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

Each result cited is universally quantified over the data in its own statement. Let $M=I_{n}-B'B$, a real $n\times n$ matrix. By claim 3 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product there are nonnegative reals $\beta,\beta',\gamma$ with $\lVert Bv\rVert\le\beta\lVert v\rVert$, $\lVert B'v\rVert\le\beta'\lVert v\rVert$ and $\lVert Mv\rVert\le\gamma\lVert v\rVert$ for every $v\in\mathbb{R}^{n}$. Put $C=\beta+\beta'+1$.

Step 1: $M$ is small on $E-x$. Let $0<\eta\le1$. By Subgradients near a Point of Twice Differentiability of a Convex Function, and Invariance of the Second-Order Expansion under Lipschitz Truncation §expansion, applied to $f$ at $x$ (first-order coefficient $y$, Hessian $B$) and to $g$ at $y$ (first-order coefficient $x$, Hessian $B'$), there are $\delta_{1},\delta_{2}>0$ such that

Put $\delta_{3}=\min\{\delta_{1},\delta_{2}(\beta+1)^{-1}\}$. Let $x'\in E$ with $\lVert x'-x\rVert<\delta_{3}$, put $h=x'-x$, and take $q\in U'$ with $q\in\partial_{U}f(x')$ and $x'\in\partial_{U'}g(q)$. The first estimate and the triangle inequality give $\lVert q-y\rVert\le(\beta+\eta)\lVert h\rVert\le(\beta+1)\lVert h\rVert<\delta_{2}$ and $\lVert q-y-Bh\rVert\le\eta\lVert h\rVert$. The second estimate with $z=x'$ gives $\lVert h-B'(q-y)\rVert\le\eta(\beta+1)\lVert h\rVert$. Since $B'(q-y)=B'Bh+B'(q-y-Bh)$ (claims 1 and 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product), we get $\lVert h-B'Bh\rVert\le\eta(\beta+1)\lVert h\rVert+\beta'\eta\lVert h\rVert$, that is, using $Mh=h-B'Bh$ (claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum),

Step 2: $M=0$. Let $w\in\mathbb{R}^{n}$ with $w\ne0_{\mathbb{R}^{n}}$, and let $0<\eta\le1$ with $\delta_{3}$ as in Step 1. Since $x$ is a density point of $E$, Near a Density Point Every Direction Meets the Set Closely §nearby with the positive number $\eta$ gives $\rho_{0}>0$ such that every $w'$ with $0<\lVert w'-x\rVert<\rho_{0}$ has some $x'\in E$ with $\lVert x'-w'\rVert\le\eta\lVert w'-x\rVert$. Choose $s>0$ with $s\lVert w\rVert<\min\{\rho_{0},\tfrac12\delta_{3}\}$ and apply this to $w'=x+sw$: there is $x'\in E$ with $\lVert x'-x-sw\rVert\le\eta s\lVert w\rVert$. Then $\lVert x'-x\rVert\le(1+\eta)s\lVert w\rVert\le2s\lVert w\rVert<\delta_{3}$, so by (1) and linearity,

Dividing by $s$, $\lVert Mw\rVert\le(2C+\gamma)\eta\lVert w\rVert$ for every $\eta\in(0,1]$. Given $\varepsilon>0$, the choice $\eta=\min\{1,\varepsilon((2C+\gamma)\lVert w\rVert+1)^{-1}\}$ gives $0\le\lVert Mw\rVert\le\varepsilon$; so $Mw=0_{\mathbb{R}^{n}}$ (Comparison of Real Numbers with Arbitrary Positive Slack §vanishing gives $\lVert Mw\rVert=0$, and claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ gives $Mw=0_{\mathbb{R}^{n}}$); also $M0_{\mathbb{R}^{n}}=0_{\mathbb{R}^{n}}$. Hence $B'Bw=w$ for all $w$. Taking $w=e_{j}$, the $j$th column of $B'B$, whose entries are the coordinates of $(B'B)e_{j}$, is $e_{j}$, the $j$th column of $I_{n}$; so $B'B=I_{n}$.

Step 3: the remaining assertions. By claims 4 and 3 of Elementary Properties of the Transpose of a Real Matrix and symmetry of $B$ and $B'$, $BB'=B^{\top}B'^{\top}=(B'B)^{\top}=I_{n}^{\top}=I_{n}$. Thus each of $B$, $B'$ is invertible with inverse matrix the other. By Alexandrov's Theorem for Semiconvex Functions on an Open Convex Set §hessian-bound with constant $0$ (convex functions being semiconvex with constant $0$ by Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §convexity), $0_{n}\preceq B$ and $0_{n}\preceq B'$, so both are positive semidefinite. If $z\cdot(Bz)=0$, then $Bz=0_{\mathbb{R}^{n}}$ by claim 3 of Cauchy-Schwarz Inequality for a Positive Semidefinite Quadratic Form on $\mathbb{R}^n$, so $z=I_{n}z=B'(Bz)=0_{\mathbb{R}^{n}}$; hence $z\cdot(Bz)>0$ for $z\ne0_{\mathbb{R}^{n}}$ and $B$ is positive definite. The same argument with $BB'=I_{n}$ shows that $B'$ is positive definite. Finally $\det B'\cdot\det B=\det(B'B)=\det I_{n}=1$ by The Determinant is Multiplicative and claim 1 of The Determinant of a Triangular Matrix is the Product of its Diagonal Entries, $I_{n}$ being lower triangular with diagonal entries $1$.