Proof of Local Lipschitz Bound and Continuity for a Semiconvex Function on an Open Convex Set

Throughout, $G:U\to\mathbb{R}$ denotes the function given by

$$G(z)=\varphi(z)+\frac{\mu}{2}\,\lVert z\rVert^{2},$$

which is convex on $U$ precisely because $\varphi$ is semiconvex on $U$ with constant $\mu$.

Proof of claim 1. Let $x_{0}\in U$. The set $U$ belongs to the topology of The Open Subsets of a Metric Space Form a Topology and satisfies $x_{0}\in U$ and $U\subseteq U$, so $x_{0}$ is an interior point of $U$ in $\mathbb{R}^{n}$ by Interior of a Subset of a Topological Space. Since $U$ is convex and $G$ is convex on $U$, A Convex Function is Lipschitz on a Ball around an Interior Point supplies $\rho,L_{0}\in\mathbb{R}$ with $0<\rho$ and $0\le L_{0}$ such that $\bar{B}_{d_{E}}(x_{0},\rho)\subseteq U$ and

$$\bigl|G(y)-G(x)\bigr|\le L_{0}\,\lVert y-x\rVert\qquad\text{for all }x,y\in\bar{B}_{d_{E}}(x_{0},\rho).$$

Put $R=\lVert x_{0}\rVert+\rho$, a nonnegative real number. If $x\in\bar{B}_{d_{E}}(x_{0},\rho)$ then $\lVert x-x_{0}\rVert=d_{E}(x_{0},x)\le\rho$ by Closed Ball in a Metric Space and claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, whence by the triangle inequality (claim 6 there)

$$\lVert x\rVert=\lVert x_{0}+(x-x_{0})\rVert\le\lVert x_{0}\rVert+\lVert x-x_{0}\rVert\le R .$$

Next, for arbitrary $x,y\in\mathbb{R}^{n}$ the triangle inequality gives $\lVert y\rVert\le\lVert x\rVert+\lVert y-x\rVert$ and $\lVert x\rVert\le\lVert y\rVert+\lVert x-y\rVert$, while $\lVert x-y\rVert=\lVert(-1)(y-x)\rVert=\lVert y-x\rVert$ by claim 5 there; hence

$$\bigl|\,\lVert y\rVert-\lVert x\rVert\,\bigr|\le\lVert y-x\rVert .$$

Consequently, for $x,y\in\bar{B}_{d_{E}}(x_{0},\rho)$, using $0\le\lVert x\rVert\le R$ and $0\le\lVert y\rVert\le R$,

$$\bigl|\,\lVert y\rVert^{2}-\lVert x\rVert^{2}\,\bigr|=\bigl|\,\lVert y\rVert-\lVert x\rVert\,\bigr|\cdot\bigl(\lVert y\rVert+\lVert x\rVert\bigr)\le 2R\,\lVert y-x\rVert .$$

Since $\varphi(z)=G(z)-\tfrac{\mu}{2}\lVert z\rVert^{2}$ for $z\in U$, the triangle inequality for the absolute value gives, for all $x,y\in\bar{B}_{d_{E}}(x_{0},\rho)$,

$$\bigl|\varphi(y)-\varphi(x)\bigr|\le\bigl|G(y)-G(x)\bigr|+\frac{\mu}{2}\bigl|\,\lVert y\rVert^{2}-\lVert x\rVert^{2}\,\bigr|\le\bigl(L_{0}+\mu R\bigr)\lVert y-x\rVert .$$

So claim 1 holds with this $\rho$ and with $L=L_{0}+\mu R$, which is nonnegative.

Proof of claim 2. Let $S\subseteq U$, let $x\in S$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Apply claim 1 at the point $x_{0}=x$ of $U$ to obtain $\rho,L\in\mathbb{R}$ with $0<\rho$, $0\le L$, $\bar{B}_{d_{E}}(x,\rho)\subseteq U$ and

$$\bigl|\varphi(w)-\varphi(z)\bigr|\le L\,\lVert w-z\rVert\qquad\text{for all }z,w\in\bar{B}_{d_{E}}(x,\rho).$$

Since $0\le L$ we have $0<L+1$, so $\varepsilon/(L+1)$ is a positive real number; let $\delta$ be the smaller of $\rho$ and $\varepsilon/(L+1)$, a positive real number.

Let $y\in S$ satisfy $\lVert y-x\rVert<\delta$. Then $d_{E}(x,y)=\lVert y-x\rVert<\delta\le\rho$, so $y\in\bar{B}_{d_{E}}(x,\rho)$; also $x\in\bar{B}_{d_{E}}(x,\rho)$ by claim 1 of Elementary Properties of the Closed Ball in a Metric Space. Therefore, using $0\le L$ and $\lVert y-x\rVert<\delta$,

$$\bigl|\varphi(y)-\varphi(x)\bigr|\le L\,\lVert y-x\rVert\le L\,\delta\le\frac{L}{L+1}\,\varepsilon<\varepsilon,$$

the last inequality because $0\le L<L+1$ gives $L/(L+1)<1$ and $0<\varepsilon$. This is the assertion of claim 2.