Strict second-order positivity on the critical cone gives a strict local minimum

Write $Q(d)=d^T{\mathrm{\nabla }}_{xx}^{2}L(x_{\ast },\nu ,\lambda )d$ and $S=\{d:\mathrm{\parallel }d{\mathrm{\parallel }}_2=1\}$. The critical cone intersected with $S$ is closed in $S$ by its linear equalities and inequalities. If this intersection is nonempty, it is compact by [F2], and [F2] gives a minimum $\gamma =\min Q(d)>0$ there. The relatively open set $N=\{u\in S:Q(u)>\gamma \mathrm{/}2\}$ contains that intersection. If the intersection is empty, put $N=\mathrm{\varnothing }$.

For $k\ge 1$, let $D_k$ be the set of unit directions $u=(x-x_{\ast })\mathrm{/}\mathrm{\parallel }x-x_{\ast }{\mathrm{\parallel }}_2$ for feasible $x\mathrm{\ne }{x}_{\ast }$ with $\mathrm{\parallel }x-x_{\ast }{\mathrm{\parallel }}_2<1\mathrm{/}k$ and $f(x)\le f(x_{\ast })$. If $D_k$ is empty for some $k$, there is no such point in that neighborhood and strict local minimality holds. Otherwise, if bad points occur arbitrarily close with directions outside $N$, the nonempty nested closed sets $\overline{D_k\setminus N}$ have the finite-intersection property in the compact sphere $S$. By [F3] choose $d\in {\bigcap }_k\overline{D_k\setminus N}$; since $N$ is relatively open, $d\mathrm{\notin }N$ and $\mathrm{\parallel }d{\mathrm{\parallel }}_2=1$.

For each $k$, the closure property in step 1.2 supplies a direction $u$ with $\mathrm{\parallel }u-d{\mathrm{\parallel }}_2<1\mathrm{/}k$ from a feasible bad point $x=x_{\ast }+tu$, where $0<t<1\mathrm{/}k$ and $f(x)\le f(x_{\ast })$. Taylor expansion [F4] and equality feasibility imply $Dh(x_{\ast })u=O(t)$; active inequality feasibility gives $D{g}_i(x_{\ast })u\le O(t)$ for every active $i$. The objective inequality gives $\mathrm{\nabla }f(x_{\ast }{)}^{T}u\le O(t)$. Letting $k$ grow in these bounds, using $\mathrm{\parallel }u-d{\mathrm{\parallel }}_2<1\mathrm{/}k$ and $t<1\mathrm{/}k$, shows $Dh(x_{\ast })d=0$, $D{g}_i(x_{\ast })d\le 0$ on the active set, and $\mathrm{\nabla }f(x_{\ast }{)}^{T}d\le 0$. KKT stationarity and [F1] also give $\mathrm{\nabla }f(x_{\ast }{)}^{T}d\ge 0$, so $d$ is a unit critical direction. This contradicts $d\mathrm{\notin }N$: if the critical unit set is empty there is no such $d$, and otherwise every such $d$ belongs to $N$. Hence for some $k_0$, every feasible bad point with $0<\mathrm{\parallel }x-x_{\ast }{\mathrm{\parallel }}_2<1\mathrm{/}{k}_0$ has direction in $N$.

If the critical unit set is empty, step 2.1 already excludes all bad points sufficiently near $x_{\ast }$, proving strict local minimality. Otherwise, let $\gamma >0$ be as in step 1.1. At a feasible $x=x_{\ast }+tu$, KKT gives $L(x_{\ast },\nu ,\lambda )=f(x_{\ast })$, ${\mathrm{\nabla }}_{x}L(x_{\ast },\nu ,\lambda )=0$, and $L(x,\nu ,\lambda )\le f(x)$. For bad points sufficiently close, step 2.1 gives $u\in N$, so $Q(u)>\gamma \mathrm{/}2$. The Taylor expansion [F4] of the $C^2$ Lagrangian has a remainder $o(t^2)$, uniformly as $x\to x_{\ast }$; after shrinking the neighborhood it yields $L(x,\nu ,\lambda )>f(x_{\ast })$. This contradicts $L(x,\nu ,\lambda )\le f(x)\le f(x_{\ast })$. Therefore no distinct feasible point sufficiently near $x_{\ast }$ has objective at most $f(x_{\ast })$, proving that $x_{\ast }$ is a strict local minimizer.