R E S E A R C H
Open Access
Estimates for the extremal sections of
complex
p
-balls
Duc Nam Lai
1, Qingzhong Huang
2and Binwu He
1**Correspondence:
1Department of Mathematics,
Shanghai University, Shanghai, 200444, China
Full list of author information is available at the end of the article
Abstract
The problem of maximal hyperplane section ofBp(Cn) withp≥1 is considered, which is the complex version of central hyperplane section problem ofBp(Rn). 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(Cn)-balls that does not depend onn andpis shown, which extends results of Oleszkiewicz and Pełczy ´nski, Koldobsky and Zymonopoulou, and Meyer and Pajor.
MSC: 52A21; 46B07
Keywords: complex isotropic constant;
p(Cn)-balls; complex slicing problem
1 Introduction
LetBp(Rn) andBp(Cn) denote the unit balls of the real and complexn-dimensionalp spaces,p(Rn) andp(Cn), respectively. We denote byλn(K) then-dimensional Lebesgue measure of a compact setK. We writern,p=λn(Bp(Rn))–/nandcn,p=λn(Bp(Cn))–/nsuch thatλn(rn,pBp(Rn)) = andλn(cn,pBp(Cn)) = , respectively.
The extremal volume of central hyperplane section ofBp(Rn) is studied by various au-thors (see,e.g., [–]). Especially, the left-hand side of the following inequalities is known due to Meyer and Pajor [] for all p≥ andp= , and Schmuckenschläger [] for all <p< . The right-hand side of the following inequalities is due to Ma and the third named author [], and it shows an upper bound estimate for the volume of central hyper-plane sections of normalizedp-balls that does not depend onnandp.
Theorem . Let n∈N,n≥,p≥,and H any central hyperplane inRn.Then
≤λn–
rn,pBp
Rn∩H≤√πe.
Moreover,the minimum occurs for B∞(Rn)ifξ has only one non-zero coordinate whereξ
is the normal vector of H.
Note that Theorem . is proved by determining the extremal value of the isotropic con-stant ofBp(Rn) 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 . (see Theorem .) can be proved by estimating the extremal value of the complex isotropic constant ofBp(Cn).
A noteworthy fact is that the extremal volume of complex central hyperplane section ofBp(Cn) has not been studied until recent years. Other results concerning convex bod-ies in a complex vector space as ambient space can be found in [–]. Especially, in [], Oleszkiewicz and Pełczyński proved that ≤λn–(cn,∞B∞(Cn)∩Hξ)≤, whereHξ is
the complex central hyperplane (see Section . 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 formej+σek, wherej=k,ejandekare standard basic vectors, andσ∈C,|σ|= . In [], Koldobsky and Zymonopoulou studied the extremal sections of Bp(Cn), for <p≤ and showed that the minimum corresponds to hyperplanes orthogonal to vectorsξ= (ξ, . . . ,ξn)∈Cn with |ξ|=· · ·=|ξn|and the maximum corresponds to hyperplanes orthogonal to vec-tors with only one non-zero coordinate. Moreover, a result of Meyer and Pajor [, Corol-lary .] states the following. Supposep≥, thenλn–(cn–,pBp(Cn)∩Hξ)≥; Suppose
≤p≤, thenλn–(cn–,pBp(Cn)∩Hξ)≤, wherecn–,p =λn–(Bp(Cn–))–/(n–). Re-cently, Koldobsky and König [] considered minimal volume of slabs for the complex cube.
The casep≤ of the following theorem follows directly from the work of Koldobsky and Zymonopoulou []. The following theorem extends their results top> , and it also shows an upper bound estimate for the volume of complex central hyperplane sections of normalized complexp-balls that does not depend onnandp.
Theorem . Let n∈N,n≥,p≥,and Hξ any complex hyperplane inRn.Then
≤λn–
cn,pBp
Cn∩H
ξ
≤e.
Moreover,the minimum occurs for B∞(Cn)ifξhas only one non-zero coordinate.
Our problem is different from the extremal volume of central hyperplane section of Bp(Rn) problem in two aspects. First, Bp(Cn)=Bp(Rn) except p= ; Secondly, we do only (n– )-dimensional sections, sections by subspaces coming from complex hyper-planes, rather than all (n– )-dimensional sections, and (n– )-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 ofxbyx=√ x,xforx,y∈Rn. Forx,y∈Cn, their complex scalar product by
x,ycand the modulus ofxbyx= √
x,xc.
