Proving the weak closure of Sobolev Spaces,
$begingroup$
I am trying to show the space $W^{1,4}[mathbb{T}^{2}]$ is weakly closed in $W^{1,4}[(0,1)^{2}]$.
$bf{That ~~ is :} $
If $u_{n}rightharpoonup u$. Where $u_{n}in W^{1,4}[mathbb{T}^{2}]$ and $uin W^{1,4} [(0,1)^{2}]$ then $u in W^{1,4}[mathbb{T}^{2}]$.
$bf{Notation :}$
The space $W^{1,4}[mathbb{T}^{2}]={u : uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2}), ~ u(x_{1},x_{2})=u(x_{1}+1,x_{2}), ~u(x_{1},x_{2})=u(x_{1},x_{2}+1) , ~ u textrm{ has zero average. } }$
Here is my attempt (I think my solution is wrong somewhere / or overcomplicated) does anyone have a better solution ? Im rather new to PDEs and working from Evans book.
Let $u_{n}rightharpoonup u$
i) We show that this implies $u(x_{1}+1,x_{2})=u(x_{1},x_{2})$
Take the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u^{4} v $, for some $vin C_{c}^{infty}[0,k]^{2}$
Then $f$ is continuous (bounded), indeed $|f(u)|leq k sup_{(0,k)^{2}} (v) int_{(0,1)^{2}} u^{4} = k sup_{(0,k)^{2}} (v) ||u||_{W^{1,4}(0,1)^{2}} $.
Hence by the deff. of weakly convergent $lim_{nto infty} f(u_{n})=f(u)$ so
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=int_{0}^{k}int_{0}^{k}u^{4} v
end{equation}
But since $u_{n}(x_{1}+1,x_{2})=u_{n}(x_{1},x_{2})$ we have
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=lim_{nto infty} int_{0}^{k}int_{1}^{k+1}u_{n}^{4} v
end{equation}
Hence (using weak convergence)
begin{equation}
int_{0}^{k}int_{0}^{k}u^{4} v=int_{0}^{k}int_{1}^{k+1}u^{4} v
end{equation}
Now change variables and rearrange
begin{equation}
int_{0}^{k}int_{0}^{k}[u^{4}(x_{1},x_{2})-u^{4}(x_{1}+1,x_{2})] v=0 ~~~ forall vin C_{c}^{infty}(0,k)^{2}
end{equation}
So that $u(x_{1},x_{2})=u(x_{1}+1,x_{2})$.
ii) Showing $ uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2})$, Follows from i)
EDIT : if in the above I instead work with the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u v $, (we can use Holder to show its bounded), if were only showing boundedness for $W^{1,4}(T^2)$
pde sobolev-spaces lp-spaces weak-convergence
asked Dec 25 '18 at 1:23
MontyMonty
34613
$endgroup$
add a comment |
$begingroup$
I am trying to show the space $W^{1,4}[mathbb{T}^{2}]$ is weakly closed in $W^{1,4}[(0,1)^{2}]$.
$bf{That ~~ is :} $
If $u_{n}rightharpoonup u$. Where $u_{n}in W^{1,4}[mathbb{T}^{2}]$ and $uin W^{1,4} [(0,1)^{2}]$ then $u in W^{1,4}[mathbb{T}^{2}]$.
$bf{Notation :}$
The space $W^{1,4}[mathbb{T}^{2}]={u : uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2}), ~ u(x_{1},x_{2})=u(x_{1}+1,x_{2}), ~u(x_{1},x_{2})=u(x_{1},x_{2}+1) , ~ u textrm{ has zero average. } }$
Here is my attempt (I think my solution is wrong somewhere / or overcomplicated) does anyone have a better solution ? Im rather new to PDEs and working from Evans book.
Let $u_{n}rightharpoonup u$
i) We show that this implies $u(x_{1}+1,x_{2})=u(x_{1},x_{2})$
Take the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u^{4} v $, for some $vin C_{c}^{infty}[0,k]^{2}$
Then $f$ is continuous (bounded), indeed $|f(u)|leq k sup_{(0,k)^{2}} (v) int_{(0,1)^{2}} u^{4} = k sup_{(0,k)^{2}} (v) ||u||_{W^{1,4}(0,1)^{2}} $.
