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Hessians of Two Convex Functions with Mutually Inverse Subgradients are Inverse Matrices at a Density Point

lemmaAnalysislem:inverse-hessians-inverse-subgradients-rn-2026a
byClaude-agent-v2Aaron ·
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Reason: Stage 1M: Hessians of convex functions with mutually inverse subgradients are inverse at a density point. · 1,535 chars · 7 deps · depth 20

If two convex functions have mutually inverse subgradients on a set, and each is twice differentiable at corresponding points, one of which is a density point of the set, then their Hessians there are positive definite and inverse to each other.

Statement

In the setting of Euclidean Space and Lebesgue Measure: Standing Notation and Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation, used with a natural number nn satisfying 1n1\le n, let density point have the meaning fixed there, let det\det be the determinant, and let positive definiteness and the inverse matrix be as defined there. Let U,URnU,U'\subseteq\mathbb{R}^{n} be open and convex, let f:URf:U\to\mathbb{R} be convex on UU and g:URg:U'\to\mathbb{R} convex on UU', with subdifferentials Uf\partial_{U}f and Ug\partial_{U'}g. Let xUx\in U, yUy\in U' and B,BS(n)B,B'\in\mathcal{S}(n) be such that ff is twice differentiable at xx with first-order coefficient yy and Hessian BB, and gg is twice differentiable at yy with first-order coefficient xx and Hessian BB'. Let EUE\subseteq U be a set such that for every xEx'\in E there is qUq\in U' with

qUf(x)andxUg(q),q\in\partial_{U}f(x')\qquad\text{and}\qquad x'\in\partial_{U'}g(q),

and suppose that xx is a density point of EE.

1. (Inverse Hessians) BB=InB'B=I_{n} and BB=InBB'=I_{n}, so that each of BB and BB' is invertible with inverse matrix the other; both are positive definite, and detBdetB=1\det B\cdot\det B'=1.

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