E extracta mathematicae Vol. 33, Núm. 1, 51 – 66 (2018)

The Geometry of L(3l2∞) and Optimal Constants in the
Bohnenblust-Hille Inequality for Multilinear Forms

and Polynomials

Sung Guen Kim ∗

Department of Mathematics, Kyungpook National University,

Daegu 702 − 701, South Korea, sgk317@knu.ac.kr

Presented by Ricardo Garćıa Received December 23, 2016

Abstract: We classify the extreme and exposed 3-linear forms of the unit ball of L(3l2∞). We
introduce optimal constants in the Bohnenblust-Hille inequality for symmetric multilinear
forms and polynomials and investigate about their relations.

Key words: Extreme points, exposed points, the optimal constants in the Bohnenblust-Hille
inequality for symmetric multilinear forms and polynomials.

AMS Subject Class. (2010): 46A22.

1. Introduction

We write BE for the closed unit ball of a real Banach space E and the
dual space of E is denoted by E∗. x ∈ BE is called an extreme point of BE if
y, z ∈ BE with x = 12(y + z) implies x = y = z. x ∈ BE is called an exposed
point of BE if there is a f ∈ E∗ so that f(x) = 1 = ∥f∥ and f(y) < 1 for
every y ∈ BE \ {x}. It is easy to see that every exposed point of BE is an
extreme point. We denote by extBE and expBE the sets of extreme and
exposed points of BE, respectively. Let n ∈ N, n ≥ 2. A mapping P : E → R
is a continuous n-homogeneous polynomial if there exists a continuous n-linear
form L on the product E × · · · × E such that P(x) = L(x, . . . , x) for every
x ∈ E. We denote by L(nE) the Banach space of all continuous n-linear forms
on E endowed with the norm ∥L∥ = sup∥xj∥=1, 1≤j≤n |L(x1, . . . , xn)|. Ls(

nE)
denotes the closed subspace of L(nE) consisting all continuous symmetric
n-linear forms on E. P(nE) denotes the Banach space of all continuous n-
homogeneous polynomials from E into R endowed with the norm ∥P∥ =

∗This research was supported by the Basic Science Research Program through the Na-
tional Research Foundation of Korea(NRF) funded by the Ministry of Education, Science
and Technology (2013R1A1A2057788).

51



52 s. g. kim

sup∥x∥=1 |P(x)|. Note that the spaces L(nE), Ls(nE), P(nE) are very different
from a geometric point of view. In particular, for integral multilinear forms
and integral polynomials one has ([2], [9], [32])

extBLI(nE) = {ϕ1ϕ2 · · · ϕn : ϕi ∈ extBE∗}

and
extBPI(nE) = {±ϕ

n : ϕ ∈ E∗, ∥ϕ∥ = 1},
where LI(nE) and PI(nE) are the spaces of integral n-linear forms and integral
n-homogeneous polynomials on E, respectively. For more details about the
theory of multilinear mappings and polynomials on a Banach space, we refer
to [10].

In 1998, Choi et al. ([4], [5]) characterized the extreme points of the unit
ball of P(2l21) and P(

2l22). Kim [15] classified the exposed 2-homogeneous poly-
nomials on P(2l2p) (1 ≤ p ≤ ∞). Kim ([17], [19], [23]) classified the extreme,
exposed, smooth points of the unit ball of P(2d∗(1, w)2), where d∗(1, w)2 = R2
with the octagonal norm of weight w.

In 2009, Kim [16] initiated extremal problems for bilinear forms on a clas-
sical finite dimensional real Banach space and classified the extreme, exposed,
smooth points of the unit ball of Ls(2l2∞). Kim ([18], [20]–[22]) classified
the extreme, exposed, smooth points of the unit balls of Ls(2d∗(1, w)2) and
L(2d∗(1, w)2).

We refer to ([1]–[9], [11]–[32] and references therein) for some recent work
about extremal properties of multilinear mappings and homogeneous polyno-
mials on some classical Banach spaces.

Let K = R or C. The Bohnenblust-Hille inequality for n-linear forms ([3]
and references therein) tells us that there exists a sequence of positive scalars
(C(n : K))∞n=1 in [1, ∞] such that(

∞∑
j1,...,jn=1

∣∣T(ej1, . . . , ejn)∣∣ 2nn+1
)n+1

2n

≤ C(n : K)∥T∥

for all continuous n-linear forms T : c0 × · · · × c0 → K. The optimal constant
in the Bohnenblust-Hille inequality for n-linear forms C(n : K) is defined by

C(n : K) := sup

{(
∞∑

j1,...,jn=1

|T(ej1, . . . , ejn)|
2n
n+1

)n+1
2n

:

T ∈ L(nc0 : K), ∥T∥ = 1

}
.



the geometry of L(3l2∞) 53

We introduce the optimal constant in the Bohnenblust-Hille inequality for
symmetric n-linear forms Cs(n : K) is defined by

Cs(n : K) := sup

{(
∞∑

j1,...,jn=1

|T(ej1, . . . , ejn)|
2n
n+1

)n+1
2n

:

T ∈ Ls(nc0 : K), ∥T∥ = 1

}
.

