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18409 Topics in Theoretical Computer Science An Algorithmists Toolkit Fall 2007

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1

18409 An Algorithmistrsquos Toolkit November 1 2007

Lecture 15 Lecturer Jonathan Kelner Scribe Ali ParandehGheibi

The Brunn-Minkowski inequality

Definition 1 For A B sube Rn the Minkowski sum A oplus B is given by

A oplus B = a + b | a isin A b isin B (1)

The Minkowski sum can be defined for any subsets of Rn but it is nicely behaved if A and B are convex Intuitively the Minkowski sum is obtained by moving one of the sets around the boundary of the other one The following figures give two simple examples of Minkowski sums for convex sets

The Brunn-Minkowski inequality which relates the volume of A oplus B to the volumes of A and B implies many important theorems in convex geometry The goal is to bound vol(A + B) in terms of vol(A) and vol(B) The following are some loose bounds that can be simply verified

Fact 1 vol(A oplus B) ge maxvol(A) vol(B)

Proof Let a isin A We have a oplus B sube A oplus B by definition Hence

vol(A oplus B) ge vol(a oplus B) = vol(B)

Similarly vol(A oplus B) ge vol(B)

Fact 2 vol(A oplus B) ge vol(A) + vol(B)

Proof By moving one of the sets around the other one (summing the extreme points) we can get disjoint copies of A and B in A oplus B

The bound given by Fact 2 is loose To see that consider the case that A = B In this case A oplus A = 2A and hence vol(A oplus A) = 2nvol(A) So the volume of A oplus A grows exponentially with n while the lower bound given in the above fact do not This suggests taking the n-th roots and still get a valid bound Let us first prove it for boxes

Lemma 1 Let A and B be boxes in Rn then

1 1 1 n n n(vol(A oplus B)) ge (vol(A)) + (vol(B))

Proof Let A have sides a1 a2 an and B have sides b1 b2 bn It directly follows from the definition of Minkowski sum that A oplus B has sides a1 + b1 a2 + b2 an + bn

We just need to show the following

1 1 n n(vol(A)) + (vol(B)) le 1 (2)1

n(vol(A oplus B))

15-1

Figure 1 Minkowski sum examples

Figure 2 A+ and B+ as defined in proof of Theorem 1

We can rewrite the left-hand side of (2) as 1 1 1 1 n n n n( n

i=1 ai) + ( ni=1

1

bi) = ( ni=1 ai)

1 + ( ni=1 bi)

1 n n nn n n( i=1(ai + bi)) ( i=1(ai + bi)) ( i=1(ai + bi)) n 1 n 1

n n ai bi = +

(ai + bi) (ai + bi)i=1 i=1 n n1 ai 1 bi le

n ai + bi +

n ai + bi = 1 (3)

i=1 i=1

where the inequality follows from Arithmetic Mean - Geometric Mean (AM-GM) inequality

The following theorem generalizes the inequality to non-convex bodies that can be obtained by adding up boxes (measurable sets)

Theorem 1 Let A B sube Rn be compact measurable sets then

1 1 1 n n n(vol(A oplus B)) ge (vol(A)) + (vol(B)) (4)

The equality holds when A is a translation of a dilation of a translation of B up to zero-measure sets An equivalent version of Brunn-Minkowski inequality is given by 1 1 1

n nvol(λA oplus (1 minus λ)B) n ge λ vol(A) + (1 minus λ) vol(B) λ isin [0 1] (5)

15-2

The equivalence of (4) and (5) follows directly from the fact that vol(λA) = λnvol(A) 1 1 1 n nvol(λA oplus (1 minus λ)B)

n ge vol(λA) + vol((1 minus λ)B) 1 1 n n= λnvol(A) + (1 minus λ)nvol(B) 1 1

n n= λ vol(A) + (1 minus λ) vol(B) (6)

The inequality in (5) implies that n-th root of the volume function is concave with respect to Minkowski sum

Here we sketch the proof idea for Theorem 1 by proving (4) for any set constructed by a finite collection of boxes The proof can be generalized to any measurable set by approximating the set by a sequence of finite collection of boxes and taking the limit We omit the real analysis details for brevity

Let A and B be finite collection of boxes in Rn We prove (4) by induction on the number of boxes in A cup B Define the following subsets of Rn

A+ = A cap x isin Rn | xn ge 0 Aminus = A cap x isin Rn | xn le 0 B+ = B cap x isin Rn | xn ge 0 Bminus = B cap x isin Rn | xn le 0 (7)

Translate A and B such that the following conditions hold

(1) A has some pair of boxes separated by the hyperplane x isin Rn | xn = 0 ie there exists a box that lies completely in the half-space x isin Rn | xn ge 0 and there is some other box that lies in its complement half-space (see figure 2)

(2) It is true thatvol(A+) vol(B+)

