THE JOHN ELLIPSOID THEOREM

RALPH HOWARD DEPARTMENT OF MATHEMATICS UNIVERSITY OF SOUTH CAROLINA COLUMBIA, S.C. 29208, USA

                             HOWARD@MATH.SC.EDU

The following is a lecture given in the functional analysis seminar at the University of South Carolina.

Contents

1. Introduction. 1 2. Proof of the theorem. 2 3. The case of centrally symmetric convex bodies. 6 4. Proof of the uniquness of the John ellipsoid. 7 4.1. Examples of where the inequalities are sharp. 10 References 10

1. Introduction.

An affine map Φ: R^{n} → R^{n} is a map of the form

Φ(x) := Ax + b

where A: R^{n} → R^{n} is linear and b ∈ R^{n} is a constant vector. This is nonsingular iff det A ≠ 0. An ellipsoid in R^{n} the image of the closed unit ball B^{n} of R^{n} under a nonsingular affine map. Our goal here is to prove the following famous result of Fritz John.

Theorem 1 (John [3]). Let K ⊂ R^{n} be a convex body (that is a compact convex set with nonempty interior). Then there is an ellipsoid E (called the John ellipsoid which will turn out to be the ellipsoid of maximal volume contained in K) so that if c is the center of E then the inclusions

E ⊆ K ⊆ c + n(E − c).

hold. (Here c + n(E − c) is the set of points {c + n(x − c) : x ∈ E}. This is the dilation of E by a factor of n with center c.)

Date: November 1997.

Remark 1.1. This result is sharp in the sense that the dilation factor n can not be replaced by a smaller factor. As an example let x_{0}, …, x_{n} be an affinely independent set of points in R^{n} and let K be the convex hull of {x_{0}, …, x_{n}}. That is K is an n dimensional simplex in R^{n}. Then if E is the ellipsoid of maximal volume in K then E ⊆ K ⊆ c + n(E − c), but there is no ellipsoid E′ ⊆ K so that E′ ⊆ K ⊆ c′ + m(E′ − c′) for any real number m < n. For the outline the proof of these claims see Section 4.1. For anther example see Remark 2.3. □

The proof here is one I came up with based loosely on ideas in the expository article [2] of Berger where he gives a similar proof in the case K is symmetric about the origin. For a different proof, which in many ways is preferable to the bare handed approach here, see the article of Keith Ball [1]. John original proof [3] (and I thank Steve Dilworth for giving me a copy of this paper) is quite different and deduces the result from a more general result on maxima and minima of functions subject to inequality constraints very much in spirit of what is now called “geometric programming”.

2. Proof of the theorem.

For the rest of this section K will be a convex body in R^{n}. The basic idea of the proof is to choose E ⊂ R^{n} to be an ellipsoid of maximal volume. Then by an affine change of variables we can assume that E is the unit ball B^{n}. The proof is completed by showing that if K contains a point p at a distance greater than n from the origin of R^{n} then the convex hull of B^{n} ∪ {p}, and thus also K, contains an ellipsoid of volume greater than B^{n} which would contradict that B^{n} has maximal volume. See Figure 1

p B^{n}  E  If ∥p∥ is large enough then conv{B^{n}, p} will contain an ellipsoid E with Vol(E) > Vol(B^{n}).

Figure 1

We start by showing that K contains an ellipsoid of maximal volume. If Φ(x) = Ax + b is an affine max then the volume of the ellipsoid E = Φ[B^{n}] is Vol(E) = | det(A)| Vol(B^{n}). This is a standard piece of affine geometry, but it can also be deduced from the change of variable formula for integrals.

Lemma 2.1. The convex body K contains an ellipsoid of maximal volume.

Proof. Let N := n + n^{2}. Then the set of ordered pairs (A, b) where A is an n × n matrix and b ∈ R^{n} is just R^{N}. Let E := {(A, b) ∈ R^{N} : AB^{n} + b ⊂ K}. Then E is a closed bounded and thus compact subset of R^{N} and (A, b) ↦ | det(A)| is a continuous function on E. Thus there is an (A_{0}, b_{0}) that maximizes this function on E. Then E := A_{0}B^{n} + b_{0} is the desired ellipsoid. □

