Weak convergence in the Sobolev space and compact embeddedness

Question

Let

• $\Omega\subseteq\mathbb R^n$ be a bounded domain
• $H:=W_0^{1,2}(\Omega)$ be the Sopolev space
• $u\in C^0\left(\overline\Omega\times [0,\infty)\right)\cap C^{2,1}\left(\Omega\times (0,\infty)\right)\cap L^\infty(\Omega)$
• $w\in L^2(\Omega)$ and $(t_k)_{k\in\mathbb N}\subseteq (1,\infty)$ with $t_k\to\infty$ $$u(\;\cdot\;,t_k)\stackrel{L^2(\Omega)}{\to}w\tag{1}$$ for $k\to\infty$

Let's assume, that $\left(u\left(\;\cdot\;,t_k\right)\right)_{k\in\mathbb N}$ is bounded in $H$. I want to show, that $w\in H\cap L^\infty(\Omega)$.

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I'm reading a proof and they state the following:

1. Since $\left(u\left(\;\cdot\;,t_k\right)\right)_{k\in\mathbb N}$ is bounded in $H$ and $H$ is compactly embedded into $L^2(\Omega)$, there exists a subsequence $(k_{\mathcal l})_{\mathcal l\in\mathbb N}$ such that $$u\left(\;\cdot\;,t_{k_{\mathcal l}}\right)\stackrel H{\to}w\;\;\;\text{for }\mathcal l\to\infty\tag{2}$$
2. Moreover, $$u\left(\;\cdot\;,t_{k_{\mathcal l}}\right)\stackrel {\mathcal l\to\infty}{\to}w\;\;\;\text{almost everywhere in }\Omega\tag{3}$$
3. Thus, since $u\in L^\infty(\Omega)$, we've got $w\in H\cap L^\infty(\Omega)$.

What I don't understand is:

1. The compact embeddedness should only yield convergence of a subsequence in $L^2(\Omega)$ instead of $(2)$. Moreover, that would be no new information, since we've already got this by $(1)$.
2. Again, this immediately follows from $(1)$. So, I don't understand why they argue this way.
3. I don't understand why we can conclude that

Answer 1

1) also can be justified by that $H^1$ is a reflexive Banach space, so from the bounded sequence you can extract a weakly convergent subsequence with a limit in $H^1(\Omega)$ and because the embedding operator is compact that means that it maps weakly convergent sequence to strongly convergent, i.e $u(.,t_{k_l})\rightarrow w$ strongly in $L_2$ (here, we already now that $w\in H^1(\Omega)$ because $w$ is the unique limit of the weakly convergent subsequence). From this subsequence you can extract a convergent pointwise almost everywhere subequence (denote it again by $u(.,t_{k_l})$). Since $u\in L_\infty(\Omega)\Rightarrow \exists M: |u(x)|\leq M$ a.e in $\Omega$. Therefore $u(x,t_{k_l})\leq M \quad \forall \,k_l$ and the poinwise a.e limit function $w$ should be also bounded by the same constant $M$ a.e in $\Omega$.

Note that $u\in L_\infty(\Omega)$ because $u\in C^0(\overline\Omega)$ for each $t$.