Proof of A Convex Function is Continuous near an Interior Point

Let $\rho$ and $L$ be as provided by A Convex Function is Lipschitz on a Ball around an Interior Point, so that $0<\rho$, $0\le L$, the closed ball $B=\bar{B}_{d_{E}}(x_{0},\rho)$ is contained in $C$, and

By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ we have $d_{E}(x,y)=\lVert x-y\rVert$, and by claim 5 of that lemma, applied with the factor $-1$, $\lVert x-y\rVert=\lVert y-x\rVert$; so the displayed bound reads $|u(y)-u(x)|\le L\,d_{E}(x,y)$.

Let $x\in B$ and let $\eta\in\mathbb{R}$ with $0<\eta$. We exhibit a $\theta\in\mathbb{R}$ with $0<\theta$ such that every $y\in B$ with $d_{E}(x,y)<\theta$ satisfies $d_{\mathbb{R}}(u(y),u(x))<\eta$; by Continuous Map Between Metric Spaces this makes the restriction of $u$ to $B$ continuous at $x$ relative to $B$, and as $x\in B$ is arbitrary, continuous on $B$.

Suppose first that $L=0$. Take $\theta=1$, which is positive by claim 6 of Elementary Order Arithmetic in an Ordered Field. For every $y\in B$ the bound above and Zero Products and Elementary Identities in a Field give

Suppose now that $L\ne0$, so that $0<L$ and $L$ has an inverse $L^{-1}$ with $0<L^{-1}$, by claim 7 of Elementary Order Arithmetic in an Ordered Field. Take $\theta=\eta\,L^{-1}$, which is positive by claim 5 of that lemma. Let $y\in B$ with $d_{E}(x,y)<\theta$. Multiplying this strict inequality by the positive number $L$, by claim 10 of Elementary Order Arithmetic in an Ordered Field,

and combining with $|u(y)-u(x)|\le L\,d_{E}(x,y)$ by the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field,