Hence by the deff. of weakly convergent $lim_{nto infty} f(u_{n})=f(u)$ so
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=int_{0}^{k}int_{0}^{k}u^{4} v
end{equation}
But since $u_{n}(x_{1}+1,x_{2})=u_{n}(x_{1},x_{2})$ we have
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=lim_{nto infty} int_{0}^{k}int_{1}^{k+1}u_{n}^{4} v
end{equation}
Hence (using weak convergence)
begin{equation}
int_{0}^{k}int_{0}^{k}u^{4} v=int_{0}^{k}int_{1}^{k+1}u^{4} v
end{equation}
Now change variables and rearrange
begin{equation}
int_{0}^{k}int_{0}^{k}[u^{4}(x_{1},x_{2})-u^{4}(x_{1}+1,x_{2})] v=0 ~~~ forall vin C_{c}^{infty}(0,k)^{2}
end{equation}
So that $u(x_{1},x_{2})=u(x_{1}+1,x_{2})$.
ii) Showing $ uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2})$, Follows from i)
EDIT : if in the above I instead work with the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u v $, (we can use Holder to show its bounded), if were only showing boundedness for $W^{1,4}(T^2)$
pde sobolev-spaces lp-spaces weak-convergence
asked Dec 25 '18 at 1:23
MontyMonty
34613
$endgroup$
$begingroup$
Edit : I have already noticed a mistake in proving boundedness I have assumed $u$ to be periodic...
$endgroup$
– Monty
Dec 25 '18 at 1:30
$begingroup$
Edit : nor is it linear. unless we add a $(~)^{1/4}$
$endgroup$
– Monty
Dec 26 '18 at 18:12
add a comment |
$begingroup$
I am trying to show the space $W^{1,4}[mathbb{T}^{2}]$ is weakly closed in $W^{1,4}[(0,1)^{2}]$.
$bf{That ~~ is :} $
If $u_{n}rightharpoonup u$. Where $u_{n}in W^{1,4}[mathbb{T}^{2}]$ and $uin W^{1,4} [(0,1)^{2}]$ then $u in W^{1,4}[mathbb{T}^{2}]$.
$bf{Notation :}$
The space $W^{1,4}[mathbb{T}^{2}]={u : uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2}), ~ u(x_{1},x_{2})=u(x_{1}+1,x_{2}), ~u(x_{1},x_{2})=u(x_{1},x_{2}+1) , ~ u textrm{ has zero average. } }$
Here is my attempt (I think my solution is wrong somewhere / or overcomplicated) does anyone have a better solution ? Im rather new to PDEs and working from Evans book.
Let $u_{n}rightharpoonup u$
i) We show that this implies $u(x_{1}+1,x_{2})=u(x_{1},x_{2})$
Take the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u^{4} v $, for some $vin C_{c}^{infty}[0,k]^{2}$
Then $f$ is continuous (bounded), indeed $|f(u)|leq k sup_{(0,k)^{2}} (v) int_{(0,1)^{2}} u^{4} = k sup_{(0,k)^{2}} (v) ||u||_{W^{1,4}(0,1)^{2}} $.
Hence by the deff. of weakly convergent $lim_{nto infty} f(u_{n})=f(u)$ so
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=int_{0}^{k}int_{0}^{k}u^{4} v
end{equation}
But since $u_{n}(x_{1}+1,x_{2})=u_{n}(x_{1},x_{2})$ we have
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=lim_{nto infty} int_{0}^{k}int_{1}^{k+1}u_{n}^{4} v
end{equation}
Hence (using weak convergence)
begin{equation}
int_{0}^{k}int_{0}^{k}u^{4} v=int_{0}^{k}int_{1}^{k+1}u^{4} v
end{equation}
Now change variables and rearrange
begin{equation}
int_{0}^{k}int_{0}^{k}[u^{4}(x_{1},x_{2})-u^{4}(x_{1}+1,x_{2})] v=0 ~~~ forall vin C_{c}^{infty}(0,k)^{2}
end{equation}
So that $u(x_{1},x_{2})=u(x_{1}+1,x_{2})$.
ii) Showing $ uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2})$, Follows from i)
EDIT : if in the above I instead work with the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u v $, (we can use Holder to show its bounded), if were only showing boundedness for $W^{1,4}(T^2)$
pde sobolev-spaces lp-spaces weak-convergence
asked Dec 25 '18 at 1:23
MontyMonty
34613
$endgroup$
I am trying to show the space $W^{1,4}[mathbb{T}^{2}]$ is weakly closed in $W^{1,4}[(0,1)^{2}]$.
$bf{That ~~ is :} $
If $u_{n}rightharpoonup u$. Where $u_{n}in W^{1,4}[mathbb{T}^{2}]$ and $uin W^{1,4} [(0,1)^{2}]$ then $u in W^{1,4}[mathbb{T}^{2}]$.
