Dentable bounded set and slice

Let $C$ be a nonempty bounded subset of a real or complex Banach space $X$. For $x^{\ast }\in X^{\ast }$ and $\alpha >0$, the slice of $C$ determined by $(x^{\ast },\alpha )$ is

$S(C,x^{\ast },\alpha ):=\{x\in C:\mathrm{Re}{x}^{\ast }(x)>{\sup }_{y\in C}\mathrm{Re}{x}^{\ast }(y)-\alpha \}\mathrm{.}$

The real part is omitted over the real field. Boundedness of $C$ and continuity of $x^{\ast }$ make the displayed supremum finite, and its defining property makes the slice nonempty.

The norm diameter of $D\subseteq X$ is $\mathrm{diam}D:=\sup \{\mathrm{\parallel }x-y\mathrm{\parallel }:x,y\in D\}$, with diameter $0$ for a singleton. The set $C$ is dentable if for every $\epsilon >0$ it has a slice $S(C,x^{\ast },\alpha )$ with diameter less than $\epsilon$.

• If $C$ has diameter zero, the zero functional gives the whole set as a slice, so $C$ is dentable. This includes the singleton unit ball of the zero Banach space.
• If $C$ has positive diameter and a slice is smaller than $C$, its defining functional is necessarily nonzero. Thus the zero functional introduces no spurious nondegenerate denting.
• Strict inequality and $\alpha >0$ ensure that endpoint nonattainment of the supremum does not make a slice empty.