Consider a convex, open, non-empty set $B \subset \mathbb{R}^k$ and a function $V: B\rightarrow \mathbb{R}$ convex (and, hence, continuous) in $B$. Consider $\epsilon \in \mathbb{R}$, $\epsilon>0$. Consider the hyper-cube $K\subset B$.
Statement: There is a $\delta>0$ such that $V(\cdot)$ varies by less than $\epsilon$ over each cube of side $2\delta$ that intersects $K$.
Could you help me to prove the statement? I don't get the following things:
(1) I know that by continuity of $V$, $\forall \tilde{u} \in B$, there exists $\delta_{\tilde{u}}>0$ such that $\forall u \in B$ with $||u-\tilde{u}||\leq \delta_{\tilde{u}}$ we have $|V(u)-V(\tilde{u})|\leq \epsilon $.
(2) I understand that we can pick $\tilde{u}$ is a way such that the $k$-sphere $\{ u \in B \text{ s.t. } ||u-\tilde{u}||\leq \delta_{\tilde{u}}\}$ intersects $K$.
(3) I don't get why the statement talks about "each cube". Is there some uniform continuity that I don't get? And why cube instead of sphere?
The continuous function $V$ is uniformly continuous on $K$ since this is a compact set. Note that some cubes of side $2\delta$ are not contained in $K$ but the closure of their union is also compact, and thus you also have uniform continuity on that closure.
Nothing changes if you replace "each cube" by "each sphere".