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Estimates for the extremal sections of complex p -balls

Abstract

The problem of maximal hyperplane section of B p ( C n ) with p1 is considered, which is the complex version of central hyperplane section problem of B p ( R n ). The relation between the complex slicing problem and the complex isotropic constant of a body is established, an upper bound estimate for the volume of complex central hyperplane sections of normalized complex p ( C n )-balls that does not depend on n and p is shown, which extends results of Oleszkiewicz and Pełczyński, Koldobsky and Zymonopoulou, and Meyer and Pajor.

MSC:52A21, 46B07.

1 Introduction

Let B p ( R n ) and B p ( C n ) denote the unit balls of the real and complex n-dimensional p spaces, p ( R n ) and p ( C n ), respectively. We denote by λ n (K) the n-dimensional Lebesgue measure of a compact set K. We write r n , p = λ n ( B p ( R n ) ) 1 / n and c n , p = λ 2 n ( B p ( C n ) ) 1 / 2 n such that λ n ( r n , p B p ( R n ))=1 and λ 2 n ( c n , p B p ( C n ))=1, respectively.

The extremal volume of central hyperplane section of B p ( R n ) is studied by various authors (see, e.g., [18]). Especially, the left-hand side of the following inequalities is known due to Meyer and Pajor [6] for all p2 and p=1, and Schmuckenschläger [8] for all 1<p<2. The right-hand side of the following inequalities is due to Ma and the third named author [7], and it shows an upper bound estimate for the volume of central hyperplane sections of normalized p -balls that does not depend on n and p.

Theorem 1.1 Let nN, n2, p1, and H any central hyperplane in R n . Then

1 λ n 1 ( r n , p B p ( R n ) H ) π e .

Moreover, the minimum occurs for B ( R n ) if ξ has only one non-zero coordinate where ξ is the normal vector of H.

Note that Theorem 1.1 is proved by determining the extremal value of the isotropic constant of B p ( R n ) together with the well-known relation between the slicing problem and the isotropic constant of a body. Motivated by this idea, we define a new quantity, called the complex isotropic constant, and establish its relation to the complex slicing problem. Thus, the complex version of Theorem 1.1 (see Theorem 1.2) can be proved by estimating the extremal value of the complex isotropic constant of B p ( C n ).

A noteworthy fact is that the extremal volume of complex central hyperplane section of B p ( C n ) has not been studied until recent years. Other results concerning convex bodies in a complex vector space as ambient space can be found in [919]. Especially, in [17], Oleszkiewicz and Pełczyński proved that 1 λ 2 n 2 ( c n , B ( C n ) H ξ )2, where H ξ is the complex central hyperplane (see Section 2.1 for the definition). Furthermore, they showed that the minimal sections are the ones orthogonal to vectors with only one non-zero coordinate, and the maximal sections are orthogonal to vectors of the form e j +σ e k , where jk, e j and e k are standard basic vectors, and σC, |σ|=1. In [16], Koldobsky and Zymonopoulou studied the extremal sections of B p ( C n ), for 0<p2 and showed that the minimum corresponds to hyperplanes orthogonal to vectors ξ=( ξ 1 ,, ξ n ) C n with | ξ 1 |==| ξ n | and the maximum corresponds to hyperplanes orthogonal to vectors with only one non-zero coordinate. Moreover, a result of Meyer and Pajor [[6], Corollary 2.5] states the following. Suppose p2, then λ 2 n 2 ( c n 1 , p B p ( C n ) H ξ )1; Suppose 1p2, then λ 2 n 2 ( c n 1 , p B p ( C n ) H ξ )1, where c n 1 , p = λ 2 n 2 ( B p ( C n 1 ) ) 1 / ( 2 n 2 ) . Recently, Koldobsky and König [13] considered minimal volume of slabs for the complex cube.

The case p2 of the following theorem follows directly from the work of Koldobsky and Zymonopoulou [16]. The following theorem extends their results to p>2, and it also shows an upper bound estimate for the volume of complex central hyperplane sections of normalized complex p -balls that does not depend on n and p.

