�
EXTRACTA

MATHEMATICAE

Volumen 33, Número 2, 2018

instituto de investigación de matemáticas de la

universidad de extremadura

EXTRACTA MATHEMATICAE

Vol. 37, Num. 2 (2022), 243 – 259

doi:10.17398/2605-5686.37.2.243

Available online March 2, 2022

Smooth 2-homogeneous polynomials on
the plane with a hexagonal norm

Sung Guen Kim

Department of Mathematics, Kyungpook National University
Daegu 702-701, South Korea

sgk317@knu.ac.kr

Received September 14, 2021 Presented by Ricardo Garćıa
Accepted February 2, 2022

Abstract: Motivated by the classifications of extreme and exposed 2-homogeneous polynomials on

the plane with the hexagonal norm ‖(x,y)‖ = max{|y|, |x| + 1
2
|y|} (see [15, 16]), we classify all

smooth 2-homogeneous polynomials on R2 with the hexagonal norm.

Key words: The Krein-Milman theorem, smooth points, extreme points, exposed points, 2-

homogeneous polynomials on the plane with the hexagonal norm.

MSC (2020): 46A22.

1. Introduction

One of the main results about smooth points is known as “the Mazur
density theorem”. Recall that the Mazur density theorem ([9, p. 71]) says that
the set of all the smooth points of a solid closed convex subset of a separable
Banach space is a residual subset of its boundary. We denote by BE the closed
unit ball of a real Banach space E and also by E∗ the dual space of E. We
recall that a point x ∈ BE is said to be an extreme point of BE if the equation
x = 1

2
(y + z) for some y,z ∈ BE implies that x = y = z. A point x ∈ BE is

called an exposed point of BE if there is an 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. A point x ∈ BE is called a smooth point
of BE if there is a unique f ∈ E∗ so that f(x) = 1 = ‖f‖. We denote by
ext BE, exp BE and sm BE the set of extreme points, the set of exposed points
and the set of smooth points of BE, respectively. For n ∈ N, we denote by
L(nE) the Banach space of all continuous n-linear forms on E endowed with
the norm ‖T‖ = sup‖xk‖=1 |T(x1, · · · ,xn)|. A n-linear form T is symmetric if
T(x1, . . . ,xn) = T(xσ(1), . . . ,xσ(n)) for every permutation σ on {1, 2, . . . ,n}.
We denote by Ls(nE) the Banach space of all continuous symmetric n-linear

ISSN: 0213-8743 (print), 2605-5686 (online)

© The author(s) - Released under a Creative Commons Attribution License (CC BY-NC 3.0)

https://doi.org/10.17398/2605-5686.37.2.243
mailto:sgk317@knu.ac.kr
https://publicaciones.unex.es/index.php/EM
https://creativecommons.org/licenses/by-nc/3.0/


244 s.g. kim

forms on E. A mapping P : E → R is a continuous n-homogeneous polynomial
if there exists a unique T ∈ Ls(nE) such that P(x) = T(x, · · · ,x) for every
x ∈ E. In this case it is convenient to write T = P̌ . We denote by P(nE)
the Banach space of all continuous n-homogeneous polynomials from E into
R endowed with the norm ‖P‖ = sup‖x‖=1 |P(x)|. For more details about
the theory of multilinear mappings and polynomials on a Banach space, we
refer to [7].

Choi et al. [2, 3, 4, 5] initiated and characterized the smooth points,
extreme points and exposed points of the unit balls of P(2l21),P(

2l22) and
P(2c0). Kim [10] and Choi and Kim [6] classified the exposed 2-homogeneous
polynomials on P(2l2p) (1 ≤ p ≤∞). Kim et al. [17] characterized the exposed
2-homogeneous polynomials on Hilbert spaces. Kim [11, 12, 14] classified
the smooth points, extreme points and exposed points of the unit ball of
P(2d∗(1,w)2), where d∗(1,w)2 = R2 with the octagonal norm of weight w.
For some applications of the classification of the extreme points of the unit
ball of P(2d∗(1,w)2), Kim [13] investigated polarization and unconditional
constants of P(2d∗(1,w)2). Thus we fully described the geometry of the unit
ball of P(2d∗(1,w)2). We refer to [1, 8, 18, 19] and references therein for some
recent work about extremal properties of homogeneous polynomials on some
classical Banach spaces.

We will denote by P(x,y) = ax2 + by2 + cxy a 2-homogeneous polynomial
on a real Banach space of dimension 2 for some a,b,c ∈ R. Let 0 < w < 1 be
fixed. We denote R2

h(w)
= R2 with the hexagonal norm of weight w by

‖(x,y)‖h(w) := max
{
|y| , |x| + (1 −w)|y|

}
.

Throughout the paper we will denote R2
h( 1

2
)

by H. Kim [15, 16] classified the
extreme and exposed points of the unit ball of P(2H) as follows:

(a) ext BP(2H) =
{
±y2,±

(
x2 +

1

4
y2 ±xy

)
, ±
(
x2 +

3

4
y2
)
,

±
[
x2 +

(c2
4
− 1
)
y2 ± cxy

]
(0 ≤ c ≤ 1) ,

±
[
ax2 +

(a + 4√1 −a
4

− 1
)
y2

± (a + 2
√

1 −a)xy
]

(0 ≤ a ≤ 1)
}

;

(b) exp BP(2H) = ext BP(2H).



smooth 2-homogeneous polynomials 245

In this paper we classify sm BP(2H) using the classifications of ext BP(2H)
and exp BP(2H).

