A local minimum with a gradient inequality is stable

Fix $\epsilon >0$. Choose an open neighborhood $W\subseteq U$ of $z$ on which $f\ge f(z)$ and the gradient inequality both hold, and choose $0<r<\epsilon$ with $\bar{B}(z,r)\subset W$. The $C^2$ hypothesis gives continuous partial derivatives, so $f$ is totally differentiable; Fermat's theorem at its interior local minimum gives $\mathrm{\nabla }f(z)=0$. Hence $z$ is an equilibrium.

The field $-\mathrm{\nabla }f$ is continuous and locally Lipschitz, as follows directly. On any closed ball $K\subset U$ of positive radius, compactness and extreme values bound all $\mathrm{\mid }{\mathrm{\partial }}_j{\mathrm{\partial }}_{i}f\mathrm{\mid }$ by one $M\ge 0$. Each ${\mathrm{\partial }}_{i}f$ is totally differentiable by [F2]. For $u,v\in K$, the chain rule on $u+t(v-u)$ bounds the derivative of ${\mathrm{\partial }}_{i}f(u+t(v-u))$ by $M{\sum }_{j<m}\mathrm{\mid }{v}_j-u_j\mathrm{\mid }\le M\sqrt{m}\mathrm{\parallel }v-u\mathrm{\parallel }$. The scalar mean value theorem on $[0,1]$ gives $\mathrm{\mid }{\mathrm{\partial }}_{i}f(v)-{\mathrm{\partial }}_{i}f(u)\mathrm{\mid }\le M\sqrt{m}\mathrm{\parallel }v-u\mathrm{\parallel }$, hence $\mathrm{\parallel }\mathrm{\nabla }f(v)-\mathrm{\nabla }f(u)\mathrm{\parallel }\le mM\mathrm{\parallel }v-u\mathrm{\parallel }$. These bounds are independent of time for the autonomous field on $\mathbb{R}\times U$, so the local and maximal ODE existence and uniqueness results apply.

Put $\alpha =1-\rho >0$ and $\eta =(c\alpha r\mathrm{/}4{)}^{1\mathrm{/}\alpha }>0$. On positive bases, the derivative formula, positivity and the mean value theorem make $s\mapsto s^{\alpha }$ strictly increasing; the iterated-power law gives ${\eta }^{\alpha }=c\alpha r\mathrm{/}4$. Continuity of $f$ at $z$ permits $0<\delta <r\mathrm{/}4$ such that $\mathrm{\parallel }y-z\mathrm{\parallel }<\delta$ implies $0\le f(y)-f(z)<\eta$. Thus $[f(y)-f(z){]}^{\alpha }\mathrm{/}(c\alpha )<r\mathrm{/}4$, including $f(y)=f(z)$ by the zero-base convention.

Let $\mathrm{\parallel }y-z\mathrm{\parallel }<\delta$ and let $x$ be the maximal solution with $x(0)=y$, with forward interval $[0,T)$, $0<T\le \mathrm{\infty }$. If $\mathrm{\parallel }x(t_1)-z\mathrm{\parallel }\ge r$ for some $t_1<T$, the set $E=\{t\in [0,t_1]:\mathrm{\parallel }x(t)-z\mathrm{\parallel }\ge r\}$ is nonempty and closed by continuity, hence compact. The minimum of the identity function on $E$ gives $\tau >0$. Continuity forces $\mathrm{\parallel }x(\tau )-z\mathrm{\parallel }=r$: a larger value would give an earlier time in $E$. On $[0,\tau ]$ the trajectory stays in $\bar{B}(z,r)\subset W$, so [F5] bounds its length and hence $\mathrm{\parallel }x(\tau )-y\mathrm{\parallel }$ by $[f(y)-f(z){]}^{\alpha }\mathrm{/}(c\alpha )<r\mathrm{/}4$. The triangle inequality gives $\mathrm{\parallel }x(\tau )-z\mathrm{\parallel }<r\mathrm{/}2$, a contradiction. Consequently $x(t)\in B(z,r)$ for $0\le t<T$; applying the same length and displacement estimates to each $[0,t]$, $0<t<T$, gives $\mathrm{\parallel }x(t)-z\mathrm{\parallel }<r\mathrm{/}2$, also true at $t=0$. These restrictions are $C^1$ in the closed-interval convention, since the ODE has continuous right-hand side.

If $T<\mathrm{\infty }$, the entire forward image lies in $K_0=\bar{B}(z,r\mathrm{/}2)\subset U$, which is compact by Heine–Borel. The autonomous compact-continuation theorem applies to this continuous locally Lipschitz field and extends the solution past $T$, contradicting maximality. Therefore $T=\mathrm{\infty }$.

For every $\epsilon >0$ we have obtained $\delta >0$ for which every maximal solution starting within $\delta$ of $z$ exists for all forward time and satisfies $\mathrm{\parallel }x(t)-z\mathrm{\parallel }<r\mathrm{/}2<\epsilon$. In particular this proves every test-ball condition of the local ODE stability definition in [F6], and hence Lyapunov stability. If a complete flow is available, it also proves its usual metric tolerance condition. No attraction to $z$ is asserted.