Origin-symmetric convex bodies inCnare the unit balls of norms onCn. We denote by · Kthe norm corresponding to the bodyK:
K=z∈Cn: · K≤
We identifyCnwithRnusing the standard mapping, that is, for
ξ= (ξ, . . .ξn) = (ξ+iξ, . . . ,ξn+iξn)∈Cn, (ξ+iξ, . . . ,ξn+iξn)
τ
→(ξ,ξ, . . . ,ξn,ξn). (.)
Since norms onCnsatisfy the equality
λz=|λ|z, ∀z∈Cn,∀λ∈C,
origin-symmetric complex convex bodies correspond to those origin-symmetric convex bodiesKinRnthat are invariant with respect to any coordinate-wise two-dimensional rotation, namely for eachθ∈[, π] and each (ξ,ξ, . . .ξn,ξn)∈Rn
ξK=Rθ(ξ,ξ), . . . ,Rθ(ξn,ξn)K, (.)
whereRθstands for the counterclockwise rotation ofRby the angleθwith respect to the
origin. We define the mapRn→Rn:
ξ= (ξ,ξ, . . . ,ξn,ξn)→
Rθ(ξ,ξ), . . . ,Rθ(ξn,ξn)
asRθfor eachθ∈[, π].
Forξ∈Cn,|ξ|= , denote by
Hξ=
z∈Cn: z,ξc= n
k=
zkξk=
the complex hyperplane through the origin, perpendicular toξ. Under the standard map-ping fromCntoRnthe hyperplaneH
ξ turns into a (n– )-dimensional subspace ofRn
orthogonal to the vectors
ξ= (ξ,ξ, . . . ,ξn,ξn) and ξ†= (–ξ,ξ, . . . , –ξn,ξn).
The orthogonal two-dimensional subspaceHξ⊥has an orthonormal basisξ,ξ†.
LetBp(Cn) be thep(Cn)-balls, when viewed as a subset ofRn:
Bp
Cn=
(x+ix, . . . ,xn+ixn)∈Cn: n
j=
xj+xjp/≤
=
(x,x, . . . ,xn,xn)∈Rn: n
j=
xj+xjp/≤
,
if <p<∞, and
B∞Cn=
(x+ix, . . . ,xn+ixn)∈Cn:max ≤j≤n
xj+xj≤
=
(x,x, . . . ,xn,xn)∈Rn:max ≤j≤n
xj+xj≤
If p≥,Bp(Cn) isRθ-invariant convex body inRn. Here a convex body is a compact
convex set with non-empty interior.
2.2 Complex isotropic bodies
First, noting that a subsetK⊂Cnis called a complex convex body means thatKis a convex body inRnunder the map (.). An important notion in asymptotic convex geometry is the quantity called isotropic constant (see,e.g., [–]).
Recall that a convex bodyKinRnis called isotropic with the isotropic constantL K> ifλn(K) = , the origin is the center of mass ofK, and
K
x,ydx=LKy (.)
for everyy∈Rn.
Inspired by the interplay of (real) slicing problem and isotropic constant (see, e.g., [, ]), we will consider the complex slicing problem via the quantity called complex isotropic constant.
Definition A bodyK⊂Rnis called complex isotropic with the complex isotropic
con-stantLK> , ifλn(K) = , the origin is the center of mass ofK, and
K
x,y+x†,ydx=LKy (.)
for ally∈Rn.
This definition is natural since we identifyCnwithRnusing the mappingτ and
x,yc
=τ(x),τ(y)+τ(x)†,τ(y)
=τ(x),τ(y)+τ(x),τ(y)†. (.)
Equivalently,
K
x⊗x+x†⊗x†dx=LKIn, (.)
whereIndenotes the identity operator onRn, andx⊗xis the rank linear operator on
Rnthat takesyto x,yx. More precisely, (.) means
K
(xixj+xixj)dx= for ≤i=j≤n, (.)
K
(xixj–xixj)dx= for ≤i,j≤n (.)
and
K
Summing (.) withi= , . . . ,n, we have
K
xdx=nLK. (.)
