Let $I = (0,1)$ i.e. a bounded interval.Prove that for bounded sequence $u_k$ in $W^{1,p}(I)$ for $p = (1,\infty]$. Exist some $u\in W^{1,p}(I)$ also.such that $u_k\to u$ in $L^\infty(I)$.
My attempt,using the Sobolev embedding theorem such that $W^{1,p}(I)\subset C(\bar{I})$ is compact inclusion.
Then we know exist subsequence of $u_{n_k} \to u$ in $L^{\infty}(I)$ such that $u\in L^{\infty}(I)\subset L^p(I)$ also.
Now we need to find the weak derivative of $u$ in order to prove that $u\in W^{1,p}$.
There are three possible ideas:
- first one is if we can prove $u'_{n_k} \to v$ in $L^p(I)$.Then we are done since it even implies that $u_{n_k}\to u$ in $W^{1,p}(I)$ (which may not be very likely to happens since we know there are only weak convergence for derivative:that is $u'_{n_k}\to u'$ weakly)
- using the criterion that if $$|\int \varphi'udx|\le \|u\|_{L^p}\|\varphi\|_{L^{p'}}$$ then we are done. (which is given in Brezis's Sobolev space book page 206 Prop 8.3)
Maybe we can insert $u_{k_n}$ into it. then the hard part is the integral $$|\int(u_{n_k}-u)\varphi' dx |\le C\|\varphi\|_{p'} ?$$ 3. If we can prove $u$ is absolute continuous,then the derivative in measure theory sense exist.
I find three of them is not very obvious,I guess may be I misinterpret something?
$u_k'$ is bounded in $L^p$, so identifying $L^p = (L^q)^*$ it has a subsequence converging weak-* to some $v \in L^p$. Let $\phi \in C^\infty_c(I)$. Then $\phi' \in L^q$ and so $\int_I u_{k_j}' \phi\,dx \to \int_I v \phi\,dx$. On the other hand $$ \int_I u_{k_j}' \phi\,dx = - \int_I u_{k_j} \phi'\,dx \to -\int_I u \phi'\,dx$$ so $\int_I v \phi\,dx = -\int_I u \phi'\,dx$ for all $\varphi \in C^\infty_c(I)$, which shows that $u' = v$.