Maximizing ratio volume/diameter^n by an affinity

Question

Suppose we have a convex compact body $D\subset \mathbb R^n$. We can try to apply affine transformation keeping the volume and decreasing the diameter of $D$.

It is clear that there is a constant $\lambda_n$ such that for any $D$ there is an affinity $F$ such that $diamater(F(D))^n\leq\lambda_n Volume(F(D))$. I'm interesting in the optimal value for $\lambda_n$.

The article "On the thinnest non-separable lattice of convex bodies" (E. Makai, p.23) gives an estimate $\lambda_n\leq (_n^{2n})n^{n/2}/k_n$ where $k_n$ is the volume of the convex hull of the unit sphere and $(\pm\sqrt{n},0,0,..)$, $\lambda_2$ is also computed.

I can not find in literature any better estimate for $\lambda_3$, but it seems that I can prove a few better estimate by school-like methods: if there is no affinity decreasing diameter, that means that in $D$ there are a lot of diameters which will increase if we apply an infinitesimal affinity, so we get an estimate. I'm sure that I'm not the first one who apply such a simple idea to this problem. Do you know any others references concerning this problem?

Answer 1

A closely related problem, considering the $(n-1)$-dimensional surface area instead of the diameter of the body, has been solved by Keith Ball, Volume ratios and a reverse isoperimetric inequality. J. London Math. Soc. (2) 44 (1991), no. 2, 351–359 [MR1136445]. The extreme case, as expected, is the $n$-dimensional simplex. Ball's proof may also work for the diameter. The article is available on the arXiv, see http://arxiv.org/abs/math/9201205 . The 2-dimensional case was solved long before, with a very short, elementary proof.

A quick estimate of the diameter can be obtained by taking the minimum volume ellipsoid containing the body $K$ and using the known (best possible, by the way) estimate of the ellipsoid's volume. An affine transformation that turns the ellipsoid into a ball yields a bound on the diameter - perhaps not the best possible, but a fairly good one.