Now, we show the relation between real isotropic bodies and complex isotropic bodies when we consider the convex body defined inRn.
Theorem . Let K⊂Rnbe a convex body withλ
n(K) = ,center of mass at the origin. Then
(i) ifKis(real)isotropic,thenKis complex isotropic;
(ii) there exists a complex isotropic convex bodyKsuch thatKis not(real)isotropic; (iii) ifKis complex isotropic andRθ-invariant for everyθ∈[, π],thenKis(real)
isotropic.
Proof (i) From (.) and the definition of isotropy (.), we have
K
x,y+x†,ydλn(x)
=
K
x,y+x,y†dλn(x)
=
K
x,ydλn(x) +
K
x,y†dλn(x) =LKy+LKy†= LKy.
From (.), we also haveL K= LK. (ii) We takeKas
K=A×B
R× · · · ×B
R
n–
,
where
A=conv{P,Q,R},
withP= (–,),Q= (–, –),R= (, ). ThenAis a right triangle and the origin is the center of mass ofA. Therefore,Kis convex body and the origin is the center of mass ofK. Moreover, it follows thatλ(A) = /,
A
xdxdx=
–
–x+
x–
xdxdx=
(.)
and
A
xdxdx=
–
–x+
x–
xdxdx=
. (.)
In order to verify (.), we divide it into two cases. For the case that ≤i≤n,
K
xi+xidλn(x)
=πn–λ(A)
B(R)
xi+xidxidxi
=πn–
·π
rdr= π
n–.
For the case thati= , together with (.), (.), we obtain
K
x+xdλn(x)
=πn–
A
x+xdxdx
=πn–·
+
= π
n–.
Therefore,K is complex isotropic. However,K is not (real) isotropic in view of (.) and (.).
(iii) From the assumption thatK is a complex isotropicRθ-invariant convex body in
Rn, we haveR
π/K=K. Note thatRπ/x=x†and|det(Rπ/)|= , together with (.), we obtain
L Ky=
K
x,y+x†,ydλn(x)
=
K
x,ydλn(x) +
K
x†,ydλn(x)
=
K
x,ydλn(x) +
Rπ/K
Rπ/x,ydλn(x)
=
K
x,ydλn(x) +
K
x,ydλn(x)
=
K
x,ydλn(x).
From (.), it follows thatL
K= LK.
By Theorem ., the class of complex isotropic bodies is larger than ones of real isotropic bodies.
3 The complex slicing problem
Observe that (K∩span{Hξ,ξ})∩(Hξ+tξ) =K∩(Hξ+tξ), together with Brunn’s theorem
(see,e.g., [, p.]), we have the following lemma.
Lemma . Let K be an origin-symmetric convex body inRn,and f(t) =λn–(K∩(Hξ+
Lemma . Let K be a Rθ-invariant body inRn,thenλn–(K∩(Hξ + (rξ +rξ†)))is constant for allr
+r=t.
Proof SinceRθHξ =Hξ andRθK=Kforθ∈[, π], we obtain
Rθ
K∩Hξ+
rξ+rξ†
=K∩Hξ+
rRθξ+rRθξ†
. (.)
Obviously, there exists a mapRθsuch that
rξ+rξ†=tRθξ
for anyr,r> satisfies
r +r=t. Together with (.), we obtain
VK∩Hξ+
rξ+rξ†
=VK∩(Hξ +tRθξ)
=V Rθ
K∩(Hξ +tξ)
=VK∩(Hξ +tξ)
for anyr,r> satisfies
r
+r=t.
Lemma . Let K be a Rθ-invariant complex isotropic body inRnwithλn(K) = ,then
λn(K) = π
∞
tf(t)dt andLK= π
∞ t
f(t)dt,where f(t) =λ
n–(K∩(Hξ+tξ)).
Proof From (.) and Lemma ., we have
L K=
K
x,ξ+x,ξ†dx
= π
∞
tλn–
K∩(Hξ+tξ)
dt= π
∞
tf(t)dt.
A similar argument showsλn(K) = π
∞
tf(t)dt.
The following lemma is given by Milman and Pajor in [, Lemma .].