2. Results

Theorem 2.1. ([15]) Let P(x,y) = ax2 + by2 + cxy ∈P(2H) with a ≥ 0,
c ≥ 0 and a2 + b2 + c2 6= 0. Then:

Case 1: c < a.

If a ≤ 4b, then

‖P‖ = max
{
a,b,

∣∣∣1
4
a + b

∣∣∣ + 1
2
c,

4ab− c2

4a
,

4ab− c2

2c + a + 4b
,

4ab− c2

|2c−a− 4b|

}
= max

{
a,b,

∣∣∣1
4
a + b

∣∣∣ + 1
2
c

}
.

If a > 4b, then ‖P‖ = max
{
a, |b|,

∣∣∣14a + b∣∣∣ + 12c, |c2−4ab|4a }.
Case 2: c ≥ a.

If a ≤ 4b, then ‖P‖ = max
{
a,b,

∣∣∣14a + b∣∣∣ + 12c, |c2−4ab|2c+a+4b}.
If a > 4b, then ‖P‖ = max

{
a, |b|,

∣∣∣14a + b∣∣∣ + 12c, c2−4ab2c−a−4b}.
Note that if ‖P‖ = 1, then |a| ≤ 1, |b| ≤ 1, |c| ≤ 2.

Theorem 2.2. ([15, 16])

ext BP(2H) = exp BP(2H)

=

{
±y2 , ±

(
x2 +

1

4
y2 ±xy

)
, ±
(
x2 +

3

4
y2
)
,

±
[
x2 +

(
c2

4
− 1
)
y2 ± cxy

]
(0 ≤ c ≤ 1) ,

±
[
ax2 +

(
a + 4

√
1 −a

4
− 1
)
y2

± (a + 2
√

1 −a)xy
]

(0 ≤ a ≤ 1)
}
.

By the Krein-Milman theorem, a convex function (like a functional norm,
for instance) defined on a convex set (like the unit ball of a finite dimen-
sional polynomial space) attains its maximum at one extreme point of the
convex set.



246 s.g. kim

Theorem 2.3. ([16]) Let f ∈ P(2H)∗ with α = f(x2), β = f(y2),
γ = f(xy). Then

‖f‖ = max
{
|β|,

∣∣∣α + 1
4
β
∣∣∣ + |γ|,∣∣∣α + 3

4
β
∣∣∣,∣∣∣∣α +

(
c2

4
− 1
)
β

∣∣∣∣ + c|γ| (0 ≤ c ≤ 1),∣∣∣∣aα +
(
a + 4

√
1 −a

4
− 1
)
β

∣∣∣∣ + (a + 2√1 −a)|γ| (0 ≤ a ≤ 1)
}
.

Proof. It follows from Theorem 2.2 and the fact that ‖f‖ = sup
P∈ext B

∣∣f(P)∣∣,
where B := BP(2H).

Note that if ‖f‖ = 1, then |α| ≤ 1, |β| ≤ 1, |γ| ≤ 1
2
.

Remark. Let P(x,y) = ax2 + by2 + cxy ∈P(2H) with ‖P‖ = 1. Then the
following are equivalent:

(1) P is smooth;

(2) −P(x,y) = −ax2 − by2 − cxy is smooth;
(3) P(x,−y) = ax2 + by2 − cxy is smooth.

As a consequence of the previous remark, our attention can be restricted
to polynomials Q(x,y) = ax2 + by2 + cxy ∈P(2H) with a ≥ 0, c ≥ 0.

We are in position to prove the main result of this paper.

Theorem 2.4. Let P(x,y) = ax2 + by2 + cxy ∈P(2H) with a ≥ 0, c ≥ 0,
‖P‖ = 1. Then P is a smooth point of the unit ball of P(2H) if and only if
one of the following mutually exclusive conditions holds:

(1) a = 0 , 0 < |b| < 1 ;
(2) a = 1 , b = −3

4
, 1

4
, c < 1 ;

(3) a = 1 , −1 < b < −3
4
, b− c

2
> −5

4
, c

2

4
− b < 1 ;

(4) a = 1 , −3
4
< b < 1

4
;

(5) a = 1 , 1
4
≤ b , b + c

2
< 3

4
;

(6) 0 < a < 1 , b = 0 ;

(7) 0 < a < 1 , c ≤ a , 0 6= 4b < a ;
(8) 0 < a < 1 , 0 < c ≤ a < 4b ;



smooth 2-homogeneous polynomials 247

(9) 0 < a < 1 , 4b = a < c ;

(10) 0 < a < 1 , 0 6= 4b < a < c , c 6= a + 2
√

1 −a ;
(11) 0 < a < 1 , a < 4b , a < c .

Proof. Let Q(x,y) = ax2 + by2 + cxy ∈ P(2H) with a ≥ 0, c ≥ 0
and ‖Q‖ = 1.

Case 1: a = 0.

Note that if b = 0 or ±1, then Q is not smooth. In fact, if b = 0, then
Q = 2xy. For j = 1, 2, let fj ∈P(2H)∗ be such that

f1
(
x2
)

=
1

4
, f1

(
y2
)

= 1 , f1(xy) =
1

2
, f2

(
x2
)

= 0 = f2
(
y2
)
, f2(xy) =

1

2
.

