[proofplan] The proof is just the equivalence between the differential equation for a constant curve and the algebraic equation defining an equilibrium. First we compute that every constant curve has zero derivative at every point where the equation is imposed, using the difference quotient at interior points and the corresponding one-sided quotient at endpoints when endpoints belong to a non-singleton interval. If $f(x^*)=0$, this derivative agrees with $f(x_J(t))$ at every such time. Conversely, if the constant curve solves the equation on any nonempty interval, evaluating the equation at a time in that interval gives $f(x^*)=0$. [/proofplan]

[step:Compute the derivative of the constant curve]Let $J\subset \mathbb{R}$ be a nonempty interval, and let $x_J:J\to U$ be defined by $x_J(t)=x^*$ for every $t\in J$. If $J$ is a singleton interval, the statement's convention imposes no differential equation on $J$, so there is no derivative to compute. Assume now that $J$ has at least two points. If $t\in J$ is an interior point of $J$, then for every sufficiently small nonzero $h\in \mathbb{R}$ with $t+h\in J$, \begin{align*} \frac{x_J(t+h)-x_J(t)}{h}=0. \end{align*} Hence $\dot{x}_J(t)=0$. If $t$ is an endpoint of $J$ belonging to $J$, the same computation with the appropriate one-sided nonzero values of $h$ gives the one-sided derivative $\dot{x}_J(t)=0$. Thus the derivative required in the [ordinary differential equation](/page/Ordinary%20Differential%20Equation) is identically zero at every point of $J$ where the equation is imposed.[/step]

[guided]We first isolate the only calculus fact needed in the proof: a constant curve has zero derivative wherever the definition of solution asks for a derivative. Fix a nonempty interval $J\subset \mathbb{R}$ and define the curve $x_J:J\to U$ by $x_J(t)=x^*$ for all $t\in J$. This is well-defined because the theorem assumes $x^*\in U$. If $J$ is a singleton interval, the convention in the theorem imposes no differential equation on $J$. Thus no derivative is required in that case. Assume now that $J$ has at least two points. Let $t\in J$ be an interior point of $J$. For every sufficiently small nonzero $h\in \mathbb{R}$ such that $t+h\in J$, both $x_J(t+h)$ and $x_J(t)$ equal $x^*$. Therefore \begin{align*} \frac{x_J(t+h)-x_J(t)}{h}=\frac{x^*-x^*}{h}=0. \end{align*} Taking the limit of this difference quotient gives $\dot{x}_J(t)=0$. If $t$ is an endpoint of $J$ and $t\in J$, the definition of solution on a non-singleton interval uses the corresponding one-sided derivative. Because $J$ has at least two points, there are admissible one-sided nonzero increments $h$ with $t+h\in J$ arbitrarily close to $0$ from the side lying inside $J$. For those increments the quotient computation is again always $0$. Hence the one-sided derivative is also $0$. Thus the constant curve has derivative $0$ at every point where the differential equation is imposed.[/guided]

[step:Use the equilibrium equation to prove the constant curve is a solution]Assume that $x^*$ is an equilibrium point of $f$, meaning that $f(x^*)=0$. For every nonempty interval $J\subset \mathbb{R}$ and every $t\in J$ where the differential equation is imposed, \begin{align*} \dot{x}_J(t)=0=f(x^*)=f(x_J(t)). \end{align*} Therefore $x_J$ solves $\dot{x}=f(x)$ on $J$. Since $J$ was arbitrary, the constant curve solves the equation on every nonempty interval.[/step]

[guided]Assume that $x^*$ is an equilibrium point of $f$. By the definition stated in the theorem, this means \begin{align*} f(x^*)=0. \end{align*} Fix a nonempty interval $J\subset \mathbb{R}$ and a point $t\in J$ at which the equation is imposed, using a one-sided derivative if $t$ is an endpoint belonging to $J$. From the derivative computation for the constant curve, $\dot{x}_J(t)=0$. Since $x_J(t)=x^*$, we have \begin{align*} \dot{x}_J(t)=0=f(x^*)=f(x_J(t)). \end{align*} Thus the differential equation holds at every required time in $J$. Because $J$ was arbitrary, the constant curve solves $\dot{x}=f(x)$ on every nonempty interval.[/guided]

[step:Evaluate a constant solution to recover the equilibrium equation]Assume conversely that $x_J$ solves $\dot{x}=f(x)$ on every nonempty interval $J\subset \mathbb{R}$. Apply this assumption to the interval $J=\mathbb{R}$. For any $t\in \mathbb{R}$, the differential equation gives \begin{align*} 0=\dot{x}_{\mathbb{R}}(t)=f(x_{\mathbb{R}}(t))=f(x^*). \end{align*} Thus $f(x^*)=0$, so $x^*$ is an equilibrium point of $f$. This proves the desired equivalence.[/step]

[guided]Assume conversely that, for every nonempty interval $J\subset \mathbb{R}$, the constant curve $x_J$ solves $\dot{x}=f(x)$ on $J$. We need to recover the algebraic equilibrium equation. Choose the particular nonempty interval $J=\mathbb{R}$, where every point is an interior point, so the usual derivative is used. For any $t\in\mathbb{R}$, the solution property gives \begin{align*} \dot{x}_{\mathbb{R}}(t)=f(x_{\mathbb{R}}(t)). \end{align*} The derivative computation gives $\dot{x}_{\mathbb{R}}(t)=0$, and the definition of the constant curve gives $x_{\mathbb{R}}(t)=x^*$. Therefore \begin{align*} 0=\dot{x}_{\mathbb{R}}(t)=f(x_{\mathbb{R}}(t))=f(x^*). \end{align*} Hence $f(x^*)=0$, so $x^*$ is an equilibrium point of $f$. This completes the equivalence.[/guided]