[proofplan] The proof is a direct comparison between [Lyapunov stability](/page/Lyapunov%20Stability) and uniform boundedness of the linear flow. If the origin is Lyapunov stable, then the stability condition keeps a sufficiently small ball inside the unit ball for all future times; linearity of $x \mapsto e^{tA}x$ then scales this unit-ball estimate to every initial condition. Conversely, a uniform operator bound for $e^{tA}$ immediately gives the $\varepsilon$-$\delta$ condition for Lyapunov stability. [/proofplan] [step:Express Lyapunov stability using the linear flow] For each initial condition $x_0 \in \mathbb{R}^n$, the solution of the linear autonomous system with $x(0)=x_0$ is the map \begin{align*} x_{x_0}: [0,\infty) &\to \mathbb{R}^n \end{align*} defined by $x_{x_0}(t)=e^{tA}x_0$. Thus the equilibrium $0$ is Lyapunov stable precisely when, for every $\varepsilon>0$, there exists $\delta>0$ such that \begin{align*} |x_0|<\delta \implies |e^{tA}x_0|<\varepsilon \quad \text{for every } t\ge 0. \end{align*} [/step] [step:Scale the Lyapunov stability estimate to obtain a uniform bound] Assume that $0$ is Lyapunov stable. Apply the stability condition with $\varepsilon=1$. Then there exists $\eta>0$ such that, whenever $y \in \mathbb{R}^n$ satisfies $|y|<\eta$, \begin{align*} |e^{tA}y|<1 \quad \text{for every } t\ge 0. \end{align*} Define $M:=2/\eta$. We prove that this $M$ works. If $x=0$, then $|e^{tA}x|=0=M|x|$ for every $t\ge 0$. Now let $x \in \mathbb{R}^n$ with $x \ne 0$, and define the scalar \begin{align*} \alpha:=\frac{\eta}{2|x|}. \end{align*} Then $\alpha>0$ and $|\alpha x|=\eta/2<\eta$. Applying the preceding stability estimate to $y=\alpha x$ gives \begin{align*} |e^{tA}(\alpha x)|<1 \quad \text{for every } t\ge 0. \end{align*} Since $e^{tA}: \mathbb{R}^n \to \mathbb{R}^n$ is linear for each fixed $t\ge 0$, we have $e^{tA}(\alpha x)=\alpha e^{tA}x$. Therefore \begin{align*} \alpha |e^{tA}x|<1. \end{align*} Dividing by $\alpha>0$ yields \begin{align*} |e^{tA}x|<\frac{1}{\alpha}=\frac{2|x|}{\eta}=M|x|. \end{align*} Hence $|e^{tA}x|\le M|x|$ for all $x \in \mathbb{R}^n$ and all $t\ge 0$. [guided] Assume that the equilibrium $0$ is Lyapunov stable. The definition says that every target radius can be protected uniformly for all future time, provided the initial condition starts close enough to $0$. We use the target radius $1$, because a bound on the image of one small ball can be rescaled to all of $\mathbb{R}^n$ using linearity. Applying Lyapunov stability with $\varepsilon=1$, there exists $\eta>0$ such that \begin{align*} |y|<\eta \implies |e^{tA}y|<1 \quad \text{for every } t\ge 0. \end{align*} Define the constant \begin{align*} M:=\frac{2}{\eta}. \end{align*} We now verify that $M$ gives the desired uniform estimate. First, if $x=0$, then $e^{tA}x=0$ because $e^{tA}: \mathbb{R}^n \to \mathbb{R}^n$ is linear. Hence \begin{align*} |e^{tA}x|=0=M|x| \end{align*} for every $t\ge 0$. Now let $x \ne 0$. We want to use the small-ball estimate, but $x$ itself need not satisfy $|x|<\eta$. The way to force it into the stable ball is to scale it down. Define \begin{align*} \alpha:=\frac{\eta}{2|x|}. \end{align*} Then $\alpha>0$, and the scaled vector $\alpha x$ satisfies \begin{align*} |\alpha x|=\alpha |x|=\frac{\eta}{2}<\eta. \end{align*} Therefore the Lyapunov stability estimate applies to $y=\alpha x$, giving \begin{align*} |e^{tA}(\alpha x)|<1 \quad \text{for every } t\ge 0. \end{align*} For each fixed $t\ge 0$, the map $e^{tA}: \mathbb{R}^n \to \mathbb{R}^n$ is linear, so \begin{align*} e^{tA}(\alpha x)=\alpha e^{tA}x. \end{align*} Since $\alpha>0$, taking Euclidean norms gives \begin{align*} |e^{tA}(\alpha x)|=|\alpha e^{tA}x|=\alpha |e^{tA}x|. \end{align*} Combining this identity with $|e^{tA}(\alpha x)|<1$, we obtain \begin{align*} \alpha |e^{tA}x|<1. \end{align*} Dividing by $\alpha$ gives \begin{align*} |e^{tA}x|<\frac{1}{\alpha}. \end{align*} By the definition of $\alpha$, \begin{align*} \frac{1}{\alpha}=\frac{2|x|}{\eta}=M|x|. \end{align*} Thus \begin{align*} |e^{tA}x|<M|x| \end{align*} for every nonzero $x \in \mathbb{R}^n$ and every $t\ge 0$. Together with the case $x=0$, this proves the required uniform bound. [/guided] [/step] [step:Use the uniform bound to verify Lyapunov stability] Conversely, assume that there exists $M>0$ such that \begin{align*} |e^{tA}x| \le M|x| \end{align*} for every $x \in \mathbb{R}^n$ and every $t\ge 0$. Let $\varepsilon>0$ be given, and define \begin{align*} \delta:=\frac{\varepsilon}{2M}. \end{align*} If $x_0 \in \mathbb{R}^n$ satisfies $|x_0|<\delta$, then for every $t\ge 0$, \begin{align*} |e^{tA}x_0| \le M|x_0|<M\delta=\frac{\varepsilon}{2}<\varepsilon. \end{align*} This is exactly the Lyapunov stability condition for the equilibrium $0$. Hence $0$ is Lyapunov stable. [/step] [step:Conclude the equivalence] The first implication shows that Lyapunov stability of $0$ forces a uniform bound on the family of linear maps $(e^{tA})_{t\ge 0}$. The reverse implication shows that such a uniform bound implies the $\varepsilon$-$\delta$ stability condition for all future times. Therefore the equilibrium $0$ is Lyapunov stable if and only if there exists $M>0$ such that \begin{align*} |e^{tA}x| \le M|x| \end{align*} for all $x \in \mathbb{R}^n$ and all $t\ge 0$. [/step]