= Solution
\b[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.] Define
$$
u(x)=\alpha+\int_a^xv(t)dt.
$$
Then $u\in W^{1,1}$ and $u'=v$. Weak $L^1$ convergence tested against $\mathbf1_{(a,x)}$ 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 $|u_j'|$ 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 particular
$$
\boxed{u_{j_k}\to u\text{ in }L^1(a,b),\qquad u'=v,\quad u(a)=\alpha,\ u(b)=\beta.}
$$
<Uniform 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.
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