The closed unit ball of c-zero is not dentable

Fix an arbitrary slice with a strict margin. Let $S=S(B,\varphi ,\alpha )$ be a slice. By [L2], write $\varphi (x)={\sum }_n{a}_n{x}_n$. One has ${\sup }_{x\in B}\mathrm{Re}\varphi (x)=\parallel \varphi \parallel$: the upper bound is the dual-norm inequality, while multiplying any almost norming vector by a scalar of modulus one makes its value real and nonnegative. Choose $x\in S$ and set

Construct two points in the slice at distance two. The ${\mathrm{\ell }}^1$ truncation convergence in [L2] implies $a_k\to 0$, so take $k$ with $2\mathrm{\mid }{a}_k\mathrm{\mid }<\delta$. Retain all coordinates of $x$ except put $y_k=1$ and $z_k=-1$. A one-coordinate change preserves convergence to zero, and $\parallel y{\parallel }_{\mathrm{\infty }},\parallel z{\parallel }_{\mathrm{\infty }}\le 1$, so $y,z\in B$. Moreover,

Compute the diameter and deduce nondentability. The triangle inequality bounds the diameter of $B$, and thus of $S$, above by two. Step 2.1 attains two, so every slice has diameter exactly two. In particular no slice has diameter below one, and [L3] says that $B$ is not dentable. The set $B$ is nonempty, bounded, closed, and convex in the Banach space from [L1].

Audit zero, strict-boundary, and complex cases. [L2, L3, step 1.1, step 2.1, step 3.1] If $\varphi =0$, then the slice is all of $B$ and $e_k,-e_k$ are direct witnesses. For nonzero $\varphi$, the positive $\delta$ and strict inequality $2\mathrm{\mid }{a}_k\mathrm{\mid }<\delta$ keep both witnesses inside the slice rather than only on its boundary; a zero remote coefficient is harmless. In the complex case [L2] uses the bilinear pairing and [L3] uses $\mathrm{Re}\varphi$, so no conjugation is inserted. A singleton zero ball is not involved: $c_0$ contains every $e_k$.