THE JOHN ELLIPSOID THEOREM RALPH HOWARD DEPARTMENT OF MATHEMATICS UNIVERSITY OF SOUTH CAROLINA COLUMBIA, S.C. 29208, USA [email protected] 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 a±ne map ©: Rn ! Rn is a map of the form ©(x):=Ax + b where A: Rn ! Rn is linear and b 2 Rn is a constant vector. This is nonsingular i® det A =0.An6 ellipsoid in Rn the image of the closed unit ball Bn of Rn under a nonsingular a±ne map. Our goal here is to prove the following famous result of Fritz John. Theorem 1 (John [3]). Let K ½ Rn 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 fc + n(x ¡ c):x2Eg. This is the dilation of E by a factor of n with center c.) Date: November 1997. 1 2 RALPH HOWARD 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 x0;:::;xn be an a±nely independent set of points in Rn and let K be the convex hull of n fx0;:::;xng. That is K is an n dimensional simplex in R . Then if E is the ellipsoid of maximal volume in K then E ⊆ K ⊆ c + n(E ¡ c), but there is no ellipsoid E0 ⊆ K so that E0 ⊆ K ⊆ c0 + m(E0 ¡ c0) 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 ex- pository article [2] of Berger where he gives a similar proof in the case K is symmetric about the origin. For a di®erent 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 di®erent 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 Rn. The basic idea of the proof is to choose E ½ Rn to be an ellipsoid of maximal volume. Then by an a±ne change of variables we can assume that E is the unit ball Bn. The proof is completed by showing that if K contains a point p at a distance greater than n from the origin of Rn then the convex hull of Bn [fpg, and thus also K, contains an ellipsoid of volume greater than Bn which would contradict that Bn has maximal volume. See Figure 1 Bn n E If kpk is large enough then convfB ;pg will contain an ellipsoid E with Vol(E) > Vol(B n ). p Figure 1 We start by showing that K contains an ellipsoid of maximal volume. If ©(x)=Ax + b is an a±ne max then the volume of the ellipsoid E =©[Bn] is Vol(E)=jdet(A)j Vol(B n ). This is a standard piece of a±ne 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 + n2. Then the set of ordered pairs (A; b) where A is an n £ n matrix and b 2 Rn is just RN . Let E := f(A; b) 2 RN : ABn + b ½ THE JOHN ELLIPSOID THEOREM 3 Kg. Then E is a closed bounded and thus compact subset of RN and (A; b) 7! j det(A)j is a continuous function on E. Thus there is an (A0;b0) n that maximizes this function on E. Then E := A0B + b0 is the desired ellipsoid. ¤ n If E = A0B +b0 is an ellipsoid of maximal volume in K then by replacing ¡1 n ¡1 K by A0 (K ¡ b0) we can assume that B = A0 (E ¡ b0) is an ellipsoid of maximal volume in K. Then to prove John's Theorem it is enough to show that if p 2 K then kpk·n(where k¢k is the standard Euclidean norm). The geometric idea of the proof is shown in Figure 1. If kpk is large then the convex hull convfBn;pg of Bn and p will contain an ellipsoid E with Vol(E) > Vol(B n ). But K is convex so E ½ convfBn;p}⊆Kwhich contradicts that Bn is an ellipsoid of maximal volume in K. What takes some work is showing the critical distance where Bn stops having maximal volume in convfBn;pg is kpk >n. x=¡1 B2 ¸ 2 ©t[B ] Figure 2 We will construct the ellipsoid E as an a±ne image of the ball Bn. This will ¯rst 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 di®erential equation to be taken seriously, we will construct our a±ne maps as flows of solutions to di®erential equations. Let ¸>0 then in the plane consider the system of equations x_ =1+x y_=¡¸y: The solution with initial condition x(0) = x0 and y(0) = y0 is t x(t)=¡1+e(1 + x0) ¡¸t y(t)=e y0: ¸ Therefore if ©t is the one parameter group of di®eomorphisms generated by this system of equations (or what is the same thing the one parameter @ @ group generated by the vector ¯eld (1 + x) @x ¡ ¸@y)is ¸ t ¡¸t t t ¡¸t ©t(x; y)=(¡1+e(1 + x);e y)=(¡1+e;0)+(ex; e y) 4 RALPH HOWARD ¸ so that for each ¯xed t the map (x; y) 7! ©t (x; y) is an a±ne map. The ¸ ¸ lines x = ¡1 and y = 0 are ¯xed (set-wise) by ©t . The e®ect of ©t on the two dimensional ball B2 for a small positive value of t is shown in Figure 2. 2 For a>1 let Ca := convfB ; (a; 0)g be the convex hull of the two dimen- sion ball B2 and the point (a; 0). See Figure 3 2 B (a; 0) 2 Ca is convex hull of B and (a; 0). Figure 3 The following is the hard step in the proof of John's theorem. 1 ¸ 2 Lemma 2.2. If ¸> then ©t [B ] ½ Ca for small positive values of t. a¡1 Proof. We ¯rst note that the tangent lines to @B2 through (a; 0) (which are C y p§1 x ¡ a part of the boundary of a) are = a2¡1 ( ). To see this consider the y p¡1 x ¡ a a; line = 2 ( ). Direct calculation shows that both the points ( 0) ³ pa ¡1´ ³ p ´ 1 a2¡1 1 a2¡1 and a ; a are on this line and the point a ; a is on the unit ³ p ´ 2 circle (see Figure 4). The slope of the radius to @B2 ending at 1 ; a ¡1 p a a a2 ¡ y p¡1 x ¡ a p¡1 is 1). The slope of = 2 ( )is 2 which is the negative a ¡1 a ¡1³ p ´ 2 1 a2¡1 reciprocal of the slope of the redius to @B ending at a ; a . Therefore y p¡1 x¡a @B2 = a2¡1 ( ) is prependicular to the radius where it meets and this is tangent. µ p ¶ 2 1 ; a ¡1 a a y = p¡1 (x ¡ a) a2¡1 (a; 0) Figure 4 Therefore to prove the lemma it is enough to show that for ¸> 1 and ³ p ´ a¡1 2 t ¸ 1 ; a ¡1 y p¡1 x ¡ a small that ©t a a lies below the line = a2¡1 ( ). (For as THE JOHN ELLIPSOID THEOREM 5 ³ p ´ 2 1 a2¡1 @B is tangent to this line at a ; a it will also be caried into Ca by ¸ ©t . See Figure 5.) µ p ¶ 2 1 ; a ¡1 a a µ p ¶ 2 ¸ 1 ; a ¡1 ©t a a Figure 5 ¸ Either directly from the de¯nition of ©t or from the system of di®erential equations de¯ning it we have ¯ µ p ¶¯ µ p ¶ d 1 a2 ¡ 1 ¯ 1 ¡¸ a2 ¡1 ¸ ; ¯ x ; y ; : dt©t a a ¯ =(_(0) _(0)) = 1+a a t=0 p ¡ 2 ¢ Therefore the slope of the tangent to the flow line t 7! ©¸ 1 ; a ¡1 at ¡ p ¢ t a a 1 ; a2¡1 a a is p y_(0) ¡¸ a2 ¡ 1 = : x_(0) a +1 p ¡ 2 ¢ ¸ 1 ; a ¡1 y p¡1 x ¡ a Therfore ©t a a will be below the line = a2¡1 ( ) provided p ¡¸ a2 ¡ ¡ 1 < p 1 : a +1 a2 ¡1 1 A little algebra shows this is equivalent to ¸> a¡1.