[proofplan] We parametrize ellipsoids by their centers and positive semidefinite shape matrices, then restrict to a compact parameter set using the compactness of $K$ and a uniform bound on all semiaxes. A continuous volume functional attains its maximum on this compact set, and the nonempty interior of $K$ guarantees that the maximizer is nondegenerate. For uniqueness, we affinely normalize one maximizer to the Euclidean unit ball and use convexity of $K$ to build a new inscribed ellipsoid from two hypothetical maximizers; determinant convexity gives a larger volume unless the shape matrices agree, and a separate translation argument rules out two equal-radius translated balls. [/proofplan]

[step:Parametrize all candidate ellipsoids by centers and shape matrices] Let $B := B(0,1) \subset \mathbb{R}^n$ denote the open Euclidean unit ball, and let $\overline{B} := \overline{B}(0,1)$ denote its closure. We write an ellipsoid in $\mathbb{R}^n$ as a set of the form \begin{align*} E(a,S) := a + S\overline{B} = \{a + Sx : x \in \overline{B}\}, \end{align*} where $a \in \mathbb{R}^n$ and $S: \mathbb{R}^n \to \mathbb{R}^n$ is represented by a symmetric positive semidefinite matrix. If $S$ is positive definite, then $E(a,S)$ has nonempty interior and \begin{align*} \mathcal{L}^n(E(a,S)) = \det(S)\,\mathcal{L}^n(\overline{B}). \end{align*} If $S$ is singular, then $E(a,S)$ is contained in a proper affine subspace of $\mathbb{R}^n$, hence $\mathcal{L}^n(E(a,S)) = 0$, which is also consistent with $\det(S)=0$. Every nondegenerate ellipsoid may be written in this form: if $E = a + T\overline{B}$ for an invertible [linear map](/page/Linear%20Map) $T: \mathbb{R}^n \to \mathbb{R}^n$, then $S := (TT^\top)^{1/2}$ is symmetric positive definite and $T\overline{B} = S\overline{B}$ because the orthogonal factor in the [polar decomposition](/theorems/3074) preserves $\overline{B}$. [/step]

[step:Build a compact parameter space for inscribed ellipsoids] Define the diameter of $K$ by \begin{align*} D := \sup\{|x-y| : x,y \in K\}. \end{align*} Since $K$ is compact, $D < \infty$. Let $\operatorname{Sym}(n)$ denote the finite-dimensional real [vector space](/page/Vector%20Space) of symmetric $n \times n$ matrices, identified with symmetric linear maps $\mathbb{R}^n \to \mathbb{R}^n$. Consider the parameter set \begin{align*} \mathcal{P} := \{(a,S) \in K \times \operatorname{Sym}(n) : S \geq 0,\ \|S\|_{\mathrm{op}} \leq D/2,\ E(a,S) \subset K\}. \end{align*} We claim that every ellipsoid contained in $K$ is represented by some element of $\mathcal{P}$. Indeed, suppose $E(a,S) \subset K$ with $S = S^\top \geq 0$. First, $a \in K$ because $a \in E(a,S)$, obtained by taking $0 \in \overline{B}$. Next, for every $u \in \mathbb{R}^n$ with $|u|=1$, both $a+Su$ and $a-Su$ lie in $K$. Therefore \begin{align*} 2|Su| = |(a+Su)-(a-Su)| \leq D. \end{align*} Taking the supremum over all unit vectors $u$ gives $\|S\|_{\mathrm{op}} \leq D/2$. The set $\mathcal{P}$ is compact. The conditions $a \in K$, $S=S^\top$, $S \geq 0$, and $\|S\|_{\mathrm{op}} \leq D/2$ define a closed and bounded subset of the finite-dimensional space $\mathbb{R}^n \times \operatorname{Sym}(n)$. The containment condition is closed: if $(a_j,S_j) \in \mathcal{P}$ and $(a_j,S_j) \to (a,S)$, then for every $x \in \overline{B}$, \begin{align*} a_j + S_jx \in K \end{align*} for all $j$, and since $K$ is closed, \begin{align*} a+Sx = \lim_{j \to \infty}(a_j+S_jx) \in K. \end{align*} Thus $E(a,S) \subset K$, so $(a,S) \in \mathcal{P}$. [/step]

[step:Maximize the volume functional on the compact parameter space] Define the volume functional \begin{align*} V: \mathcal{P} &\to [0,\infty) \\ (a,S) &\mapsto \mathcal{L}^n(E(a,S)) = \det(S)\,\mathcal{L}^n(\overline{B}). \end{align*} The map $S \mapsto \det(S)$ is continuous on $\operatorname{Sym}(n)$, so $V$ is continuous on the compact set $\mathcal{P}$. Hence there exists $(a_0,S_0) \in \mathcal{P}$ such that \begin{align*} V(a_0,S_0)=\max_{(a,S)\in\mathcal{P}} V(a,S). \end{align*} Because $K$ has nonempty interior, there exist $x_0 \in K$ and $r>0$ such that $\overline{B}(x_0,r) \subset K$. Let $I_n: \mathbb{R}^n \to \mathbb{R}^n$ denote the identity linear map, represented by the $n \times n$ identity matrix. Then $E(x_0,rI_n) \subset K$, and \begin{align*} V(x_0,rI_n)=r^n\mathcal{L}^n(\overline{B})>0. \end{align*} Therefore $V(a_0,S_0)>0$, so $\det(S_0)>0$. Since $S_0$ is symmetric positive semidefinite, this implies that $S_0$ is positive definite. Thus $E(a_0,S_0)$ is a genuine ellipsoid of maximal volume contained in $K$. [/step]