Theorem 1.2 Let nN, n2, p1, and H ξ any complex hyperplane in R 2 n . Then

1 λ 2 n 2 ( c n , p B p ( C n ) H ξ ) 3e.

Moreover, the minimum occurs for B ( C n ) if ξ has only one non-zero coordinate.

Our problem is different from the extremal volume of central hyperplane section of B p ( R 2 n ) problem in two aspects. First, B p ( C n ) B p ( R 2 n ) except p=2; Secondly, we do only (2n2)-dimensional sections, sections by subspaces coming from complex hyperplanes, rather than all (2n2)-dimensional sections, and (2n1)-dimensional sections in real case.

2 Notations and preliminaries

2.1 Background on complex vector space

Throughout this paper, we denote their real scalar product by x,y and the Euclidean norm of x by x= x , x for x,y R n . For x,y C n , their complex scalar product by x , y c and the modulus of x by x= x , x c .

Origin-symmetric convex bodies in C n are the unit balls of norms on C n . We denote by K the norm corresponding to the body K:

K= { z C n : K 1 } .

We identify C n with R 2 n using the standard mapping, that is, for

ξ = ( ξ 1 , ξ n ) = ( ξ 11 + i ξ 12 , , ξ n 1 + i ξ n 2 ) C n , ( ξ 11 + i ξ 12 , , ξ n 1 + i ξ n 2 ) τ ( ξ 11 , ξ 12 , , ξ n 1 , ξ n 2 ) .
(2.1)

Since norms on C n satisfy the equality

λz=|λ|z,z C n ,λC,

origin-symmetric complex convex bodies correspond to those origin-symmetric convex bodies K in R 2 n that are invariant with respect to any coordinate-wise two-dimensional rotation, namely for each θ[0,2π] and each ( ξ 11 , ξ 12 , ξ n 1 , ξ n 2 ) R 2 n

ξ K = R θ ( ξ 11 , ξ 12 ) , , R θ ( ξ n 1 , ξ n 2 ) K ,
(2.2)

where R θ stands for the counterclockwise rotation of R 2 by the angle θ with respect to the origin. We define the map R 2 n R 2 n :

ξ=( ξ 11 , ξ 12 ,, ξ n 1 , ξ n 1 ) ( R θ ( ξ 11 , ξ 12 ) , , R θ ( ξ n 1 , ξ n 2 ) )

as R θ ˜ for each θ[0,2π].

For ξ C n , |ξ|=1, denote by

H ξ = { z C n : z , ξ c = k = 1 n z k ξ k ¯ = 0 }

the complex hyperplane through the origin, perpendicular to ξ. Under the standard mapping from C n to R 2 n the hyperplane H ξ turns into a (2n2)-dimensional subspace of R 2 n orthogonal to the vectors

ξ=( ξ 11 , ξ 12 ,, ξ n 1 , ξ n 2 )and ξ =( ξ 12 , ξ 11 ,, ξ n 2 , ξ n 1 ).

The orthogonal two-dimensional subspace H ξ has an orthonormal basis ξ, ξ .

Let B p ( C n ) be the p ( C n )-balls, when viewed as a subset of R 2 n :

B p ( C n ) = { ( x 11 + i x 12 , , x n 1 + i x n 2 ) C n : j = 1 n ( x j 1 2 + x j 2 2 ) p / 2 1 } = { ( x 11 , x 12 , , x n 1 , x n 2 ) R 2 n : j = 1 n ( x j 1 2 + x j 2 2 ) p / 2 1 } ,

if 0<p<, and

B ( C n ) = { ( x 11 + i x 12 , , x n 1 + i x n 2 ) C n : max 1 j n ( x j 1 2 + x j 2 2 ) 1 } = { ( x 11 , x 12 , , x n 1 , x n 2 ) R 2 n : max 1 j n ( x j 1 2 + x j 2 2 ) 1 } .

If p1, B p ( C n ) is R θ -invariant convex body in R 2 n . Here a convex body is a compact convex set with non-empty interior.