If E = A_{0}B^{n}+b_{0} is an ellipsoid of maximal volume in K then by replacing K by A_{0}^{−1}(K − b_{0}) we can assume that B^{n} = A_{0}^{−1}(E − b_{0}) is an ellipsoid of maximal volume in K. Then to prove John’s Theorem it is enough to show that if p ∈ K then ∥p∥ ≤ n (where ∥ · ∥ is the standard Euclidean norm). The geometric idea of the proof is shown in Figure 1. If ∥p∥ is large then the convex hull conv{B^{n}, p} of B^{n} and p will contain an ellipsoid E with Vol(E) > Vol(B^{n}). But K is convex so E ⊂ conv{B^{n}, p} ⊆ K which contradicts that B^{n} is an ellipsoid of maximal volume in K. What takes some work is showing the critical distance where B^{n} stops having maximal volume in conv{B^{n}, p} is ∥p∥ > n.

x = −1 B^{2} Φ_{t}^{λ}[B^{2}] Figure 2

We will construct the ellipsoid E as an affine image of the ball B^{n}. This will first be done in two dimensions and then extended to higher dimensions by symmetry. As it is a truth universally acknowledged that any problem in analysis needs a differential equation to be taken seriously, we will construct our affine maps as flows of solutions to differential equations. Let λ > 0 then in the plane consider the system of equations

ẋ = 1 + x
ẏ = −λy.

The solution with initial condition x(0) = x_{0} and y(0) = y_{0} is

x(t) = −1 + e^{t}(1 + x_{0})
y(t) = e^{−λt}y_{0}.

Therefore if Φ_{t}^{λ} is the one parameter group of diffeomorphisms generated by this system of equations (or what is the same thing the one parameter group generated by the vector field (1 + x)\frac{∂}{∂x} − λ\frac{∂}{∂y}) is

Φ_{t}^{λ}(x, y) = (−1 + e^{t}(1 + x), e^{−λt}y) = (−1 + e^{t}, 0) + (e^{t}x, e^{−λt}y)

so that for each fixed t the map (x, y) ↦ Φ_{t}^{λ}(x, y) is an affine map. The lines x = −1 and y = 0 are fixed (set-wise) by Φ_{t}^{λ}. The effect of Φ_{t}^{λ} on the two dimensional ball B^{2} for a small positive value of t is shown in Figure 2.

For a > 1 let C_{a} := conv{B^{2}, (a, 0)} be the convex hull of the two dimension ball B^{2} and the point (a, 0). See Figure 3

(a, 0) B^{2}  C_{a} is convex hull of B^{2} and (a, 0).

Figure 3

The following is the hard step in the proof of John’s theorem.

Lemma 2.2. If λ > \frac{1}{a − 1} then Φ_{t}^{λ}[B^{2}] ⊂ C_{a} for small positive values of t.

Proof. We first note that the tangent lines to ∂B^{2} through (a, 0) (which are part of the boundary of C_{a}) are y = \frac{±1}{\sqrt{a^{2}−1}}(x − a). To see this consider the line y = \frac{−1}{\sqrt{a^{2}−1}}(x − a). Direct calculation shows that both the points (a, 0)

and (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a}) are on this line and the point (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a}) is on the unit

circle (see Figure 4). The slope of the radius to ∂B^{2} ending at (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a})

is \sqrt{a^{2} − 1}). The slope of y = \frac{−1}{\sqrt{a^{2}−1}}(x − a) is \frac{−1}{\sqrt{a^{2}−1}} which is the negative

reciprocal of the slope of the redius to ∂B^{2} ending at (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a}). Therefore

y = \frac{−1}{\sqrt{a^{2}−1}}(x−a) is prependicular to the radius where it meets ∂B^{2} and this is tangent.

( \frac{1}{a} , ^{\frac{\sqrt{a^{2}−1}}{a}} )  y = \frac{−1}{\sqrt{a^{2}−1}} (x − a)  (a, 0) Figure 4

Therefore to prove the lemma it is enough to show that for λ > \frac{1}{a−1} and

small t that Φ_{t}^{λ} (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a}) lies below the line y = \frac{−1}{\sqrt{a^{2}−1}}(x − a). (For as ∂B^{2} is tangent to this line at (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a}) it will also be caried into C_{a} by

Φ_{t}^{λ}. See Figure 5.)

\begin{matrix}( \frac{1}{a} , ^{\frac{\sqrt{a^{2}−1}}{a}} ) \\ Φ_{t}^{λ} ( \frac{1}{a} , ^{\frac{\sqrt{a^{2}−1}}{a}} ) \end{matrix} Figure 5

Either directly from the definition of Φ_{t}^{λ} or from the system of differential equations defining it we have

\frac{d}{dt}Φ_{t}^{λ}(\frac{1}{a}, \frac{\sqrt{a^{2} − 1}}{a} )|_{t=0} = (ẋ(0), ẏ(0)) = (1 + \frac{1}{a}, \frac{−λ\sqrt{a^{2} − 1}}{a}).

