 Solution

A minimizing sequence has an energy upper bound $C$. Set $H(t)=\Phi (t)-M\ge 0$. Then $\int H(u_j^{\prime })\le C-M(b-a)=C_1$. Since $H(t)/|t|\to \infty$, for every $A>0$ there is $R_A$ such that $H(t)\ge A|t|$ for $|t|>R_A$. Consequently, for every measurable $E$,Taking first $A$ large and then $|E|$ small proves uniform integrability; taking $E=(a,b)$ also gives a uniform $L^1$ bound. The Dunford-Pettis theorem on a finite-measure space identifies bounded uniformly integrable subsets of $L^1$ with relatively weakly compact ones. Thus a subsequence satisfies