John ellipsoid ($\mathcal{E}$)

The John ellipsoid of a convex body is its inscribed ellipsoid of largest volume. For the unit ball $K$ of a $d$-dimensional normed vector space, change coordinates so that this ellipsoid is the Euclidean unit ball. The resulting coordinates are called John position. In these coordinates the John contact decomposition implies $B_2^d\subseteq K\subseteq \sqrt{d}{B}_2^d$, or $\Vert x{\Vert }_E\le |x|\le \sqrt{d}\Vert x{\Vert }_E$.

For a centrally symmetric convex body in John position, there are contact points $u_j\in \partial K\cap S^{d-1}$ and positive weights $c_j$ such that ${\sum }_j{c}_j{u}_j\otimes u_j=I$. Taking the trace gives ${\sum }_j{c}_j=d$. A supporting hyperplane at $u_j$ shows $|⟨u_j,x⟩|\le \Vert x{\Vert }_K$.

Here is the maximum-volume argument for the decomposition. If $I$ were outside the convex cone generated by the contact matrices, a separating hyperplane would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract a small positive multiple of $I$ so these inequalities become strict while the trace remains positive. The support function of $(I+tA)B_2^d$ is $|(I+tA)v|$. It decreases below $1$ near all contact directions for small positive $t$; away from them, the strict gap $h_K(v)>1$ preserves containment. Its volume increases, since $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$, a contradiction. The cone is closed because its generators have trace $1$ and form a compact set; finite-dimensional convexity gives a finite decomposition.

A centrally symmetric convex body is in John position when its John ellipsoid is the Euclidean unit ball.