[proofplan] We first use reflexivity to extract a subsequence that converges weakly in $X$. Sequential weak closedness of $\mathcal A$ then places the weak limit back in the admissible class. Compactness of the embedding gives strong convergence in $Y$ along a further subsequence, and continuity of the embedding shows that this same subsequence also converges weakly in $Y$ to the embedded weak limit. Finally, uniqueness of weak limits in the [Banach space](/page/Banach%20Space) $Y$ identifies the strong $Y$-limit with the embedded $X$-limit. [/proofplan]

[step:Extract a weakly convergent subsequence in $X$] Since $(u_k)_{k=1}^{\infty}$ is bounded in the reflexive Banach space $X$, the [weak sequential compactness](/theorems/214) of bounded sets in reflexive Banach spaces, equivalently the [Eberlein-Smulian theorem](/theorems/987), gives a point $u\in X$ and a strictly increasing map \begin{align*} k:\mathbb N\to\mathbb N \end{align*} such that the subsequence $(u_{k_j})_{j=1}^{\infty}$ satisfies \begin{align*} u_{k_j}\rightharpoonup u \quad \text{in } X. \end{align*} Here $k_j$ denotes $k(j)$. The cited compactness input is: bounded sequences in reflexive Banach spaces admit weakly convergent subsequences. [/step]

[step:Use sequential weak closedness to keep the limit admissible] For every $j\in\mathbb N$, $u_{k_j}\in\mathcal A$ because the original sequence takes values in $\mathcal A$. Since $\mathcal A$ is sequentially weakly closed in $X$ and \begin{align*} u_{k_j}\rightharpoonup u \quad \text{in } X, \end{align*} the definition of sequential weak closedness gives \begin{align*} u\in\mathcal A. \end{align*} [/step]

[step:Apply compactness of the embedding to obtain strong convergence in $Y$] The subsequence $(u_{k_j})_{j=1}^{\infty}$ is bounded in $X$, because it is a subsequence of the bounded sequence $(u_k)_{k=1}^{\infty}$. Since \begin{align*} j:X\to Y \end{align*} is compact, the sequence $(j(u_{k_j}))_{j=1}^{\infty}$ has a strongly convergent subsequence in $Y$. Thus there exist a point $v\in Y$ and a strictly increasing map \begin{align*} \ell:\mathbb N\to\mathbb N \end{align*} such that \begin{align*} j(u_{k_{\ell_r}})\to v \quad \text{in } Y. \end{align*} Here $\ell_r$ denotes $\ell(r)$. [/step]

[step:Identify the strong $Y$-limit with the embedded weak $X$-limit]We prove that $v=j(u)$. Let $\varphi\in Y^*$ be arbitrary. Since $j:X\to Y$ is continuous and linear, the composition \begin{align*} \varphi\circ j:X\to\mathbb R \end{align*} is a bounded linear functional on $X$. The [weak convergence](/page/Weak%20Convergence) $u_{k_{\ell_r}}\rightharpoonup u$ in $X$ therefore gives \begin{align*} \varphi(j(u_{k_{\ell_r}}))\to \varphi(j(u)). \end{align*} On the other hand, the strong convergence $j(u_{k_{\ell_r}})\to v$ in $Y$ implies weak convergence in $Y$, because \begin{align*} |\varphi(j(u_{k_{\ell_r}}))-\varphi(v)|\le \|\varphi\|_{Y^*}\|j(u_{k_{\ell_r}})-v\|_Y \end{align*} and the right-hand side tends to $0$. Hence \begin{align*} \varphi(j(u))=\varphi(v) \end{align*} for every $\varphi\in Y^*$. By uniqueness of weak limits in Banach spaces, equivalently by separation of points by bounded linear functionals, we conclude that \begin{align*} v=j(u). \end{align*}[/step]

[guided]The compact embedding gives a strong limit in $Y$, but at first it only gives a point $v\in Y$. We must show that this point is the same as the image of the weak $X$-limit $u$ under the embedding map $j:X\to Y$. Let $\varphi\in Y^*$ be an arbitrary bounded linear functional on $Y$. Since $j:X\to Y$ is continuous and linear, the composition \begin{align*} \varphi\circ j:X\to\mathbb R \end{align*} is a bounded linear functional on $X$. This is exactly the kind of test functional allowed in the definition of weak convergence in $X$. Because the further subsequence $(u_{k_{\ell_r}})_{r=1}^{\infty}$ is still a subsequence of the weakly convergent sequence $(u_{k_j})_{j=1}^{\infty}$, we have \begin{align*} u_{k_{\ell_r}}\rightharpoonup u \quad \text{in } X. \end{align*} Testing this weak convergence against $\varphi\circ j$ gives \begin{align*} \varphi(j(u_{k_{\ell_r}}))\to \varphi(j(u)). \end{align*} Now we use the strong convergence obtained from compactness. Since \begin{align*} j(u_{k_{\ell_r}})\to v \quad \text{in } Y, \end{align*} the boundedness of $\varphi$ gives \begin{align*} |\varphi(j(u_{k_{\ell_r}}))-\varphi(v)|\le \|\varphi\|_{Y^*}\|j(u_{k_{\ell_r}})-v\|_Y. \end{align*} The norm on the right tends to $0$, so \begin{align*} \varphi(j(u_{k_{\ell_r}}))\to \varphi(v). \end{align*} The same scalar sequence therefore has two limits: \begin{align*} \varphi(j(u_{k_{\ell_r}}))\to \varphi(j(u)) \end{align*} and \begin{align*} \varphi(j(u_{k_{\ell_r}}))\to \varphi(v). \end{align*} [Uniqueness of limits](/theorems/625) in $\mathbb R$ gives \begin{align*} \varphi(j(u))=\varphi(v). \end{align*} Because this equality holds for every $\varphi\in Y^*$, uniqueness of weak limits in Banach spaces, equivalently separation of points by bounded linear functionals, implies \begin{align*} v=j(u). \end{align*} This is the point where the strong compactness conclusion in $Y$ is synchronized with the weak compactness conclusion in $X$.[/guided]

[step:Relabel the final subsequence and record both convergences] Define a final strictly increasing index map \begin{align*} m:\mathbb N\to\mathbb N \end{align*} by \begin{align*} m_r=k_{\ell_r}. \end{align*} The sequence $(u_{m_r})_{r=1}^{\infty}$ is a subsequence of the original sequence. Since it is a subsequence of $(u_{k_j})_{j=1}^{\infty}$, it satisfies \begin{align*} u_{m_r}\rightharpoonup u \quad \text{in } X. \end{align*} Since $j(u_{m_r})=j(u_{k_{\ell_r}})$ and $v=j(u)$, it also satisfies \begin{align*} j(u_{m_r})\to j(u) \quad \text{in } Y. \end{align*} Under the standard identification of $X$ with its image under $j$, this is \begin{align*} u_{m_r}\to u \quad \text{in } Y. \end{align*} Renaming this subsequence as $(u_{k_j})_{j=1}^{\infty}$ gives the asserted subsequence with both convergences, and the limit $u$ belongs to $\mathcal A$. [/step]