By Theorem 2.3, fj(Q) = 1 = ‖fj‖ for j = 1, 2. Thus Q is not smooth. If
b = ±1, then P = ±y2. For j = 1, 2, let fj ∈P(2H)∗ be such that

f1
(
x2
)

= ±
1

4
, f1

(
y2
)

= ±1 , f1(xy) = ±
1

2
,

f2
(
x2
)

= 0 = f2(xy) , f2
(
y2
)

= ±1 .

By Theorem 2.3, fj(Q) = 1 = ‖fj‖ for j = 1, 2. Thus Q is not smooth.

Claim: if a = 0, 0 < |b| < 1, then Q is smooth.

Without loss of generality, we may assume that 0 < b < 1. By Theorem
2.1, 1 = ‖Q‖ = b + 1

2
c. Thus c = 2(1 − b), so 0 < c < 2. Let f ∈ P(2H)∗

be such that f(Q) = 1 = ‖f‖. Notice that 1 = bβ + cγ. We will show that
α = 1

4
, β = 1, γ = 1

2
. Since 0 < b < 1, 0 < c < 2, we can choose δ > 0 such

that

0 < 2(1 − b) + t = c + t < 2 , 0 < b−
1

2
t < 1 ,

for all t ∈ (−δ,δ). Let Qt(x,y) =
(
b − 1

2
t
)
y2 + (c + t)xy for all t ∈ (−δ,δ).

By Theorem 2.1, ‖Qt‖ = 1 for all t ∈ (−δ,δ). For all t ∈ (−δ,δ),

1 = bβ + cγ ≥ f(Qt) =
(
b−

1

2
t

)
β + (c + t)γ ,

which shows that t
(
γ − 1

2
β
)
≤ 0, for all t ∈ (−δ,δ). Thus γ = 1

2
β. Since

1 = f(Q) = bβ + cγ = 2γ, we have β = 1, γ = 1
2
. By Theorem 2.3, 1 ≥



248 s.g. kim

∣∣∣α + 14β∣∣∣ + |γ| = ∣∣∣α + 14∣∣∣ + 12, so
−

3

4
≤ α ≤

1

4
. (1)

By Theorem 2.3, for 0 ≤ c̃ ≤ 1,

1 ≥
∣∣∣∣α +

(
c̃2

4
− 1
)∣∣∣∣ + c̃2 = −

(
α +

(
c̃2

4
− 1
))

+
c̃

2
,

which implies that
4α ≥ sup

0≤c̃≤1
(2c̃− c̃2) = 1 . (2)

By (1) and (2), α = 1
4
. Therefore, Q is smooth.

Case 2: a = 1.

If b = −1, then Q = x2 −y2. For j = 1, 2, let fj ∈P(2H)∗ be such that

f1(x
2) = 1 , f1(y

2) = 0 = f1(xy) ,

f2(x
2) = 0 = f2(xy) , f2(y

2) = −1 .

By Theorem 2.3, fj(Q) = 1 = ‖fj‖ for j = 1, 2. Hence, Q is not smooth.

Claim: if
(
a = 1, b = −3

4
, 1

4
, c < 1

)
,
(
a = 1, −1 < b < 1

4
, b 6= −3

4

)
or(

a = 1, 1
4
≤ b, b + c

2
< 3

4

)
, then Q is smooth.

Note that if a = 1,b = −3
4
, then c ≤ 1. Note also that if a = 1,b = −3

4
,c =

1, then Q is not smooth.
Suppose that a = 1, b = −3

4
, c < 1. Let f ∈ P(2H)∗ be such that

f(Q) = 1 = ‖f‖. Then 1 = α − 3
4
β + cγ. We will show that α = 1,

β = γ = 0. Since 0 ≤ c < 1 and by Theorem 2.1, we can choose δ > 0
such that ‖Ru‖ = ‖Sv‖ = 1 for all u,v ∈ (−δ,δ), where

Ru(x,y) = x
2 −

3

4
y2 + (c + u)xy ,

Sv(x,y) = x
2 −

(
3

4
+ v

)
y2 + cxy ∈P(2H) .

It follows that, for all u,v ∈ (−δ,δ),

1 = α−
3

4
β + cγ ≥ f(Ru) = α−

3

4
β + (c + u)γ ,

1 = α−
3

4
β + cγ ≥ f(Sv) = α−

(
3

4
+ v

)
β + cγ ,



smooth 2-homogeneous polynomials 249

which shows that α = 1, β = γ = 0. Therefore, Q is smooth. By a similar
argument, if a = 1, b = 1

4
, c < 1, then Q is smooth.

Suppose that a = 1, −1 < b < 1
4
, b 6= −3

4
. Let a = 1, −1 < b < −3

4
. We

will show that c < 1. If not, then 1 ≤ c ≤ 2. By Theorem 2.1, b− c
2
≥ −5

4
,

c2−4b
2c−1−4b ≤ 1, which shows that c = 1, b ≥−

3
4
. This is a contradiction. Hence,

by Theorem 2.1, b− c
2
≥−5

4
, c

2

4
− b ≤ 1. We claim that if

a = 1 , −1 < b < −
3

4
, b−

c

2
> −

5

4
,

c2

4
− b < 1 ,

then Q is smooth. Let f ∈ P(2H)∗ be such that f(Q) = 1 = ‖f‖. Then,
1 = α + bβ + cγ. We will show that α = 1, β = γ = 0. By Theorem 2.1, we
can choose δ > 0 such that ‖Ru‖ = ‖Sv‖ = 1 for all u,v ∈ (−δ,δ), where

Ru(x,y) = x
2 + by2 + (c + u)xy ,

Sv(x,y) = x
2 + (b + v)y2 + cxy ∈P(2H) .