= (8)vol(A) vol(B)

Note that translation of A or B just translates A oplus B so any statement about the translated sets is valid for the original ones

Since A+ and Aminus are strict subsets of A A+ cup B+ and Aminus cup Bminus have fewer boxes than A cup B Therefore (4) is true for them by induction hypothesis Moreover A+ oplusB+ and Aminus oplusBminus are disjoint because the former just contains points with non-negative xn coordinate and the latter lies in the half-space x isin Rn | xn lt 0 Hence we have

vol(A oplus B) ge vol( A+ oplus B+) + vol(Aminus oplus B

minus) n n n nvol(A+) 1

+ vol(B+) 1 n + vol(Aminus)

1 + vol(Bminus)

1 n ge vol(B+) 1 n vol(Bminus) 1 n

= vol(A+) 1 + n

+ vol(Aminus) 1 + n

vol(A+) vol(Aminus) vol(B) 1 n

= vol(A+) + vol(Aminus) 1 + n

vol(A) 1 n1 n n= vol(A) + vol(B) (9)

where the second inequality follows from the induction hypothesis and the second equality is implied by (8)

Applications of Brunn-Minkowski Inequality

In this section we demonstrate the power of Brunn-Minkowski inequality by proving some important theorems in convex geometry

15-3

2

21 Volumes of Parallel Slices

Let K sube Rn be a convex body A parallel slicedenoted by Kt is defined as an intersection of the body with a specific hyperplane ie

Kt = K cap x isin Rn | xn = t

Denote the volume of the parallel slice by vK (t) as (n minus 1)-dimensional volume of Kt ie

vK (t) = volnminus1(Kt) (10)

We are interested in the behavior of the function vK (t) and whether it is concave or not Consider the Euclidean ball in Rn The following plots of vK (t) for different n suggest that except for

n = 2 the function vK (t) is not concave in t As another example consider a circular cone in R3 the volume of parallel slices is proportional to t2 so

vK (t) is not a concave function More generally vK (t) is proportional to tnminus1 for a circular cone in Rn This suggests the (n minus 1)-th root of vK being a concave function This guess is verified by Brunnrsquos theorem

1Theorem 2 (Brunnrsquos Theorem) Let vK (t) be defined as in (10) then the function vK (t) nminus1

Proof Let K be the convex body Also let s r t isin R with s = (1 minus λ)r + λt for some λ isin [0 1] Define the (n minus 1)-dimensional slices Ar As At as follows

Ar = K cap x isin Rn | x1 = r

As = K cap x isin Rn | x1 = s

At = K cap x isin Rn | x1 = t

First we claim that (1 minus λ)Ar oplus λAt sube As (11)

To prove that we need to show the following

z = (1 minus λ)x oplus λy isin As for all x isin Ar y isin At

To show that connect the points (r x) and (t y) with a straight line (see figure 3) By convexity of K the line lies completely in the body In particular the point (s z) which is a convex combination of (r x) and (t y) lies in As Therefore z isin As and the claim in (11) is true Now by applying the equivalent version of Brunn-Minkowski inequality in (5) we have

1 1 1 nminus1vol(As) ge (1 minus λ)vol(Ar) nminus1 + λvol(At) nminus1

1 1 1

rArr vK (s) nminus1 ge (1 minus λ)vK (r) nminus1 + λvK (t)nminus1 (12)

22 Isoperimetric Inequality

A few lectures ago we asked the question of finding the body with the smallest surface area given a volume It turns out that the Euclidean balls have the smallest surface area This face is direct consequence of the Isoperimetric inequality Before stating the theorem let us define the surface area of a body using Minkowski sum

Let K be a body The surface area ((n-1)-dimensional volume) of K is defined as the rate of volume increase as we add a small Euclidean ball to the body ie

2S(K) = vol(partK) = lim vol(K oplus Bn) minus vol(K)

(13) 0 rarr

15-4

Figure by MIT OpenCourseWare

Figure 3 n-dimensional convex body K in Theorem 2

Theorem 3 (Isoperimetric inequality) For any convex body K with n-dimensional volume denoted by V (K)

22

22

2

2 (15)

2

2 (16)

22

2n nnV (K) nminus1

V (Bn) 1

2

22

2

2

1 nV (K)

nminus1 V (Bn)

1 1 n n

and surface area S(K) 1 1n

V

V ((B

Kn

) le S

S

((B

Kn

) nminus1

(14)) )

Proof By employing Brunn-Minkowski inequality we have the following n1 1V (K oplus Bn + V (Bn) ge V (K) )n n 1 n V (Bn) n

= V (K) 1 + V (K) V (Bn) 1

n V (K) 1 + n ge

V (K)

where the second inequality is obtained by keeping the first two terms of the Taylor expansion of (1 + x)n Now the definition of the surface area in (13) implies 1

