Is a ball the hardest body to approximate by polytopes (in the Banach–Mazur metric)?

Question

$\DeclareMathOperator\conv{conv}\DeclareMathOperator\Vol{Vol}$In the paper "An extremal property of the hypersphere" by Macbeath, the following functionals were introduced (here $n$ is fixed, $N \geq n+1$ and $K \subset \mathbb{R}^{n}$ is a convex body)

$$\psi_N (K) = \sup \Big\{ \frac{\Vol(\conv(x_{i})_{i=1}^{N})}{\Vol(K)}: x_1 ,\dotsc, x_N \in K\Big\}$$

and it was proven that they are all minimized for $K = B$, where $B$ is an Euclidean ball; that is, balls are the hardest to approximate by polytopes, in a volumetric sense, in any dimension. His 1 page argument uses Steiner symmetrizations: this functional decreases under these symmetrizations, and is continuous with respect to the Hausdorff metric. Since one can always choose a sequence of symmetrizations that make $K$ converge to an Euclidean ball, we are done.

I define similar functionals (here I assume that $K$ is centered at the origin):

$$\phi_N (K) = \sup \big\{ r>0: \exists x_1 ,\dotsc, x_N \in K,\quad rK \subset \conv(x_i)_{i=1}^{N} \big\}$$

and ask: do these functionals all minimize for $K=B$ as well? i.e, are balls also the hardest to approximate in the Banach–Mazur distance as well? I expect there to be a silly counterexample to this overly ambitious conjecture, but I couldn't find it.

Answer 1

Let me give a proof of rozu's guess that $\phi_3(D)=1/2$ and $\phi_3(S)=\sqrt{2}-1$ if $D$, $S$ are a disk and a square. This disproves your conjecture.

First of all, for every $K$ and every $N$ the convex hull of every $N$ points in $K$ is contained in the convex hull of $N$ points on the boundary of $K$. Thus, we may suppose that the points $x_1,\dotsc,x_N$ lie on the boundary of $K$.

For a disk $D$, if $x_1x_2x_3$ is a regular triangle, the maximal $r$ such that $rD\subset \triangle x_1x_2x_3$ equals $1/2$. Otherwise, either 0 is not inside $\triangle x_1x_2x_3$ (in this case there is no $r$ for which $rD\subset \triangle x_1x_2x_3$), or there exist indices $i<j$ such that $\angle x_i0x_j>2\pi/3$, and the distance from 0 to $x_ix_j$ is less then $1/2$, so, the maximal $r$ for which $rD\subset \triangle x_1x_2x_3$ is less then $1/2$.

For a square $S=ABCD$, choose Cartesian coordinates so that the vertices of $S$ have coordinates $(\pm 1,\pm 1)$: $A=(-1,-1)$, $B=(1,-1)$ etc. Consider the square $S_1=A_1B_1C_1D_1$, where $A_1=(\sqrt{2}-1)A=(1-\sqrt{2},1-\sqrt{2})$ etc. I claim that $S_1$ is contained in a triangle with the vertices on the boundary of $S$, but is not contained in the interior of any such triangle. This is equivalent to the desired relation $\phi_3(S)=\sqrt{2}-1$.

Consider the points $K=(3-2\sqrt{2},-1)\in AB$, $L=(-1,3-2\sqrt{2})\in AD$. Then $A_1$ is the midpoint of $KL$, the vectors $\overline{KC}=(2\sqrt{2}-2,2)$, $\overline{B_1C}=(2-\sqrt{2},\sqrt{2})$ are collinear, thus, $B_1$ lies on $KC$, analogously $D_1$ lies on $CL$. This shows that $S_1$ is inscribed to the triangle $\triangle CKL$ (three vertices of the square $S_1$ lie on the sides of $\triangle CKL$ and $C_1$ lies inside it).

Now assume that $S_1$ lies in the interior of a triangle $\triangle PQR$ with $P,Q,R$ lying on the boundary of $S$. One of open intervals $AB,BC,CD,DA$ contains none of $P,Q,R$. Without loss of generality, this is $BC$. Then both closed segments $AB$ and $CD$ contain at least one vertex of $\triangle PQR$, otherwise $\triangle PQR$ clearly does not contain $S_1$. Without loss of generality, $P\in CD$, $Q\in AB$. If $R$ belongs to $AB\cup CD$, without loss of generality $R\in CD$, then $\triangle PQR\subset \triangle CQD$. Without loss of generality, $QA\leqslant QB$, then $Q\in [AK]$, and $CQ$ intersects $S_1$, a contradiction.

So, $R\in AD$. Without loss of generality, $CP\leqslant BQ$. Then $P$ lies between the vertical lines $BC$ and $B_1C_1$, as otherwise $B_1$ and $R$ would be on different sides of $PQ$. Let $PQ$, $PR$ intersect the diagonal $BD$ at $U,X$ respectively. The order of points on the line $BD$ is $BUB_1D_1XD$. We have $$DP/QB=DU/UB>DB_1/B_1B=\sqrt{2}+1.$$ Next, if $\overline{DX}=(x,-x)$, then $x<2-\sqrt{2}$ and $x/DP+x/DR=RX/RP+XP/RP=1$, thus $$1/DP+1/DR=1/x>\frac1{2-\sqrt{2}}=\frac{2+\sqrt{2}}2.$$ Analogously, $$1/AQ+1/AR>\frac{2+\sqrt{2}}2.$$ Denote $DP=t$, then $t\in [\sqrt{2},2]$ since $P$ lies between the vertical lines $BC$ and $B_1C_1$. We get $1/DR>\frac{2+\sqrt{2}}2-1/t=\frac{t(2+\sqrt{2})-2}{2t}$. Also, $QB<t/(\sqrt{2}+1)=t(\sqrt{2}-1)$, thus $AQ>2-t(\sqrt{2}-1)$ and $1/AR>\frac{2+\sqrt{2}}2-1/AQ=\frac{2+\sqrt{2}}2-\frac1{2-t(\sqrt{2}-1)}=\frac{2+2\sqrt{2}-\sqrt{2}t}{2(2-t(\sqrt{2}-1))}$. Therefore, $$ 2=AR+DR<\frac{2t}{t(2+\sqrt{2})-2}+\frac{2(2-t(\sqrt{2}-1))} {2+2\sqrt{2}-\sqrt{2}t}, $$ but after simplifications (that's quadratic inequality in $t$ after all) we get that we have equality for $t=\sqrt{2},t=2$ and the opposite inequality for $\sqrt{2}<t<2$, a contradiction.