& \mbox{if } \frac{1}{K}+\eta\leq t\leq  1-\frac{1}{K}-\eta\\
   \varphi_1^i(f_{ss'}^j)\dfrac{\eta-dist(t,[1-\frac{1}{K},1]_i)}{\eta}, & \mbox{if } 1-\frac{1}{K}-\eta\leq t\leq 1
   \end{cases}$$ Obviously, we have $$\phi_x'(g_{ss'}^j)=\phi_x(g_{ss'}^j).$$ holds for each $x\in [0,1]\subset Sp(B)$ and $1\leq j\leq p,1\leq s,s'\leq k_j$.

Define $$\overline{f}_{ss'}^j(x):=E_{\widetilde{W_j}}(x)\cdot\phi_x'(g_{ss'}^j)\cdot E_{\widetilde{W_j}}(x).$$ We also have $$\overline{f}_{s_1s_1'}^j(x)\cdot\overline{f}_{s_2s_2'}^j(x)=\delta_{s_1's_2}\cdot\overline{f}_{s_1s_2'}^j(x)$$ for all $1\leq s_1,s_1',s_2,s_2'\leq k_j$.

Let $h_j$ be the test function corresponding to $\widetilde{V_j}$ and $X_j$. In a similar way, from the pairing results, for any $x_1,x_2,x_3,x_4\in [0,1]$, we have $$\begin{aligned}
      \#_{X_j}\big(Sp\phi_{x_1}\cap \widetilde{V_j}\big) &\leq& \#(\Lambda_1({h_j})) \\
   &\leq &\#_{X_j}\big(Sp\phi_{x_2}{\cap} B(\widetilde{V_j},\frac{1}{3}\eta)\big) \\
   &\leq& \#(\Lambda_1({h_j})\cup\Lambda_2({h_j})) \\
   &\leq& \#_{X_j}\big(Sp\phi_{x_3}{\cap} B(\widetilde{V_j},\frac{2}{3}\eta)\big) \\
   &\leq& \#\big(\Lambda_1({h_j})\cup\Lambda_2({h_j})\cup\Lambda_3({h_j})\big) \\
   &\leq& \#_{X_j}\big(Sp\phi_{x_4}\cap B(\widetilde{V_j},\eta)\big).
   \end{aligned}$$

We still consider the following two cases:

Case 1.$\quad$ $\Lambda_2({h_j})=\Lambda_3({h_j})=\varnothing$

This case is just similar to case 1 in step 2, we have $$\#_{X_j}\big(Sp\phi_x\cap B(\widetilde{V_j}, \frac{1}{3}\eta)\big)=\#_{X_j}\big(Sp\phi_x\cap B(\widetilde{V_j}, \frac{2}{3}\eta)\big).$$ Therefore, $$Sp\phi_x\cap B(\widetilde{V_j},\frac{2}{3}\eta)\backslash Sp\phi_x\cap B(\widetilde{V_j},\frac{1}{3}\eta)=\varnothing.$$ Recall that $$\widetilde{V_j}=\bigcup_{\alpha_{ij}\neq0}V_0^i\cup\bigcup_{\beta_{ij}\neq0}V_K^i$$ Then, if $i$ satisfies $\alpha_{ij}\neq 0$, we will have $ Sp\phi_x{\cap}(B(V_0^i,\frac{2}{3}\eta)\backslash B(V_0^i,\frac{1}{3}\eta))=\varnothing,$ and if $i$ satisfies $\beta_{ij}\neq 0$, we will have $Sp\phi_x{\cap}\big(B(V_K^i,\frac{2}{3}\eta)\backslash B(V_K^i,\frac{1}{3}\eta)\big)=\varnothing.$

Define $\chi_j$ as follows: let $a=(a(\theta_1),a(\theta_2),\cdots,a(\theta_p))\in F_1$, where $a(\theta_j)=I_{k_j}$ and $a(\theta_s)=0_{k_s}$, if $s\neq j$. Set $$\chi_j(t)=(\chi_j(t,1),\chi_j(t,2),\cdots,\chi_j(t,l)),$$ where $$\chi_j(t,i)=
   \begin{cases}
    {\rm sgn}(\alpha_{ij})\cdot\varphi_0^i(a),\,\, & \mbox{if }\,\,t\in \overline{ B(V_0^i,\frac{1}{3}\eta)},  \\
    linear, & \mbox{if }\,\, t\in \overline{B(V_0^i,\frac{2}{3}\eta)}\backslash B(V_0^i,\frac{1}{3}\eta),\\
    0_{l_i}, & \mbox{if }\,\, t\in [0,1]_i\backslash\big(B(V_0^i,\frac{2}{3}\eta){\cup} B(V_K^i,\frac{2}{3}\eta)\big), \\
    linear, & \mbox{if }\,\, t\in \overline{ B(V_K^i,\frac{2}{3}\eta)}\backslash B(V_K^i,\frac{1}{3}\eta),\\
    