Stability Proof is Wrong?

Question

I read the book "A Linear Systems Primer" [1] and confused about a proof for a theorem:

In this section we first study the stability properties of the equilibrium $x = 0$ of linear autonomous homogeneous systems: \begin{equation} \dot{x}=A x,~ t\ge 0 \tag{4.16} \end{equation} Recall that $x = 0$ is always an equilibrium of (4.16). Recall also that the solution of (4.16) for $x(0) = x_0$ is given by \begin{equation} \phi(t,x_0)=\varPhi(t,0)x_0=\varPhi(t-0,0)x_0 \\ \triangleq \varPhi(t)x_0=e^{At}x_0 \end{equation} Theorem 4.12. The equilibrium $x=0$ of (4.16) is stable if and only if the solution of (4.16) are bounded, i.e., if and only if \begin{equation} \sup_{t \ge t_0} || \varPhi(t) || \triangleq k < \infty, \end{equation} where $||\varPhi(t)||$ denotes the matrix norm induced by the vector norm used on $\mathbb{R}^n $ and $k$ denotes a constant.
Proof. Assume that the equilibrium $x=0$ of (4.16) is stable. Then for $\epsilon=1$ there is a $\delta=\delta(1)>0$ such that $||\phi(t,x_0)||<1$ for all $t\ge 0$ and all $x_0$ with $||x_0|| \le \delta$. In this case \begin{equation} ||\phi(t,x_0)||=||\varPhi(t)x_0||=||~[\varPhi(t)(x_0\delta)/||x_0||]~||(||x_0||/\delta) < ||x_0||/\delta \tag{*} \end{equation} for all $x_0 \ne 0$ and all $t > 0$. Using the definition of matrix norm, it follows that \begin{equation} ||\varPhi(t)|| \le \delta^{-1}, ~ t \ge 0. \tag{**} \end{equation} We have proved that if the equilibrium $x=0$ of (4.16) is stable, then the solutions of (4.16) are bounded.
.......(proof of necessity)...

I don't understand how the inequality $(**)$ is justified. The definition of a matrix norm is given by $||\varPhi(t)||=\sup_{||x||=1}||\varPhi(t)x||$ where $x$ should be an arbitrary vector. But according to the formula $(*)$ , \begin{equation} ||\varPhi(t)x_0||<||x_0||/\delta \iff ||\varPhi(t) \frac{x_0}{||x_0||}|| < \delta^{-1} \end{equation} It seems correct since $ || \frac{x_0}{||x_0||} || = 1$ and it seems consistent with the definition of a matrix norm. However, we should also notice that there is a restriction on $||x_0||$, i.e., $||x_0|| \le \delta $ as stated in the proof. So $ \frac{x_0}{||x_0||} $ is not arbitrary anymore then it is NOT consistent with the definition of a matrix norm. Therefore I think $(**)$ is NOT correct.

So what is the problem here? Is there anything wrong in my thought?

References:
[1] Antsaklis, Panos J., and Anthony N. Michel. A linear systems primer. Vol. 1. Boston: Birkhäuser, 2007.

Answer 1

There is no issue in the part of the proof you have provided. Here is a clearer explanation of the step marked (**).

We wish to show that $\sup_{\|x\|\leq 1}\|\Phi(t)x\|\leq \delta^{-1}$ (the left side is one of the many equivalent definitions of the operator norm, note the slight difference in the domain of the supremum when compared to the definition you are using).

To show this amounts to proving that for all $\|x\|\leq 1$, we necessarily have $\|\Phi(t)x\|\leq \delta^{-1}$. Thus, let $x$ be arbitrary such that $\|x\|\leq 1$. Form the new vector $x_0=\delta x$. Then $\|x_0\|\leq \delta$. Hence, by construction of $\delta$, it follows that $\|\Phi(t)x_0\|<1$, which implies that $\|\Phi(t)\delta x\|\leq 1$ and therefore $\|\Phi(t) x\|< \delta^{-1}$. Since $x$ was arbitrary subject to the constraint $\|x\|\leq 1$, we may take the supremum on both sides of the last equation to deduce that $$ \sup_{\|x\|\leq 1}\|\Phi(t) x\|\leq \delta^{-1}, $$ and therefore $\|\Phi(t)\|\leq \delta^{-1}$, as desired.