Proving the weak closure of Sobolev Spaces,

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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 $u\in W^{1,4} [(0,1)^{2}]$ then $u \in W^{1,4}[\mathbb{T}^{2}]$.


$\bf{Notation :}$

The space $W^{1,4}[\mathbb{T}^{2}]=\{u : u\in L^{4}_{loc}(\mathbb{R}^{2}) , ~ u_{x_{i}}\in L^{4}_{loc}(\mathbb{R}^{2}), ~ u(x_{1},x_{2})=u(x_{1}+1,x_{2}), ~u(x_{1},x_{2})=u(x_{1},x_{2}+1) , ~ u \textrm{ has zero average. } \}$


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 $v\in C_{c}^{\infty}[0,k]^{2}$

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 $\lim_{n\to \infty} f(u_{n})=f(u)$ so

\begin{equation} \lim_{n\to \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_{n\to \infty} \int_{0}^{k}\int_{0}^{k}u_{n}^{4} v=\lim_{n\to \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 v\in C_{c}^{\infty}(0,k)^{2} \end{equation}

So that $u(x_{1},x_{2})=u(x_{1}+1,x_{2})$.

ii) Showing $ u\in 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 $W^{1,4}(T^2)$