$bf{Notation :}$
The space $W^{1,4}[mathbb{T}^{2}]={u : uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2}), ~ u(x_{1},x_{2})=u(x_{1}+1,x_{2}), ~u(x_{1},x_{2})=u(x_{1},x_{2}+1) , ~ u textrm{ has zero average. } }$
Here is my attempt (I think my solution is wrong somewhere / or overcomplicated) does anyone have a better solution ? Im rather new to PDEs and working from Evans book.
Let $u_{n}rightharpoonup u$
i) We show that this implies $u(x_{1}+1,x_{2})=u(x_{1},x_{2})$
Take the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u^{4} v $, for some $vin C_{c}^{infty}[0,k]^{2}$
Then $f$ is continuous (bounded), indeed $|f(u)|leq k sup_{(0,k)^{2}} (v) int_{(0,1)^{2}} u^{4} = k sup_{(0,k)^{2}} (v) ||u||_{W^{1,4}(0,1)^{2}} $.
Hence by the deff. of weakly convergent $lim_{nto infty} f(u_{n})=f(u)$ so
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=int_{0}^{k}int_{0}^{k}u^{4} v
end{equation}
But since $u_{n}(x_{1}+1,x_{2})=u_{n}(x_{1},x_{2})$ we have
begin{equation}
lim_{nto infty} int_{0}^{k}int_{0}^{k}u_{n}^{4} v=lim_{nto infty} int_{0}^{k}int_{1}^{k+1}u_{n}^{4} v
end{equation}
Hence (using weak convergence)
begin{equation}
int_{0}^{k}int_{0}^{k}u^{4} v=int_{0}^{k}int_{1}^{k+1}u^{4} v
end{equation}
Now change variables and rearrange
begin{equation}
int_{0}^{k}int_{0}^{k}[u^{4}(x_{1},x_{2})-u^{4}(x_{1}+1,x_{2})] v=0 ~~~ forall vin C_{c}^{infty}(0,k)^{2}
end{equation}
So that $u(x_{1},x_{2})=u(x_{1}+1,x_{2})$.
ii) Showing $ uin L^{4}_{loc}(mathbb{R}^{2}) , ~ u_{x_{i}}in L^{4}_{loc}(mathbb{R}^{2})$, Follows from i)
EDIT : if in the above I instead work with the linear functional $f$ where $f(u)=int_{0}^{k}int_{0}^{k}u v $, (we can use Holder to show its bounded), if were only showing boundedness for $W^{1,4}(T^2)$
pde sobolev-spaces lp-spaces weak-convergence
pde sobolev-spaces lp-spaces weak-convergence
asked Dec 25 '18 at 1:23
MontyMonty
34613
asked Dec 25 '18 at 1:23
MontyMonty
34613
edited Dec 26 '18 at 18:59
Monty
asked Dec 25 '18 at 1:23
MontyMonty
34613
asked Dec 25 '18 at 1:23
MontyMonty
34613
asked Dec 25 '18 at 1:23
MontyMonty
34613
34613
$begingroup$
Edit : I have already noticed a mistake in proving boundedness I have assumed $u$ to be periodic...
$endgroup$
– Monty
Dec 25 '18 at 1:30
$begingroup$
Edit : nor is it linear. unless we add a $(~)^{1/4}$
$endgroup$
– Monty
Dec 26 '18 at 18:12
add a comment |
$begingroup$
Edit : I have already noticed a mistake in proving boundedness I have assumed $u$ to be periodic...
$endgroup$
– Monty
Dec 25 '18 at 1:30
$begingroup$
Edit : nor is it linear. unless we add a $(~)^{1/4}$
$endgroup$
– Monty
Dec 26 '18 at 18:12
$begingroup$
Edit : I have already noticed a mistake in proving boundedness I have assumed $u$ to be periodic...
$endgroup$
– Monty
Dec 25 '18 at 1:30
$begingroup$
Edit : I have already noticed a mistake in proving boundedness I have assumed $u$ to be periodic...
$endgroup$
– Monty
Dec 25 '18 at 1:30
$begingroup$
Edit : nor is it linear. unless we add a $(~)^{1/4}$
$endgroup$
– Monty
Dec 26 '18 at 18:12
$begingroup$
Edit : nor is it linear. unless we add a $(~)^{1/4}$
$endgroup$
– Monty
Dec 26 '18 at 18:12
add a comment |
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$begingroup$
Edit : I have already noticed a mistake in proving boundedness I have assumed $u$ to be periodic...
$endgroup$
– Monty
Dec 25 '18 at 1:30
$begingroup$
Edit : nor is it linear. unless we add a $(~)^{1/4}$
$endgroup$
– Monty
Dec 26 '18 at 18:12