2.2 Complex isotropic bodies

First, noting that a subset K C n is called a complex convex body means that K is a convex body in R 2 n under the map (2.1). An important notion in asymptotic convex geometry is the quantity called isotropic constant (see, e.g., [2028]).

Recall that a convex body K in R n is called isotropic with the isotropic constant L K >0 if λ n (K)=1, the origin is the center of mass of K, and

K x , y 2 dx= L K 2 y 2
(2.3)

for every y R n .

Inspired by the interplay of (real) slicing problem and isotropic constant (see, e.g., [23, 27]), we will consider the complex slicing problem via the quantity called complex isotropic constant.

Definition A body K R 2 n is called complex isotropic with the complex isotropic constant L K >0, if λ 2 n (K)=1, the origin is the center of mass of K, and

K ( x , y 2 + x , y 2 ) dx= L K 2 y 2
(2.4)

for all y R 2 n .

This definition is natural since we identify C n with R 2 n using the mapping τ and

| x , y c | 2 = τ ( x ) , τ ( y ) 2 + ( τ ( x ) ) , τ ( y ) 2 = τ ( x ) , τ ( y ) 2 + ( τ ( x ) ) , τ ( y ) 2 .
(2.5)

Equivalently,

K ( x x + x x ) dx= L K 2 I 2 n ,
(2.6)

where I 2 n denotes the identity operator on R 2 n , and xx is the rank 1 linear operator on R 2 n that takes y to x,yx. More precisely, (2.6) means

K ( x i 1 x j 1 + x i 2 x j 2 )dx=0for 1ijn,
(2.7)
K ( x i 1 x j 2 x i 2 x j 1 )dx=0for 1i,jn
(2.8)

and

K ( x i 1 2 + x i 2 2 ) dx= L K 2 for 1in.
(2.9)

Summing (2.9) with i=1,,n, we have

K x 2 dx=n L K 2 .
(2.10)

Now, we show the relation between real isotropic bodies and complex isotropic bodies when we consider the convex body defined in R 2 n .

Theorem 2.1 Let K R 2 n be a convex body with λ 2 n (K)=1, center of mass at the origin. Then

  1. (i)

    if K is (real) isotropic, then K is complex isotropic;

  2. (ii)

    there exists a complex isotropic convex body K such that K is not (real) isotropic;

  3. (iii)

    if K is complex isotropic and R θ -invariant for every θ[0,2π], then K is (real) isotropic.

Proof (i) From (2.5) and the definition of isotropy (2.3), we have

K ( x , y 2 + x , y 2 ) d λ 2 n ( x ) = K ( x , y 2 + x , y 2 ) d λ 2 n ( x ) = K x , y 2 d λ 2 n ( x ) + K x , y 2 d λ 2 n ( x ) = L K 2 y 2 + L K 2 y 2 = 2 L K 2 y 2 .

From (2.4), we also have L K 2 =2 L K 2 .

(ii) We take K as

K=A× B 2 ( R 2 ) × × B 2 ( R 2 ) n 1 ,

where

A=conv{P,Q,R},

with P=( 1 2 , 3 2 ), Q=( 1 2 , 3 2 ), R=(1,0). Then A is a right triangle and the origin is the center of mass of A. Therefore, K is convex body and the origin is the center of mass of K. Moreover, it follows that λ 2 (A)=9/4,

A x 11 2 d x 11 d x 12 = 1 2 1 x 11 1 x 11 + 1 x 11 2 d x 12 d x 11 = 9 32
(2.11)

and

A x 12 2 d x 11 d x 12 = 1 2 1 x 11 1 x 11 + 1 x 12 2 d x 12 d x 11 = 27 32 .
(2.12)

In this example, it is easy to verify that (2.7) and (2.8) are true.

In order to verify (2.9), we divide it into two cases. For the case that 2in,

K ( x i 1 2 + x i 2 2 ) d λ 2 n ( x ) = π n 2 λ 2 ( A ) B 2 ( R 2 ) ( x i 1 2 + x i 2 2 ) d x i 1 d x i 2 = π n 2 9 4 2 π 0 1 r 3 d r = 9 8 π n 1 .

