2\delta \int_{B_\delta(x)}| u'_{\varepsilon}|^2\dx\leq (2\delta)(2{\varepsilon}) 4{\left}(\frac{1}{2{\varepsilon}}\int_{B_\delta(x)}\phi_{\varepsilon}|^2 u'_{\varepsilon}|^2\dx{\right})\, \leq\, 16c_0{\varepsilon}\delta \,\stackrel{{\varepsilon}{\downarrow}0}\longrightarrow\,0.$$ By lower semicontinuity of the total variation on open sets we conclude that $|Du|(B_\delta(x))=0$. This establishes $u\in \P(O)$ with $J_u\subset \{m_1,\cdots,m_N\}$.

*Step 2.2. Proof of the lower bound.* Without loss of generality, we prove the lower bound in the case $J_u=\{0\}$ and $ D:=u(0^+)=-u(0^-)>0$. Using the same argument as in [@Iur Pag. 7] for any $0<d<D$ there exist six points $y_1<x^1_{\varepsilon}\leq\tilde{x}^1_{\varepsilon}<\tilde{x}^2_{\varepsilon}\leq x^2_{\varepsilon}<y_2$ such that $$\begin{gathered}
\lim_{{\varepsilon}{\rightarrow}0}\phi_{\varepsilon}(y_1)= \lim_{{\varepsilon}{\rightarrow}0} \phi_{\varepsilon}(y_2) = 1,\\
\lim_{{\varepsilon}{\rightarrow}0}\phi_{\varepsilon}(x^1_{\varepsilon})= \lim_{{\varepsilon}{\rightarrow}0} \phi_{\varepsilon}(x^2_{\varepsilon}) =0,\\
u_{\varepsilon}(\tilde{x}^1_{\varepsilon})= -D+d,\quad\quad \; u_{\varepsilon}(\tilde{x}^2_{\varepsilon})= D-d.\end{gathered}$$ Using the Modica-Mortola trick in the intervals $(y_1,x^1_{\varepsilon})$ and $(x^2_{\varepsilon},y_2)$ as above, we compute: $$\label{dis1:liminf} 
\liminf_{{\varepsilon}{\downarrow}0}\LF_{\varepsilon}(u_{\varepsilon},\phi_{\varepsilon};(y_1,x^1_{\varepsilon})\cup(x^2_{\varepsilon},y_2))\geq\liminf_{{\varepsilon}{\downarrow}0}\int^{x^1_{\varepsilon}}_{y_1}(1-\phi_{\varepsilon})|\phi'_{\varepsilon}| \dx+\int_{x^2_{\varepsilon}}^{y_2}(1-\phi_{\varepsilon})|\phi'_{\varepsilon}| \dx\,\geq\,1.$$ For the estimation on the interval $I_{\varepsilon}=(\txk^1,\txk^2)$ let us introduce: $$\begin{gathered}
G_{\varepsilon}:= {\left}\{ w\in H^1(I_{\varepsilon}) : w(\txk^1)=-D+d,\, w(\txk^2)=D-d {\right}\},\\
Z_{\varepsilon}:= {\left}\{ z\in H^1(I_{\varepsilon}) :\eta\leq z\leq1 \text{ a.e. on } I_{\varepsilon}{\right}\},\\
H_{\varepsilon}(w,z):=\int_{I_{\varepsilon}}{\left}(\frac{1}{2{\varepsilon}}z^2|w'|^2+\frac{(1-z)^2}{2{\varepsilon}}{\right})\dx,\\
h_{\varepsilon}(z)= \inf_{w\in W_{\varepsilon}}H_{\varepsilon}(w,z) \text{ for } z\in Z_{\varepsilon}.\end{gathered}$$ Note that by Cauchy-Schwarz inequality, we have for $w\in G_{\varepsilon}$ and $z\in Z_{\varepsilon}$, $$\int_{I_{\varepsilon}}z^2|w'|^2\, \geq \, \left(\int_{I_{\varepsilon}}|w'|\dx\right)^2{\left}(\int_{I_{\varepsilon}}\frac{1}{z^2}{\right})^{-1}\, \geq\, 4(D-d)^2{\left}(\int_{I_{\varepsilon}}\frac{1}{z^2}{\right})^{-1}.$$ We deduce the lower bound $$\label{eq:minimo}
h_{\varepsilon}(z) \, \geq  \, 4(D-d)^2 {\left}(2{\varepsilon}\int_{I_{\varepsilon}}\frac{1}{z^2}\dx{\right})^{-1}+ \int_{I_{\varepsilon}}{\left}(\frac{(1-z)^2}{2{\varepsilon}}{\right})\dx.$$ Let us remark that optimizing $H_{\varepsilon}(w,z)$ with respect to $w\in G_{\varepsilon}$ we see that this inequality is actually an equality.\
Consider for $0 