[proofplan] We first compare $I$ and its sequential weak relaxation directly from the definition: constant recovery sequences give $\overline I\le I$, while every admissible recovery sequence has $\liminf$ at least $\inf_{\mathcal A}I$. This proves that $I$ and $\overline I$ have the same infimum. We then take a minimizing sequence for $\overline I$, use the assumed boundedness and [weak sequential compactness](/theorems/214) in reflexive Banach spaces to extract a weakly convergent subsequence, use sequential weak closedness of $\mathcal A$ to keep the limit admissible, and finally apply sequential weak lower semicontinuity to show that the limit attains the common infimum. [/proofplan]

[step:Compare the relaxation with the original functional by constant recovery sequences]Let $u\in\mathcal A$. Define the constant sequence \begin{align*} u_k:=u\qquad\text{for every }k\in\mathbb N. \end{align*} Then $(u_k)_{k=1}^{\infty}\subset\mathcal A$ and $u_k\rightharpoonup u$ in $X$, because every norm-convergent constant sequence is weakly convergent to the same point. This sequence is admissible in the defining infimum for $\overline I[u]$, hence \begin{align*} \overline I[u]\le \liminf_{k\to\infty} I[u_k]=I[u]. \end{align*} Taking the infimum over $u\in\mathcal A$ gives \begin{align*} \inf_{u\in\mathcal A}\overline I[u]\le \inf_{u\in\mathcal A}I[u]=\alpha. \end{align*}[/step]

[guided]Fix an arbitrary point $u\in\mathcal A$. The relaxation $\overline I[u]$ is defined by minimizing over all sequences in $\mathcal A$ that converge weakly to $u$. The simplest such sequence is the constant sequence \begin{align*} u_k:=u\qquad\text{for every }k\in\mathbb N. \end{align*} This sequence lies in $\mathcal A$ because $u\in\mathcal A$. It converges weakly to $u$ in $X$ because, for every bounded linear functional $\ell\in X^*$, one has \begin{align*} \ell(u_k)=\ell(u)\qquad\text{for every }k\in\mathbb N. \end{align*} Thus the constant sequence is one of the competitors in the infimum defining $\overline I[u]$. Therefore \begin{align*} \overline I[u]\le \liminf_{k\to\infty} I[u_k]. \end{align*} Since $I[u_k]=I[u]$ for every $k$, the right-hand side equals $I[u]$, and so \begin{align*} \overline I[u]\le I[u]. \end{align*} Because this inequality holds for every $u\in\mathcal A$, taking infima over $\mathcal A$ yields \begin{align*} \inf_{u\in\mathcal A}\overline I[u]\le \inf_{u\in\mathcal A}I[u]=\alpha. \end{align*}[/guided]

[step:Bound the relaxed functional from below by the original infimum] Let $u\in\mathcal A$, and let $(u_k)_{k=1}^{\infty}\subset\mathcal A$ be any sequence such that $u_k\rightharpoonup u$ in $X$. Since $\alpha=\inf_{v\in\mathcal A} I[v]$, we have \begin{align*} I[u_k]\ge \alpha \end{align*} for every $k\in\mathbb N$. Hence \begin{align*} \liminf_{k\to\infty} I[u_k]\ge \alpha. \end{align*} Taking the infimum over all admissible weak recovery sequences for $u$ gives \begin{align*} \overline I[u]\ge \alpha. \end{align*} Since $u\in\mathcal A$ was arbitrary, \begin{align*} \inf_{u\in\mathcal A}\overline I[u]\ge \alpha. \end{align*} Together with the previous step, \begin{align*} \inf_{u\in\mathcal A}\overline I[u]=\alpha=\inf_{u\in\mathcal A}I[u]. \end{align*} [/step]

[step:Choose a bounded minimizing sequence for the relaxed functional]By the equality of infima just proved, define the common level \begin{align*} m:=\inf_{u\in\mathcal A}\overline I[u]=\alpha\in\mathbb R. \end{align*} By the definition of infimum, there exists a sequence $(v_k)_{k=1}^{\infty}\subset\mathcal A$ such that \begin{align*} \overline I[v_k]\to m \end{align*} as $k\to\infty$. Thus $(v_k)_{k=1}^{\infty}$ is a minimizing sequence for $\overline I$ on $\mathcal A$. By hypothesis, this sequence is bounded in $X$.[/step]

[guided]The previous step proved that the relaxed infimum is the real number $m=\alpha$. Since $m$ is an infimum over the nonempty set $\mathcal A$, for each $k\in\mathbb N$ there exists $v_k\in\mathcal A$ such that \begin{align*} \overline I[v_k]\le m+\frac{1}{k}. \end{align*} The opposite inequality $m\le \overline I[v_k]$ holds because $m$ is the infimum of $\overline I$ over $\mathcal A$. Hence \begin{align*} m\le \overline I[v_k]\le m+\frac{1}{k} \end{align*} for every $k\in\mathbb N$, and the [squeeze theorem](/theorems/627) gives \begin{align*} \overline I[v_k]\to m \end{align*} as $k\to\infty$. Therefore $(v_k)_{k=1}^{\infty}$ is a minimizing sequence for $\overline I$ on $\mathcal A$. The boundedness assumption in the theorem applies exactly to such sequences, so $(v_k)_{k=1}^{\infty}$ is bounded in $X$.[/guided]