V (K) + nV (K) V (B2 n) n minus V (K)

S(K) = V (partK) ge

V (K)

V (Bn) 1n

= nV (K) V (K)

nminus1 1 V (Bn= nV (K) )n n

It is straightforward to show that for a n-dimensional unit ball it holds that

S(Bn) = nV (Bn

Therefore

)

S(K) geS(Bn nV (Bn

S(Bn nV (Bn

) )

S(K) nminus1 nminus1

gerArr ) )

2

1n

= V

V ((B

Kn

) (17)

)

15-5

Figure 4 n-dimensional circular cone

23 Grunbaumrsquos Theorem

We would like to a pick a point x in a high-dimensional convex body so that if we cut the convex body with any hyperplane the piece containing x is big A reasonable choice for x is the centroid ie

1 x = ydy

vol(K) yisinK

This choice guarantees to get at least half of the volume for any origin symmetric body like a cube or a ball The question is how much we are guaranteed to get for general convex body and what is the worst case Do we get a constant fraction of the body or it grows or shrinks with the dimension

Let us first consider the simple example for a circular n-dimensional cone Suppose we cut the cone C by the hyperplane x1 = x1 at its centroid where

1 h

tR nminus1

n x1 =

vol(C) t=0 tvolnminus1

h dt =

n + 1 h (18)

Now we have vol(left cone) vol(base-rsquoleft conersquo) n 1

vol(C) =

vol(base minus C) n + 1 rarr

e as n rarrinfin

The Grunbaumrsquos theorem states that the circular cone is indeed the worst case if we choose the centroid

Theorem 4 (Grunbaumrsquos theorem) Let K be a convex body and divide it into K1 and K2 using a hyperplane If K1 contains the centroid of K then

vol(K1) 1vol(K)

ge e (19)

In particular the hyperplane through the centroid divides the volume into almost equal pieces and the worst case ratio is approximately 037063

15-6

Figure 1 Minkowski sum examples

Figure 2 A+ and B+ as defined in proof of Theorem 1

We can rewrite the left-hand side of (2) as 1 1 1 1 n n n n( n

i=1 ai) + ( ni=1

1

bi) = ( ni=1 ai)

1 + ( ni=1 bi)

1 n n nn n n( i=1(ai + bi)) ( i=1(ai + bi)) ( i=1(ai + bi)) n 1 n 1

n n ai bi = +

(ai + bi) (ai + bi)i=1 i=1 n n1 ai 1 bi le

n ai + bi +

n ai + bi = 1 (3)

i=1 i=1

where the inequality follows from Arithmetic Mean - Geometric Mean (AM-GM) inequality

The following theorem generalizes the inequality to non-convex bodies that can be obtained by adding up boxes (measurable sets)

Theorem 1 Let A B sube Rn be compact measurable sets then

1 1 1 n n n(vol(A oplus B)) ge (vol(A)) + (vol(B)) (4)

The equality holds when A is a translation of a dilation of a translation of B up to zero-measure sets An equivalent version of Brunn-Minkowski inequality is given by 1 1 1

n nvol(λA oplus (1 minus λ)B) n ge λ vol(A) + (1 minus λ) vol(B) λ isin [0 1] (5)

15-2

The equivalence of (4) and (5) follows directly from the fact that vol(λA) = λnvol(A) 1 1 1 n nvol(λA oplus (1 minus λ)B)

n ge vol(λA) + vol((1 minus λ)B) 1 1 n n= λnvol(A) + (1 minus λ)nvol(B) 1 1

n n= λ vol(A) + (1 minus λ) vol(B) (6)

The inequality in (5) implies that n-th root of the volume function is concave with respect to Minkowski sum

Here we sketch the proof idea for Theorem 1 by proving (4) for any set constructed by a finite collection of boxes The proof can be generalized to any measurable set by approximating the set by a sequence of finite collection of boxes and taking the limit We omit the real analysis details for brevity

Let A and B be finite collection of boxes in Rn We prove (4) by induction on the number of boxes in A cup B Define the following subsets of Rn

A+ = A cap x isin Rn | xn ge 0 Aminus = A cap x isin Rn | xn le 0 B+ = B cap x isin Rn | xn ge 0 Bminus = B cap x isin Rn | xn le 0 (7)

Translate A and B such that the following conditions hold

(1) A has some pair of boxes separated by the hyperplane x isin Rn | xn = 0 ie there exists a box that lies completely in the half-space x isin Rn | xn ge 0 and there is some other box that lies in its complement half-space (see figure 2)

(2) It is true thatvol(A+) vol(B+)

= (8)vol(A) vol(B)

Note that translation of A or B just translates A oplus B so any statement about the translated sets is valid for the original ones

