Past exam of the partial differential equations for pure and applied mathematicians and calculus of variations course of the University of Oxford / 2008 / B / 8 / a / iii / Solution

The compactness conclusion justified by the hypotheses is $u\in W^{1,1}$ with the prescribed traces; the printed claim $u\in W^{1,\infty }$ needs additional regularity. DefineThen $u\in W^{1,1}$ and $u^{\prime }=v$. Weak $L^1$ convergence tested against $1_{(a,x)}$ gives $u_{j_k}(x)\to u(x)$ for every $x$, including the endpoint trace at $b$. Uniform integrability of the derivatives makes the functions equicontinuous: their increment over an interval is bounded by the integral of $|u_j^{\prime }|$ on that interval, uniformly small for short intervals. Their fixed initial trace and common derivative $L^1$ bound also make them uniformly bounded. The Arzelà-Ascoli theorem therefore upgrades the convergence to uniform convergence, and in particularUniform integrability does not bound the derivative in $L^{\infty }$. The natural compact class here is absolutely continuous functions, rather than Lipschitz functions. This distinction can affect attainment: smooth strictly convex superlinear integrands can have non-Lipschitz minimizing limits, so it is not repaired merely by adding convexity in the last variable. If one additionally has a uniform Lipschitz bound on the minimizing sequence, weak testing on measurable subsets passes that bound to $v$ and proves the printed $u\in \mathcal{A}$. Alternatively one can formulate the direct method from the outset in the fixed-trace $W^{1,1}$ class. The proof above gives all the asserted convergence and trace conclusions without assuming the missing bound.