Thus α = 1, β = γ = 0. Therefore, Q is smooth.

Note that if

a = 1 , −1 < b < −
3

4
, b−

c

2
≥−

5

4
,

c2

4
− b = 1 ,

then Q is not smooth letting fj ∈P(2H)∗ be such that

f1(x
2) = 1 , f1(y

2) = 0 = f1(xy) ,

f2(x
2) = −

c2

4
, f2(y

2) = −1 , f2(xy) =
c

2
.

Thus x2 + (c
2

4
− 1)y2 + cxy (0 ≤ c ≤ 1) is not smooth.

Note also that if

a = 1 , −1 < b < −
3

4
, b−

c

2
= −

5

4
,

c2

4
− b ≤ 1 ,

then Q is not smooth letting fj ∈P(2H)∗ be such that

f1(x
2) = 1 , f1(y

2) = 0 = f1(xy) ,

f2(x
2) = −

1

4
, f2(y

2) = −1 , f2(xy) =
1

2
.



250 s.g. kim

Let a = 1, −3
4
< b < 1

4
. We will show that Q is smooth. First, suppose

that −3
4
< b < 0. Since ‖Q‖ = 1, by Theorem 2.1, we have 0 ≤ c ≤ 1. Let

f ∈ P(2H)∗ be such that f(Q) = 1 = ‖f‖. Then 1 = α + bβ + cγ. We will
show that α = 1, β = 0 = γ. Since −3

4
< b < 0, By Theorem 2.1, we can

choose δ > 0 such that ‖Ru‖ = ‖Sv‖ = 1 for all u,v ∈ (−δ,δ), where

Ru(x,y) = x
2 + (b + u)y2 + cxy ,

Sv(x,y) = x
2 + by2 + (c + v)xy ∈P(2H) .

Thus α = 1, β = 0 = γ. Hence, Q is smooth.
Suppose that c = 1. Then 1 = α + γ, α ≥ 0, γ ≥ 0. By Theorem 2.3,

1 ≥ sup
0≤ã≤1

ãα + (ã + 2
√

1 − ã)γ

= sup
0≤ã≤1

2
√

1 − ã(1 −α) + ã = 1 + (1 −α)2 ,

which implies that α = 1. Therefore, α = 1, β = 0 = γ. We have shown that
if 0 < c ≤ 1, then Q is smooth. Suppose that c = 0. Since 1 = α + bβ, β = 0,
we have α = 1. By Theorem 2.3, 1 ≥

∣∣∣α + 14β∣∣∣+ |γ| = 1 + γ, which shows that
γ = 0. Hence, Q is smooth.

Suppose that 0 ≤ b < 1
4
. Since ‖Q‖ = 1, by Theorem 2.1, 0 ≤ c ≤ 1.

Let f ∈ P(2H)∗ be such that f(Q) = 1 = ‖f‖. We will show that α = 1,
β = 0 = γ. Since 1 = f(Q) = α + bβ + cγ, we have α > 0. Indeed, if α ≤ 0,
then

1 ≤ bβ + cγ ≤ b|β| + c|γ| <
1

4
+

1

2
=

3

4
,

which is a contradiction. We also claim that α + 1
4
β ≥ 0. If not, then

α < 1
4
|β| ≤ 1

4
, which implies that

3

4
< 1 −α = bβ + cγ ≤ b|β| + c|γ| <

3

4
,

which is a contradiction. Note that

α + bβ = 1 − cγ ≥ 1 − c|γ| ≥ 1 −
c

2
≥

1

2
.

By Theorem 2.3,

α +
1

4
β + |γ| =

∣∣∣α + 1
4
β
∣∣∣ + |γ| ≤ 1 = α + bβ + cγ ≤ α + bβ + c|γ| ,



smooth 2-homogeneous polynomials 251

which shows that (
1

4
− b
)
β ≤ (c− 1)|γ| ≤ 0 .

Hence, β ≤ 0. By Theorem 2.3, for all 0 ≤ c̃ ≤ 1, it follows that

α +

(
1 −

c̃2

4

)
|β| + c̃|γ| =

∣∣∣∣α +
(
c̃2

4
− 1
)
β

∣∣∣∣ + c̃|γ|
≤ 1 = α + bβ + cγ
≤ α + bβ + c|γ| = α− b|β| + c|γ| ,

which implies that(
1 −

c̃2

4
+ b

)
|β| ≤ (c− c̃)|γ| (0 ≤ c̃ ≤ 1) .