Since A+ and Aminus are strict subsets of A A+ cup B+ and Aminus cup Bminus have fewer boxes than A cup B Therefore (4) is true for them by induction hypothesis Moreover A+ oplusB+ and Aminus oplusBminus are disjoint because the former just contains points with non-negative xn coordinate and the latter lies in the half-space x isin Rn | xn lt 0 Hence we have

vol(A oplus B) ge vol( A+ oplus B+) + vol(Aminus oplus B

minus) n n n nvol(A+) 1

+ vol(B+) 1 n + vol(Aminus)

1 + vol(Bminus)

1 n ge vol(B+) 1 n vol(Bminus) 1 n

= vol(A+) 1 + n

+ vol(Aminus) 1 + n

vol(A+) vol(Aminus) vol(B) 1 n

= vol(A+) + vol(Aminus) 1 + n

vol(A) 1 n1 n n= vol(A) + vol(B) (9)

where the second inequality follows from the induction hypothesis and the second equality is implied by (8)

Applications of Brunn-Minkowski Inequality

In this section we demonstrate the power of Brunn-Minkowski inequality by proving some important theorems in convex geometry

15-3

2

21 Volumes of Parallel Slices

Let K sube Rn be a convex body A parallel slicedenoted by Kt is defined as an intersection of the body with a specific hyperplane ie

Kt = K cap x isin Rn | xn = t

Denote the volume of the parallel slice by vK (t) as (n minus 1)-dimensional volume of Kt ie

vK (t) = volnminus1(Kt) (10)

We are interested in the behavior of the function vK (t) and whether it is concave or not Consider the Euclidean ball in Rn The following plots of vK (t) for different n suggest that except for

n = 2 the function vK (t) is not concave in t As another example consider a circular cone in R3 the volume of parallel slices is proportional to t2 so

vK (t) is not a concave function More generally vK (t) is proportional to tnminus1 for a circular cone in Rn This suggests the (n minus 1)-th root of vK being a concave function This guess is verified by Brunnrsquos theorem

1Theorem 2 (Brunnrsquos Theorem) Let vK (t) be defined as in (10) then the function vK (t) nminus1

Proof Let K be the convex body Also let s r t isin R with s = (1 minus λ)r + λt for some λ isin [0 1] Define the (n minus 1)-dimensional slices Ar As At as follows

Ar = K cap x isin Rn | x1 = r

As = K cap x isin Rn | x1 = s

At = K cap x isin Rn | x1 = t

First we claim that (1 minus λ)Ar oplus λAt sube As (11)

To prove that we need to show the following

z = (1 minus λ)x oplus λy isin As for all x isin Ar y isin At

To show that connect the points (r x) and (t y) with a straight line (see figure 3) By convexity of K the line lies completely in the body In particular the point (s z) which is a convex combination of (r x) and (t y) lies in As Therefore z isin As and the claim in (11) is true Now by applying the equivalent version of Brunn-Minkowski inequality in (5) we have

1 1 1 nminus1vol(As) ge (1 minus λ)vol(Ar) nminus1 + λvol(At) nminus1

1 1 1

rArr vK (s) nminus1 ge (1 minus λ)vK (r) nminus1 + λvK (t)nminus1 (12)

22 Isoperimetric Inequality

A few lectures ago we asked the question of finding the body with the smallest surface area given a volume It turns out that the Euclidean balls have the smallest surface area This face is direct consequence of the Isoperimetric inequality Before stating the theorem let us define the surface area of a body using Minkowski sum

Let K be a body The surface area ((n-1)-dimensional volume) of K is defined as the rate of volume increase as we add a small Euclidean ball to the body ie

2S(K) = vol(partK) = lim vol(K oplus Bn) minus vol(K)

(13) 0 rarr

15-4

Figure by MIT OpenCourseWare

Figure 3 n-dimensional convex body K in Theorem 2

Theorem 3 (Isoperimetric inequality) For any convex body K with n-dimensional volume denoted by V (K)

22

22

2

2 (15)

2

2 (16)

22

2n nnV (K) nminus1

V (Bn) 1

2

22

2

2

1 nV (K)

nminus1 V (Bn)

1 1 n n

and surface area S(K) 1 1n

V

V ((B

Kn

) le S

S

((B

Kn

) nminus1

(14)) )

Proof By employing Brunn-Minkowski inequality we have the following n1 1V (K oplus Bn + V (Bn) ge V (K) )n n 1 n V (Bn) n

= V (K) 1 + V (K) V (Bn) 1

n V (K) 1 + n ge

V (K)

where the second inequality is obtained by keeping the first two terms of the Taylor expansion of (1 + x)n Now the definition of the surface area in (13) implies 1

V (K) + nV (K) V (B2 n) n minus V (K)

S(K) = V (partK) ge

V (K)

V (Bn) 1n

= nV (K) V (K)

nminus1 1 V (Bn= nV (K) )n n

It is straightforward to show that for a n-dimensional unit ball it holds that

S(Bn) = nV (Bn

Therefore

)