It is obvious that Cs(n : K) ≤ C(n : K). We also introduce the optimal
constant in the Bohnenblust-Hille inequality for n-homogeneous polynomials
Cp(n : K) is defined by

Cp(n : K) := sup

{(
∞∑
j=1

|P(ej)|
2n
n+1

)n+1
2n

: P ∈ P(nc0 : K), ∥P∥ = 1

}
.

Recently, Diniz et al. [12] showed that C(2 : R) =
√
2.

In this paper, we classify the extreme and exposed 3-linear forms of the
unit ball of L(3l2∞). We introduce optimal constants in the Bohnenblust-Hille
inequality for symmetric multilinear forms and polynomials and investigate
about their relations.

2. The extreme points of the unit ball of L(3l2∞)

Let T ∈ L(3l2∞) be given by

T
(
(x1, x2), (y1, y2), (z1, z2)

)
= ax1y1z1 + bx2y2z2 + c1x2y1z1 + c2x1y2z1

+ c3x1y1z2 + d1x1y2z2 + d2x2y1z2 + d3x2y2z1

for some a, b, cj, dj ∈ R and for j = 1, 2, 3. For simplicity, we will denote
T = (a, b, c1, c2, c3, d1, d2, d3).

Theorem 2.1. Let T = (a, b, c1, c2, c3, d1, d2, d3) ∈ L(3l2∞). Then

∥T∥ = max
{
|a + c1 + c2 + d3| + |b + c3 + d1 + d2|,
|a − c2 − c3 + d1| + |b + c1 − d2 − d3|,
|a − b + c3 − d3| + |c1 − c2 − d1 + d2|,
|a + b − c1 − d1| + |c2 − c3 + d2 − d3|

}
.



54 s. g. kim

Proof. Note that extBl2∞ = {(1, 1), (1, −1), (−1, 1), (−1, −1)}. By the Krein-
Milman Theorem, Bl2∞ = co(extBl2∞). By the continuity and trilinearity of
T ,

∥T∥ = max
{
|T((1, 1), (1, 1), (1, 1))|, |T((1, −1), (1, 1), (1, 1))|
|T((1, 1), (1, −1), (1, 1))|, |T((1, 1), (1, 1), (1, −1))|,
|T((1, −1), (1, −1), (1, 1))|, |T((1, −1), (1, 1), (1, −1))|,
|T((1, 1), (1, −1), (1, −1))|, |T((1, −1), (1, −1), (1, −1))|

}
= max

{
|a + c1 + c2 + d3| + |b + c3 + d1 + d2|,
|a − c2 − c3 + d1| + |b + c1 − d2 − d3|,
|a − b + c3 − d3| + |c1 − c2 − d1 + d2|,
|a + b − c1 − d1| + |c2 − c3 + d2 − d3|

}
.

Note that if ∥T∥ = 1, then |a| ≤ 1, |b| ≤ 1, |cj| ≤ 1, |dj| ≤ 1, for j = 1, 2, 3.

Theorem 2.2.

extBL(3l2∞) =
{
(a, b, c1, c2, c3, d1, d2, d3) :

a =
1

8
(ϵ1 + ϵ2 + ϵ3 + ϵ4 + ϵ5 + ϵ6 + ϵ7 + ϵ8),

b =
1

8
(ϵ1 − ϵ2 − ϵ3 − ϵ4 + ϵ5 + ϵ6 + ϵ7 − ϵ8),

c1 =
1

8
(ϵ1 − ϵ2 + ϵ3 + ϵ4 − ϵ5 − ϵ6 + ϵ7 − ϵ8),

c2 =
1

8
(ϵ1 + ϵ2 − ϵ3 + ϵ4 − ϵ5 + ϵ6 − ϵ7 − ϵ8),

c3 =
1

8
(ϵ1 + ϵ2 + ϵ3 − ϵ4 + ϵ5 − ϵ6 − ϵ7 − ϵ8),

d1 =
1

8
(ϵ1 + ϵ2 − ϵ3 − ϵ4 − ϵ5 − ϵ6 + ϵ7 + ϵ8),

d2 =
1

8
(ϵ1 − ϵ2 + ϵ3 − ϵ4 − ϵ5 + ϵ6 − ϵ7 + ϵ8),

d3 =
1

8
(ϵ1 − ϵ2 − ϵ3 + ϵ4 + ϵ5 − ϵ6 − ϵ7 + ϵ8),

ϵj = ±1, for j = 1, 2, . . . , 8
}
.



