# Maximal Ellipsoid

John's Theorem can be stated as "To every compact, convex body, there is a unique inscribed ellipsoid, whose volume is maximal among all inscribed ellipsoids." It goes on to classify this maximal ellipsoid. By using this theorem, one can prove that the ellipsoid of maximal volume which is contained in a square is a circle.

This strikes me as a problem which was probably studied well before Fritz John, and yet I have been unable to prove the statement about squares and circles in an elegant, but low-brow manner. Any thoughts?

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Here's an attempt at a low-brow proof. Take a max area ellipse. Apply an affine transform to make it a circle; then the problem becomes to show that a minimal area parallelogram containing a circle is a square. It is easy to see that both the height of the parallelogram and its base are at least the diameter. Q.E.D.

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By the way, the argument seems easy to adapt for cubes of any dimension. – t3suji Jan 20 '10 at 22:17
When you go to higher dimensions, you might as well prove Auerbach's lemma in general, since the proof is elementary and simple, and remark that Auerbach's lemma for the Euclidean ball gives John's theorem for the cube. – Bill Johnson Jan 21 '10 at 0:26
Thank you both, this is very elegant. – Ben Weiss Jan 21 '10 at 4:05

Ben, John's theorem is easier for unit balls of spaces which have a one symmetric basis because you can argue that the basis vectors must be orthogonal with respect to the Euclidean norm determined by the maximal volume ellipsoid and all the basis vectors must have the same Euclidean norm. At UMich you have a top expert in this direction (Vershynin); talk to him.

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Yes, I learned about it from Keith Ball (when he was briefly here), and have discussed this subject with Vershynin (truly an expert). Perhaps I wasn't clear enough, in that what I'm looking for is a proof about the maximal ellipsoid of a square without appealing to "heavy machinery". I'm was searching for a low-brow manner to bring this to the attention of undergraduates, and discovered that I could not (without proving John's Theorem itself). – Ben Weiss Jan 20 '10 at 21:59