For the case that i=1, together with (2.11), (2.12), we obtain

K ( x 11 2 + x 12 2 ) d λ 2 n ( x ) = π n 1 A ( x 11 2 + x 12 2 ) d x 11 d x 12 = π n 1 ( 9 32 + 27 32 ) = 9 8 π n 1 .

Therefore, K is complex isotropic. However, K is not (real) isotropic in view of (2.11) and (2.12).

(iii) From the assumption that K is a complex isotropic R θ -invariant convex body in R 2 n , we have R π / 2 ˜ K=K. Note that R π / 2 ˜ x= x and |det( R π / 2 ˜ )|=1, together with (2.4), we obtain

L K 2 y 2 = K ( x , y 2 + x , y 2 ) d λ 2 n ( x ) = K x , y 2 d λ 2 n ( x ) + K x , y 2 d λ 2 n ( x ) = K x , y 2 d λ 2 n ( x ) + R π / 2 ˜ K R π / 2 ˜ x , y 2 d λ 2 n ( x ) = K x , y 2 d λ 2 n ( x ) + K x , y 2 d λ 2 n ( x ) = 2 K x , y 2 d λ 2 n ( x ) .

From (2.3), it follows that L K 2 =2 L K 2 . □

By Theorem 2.1, the class of complex isotropic bodies is larger than ones of real isotropic bodies.

3 The complex slicing problem

Observe that (Kspan{ H ξ ,ξ})( H ξ +tξ)=K( H ξ +tξ), together with Brunn’s theorem (see, e.g., [[29], p.18]), we have the following lemma.

Lemma 3.1 Let K be an origin-symmetric convex body in R 2 n , and f(t)= λ 2 n 2 (K( H ξ +tξ)) for tR. Then f ( t ) 1 / ( 2 n 2 ) is concave and f(t) is decreasing for t0. Moreover, f(t)f(0).

Lemma 3.2 Let K be a R θ -invariant body in R 2 n , then λ 2 n 2 (K( H ξ +( r 1 ξ+ r 2 ξ ))) is constant for all r 1 2 + r 2 2 =t.

Proof Since R θ ˜ H ξ = H ξ and R θ ˜ K=K for θ[0,2π], we obtain

R θ ˜ [ K ( H ξ + ( r 1 ξ + r 2 ξ ) ) ] =K ( H ξ + ( r 1 R θ ˜ ξ + r 2 R θ ˜ ξ ) ) .
(3.1)

Obviously, there exists a map R θ ˜ such that

r 1 ξ+ r 2 ξ =t R θ ˜ ξ

for any r 1 , r 2 >0 satisfies r 1 2 + r 2 2 =t.

Together with (3.1), we obtain

V ( K ( H ξ + ( r 1 ξ + r 2 ξ ) ) ) = V ( K ( H ξ + t R θ ˜ ξ ) ) = V ( R θ ˜ ( K ( H ξ + t ξ ) ) ) = V ( K ( H ξ + t ξ ) )

for any r 1 , r 2 >0 satisfies r 1 2 + r 2 2 =t. □

Lemma 3.3 Let K be a R θ -invariant complex isotropic body in R 2 n with λ 2 n (K)=1, then λ 2 n (K)=2π 0 tf(t)dt and L K =2π 0 t 3 f(t)dt, where f(t)= λ 2 n 2 (K( H ξ +tξ)).

Proof From (2.4) and Lemma 3.2, we have

L K 2 = K ( x , ξ 2 + x , ξ 2 ) d x = 2 π 0 t 3 λ 2 n 2 ( K ( H ξ + t ξ ) ) d t = 2 π 0 t 3 f ( t ) d t .

A similar argument shows λ 2 n (K)=2π 0 tf(t)dt. □

The following lemma is given by Milman and Pajor in [[27], Lemma 2.1].

Lemma 3.4 Let φ: R n R + be a measurable function such that φ =1 and let L be a symmetric convex body in R n . Then the function

F(p)= ( R n x L p φ ( x ) d x / L x L p d x ) 1 / ( n + p )

is an increasing function of p on (n,+).

