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The displayed spikes have $L^p$ norm one. Pairing with $g\in L^{p^{\prime }}$ and applying Holder inequality bounds the pairing by $({\int }_0^{1/j}|g{|}^{p^{\prime }}{)}^{1/p^{\prime }}$, which tends to zero. At exponent one, however, $j{1}_{(0,1/j)}$ cannot converge weakly in $L^1$: every interval away from zero forces any limit to vanish there, while the constant test function preserves integral one. Their failure of uniform integrability records exactly the obstruction in the Dunford-Pettis theorem.

On a finite measure space, a family is uniformly integrable exactly when it admits the displayed bound for some nonnegative increasing convex superlinear function $\Psi$. By the Dunford-Pettis theorem this ensures relative weak compactness in $L^1$. For the sufficient direction, choose $R_A$ so $\Psi (s)\ge As$ for $s>R_A$; then ${\int }_E|v_j|\le R_A\mu (E)+A^{-1}{\sup }_j\int \Psi (|v_j|)$ gives uniform integrability. Conversely choose increasing thresholds $R_m\to \infty$ so ${\sup }_j{\int }_{|v_j|>R_m}|v_j|\le 2^{-m}$. The convex function $\Psi (s)={\sum }_{m\ge 1}(s-R_m{)}_{+}$ is locally a finite sum, has uniformly bounded integrals by summing these tail estimates, and is superlinear because for $s\ge 2R_N$ its first $N$ terms sum to at least $Ns/2$. This proves the criterion in both directions.

The De la Vallée-Poussin criterion characterizes uniform integrability by existence of a nonnegative increasing convex function $\Psi$ with $\Psi (s)/s\to \infty$ and a uniform bound on $\int \Psi (|v_j|)$. On a finite interval this gives a weakly convergent subsequence in $L^1$ by the Dunford-Pettis theorem.

After discarding a finite prefix of the minimizing sequence if needed, put $H(t)=\Phi (t)-M\ge 0$. Its derivative sequence has $\int H(u_j^{\prime })\le C$. The superlinear bound directly gives uniform integrability: for $H(t)\ge A|t|$ outside $[-R_A,R_A]$,If the convex version of the criterion is used, choose increasing $R_m\to \infty$ so $H(t)\ge m^2|t|$ for $|t|\ge R_m$, and set $\Psi (s)={\sum }_{m\ge 1}(s-R_m{)}_{+}$. This is locally a finite sum, convex and increasing. It is superlinear because for fixed $N$ and $s\ge 2R_N$, $\Psi (s)\ge Ns/2$. If exactly $m$ terms are active, $\Psi (s)\le ms\le H(t)$ for $s=|t|$, so its integrals have the same bound. The criterion therefore supplies $u_{j_k}^{\prime }\rightharpoonup v$ in $L^1$.

Define $u(x)=\alpha +{\int }_a^{x}v$. Interval-indicator tests give $u_{j_k}(x)\to u(x)$ for every $x$, including $b$, so its endpoint values are correct and $u\in W^{1,1}$. The common derivative $L^1$ bound makes these functions uniformly bounded. Uniform integrability also gives common equicontinuity, since an increment is bounded by the derivative integral on the intervening interval. The Arzelà-Ascoli theorem yields uniform convergence to the same identified limit, and in particularHere the printed admissible space is $W^{1,1}$, so this compactness argument gives precisely the needed membership.

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