Let $X$ be a reflexive [Banach space](/page/Banach%20Space), let $\mathcal A\subset X$ be nonempty and sequentially weakly closed for the [weak topology](/page/Weak%20Topology) of $X$, and let $I:\mathcal A\to(-\infty,\infty]$ be a functional with finite infimum $\alpha:=\inf_{u\in\mathcal A} I[u]\in\mathbb R$. For each $u\in\mathcal A$, define the sequential weak relaxation of $I$ on $\mathcal A$ by $\overline I[u]:=\inf\{\liminf_{k\to\infty} I[u_k]:(u_k)_{k=1}^{\infty}\subset\mathcal A,\ u_k\rightharpoonup u\text{ in }X\}$. Assume that $\overline I$ is sequentially weakly lower semicontinuous on $\mathcal A$, meaning that whenever $(u_k)_{k=1}^{\infty}\subset\mathcal A$ and $u_k\rightharpoonup u\in\mathcal A$ in $X$, one has $\overline I[u]\le \liminf_{k\to\infty}\overline I[u_k]$. Assume also that every sequence $(v_k)_{k=1}^{\infty}\subset\mathcal A$ satisfying $\overline I[v_k]\to\inf_{u\in\mathcal A}\overline I[u]$ is bounded in $X$. Then there exists $u_*\in\mathcal A$ such that $\overline I[u_*]=\min_{u\in\mathcal A}\overline I[u]$. Moreover, $\min_{u\in\mathcal A}\overline I[u]=\inf_{u\in\mathcal A} I[u]$.