Therefore the slope of the tangent to the flow line t ↦ Φ_{t}^{λ}(\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a} ) at (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a} ) is

\frac{ẏ(0)}{ẋ(0)} = \frac{−λ\sqrt{a^{2} − 1}}{a + 1}.

Therfore Φ_{t}^{λ}(\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a} ) will be below the line y = \frac{−1}{\sqrt{a^{2}−1}}(x − a) provided

\frac{−λ\sqrt{a^{2} − 1}}{a + 1} < \frac{−1}{\sqrt{a^{2} − 1}}.

A little algebra shows this is equivalent to λ > \frac{1}{a−1}. This completes the proof of the lemma □

Proof of Thoerem 1. Assume that B^{n} ⊂ K is an ellipsoid of maximal volume in K. Then we need to show that then is not a point p of K with ∥p∥ > n. Assume, toward a contradiction, that there is such a point. Choose orthogonal coordinates (x, y^{1}, …, y^{n−1}) on R^{n} so that p is on the positive x axis. Then if a = ∥p∥ then in these coordinates p = (a, 0, …, 0). The higher dimensional version of C_{a} used above it C_{a}^{n} := conv{B^{n}, p} = conv{B^{n}, (a, 0, …, 0)}. Define the higher dimensional analogue Φ_{t}^{λ}, that is

Ψ_{t}^{λ}(x, y^{1}, …, y^{n−1}) = (−1 + e^{t}(x + 1), e^{−λt}y^{1}, …, e^{−λt}y^{n−1}).

As before this is an affine map. From Lemma 2.2 and symmetry we have that Ψ_{t}^{λ}[B^{n}] ⊂ C_{a}^{n} ⊂ K for small t provided λ > \frac{1}{a−1}. The volume of the ellipsoid Ψ_{t}^{λ}[B_{a}^{n}] is

Vol(Ψ_{t}^{λ}[B^{n}]) = e^{(1−(n−1)λ)t} Vol(B^{n}).

Now if a > n then it is possible to choose λ with λ > \frac{1}{a−1} and 1−(n−1)λ > 0. This implies for small positive t that Ψ_{t}^{λ}[B^{n}] ⊂ C_{a}^{n} ⊆ K and Vol(Ψ_{t}^{λ}[B^{n}]) > Vol(B^{n}). This contradiction completes the proof of John’s theorem. □

Remark 2.3. The proof of the theorem gives anther example showing that the dilation factor n is sharp in John’s theorem. If a = n then B^{n} will be the ellipsoid of maximal volume in C_{a} and C_{a} ⊂ αB^{n} only for α ≥ n. □

3. The case of centrally symmetric convex bodies.

When K is symmetric about the origin, that is −K = K, then the dilation factor in John’s Theorem can be improved.

Theorem 2. Let K be a convex body in R^{n} which is symmetric about the origin. Then there is an ellipsoid E, also symmetric about the origin, so that

E ⊆ K ⊆ \sqrt{n}E.

Remark 3.1. The factor \sqrt{n} is this result is sharp. As an example consider K := {(x^{1}, …, x^{n}) ∈ R^{n} : |x^{i}| ≤ 1} be the unit ball of ℓ_{n}^{∞}. Then the unit ball B^{n} is the ellipsoid of maximal volume in K and the point P := (1, , …, 1) ∈ K has ∥p∥ = \sqrt{n} which shows that \sqrt{n} is best possible. □

Remark 3.2. Let (X, | · |_{X}) and (Y, | · |_{Y}) be two finite n-dimensional vector spaces. Then define the Banach-Mazur distance between the two spaces as

dist^{BM}(X, Y) := inf_{T : X→Y} ∥T ∥∥T^{−1}∥

where T ranges over all linear isomorphisms of T: X → Y and ∥ · ∥ is the operator norm. Then a restatement of Theorem 2 is that for and n dimensional X that dist BM(X, ℓ_{n}^{2}) ≤ \sqrt{n} and Remark 3.1 implies that dist^{BM}(ℓ_{n}^{∞}, ℓ_{n}^{2}) = dist(ℓ_{n}^{1}, ℓ_{n}^{2}) = \sqrt{n} so that this is the sharp bound. □

Proof. The proof is very much like the general case. A variant on Lemma 2.1 shows there as a symmetric ellipsoid E ⊂ K that has maximal volume. By an affine change of coordinates we can assume that the ball B^{n} is the ellipsoid of maximal volume in K.