Theorem 3.5 Let K be a R θ -invariant complex isotropic convex body in R 2 n with λ 2 n (K)=1, then

L K 1 2 π 1 λ 2 n 2 ( K H ξ ) 1 / 2 .
(3.2)

Proof Let f(t)= λ 2 n 2 (K( H ξ +tξ)). From Lemma 3.2, it follows that f(t) is an even function. Note that K is R θ -invariant implies K is origin symmetric. From Lemma 3.1, we have f ( t ) =f(0). Taking φ(t):=f(t)/f(0) and L:=[1,1]R in Lemma 3.4, combining with Lemma 3.3, we get

F(3)= ( R | t | 3 f ( t ) / f ( 0 ) d t 1 1 | t | 3 d t ) 1 / 4 = ( 2 L K 2 π λ 2 n 2 ( K H ξ ) ) 1 / 4

and

F(1)= ( R | t | f ( t ) / f ( 0 ) d t 1 1 | t | d t ) 1 / 2 = ( 1 π λ 2 n 2 ( K H ξ ) ) 1 / 2 .

From the comparison F(3)F(1), we have

L K 2 1 2 π λ 2 n 2 ( K H ξ ) .
(3.3)

 □

Remark If

f(t)={ f ( 0 ) , if  a t a  for some  a > 0 , 0 , otherwise ,

we have equality in (3.2).

The following lemma is due to Marshall et al. [30]. A simple proof is given in [[27], Lemma 2.6].

Lemma 3.6 Let h: R + R + be a decreasing function and let Φ: R + R + satisfying Φ(0)=0 and such that Φ and Φ(t)/t are increasing. Then

G(p)= ( 0 h ( Φ ( t ) ) t p d t 0 h ( t ) t p d t ) 1 / ( p + 1 )

is a decreasing function of p on (1,) (provided the integrals in G(p) are well defined).

Theorem 3.7 Let K be a R θ -invariant complex isotropic convex body in R 2 n with λ 2 n (K)=1, then

L K 3 π 1 λ 2 n 2 ( K H ξ ) 1 / 2 .

Proof Let f(t)= λ 2 n 2 (K( H ξ +tξ)), t0. Let h(t)= ( 1 t ) 2 n 2 for 0t1, h(t)=0 for t>1 and set Φ(t)=1 ( f ( t ) / f ( 0 ) ) 1 / ( 2 n 2 ) . Note that K is R θ -invariant implies K is origin symmetric. From Lemma 3.1, Φ(t) is convex and all hypotheses of Lemma 3.6 are satisfied, so by Lemma 3.3 we get

G(3)= ( 0 t 3 f ( t ) / f ( 0 ) d t 0 1 t 3 ( 1 t ) 2 n 2 d t ) 1 / 4 = ( n ( n + 1 ) ( 2 n + 1 ) ( 2 n 1 ) L K 2 3 π λ 2 n 2 ( K H ξ ) ) 1 / 4

and

G(1)= ( 0 t f ( t ) / f ( 0 ) d t 0 1 t ( 1 t ) 2 n 2 d t ) 1 / 2 = ( n ( 2 n 1 ) π λ 2 n 2 ( K H ξ ) ) 1 / 2 .

From the comparison G(3)G(1), we have

L K 2 3 n ( 2 n 1 ) π ( n + 1 ) ( 2 n + 1 ) λ 2 n 2 ( K H ξ ) .

Note that

n ( 2 n 1 ) ( n + 1 ) ( 2 n + 1 )
(3.4)

is increasing when the positive integer n is increasing. Thus, the maximum of (3.4) occurs as n tends to infinity. □

4 Extremum of the complex isotropic constant of B p ( C n )

The following lemma is proved in the spirit of the real counterpart (see, e.g., [[29], p.32]).

Lemma 4.1 Let 0<p, then

λ 2 n ( B p ( C n ) ) = π n ( Γ ( 1 + 2 p ) ) n Γ ( 1 + 2 n p ) .