Thus (
1 −

c2

4
+ b

)
|β| = lim

c̃→c−

(
1 −

c̃2

4
+ b

)
|β| ≤ lim

c̃→c−
(c− c̃)|γ| = 0 ,

so β = 0. Since 1 = f(Q) = α + cγ, we have γ ≥ 0. By Theorem 2.3,

ãα +
(
ã + 2

√
1 − ã

)
γ ≤ 1 = α + cγ (0 ≤ ã ≤ 1) ,

which implies that

(ã− c + 2
√

1 − ã)γ ≤ (1 − ã)α (0 ≤ ã ≤ 1) . (3)

If c < 1, then

(1 − c)γ = lim
ã→1−

(
ã− c + 2

√
1 − ã

)
γ ≤ lim

ã→1−
(1 − ã)α = 0 ,

so γ = 0. Therefore, α = 1, β = 0. Suppose that c = 1. By (3),

(ã− 1 + 2
√

1 − ã)γ ≤ (1 − ã)α (0 ≤ ã ≤ 1) ,

which implies that

2γ = lim
ã→1−

(
2 −
√

1 − ã
)
γ ≤

(
lim
ã→1−

√
1 − ã

)
α = 0 ,

so γ = 0. Therefore, α = 1, β = 0 = γ. Hence, Q is smooth.



252 s.g. kim

Suppose that a = 1, 1
4
≤ b. Since ‖Q‖ = 1, we have b+ c

2
≤ 3

4
. If b+ c

2
= 3

4
,

then Q is not smooth letting fj ∈P(2H)∗ be such that

f1(x
2) =

1

4
, f1(y

2) = 1 , f1(xy) =
1

2
,

f2(x
2) = 1 , f2(y

2) = 0 = f2(xy) .

Let b + c
2
< 3

4
. Note that if b = 1

4
, then Q = x2 + 1

4
y2 + cxy for 0 ≤ c < 1.

Let f ∈P(2H)∗ be such that f(Q) = 1 = ‖f‖. Then α = 1, β = 0 = γ. Thus
Q is smooth.

Suppose that a = 1, 1
4
< b. Let f ∈P(2H)∗ be such that f(Q) = 1 = ‖f‖.

Then α = 1, β = 0 = γ. Thus Q is smooth.

Case 3: 0 < a < 1.

Suppose that b = 0. We will show that c > a. If not, then ‖Q‖ < 1,
which is a contradiction. Hence, c > a. We claim that Q is smooth. Let
f ∈ P(2H)∗ be such that f(Q) = 1 = ‖f‖. We will show that α = 1

c2
,

β =
4(1−a)
c2

, γ =
2(c−1)
c2

. Note that 1
4
a + 1

2
c < 1, 0 < c < 2. We may choose

δ > 0 such that ‖Ru‖ = ‖Sv‖ = 1 for all u,v ∈ (−δ,δ), where

Ru(x,y) =
(
a + u(2 − 2c−u)

)
x2 + (c + u)xy ,

Sv(x,y) =

(
a +

4(a− 1)v
1 − 4v

)
x2 + vy2 + cxy ∈P(2H) .

Then γ = 2(c− 1)α, β = 4(1 −a)α. It follows that

1 = aα + cγ = c(2 − c)α + c(2c− 2)α = c2α,

proving that α = 1
c2

, β =
4(1−a)
c2

, γ =
2(c−1)
c2

. Thus Q is smooth.
Suppose that b 6= 0. Let c ≤ a. Suppose that c ≤ a ≤ 4b. Notice that if

a = 4b, then ‖Q‖ < 1. Hence, Q is not smooth.
Suppose that a < 4b. Then, 0 < b ≤ 1. If b = 1, then ‖Q‖ > 1, which

is impossible. We claim that if c = a, 0 < b < 1, then Q is smooth. Let
0 < b < 1. By Theorem 2.1, 1 = ‖Q‖ = 3

4
a + b. Therefore,

Q = ax2 +

(
1 −

3

4
a

)
y2 + axy

for 0 < a < 1. Let f ∈ P(2H)∗ be such that f(Q) = 1 = ‖f‖. Then
1 = aα +

(
1 − 3

4
a
)
β + aγ. We will show that α = 1

4
, β = 1, γ = 1

2
. We can



smooth 2-homogeneous polynomials 253

choose δ > 0 such that ‖Ru‖ = ‖Sv‖ = 1 for all u,v ∈ (−δ,δ), where

Ru(x,y) = ax
2 +

(
1 −

3a

4
+ u

)
y2 + (a− 2u)xy ,

Sv(x,y) = (a− 2v)x2 +
(

1 −
3a

4

)
y2 + (a + v)xy ∈P(2H) .

Then β = 2γ, γ = 2α. Therefore, α = 1
4
, β = 1, γ = 1

2
. Thus Q is smooth.

Notice that if 0 = c < a < 4b, then Q is not smooth letting fj ∈ P(2H)∗
be such that

f1(x
2) =

1

4
= f2(x

2) , f1(y
2) = 1 = f2(y

2) ,

f1(xy) =
1

2
, f2(xy) = 0 .

Claim: if 0 < c < a < 4b, then Q is smooth.

By Theorem 2.1, 1 = ‖Q‖ = 1
4
a + b + 1

2
c. Thus 0 < b < 1. Let f ∈P(2H)∗

be such that f(Q) = 1 = ‖f‖. We will show that α = 1
4
,β = 1,γ = 1

2
. We

choose δ > 0 such that ‖Ru,v‖ = 1 for all u,v ∈ (−δ,δ), where

Ru,v(x,y) = (a + u)x
2 + (b + v)y2 +

(
c−

1

2
u− 2v

)
xy ∈P(2H) .

Thus α = 1
4
, β = 1, γ = 1

2
. Therefore, Q is smooth.

