For a positive integer $n$ and $p \in [1,\infty]$, let $\mathbb B_{n,p} := \{x \in \mathbb R^n \mid \|x\|_p \le 1\}$ be the $\ell_p$ unit-ball in $\mathbb R^n$. Fix $h \in [0, 1]$.
Question. Of all convex subsets $A$ of $\mathbb B_{n,p}$ with $d(0,A) = h$, which shape maximizes the volume ?
N.B.: $d(0,A) := \inf_{a \in A}\|a\|_p$ is the distance of $A$ from the origin.
Intuitively, the answer to the above question for $p \in \{2,\infty\}$ should be spherical caps, i.e $A=\{x \in \mathbb B_{n,p} \mid x_1 \ge h\}$. I think the question should be solvable using isoperimetric inequalities (e.g Brunn-Minkowski), but I'm not sure how to proceed.
Observation
Without the convexity constraint, a simple symmetry-based argument yields $vol(A) \le vol(Annulus(h,1)) = 1-h^2$ whenever $d(0,A)=h$, where $$ Annulus(h,1) := \{x \in \mathbb R^n \mid h \le \|x\|_p \le 1\} $$
I answered taking $d(0, A):= \sup_{a\in A} \Vert a \Vert_p$. Otherwise, as pointed out in the comments by copper.hat, every set that contains the origin in its closure has zero distance.
The shape $A$ that minimizes the distance to the origin for a given volume is (up to sets of measure zero) a ball. This follows from the observation that if $A$ is not (up to sets of measure zero) a ball, then the ball with radius equal to the $d(0,A)$ has the same distance to the origin, but larger volume as $d(0,A)$ (as $A$ is contained in the ball of radius $d(0,A)$). Thus, I can shrink the radius to get the volume of $A$ and get smaller distance to the origin. However, then $A$ did not have minimal distance to the origin.