If I define the approximation of the second derivative as $$\delta^2_xV_{i}=\dfrac{D^+_xV_{i}-D^-_xV_{i}}{(x_{i+1}-x_{i-1})/2}$$ where $$D^+_xV_{i}=\dfrac{V_{i+1}-V_i}{x_{i+1}-x_i}, D^-_xV_{i}=\dfrac{V_{i}-V_{i-1}}{x_i-x_{i-1}},$$
and define a function $$ \phi(x_i)=CN^{-2}+CN^{-2}\dfrac{\tau}{\sqrt{\epsilon}}\begin{cases} \dfrac{x_i}{\tau}& 0\leq x_i\leq\tau\\ 1&\tau\leq x_i\leq1-\tau\\ \dfrac{1-x_i}{\tau}&1-\tau\leq x_i\leq1. \end{cases}$$ where $\tau=\min\{1/4,2\sqrt{\epsilon}\ln N_x\}$. If I have an operator $L\psi=-\epsilon\delta^2\psi+b_i\psi$, with $b_i\geq\beta$.
How can I show that $$ L\phi(x_i)\geq \begin{cases} C\sqrt{\epsilon}N^{-1}+N^{-2}& ,x_i=\tau\text{ or }x_i=1-\tau\\ CN^{-2}&, \text{otherwise}. \end{cases}$$ I can see that $Lx_i=b_ix_i\geq\beta x_i$. So this is what I have done $$ L\phi(x_i)=L(CN^{-2})+CN^{-2}\dfrac{\tau}{\sqrt{\epsilon}}\begin{cases} \dfrac{L(x_i)}{\tau}& 0\leq x_i\leq\tau\\ L(1)&\tau\leq x_i\leq1-\tau\\ \dfrac{L(1-x_i)}{\tau}&1-\tau\leq x_i\leq1. \end{cases}$$ $$ =b_iCN^{-2}+CN^{-2}\dfrac{\tau}{\sqrt{\epsilon}}\begin{cases} \dfrac{b_i x_i}{\tau}& 0\leq x_i\leq\tau\\ b_i&\tau\leq x_i\leq1-\tau\\ \dfrac{b_i(1-x_i)}{\tau}&1-\tau\leq x_i\leq1. \end{cases}$$ $$ \geq \beta CN^{-2}+\beta CN^{-2}\dfrac{\tau}{\sqrt{\epsilon}}\begin{cases} \dfrac{x_i}{\tau}& 0\leq x_i\leq\tau\\ 1&\tau\leq x_i\leq1-\tau\\ \dfrac{1-x_i}{\tau}&1-\tau\leq x_i\leq1. \end{cases}$$ since $b_i\geq\beta$. How do I proceed from here