Extending a Convex Function from a Closed Ball to All of $\mathbb{R}^n$

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 with a natural number $n$ satisfying $1\le n$: the real numbers and least upper bounds, the Euclidean norm $\lVert\,\cdot\,\rVert$, dot product, distance $d_{E}$, notion of openness and closed balls $\bar{B}(x,r)$, and the convention that Lipschitz maps are understood for the restricted Euclidean distances, are as fixed there.

Let $U\subseteq\mathbb{R}^{n}$ be open and convex, let $f:U\to\mathbb{R}$ be convex on $U$ with subdifferential $\partial_{U}f$, let $y_{0}\in U$, and let $r,M\in\mathbb{R}$ with $0<r$ and $0\le M$ be such that $\bar{B}(y_{0},2r)\subseteq U$ and

$$|f(z)-f(w)|\le M\,\lVert z-w\rVert\qquad\text{for all }z,w\in\bar{B}(y_{0},2r).$$

Let

$$S=\bigl\{\,(z,q)\;:\;z\in\bar{B}(y_{0},r)\ \text{ and }\ q\in\partial_{U}f(z)\,\bigr\},$$

and for $x\in\mathbb{R}^{n}$ let $A(x)=\{\,f(z)+q\cdot(x-z)\;:\;(z,q)\in S\,\}\subseteq\mathbb{R}$. Then the following hold.

1. (The extension is well defined) (i) The set $S$ is nonempty, and for every $x\in\mathbb{R}^{n}$ the set $A(x)$ is nonempty and bounded above. We may therefore define $F:\mathbb{R}^{n}\to\mathbb{R}$ by letting $F(x)$ be the least upper bound of $A(x)$.

2. (Convexity) (ii) $F$ is convex on $\mathbb{R}^{n}$.

3. (Agreement on the ball) (iii) $F(x)=f(x)$ for every $x\in\bar{B}(y_{0},r)$.

4. (Lipschitz bound) (iv) $F$ is Lipschitz with constant $M$ on $\mathbb{R}^{n}$; that is, $|F(x)-F(x')|\le M\lVert x-x'\rVert$ for all $x,x'\in\mathbb{R}^{n}$.