Again we start by looking at a two dimensional problem. This time let S_{a} := conv(B^{2} ∪ {(a, 0), (−a, 0)} (See Figure 6.) As we wish to preserve the symmetry of figures this time we look at the flow of the equations

ẋ = x
ẏ = −λy

where λ > 0. This has as flow

Φ_{t}^{λ}(x, y) = (e^{t}x, e^{−λt}y)

which is linear. Again we want to know when Φ_{t}^{λ}[B^{2}] ⊂ S_{a} for small t. As in the proof of Lemma 2.2 looking at what happens to the point (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a})

( \frac{1}{a} , ^{\frac{\sqrt{a^{2}−1}}{a}} ) ( \frac{1}{a} , ^{\frac{−\sqrt{a^{2}−1}}{a}} )  (a, 0) (−a, 0)  ( \frac{−1}{a} , ^{\frac{\sqrt{a^{2}−1}}{a}} ) ( \frac{−1}{a} , ^{\frac{−\sqrt{a^{2}−1}}{a}} )B^{2}  y = \frac{−1}{\sqrt{a^{2}−1}} (x − a) Figure 6

is the key. We have

\frac{d}{dt}Φ_{t}^{λ}(\frac{1}{a}, \frac{\sqrt{a^{2} − 1}}{a} )|_{t=0} = (ẋ(0), ẏ(0)) = (\frac{1}{a}, \frac{−λ\sqrt{a^{2} − 1}}{a}).

Thus the condition for Φ_{t}^{λ} (\frac{1}{a}, \frac{\sqrt{a^{2}−1}}{a}) to be below the line y = \frac{−1}{\sqrt{a^{2}−1}}(x − a) for small t is that

\frac{ẏ(0)}{ẏ(0)} = −λ\sqrt{a^{2} − 1} < \frac{−1}{\sqrt{a^{2} − 1}}.

Some algebra then shows that Φ_{t}^{λ}[B^{2}] ⊂ S_{a} for small t if λ?\frac{1}{a^{2}−1}.

To complete the proof if K is a symmetric convex body so that B^{n} is the ellipsoid of maximum volume contained in K then we wish to show that any p ∈ K satisfies ∥p∥ ≤ \sqrt{n}. We can choose coordinates x, y^{1}, …, y^{n−1} so that p is on the positive x-axis. Then p = (a, 0, …, 0). By symmetry −p = (a, 0, …, 0) is also in K. Let S_{a}^{n} = conv(B^{n} ∪ {p, −p}) and define a one parameter group of linear maps acting on R^{n} by

Ψ_{t}^{λ}(x, y^{1}, …, y^{n−1}) = (e^{t}x, e^{−λt}y^{1}, …, e^{−λt}y^{n−1}).

Then by symmetry Ψ_{t}^{λ}[B^{n}] ⊂ S_{a} ⊆ K for small t if λ > \frac{1}{a^{2}−1}. The volume of Ψ_{t}^{λ}[B^{n}] is

Vol(Ψ_{t}^{λ}[B^{n}]) = e^{(1−(n−1)λ)t} Vol(B^{n}).

Now if a = ∥p∥ > \sqrt{n} then it is possible to choose λ so that λ > \frac{1}{a^{2}−1} (so that Ψ_{t}^{λ}[B^{n}] ⊂ S_{a} ⊆ K for small t) and 1 − (n − 1)λ > 0 (so that Vol(Ψ_{t}^{λ}[B^{n}]) > Vol(B^{n}). This is a contradiction and completes the proof of Theorem 2. □

4. Proof of the uniquness of the John ellipsoid.

Theorem 3. If K ⊂ R^{n} is a compact body then the ellipsoid of maximal volume is unique.

Lemma 4.1. Let E be an ellipsoind in R^{n}, so that E = AB^{n} + b with A linear and b ∈ R^{n}. Then in this representation we can take A to be positive definite.