the geometry of L(3l2∞) 55

Proof. Let T = (a, b, c1, c2, c3, d1, d2, d3) ∈ L(3l2∞) with ∥T∥ = 1. Note that

T
(
(1, 1), (1, 1), (1, 1)

)
= a + b + c1 + c2 + c3 + d1 + d2 + d3,

T
(
(1, −1), (1, 1), (1, 1)

)
= a − b − c1 + c2 + c3 + d1 − d2 − d3,

T
(
(1, 1), (1, −1), (1, 1)

)
= a − b + c1 − c2 + c3 − d1 + d2 − d3,

T
(
(1, 1), (1, 1), (1, −1)

)
= a − b + c1 + c2 − c3 − d1 − d2 + d3,

T
(
(1, −1), (1, −1), (1, 1)

)
= a + b − c1 − c2 + c3 − d1 − d2 + d3,

T
(
(1, −1), (1, 1), (1, −1)

)
= a + b − c1 + c2 − c3 − d1 + d2 − d3,

T
(
(1, 1), (1, −1), (1, −1)

)
= a + b + c1 − c2 − c3 + d1 − d2 − d3,

T
(
(1, −1), (1, −1), (1, −1)

)
= a − b − c1 − c2 − c3 + d1 + d2 + d3.

Let A = (aij)1≤i,j≤8 be the 8 × 8 matrix such that

ai1 = 1 (i = 1, . . . , 8), ai2 = 1 (i = 1, 5, 6, 7), ak2 = −1 (k = 2, 3, 4, 8),
ai3 = 1 (i = 1, 3, 4, 7), ak3 = −1 (k = 2, 5, 6, 8), ai4 = 1 (i = 1, 2, 4, 6),
ak4 = −1 (k = 3, 5, 7, 8), ai5 = 1 (i = 1, 2, 3, 5), ak5 = −1 (k = 4, 6, 7, 8),
ai6 = 1 (i = 1, 2, 7, 8), ak6 = −1 (k = 3, 4, 5, 6), ai7 = 1 (i = 1, 3, 6, 8),
ak7 = −1 (k = 2, 4, 5, 7), ai8 = 1 (i = 1, 4, 5, 8), ak8 = −1 (k = 2, 3, 6, 7).

By calculation, det(A) = −212, so A is invertible. Note that

AT =
(
T
(
(1, 1), (1, 1), (1, 1)

)
, T
(
(1, −1), (1, 1), (1, 1)

)
,

T
(
(1, 1), (1, −1), (1, 1)

)
, T
(
(1, 1), (1, 1), (1, −1)

)
,

T
(
(1, −1), (1, −1), (1, 1)

)
, T
(
(1, −1), (1, 1), (1, −1)

)
,

T
(
(1, 1), (1, −1), (1, −1)

)
, T
(
(1, −1), (1, −1), (1, −1)

))t
and ∥AT∥∞ = ∥T∥. Note also that

T = A−1
(
T
(
(1, 1), (1, 1), (1, 1)

)
, T
(
(1, −1), (1, 1), (1, 1)

)
,

T
(
(1, 1), (1, −1), (1, 1)

)
, T
(
(1, 1), (1, 1), (1, −1)

)
,

T
(
(1, −1), (1, −1), (1, 1)

)
, T
(
(1, −1), (1, 1), (1, −1)

)
,

T
(
(1, 1), (1, −1), (1, −1)

)
, T
(
(1, −1), (1, −1), (1, −1)

))t
.



56 s. g. kim

We claim that T ∈ extBL(3l2∞) if and only if

1 =
∣∣T((1, 1), (1, 1), (1, 1))∣∣ = ∣∣T((1, −1), (1, 1), (1, 1))∣∣

=
∣∣T((1, 1), (1, −1), (1, 1))∣∣ = ∣∣T((1, 1), (1, 1), (1, −1))∣∣

=
∣∣T((1, −1), (1, −1), (1, 1))∣∣ = ∣∣T((1, −1), (1, 1), (1, −1))∣∣

=
∣∣T((1, 1), (1, −1), (1, −1))∣∣ = ∣∣T((1, −1), (1, −1), (1, −1))∣∣.