Claim: if c ≤ a, 4b < a, then Q is smooth.

Suppose that c = a, 4b < a. By Theorem 2.1, 1 = ‖Q‖ =
∣∣∣14a + b∣∣∣ + 12a.

Notice that 1
4
a + b < 0. Thus

Q = ax2 +

(
1

4
a− 1

)
y2 + axy

for 0 < a < 1. We will show that Q is smooth. Let f ∈ P(2H)∗ be such
that f(Q) = 1 = ‖f‖. We will show that α = −1

4
, β = −1, γ = 1

2
. Choose

0 < δ < 1 such that

0 < a + 2v < a + v < 1 ,
(a + v)2 − 4a

(
1
4
a− 1

)
2(a + v) −a− 4

(
1
4
a− 1

) < 1



254 s.g. kim

for all v ∈ (−δ, 0). Let

Rv = (a + 2v)x
2 +

(
1

4
a− 1

)
y2 + (a + v)xy

for v ∈ (−δ, 0). By Theorem 2.1, 1 = ‖Rv‖. Thus γ ≥ −2α. Choose
0 < δ1 < 1 such that

0 < a + v < 1 ,
a2 − 4(a + v)

(
1
4
a− 1 − 1

4
v
)

2a− (a + v) − 4
(

1
4
a− 1 − 1

4
v
) < 1

for all v ∈ (−δ1, 0). Let

Sv = (a + v)x
2 +

(
1

4
a− 1 −

1

4
v

)
y2 + axy

for v ∈ (−δ1, 0). By Theorem 2.1, 1 = ‖Sv‖. Thus α ≥ 14β. Choose 0 < δ2 < 1
such that

(a + 2v)2 − 4a
(

1
4
a− 1 + v

)
2(a + 2v) −a− 4

(
1
4
a− 1 + v

) < 1
for all v ∈ (0,δ2). Let

Wu = ax
2 +

(
1

4
a− 1 + u

)
y2 + (a + 2u)xy

for u ∈ (0,δ2). By Theorem 2.1, 1 = ‖Wu‖. Thus β ≤ −2γ. Let β = −1 + �
for some 0 ≤ � < 1. By Theorem 2.3, it follows that

1 ≥ sup
0≤c≤1

∣∣∣∣α +
(
c2

4
− 1
)

(−1 + �)
∣∣∣∣ + cγ

= sup
0≤c≤1

−
1

4
(c− 2γ)2 + γ2 −γ +

5

4
+ �

(
1

a
−

5

4
+
c2

4

)
≥ max

{
γ2 −γ +

5

4
+ �

(
1

a
−

5

4
+

(2γ)2

4

)
,

−
1

4
(1 − 2γ)2 + γ2 −γ +

5

4
+ �

(
1

a
− 1
)}

= max

{(
γ −

1

2

)2
+ 1 + �

(
1

a
−

5

4
+ γ2

)
, 1 + �

(
1

a
− 1
)}

≥ 1 + �
(

1

a
− 1
)
≥ 1 ,



smooth 2-homogeneous polynomials 255

which shows that � = 0 = (γ − 1
2
)2. Thus α = −1

4
, β = −1, γ = 1

2
. Hence, Q

is smooth.
Suppose that c < a, 4b < a. Note that −1 ≤ b < 0. If b = −1, then

Q = ax2 −y2. We will show that it is smooth. Let f ∈P(2H)∗ be such that
f(Q) = 1 = ‖f‖. Notice that α = 0, β = −1, γ = 0. Hence, Q is smooth.

Let −1 < b < 0. Then c > 0.

Claim: 1 =
|c2−4ab|

4a
= c

2−4ab
4a

.

First, suppose that 1
4
a ≥ |b|. Then

∣∣1
4
a+b

∣∣+ 1
2
c = 1

4
a+b+ 1

2
c < a < 1. By

Theorem 2.1, 1 = ‖Q‖ = |c
2−4ab|

4a
. Let 1

4
a < |b|. Notice that

∣∣1
4
a + b

∣∣ + 1
2
c <

c2+4a|b|
4a

, so 1 =
|c2−4ab|

4a
= c

2−4ab
4a

. Suppose that 0 < c < 1. We will show that
Q is smooth. Let f ∈ P(2H)∗ be such that f(Q) = 1 = ‖f‖. We will show
that α = − c

2

4a2
, β = −1, γ = c

2a
. We choose δ > 0 such that ‖Rv‖ = ‖Sw‖ = 1

for all v,w ∈ (−δ,δ), where

Rv(x,y) =

(
a−

av

1 + b + v

)
x2 + (b + v)y2 + cxy ,

Sw(x,y) = ax
2 +

(
b +

w(2c + w)

4a

)
y2 + (c + w)xy ∈P(2H) .

Notice that β = a
1+b

α, γ = − c
2a
β. Therefore, α = − c

2

4a2
, β = −1, γ = c

2a
.

Hence, Q is smooth.
Suppose that c = 0. Then Q = ax2 −y2 for 0 < a < 1, which is smooth.
Suppose that c > a.

Claim: if c > a = 4b, then Q is smooth.

Notice that Q = ax2 + a
4
y2 + (2 − a)xy. Let f ∈ P(2H)∗ be such

that f(Q) = 1 = ‖f‖. By the previous arguments, α = 1
4
,β = 1,γ = 1

2
.

