The Subdifferential of a Convex Function on an Open Convex Set is Nonempty

We work in the setting of Euclidean Space and Lebesgue Measure: Standing Notation, whose notation is fixed for every dimension and is used here in the dimensions $n$ and $n+1$ for a natural number $n$ with $1\le n$: the real numbers and sequences, and the Euclidean norm $\lVert\,\cdot\,\rVert$, dot product, distance $d_{E}$, topology, the notions of open, closed, bounded and compact subsets, and the closed balls $\bar{B}(x,r)$, are as fixed there. Let $\iota:\mathbb{R}^{n}\times\mathbb{R}^{1}\to\mathbb{R}^{n+1}$ be the concatenation map, a linear bijection under which the dot product and the norm split coordinatewise; as in A Lipschitz Function on an Open Interval is Differentiable Almost Everywhere we identify a point of $\mathbb{R}^{1}$ with its single coordinate, so that we may write $\iota(z,t)$ for $z\in\mathbb{R}^{n}$ and $t\in\mathbb{R}$.

Let $U\subseteq\mathbb{R}^{n}$ be open and convex, let $f:U\to\mathbb{R}$ be convex on $U$, and let $y\in U$. The subdifferential $\partial_{U}f(y)$ is as defined there. Then the following hold.

1. (A subgradient inequality on one ball holds on all of $U$) ¶ Let $p\in\mathbb{R}^{n}$ and let $r\in\mathbb{R}$ with $0<r$ and $\bar{B}(y,r)\subseteq U$, and suppose that

Then $p\in\partial_{U}f(y)$.

2. (Nonemptiness) ¶ $\partial_{U}f(y)$ is nonempty.