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

We use the notation of the statement. Algebraic manipulations of dot products use The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite. For $(z,q)\in S$ let $\ell_{z,q}:\mathbb{R}^{n}\to\mathbb{R}$ be given by $\ell_{z,q}(x)=f(z)+q\cdot(x-z)$, and let $\mathcal{F}$ be the set of all functions $\ell_{z,q}$ with $(z,q)\in S$, so that $A(x)=\{g(x):g\in\mathcal{F}\}$ for every $x\in\mathbb{R}^{n}$.

The subgradients are bounded. Since $\bar{B}(y_{0},2r)\subseteq U$ and $f$ satisfies the stated Lipschitz bound on $\bar{B}(y_{0},2r)$, claim 2 of Elementary Calculus of the Subdifferential of a Convex Function gives

$$\lVert q\rVert\le M\qquad\text{for every }(z,q)\in S. \tag{B}$$

Claim 1. The set $U$ is open and convex and $y_{0}\in\bar{B}(y_{0},r)\subseteq U$, so The Subdifferential of a Convex Function on an Open Convex Set is Nonempty (ii) provides some $q\in\partial_{U}f(y_{0})$; hence $(y_{0},q)\in S$ and $S$ is nonempty, and therefore so are $\mathcal{F}$ and each $A(x)$.

Let $x\in\mathbb{R}^{n}$ and $(z,q)\in S$. By The Cauchy-Schwarz Inequality on $\mathbb{R}^n$, claim 3 of Properties of the Absolute Value in an Ordered Field and (B),

$$q\cdot(x-z)\le|q\cdot(x-z)|\le\lVert q\rVert\,\lVert x-z\rVert\le M\bigl(\lVert x-y_{0}\rVert+r\bigr),$$

using claim 6 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and $\lVert z-y_{0}\rVert\le r$. The Lipschitz hypothesis gives $f(z)\le f(y_{0})+M\lVert z-y_{0}\rVert\le f(y_{0})+Mr$. Hence every element of $A(x)$ is at most $f(y_{0})+Mr+M(\lVert x-y_{0}\rVert+r)$, so $A(x)$ is bounded above.

Each $\ell_{z,q}$ is convex on $\mathbb{R}^{n}$: for $x_{1},x_{2}\in\mathbb{R}^{n}$ and $\theta\in\mathbb{R}$ with $0\le\theta\le1$, The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite gives

$$\ell_{z,q}\bigl(\theta x_{1}+(1-\theta)x_{2}\bigr)=f(z)+q\cdot\bigl(\theta x_{1}+(1-\theta)x_{2}-z\bigr)=\theta\,\ell_{z,q}(x_{1})+(1-\theta)\,\ell_{z,q}(x_{2}),$$

so the defining inequality of Convex Real-Valued Function on a Convex Subset of $\mathbb{R}^n$ holds with equality. Claim 4 of Affine Functions, Sums, Nonnegative Multiples and Pointwise Suprema of Convex Functions, applied to the nonempty set $\mathcal{F}$ of convex functions on the convex set $\mathbb{R}^{n}$, therefore shows both that $F$ is well defined, its value at $x$ being the least upper bound of $A(x)$, and that $F$ is convex on $\mathbb{R}^{n}$. This proves claims 1 and 2.

Claim 3. Let $x\in\bar{B}(y_{0},r)$. For $(z,q)\in S$ we have $q\in\partial_{U}f(z)$ and $x\in\bar{B}(y_{0},r)\subseteq U$, so Subdifferential of a Real-Valued Function on a Convex Subset of $\mathbb{R}^n$ (i) gives $f(x)\ge f(z)+q\cdot(x-z)=\ell_{z,q}(x)$. Hence $f(x)$ is an upper bound of $A(x)$ and $F(x)\le f(x)$.

Conversely, $x\in U$ and $U$ is open and convex, so The Subdifferential of a Convex Function on an Open Convex Set is Nonempty (ii) provides $q_{x}\in\partial_{U}f(x)$; then $(x,q_{x})\in S$ and $\ell_{x,q_{x}}(x)=f(x)+q_{x}\cdot(x-x)=f(x)$, so $f(x)\in A(x)$ and $F(x)\ge f(x)$. Therefore $F(x)=f(x)$.

Claim 4. Let $x,x'\in\mathbb{R}^{n}$ and $(z,q)\in S$. By The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, The Cauchy-Schwarz Inequality on $\mathbb{R}^n$, claim 3 of Properties of the Absolute Value in an Ordered Field and (B),

$$\ell_{z,q}(x)=\ell_{z,q}(x')+q\cdot(x-x')\le\ell_{z,q}(x')+M\lVert x-x'\rVert\le F(x')+M\lVert x-x'\rVert .$$

Thus $F(x')+M\lVert x-x'\rVert$ is an upper bound of $A(x)$, so $F(x)\le F(x')+M\lVert x-x'\rVert$. Exchanging $x$ and $x'$ and using $\lVert x'-x\rVert=\lVert x-x'\rVert$, which is claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square with the scalar $-1$, gives $F(x')-F(x)\le M\lVert x-x'\rVert$. By claim 6 of Properties of the Absolute Value in an Ordered Field the two inequalities give $|F(x)-F(x')|\le M\lVert x-x'\rVert$, so $F$ is Lipschitz with constant $M$.