Let $K \subset \mathbb{R}^d$ be a convex body, symmetric about the origin and with nonempty interior. Then John's theorem asserts that there exists a unique ellipsoid $E$ of minimal volume such that $K \subset E$.

My question: let $K, K'$ be two such convex sets (that is, symmetric with nonempty interior), and apply John's theorem to get the John's ellipsoids $E, E'$ respectively. Suppose that $K \subset K'$. Is it true, then, that $E \subset E'$?

I have been concerned with how John's ellipsoid varies as you vary the convex set $K$, in particular how the gauge function of $E$ varies. Suppose my claim is true: then it would follow that for any two convex bodies $K_1, K_2$ (again, symmetric about the origin and with nonempty interior) with John's ellipsoids $E_1, E_2$, we'd have $$ \sup_{v \in \mathbb{R}^d} \left| 1 - \frac{n_{E_1}(v)}{n_{E_2}(v)} \right| \leq \sup_{v \in \mathbb{R}^d} \left| 1 - \frac{n_{K_1}(v)}{n_{K_2}(v)} \right| $$ where $n_K$ is the gauge function of a convex body $K$.

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    $\begingroup$ you may wish to check out: mathoverflow.net/questions/127114/… $\endgroup$ – Suvrit Nov 6 '13 at 16:06
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    $\begingroup$ Let's see. Specialize to the case where $K'$ is itself an ellipsoid. If what you want to be true were indeed true, you would get that the John's ellipsoid for $K$ must be contained in $K'$. In other words, the John's ellipsoid for $K$ would be contained in the intersection of all ellipsoids containing $K$. $\endgroup$ – Bill Johnson Nov 6 '13 at 16:13
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    $\begingroup$ By the way, the minimal ellipsoid containing $K$ is more usually called Löwner's ellipsoid. John's ellipsoid is the maximal volume ellipsoid contained in $K$. $\endgroup$ – Douglas Zare Nov 6 '13 at 19:57
  • $\begingroup$ With the correct terminology, this question is contained within the question to which Suvrit linked, so I'll vote to close as an exact duplicate. $\endgroup$ – Douglas Zare Nov 6 '13 at 21:22

Suvrit's and Bill Johnson's comments say that the answer is no. Here is an explicit example. Let $K'$ be the unit disk, so it equals its minimal area containing ellipsoid. Let $K$ be the rectangle with vertices $(\pm \frac{3}{5},\pm \frac{4}{5})$ so it is inscribed in the unit disk. Affine transformations commute with the "minimum volume ellipsoid" operator. $K$ is the image of the square $S$ with vertices $(\pm \frac{3}{5}, \pm \frac{3}{5})$ under the map $(x,y) \mapsto (x,\frac{4}{3} y)$. The minimal ellipsoid containing the square $S$ is the circle of radius $\frac{3}{5}\sqrt{2}$, and the image of this under the affine transformation is an ellipse with semimajor axis $\frac{4}{3}\frac{3}{5}\sqrt{2} = \frac{4}{5}\sqrt{2} = 1.131 \gt 1.$ This is not contained in $K'$ even though its area $\frac{24}{25}\pi$ is slightly smaller than the unit disk's area $\pi$.

  • $\begingroup$ Hey, guys; Alex is a student and was supposed to work this out for himself. $\endgroup$ – Bill Johnson Nov 6 '13 at 20:25
  • $\begingroup$ It wasn't a HW problem, Wlodek--Alex is a grad student at Courant and I am at Texas A&M. $\endgroup$ – Bill Johnson Nov 6 '13 at 20:43
  • $\begingroup$ @Bill Johnson: If you feel this question is too elementary, despite the fact that a similar question was asked before and had 4 upvotes and an answer, then maybe you should vote to close, or state explicitly in your comment that you think this is too easy to answer. $\endgroup$ – Douglas Zare Nov 6 '13 at 21:11

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