## 3 Main Results
Substituting \(y_{j}(k)=y^{+}(k)\) into (22) gives  
\[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]3Otherwise, if \(s_{i}(k)\
eq 1\), the following results can be concluded from Equation (22):  
\[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]4 \[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]5 \[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]6 \[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]7  
Combining (24) and (25) gives  
\[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]8  
It is clear that \(\Omega^{+}(k+1)+\Omega^{+}(k)-2w^{+}>0\). Therefore, \(\Delta V_{1}(k+1)\leq 0\) holds. Now, one only needs to prove that \(\Delta V_{1}(k+1)\
ot\equiv 0\). Assume that there exists a time instant \(K_{1}\) such that \(\Delta V_{1}(k+1)\equiv 0\) when \(k>K_{1}\). Define \(\mathcal{S}_{1}(h)=\{i|y_{i}(K_{1}+h)=y^{+}(K_{1})\}\) and \(N_{1}=\text{card}(\mathcal{S}_{1}(0))\). Since \(\Delta V_{1}(k+1)\equiv 0\), it means \(y^{+}(K_{1}+h)=y^{+}(K_{1})\) for all \(h>0\), i.e., \(\text{card}(\mathcal{S}_{1}(h))>0\) must hold for all \(h>0\). According to Assumption 4, at least one agent in \(\mathcal{S}_{1}(h)\) whose neighbor's output is less than \(y^{+}(K_{1})\), i.e., there exist a certain follower \(y_{p}(K_{1}+h)\) satisfying \(y_{p}(K_{1}+h)\in\mathcal{S}_{1}(h)\) and \(s_{p}(K_{1}+h)\
eq 1\). From (25), one derives \(y_{p}(K_{1}+h+1)<y^{+}(K_{1})\), which implies  
\[\gamma_{ii}(k)=1-d_{i}\cdot\rho_{i}\cdot\frac{\phi_{i}(k)\widehat{\phi}_{i}(k )}{\lambda+|\widehat{\phi}_{i}(k)|^{2}}=1-\frac{d_{i}\rho_{i}\phi_{i}(k) \widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2}} \tag{19}\]9  
Repeating the process in (27), one gets  
\[0<\frac{\phi_{i}(k)\widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2} }\leq\frac{b|\widehat{\phi}_{i}(k)|}{2\sqrt{\lambda}|\widehat{\phi}_{i}(k)|}< \frac{b}{2\sqrt{\lambda_{min}}}<1 \tag{20}\]0 \[0<\frac{\phi_{i}(k)\widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2} }\leq\frac{b|\widehat{\phi}_{i}(k)|}{2\sqrt{\lambda}|\widehat{\phi}_{i}(k)|}< \frac{b}{2\sqrt{\lambda_{min}}}<1 \tag{20}\]1 \[0<\frac{\phi_{i}(k)\widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2} }\leq\frac{b|\widehat{\phi}_{i}(k)|}{2\sqrt{\lambda}|\widehat{\phi}_{i}(k)|}< \frac{b}{2\sqrt{\lambda_{min}}}<1 \tag{20}\]2 \[0<\frac{\phi_{i}(k)\widehat{\phi}_{i}(k)}{\lambda+|\widehat{\phi}_{i}(k)|^{2} }\leq\frac{b|\widehat{\phi}_{i}(k)|}{2\sqrt{\lambda}|\widehat{\phi}_{i}(k)|}< \frac{b}{2\sqrt{\lambda_{min}}}<1 \tag{20}\]3  
which contracts to \(\text{card}(\mathcal{S}_{1}(h))>0\). Therefore, the assumption is violated. According to Lemma 2, one concludes that \(e^{+}(k)\) will converge to 0. This completes the proof. Remark 6: _According to the analysis in Theorem 1, if there is only one leader, all followers will converge to the leader's output. In other words, consensus is a special case of the containment control discussed in this paper._