Are the translates of a function in $H^1(\mathbb{R}^2)$ bounded by a function with finite integral over some compact set? Suppose that $u\in H^1(\mathbb{R}^2)$ and let $u_n(\cdot,\cdot)=u(\cdot,\cdot+n)$. Let $E\subset\mathbb{R}^2$ be compact. Can I find a non-negative function $f\in L^1(\mathbb{R}^2)$ such that $|u_n|\leq f$ on $E$ for all $n\in\mathbb{N}$?
My Try: In the case of $u\in H^1(\mathbb{R}^2)\cap L^\infty(\mathbb{R}^2)$, one can choose $f\in L^1(\mathbb{R}^2)$ to be
$$f=\begin{cases}\|u\|_{\infty} & \text{ on E},\\
0 & \text{ on } E^c.\end{cases} $$
However, $H^1\not\subset L^\infty$ so this does not work in that case. Is this result even true for $u\in H^1(\mathbb{R}^2)$?
Up the ante. In case the above is false, can one find a sequence $\{a_n\}$ in $\mathbb{R}^2$ with $a_n\to\infty$ and a function $f\in L^1(\mathbb{R}^2)$ such that $|u_n(\cdot)|:=|u(\cdot+a_n)|\leq f$ on $E$ for all $n\in\mathbb{N}$?
 A: I think the claim is true if $E \subset \mathbb R \times (-1/2,+1/2)$.
Define $v_n:E \to \mathbb R$ by
$$
v_n(x) = \max(|u_n(x)|, v_{n-1}(x)),
$$
where $v_0=0$. This implies $\nabla v_n(x) \in \{ \pm \nabla |u_n(x)|, \nabla v_{n-1}(x)\}$.
Then there are disjoint subsets $E_1\dots E_n\subset E$ such that
$$
v_n = \sum_{i=1}^n \chi_{E_i} |u_i|,
$$
which implies
$$
|\nabla v_n| = \sum_{i=1}^n \chi_{E_i} |\nabla u_i|.
$$
Due to the assumption on $E$, the sets $E_i - (0,i)$ are disjoint, so
$$
\|v_n\|_{L^2(E)}^2 = \sum_i \int_{E_i} |u_i|^2 = \sum_i \int_{E_i-(0,i)}|u|^2 
= \int_{\cup_i (E_i-(0,i))}|u|^2 
\le \|u\|_{L^2}^2
$$
and likewise $\|\nabla v_n\|_{L^2(E)}^2 \le \|\nabla u\|_{L^2}^2$.
So $(v_n)$ is bounded in $H^1(E)$, hence admits a weakly converging subsequence $v_{n_k} \rightharpoonup v$ in $H^1(E)$.
Clearly $|u_n| \le v_m$ for all $n<m$, which carries over to $|u_n|\le v$.
And the claim is true with $f:=\chi_E v$.
This proof also shows that the weaker statement of the second question is true: choose $a_n:=n\cdot w$, where $w$ is such that $E\subset \mathbb R \times(-w/2,+w/2)$.