S(K) geS(Bn nV (Bn

S(Bn nV (Bn

) )

S(K) nminus1 nminus1

gerArr ) )

2

1n

= V

V ((B

Kn

) (17)

)

15-5

Figure 4 n-dimensional circular cone

23 Grunbaumrsquos Theorem

We would like to a pick a point x in a high-dimensional convex body so that if we cut the convex body with any hyperplane the piece containing x is big A reasonable choice for x is the centroid ie

1 x = ydy

vol(K) yisinK

This choice guarantees to get at least half of the volume for any origin symmetric body like a cube or a ball The question is how much we are guaranteed to get for general convex body and what is the worst case Do we get a constant fraction of the body or it grows or shrinks with the dimension

Let us first consider the simple example for a circular n-dimensional cone Suppose we cut the cone C by the hyperplane x1 = x1 at its centroid where

1 h

tR nminus1

n x1 =

vol(C) t=0 tvolnminus1

h dt =

n + 1 h (18)

Now we have vol(left cone) vol(base-rsquoleft conersquo) n 1

vol(C) =

vol(base minus C) n + 1 rarr

e as n rarrinfin

The Grunbaumrsquos theorem states that the circular cone is indeed the worst case if we choose the centroid

Theorem 4 (Grunbaumrsquos theorem) Let K be a convex body and divide it into K1 and K2 using a hyperplane If K1 contains the centroid of K then

vol(K1) 1vol(K)

ge e (19)

In particular the hyperplane through the centroid divides the volume into almost equal pieces and the worst case ratio is approximately 037063

15-6

The equivalence of (4) and (5) follows directly from the fact that vol(λA) = λnvol(A) 1 1 1 n nvol(λA oplus (1 minus λ)B)

n ge vol(λA) + vol((1 minus λ)B) 1 1 n n= λnvol(A) + (1 minus λ)nvol(B) 1 1

n n= λ vol(A) + (1 minus λ) vol(B) (6)

The inequality in (5) implies that n-th root of the volume function is concave with respect to Minkowski sum

Here we sketch the proof idea for Theorem 1 by proving (4) for any set constructed by a finite collection of boxes The proof can be generalized to any measurable set by approximating the set by a sequence of finite collection of boxes and taking the limit We omit the real analysis details for brevity

Let A and B be finite collection of boxes in Rn We prove (4) by induction on the number of boxes in A cup B Define the following subsets of Rn

A+ = A cap x isin Rn | xn ge 0 Aminus = A cap x isin Rn | xn le 0 B+ = B cap x isin Rn | xn ge 0 Bminus = B cap x isin Rn | xn le 0 (7)

Translate A and B such that the following conditions hold

(1) A has some pair of boxes separated by the hyperplane x isin Rn | xn = 0 ie there exists a box that lies completely in the half-space x isin Rn | xn ge 0 and there is some other box that lies in its complement half-space (see figure 2)

(2) It is true thatvol(A+) vol(B+)

= (8)vol(A) vol(B)

Note that translation of A or B just translates A oplus B so any statement about the translated sets is valid for the original ones

Since A+ and Aminus are strict subsets of A A+ cup B+ and Aminus cup Bminus have fewer boxes than A cup B Therefore (4) is true for them by induction hypothesis Moreover A+ oplusB+ and Aminus oplusBminus are disjoint because the former just contains points with non-negative xn coordinate and the latter lies in the half-space x isin Rn | xn lt 0 Hence we have

vol(A oplus B) ge vol( A+ oplus B+) + vol(Aminus oplus B

minus) n n n nvol(A+) 1

+ vol(B+) 1 n + vol(Aminus)

1 + vol(Bminus)

1 n ge vol(B+) 1 n vol(Bminus) 1 n

= vol(A+) 1 + n

+ vol(Aminus) 1 + n

vol(A+) vol(Aminus) vol(B) 1 n

= vol(A+) + vol(Aminus) 1 + n

vol(A) 1 n1 n n= vol(A) + vol(B) (9)

where the second inequality follows from the induction hypothesis and the second equality is implied by (8)

Applications of Brunn-Minkowski Inequality

In this section we demonstrate the power of Brunn-Minkowski inequality by proving some important theorems in convex geometry

15-3

2

21 Volumes of Parallel Slices

Let K sube Rn be a convex body A parallel slicedenoted by Kt is defined as an intersection of the body with a specific hyperplane ie

Kt = K cap x isin Rn | xn = t

Denote the volume of the parallel slice by vK (t) as (n minus 1)-dimensional volume of Kt ie

vK (t) = volnminus1(Kt) (10)

