Proof. (i) Put \(u = x - x^*\), so \(u' = Au + g(u)\) with \(g(u) = f(x^* + u) - Au\). Since \(f\) is \(C^1\), \(\|g(u)\| \leq \varepsilon (r)\|u\|\) for \(\|u\| \leq r\), with \(\varepsilon (r) \to 0\) as \(r \to 0\). By Theorem 5.3.2(i) there are \(K \geq 1\) and \(\alpha > 0\) with \(\|e^{At}\| \leq Ke^{-2\alpha t}\). Choose \(r\) so small that \(K\varepsilon (r) \leq \alpha \). While \(\|u(s)\| \leq r\) on \([0, t]\), variation of parameters gives \[ u(t) = e^{At}u_0 + \int _0^te^{A(t - s)}g(u(s))\,ds, \qquad e^{2\alpha t}\|u(t)\| \leq K\|u_0\| + K\varepsilon (r)\int _0^te^{2\alpha s}\|u(s)\|\,ds , \] and Gronwall’s inequality applied to \(e^{2\alpha t}\|u(t)\|\) gives \(\|u(t)\| \leq K\|u_0\|e^{(K\varepsilon (r) - 2\alpha )t} \leq K\|u_0\|e^{-\alpha t}\). If \(\|u_0\| < r/K\), this bound is less than \(r\), so the solution never reaches \(\|u\| = r\), the estimate holds for all \(t \geq 0\), and \(u(t) \to 0\) exponentially.

(ii) We sketch the proof. Decompose \(\mathbb R^n\) into the sum of the generalised eigenspaces of eigenvalues with positive real part and the rest, and define a quadratic form that grows along solutions in the first part and is controlled in the second; a solution starting arbitrarily close to \(x^*\) in the first part is driven out of a fixed ball. A complete proof uses the Lyapunov functions of the next section. □