Thus Q is smooth.

Claim: if c > a > 4b, c 6= a + 2
√

1 −a, then Q is smooth.

By Theorem 2.1, −1 < b < 1
4
, 0 < c < 2. Notice that∣∣∣∣14a + b

∣∣∣∣ + 12c < 1 and c
2 − 4ab

2c−a− 4b
= 1 ,

or

c2 − 4ab
2c−a− 4b

< 1 and

∣∣∣∣14a + b
∣∣∣∣ + 12c = 1 .



256 s.g. kim

First, suppose that
∣∣1

4
a + b

∣∣ + 1
2
c < 1, c

2−4ab
2c−a−4b = 1. Let f ∈ P(

2H)∗ be such
that f(Q) = 1 = ‖f‖. We will show that α = (c−4b)

2

(2c−a−4b)2 , β =
4(c−a)2

(2c−a−4b)2 ,

γ =
2(c−a)(c−4b)
(2c−a−4b)2 . We may choose δ > 0 such that

0 < 1 − 4b− 4v , 0 < a +
4(a− 1)v

1 − 4b− 4v
< 1 , −1 < b + v <

1

4
,

4(b + v) < a +
4(a− 1)v

1 − 4b− 4v
< c,

∣∣∣∣14
(
a +

4(a− 1)v
1 − 4b− 4v

)
+ b + v

∣∣∣∣ + 12c < 1
for all v ∈ (−δ,δ). Let

Rv(x,y) =

(
a +

4(a− 1)v
1 − 4b− 4v

)
x2 + (b + v)y2 + cxy

for all v ∈ (−δ,δ). By Theorem 2.1,

‖Rv‖ =
c2 − 4

(
a +

4(a−1)v
1−4b−4v

)
(b + v)

2c−
(
a +

4(a−1)v
1−4b−4v

)
− 4(b + v)

= 1

for all v ∈ (−δ,δ). Notice that

β =
4(1 −a)
1 − 4b

α. (4)

We may choose � > 0 such that

−1 < b +
w(2c− 2 + w)

4(a− 1)
<

1

4
, 4

(
b +

w(2c− 2 + w)
4(a− 1)

)
< a < c + w < 2 ,∣∣∣∣14a + b + w(2c− 2 + w)4(a− 1)

∣∣∣∣ + 12 (c + w) < 1
for all w ∈ (−�,�). Let

Sw(x,y) = ax
2 +

(
b +

w(2c− 2 + w)
4(a− 1)

)
y2 + (c + w)xy

for all w ∈ (−�,�). By Theorem 2.1,

‖Sw‖ =
(c + w)2 − 4a

(
b +

w(2c−2+w)
4(a−1)

)
2(c + w) −a− 4

(
b +

w(2c−2+w)
4(a−1)

) = 1



smooth 2-homogeneous polynomials 257

for all w ∈ (−�,�). Notice that γ = (c−1)
2(1−a)β and by (4), γ =

2(c−1)
1−4b α. It follows

that

1 = aα + bβ + cγ = α

(
a +

4b(1 −a)
1 − 4b

+
2c(c− 1)

1 − 4b

)
= α

(
2c−a− 4b

1 − 4b

)
,

which implies that α = 1−4b
2c−a−4b and

1−4b
2c−a−4b =

(c−4b)2
(2c−a−4b)2 . Therefore,

α =
(c− 4b)2

(2c−a− 4b)2
, β =

4(c−a)2

(2c−a− 4b)2
, γ =

2(c−a)(c− 4b)
(2c−a− 4b)2

.

Thus Q is smooth.
Suppose that c

2−4ab
2c−a−4b < 1,

∣∣1
4
a + b

∣∣ + 1
2
c = 1. Note that 1

4
a + b 6= 0. First,

suppose that 1
4
a + b > 0. Let f ∈P(2H)∗ be such that f(Q) = 1 = ‖f‖. We

will show that α = 1
4
, β = 1, γ = 1

2
. We choose δ > 0 such that

Ru(x,y) = (a + u)x
2 +

(
b−

1

4
u

)
y2 + cxy ,

Sv(x,y) = ax
2 +

(
b−

v

2

)
y2 + (c + v)xy ∈P(2H)

for all u,v ∈ (−δ,δ). Notice that γ = 1
2
β, γ = 2α. Thus α = 1

4
, β = 1, γ = 1

2
.

Hence, Q is smooth.
Next, suppose that 1

4
a + b < 0. Let f ∈P(2H)∗ be such that f(Q) = 1 =

‖f‖. By the previous argument, α = −1
4
, β = −1, γ = 1

2
. Thus Q is smooth.

Suppose that c > a > 4b, c = a + 2
√

1 −a. We will show that Q is not

smooth. By Theorem 2.1, 1 = ‖Q‖ ≥
(
a+2
√

1−a
)2
−4ab

2
(
a+2
√

1−a
)
−a−4b

. Thus −1 < b ≤

a+4
√

1−a
4

− 1 < 0, so 1
4
a + b < 0. Since

1 ≥
∣∣∣∣14a + b

∣∣∣∣ + 12c = −
(

1

4
a + b

)
+

1

2
c,

which implies that b ≥ a+4
√

1−a
4

− 1, so b = a+4
√

1−a
4

− 1 and

Q = ax2 +

(
a + 4

√
1 −a

4
− 1
)
y2 +

(
a + 2

√
1 −a

)
xy (0 < a < 1) .