(⇒): Otherwise. Then we have 8 cases as follows:

Case 1 :
∣∣T((1, 1), (1, 1), (1, 1))∣∣ < 1 or

Case 2 :
∣∣T((1, −1), (1, 1), (1, 1))∣∣ < 1 or

Case 3 :
∣∣T((1, 1), (1, −1), (1, 1))∣∣ < 1 or

Case 4 :
∣∣T((1, 1), (1, 1), (1, −1))∣∣ < 1 or

Case 5 :
∣∣T((1, −1), (1, −1), (1, 1))∣∣ < 1 or

Case 6 :
∣∣T((1, −1), (1, 1), (1, −1))∣∣ < 1 or

Case 7 :
∣∣T((1, 1), (1, −1), (1, −1))∣∣ < 1 or

Case 8 :
∣∣T((1, −1), (1, −1), (1, −1))∣∣ < 1.

Case 1:
∣∣T((1, 1), (1, 1), (1, 1))∣∣ < 1. Let

ϵ1 := T
(
(1, 1), (1, 1), (1, 1)

)
,

ϵ2 := T
(
(1, −1), (1, 1), (1, 1)

)
,

ϵ3 := T
(
(1, 1), (1, −1), (1, 1)

)
,

ϵ4 := T
(
(1, 1), (1, 1), (1, −1)

)
,

ϵ5 := T
(
(1, −1), (1, −1), (1, 1)

)
,

ϵ6 := T
(
(1, −1), (1, 1), (1, −1)

)
,

ϵ7 := T
(
(1, 1), (1, −1), (1, −1)

)
,

ϵ8 := T
(
(1, −1), (1, −1), (1, −1)

)
.

Then,
AT = (ϵ1, ϵ2, ϵ3, ϵ4, ϵ5, ϵ6, ϵ7, ϵ8)

t.

Let n0 ∈ N such that |ϵ1| + 1n0 < 1. Let T1, T2 ∈ L(
3l2∞) be the solutions of

AT1 =
(
ϵ1+

1

n0
, ϵ2, ϵ3, ϵ4, ϵ5, ϵ6, ϵ7, ϵ8

)t
, AT2 =

(
ϵ1−

1

n0
, ϵ2, ϵ3, ϵ4, ϵ5, ϵ6, ϵ7, ϵ8

)t
.



the geometry of L(3l2∞) 57

Note that Tj ̸= T, ∥Tj∥ = ∥ATj∥∞ = 1 for j = 1, 2. It follows that

A
(1
2
(T1 + T2)

)
= (ϵ1, ϵ2, ϵ3, ϵ4, ϵ5, ϵ6, ϵ7, ϵ8)

t = AT,

which shows that
1

2
(T1 + T2) = A

−1(ϵ1, ϵ2, ϵ3, ϵ4, ϵ5, ϵ6, ϵ7, ϵ8)
t = T,

so T is not extreme. By the similar argument in the Case 1, if any of Cases
2–8 holds, then we may reach to a contradiction.

(⇐): Let ϵj ∈ R be given for j = 1, 2, . . . , 8. Consider the following system
of 8 simultaneous linear equations: AT = (ϵ1, ϵ2, ϵ3, ϵ4, ϵ5, ϵ6, ϵ7, ϵ8)

t, ie,

a + b + c1 + c2 + c3 + d1 + d2 + d3 = ϵ1,

a − b − c1 + c2 + c3 + d1 − d2 − d3 = ϵ2,
a − b + c1 − c2 + c3 − d1 + d2 − d3 = ϵ3,
a − b + c1 + c2 − c3 − d1 − d2 + d3 = ϵ4,
a + b − c1 − c2 + c3 − d1 − d2 + d3 = ϵ5,
a + b − c1 + c2 − c3 − d1 + d2 − d3 = ϵ6,
a + b + c1 − c2 − c3 + d1 − d2 − d3 = ϵ7,
a − b − c1 − c2 − c3 + d1 + d2 + d3 = ϵ8.

(∗)

We get the unique solution of (∗) as follows: T = A−1(ϵ1, ϵ2, ϵ3, ϵ4, ϵ5, ϵ6, ϵ7, ϵ8)t,
ie,

a =
1

8
(ϵ1 + ϵ2 + ϵ3 + ϵ4 + ϵ5 + ϵ6 + ϵ7 + ϵ8),

b =
1

8
(ϵ1 − ϵ2 − ϵ3 − ϵ4 + ϵ5 + ϵ6 + ϵ7 − ϵ8),

c1 =
1

8
(ϵ1 − ϵ2 + ϵ3 + ϵ4 − ϵ5 − ϵ6 + ϵ7 − ϵ8),

c2 =
1

8
(ϵ1 + ϵ2 − ϵ3 + ϵ4 − ϵ5 + ϵ6 − ϵ7 − ϵ8),

c3 =
1

8
(ϵ1 + ϵ2 + ϵ3 − ϵ4 + ϵ5 − ϵ6 − ϵ7 − ϵ8),

d1 =
1

8
(ϵ1 + ϵ2 − ϵ3 − ϵ4 − ϵ5 − ϵ6 + ϵ7 + ϵ8),

d2 =
1

8
(ϵ1 − ϵ2 + ϵ3 − ϵ4 − ϵ5 + ϵ6 − ϵ7 + ϵ8),

d3 =
1

8
(ϵ1 − ϵ2 − ϵ3 + ϵ4 + ϵ5 − ϵ6 − ϵ7 + ϵ8).

