Assumption:
the expressions are all meaningful, i.e., there exists no case that division by zero.

Proof.
If $\{s(t)\}$ is a bounded sequence, then by (3), $\{\Vert \sigma (t)\Vert \}$ is a bounded sequence. Then by $(1)$ and $(2)$ it follows that
$$\underset{t\to \infty }{\lim }\Vert s(t)\Vert =0$$

Now assume that $\{\Vert s(t)\Vert \}$ is unbounded. It follows that there exists a subsequence $\left\{t_n\right\}$ such that
$$\underset{t_n\to \infty }{\lim }\Vert s\left(t_n\right)\Vert =\infty$$and
$$\Vert s(t)\Vert \le \Vert s\left(t_n\right)\Vert \text{?for?}t\le t_n$$
Now along the subsequence $\left\{t_n\right\}$
$$\begin{array}{rl}\frac{{\left[s{\left(t_n\right)}^{T}s\left(t_n\right)\right]}^{1/2}}{{\left[b_1\left(t_n\right)+b_2\left(t_n\right)\sigma {\left(t_n\right)}^T\sigma \left(t_n\right)\right]}^{1/2}}& \ge \frac{\Vert s\left(t_n\right)\Vert }{{\left[K+K\Vert \sigma \left(t_n\right){\Vert }^2\right]}^{1/2}}\\ & \ge \frac{\Vert s\left(t_n\right)\Vert }{K^{1/2}+K^{1/2}\Vert \sigma \left(t_n\right)\Vert }\\ & \ge \frac{\Vert s\left(t_n\right)\Vert }{K^{1/2}+K^{1/2}\left[C_1+C_2\Vert s\left(t_n\right)\Vert \right]}\end{array}$$
Hence, when $t_n$ approaches infinity, we have
$$\frac{{\left[s{\left(t_n\right)}^{T}s\left(t_n\right)\right]}^{1/2}}{{\left[b_1\left(t_n\right)+b_2\left(t_n\right)\sigma {\left(t_n\right)}^T\sigma \left(t_n\right)\right]}^{1/2}}\ge \frac{1}{K^{1/2}{C}_2}>0$$
but this contradicts (1) and hence the assumption that $\{\Vert s(t)\Vert \}$ is unbounded is false and the result follows.

The proof is complete.

總結

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