Proof. It is a standard result from linear algebra (the “polar decomposition”) that any nonsingular matrix can be expresses as a product A = P U where P is positive definite and U is orthogonal. For any orthogonal matrix we have UB^{n} = B^{n}. Thus AB^{n} = P UB^{n} = P B^{n}. Thus we can replace A by P in the representation E = AB^{n} + b and assume that A is positive definite. □

Proposition 4.2. Let A and B be n × n positive definite matrices. Then

det(A + B)^{\frac{1}{n}} ≥ det(A)^{\frac{1}{n}} + det(B)^{\frac{1}{n}}

with equality iff there is a positive constant c so that B = cA. In particular if equality holds and det(A) = det(B) then A = B.

Proof. As A is positive definite it has a positive definite square root P. That is A = P^{2}. Then

det(A + B)^{\frac{1}{n}} = det(P^{2} + B)^{\frac{1}{n}} = det(P)^{\frac{2}{n}} det(I + P^{−1}BP^{−1})^{\frac{1}{n}}

and

det(A)^{\frac{1}{n}} + det(B)^{\frac{1}{n}} = det(P)^{\frac{2}{n}} (1 + det(P^{−1}AP^{−1})^{\frac{1}{n}}).

Thus letting C = P^{−1}BP^{−1} it is enough to show det(I +C)^{\frac{1}{n}} ≥ 1+det(C)^{\frac{1}{n}} with equality iff C = cI. (If C = cI then P^{−1}BP^{−1} = cI so that B = cP^{2} = cA.) Now let λ_{1}, …, λ_{n} be the eigenvalues of C. Then det(I + C)^{\frac{1}{n}} ≥ 1 + det(C)^{\frac{1}{b}} is equivalent to

(4.1)  (∏_{k=1}^{n}(1 + λ_{k}))^{\frac{1}{n}} ≥ 1 + (λ_{1} ⋯ λ_{n})^{\frac{1}{n}}

with equality iff λ_{k} = c for all k and some positive c. Let σ_{k} = σ_{k}(λ_{1}, …, λ_{n}) be the k elementary symmetric function of λ_{1}, …, λ_{n}. That is

σ_{k}(λ_{1}, …, λ_{n}) := ∑_{1≤i_{1}<i_{2}<···<i_{k}≤n} λ_{i_{1}}λ_{i_{2}} ⋯ λ_{i_{k}}.

Then

∏_{k=1}^{n}(1 + λ_{k}) = 1 + σ_{1} + σ_{2} + ⋯ + σ_{n}.

(Note σ_{n} = λ_{1} ⋯ λ_{n}.) By the inequality between the arithmetic and geometric means we have

σ_{k}(λ_{1}, …, λ_{n}) = \binom{n}{k} ∑_{1≤i_{1}<i_{2}<···<i_{k}≤n} \frac{λ_{i_{1}}λ_{i_{2}} ⋯ λ_{i_{k}}}{\binom{n}{k}}

≥ \binom{n}{k} ∏_{1≤i_{1}<i_{2}<···<i_{k}≤n}(λ_{i_{1}}λ_{i_{2}} ⋯ λ_{i_{k}})\frac{1}{\binom{n}{k}}

= \binom{n}{k}(λ_{1}λ_{2} ⋯ λ_{n})\begin{matrix}( & {n−1 \atop k−1} & ) \\ ( & ^{n}_{k} & ) \end{matrix}

= \binom{n}{k}(λ_{1}λ_{2} ⋯ λ_{n})^{\frac{k}{n}}

and equality will hold iff λ_{i_{1}}λ_{i_{2}} ⋯ λ_{i_{k}} = λ_{j_{1}}λ_{j_{2}} ⋯ λ_{j_{k}} for all subsets {i_{1}, …, i_{k}} and {i_{1}, …, i_{k}} of {1, …, n} of size k. This is the case iff there is a c so that λ_{i} = c for all i.

