Let $I:=(-1,1)$ and function $u$ is defined on $I$. Assume function $u$ is continuous on $(-1,0)$ and $(0,1)$, and we define $u^-(0)=\lim_{x\to 0^-}u(x)$ and $u^+(0)=\lim_{x\to 0^+}u(x)$. We also assume that $u(0^-)\leq u(0)\leq u(0^+)$.
Now given two sequences $\{x_n\}\subset I$ and $\{y_n\}\subset I$ such that for each $n\in \mathbb N$, $-1<x_n<y_n<1$, and $\lim_{n\to\infty} x_n=\lim_{n\to\infty} y_n=0$. I wish to prove that $$ \liminf_{n\to\infty}\,\inf_{x\in(x_n,y_n)}u(x)\geq u^-(0). $$ Here $(x_n,y_n)$ denotes an interval with endpoints $x_n$ and $y_n$. Note that we don't have $x_n<0<y_n$ or $x_n<y_n<0$. To prove, I tried with contradiction but it does not work... Any help is really welcome!
If $u^-(0)$ is minus infinity then also the left side of the inequality is minus infinity. So we may assume that $u^-(0)$ is finite. Fix $\epsilon >0$. If we can prove that for all $k\geq n$ and n sufficiently large we have $u(x)\geq u^-(0)-\epsilon$ for all $x\in (x_k,y_k)$ then $\liminf_k \inf_{x\in (x_k,y_k)} u(x) \geq u^-(0)-\epsilon$.
Choose $\delta_1 >0$ such that for all $x\in (-\delta_1,0)$ we have $|u(x)-u^-(0)|<\epsilon$. Similarly choose $\delta_2>0$ such that for all $x\in (0,\delta_2)$ we have $|u(x)-u^+(x)|<\epsilon$. Set $\delta =\min \{\delta_1, \delta_2\}$. Then for all $x\in B_\delta(0)$ we have $u(x)\geq u^-(0)-\epsilon$. Note that for $n$ sufficiently large we have $(x_k,y_k)\subset B_{\delta}(0)$ for all $k\geq n$, so the claim follows.