Proof Obviously, λ 2 n ( B ( C n ))= π n . We only need to consider the case that 0<p<. Note that we identify p ( C n ) with the real 2n-dimensional space equipped with the norm

x p = [ ( x 11 2 + x 12 2 ) p / 2 + + ( x n 1 2 + x n 2 2 ) p / 2 ] 1 / p .

Then

R 2 n e x p p d λ 2 n ( x ) = i = 1 n ( R 2 e ( x i 1 2 + x i 2 2 ) p / 2 d x i 1 d x i 2 ) = ( π Γ ( 1 + 2 p ) ) n .

On the other hand, we compute the same integral in polar coordinates and the polar formula for the volume:

R 2 n e x p p d λ 2 n ( x ) = S 2 n 1 0 e r p θ p p r 2 n 1 d r d θ = Γ ( 2 n / p ) p S 2 n 1 θ p 2 n d θ = Γ ( 1 + 2 n p ) λ 2 n ( B p ( C n ) ) .

Comparing these two expressions for the same integral, we get the result. □

Theorem 4.2 Let 1p, then c n , p B p ( C n ) is a complex isotropic convex body in R 2 n . Furthermore, its complex isotropic constant is

L B p ( C n ) 2 = Γ ( 1 + 2 n p ) 1 + 1 / n Γ ( 1 + 4 p ) 2 π Γ ( 1 + 2 n + 2 p ) Γ ( 1 + 2 p ) 2 .

Proof Assume that 1p<. It is easy to verify that (2.7) and (2.8) are true for B p ( C n ). Now, we only need to verify (2.9) for B p ( C n ). Actually, we can explicitly calculate L B p ( C n ) by using Lemma 4.1, to find that

L B p ( C n ) 2 = 1 λ 2 n ( B p ( C n ) ) 1 + 1 n B p ( C n ) ( x 11 2 + x 12 2 ) d x = 2 π λ 2 n ( B p ( C n ) ) 1 + 1 n 0 1 r 3 ( 1 r p ) 2 n 2 p λ 2 n 2 ( B p ( C n 1 ) ) d r = Γ ( 1 + 2 n p ) 1 + 1 / n Γ ( 1 + 4 p ) 2 π Γ ( 1 + 2 n + 2 p ) Γ ( 1 + 2 p ) 2 .

Thus, c n , p B p ( C n ) (1p<) is complex isotropic.

Similarly, π 1 / 2 B ( C n ) is complex isotropic and L B ( C n ) 2 =1/2π. □

Remark From Theorem 2.1(iii), c n , p B p ( C n ) is in fact isotropic since it is R θ -invariant. However, the complex isotropic constant of B p ( C n ) is more useful as the argument in Section 3.

Next, we determine the extreme value of L B p ( C n ) for p1. The approach that we adopted is similar to the real case due to Ma and the third named author [7].

Lemma 4.3 Let p>0. Then, for each given positive integer n,

F(p)= Γ ( 1 + 2 n p ) 1 + 1 / n Γ ( 1 + 4 p ) Γ ( 1 + 2 n + 2 p ) Γ ( 1 + 2 p ) 2

is a decreasing function for 0<p<2 and an increasing function for p2.

Proof Making the change of variables q=2/p, we have

G(q)=F ( 2 q ) = Γ ( 1 + n q ) 1 + 1 / n Γ ( 1 + 2 q ) Γ ( 1 + ( n + 1 ) q ) Γ ( 1 + q ) 2 .
(4.1)

It follows that

d ln G ( q ) d q =2 ( ψ ( 1 + 2 q ) ψ ( 1 + q ) ) (n+1) ( ψ ( 1 + ( n + 1 ) q ) ψ ( 1 + n q ) ) ,

where ψ(x)= Γ (x)/Γ(x). Now,

ψ ( 1 + ( n + 1 ) q ) ψ ( 1 + n q ) = 0 1 z ( 1 ( 1 + z ) 1 + n q 1 ( 1 + z ) 1 + ( n + 1 ) q ) d z = 1 ( n + 1 ) q 1 t 1 t 1 q 1 d 1 t n + 1 ,
(4.2)

where we use the following integral representation for the function ψ:

ψ(x)= 0 1 z ( e z 1 ( 1 + z ) x ) dz,

and a change of variable t= ( 1 + z ) q .

