[proofplan]
We use the explicit exponential stability hypothesis and verify the two clauses in the definition of [asymptotic stability](/page/Asymptotic%20Stability): [Lyapunov stability](/page/Lyapunov%20Stability) and local attractivity. The hypothesis $f(x^*)=0$ identifies $x^*$ as an equilibrium because the constant map at $x^*$ solves the differential equation. Exponential stability gives a neighbourhood of $x^*$ on which every forward solution exists for all $t\geq 0$ and satisfies an exponentially decaying estimate relative to its initial distance from $x^*$. The estimate at arbitrary time gives Lyapunov stability after choosing the initial radius small enough, and letting $t\to\infty$ in the same estimate gives convergence to $x^*$.
[/proofplan]
[step:Unpack the exponential stability estimate near $x^*$]
By exponential stability of the equilibrium $x^*$, there exist constants $r>0$, $C>0$, and $\alpha>0$ such that for every initial point $x_0\in U$ with $|x_0-x^*|<r$, there exists a unique forward solution $\gamma_{x_0}\in C^1([0,\infty);U)$ satisfying $\gamma_{x_0}'(t)=f(\gamma_{x_0}(t))$ for every $t\geq 0$ and $\gamma_{x_0}(0)=x_0$, and this solution satisfies
\begin{align*}
|\gamma_{x_0}(t)-x^*|\leq C e^{-\alpha t}|x_0-x^*|
\end{align*}
for every $t\geq 0$. Since $f(x^*)=0$, the constant map $\gamma_{x^*}:[0,\infty)\to U$ defined by $\gamma_{x^*}(t)=x^*$ satisfies $\gamma_{x^*}'(t)=0=f(x^*)$ for every $t\geq 0$, so $x^*$ is an equilibrium. Since $f$ is locally Lipschitz, the forward solution with initial value $x_0$ is unique on every interval on which it is defined, so this estimate applies to the trajectory determined by $x_0$.
[guided]
We prove the two parts of asymptotic stability directly from the exponential estimate. The hypothesis supplies constants $r>0$, $C>0$, and $\alpha>0$ such that whenever $x_0\in U$ satisfies $|x_0-x^*|<r$, there exists a unique forward solution $\gamma_{x_0}\in C^1([0,\infty);U)$ satisfying $\gamma_{x_0}'(t)=f(\gamma_{x_0}(t))$ for every $t\geq 0$ and $\gamma_{x_0}(0)=x_0$, and this solution satisfies
\begin{align*}
|\gamma_{x_0}(t)-x^*|\leq C e^{-\alpha t}|x_0-x^*|
\end{align*}
for every $t\geq 0$.
First, $x^*$ is an equilibrium. Since $f(x^*)=0$, the constant map $\gamma_{x^*}:[0,\infty)\to U$ defined by $\gamma_{x^*}(t)=x^*$ satisfies $\gamma_{x^*}'(t)=0=f(x^*)$ for every $t\geq 0$.
Now we prove Lyapunov stability. Let $\varepsilon>0$ be given, and define
\begin{align*}
\delta:=\min\left\{r,\frac{\varepsilon}{C}\right\}.
\end{align*}
If $x_0\in U$ and $|x_0-x^*|<\delta$, then $|x_0-x^*|<r$, so the exponential estimate applies. Also $e^{-\alpha t}\leq 1$ for every $t\geq 0$, because $\alpha>0$. Hence
\begin{align*}
|\gamma_{x_0}(t)-x^*|\leq C e^{-\alpha t}|x_0-x^*|\leq C|x_0-x^*|<C\delta\leq \varepsilon
\end{align*}
for every $t\geq 0$. This is the Lyapunov stability clause.
Finally, we prove local attractivity. Take any $x_0\in U$ with $|x_0-x^*|<r$. The exponential estimate gives
\begin{align*}
0\leq |\gamma_{x_0}(t)-x^*|\leq C e^{-\alpha t}|x_0-x^*|
\end{align*}
for every $t\geq 0$. Since $\alpha>0$, the scalar function $t\mapsto C e^{-\alpha t}|x_0-x^*|$ tends to $0$ as $t\to\infty$. The non-negative function $t\mapsto |\gamma_{x_0}(t)-x^*|$ is bounded above by this scalar function, so for every $\eta>0$ there exists $T\geq 0$ such that $0\leq |\gamma_{x_0}(t)-x^*|<\eta$ whenever $t\geq T$. Therefore
\begin{align*}
\lim_{t\to\infty}|\gamma_{x_0}(t)-x^*|=0.
\end{align*}
Thus the same radius $r$ gives a neighbourhood of attraction. Together with the equilibrium verification, this proves that $x^*$ is asymptotically stable.
[/guided]
[/step]
[step:Choose the initial radius to prove Lyapunov stability]
Let $\varepsilon>0$ be given. Define
\begin{align*}
\delta:=\min\left\{r,\frac{\varepsilon}{C}\right\}.
\end{align*}
If $x_0\in U$ satisfies $|x_0-x^*|<\delta$, then $|x_0-x^*|<r$, so the exponential estimate applies. Since $e^{-\alpha t}\leq 1$ for every $t\geq 0$, we obtain
\begin{align*}
|\gamma_{x_0}(t)-x^*|\leq C e^{-\alpha t}|x_0-x^*|\leq C|x_0-x^*|<C\delta\leq \varepsilon
\end{align*}
for every $t\geq 0$. This proves Lyapunov stability of $x^*$.
[/step]
[step:Let the exponential factor tend to zero to prove attractivity]
Take any $x_0\in U$ with $|x_0-x^*|<r$. The same exponential estimate gives
\begin{align*}
0\leq |\gamma_{x_0}(t)-x^*|\leq C e^{-\alpha t}|x_0-x^*|
\end{align*}
for every $t\geq 0$. Because $\alpha>0$, the scalar function
\begin{align*}
t\mapsto C e^{-\alpha t}|x_0-x^*|
\end{align*}
from $[0,\infty)$ to $[0,\infty)$ tends to $0$ as $t\to\infty$. Let $\eta>0$. There exists $T\geq 0$ such that $C e^{-\alpha t}|x_0-x^*|<\eta$ whenever $t\geq T$. Hence $0\leq |\gamma_{x_0}(t)-x^*|<\eta$ whenever $t\geq T$, and therefore
\begin{align*}
\lim_{t\to\infty}|\gamma_{x_0}(t)-x^*|=0.
\end{align*}
Thus every trajectory starting from an initial condition $x_0\in U$ with $|x_0-x^*|<r$ converges to $x^*$ as $t\to\infty$.
[/step]
[step:Combine stability and attractivity]
The previous two steps show that $x^*$ is Lyapunov stable and that choosing $\rho:=r$ gives a radius such that every solution starting from an initial condition $x_0\in U$ with $|x_0-x^*|<\rho$ converges to $x^*$ as $t\to\infty$. Together with the equilibrium verification from the first step, these are precisely the requirements in the definition of asymptotic stability of an equilibrium. Therefore $x^*$ is asymptotically stable.
[/step]
