How to show convergence of this sequence in $L^2?$ $\newcommand\normo[1]{\left\lVert#1\right\rVert_{H^1}}$
I am reading a paper where the author considers a sequence $\normo{u_n(0)} < C$ for some constant $C>0,$ where $u_n(t,x):\mathbb{R}\times \mathbb{R}^n\to\mathbb{C}$ and $H^1(\mathbb{R}^n)$ is the usual Sobolev Space. 
Then the author says that this implies that there exists a subsequence $u_{\phi(n)}(0)$ and $U_0\in H^1(\mathbb{R}^n)$ such that,
$$u_{\phi(n)}(0)\to U_0 \quad (*)$$
in $L^2_{\text{loc}}(\mathbb{R}^n)$ as $n\to +\infty.$ 
Furthermore, previously in the paper, it was shown that,
$$\forall \epsilon>0, \exists K_0,\text{ such that }\forall n\geq 1,\quad \int_{|x|>K_0}|u_n(0,x)|^2 < \epsilon$$
and so then the author says that we can conclude that,
$$u_{\phi(n)}(0)\to U_0 \quad (**)$$
in $L^2(\mathbb{R}^n)$ as $n\to +\infty.$ 
I do not understand this argument, in particular I do not know how being a bounded sequence in $H^1$ implies the existence of a subsequence that converges in $L^2_{\text{loc}}$ and how using the result in the paper we can derive that this subsequence converges in $L^2.$ Any references/hints/comments related to this will be much appreciated.
 A: The first part of your question is just a consequence of Rellich's Theorem, which gives you the compact embedding of $H^1(\Omega)\hookrightarrow L^2(\Omega)$ for bounded sets. Since this compact embedding holds for any bounded set you immediately get your first convergence.
For your second question, this is just an application of the following standard functional analysis lemma: If a bounded sequence $\{x_n\}\subset X$ weakly converges $x_n\rightharpoonup x\in X$ and $$
\limsup_{n\to+\infty}\Vert x_n\Vert_X\leq \Vert x\Vert_X,
$$ 
then $x_n\to x$ strongly in $X$.
So, with this lemma together with your estimate $$
\forall\varepsilon>0, \ \exists K_0>0, \ \forall n\geq 1, \ \ \int_{\vert x\vert\geq K_0}\vert u_n(0,x)\vert^2<\varepsilon.
$$
and the weakly convergence in $L^2(\mathbb{R})$ you can easily conclude. I let you the details. 
PS: Let me be a little more specific on the second part of your question: Fix any $\varepsilon>0$. By using your local $L^2$-strong convergence and the previous estimate prove that $$
\limsup_{n\to+\infty}\Vert u_n(0,\cdot)\Vert_{L^2}\leq \Vert U_0\Vert_{L^2}+\varepsilon.
$$
Since $\varepsilon>0$ was arbitrary, you can conclude by using the lemma above and the weak convergence. I hope my answer helps you. 
