Bounding a convex set by ellipses

Question

Given any compact convex set $K$ in $\mathbb{R}^d$ with non-empty interior, does there exists an affine transformation $T$ such that:

$$\overline{B} (0, 1) \subset T(K) \subset d \cdot \overline{B} (0, 1)?$$

The constant $d$ comes from the case of a regular tetrahedron, which intuitively seems to be the worst possible case. If this is false, can we replace $d$ by some function of $d$?

I don't quite know how to proceed or which keywords to use to find references.

Answer 1

Yes, such a $T$ exists, and $d$ is optimal. A proof is in Theorem 4.1 of Kannan, Ravi; Lovász, László; Simonovits, Miklos, Isoperimetric problems for convex bodies and a localization lemma, Discrete Comput. Geom. 13, No. 3-4, 541-559 (1995). ZBL0824.52012. It is described as "folklore".

Theorem. If $K$ is in isotropic position and $B$ is the unit ball about $0,$ then $$\sqrt{\frac{n+2}n}B\subseteq K\subseteq \sqrt{n(n+2)}B$$

(Note the slightly weaker but simpler inequalities $B \subseteq K \subseteq (n + 1)B.$ The inequalities as stated are tight for the regular simplex. [...])

Their $n$ is your $d.$ "Isotropic position" means that $\frac{1}{\operatorname{vol} K}\int_K xdx$ is the zero vector and $\frac{1}{\operatorname{vol} K}\int_{K} xx^Tdx$ is the $n\times n$ identity matrix. It is (mathematically) easy to get your $K$ into isotropic position by an affine transformation. Scaling then gives $B\subseteq \sqrt{\frac n{n+2}} K\subseteq nB.$