(∗∗)



58 s. g. kim

Let T1 = (a+ϵ, b+δ, c1 +γ1, c2 +γ2, c3 +γ3, d1 +ρ1, d2 +ρ2, d3 +ρ3) ∈ L(3l2∞)
and T2 = (a−ϵ, b−δ, c1 −γ1, c2 −γ2, c3 −γ3, d1 −ρ1, d2 −ρ2, d3 −ρ3) ∈ L(3l2∞)
be such that 1 = ∥T1∥ = ∥T2∥ for some ϵ, δ, γj, ρj for j = 1, 2, 3. Then, for
k = 1, 2,

1 ≥
∣∣Tk((1, 1), (1, 1), (1, 1))∣∣ = 1 + |ϵ + δ + γ1 + γ2 + γ3 + ρ1 + ρ2 + ρ3|,

1 ≥
∣∣Tk((1, −1), (1, 1), (1, 1))∣∣ = 1 + |ϵ − δ − γ1 + γ2 + γ3 + ρ1 − ρ2 − ρ3|,

1 ≥
∣∣Tk((1, 1), (1, −1), (1, 1))∣∣ = 1 + |ϵ − δ + γ1 − γ2 + γ3 − ρ1 + ρ2 − ρ3|,

1 ≥
∣∣Tk((1, 1), (1, 1), (1, −1))∣∣ = 1 + |ϵ − δ + γ1 + γ2 − γ3 − ρ1 − ρ2 + ρ3|,

1 ≥
∣∣Tk((1, −1), (1, −1), (1, 1))∣∣ = 1 + |ϵ + δ − γ1 − γ2 + γ3 − ρ1 − ρ2 + ρ3|,

1 ≥
∣∣Tk((1, −1), (1, 1), (1, −1))∣∣ = 1 + |ϵ + δ − γ1 + γ2 − γ3 − ρ1 + ρ2 − ρ3|,

1 ≥
∣∣Tk((1, 1), (1, −1), (1, −1))∣∣ = 1 + |ϵ + δ + γ1 − γ2 − γ3 + ρ1 − ρ2 − ρ3|,

1 ≥
∣∣Tk((1, −1), (1, −1), (1, −1))∣∣ = 1 + |ϵ − δ − γ1 − γ2 − γ3 + ρ1 + ρ2 + ρ3|.

Therefore, we have

0 = ϵ + δ + γ1 + γ2 + γ3 + ρ1 + ρ2 + ρ3,

0 = ϵ − δ − γ1 + γ2 + γ3 + ρ1 − ρ2 − ρ3,

0 = ϵ − δ + γ1 − γ2 + γ3 − ρ1 + ρ2 − ρ3,

0 = ϵ − δ + γ1 + γ2 − γ3 − ρ1 − ρ2 + ρ3,

0 = ϵ + δ − γ1 − γ2 + γ3 − ρ1 − ρ2 + ρ3,

0 = ϵ + δ − γ1 + γ2 − γ3 − ρ1 + ρ2 − ρ3,

0 = ϵ + δ + γ1 − γ2 − γ3 + ρ1 − ρ2 − ρ3,

0 = ϵ − δ − γ1 − γ2 − γ3 + ρ1 + ρ2 + ρ3.

Hence, A(ϵ, δ, γ1, γ2, γ3, ρ1, ρ2, ρ3)
t = 0. By (∗∗), 0 = ϵ = δ = γ1 = γ2 = γ3 =

ρ1 = ρ2 = ρ3. Hence, T is extreme. Therefore, we complete the proof.