For j = 1, 2, let fj ∈P(2H)∗ be such that

f1
(
x2
)

= −
1

4
, f1

(
y2
)

= −1 , f1(xy) =
1

2
, f2

(
x2
)

=

(
2 −
√

1 −a
)2

4
,

f2
(
y2
)

= 1 −a, f2(xy) =
√

1 −a
(
2 −
√

1 −a
)

2
.



258 s.g. kim

Clearly fj(Q) = 1 = ‖f1‖ for j = 1, 2. We claim that ‖f2‖ = 1. Indeed, for
P = a′x2 + b′y2 + c′xy ∈P(2H), we have

δ(2−√1−a
2

,
√

1−a
)(P) = P(2 −√1 −a

2
,
√

1 −a
)

= a′
(

2 −
√

1 −a
2

)2
+ b′

(√
1 −a

)2
+ c′

(
2 −
√

1 −a
2

)√
1 −a

= f2(P) ,

which implies that f2 = δ
(

2−
√
1−a
2

,
√

1−a
). Thus

‖f2‖ =
∥∥∥∥δ(2−√1−a

2
,
√

1−a
)∥∥∥∥ ≤

∥∥∥∥
(

2 −
√

1 −a
2

,
√

1 −a
)∥∥∥∥

h( 1
2

)

= 1 .

Since f2(Q) = 1, ‖f2‖ = 1. Therefore, Q is not smooth.

Claim: if c > a, a < 4b, then Q is smooth.

By Theorem 2.1, 0 < b < 1, 0 < c < 2. Let f ∈ P(2H)∗ be such
that f(Q) = 1 = ‖f‖. By the previous arguments, α = 1

4
, β = 1, γ = 1

2
. Thus

Q is smooth.
Therefore, we complete the proof.

Acknowledgements

The author is thankful to the referee for the careful reading and
considered suggestions leading to a better presented paper.

References

[1] R.M. Aron, M. Klimek, Supremum norms for quadratic polynomials, Arch.
Math. (Basel) 76 (2001), 73 – 80.

[2] Y.S. Choi, H. Ki, S.G. Kim, Extreme polynomials and multilinear forms
on l1, J. Math. Anal. Appl. 228 (1998), 467 – 482.

[3] Y.S. Choi, S.G. Kim, The unit ball of P(2l22), Arch. Math. (Basel) 71
(1998), 472 – 480.

[4] Y.S. Choi, S.G. Kim, Extreme polynomials on c0, Indian J. Pure Appl.
Math. 29 (1998), 983 – 989.

[5] Y.S. Choi, S.G. Kim, Smooth points of the unit ball of the space P(2l1),
Results Math. 36 (1999), 26 – 33.

[6] Y.S. Choi, S.G. Kim, Exposed points of the unit balls of the spaces P(2l2p)
(p = 1, 2,∞), Indian J. Pure Appl. Math. 35 (2004), 37 – 41.



smooth 2-homogeneous polynomials 259

[7] S. Dineen, “Complex Analysis on Infinite Dimensional Spaces”, Springer-
Verlag, London, 1999.

[8] B.C. Grecu, G.A. Muñoz-Fernández, J.B. Seoane-Sepúlveda,
The unit ball of the complex P(3H), Math. Z. 263 (2009), 775 – 785.

[9] R.B. Holmes, “Geometric Functional Analysis and its Applications”,
Springer-Verlag, New York-Heidelberg, 1975.

[10] S.G. Kim, Exposed 2-homogeneous polynomials on P(2l2p) for 1 ≤ p ≤ ∞),
Math. Proc. R. Ir. Acad. 107 (2007), 123 – 129.

[11] S.G. Kim, The unit ball of P(2D∗(1,W)2), Math. Proc. R. Ir. Acad. 111A (2)
(2011), 79 – 94.

[12] S.G. Kim, Smooth polynomials of P(2d∗(1,w)2), Math. Proc. R. Ir. Acad.
113A (1) (2013), 45 – 58.

[13] S.G. Kim, Polarization and unconditional constants of P(2d∗(1,w)2), Com-
mun. Korean Math. Soc. 29 (2014), 421 – 428.

[14] S.G. Kim, Exposed 2-homogeneous polynomials on the two-dimensional real
predual of Lorentz sequence space, Mediterr. J. Math. 13 (2016), 2827 – 2839.

[15] S.G. Kim, Extreme 2-homogeneous polynomials on the plane with a hexago-
nal norm and applications to the polarization and unconditional constants,
Studia Sci. Math. Hungar. 54 (2017), 362 – 393.

[16] S.G. Kim, Exposed polynomials of P(2R2
h( 1

2
)
), Extracta Math. 33(2) (2018),

127 – 143.
[17] S.G. Kim, S.H. Lee, Exposed 2-homogeneous polynomials on Hilbert spaces,

Proc. Amer. Math. Soc. 131 (2003), 449 – 453.

[18] G.A. Muñoz-Fernández, S.Gy. Révész, J.B. Seoane-Sepúlveda,
Geometry of homogeneous polynomials on non symmetric convex bodies,
Math. Scand. 105 (2009), 147 – 160.

[19] R.A. Ryan, B. Turett, Geometry of spaces of polynomials, J. Math. Anal.
Appl. 221 (1998), 698 – 711.


	Introduction
	Results