Proof. [04U0]
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Proof.
By translating and subtracting a linear function assume that . Assume by way of contradiction that we can find and some such that
for all . We first show that is trapped by two tangent planes at .
Let and be the points on where the hyperplanes perpendicular to the shortest axis of the John ellipsoid become tangent to , and let and denote subgradients at these points. Since
we have that for all . By this observation and convexity we can rotate and pass to a subsequence such that
Then is trapped by the planes . We conclude that
To complete the proof, we show that the volumes of sections obtained with tilted supporting planes are too large. Take the largest such that and consider the sections
Then engulf . Furthermore,
where as . Indeed, if not, then for some small and a sequence we would have for all . Convexity and imply that for all , which in turn implies that
contradicting the definition of .
Finally, let be the point in furthest in the direction. Since grows at least quadratically, we have
Recall that . Since for all , contains the cone with vertex and base given by a ball of radius on the hyperplane . We conclude that
contradicting our definition of for large. ∎