## 3 Main Results
Since \(\widehat{\phi}_{i}(k)\) and \(\phi_{i}(k)\) are bounded, we choose \(\lambda>\lambda_{min}\) (\(\lambda_{min}>b^{2}/4\)) to satisfy the following inequality:  
\[0<\frac{\phi_{i}(k)\widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2} }\leq\frac{b|\widehat{\phi}_{i}(k)|}{2\sqrt{\lambda}|\widehat{\phi}_{i}(k)|}< \frac{b}{2\sqrt{\lambda_{min}}}<1 \tag{20}\]  
which means \(0<\gamma_{ii}(k)<1\) when \(0<\rho_{i}<\frac{1}{d_{i}}\) and \(\lambda>\lambda_{min}\). By definition of the Laplacian matrix, one has \(\mathcal{L}_{1}\mathbf{1}_{N}+\mathcal{L}_{2}\mathbf{1}_{M}=0\). Note the fact that \(\boldsymbol{\Gamma}(k)\geq 0\) and \(\boldsymbol{\Gamma}(k)\mathbf{1}_{N+M}=\mathbf{1}_{N+M}\), i.e., \(\boldsymbol{\Gamma}(k)\) is a stochastic matrix. _Step 3_ Prove the convergence of the global containment error. Define the upper bounds of the followers, leaders, and the MASs as follows:  
\[y^{+}(k)=\max_{i}y_{i}(k)\] \[w^{+}=\max_{l}w_{l}\] \[\Omega^{+}(k)=\max\{y^{+}(k),w^{+}\}\]  
Note that one can then define \(y^{-}(k)=\min_{i}y_{i}(k)\), \(w^{-}=\min_{l}w_{l}\), and \(\Omega^{-}(k)=\min\{y^{-}(k),w^{-}\}\) for the lower bound. For the upper bound, \(\Omega^{+}(k)=w^{+}\) means all the followers goes into the convex hull of the leaders. Therefore, define the upper global containment error as follows:  
\[e^{+}(k)=\Omega^{+}(k)-w^{+}\]  
(Define \(e^{-}(k)=w^{-}-\Omega^{-}(k)\) for the lower global containment error). Then, one considers the following Lyapunov function \(V_{1}(k)=e^{+}(k)^{2}\) for further analysis. The time difference can be written as  
\[\Delta V_{1}(k+1) =V_{1}(k+1)-V_{1}(k)\] \[=(e^{+}(k+1)-e^{+}(k))(e^{+}(k+1)+e^{+}(k))\] \[=\big{(}\Omega^{+}(k+1)-\Omega^{+}(k)\big{)}\big{(}\Omega^{+}(k+1 )+\Omega^{+}(k)-2w^{+}\big{)} \tag{21}\]  
Rewrite \(y_{i}(k+1)\) as  
\[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]0 \[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]1  
where \(\gamma_{ij}(k)\) is the entries of \(\Gamma(k)\). According to Lemma 2, two parts need to be proved: (1) \(e^{+}(k)=0\) is an equilibrium point; (2) For any \(e^{+}(k)\
eq 0\), \(\Delta V_{1}(k)\leq 0\) and \(\Delta V_{1}(k)\
ot\equiv 0\). (1) \(e^{+}(k)=0\), i.e., \(y^{+}(k)\leq w^{+}\). Substituting \(y_{j}(k)\leq w^{+}\) and \(w_{l}\leq w^{+}\) into (22), one derives  
\[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]2  
Therefore, \(\Omega^{+}(k+1)\!=\!\max\{y^{+}(k+1),w^{+}\}=w^{+}=\Omega^{+}(k)\), i.e., \(\Delta V_{1}\ (k+1)=0\) in this case. In other words, the \(e^{+}(k)=0\) is the equilibrium point. (2) \(e^{+}(k)\
eq 0\), i.e., \(y^{+}(k)>w^{+}\). Denote \(s_{i}(k)=\sum_{j:y_{j}(k)=y^{+}(k)}\gamma_{ij}(k)\). For the \(i\)th follower, \(s_{i}(k)=1\) means the outputs of the \(i\)th follower and its neighbors are all equal to \(y^{+}(k)\), which also means only the first term in the right side of (22) exists.