Lemma . Letϕ:Rn→R
+be a measurable function such thatϕ∞= and let L be a symmetric convex body inRn.Then the function
F(p) =
Rnx p Lϕ(x)dx
L xpLdx
/(n+p)
is an increasing function of p on(–n, +∞).
Theorem . Let K be a Rθ-invariant complex isotropic convex body inRnwithλn(K) = ,then
LK≥
√
π
λn–(K∩Hξ)/
Proof Letf(t) =λn–(K∩(Hξ+tξ)). From Lemma ., it follows thatf(t) is an even
func-tion. Note thatKisRθ-invariant impliesKis origin symmetric. From Lemma ., we have
f(t)∞=f(). Takingϕ(t) :=f(t)/f() andL:= [–, ]⊂Rin Lemma ., combining with Lemma ., we get
F() =
R|t|f(t)/f()dt
–|t|dt
/ =
L K
π λn–(K∩Hξ)
/
and
F() =
R|t|f(t)/f()dt
–|t|dt
/ =
π λn–(K∩Hξ)
/ .
From the comparisonF()≥F(), we have
L
K≥
π λn–(K∩Hξ)
. (.)
Remark If
f(t) =
⎧ ⎨ ⎩
f(), if –a≤t≤afor somea> ,
, otherwise,
we have equality in (.).
The following lemma is due to Marshallet al.[]. A simple proof is given in [, Lem-ma .].
Lemma . Let h:R+→R+ be a decreasing function and let:R+→R+ satisfying
() = and such thatand(t)/t are increasing.Then
G(p) =
∞
h((t))tpdt ∞
h(t)tpdt
/(p+)
is a decreasing function of p on(–,∞) (provided the integrals in G(p)are well defined).
Theorem . Let K be a Rθ-invariant complex isotropic convex body inRnwithλn(K) = ,then
LK≤
π
λn–(K∩Hξ)/
.
Proof Letf(t) =λn–(K∩(Hξ +tξ)),t≥. Leth(t) = ( –t)n– for ≤t≤,h(t) =
fort> and set(t) = – (f(t)/f())/(n–). Note thatKisR
θ-invariant impliesKis origin
symmetric. From Lemma .,(t) is convex and all hypotheses of Lemma . are satisfied, so by Lemma . we get
G() = ∞
tf(t)/f()dt
t( –t)n–dt
/ =
n(n+ )(n+ )(n– )LK π λn–(K∩Hξ)
and
G() = ∞
tf(t)/f()dt
t( –t)n–dt
/ =
n(n– )
π λn–(K∩Hξ)
/ .
From the comparisonG()≤G(), we have
L
K≤
n(n– )
π(n+ )(n+ )λn–(K∩Hξ)
.
Note that
n(n– )
(n+ )(n+ ) (.)
is increasing when the positive integernis increasing. Thus, the maximum of (.) occurs
asntends to infinity.
4 Extremum of the complex isotropic constant ofBp(Cn)
The following lemma is proved in the spirit of the real counterpart (see,e.g., [, p.]).
Lemma . Let <p≤ ∞,then
λn
Bp
Cn=π
n(( + p))
n
( + np) .
Proof Obviously,λn(B∞(Cn)) =πn. We only need to consider the case that <p<∞. Note that we identifyp(Cn) with the real n-dimensional space equipped with the norm
xp=
x+xp/+· · ·+xn+xnp//p.
Then
Rne –xppdλ
n(x) = n
!
i=
Re
–(xi+xi)p/dx idxi
=
π
+ p
n .
On the other hand, we compute the same integral in polar coordinates and the polar for-mula for the volume:
Rne –xppdλ
n(x) =
Sn– ∞
e–rpθpprn–dr dθ
=(n/p) p
Sn–
θ–np dθ
=
+n p
λn
Bp
Cn.
Theorem . Let≤p≤ ∞,then cn,pBp(Cn)is a complex isotropic convex body inRn. Furthermore,its complex isotropic constant is
L Bp(Cn)=
( +np)+/n( + p)
π ( +n+p )( +p).
Proof Assume that ≤p<∞. It is easy to verify that (.) and (.) are true forBp(Cn). Now, we only need to verify (.) forBp(Cn). Actually, we can explicitly calculateLBp(Cn)
by using Lemma ., to find that
L Bp(Cn)=
λn(Bp(Cn))+
n
Bp(Cn)
x+xdx
= π
λn(Bp(Cn))+
n
r –rpnp–λ n–
Bp
Cn–dr
= ( + n
p)
+/n( + p) π ( +n+p )( +p).
