroup$
This is the way it is written in my exercise. So I think it is intended to be like this @LorenzoQuarisa
$endgroup$
– user3766553
Jan 5 at 16:55
$begingroup$
$Omega$ is bounded, right? Because otherwise for $h(x)=1$ the integral defining $left langle B(w,u),vrightrangle$ might fail to converge
$endgroup$
– Lorenzo Quarisa
Jan 5 at 20:39
$begingroup$
Actually there is no information about $Omega$ in my exercise. But is my idea right? When $B(u_n, v)in (W^{1,p}_0(Omega ))^*$ $forall nin mathbb{N}$ it holds that $langle B(u_n, v), u_n-urangle to 0$?
$endgroup$
– user3766553
Jan 6 at 15:54
$begingroup$
No, it does not. You need some information on the map $u_nmapsto B(u_n,v)$. A priori, it could be that this dependance makes the evaluation $left langle B(u_n,v),u_n-urightrangle$ blow-up even if $left langle F,u_n-urightrangle$ for all $Fin W_0^{1,p}(Omega)^*$. I think you need to work on the specific expression of $B(w,v)$ here. I tried myself and was not able to solve it, probably because I don't know how to use the condition that $u,vin W_0^{1,p}$ rather than just $W^{1,p}$. Any ideas on that?
$endgroup$
– Lorenzo Quarisa
Jan 6 at 16:00