A First-Order Expansion of the Subdifferential Gives a Second-Order Expansion of the Function

We work in the setting of Euclidean Space and Lebesgue Measure: Standing Notation, whose notation is fixed for every dimension and is used here with a natural number $n$ satisfying $1\le n$: the real numbers, natural numbers and finite index sets, and the Euclidean norm $\lVert\,\cdot\,\rVert$, dot product and notion of openness, are as fixed there. Write $\mathcal{S}(n)$ for the set of symmetric real $n\times n$ matrices, $Mh$ for the matrix-vector product, $M^{\top}$ for the transpose, and $\tfrac{1}{2}(M+M^{\top})$ for the corresponding scalar multiple of the sum.

Let $f:\mathbb{R}^{n}\to\mathbb{R}$ be convex on $\mathbb{R}^{n}$, which is a convex subset of itself, and let $\partial f=\partial_{\mathbb{R}^{n}}f$ be its subdifferential. Let $y\in\mathbb{R}^{n}$, let $p\in\partial f(y)$, and let $M$ be a real matrix with $n$ rows and $n$ columns. Put $S=\tfrac{1}{2}(M+M^{\top})$, which belongs to $\mathcal{S}(n)$. Then the following hold.

1. (Telescoped expansion) ¶ Let $\varepsilon,\delta\in\mathbb{R}$ with $0<\varepsilon$ and $0<\delta$, and suppose that

for every $y'\in\mathbb{R}^{n}$ with $\lVert y'-y\rVert<\delta$ and every $q\in\partial f(y')$. Then every $h\in\mathbb{R}^{n}$ with $\lVert h\rVert<\delta$ satisfies

2. (Twice differentiability) ¶ Suppose that for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\delta\in\mathbb{R}$ with $0<\delta$ such that the hypothesis of claim 1 holds for that pair $\varepsilon,\delta$. Then $f$ is twice differentiable at $y$ with first-order coefficient $p$ and Hessian $S$.