TheoremBase

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

A convex function on an open convex subset of Euclidean space has a subgradient at every point, obtained from a supporting hyperplane to a bounded piece of its epigraph; and a subgradient inequality valid on one closed ball around the point is automatically valid on the whole set.

Statement

We work in the setting of Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background, whose notation is fixed for every dimension and is used here in the dimensions nn and n+1n+1 for a natural number nn with 1≤n1\le n: the real numbers and sequences, and the Euclidean norm ∥ ⋅ ∥\lVert\,\cdot\,\rVert, dot product, distance dEd_{E}, topology, the notions of open, closed, bounded and compact subsets, and the closed balls Bˉ(x,r)\bar{B}(x,r), are as fixed there. Let ι:Rn×R1→Rn+1\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 R1\mathbb{R}^{1} with its single coordinate, so that we may write ι(z,t)\iota(z,t) for z∈Rnz\in\mathbb{R}^{n} and t∈Rt\in\mathbb{R}.

Let U⊆RnU\subseteq\mathbb{R}^{n} be open and convex, let f:U→Rf:U\to\mathbb{R} be convex on UU, and let y∈Uy\in U. The subdifferential ∂Uf(y)\partial_{U}f(y) is as defined there. Then the following hold.

1. (A subgradient inequality on one ball holds on all of UU) Let p∈Rnp\in\mathbb{R}^{n} and let r∈Rr\in\mathbb{R} with 0<r0<r and Bˉ(y,r)⊆U\bar{B}(y,r)\subseteq U, and suppose that

f(z)≥f(y)+p⋅(z−y)for every z∈Bˉ(y,r).f(z)\ge f(y)+p\cdot(z-y)\qquad\text{for every }z\in\bar{B}(y,r).

Then p∈∂Uf(y)p\in\partial_{U}f(y).

2. (Nonemptiness) ∂Uf(y)\partial_{U}f(y) is nonempty.

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