13.5 Norms and Symmetric Convex Bodies in $\R^{n}$

Definition 13.5.1 (Origin-Symmetric Convex Body).label A convex body $K\suf\R^{n}$ is origin-symmetric if $K=-K$, i.e. $x\in K\iff -x\in K$.

Proposition 13.5.2.label If $K$ is an origin-symmetric convex body, then $0\in int(K)$

Proof. Since $K$ is convex, pick $y\in int(K)$ so $-y\in int(K)$. But $int(K)$ is convexx so $0=\frac{1}{2}y+\frac{1}{2}(-y)\in int(K)$.$\square$

Definition 13.5.3 (Minkowski Gauge).label Let $K$ be an origin-symmetric convex body. Its gauge/Minkowski functional

\begin{align*}\norm{x}{K}\define \inf\curl{t>0:\frac{x}{t}\in K}\end{align*}

is a norm on $\R^{n}$ and $K=\curl{x:\norm{x}{K}\leq 1}$

Proof. Clearly $\norm{0}{K}=0$ and $\norm{x}{K}\geq 0$ for any $x\in K$. Let $\e>0$ such that $B(0,\e)\suf K$ so $\fall x\neq 0,\exists t^{*}>0$ such that $\norm{\frac{x}{t^{*}}}{2}\leq \e$ then $\frac{x}{t^{*}}\in B(0,\e)\suf K$. Thus $\norm{x}{K}<\infty$. By boundedness of $K$, $\norm{x}{K}=0\implies x=\vec{0}$.

For $\lam\neq 0$ set $\lam=\e\abs{\lam}$ then

\begin{align*}\frac{\lam x}{t}=\frac{\e\abs{\lam}x}{t}\in K\iff\frac{\abs{\lam}x}{t}\in K\end{align*}

hence

\begin{align*}\norm{\lam x}{K}=\inf\curl{t>0:\frac{\abs{\lam}x}{t}\in K}=\inf \curl{\abs{\lam}s:s>0,\frac{x}{s}\in K}=\abs{\lam}\norm{x}{K}\end{align*}

Let $x,y\in\R^{,}\e>0$. By definition there are $s\leq \norm{x}{K}+\e,t\leq \norm{y}{K}+\e$ with $\frac{x}{s},\frac{y}{t}\in K$. Since $K$ is convex

\begin{align*}\frac{x+y}{s+t}=\frac{s}{s+t}\frac{x}{s}+\frac{t}{s+t}\frac{y}{t}\in K\end{align*}

so $\norm{x+y}{K}\leq s+t\leq\norm{x}{K}+\norm{y}{K}+2\e$.

If $x\in K$ then $x/1\in K$ so $\norm{x}{K}\leq 1$. Conversely if $\norm{x}{K}\leq 1$ then pick a non-increasing sequence $t_{j}\to \norm{x}{K}$ with $x/t_{j}\in K$. If $\norm{x}{K}<1$ we may assume $t_{j}\leq 1$ so $x=t_{j}(x/t_{j})+(1-t_{j})0\in K$ while $\norm{x}{K}=1\implies x/t_{j}\to x\in K$ by closedness.$\square$

Definition 13.5.4 (Banach-Mazur Distance).label Let $K,L\suf\R^{n}$ be origin-symmetric convex bodies. The Banach-Mazur distance between $K$ and $L$ is

\begin{align*}d_{BM}(K,L)\define \inf \curl{d\geq 1:\exists T\in GL(n,\R),K\suf TL\suf dK}\end{align*}

Equivalently $d_{BM}(K,L)\leq d$ if and only if there is a linear isomorphism $T$ such that

\begin{align*}\frac{1}{d}\norm{x}{K}\leq\norm{T\inv x}{L}\leq\norm{x}{K}&&\fall x\in\R^{n}\end{align*}

Additionally $\norm{T}{}\norm{T\inv}{}\leq d$. We have several properties of $d_{BM}$. Let $K,L,M$ be origin-symmetric convex bodies in $\R^{n}$

(1)
$d_{BM}(K,L)\geq 1$ with equality if and only if $K$ and $L$ are linearly equivalent, i.e. $L=TK$ for some $T\in GL(n, R)$
(2)
$d_{BM}(K,L)=d_{BM}(L,K)$
(3)
$d_{BM}(K,M)\leq d_{BM}(K,L)d_{BM}(L,M)$

Consequently $\log d_{BM}$ defines a metric on the set of origin-symmetric convex bodies in $\R^{n}$ modulo the equivalence relation of being linearly equivalent.

Theorem 13.5.5 (John’s Proposition).label Let $C$ be an origin symmetric convex body then $\exists!$ centered ellipsoid $E$ of maximal volume in $C$ with $E\suf C\suf \sqrt{n}E$ called John’s ellipsoid^[1]

Remark 13.5.6.label

(1)
$E^{o}\supf C^{o}\supf \frac{1}{\sqrt{n}}E^{o}$
(2)
$d_{BM}(K,B_{2}^{n})\leq \sqrt{n}$
(3)
$d_{BM}(K,L)\leq d_{BM}(K,B_{2}^{n})d_{BM}(L,B_{2}^{n})\leq \sqrt{n}\sqrt{n}= n$
(4)
If $C$ is a convex body then $\exists!$ ellipsoid of maximal volume included in $C$. Moreover $x_{0}+E\suf C\suf x_{0}+n E$

which is a sharp bound due to the cube geometricallykeyboard_return

The Eigenspace

13.5 Norms and Symmetric Convex Bodies in $\R^{n}$

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The Eigenspace