We are interested in the behavior of the function vK (t) and whether it is concave or not Consider the Euclidean ball in Rn The following plots of vK (t) for different n suggest that except for

n = 2 the function vK (t) is not concave in t As another example consider a circular cone in R3 the volume of parallel slices is proportional to t2 so

vK (t) is not a concave function More generally vK (t) is proportional to tnminus1 for a circular cone in Rn This suggests the (n minus 1)-th root of vK being a concave function This guess is verified by Brunnrsquos theorem

1Theorem 2 (Brunnrsquos Theorem) Let vK (t) be defined as in (10) then the function vK (t) nminus1

Proof Let K be the convex body Also let s r t isin R with s = (1 minus λ)r + λt for some λ isin [0 1] Define the (n minus 1)-dimensional slices Ar As At as follows

Ar = K cap x isin Rn | x1 = r

As = K cap x isin Rn | x1 = s

At = K cap x isin Rn | x1 = t

First we claim that (1 minus λ)Ar oplus λAt sube As (11)

To prove that we need to show the following

z = (1 minus λ)x oplus λy isin As for all x isin Ar y isin At

To show that connect the points (r x) and (t y) with a straight line (see figure 3) By convexity of K the line lies completely in the body In particular the point (s z) which is a convex combination of (r x) and (t y) lies in As Therefore z isin As and the claim in (11) is true Now by applying the equivalent version of Brunn-Minkowski inequality in (5) we have

1 1 1 nminus1vol(As) ge (1 minus λ)vol(Ar) nminus1 + λvol(At) nminus1

1 1 1

rArr vK (s) nminus1 ge (1 minus λ)vK (r) nminus1 + λvK (t)nminus1 (12)

22 Isoperimetric Inequality

A few lectures ago we asked the question of finding the body with the smallest surface area given a volume It turns out that the Euclidean balls have the smallest surface area This face is direct consequence of the Isoperimetric inequality Before stating the theorem let us define the surface area of a body using Minkowski sum

Let K be a body The surface area ((n-1)-dimensional volume) of K is defined as the rate of volume increase as we add a small Euclidean ball to the body ie

2S(K) = vol(partK) = lim vol(K oplus Bn) minus vol(K)

(13) 0 rarr

15-4

Figure by MIT OpenCourseWare

Figure 3 n-dimensional convex body K in Theorem 2

Theorem 3 (Isoperimetric inequality) For any convex body K with n-dimensional volume denoted by V (K)

22

22

2

2 (15)

2

2 (16)

22

2n nnV (K) nminus1

V (Bn) 1

2

22

2

2

1 nV (K)

nminus1 V (Bn)

1 1 n n

and surface area S(K) 1 1n

V

V ((B

Kn

) le S

S

((B

Kn

) nminus1

(14)) )

Proof By employing Brunn-Minkowski inequality we have the following n1 1V (K oplus Bn + V (Bn) ge V (K) )n n 1 n V (Bn) n

= V (K) 1 + V (K) V (Bn) 1

n V (K) 1 + n ge

V (K)

where the second inequality is obtained by keeping the first two terms of the Taylor expansion of (1 + x)n Now the definition of the surface area in (13) implies 1

V (K) + nV (K) V (B2 n) n minus V (K)

S(K) = V (partK) ge

V (K)

V (Bn) 1n

= nV (K) V (K)

nminus1 1 V (Bn= nV (K) )n n

It is straightforward to show that for a n-dimensional unit ball it holds that

S(Bn) = nV (Bn

Therefore

)

S(K) geS(Bn nV (Bn

S(Bn nV (Bn

) )

S(K) nminus1 nminus1

gerArr ) )

2

1n

= V

V ((B

Kn

) (17)

)

15-5

Figure 4 n-dimensional circular cone

23 Grunbaumrsquos Theorem

We would like to a pick a point x in a high-dimensional convex body so that if we cut the convex body with any hyperplane the piece containing x is big A reasonable choice for x is the centroid ie

1 x = ydy

vol(K) yisinK

This choice guarantees to get at least half of the volume for any origin symmetric body like a cube or a ball The question is how much we are guaranteed to get for general convex body and what is the worst case Do we get a constant fraction of the body or it grows or shrinks with the dimension

Let us first consider the simple example for a circular n-dimensional cone Suppose we cut the cone C by the hyperplane x1 = x1 at its centroid where

1 h

tR nminus1

n x1 =

vol(C) t=0 tvolnminus1

h dt =

n + 1 h (18)

Now we have vol(left cone) vol(base-rsquoleft conersquo) n 1

vol(C) =

vol(base minus C) n + 1 rarr

e as n rarrinfin

The Grunbaumrsquos theorem states that the circular cone is indeed the worst case if we choose the centroid

Theorem 4 (Grunbaumrsquos theorem) Let K be a convex body and divide it into K1 and K2 using a hyperplane If K1 contains the centroid of K then

vol(K1) 1vol(K)

ge e (19)

In particular the hyperplane through the centroid divides the volume into almost equal pieces and the worst case ratio is approximately 037063