Corollary 2.3. If T = (a, b, c1, c2, c3, d1, d2, d3) ∈ extBL(3l2∞), then |a|,
|b|, |cj|, |dj| ∈ {0, 14,

1
2
, 3
4
, 1} for j = 1, 2, 3.



the geometry of L(3l2∞) 59

Theorem 2.4. ([26])

extBLs(3l2∞) =
{
± (1, 0, 0, 0, 0, 0, 0, 0), ±(0, 1, 0, 0, 0, 0, 0, 0, ),

±
(1
2
, 0, 0, 0, 0, −

1

2
, −

1

2
, −

1

2

)
, ±
(
0,

1

2
, −

1

2
, −

1

2
, −

1

2
, 0, 0, 0

)
,

±
(1
4
, −

3

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4

)
, ±
(
−

3

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4

)
,

±
(3
4
,
1

4
,
1

4
,
1

4
,
1

4
, −

1

4
, −

1

4
, −

1

4

)
,

±
(1
4
,
3

4
, −

1

4
, −

1

4
, −

1

4
,
1

4
,
1

4
,
1

4

)}
.

Theorem 2.5. extBLs(3l2∞) = extBL(3l2∞) ∩ Ls(
3l2∞).

Proof. It follows from Theorems 2.2 and 2.4.

Remarks. (1) 24 = |extBLs(3l2∞)| < |extBL(3l2∞)| = 2
8.

(2) Let T = (a, b, c1, c2, c3, d1, d2, d3) ∈ L(3l2∞). Then, by scaling, we may
assume that dj ≥ 0 for j = 1, 2, 3.

Proof. Let T1((x1, x2), (y1, y2), (z1, z2)) := T((ϵ1x1, x2), (ϵ2y1, y2)), (ϵ3z1, z2),
where ϵk = ±1 be given satisfying ϵjdj ≥ 0 for j = 1, 2, 3.

Question. Is it true that extBLs(nl2∞) = extBL(nl2∞) ∩ Ls(
nl2∞) for n ≥ 4?

3. The exposed points of the unit ball of L(3l2∞)

Theorem 3.1. Let f ∈ L(3l2∞)∗ with

α = f(x1y1z1), β = f(x2y2z2), γ1 = f(x2y1z1), γ2 = f(x1y2z1),

γ3 = f(x1y1z2), δ1 = f(x1y2z2), δ2 = f(x2y1z2), δ3 = f(x2y2z1).



60 s. g. kim

Then,

∥f∥ = max
{∣∣∣∣aα + bβ + ∑

j=1,2,3

cjγj +
∑

j=1,2,3

djδj

∣∣∣∣ :
a =

1

8
(ϵ1 + ϵ2 + ϵ3 + ϵ4 + ϵ5 + ϵ6 + ϵ7 + ϵ8),

b =
1

8
(ϵ1 − ϵ2 − ϵ3 − ϵ4 + ϵ5 + ϵ6 + ϵ7 − ϵ8),

c1 =
1

8
(ϵ1 − ϵ2 + ϵ3 + ϵ4 − ϵ5 − ϵ6 + ϵ7 − ϵ8),

c2 =
1

8
(ϵ1 + ϵ2 − ϵ3 + ϵ4 − ϵ5 + ϵ6 − ϵ7 − ϵ8),

c3 =
1

8
(ϵ1 + ϵ2 + ϵ3 − ϵ4 + ϵ5 − ϵ6 − ϵ7 − ϵ8),

d1 =
1

8
(ϵ1 + ϵ2 − ϵ3 − ϵ4 − ϵ5 − ϵ6 + ϵ7 + ϵ8),

d2 =
1

8
(ϵ1 − ϵ2 + ϵ3 − ϵ4 − ϵ5 + ϵ6 − ϵ7 + ϵ8),

d3 =
1

8
(ϵ1 − ϵ2 − ϵ3 + ϵ4 + ϵ5 − ϵ6 − ϵ7 + ϵ8),

ϵj = ±1, for j = 1, 2, . . . , 8
}

Proof. Proof. It follows from Theorem 2.2 and the Krein-Milman Theo-
rem.

Theorem 3.2. expBL(3l2∞) = extBL(3l2∞).

Proof. We will show that extBL(3l2∞) ⊂ expBL(3l2∞). By Theorem 2.2,
Corollary 2.3 and Remarks(2), it suffices to show that if

T =(1, 0, 0, 0, 0, 0, 0, 0),
(
−

1

2
,
1

2
,
1

2
,
1

2
, 0, 0, 0, 0

)
(
−

3

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4
,
1

4

)
,
(3
4
, −

1

4
, −

1

4
, −

1

4
,
1

4
,
1

4
,
1

4
,
1

4

)
,

then T ∈ expBL(3l2∞).
Claim: T = (1, 0, 0, 0, 0, 0, 0, 0) ∈ expBL(3l2∞).
Let f ∈ L(3l2∞)∗ with α = 1, 0 = β = γj = δj for j = 1, 2, 3. Note that, by

Corollary 2.3 and Theorems 2.2 and 3.1, ∥f∥ = 1 = f(T) and |f(S)| < 1 for
all S ∈ extBL(3l2∞)\{±T}. The claim follows from Theorem 2.3 of [21].



the geometry of L(3l2∞) 61

Claim: T =
(
− 1

2
, 1
2
, 1
2
, 1
2
, 0, 0, 0, 0

)
∈ expBL(3l2∞).

