Let $f:\mathbb R \rightarrow [0,+\infty)\;$ be a smooth function that vanishes on a finite set $A$ where $\vert A \vert \ge 2\;$. If $a_1\;,a_2\;\in A\;$ consider:
$\mathcal M= \{\;v\in W^{1,2}_{loc} (l^{-},l^{+});\;-\infty \le l^{-} \lt l^{+} \le +\infty\;,\;\lim_{x \to l^{-}} v(x)=a_1\;,\;\lim_{x \to l^{+}} v(x)=a_2 \neq a_1\;\;,\;v((l^{-},l^{+}))\subseteq \mathbb R \setminus A \}$
It also holds $\int_{l^{-}}^{l^{+}} f(v(x))\;dx\lt \infty\;$
Question: Assume a sequence $\{v_n\}\;\subset \mathcal M$ for which $\;\limsup_{\vert v_n \vert \to \infty} f(v_n) \gt 0\;$. Is this sufficient in order to conclude $\;\{v_n\}\;$ is a bounded sequence? It holds $\int_{l^{-}}^{l^{+}} f(v_n(x))\;dx\lt \infty\;$ but somehow this is not enough for the sequence to be bounded.
If it's true, I would appreciate if somebody could explain why...and if it's not I would like to know by which assumption(s) on $f$, $\;\{v_n\}\;$ is bounded.
Unfortunately, I can't come up with any good idea. Any help would be valuable.
Thanks in advance