15-6

21 Volumes of Parallel Slices

Let K sube Rn be a convex body A parallel slicedenoted by Kt is defined as an intersection of the body with a specific hyperplane ie

Kt = K cap x isin Rn | xn = t

Denote the volume of the parallel slice by vK (t) as (n minus 1)-dimensional volume of Kt ie

vK (t) = volnminus1(Kt) (10)

We are interested in the behavior of the function vK (t) and whether it is concave or not Consider the Euclidean ball in Rn The following plots of vK (t) for different n suggest that except for

n = 2 the function vK (t) is not concave in t As another example consider a circular cone in R3 the volume of parallel slices is proportional to t2 so

vK (t) is not a concave function More generally vK (t) is proportional to tnminus1 for a circular cone in Rn This suggests the (n minus 1)-th root of vK being a concave function This guess is verified by Brunnrsquos theorem

1Theorem 2 (Brunnrsquos Theorem) Let vK (t) be defined as in (10) then the function vK (t) nminus1

Proof Let K be the convex body Also let s r t isin R with s = (1 minus λ)r + λt for some λ isin [0 1] Define the (n minus 1)-dimensional slices Ar As At as follows

Ar = K cap x isin Rn | x1 = r

As = K cap x isin Rn | x1 = s

At = K cap x isin Rn | x1 = t

First we claim that (1 minus λ)Ar oplus λAt sube As (11)

To prove that we need to show the following

z = (1 minus λ)x oplus λy isin As for all x isin Ar y isin At

To show that connect the points (r x) and (t y) with a straight line (see figure 3) By convexity of K the line lies completely in the body In particular the point (s z) which is a convex combination of (r x) and (t y) lies in As Therefore z isin As and the claim in (11) is true Now by applying the equivalent version of Brunn-Minkowski inequality in (5) we have

1 1 1 nminus1vol(As) ge (1 minus λ)vol(Ar) nminus1 + λvol(At) nminus1

1 1 1

rArr vK (s) nminus1 ge (1 minus λ)vK (r) nminus1 + λvK (t)nminus1 (12)

22 Isoperimetric Inequality

A few lectures ago we asked the question of finding the body with the smallest surface area given a volume It turns out that the Euclidean balls have the smallest surface area This face is direct consequence of the Isoperimetric inequality Before stating the theorem let us define the surface area of a body using Minkowski sum

Let K be a body The surface area ((n-1)-dimensional volume) of K is defined as the rate of volume increase as we add a small Euclidean ball to the body ie

2S(K) = vol(partK) = lim vol(K oplus Bn) minus vol(K)

(13) 0 rarr

15-4

Figure by MIT OpenCourseWare

Figure 3 n-dimensional convex body K in Theorem 2

Theorem 3 (Isoperimetric inequality) For any convex body K with n-dimensional volume denoted by V (K)

22

22

2

2 (15)

2

2 (16)

22

2n nnV (K) nminus1

V (Bn) 1

2

22

2

2

1 nV (K)

nminus1 V (Bn)

1 1 n n

and surface area S(K) 1 1n

V

V ((B

Kn

) le S

S

((B

Kn

) nminus1

(14)) )

Proof By employing Brunn-Minkowski inequality we have the following n1 1V (K oplus Bn + V (Bn) ge V (K) )n n 1 n V (Bn) n

= V (K) 1 + V (K) V (Bn) 1

n V (K) 1 + n ge

V (K)

where the second inequality is obtained by keeping the first two terms of the Taylor expansion of (1 + x)n Now the definition of the surface area in (13) implies 1

V (K) + nV (K) V (B2 n) n minus V (K)

S(K) = V (partK) ge

V (K)

V (Bn) 1n

= nV (K) V (K)

nminus1 1 V (Bn= nV (K) )n n

It is straightforward to show that for a n-dimensional unit ball it holds that

S(Bn) = nV (Bn

Therefore

)

S(K) geS(Bn nV (Bn

S(Bn nV (Bn

) )

S(K) nminus1 nminus1

gerArr ) )

2

1n

= V

V ((B

Kn

) (17)

)

15-5

Figure 4 n-dimensional circular cone

23 Grunbaumrsquos Theorem

We would like to a pick a point x in a high-dimensional convex body so that if we cut the convex body with any hyperplane the piece containing x is big A reasonable choice for x is the centroid ie

1 x = ydy

vol(K) yisinK

This choice guarantees to get at least half of the volume for any origin symmetric body like a cube or a ball The question is how much we are guaranteed to get for general convex body and what is the worst case Do we get a constant fraction of the body or it grows or shrinks with the dimension

Let us first consider the simple example for a circular n-dimensional cone Suppose we cut the cone C by the hyperplane x1 = x1 at its centroid where

1 h

tR nminus1

n x1 =

vol(C) t=0 tvolnminus1

h dt =

n + 1 h (18)