Let f ∈ L(3l2∞)∗ with −α =
1
2
= β = γ1 = γ2, 0 = γ3 = δj for j = 1, 2, 3.

Note that, by Corollary 2.3 and Theorems 2.2 and 3.1, ∥f∥ = 1 = f(T) and
|f(S)| < 1 for all S ∈ extBL(3l2∞)\{±T}. The claim follows from Theorem 2.3
of [21].

Claim: T =
(
− 3

4
, 1
4
, 1
4
, 1
4
, 1
4
, 1
4
, 1
4
, 1
4

)
∈ expBL(3l2∞).

Let f ∈ L(3l2∞)∗ with α = −
1
2
, 5
14

= β = γj = δj for j = 1, 2, 3. Note that,
by Corollary 2.3 and Theorems 2.2 and 3.1, ∥f∥ = 1 = f(T) and |f(S)| < 1
for all S ∈ extBL(3l2∞)\{±T}. The claim follows from Theorem 2.3 of [21].

Claim: T =
(
3
4
, −1

4
, −1

4
, −1

4
, 1
4
, 1
4
, 1
4
, 1
4

)
∈ expBL(3l2∞).

Let f ∈ L(3l2∞)∗ with α =
1
2
, − 5

14
= β = γ1 = γ2 = −γ3 = −δj for j =

1, 2, 3. Note that, by Corollary 2.3 and Theorems 2.2 and 3.1, ∥f∥ = 1 = f(T)
and |f(S)| < 1 for all S ∈ extBL(3l2∞)\{±T}. The claim follows from Theorem
2.3 of [21]. We complete the proof.

Theorem 3.3. ([26]) expBLs(3l2∞) = extBLs(3l2∞).

Theorem 3.4. expBLs(3l2∞) = expBL(3l2∞) ∩ Ls(
3l2∞).

Proof. It follows from Theorems 3.2 and 3.3.

Question. Is it true that expBLs(nl2∞) = expBL(nl2∞) ∩Ls(
nl2∞) for n ≥ 4?

4. Optimal constants in the Bohnenblust-Hille inequality for
symmetric multilinear forms and polynomials

Theorem 4.1. 1 ≤ Cp(n : K) ≤ n
n

n!
Cs(n : K) ≤ n

n

n!
C(n : K) for all n ≥ 2.

Proof. It is enough to show that Cp(n : K) ≤ n
n

n!
Cs(n : K). Let P ∈ P(nc0 :

K), ∥P∥ = 1. By the polarization formula, ∥P̌∥ ≤ n
n

n!
∥P∥ = n

n

n!
, where P̌ is

the corresponding symmetric n-linear form to P . Hence,(
∞∑
j=1

|
n!

nn
P(ej)|

2n
n+1

)n+1
2n

≤

(
∞∑

j1,··· ,jn=1

∣∣∣∣ n!nn P̌(ej1, · · · , ejn)
∣∣∣∣

2n
n+1

)n+1
2n

≤ Cs(n : K).



62 s. g. kim

Theorem 4.2. Cs(2 : R) = C(2 : R) =
√
2.

Proof. It is enough to show that Cs(2 : R) =
√
2. Let

T
(
(x1, x2), (y1, y2)

)
=

1

2
x1y1 −

1

2
x2y2 +

1

2
x1y2 +

1

2
x2y1.

Then T ∈ Ls(2l2∞), ∥T∥ = 1. By a result of [12],
√
2 ≤ (

∑2
i,j=1 |T(ei, ej)|

4
3 )

3
4 ≤

Cs(2 : R) ≤ C(2 : R) =
√
2.

Theorem 4.3. ([16])

extBLs(2l2∞) =

{
±(1, 0, 0, 0), ±(0, 1, 0, 0), ±

1

2
(1, −1, 1, 1), ±

1

2
(1, −1, −1, −1)

}
.

Theorem 4.4.

sup

{( 2∑
i,j=1

|T(ei, ej)|
4
3

)3
4

: T ∈ Ls(2l2∞), ∥T∥ = 1, T /∈ extBLs(2l2∞)

}

= Cs(2 : R).

Proof. Let

l := sup

{( 2∑
i,j=1

|T(ei, ej)|
4
3

)3
4

: T ∈ Ls(2l2∞), ∥T∥ = 1, T /∈ extBLs(2l2∞)

}
.

For |c| < 1
2
, let

Tc
(
(x1, x2), (y1, y2)

)
=

1

2
x1y1 −

1

2
x2y2 + cx1y2 + cx2y1.