Using this inequality we have

(1 + (λ_{1} ⋯ λ_{n})^{\frac{1}{n}})^{n} = ∑_{k=0}^{n} \binom{n}{k}(λ_{1} ⋯ λ_{n})^{\frac{k}{n}}
≤ 1 + σ_{1} + σ_{2} + ⋯ + σ_{n}

= ∏_{k=1}^{n}(1 + λ_{k})

with equality iff λ_{k} = c for some c > 0. But this inequality is equivalent to the inequality (4.1). This completes the proof. □

Proof of Theorem 3. As above let P_{n} be the set of n × n positive definite matrices and let K := P_{n} × R^{n}. Let E := {(A, b) ∈ K : AB^{n} + b ⊆ K}. By Lemma 4.1 every ellipsoid contained in K is of the form AB^{n} + b for some (A, b) ∈ K. Also as K is convex the set E is also convex. Let E_{1} = A_{1}B^{n} +b_{1} and E_{2} = A_{2}B^{n} + b_{2} be two ellipsoids of maximal volume contained in K. The volume of the ellipsoid AB^{n} + b is det(A) Vol(B^{n}), so as E_{1} and E_{2} are both maximal we have det(A_{1}) = det(A_{2}). Let A_{3} = \frac{1}{2}(A_{1} + A_{2}) and b_{3} = \frac{1}{2}(b_{1} + b_{2}). Then as E is convex the ellipsoid E_{3} = A_{3}B^{n} + b_{3} ⊆ K. But from Proposition 4.2

det(A_{3})^{\frac{1}{n}} = \frac{1}{2} det(A_{1} + A_{2})^{\frac{1}{n}} ≥ \frac{1}{2}(det(A_{1})^{\frac{1}{n}} + det(A_{2})^{\frac{1}{n}}) = det(A_{1})^{\frac{1}{n}} .

But as E_{1} has maximal volume this implies det(A_{3}) = det(A_{1}) = det(A_{2}) so that equality holds in the inequality. By Proposition 4.2 this implies A_{1} = A_{2}. That is E_{2} is a translate of E_{1}.

If only remains to show b_{1} = b_{2}. If b_{1} ≠ b_{2} then K will contain the convex hull of E_{1} ∪ E_{2}. Let b_{3} = \frac{1}{2}(b_{1} + b_{2}). Then conv(E_{1} ∪ E_{2}) will contain an ellipsoid E_{3} centered and b_{3} that has volume Vol(E_{3}) > Vol(E_{1}) as it will

.  . . b_{3} b_{1}  b_{2}  E_{3} E_{2} E_{1}  Translate of E_{1} centered at b_{3}. E_{3} contains a translate of E_{1} and thus has greater volume.

Figure 7

contain a translate of E_{1} centered at b_{3} (see Figure 7 where we have done an affine transformation so that we can assume that E_{1} and E_{2} are balls.)

But this contradicts that E_{1} had maximal volume and so b_{1} = b_{2}. This completes the proof. □

4.1. Examples of where the inequalities are sharp. Above we we mentioned some examples there the inequalities are sharp. However to show that this is the case by direct calculation is not easy so we show here that the uniqueness result Theorem 3 can be used to find the ellipsoid of maximum volume in case where symmetry is present. Consider the case of a the standard regular simplex σ^{n}. The easiest realization of this is to view the affine space R^{n} as the hyperplane in R^{n+1} defined by x^{1}+x^{2}+⋯+x^{n+1} = 1. Then let e_{1}, …, e_{n+1} be the standard basis of R^{n+1}. Then σ^{n} is the convex hull of e_{1}, …, e_{n+1}. That is σ is the subset of R^{n+1} defined by x^{1} + ⋯ + x^{n+1} = 1 and x^{i} ≥ 0. Let S_{n+1} be the permutation group on n + 1 objects. Then S_{n+1} acts on σ_{n} ⊂ R^{n} by permuting its vertices e_{1}, …, e_{n+1}. This action is by affine maps and in fact by isometries. This implies that this action maps ellipsoids to ellipsoids and that it preserves their volume. As the ellipsoid of maximal volume in σ^{n} is unique this implies that it is left fixed by all elements of S_{n+1}. However it is not hard to check that the only ellipsoids in R^{n} invariant under all elements are the balls centered at (\frac{1}{n+1}, …, \frac{1}{n+1}). Using this and that all simplexes are affinely equivalent it is not hard to verify the claims in Remark 1.1.

The claims of Remark 3.1 can be checked in a similar manner.

References

1. Keith Ball, Ellipsoids of maximal volume in convex bodies, Geom. Dedicata 41 (1992), no. 2, 241–250.

2. Marcel Berger, Convexity, Amer. Math. Monthly 97 (1990), no. 8, 650–678.

3. Fritz John, Extremum problems with inequalities as subsidiary conditions, Studies and Essays Presented to R. Courant on his 60th Birthday, January 8, 1948, Interscience Publishers, Inc., New York, N. Y., 1948, pp. 187–204.