Let $\{K_i\}$ be a family of smooth, origin-symmetric, strictly convex bodies such that $K_i$ converge in the Hausdorff distance (or you may assume $\partial K_i\to \partial K$ smoothly, in the sense of differential geometry, where $\partial K$ stands for the boundary hypersurface of $K$) to a smooth, strictly convex body $K.$

The Banach-Mazur distance of an origin-symmetric convex body $L$ from the unit ball of $\mathbb{R}^n$, $B$, is defined as follows $$d(L,B)=\min\{r; B\subseteq\Phi L\subseteq rB, \Phi\in GL(n)\}.$$

Here, $GL(n)$ is the group of linear transformations.

Is there a family of linear transformations $A_i\in GL(n)~$ for $i=1,2,\cdots,\infty$ (including $i=\infty$), such that for each $A_i$, $B\subseteq A_i K_i\subseteq d(K_i,B)B$ and $A_i\to A_{\infty}$ uniformly, or at least $\det A_i\to \det A_{\infty}$.Each of these claim implies that the inradius of $(A_iK_i) 1/(detA_i)$ converge to the inradius of$ (A_{\infty}K_{\infty})1/(detA_{\infty})$.

  • $\begingroup$ There is a problem with the definition of BM distance: as stated, the distance from $B$ to itself is 1. Should $r$ in the definition be replaced with $1+r$ or $e^r$? $\endgroup$ Sep 27, 2015 at 5:47
  • $\begingroup$ @VictorProtsak: $\log r$ is a distance in the usual sense, however for this case it is usual to leave the $r$ as it is, so for example, the triangle inequality is $d(K,L)\leq d(K,J)\,d(J,L)$. $\endgroup$ Sep 27, 2015 at 12:59

1 Answer 1


This got more complicated than what I thought, and the type of convergence (Hausdorff versus smooth) may be important. I have a non-smooth counter-example, did you already find that? It seems that it can be modified to obtain smooth sets, but the convergence is only in the Hausdorff distance. Perhaps a stronger convergence, say, $\partial K_i\to \partial K$ as $C^2$ hypersurfaces, may be enough to establish what you want, but that may be complicated to do.

Definition: We say that a transformation $A\in GL(n)$ places $K$ in optimum (or Banach-Mazur) position (or does the job) iff $$B\subseteq AK\subseteq d(K,B)B.$$

Remark: Observe that this fact depends on $A$ only up to left multiplication by orthogonal transformations, that is, if $A$ does the job, then so does any $UA$ where $U$ is orthogonal. (This is a consequence of the fact that $B\subseteq K$ iff $UB\subseteq UK$ iff $B\subseteq UK$ because $UB=B$. Similarly, we have $K\subseteq rB$ iff $UK\subseteq rB$.) Since any $A$ has a unique polar decomposition $A=UP$ where $U$ is orthogonal and $P$ is symmetric and positive semidefinite, we can discard the $U$ and consider only operators $A=P$ that are symmetric and positive.

Now, an operator $P$ that places $K$ in Banach-Mazur position is not uniquely determined by $K$ as shows the following

Example: In $\mathbb R^3$ take $K$ the convex hull of $B\cup\{(\pm 1,\pm 1, 0)\}$. This set has $d(K,B)=\sqrt 2$, and any $P_t=\mathrm{diag}(1,1,t)$ for $t\in[1,\sqrt 2]$ will place $K$ in optimum position. The fact that $d(K,B)\geq\sqrt 2$ follows by looking at the intersection of $K$ with the $xy$-plane, which is a square. Here there's a gap: I'm not proving that the operators $P_t$ are the only positive operators $P$ that do the job.

Now, it is possible to slightly grow $K$ to obtain a body $K_{+,\epsilon}$ that has the same Banach-Mazur distance to $B$, in a way that blocks all the operators $P_t$ but the lowest $P_1$ from doing their job, and there is another way to slightly shrink $K$ to a body $K_{-,\epsilon}$ so that all the $P_t$ but the highest $P_{\sqrt 2}$ are blocked. Our modifications won't affect the intersection of $K$ with the $xy$ plane, and that's why the Banach-Mazur distance to $B$ is not decreased. The fact that some of the $P_t$ still place $K$ between $B$ and $\sqrt 2B$ show that the Banach-Mazur is not increased either.

The modifications are as follows: to obtain $K_{+,\epsilon}$ for some $\epsilon>0$, consider the circle $C=\{(x,y,z):x^2+y^2+z^2=\sqrt 2\textrm{ and }x+y=0\}$ and let $C_\epsilon=\{(x,y,z)\in C:|z|\leq\epsilon\}$. Let $K_{+,\epsilon}$ be the convex hull of $K\cup C_\epsilon$. Observe that $P_t(C_\epsilon)\subseteq\sqrt B$ iff $t=1$, as we wanted.

To obtain $K_{-,\epsilon}$, let $p_\epsilon=(\cos\epsilon, 0,\sin\epsilon)$, and let $$K_{-,\epsilon}=\{(x,y,z)\in K:|\langle P_{\sqrt 2}(x,y,z),p_\epsilon\rangle|\leq 1\}.$$ Observe that $P_tK_{-,\epsilon}\supseteq B$ iff $t=\sqrt 2$, since the point $P_{\sqrt 2}^{-1}p_\epsilon$ is on the border of $K_{-,\epsilon}$ and is only thrown outside the interior of $B$ by $P_t$ when $t=\sqrt 2$.

Finally, we can construct the sequence $K_i:=K_{(-1)^i,\epsilon_i}$ for some sequence $0<\epsilon_i\to 0$, so that $K_i\to K$ in the Hausdorff distance. The respective unique positive operators $P_i$ that place $K_i$ in optimum position alternate between $P_1$ and $P_\sqrt 2$, with respective determinant 1 and $\sqrt 2$. Allowing general operators $A_i=U_iP_i$ with $U_i$ orthogonal doesn't change the picture, since the determinant is not altered.

The bodies involved in the counterexample can probably be made smooth and strictly convex, but convergence in the $C^2$ sense won't take place.


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