Proof of Subgradients near a Point of Twice Differentiability of a Convex Function, and Invariance of the Second-Order Expansion under Lipschitz Truncation

Each result cited is universally quantified over the data in its own statement. Throughout, $\beta=\lVert B\rVert$ is the norm of $B\in\mathcal{S}(n)$, so that $|k\cdot(Bk)|\le\beta\lVert k\rVert^{2}$ for every $k\in\mathbb{R}^{n}$ by claim 2 of Properties of the Norm of a Symmetric Real Matrix, and $0\le\beta$ by claim 1 there.

Claim 1. By Basic Properties of Twice Differentiability at a Point §gradient, $f$ is differentiable at $y$ with derivative matrix the one-row matrix whose entry in column $i$ is the $i$th coordinate of $p$. Since $U$ is open and convex and $f$ is convex on $U$, Elementary Calculus of the Subdifferential of a Convex Function §gradient gives $\partial_{U}f(y)=\{g\}$ for the point $g$ whose $i$th coordinate is that entry, that is, $g=p$ (points of $\mathbb{R}^{n}$ are equal when their coordinates are). Hence $\partial_{U}f(y)=\{p\}$.

Claim 2. Let $\varepsilon>0$. Put $t=\min\{1,\varepsilon(\beta+1)^{-1}\}$, which satisfies $0<t\le1$ and $\beta t\le\varepsilon$, and $\eta=\tfrac{1}{10}\varepsilon t>0$. By the definition of twice differentiability applied with $\eta$ there is $\delta_{0}>0$ such that every $h$ with $\lVert h\rVert<\delta_{0}$ satisfies $y+h\in U$ and

Put $\delta=\tfrac12\delta_{0}$. Let $y'\in\mathbb{R}^{n}$ with $\lVert y'-y\rVert<\delta$ and put $h=y'-y$; then $y'=y+h\in U$ since $\lVert h\rVert<\delta_{0}$. Let $q\in\partial_{U}f(y')$ and $w=q-p-Bh$. If $h=0_{\mathbb{R}^{n}}$, then $y'=y$ and $q=p$ by claim 1, so $w=0_{\mathbb{R}^{n}}$ (as $B0_{\mathbb{R}^{n}}=0_{\mathbb{R}^{n}}$ by claim 1 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product) and the inequality holds. If $w=0_{\mathbb{R}^{n}}$ there is nothing to prove. Otherwise $h\ne0_{\mathbb{R}^{n}}$ and $w\ne0_{\mathbb{R}^{n}}$; put $e=\lVert w\rVert^{-1}w$ and $k=t\lVert h\rVert e$, so that $\lVert k\rVert=t\lVert h\rVert$ and $w\cdot k=t\lVert h\rVert\lVert w\rVert$. Then $\lVert h+k\rVert\le(1+t)\lVert h\rVert\le2\lVert h\rVert<\delta_{0}$, so $y+h+k\in U$, and the subgradient inequality of Subdifferential of a Real-Valued Function on a Convex Subset of $\mathbb{R}^n$ §subdifferential at $y'$ gives $f(y+h+k)\ge f(y+h)+q\cdot k$. Applying (1) at $h+k$ (upper bound) and at $h$ (lower bound), and expanding $(h+k)\cdot(B(h+k))=h\cdot(Bh)+2(Bh)\cdot k+k\cdot(Bk)$ by linearity of the matrix-vector product (claim 1 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product), bilinearity and symmetry of the dot product (Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$), and the identity $h\cdot(Bk)=k\cdot(Bh)$, which holds because both sides equal $\sum_{i,j\in[n]}h_{i}B_{ij}k_{j}$ by the double-sum formula of claim 4 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and $B_{ij}=B_{ji}$, we get

Hence $w\cdot k\le\tfrac12\beta t^{2}\lVert h\rVert^{2}+5\eta\lVert h\rVert^{2}$, that is, $t\lVert h\rVert\lVert w\rVert\le(\tfrac12\beta t^{2}+5\eta)\lVert h\rVert^{2}$. Dividing by the positive number $t\lVert h\rVert$,

Claim 3. By claim 1, $\partial_{U}f(y)=\{p\}$, so Elementary Calculus of the Subdifferential of a Convex Function §continuity applied with the positive number $L-\lVert p\rVert$ gives $\delta_{1}>0$ such that every $y'\in U$ with $\lVert y'-y\rVert<\delta_{1}$ and every $q\in\partial_{U}f(y')$ satisfy $\lVert q-p\rVert<L-\lVert p\rVert$, hence $\lVert q\rVert\le\lVert q-p\rVert+\lVert p\rVert<L$ by the triangle inequality (claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$). Since $U$ is open there is $r_{0}>0$ with every $y'$ satisfying $\lVert y'-y\rVert<r_{0}$ in $U$; put $r=\min\{r_{0},\delta_{1}\}$. Let $\lVert y'-y\rVert<r$. Then $y'\in U$, $\partial_{U}f(y')$ is nonempty by The Subdifferential of a Convex Function on an Open Convex Set is Nonempty, and any $q$ in it satisfies $\lVert q\rVert<L$, so $f^{L}(y')=f(y')$ by The Lipschitz Truncation of a Convex Function: a Global Lipschitz Convex Minorant Agreeing with It Where the Slope is Small §agreement.

Finally let $\varepsilon>0$, let $\delta$ be as in the definition of twice differentiability of $f$ at $y$ with this $\varepsilon$, and put $\delta'=\min\{\delta,r\}$. For $\lVert h\rVert<\delta'$ we have $y+h\in\mathbb{R}^{n}$, $f^{L}(y+h)=f(y+h)$ and $f^{L}(y)=f(y)$, so

Since $\mathbb{R}^{n}$ is open and convex and $f^{L}$ is defined on it, $f^{L}$ is twice differentiable at $y$ with first-order coefficient $p$ and Hessian $B$.