so that for any $\nu \ge \nu_{\kappa, \delta}$, $A_\nu := [-\frac{1}{\nu}, \frac{1}{\nu}]$ satisfies, $$\frac{|A_\nu|}{|\Omega|} \le  \delta\, , \qquad
\frac{\int_{A_\nu} v_\nu}{\| v_\nu \|_1} \ge \kappa.$$

Since the potential $V_\nu$ is even, so is the torsion function $v_\nu$ and hence it suffices to determine $v_\nu$ on $[0, 1]$. On the interval $[ 0,\frac{1}{\nu}]$, the function $v_\nu$ is of the form $$v_{\nu, 1}(x) := - \frac{1}{2}x^2 + \gamma_\nu\, ,
\quad 0 \le x \le \frac{1}{\nu}\, ,$$ whereas on the interval $[\frac{1}{\nu}, 1]$, we make the following Ansatz $$v_{\nu, 2}(x) := \frac{1}{\nu^4}
- \alpha_\nu e^{\nu^2 x}
+ \beta_\nu e^{-\nu^2 x}\, ,
\quad \frac{1}{\nu}  \le x \le 1\, .$$ It is straightforward to verify that $(-\Delta + V_{\nu}) v_{\nu, 1} = 1$ on $[0, \frac{1}{\nu}]$ and $(-\Delta + V_{\nu}) v_{\nu, 2} = 1$ on $[\frac{1}{\nu}, 1]$. The constants $\gamma_\nu,$ $\alpha_\nu$, and $\beta_\nu$ are determined by the boundary condition $v_{\nu, 2}(1) = 0$ and the matching conditions $$v_{\nu, 1}(\nu^{-1}) = v_{\nu, 2}(\nu^{-1})\, ,\quad
v_{\nu, 1}'(\nu^{-1}) = v_{\nu, 2}'(\nu^{-1})\, ,$$ where $'$ denotes the derivative with respect to the variable $x$. The boundary condition $v_{\nu, 2}(1) = 0$, or explicitly, $$\frac{1}{\nu^4} - \alpha_\nu e^{\nu^2} + \beta_\nu e^{-\nu^2} =0\, ,$$ yields $$\label{first formula beta nu}
\beta_\nu = \alpha_\nu 