[step:Extract a weakly convergent subsequence and keep its limit admissible]Since $X$ is reflexive and $(v_k)_{k=1}^{\infty}$ is bounded in $X$, the weak [sequential compactness](/page/Sequential%20Compactness) theorem for bounded sequences in reflexive Banach spaces gives a subsequence $(v_{k_j})_{j=1}^{\infty}$ and a point $u_*\in X$ such that \begin{align*} v_{k_j}\rightharpoonup u_*\qquad\text{in }X. \end{align*} Because each $v_{k_j}$ belongs to $\mathcal A$ and $\mathcal A$ is sequentially weakly closed, the weak limit also belongs to $\mathcal A$: \begin{align*} u_*\in\mathcal A. \end{align*}[/step]

[guided]The boundedness assumption is used exactly here. We have a minimizing sequence $(v_k)_{k=1}^{\infty}$ in the reflexive [Banach space](/page/Banach%20Space) $X$, and the hypothesis says that this sequence is bounded in the norm $\|\cdot\|_X$. The weak sequential [compactness theorem](/theorems/2748) for bounded sequences in reflexive Banach spaces states that every bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Applying that theorem to $(v_k)_{k=1}^{\infty}$, we obtain a subsequence $(v_{k_j})_{j=1}^{\infty}$ and a point $u_*\in X$ such that \begin{align*} v_{k_j}\rightharpoonup u_*\qquad\text{in }X. \end{align*} This gives a candidate minimizer in the ambient space $X$, but we still need admissibility. The admissible set $\mathcal A$ is sequentially weakly closed, meaning that whenever a sequence in $\mathcal A$ converges weakly in $X$, its weak limit remains in $\mathcal A$. Since $v_{k_j}\in\mathcal A$ for every $j\in\mathbb N$ and $v_{k_j}\rightharpoonup u_*$ in $X$, sequential weak closedness gives \begin{align*} u_*\in\mathcal A. \end{align*} Thus the weak compactness step produces not merely a limit in $X$, but an admissible limit.[/guided]

[step:Apply sequential weak lower semicontinuity to identify the minimizer]Since $\overline I$ is sequentially weakly lower semicontinuous on $\mathcal A$ and $v_{k_j}\rightharpoonup u_*$ in $X$ with $v_{k_j},u_*\in\mathcal A$, we have \begin{align*} \overline I[u_*]\le \liminf_{j\to\infty}\overline I[v_{k_j}]. \end{align*} Because $(v_k)_{k=1}^{\infty}$ is a minimizing sequence with $\overline I[v_k]\to m$, every subsequence has the same limit, so \begin{align*} \liminf_{j\to\infty}\overline I[v_{k_j}]=m. \end{align*} Therefore \begin{align*} \overline I[u_*]\le m. \end{align*} On the other hand, $m=\inf_{u\in\mathcal A}\overline I[u]$ and $u_*\in\mathcal A$, so \begin{align*} m\le \overline I[u_*]. \end{align*} Combining the two inequalities gives \begin{align*} \overline I[u_*]=m. \end{align*} Hence $\overline I$ attains its minimum on $\mathcal A$, and using the equality of infima from above, \begin{align*} \min_{u\in\mathcal A}\overline I[u]=\overline I[u_*]=m=\inf_{u\in\mathcal A}I[u]. \end{align*} This proves both asserted conclusions.[/step]

[guided]The subsequence $(v_{k_j})_{j=1}^{\infty}$ still minimizes $\overline I$ because it is taken from a sequence satisfying $\overline I[v_k]\to m$. Since $v_{k_j}\rightharpoonup u_*$ in $X$ and both $v_{k_j}$ and $u_*$ belong to $\mathcal A$, the sequential weak lower semicontinuity hypothesis applies and gives \begin{align*} \overline I[u_*]\le \liminf_{j\to\infty}\overline I[v_{k_j}]. \end{align*} Because convergence of a sequence implies convergence of every subsequence to the same limit, we have \begin{align*} \liminf_{j\to\infty}\overline I[v_{k_j}]=m. \end{align*} Thus \begin{align*} \overline I[u_*]\le m. \end{align*} The reverse inequality follows from the definition of $m$ as the infimum of $\overline I$ on $\mathcal A$: since $u_*\in\mathcal A$, \begin{align*} m\le \overline I[u_*]. \end{align*} Combining the two inequalities yields \begin{align*} \overline I[u_*]=m. \end{align*} Therefore $u_*$ is a minimizer of $\overline I$ on $\mathcal A$. Finally, the earlier comparison of infima identified $m$ with $\inf_{u\in\mathcal A}I[u]$, so \begin{align*} \min_{u\in\mathcal A}\overline I[u]=\overline I[u_*]=m=\inf_{u\in\mathcal A}I[u]. \end{align*} This proves both the existence of a relaxed minimizer and the equality of the relaxed and original infima.[/guided]