Now we have vol(left cone) vol(base-rsquoleft conersquo) n 1

vol(C) =

vol(base minus C) n + 1 rarr

e as n rarrinfin

The Grunbaumrsquos theorem states that the circular cone is indeed the worst case if we choose the centroid

Theorem 4 (Grunbaumrsquos theorem) Let K be a convex body and divide it into K1 and K2 using a hyperplane If K1 contains the centroid of K then

vol(K1) 1vol(K)

ge e (19)

In particular the hyperplane through the centroid divides the volume into almost equal pieces and the worst case ratio is approximately 037063

15-6

Figure by MIT OpenCourseWare

Figure 3 n-dimensional convex body K in Theorem 2

Theorem 3 (Isoperimetric inequality) For any convex body K with n-dimensional volume denoted by V (K)

22

22

2

2 (15)

2

2 (16)

22

2n nnV (K) nminus1

V (Bn) 1

2

22

2

2

1 nV (K)

nminus1 V (Bn)

1 1 n n

and surface area S(K) 1 1n

V

V ((B

Kn

) le S

S

((B

Kn

) nminus1

(14)) )

Proof By employing Brunn-Minkowski inequality we have the following n1 1V (K oplus Bn + V (Bn) ge V (K) )n n 1 n V (Bn) n

= V (K) 1 + V (K) V (Bn) 1

n V (K) 1 + n ge

V (K)

where the second inequality is obtained by keeping the first two terms of the Taylor expansion of (1 + x)n Now the definition of the surface area in (13) implies 1

V (K) + nV (K) V (B2 n) n minus V (K)

S(K) = V (partK) ge

V (K)

V (Bn) 1n

= nV (K) V (K)

nminus1 1 V (Bn= nV (K) )n n

It is straightforward to show that for a n-dimensional unit ball it holds that

S(Bn) = nV (Bn

Therefore

)

S(K) geS(Bn nV (Bn

S(Bn nV (Bn

) )

S(K) nminus1 nminus1

gerArr ) )

2

1n

= V

V ((B

Kn

) (17)

)

15-5

Figure 4 n-dimensional circular cone

23 Grunbaumrsquos Theorem

We would like to a pick a point x in a high-dimensional convex body so that if we cut the convex body with any hyperplane the piece containing x is big A reasonable choice for x is the centroid ie

1 x = ydy

vol(K) yisinK

This choice guarantees to get at least half of the volume for any origin symmetric body like a cube or a ball The question is how much we are guaranteed to get for general convex body and what is the worst case Do we get a constant fraction of the body or it grows or shrinks with the dimension

Let us first consider the simple example for a circular n-dimensional cone Suppose we cut the cone C by the hyperplane x1 = x1 at its centroid where

1 h

tR nminus1

n x1 =

vol(C) t=0 tvolnminus1

h dt =

n + 1 h (18)

Now we have vol(left cone) vol(base-rsquoleft conersquo) n 1

vol(C) =

vol(base minus C) n + 1 rarr

e as n rarrinfin

The Grunbaumrsquos theorem states that the circular cone is indeed the worst case if we choose the centroid

Theorem 4 (Grunbaumrsquos theorem) Let K be a convex body and divide it into K1 and K2 using a hyperplane If K1 contains the centroid of K then

vol(K1) 1vol(K)

ge e (19)

In particular the hyperplane through the centroid divides the volume into almost equal pieces and the worst case ratio is approximately 037063

15-6

Figure 4 n-dimensional circular cone

23 Grunbaumrsquos Theorem

We would like to a pick a point x in a high-dimensional convex body so that if we cut the convex body with any hyperplane the piece containing x is big A reasonable choice for x is the centroid ie

1 x = ydy

vol(K) yisinK

This choice guarantees to get at least half of the volume for any origin symmetric body like a cube or a ball The question is how much we are guaranteed to get for general convex body and what is the worst case Do we get a constant fraction of the body or it grows or shrinks with the dimension

Let us first consider the simple example for a circular n-dimensional cone Suppose we cut the cone C by the hyperplane x1 = x1 at its centroid where

1 h

tR nminus1

n x1 =

vol(C) t=0 tvolnminus1

h dt =

n + 1 h (18)

Now we have vol(left cone) vol(base-rsquoleft conersquo) n 1

vol(C) =

vol(base minus C) n + 1 rarr

e as n rarrinfin

The Grunbaumrsquos theorem states that the circular cone is indeed the worst case if we choose the centroid

Theorem 4 (Grunbaumrsquos theorem) Let K be a convex body and divide it into K1 and K2 using a hyperplane If K1 contains the centroid of K then

vol(K1) 1vol(K)

ge e (19)

In particular the hyperplane through the centroid divides the volume into almost equal pieces and the worst case ratio is approximately 037063

15-6