Thus,cn,pBp(Cn) (≤p<∞) is complex isotropic.
Similarly,π–/B∞(Cn) is complex isotropic andLB∞(Cn)= /π. Remark From Theorem .(iii), cn,pBp(Cn) is in fact isotropic since it is Rθ-invariant.
However, the complex isotropic constant ofBp(Cn) is more useful as the argument in Sec-tion .
Next, we determine the extreme value ofLBp(Cn)forp≥. The approach that we adopted
is similar to the real case due to Ma and the third named author [].
Lemma . Let p> .Then,for each given positive integer n,
F(p) =( + n
p)
+/n( + p)
( +n+p )( +p)
is a decreasing function for <p< and an increasing function for p≥.
Proof Making the change of variablesq= /p, we have
G(q) =F
q
= ( +nq)
+/n( + q)
( + (n+ )q)( +q). (.)
It follows that
dlnG(q)
dq =
ψ( + q) –ψ( +q)– (n+ )ψ + (n+ )q–ψ( +nq),
whereψ(x) =(x)/(x). Now,
ψ + (n+ )q–ψ( +nq) =
∞
z
( +z)+nq–
( +z)+(n+)q
dz
= –
(n+ )q
∞
t–
tq –
d
where we use the following integral representation for the functionψ:
ψ(x) =
∞
z
e–z–
( +z)x
dz,
and a change of variablet= ( +z)q. Letn= in (.), then
ψ( + q) –ψ( +q) = – q
∞
t–
tq–
d t.
Thus,
dlnG(q)
dq =
q
∞
t–
tq –
d
tn+ –
t
.
Then it follows that
dlnG(q) dq q= = ∞ d tn+ –
t
=
and, if <q< ,
dlnG(q)
dq =
q
∞
t–
tq –
d
tn+ –
t = ∞
( –q)tq+qt–tq–
q(tq– )
tn+ –
t
dt
< ,
where we use the arithmetic-geometric mean inequality,i.e., ( –λ)x+λy≥x–λyλ.
Similarly, ifq> , we have
dlnG(q)
dq =
q
∞
t–
tq –
d
tn+ –
t = ∞
( –q)tq + qt
q––
q(tq – )
t –
tn+
dt
> .
The result follows from (.).
The following lemma can be simply deduced as in [, p.].
Lemma . For every complex isotropic convex body K inRn,
LK≥LB(Cn).
Recall that
L Bp(Cn)=
( +np)+/n( + p)
forp≥. Thus, from Lemma ., we have
LBp(Cn)≥LB
(Cn)=
(n!)n
π(n+ )
≥√
πe. (.)
From Lemma . and (.), it follows that
LBp(Cn)≤max{LB(Cn),LB∞(Cn)} (.)
forp≥. The following lemma is given by Gao [].
Lemma . Let≤x≤and y> .For fixed x,the function
f(x,y) =
+ y
ln( +xy) –ln + (y+ )x (.)
is a decreasing function of y for y> when≤x≤/and for y≥when/ <x≤.
If we setx= /pandy= nin (.) forp≥,n≥, by (.), it follows that
LBp(Cn)≤LBp(C).
Combining with (.), we have
LBp(Cn)≤LB(C)=LB∞(Cn)=√
π. (.)
Together with (.) and (.), we have the following theorem.
Theorem . Let≤p≤ ∞,then
√
πe≤LBp(Cn)≤
√
π.
Now we complete the proof of our main result.
Proof of Theorem. Combining with Theorem ., Theorem ., and Theorem ., we
obtain
≤√ π
LBp(Cn) ≤λn–
cn,pBp
Cn∩H
ξ
/
≤
π
LBp(Cn) ≤√e.
The equality of the left inequality in this theorem holds forB∞(Cn) from the remark after
Theorem . and (.).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Author details
1Department of Mathematics, Shanghai University, Shanghai, 200444, China.2College of Mathematics, Physics and
Information Engineering, Jiaxing University, Jiaxing, 314001, China.
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).
Received: 28 November 2013 Accepted: 29 May 2014 Published:06 Jun 2014 References
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