Let n=1 in (4.2), then

ψ(1+2q)ψ(1+q)= 1 2 q 1 t 1 t 1 q 1 d 1 t 2 .

Thus,

d ln G ( q ) d q = 1 q 1 t 1 t 1 q 1 d ( 1 t n + 1 1 t 2 ) .

Then it follows that

d ln G ( q ) d q | q = 1 = 1 d ( 1 t n + 1 1 t 2 ) =0

and, if 0<q<1,

d ln G ( q ) d q = 1 q 1 t 1 t 1 q 1 d ( 1 t n + 1 1 t 2 ) = 1 ( 1 q ) t 1 q + q t 0 t 1 q 1 q 2 ( t 1 q 1 ) 2 ( 1 t n + 1 1 t 2 ) d t < 0 ,

where we use the arithmetic-geometric mean inequality, i.e., (1λ)x+λy x 1 λ y λ .

Similarly, if q>1, we have

d ln G ( q ) d q = 1 q 1 t 1 t 1 q 1 d ( 1 t n + 1 1 t 2 ) = 1 ( 1 1 q ) t 1 q + 1 q t 1 q 1 1 q ( t 1 q 1 ) 2 ( 1 t 2 1 t n + 1 ) d t > 0 .

The result follows from (4.1). □

The following lemma can be simply deduced as in [[23], p.4].

Lemma 4.4 For every complex isotropic convex body K in R 2 n ,

L K L B 2 ( C n ) .

Recall that

L B p ( C n ) 2 = Γ ( 1 + 2 n p ) 1 + 1 / n Γ ( 1 + 4 p ) 2 π Γ ( 1 + 2 n + 2 p ) Γ ( 1 + 2 p ) 2
(4.3)

for p1. Thus, from Lemma 4.4, we have

L B p ( C n ) L B 2 ( C n ) = ( ( n ! ) 1 n π ( n + 1 ) ) 1 2 1 π e .
(4.4)

From Lemma 4.3 and (4.3), it follows that

L B p ( C n ) max{ L B 1 ( C n ) , L B ( C n ) }
(4.5)

for p1. The following lemma is given by Gao [3].

Lemma 4.5 Let 0x1 and y>0. For fixed x, the function

f(x,y)= ( 1 + 2 y ) lnΓ(1+xy)lnΓ ( 1 + ( y + 2 ) x )
(4.6)

is a decreasing function of y for y>0 when 0x1/2 and for y2 when 1/2<x1.

If we set x=1/p and y=2n in (4.6) for p1, n1, by (4.3), it follows that

L B p ( C n ) L B p ( C ) .

Combining with (4.5), we have

L B p ( C n ) L B 1 ( C ) = L B ( C n ) = 1 2 π .
(4.7)

Together with (4.4) and (4.7), we have the following theorem.

Theorem 4.6 Let 1p, then

1 π e L B p ( C n ) 1 2 π .

Now we complete the proof of our main result.

Proof of Theorem 1.2 Combining with Theorem 3.5, Theorem 3.7, and Theorem 4.6, we obtain

1 1 2 π 1 L B p ( C n ) λ 2 n 2 ( c n , p B p ( C n ) H ξ ) 1 / 2 3 π 1 L B p ( C n ) 3 e .

The equality of the left inequality in this theorem holds for B ( C n ) from the remark after Theorem 3.5 and (4.7). □

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Acknowledgements

The authors thank the referee for useful comments that helped to improve the presentation of the manuscript. This work is supported by the National Natural Science Foundation of China (Grant No. 11371239), Shanghai Leading Academic Discipline Project (Project Number: J50101), and the Research Fund for the Doctoral Programs of Higher Education of China (20123108110001).

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Correspondence to Binwu He.

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Lai, D.N., Huang, Q. & He, B. Estimates for the extremal sections of complex p -balls. J Inequal Appl 2014, 235 (2014). https://doi.org/10.1186/1029-242X-2014-235

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Keywords

  • complex isotropic constant
  • p ( C n )-balls
  • complex slicing problem