In a problem I'm working on I want to show that a bounded sequence $\{u_n\} \subset W_0^{1,1}((0,1))$ converges weakly. Of course, since $W^{1,1}$ is not reflexive I don't get this for free.
The reason, I think, that bounded sequences don't converge is that $W^{1,1}$ can't detect the formation of step functions. So I think that with additional assumptions on $u_n$, I might be able to show that my sequence does converge weakly. Specifically, I think that I can prove that each $u_n$ is concave, which should control the formation of step functions except at the boundary. However, I think it's plausible that I can control the growth of $Du_n$ (perhaps in some $L^p$ norm) in some small region near the endpoints. Intuitively, this should allow me to globally control the growth of $Du_n$, which should buy me weak convergence in some $W^{1,p}$ space. (This is actually what I need).
So my question is how the previous discussion can be made precise. It boils down to the following:
What assumptions (in addition to the concavity of each $u_n$) could I use in order to show that $u_n$ converges weakly in some $W^{1,p}$?