Then Tc ∈ Ls(2l2∞), ∥Tc∥ = 1. By Theorem 4.3, Tc /∈ extBLs(2l2∞). It follows
that

Cs(2 : R) ≥ l ≥ sup

{( 2∑
i,j=1

|Tc(ei, ej)|
4
3

)3
4

: |c| <
1

2

}

= sup
|c|< 1

2

(
2
(1
2

)4
3
+ 2|c|

4
3

)3
4

=
√
2 = Cs(2 : R).



the geometry of L(3l2∞) 63

Theorem 4.5. Let n ≥ 2. Then, 2
n+1
2n ≤ Cp(n : R).

Proof. Let

w := sup

{( ∞∑
j=1

|P(ej)|
2n
n+1

)n+1
2n

: P ∈ P(nc0), ∥P∥ = 1,

P
(
(xm)

∞
m=1

)
=

∞∑
j=1

ajx
n
j for some aj ∈ R

}
.

Claim: w = 2
n+1
2n .

Let P ∈ P(nc0), ∥P∥ = 1, P((xm)∞m=1) =
∑∞

j=1 ajx
n
j for some aj ∈ R. Let

A := {j ∈ N : aj ≥ 0} and B := N\A. Note that, for every k ∈ N,

1 ≥

∣∣∣∣∣P
( ∑

j∈A, j≤k
ej

)∣∣∣∣∣ = ∑
j∈A, j≤k

|aj|

and

1 ≥

∣∣∣∣∣P
( ∑

j∈B, j≤k
ej

)∣∣∣∣∣ = ∑
j∈B, j≤k

|aj|.

Hence, ∑
j∈A

|aj| ≤ 1,
∑
j∈B

|aj| ≤ 1.

It follows that( ∞∑
j=1

|P(ej)|
2n
n+1

)n+1
2n

=

(∑
j∈A

|aj|
2n
n+1 +

∑
j∈B

|aj|
2n
n+1

)n+1
2n

≤
(∑

j∈A
|aj| +

∑
j∈B

|aj|
)n+1

2n

≤ 2
n+1
2n ,

which shows that w ≤ 2
n+1
2n . Let P0

(
(xm)

∞
m=1

)
= xn1 − x

n
2 ∈ P(

nc0) for
(xm)

∞
m=1 ∈ c0. Then ∥P0∥ = 1. Hence,

2
n+1
2n =

( ∞∑
j=1

|P0(ej)|
2n
n+1

)n+1
2n

≤ w ≤ 2
n+1
2n .

Therefore, 2
n+1
2n ≤ Cp(n : R). We complete the proof.



64 s. g. kim

Corollary 4.6. C(2 : R) < 2
3
4 ≤ Cp(2 : R) ≤ 2

√
2.

Theorem 4.7. Let T : l2∞(R) × l2∞(R) → R be given by T(x, y) =∑2
i,j=1 aijxiyj, with aij ∈ R, a12 = a21. Then the symmetric bilinear forms

satisfying ( 2∑
i,j=1

|T(ei, ej)|
4
3

)3
4

=
√
2∥T∥

are given by T(x, y) = a(x1y1 − x2y2 + x1y2 + x2y1) or T(x, y) = a(−x1y1 +
x2y2 + x1y2 + x2y1) for all a ∈ R\{0}.

Proof. It follows from Theorem 4.1 of [3].

Theorem 4.8. Let T ∈ Ls(2l2∞), ∥T∥ = 1, T(ej, ej) ̸= 0 for j = 1, 2.
Then, ( 2∑

i,j=1

|T(ei, ej)|
4
3

)3
4

= Cs(2 : R)

if and only if T ∈ extBLs(2l2∞).

Proof. It follows from Theorems 4.2, 4.3 and 4.7.

Theorem 4.9.

sup{(
2∑

i,j=1

|T(ei, ej)|
6
4 )

4
6 : T ∈ extBL(3l2∞)} =

(7 + 3
3
2 )

2
3

4
< C(3 : R).

Proof. Diniz et al. [12] showed that 2
2
3 ≤ C(3 : R) ≤ 1.782. By Theo-

rem 2.2 and Corollary 2.3,

sup

{( 2∑
i,j=1

|T(ei, ej)|
6
4

)4
6

: T ∈ extBL(3l2∞)
}

=
(7 + 3

3
2 )

2
3

4
< 2

2
3 ≤ C(3 : R).

Questions. (1) Is it true that Cs(n : R) = C(n : R) for all n ≥ 3?
(2) Is it true that Cp(2 : R) = 2

3
4 ?



the geometry of L(3l2∞) 65

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