= Solution
Take the <weakly convergent> <subsequence> in (i), with limit $u\in H_0^1(B)$. For each $\phi\in H_0^1(B)$, the <weak formulation> gives
$$
\int_B\nabla u_{m_j}\cdot\nabla\phi\,dx=\int_Bf_{m_j}\phi\,dx.
$$
The left side tends to $\int\nabla u\cdot\nabla\phi$ by <weak convergence> in $H^1$. The zero-boundary <Sobolev inequality> gives $\phi\in L^6(B)$, so (ii) makes the right side tend to zero. Taking $\phi=u$ in the limiting identity yields $\int_B|\nabla u|^2=0$. The zero-boundary <Poincaré inequality> then gives $u=0$.
The same argument applies to a <weakly convergent> <subsequence> of any <subsequence> of $(u_m)$, because all remain bounded as in (i). To see why this proves convergence of the full sequence, suppose a continuous <linear functional> $F$ on $H^1(B)$ did not satisfy $F(u_m)\to0$. Some <subsequence> would have $|F(u_{m_j})|\geq\varepsilon>0$. Its boundedness gives a further <weakly convergent> <subsequence>, whose limit has just been proved to be zero; then $F$ tends to zero along it, a contradiction. Therefore
$$
\boxed{u_m\rightharpoonup0\quad\text{in }H^1(B)\text{ for the entire sequence}.}
$$
Uniqueness of a possible limit alone is not enough without the <weak sequential compactness in a Hilbert space> used to extract a further <subsequence>.
Back to article page
