[proofplan] We first record the uniqueness consequence of local Lipschitz continuity: the equilibrium solution $t\mapsto x^*$ is the only solution that can pass through $x^*$. Therefore a non-equilibrium solution cannot hit or cross $x^*$ at any positive time. The sign of $f$ on each side of $x^*$ then gives monotonicity directly from $\dot{x}=f(x)$. In the attracting case, a global trajectory that remains in the neighbourhood is monotone and bounded, hence has a limit; continuity and the non-vanishing of $f$ away from $x^*$ force that limit to be $x^*$. [/proofplan]

[step:Use uniqueness to prevent hitting the equilibrium]Let \begin{align*} V=(x^*-\eta,x^*+\eta). \end{align*} Define the autonomous vector field \begin{align*} F: \mathbb{R}\times U \to \mathbb{R}, \qquad (t,y)\mapsto f(y). \end{align*} Since $f$ is locally Lipschitz on $U$, the map $F$ is continuous and locally Lipschitz in the state variable. By the [Existence and Uniqueness of Maximal Solutions](/theorems/8313) theorem [citetheorem:8313], applied with ambient time interval $\mathbb{R}$, state space $U$, and vector field $F$, the initial value problem for $\dot{y}=F(t,y)$ has a unique maximal solution through each prescribed time and state, and every other solution with the same initial data agrees with it on the connected component of the common domain containing the initial time. The constant map \begin{align*} e: \mathbb{R} \to U, \qquad t\mapsto x^* \end{align*} solves $\dot{e}=f(e)$ because $f(x^*)=0$. Let $x:J\to U$ be a solution with $0\in J$ and $x(0)\ne x^*$. If there existed $\tau\in J$ with $\tau\ge 0$ and $x(\tau)=x^*$, then uniqueness applied to the initial value problem at time $\tau$ with initial value $x^*$ would give \begin{align*} x(t)=x^* \end{align*} for every $t$ in the connected component of the common domain $J\cap\mathbb{R}=J$ containing $\tau$. Since $J$ is an interval and $0,\tau\in J$, this connected component contains both $0$ and $\tau$. In particular $x(0)=x^*$, contradicting the hypothesis. Hence \begin{align*} x(t)\ne x^* \end{align*} for every $t\in J\cap[0,\infty)$.[/step]

[guided]The only subtle point in a one-dimensional phase-line argument is that a solution should not be allowed to arrive at the equilibrium in finite time and then remain there. Local Lipschitz continuity rules this out through uniqueness. Set \begin{align*} V=(x^*-\eta,x^*+\eta). \end{align*} To place the autonomous equation in the standard initial-value framework, define \begin{align*} F:\mathbb{R}\times U&\to \mathbb{R} \end{align*} \begin{align*} (t,y)&\mapsto f(y). \end{align*} The function $f$ is locally Lipschitz on $U$, so $F$ is locally Lipschitz in the state variable $y$, locally uniformly in $t$. It is also continuous. Therefore the Existence and Uniqueness of Maximal Solutions theorem [citetheorem:8313] applies to the differential equation $\dot{y}=F(t,y)$ with ambient time interval $\mathbb{R}$ and state space $U$: through each prescribed time and state there is a unique maximal solution, and any other solution with the same initial data agrees with it on the connected component of the common domain containing that initial time. Now define the constant equilibrium solution \begin{align*} e:\mathbb{R}&\to U \end{align*} \begin{align*} t&\mapsto x^*. \end{align*} Since $f(x^*)=0$, we have $\dot{e}(t)=0=f(e(t))$ for every $t\in\mathbb{R}$, so $e$ is a solution. Suppose, toward a contradiction, that a solution $x:J\to U$ with $x(0)\ne x^*$ reaches the equilibrium at some positive time $\tau\in J$: \begin{align*} x(\tau)=x^*. \end{align*} At time $\tau$, both $x$ and $e$ solve the same initial value problem with initial value $x^*$. The common domain of these two solutions is $J\cap\mathbb{R}=J$. Since $J$ is an interval and $0,\tau\in J$, the connected component of the common domain containing $\tau$ also contains $0$. The uniqueness conclusion from [citetheorem:8313] therefore gives $x(t)=e(t)$ for every $t\in J$, and in particular \begin{align*} x(0)=e(0)=x^*, \end{align*} contradicting the initial condition $x(0)\ne x^*$. Thus the trajectory cannot hit $x^*$ at any positive time. Since crossing $x^*$ would require hitting it by continuity of $x$, crossing is also impossible.[/guided]

[step:Derive monotonic motion toward the equilibrium in the attracting case] Assume \begin{align*} f(y)>0 \quad \text{for } y\in(x^*-\eta,x^*) \end{align*} and \begin{align*} f(y)<0 \quad \text{for } y\in(x^*,x^*+\eta). \end{align*} Let $x:J\to U$ be a solution with $x(0)\in V\setminus\{x^*\}$. Let $T\in J\cap[0,\infty)$ satisfy \begin{align*} x([0,T])\subset V. \end{align*} By the previous step, $x(t)\ne x^*$ for every $t\in[0,T]$. Since $x([0,T])$ is connected and contained in $V\setminus\{x^*\}$, it is contained entirely in one of the two intervals $(x^*-\eta,x^*)$ or $(x^*,x^*+\eta)$. If $x(0)<x^*$, then \begin{align*} x(t)\in(x^*-\eta,x^*) \quad \text{for every } t\in[0,T]. \end{align*} Hence \begin{align*} \dot{x}(t)=f(x(t))>0 \quad \text{for every interior point } t\in[0,T]. \end{align*} By the [mean value theorem](/theorems/186) applied on each subinterval $[a,b]\subset[0,T]$ with $a<b$, there exists $c\in(a,b)$ such that $x(b)-x(a)=\dot{x}(c)(b-a)>0$. Thus $x$ is strictly increasing on $[0,T]$, and because $x(t)<x^*$ throughout this interval, it moves monotonically toward $x^*$. If $x(0)>x^*$, the same argument gives \begin{align*} x(t)\in(x^*,x^*+\eta) \quad \text{for every } t\in[0,T], \end{align*} and therefore \begin{align*} \dot{x}(t)=f(x(t))<0 \quad \text{for every interior point } t\in[0,T]. \end{align*} By the mean value theorem applied on each subinterval $[a,b]\subset[0,T]$ with $a<b$, there exists $c\in(a,b)$ such that $x(b)-x(a)=\dot{x}(c)(b-a)<0$. Thus $x$ is strictly decreasing on $[0,T]$, and because $x(t)>x^*$ throughout this interval, it moves monotonically toward $x^*$. [/step]

