## 5 Concluding Remarks
Then,  
\[\|x(t)\|\leq\sup_{0\leq t\leq t}\|x(\tau)\|=\sup_{0\leq t\leq t}\|x(\tau)\|=\|x (t\prime)\|\leq\,k_{0}e^{-\rho_{0}t\prime}\|x_{0}\|+\frac{k_{0}}{\rho_{0}} \sup_{0\leq t\leq t}\|\overline{A}(\tau)\|\|x(t\prime)\|;\forall t\in\mathbf{R}_{0+}\] (A4)  
Since \(1>\frac{k_{0}}{\rho_{0}}\sup_{0\leq t\leq t}\|\overline{A}(\tau)\|;\forall t \in\mathbf{R}_{0+}\), one gets:  
\[ \begin{split}&\|x(t)\|\leq\|x(t)\|=\sup_{0\leq t\leq t}\|x( \tau)\|\leq\left(1-\frac{k_{0}}{\rho_{0}}\sup_{0\leq t\leq\infty}\|\overline{ A}(\tau)\|\right)^{-1}\!\!\!k_{0}e^{-\rho_{0}t\prime}\|x_{0}\|\\ &\leq M=k_{0}\!\!\left(\!1-\frac{k_{0}}{\rho_{0}}\sup_{0\leq t< \infty}\|\overline{A}(\tau)\|\right)^{-1}\!\!\!\|x_{0}\|\,;\,\forall t\in\mathbf{R} _{0+}\end{split} \] (A5)  
Therefore, \(x(t)\) is bounded for any \(t\in\mathbf{R}_{0+}\) if \(x_{0}\) is finite. Now, assume the following cases. _Case a_: For any \(T_{s}\in\mathbf{R}_{+}\), the sequence \(\left\{\sup_{nT_{s}\leq t<(n+m_{n})T_{s}}\|x(t)\|\right\}_{n=k}^{\infty}\) is strictly decreasing for some finite positive integer \(k=k(T_{s})\) and some positive sequence \(\{m_{n}\}\) of bounded integer numbers which satisfies \(m_{n+1}>m_{n}-1\) for \(n\geq 0\). As a result, \(\left\{\sup_{nT_{s}\leq t<(n+m_{n})T_{s}}\|x(t)\|\right\}_{n=k}^{\infty}\to 0\) as \(n\to\infty\) for any given \(T_{s}>0\). Then, one gets from Equation (A3) that \(\|x(t)\|\to 0\), as \(t\to\infty\) since \(\|x(nT_{s}+t)\|\leq k_{0}\!\!\left(1+\frac{1}{\rho_{0}}\sup_{nT_{s}\leq t<t}\|x (\tau)\|\right)\); \(\forall t\in(nT_{s}\,,\,(n+1)T_{s})\). The result is proved for this case. _Case b_: For some \(T_{s}\in\mathbf{R}_{+}\), a sequence \(\left\{\sup_{nT_{s}\leq t<(n+m_{n})T_{s}}\|x(t)\|\right\}_{n=0}^{\infty}\) can be built, with \(\{m_{n}\}_{n=0}^{\infty}\to\infty\) as \(m\to\infty\) satisfying \(\sup_{(n+m_{n})T_{s}\leq t<(n+m_{n})T_{s}}\|x(t)\|=\sup_{nT_{s}\leq t<(n+m_{n}+ m_{n}+m_{n})T_{s}}\|x(t)\|\geq\sup_{nT_{s}\leq t<(n+m_{n})T_{s}}\|x(t)\|\,;n\geq 0\) (note that the above inequality cannot be strict as \(n\to\infty\) since it has already been proven that \(\|x(t)\|<+\infty\); \(\forall t\in\mathbf{R}_{0+}\)). However, then one gets from Equation (A3) for some \(t_{n+1}\in[n+m_{n}\,,\,n+m_{n}+m_{n+m_{n}})\) since \(1>\frac{k_{0}}{\rho_{0}}\sup_{0\leq t\leq t}\|\overline{A}(\tau)\|;\forall t \in\mathbf{R}_{0+}\):  
\[ \begin{split}&\sup_{nT_{s}\leq t<(n+m_{n})T_{s}}\|x(t)\|\leq \sup_{nT_{s}\leq t<(n+m_{n}+m_{n}+m_{n})T_{s}}\|x(t)\|\leq K_{0}e^{-\rho_{0}(t_{ n+1}-nT_{s})}\|x(nT_{s})\|\\ &+\frac{K_{0}}{\rho_{0}}\sup_{nT_{s}\leq t<(n+m_{n}+m_{n}+m_{n})T _{s}}\left(\|\overline{A}(\tau)\|\right)\sup_{nT_{s}\leq t<(n+m_{n}+m_{n}+m_{n} )T}\|x(t)\|\\ &<K_{0}e^{-\rho_{0}(t_{n+1}-nT_{s})}\|x(nT_{s})\|+\sup_{nT_{s} \leq t<(n+m_{n}+m_{n}+m_{n})T}\|x(t)\|\end{split} \] (A6)  
If \(n\to\infty\), \(m_{n}\to\infty\), then \((t_{n+1}-m_{n+m_{n}})\to\infty\), thus the following contradiction arises:  
\[0=\limsup_{n\to\infty}\left(\sup_{nT_{s}\leq t<(n+m_{n}+m_{n}+m_{n})T_{s}}\|x(t )\|-\sup_{nT_{s}\leq t<(n+m_{n}+m_{n}+m_{n})T_{s}}\|x(t)\|\right)<0.\]Thus, Case b is not possible and the whole result follows from Case a. \(\Box\)  
Note that the stability abscissa of \(A_{0}\), that is, \((-\rho_{0})<0\) is not smaller than the dominant eigenvalue abscissa.