This problem needs more details but I can't write everything here. Please close this problem!
Let $(v_\epsilon)\subset W^{1,2}(I)$ where $\epsilon>0$ and $v_\epsilon\to 1$ in $L^1$ and $0\leq v_\epsilon\leq 1$ for all $\epsilon>0$.
My question: Does there exist $a$, $b>0$ such that if $$ \limsup_{\epsilon\to 0}\frac{1}{\epsilon^a}\int_I (1-v_\epsilon)^bdx<+\infty, $$ then we have $$ \limsup_{\epsilon\to 0}\frac{1}{(2t_\epsilon)^2}\int_{-t_\epsilon}^{t_\epsilon}(v_\epsilon(s))^2ds=0 $$ where $\lim_{\epsilon\to 0}t_\epsilon=0$, $\lim_{\epsilon\to 0}v(t_\epsilon)=\lim_{\epsilon\to 0}v(-t_\epsilon)=1$, and there exists $-t_\epsilon<s_\epsilon<t_\epsilon$ such that $v_\epsilon(s_\epsilon)\to 0$.
Take $v_\epsilon \equiv 1$. Then, your assumption is satisfied for all $a,b > 0$.
But $$\frac1{(2t(\epsilon))^2} \int_{-t(\epsilon)}^{t(\epsilon)}1 \mathrm{d}s = \frac1{2t(\epsilon)} \to \infty$$ as $\epsilon \to 0$.
After the question has completely changed, still a similar construction is possible: $$v_\epsilon(t) = \max\{1, n_\epsilon \, |t|\}$$ with carefully chosen $n_\epsilon$ (depending on $a$ and $b$).