[step:Show that a global attracting trajectory remaining in the neighbourhood converges to the equilibrium] Assume the attracting sign conditions from the previous step. Let $x:J\to U$ be a solution such that $[0,\infty)\subset J$, $x([0,\infty))\subset V$, and $x(0)\ne x^*$. If $x(0)<x^*$, then the previous step shows that \begin{align*} x(t)<x^* \end{align*} for every $t\ge 0$, and that $x:[0,\infty)\to \mathbb{R}$ is increasing. Hence \begin{align*} \ell=\sup_{t\ge 0} x(t) \end{align*} exists and satisfies $\ell\le x^*$. Monotonicity implies \begin{align*} \lim_{t\to\infty}x(t)=\ell. \end{align*} Suppose that $\ell<x^*$. Since $f(\ell)>0$ and $f$ is continuous at $\ell$, there exists $\delta>0$ such that $(\ell-\delta,\ell+\delta)\subset(x^*-\eta,x^*)$ and \begin{align*} f(y)\ge \frac{1}{2}f(\ell)>0 \end{align*} for every $y\in(\ell-\delta,\ell+\delta)$. Since $x(t)\to\ell$, there exists $T_0\ge 0$ such that $x(t)\in(\ell-\delta,\ell+\delta)$ for every $t\ge T_0$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Since $f$ is locally Lipschitz, it is continuous, and since $x$ is continuous, the map $s\mapsto f(x(s))$ is continuous on $[T_0,t]$. For every $t\ge T_0$, the [fundamental theorem of calculus](/theorems/632) for the differentiable map $x|_{[T_0,t]}:[T_0,t]\to\mathbb{R}$ gives \begin{align*} x(t)-x(T_0)=\int_{T_0}^{t} f(x(s))\,d\mathcal{L}^1(s). \end{align*} Using the lower bound on $f(x(s))$ on $[T_0,t]$, we obtain \begin{align*} x(t)-x(T_0)\ge \frac{1}{2}f(\ell)(t-T_0). \end{align*} The right-hand side tends to $+\infty$ as $t\to\infty$, contradicting the boundedness $x(t)\le x^*$. Therefore $\ell=x^*$. If $x(0)>x^*$, then $x$ is decreasing and bounded below by $x^*$. Define \begin{align*} \ell=\inf_{t\ge 0} x(t). \end{align*} Then $\lim_{t\to\infty}x(t)=\ell$ and $\ell\ge x^*$. If $\ell>x^*$, then $f(\ell)<0$. By continuity there exists $\delta>0$ such that $(\ell-\delta,\ell+\delta)\subset(x^*,x^*+\eta)$ and \begin{align*} f(y)\le \frac{1}{2}f(\ell)<0 \end{align*} for every $y\in(\ell-\delta,\ell+\delta)$. For all sufficiently large $t$, $x(t)$ lies in this interval, and the same integral identity gives \begin{align*} x(t)-x(T_0)\le \frac{1}{2}f(\ell)(t-T_0), \end{align*} which tends to $-\infty$ as $t\to\infty$, contradicting $x(t)\ge x^*$. Hence $\ell=x^*$. Thus every global solution remaining in $V$ converges to $x^*$ as $t\to\infty$. [/step]

[step:Reverse the signs to obtain monotonic motion away from the equilibrium] Assume instead that \begin{align*} f(y)<0 \quad \text{for } y\in(x^*-\eta,x^*) \end{align*} and \begin{align*} f(y)>0 \quad \text{for } y\in(x^*,x^*+\eta). \end{align*} Let $x:J\to U$ be a solution with $x(0)\in V\setminus\{x^*\}$, and let $T\in J\cap[0,\infty)$ satisfy $x([0,T])\subset V$. By the first step, $x(t)\ne x^*$ on $[0,T]$, so the trajectory remains on its initial side of $x^*$ while it remains in $V$. If $x(0)<x^*$, then $x(t)\in(x^*-\eta,x^*)$ on $[0,T]$, and \begin{align*} \dot{x}(t)=f(x(t))<0 \end{align*} at every interior point of $[0,T]$. Hence $x$ is strictly decreasing on $[0,T]$, so it moves monotonically away from $x^*$. If $x(0)>x^*$, then $x(t)\in(x^*,x^*+\eta)$ on $[0,T]$, and \begin{align*} \dot{x}(t)=f(x(t))>0 \end{align*} at every interior point of $[0,T]$. Hence $x$ is strictly increasing on $[0,T]$, so it moves monotonically away from $x^*$. This proves the reversed-sign assertion. [/step]