[step:Normalize two hypothetical maximizers by an affine change of variables] Suppose that $E_1$ and $E_2$ are two maximal-volume ellipsoids contained in $K$. Write \begin{align*} E_1 &= E(a_1,S_1), & E_2 &= E(a_2,S_2), \end{align*} where $S_1$ and $S_2$ are symmetric positive definite. Define the affine isomorphism \begin{align*} A: \mathbb{R}^n &\to \mathbb{R}^n \\ x &\mapsto S_1^{-1}(x-a_1), \end{align*} and define the normalized convex body \begin{align*} K' := A(K). \end{align*} Affine maps preserve convexity, compactness, and containment, and they multiply every $n$-dimensional volume by the same factor $|\det(S_1^{-1})|$. Therefore $A(E_1)=\overline{B}$ and $A(E_2)$ is another maximal-volume ellipsoid in $K'$. Write \begin{align*} A(E_2)=c+M\overline{B}, \end{align*} where $c \in \mathbb{R}^n$ and $M$ is symmetric positive definite. Since $A(E_2)$ and $\overline{B}$ have the same maximal volume, \begin{align*} \det(M)=1. \end{align*} It remains to prove that $c=0$ and $M=I_n$. [/step]

[step:Use the arithmetic-geometric mean inequality to force the normalized shape matrix to be the identity] Assume first that $M \neq I_n$. Define the symmetric positive definite matrix \begin{align*} N := \frac{I_n+M}{2}. \end{align*} For every $x \in \overline{B}$, the point \begin{align*} \frac{1}{2}x+\frac{1}{2}(c+Mx) = \frac{c}{2}+Nx \end{align*} belongs to $K'$, because $x \in \overline{B} \subset K'$, $c+Mx \in c+M\overline{B} \subset K'$, and $K'$ is convex. Hence \begin{align*} \frac{c}{2}+N\overline{B} \subset K'. \end{align*} Let $\lambda_1,\dots,\lambda_n>0$ be the eigenvalues of $M$, counted with multiplicity. Since $\det(M)=1$, \begin{align*} \prod_{i=1}^n \lambda_i = 1. \end{align*} The eigenvalues of $N$ are $(1+\lambda_i)/2$, so \begin{align*} \det(N) = \prod_{i=1}^n \frac{1+\lambda_i}{2} \geq \prod_{i=1}^n \sqrt{\lambda_i} = 1, \end{align*} where the inequality is the arithmetic-geometric mean inequality applied to each pair $1,\lambda_i$. Equality holds only when $\lambda_i=1$ for every $i$. Since $M \neq I_n$, at least one $\lambda_i \neq 1$, so the inequality is strict: \begin{align*} \det(N)>1. \end{align*} Therefore \begin{align*} \mathcal{L}^n\left(\frac{c}{2}+N\overline{B}\right) = \det(N)\mathcal{L}^n(\overline{B}) > \mathcal{L}^n(\overline{B}), \end{align*} contradicting maximality of $\overline{B}$ in $K'$. Hence $M=I_n$. [/step]

[step:Rule out a nonzero translation of the normalized unit ball] It remains to exclude the case $M=I_n$ and $c \neq 0$. Suppose $c \neq 0$. Define \begin{align*} \alpha := \frac{|c|}{2}, \qquad e := \frac{c}{|c|}. \end{align*} After translating by $-c/2$, the two balls $\overline{B}$ and $c+\overline{B}$ become the balls centered at $-\alpha e$ and $\alpha e$. Since both are contained in $K'$, their convex hull is contained in $K'$ after translating back by $c/2$. Define the linear map \begin{align*} L: \mathbb{R}^n &\to \mathbb{R}^n \\ te+w &\mapsto \sqrt{1+\alpha^2}\,te+w, \end{align*} where $t \in \mathbb{R}$ and $w \in e^\perp$. We claim that $L\overline{B}$ is contained in the convex hull of the two balls centered at $-\alpha e$ and $\alpha e$. Let $z=te+w \in \overline{B}$ with $w \in e^\perp$. Then $t^2+|w|^2 \leq 1$. The point $Lz$ equals $\sqrt{1+\alpha^2}\,te+w$. The section of the convex hull of $\overline{B}(-\alpha e,1)$ and $\overline{B}(\alpha e,1)$ at height $w$ in the $e^\perp$ direction contains all points $\tau e+w$ with \begin{align*} |\tau| \leq \alpha + \sqrt{1-|w|^2}. \end{align*} Since $\sqrt{1-|w|^2} \geq |t|$, we have \begin{align*} \alpha+\sqrt{1-|w|^2} \geq \alpha+|t| \geq \sqrt{1+\alpha^2}\,|t|, \end{align*} because $0 \leq |t| \leq 1$ and $\sqrt{1+\alpha^2}-1 \leq \alpha$. Thus $Lz$ belongs to that convex hull. Hence \begin{align*} \frac{c}{2}+L\overline{B} \subset K'. \end{align*} The determinant of $L$ is \begin{align*} \det(L)=\sqrt{1+\alpha^2}>1. \end{align*} Therefore \begin{align*} \mathcal{L}^n\left(\frac{c}{2}+L\overline{B}\right) = \det(L)\mathcal{L}^n(\overline{B}) > \mathcal{L}^n(\overline{B}), \end{align*} again contradicting maximality of $\overline{B}$ in $K'$. Thus $c=0$. [/step]

[step:Transfer uniqueness back to the original convex body] The normalized argument proves that any maximal ellipsoid in $K'$ must equal $\overline{B}$. Therefore $A(E_2)=A(E_1)$, and since $A$ is injective, \begin{align*} E_2=E_1. \end{align*} Thus the maximal-volume ellipsoid contained in $K$ is unique. Together with the existence already proved, this establishes the existence and uniqueness of the John ellipsoid of $K$. [/step]