E extracta mathematicae Vol. 33, Núm. 2, 127 – 143 (2018)

Exposed Polynomials of P
(
2R2

h(1
2
)

)
Sung Guen Kim ∗

Department of Mathematics, Kyungpook National University

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

Presented by Ricardo Garćıa Received February 2, 2018

Abstract: We show that every extreme polynomials of P
(
2R2

h( 1
2
)

)
is exposed.

Key words: The Krein-Milman Theorem, extreme polynomials, exposed polynomials, the
plane with a hexagonal norm.

AMS Subject Class. (2000): 46A22.

1. Introduction

According to the Krein-Milman Theorem, every nonempty convex set in
a Banach space is fully described by the set of its extreme points. Let n ∈ N.
We write BE for the closed unit ball of a real Banach space E and the dual
space of E is denoted by E∗. We recall that if x ∈ BE is said to be an extreme
point of BE if y, z ∈ BE and x = λy + (1 − λ)z for some 0 < λ < 1 implies
that x = y = z. 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. We denote by extBE and
expBE the sets of extreme and exposed points of BE, respectively. 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
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)|. Note that the spaces L(nE),

∗ 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).

127



128 s. g. kim

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], [42])

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

extBPI(nE) = {±ϕ
n : ϕ ∈ E∗, ∥ϕ∥ = 1},

where LI(nE) and PI(nE) are the spaces of integral n-linear forms and inte-
gral 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].

Let us say about the stories of the classification problems of extBX and
expBX if X = P(nE). Choi et al. ([4], [5]) initiated the classification problems
and classified extBX if X = P

(
2l2p

)
for p = 1, 2, where l2p = R

2 with the
lp-norm. B. Grecu [14] classified extBX if X = P

(
2l2p

)
for 1 < p < 2 or

2 < p < ∞. Kim [18] classified expBX if X = P
(
2l2p

)
for 1 ≤ p ≤ ∞. Kim

et al. [34] showed that every extreme 2-homogeneous polynomials on a real
separable Hilbert space is also exposed. Kim ([20], [26]) characterized extBX
and expBX for X = P

(
2d∗(1, w)

2
)
, where d∗(1, w)

2 = R2 with the octagonal
norm

∥(x, y)∥d∗ = max
{
|x|, |y|, |x|+|y|

1+w
: 0 < w < 1

}
.

He showed [26] that extBP(2d∗(1,w)2) ̸= expBP(2d∗(1,w)2). In [31], Kim classified
extBX and using the classification of extBX, Kim computed the polarization
and unconditional constants of the space X if X = P

(
2R2

h( 1
2
)

)
, where R2

h(w)

denotes the space R2 endowed with the hexagonal norm

∥(x, y)∥h(w) := max{|y|, |x| + (1 − w)|y|}.

We refer to ([1]–[9], [11]–[43]) and references therein for some recent work
about extremal properties of multilinear mappings and homogeneous polyno-
mials on some classical Banach spaces.

We will denote by T((x1, y1), (x2, y2)) = ax1x2 +by1y2 +c(x1y2 +x2y1) and
P(x, y) = ax2 + by2 + cxy a symmetric bilinear form and a 2-homogeneous
polynomial on a real Banach space of dimension 2, respectively. Recently,
Kim [31] classified the extreme points of the unit ball of P

(
2R2

h( 1
2
)

)
as follows:

extB
P
(
2R2

h( 1
2
)

) ={± y2, ±(x2 + 1
4
y2 ± xy

)
, ±

(
x2 + 3

4
y2
)
,

±
[
x2 +

(
c2

4
− 1

)
y2 ± cxy

]
,

±
[
cx2 +

(
c+4

√
1−c

4
− 1

)
y2 ±

(
c + 2

√
1 − c

)
xy

]
(0 ≤ c ≤ 1)

}
.



exposed polynomials of P
(
2R2

h( 1
2
)

)
129

In this paper, we show that that every extreme polynomials of P
(
2R2

h( 1
2
)

)
is

exposed.

2. Results

Theorem 2.1. ([31]) Let P(x, y) = ax2 + by2 + cxy ∈ P
(
2R2

h( 1
2
)

)
with

a ≥ 0, c ≥ 0 and a2 + b2 + c2 ̸= 0. Then:

Case 1 : c < a.
If a ≤ 4b, then

∥P∥ = max
{
a, b,

∣∣1
4
a + b

∣∣ + 1
2
c, 4ab−c

2

4a
, 4ab−c

2

2c+a+4b
, 4ab−c

2

|2c−a−4b|

}
= max

{
a, b,

∣∣1
4
a + b

∣∣ + 1
2
c
}
.

If a > 4b, then ∥P∥ = max
{
a, |b|,

∣∣1
4
a + b

∣∣ + 1
2
c,

|c2−4ab|
4a

}
.

Case 2 : c ≥ a.
If a ≤ 4b, then ∥P∥ = max

{
a, b,

∣∣1
4
a + b

∣∣ + 1
2
c,

|c2−4ab|
2c+a+4b

}
.

If a > 4b, then ∥P∥ = max
{
a, |b|,

∣∣1
4
a + b

∣∣ + 1
2
c, c

2−4ab
2c−a−4b

}
.

Theorem 2.2. ([31])

extB
P
(
2R2

h( 1
2
)

) ={± y2, ±(x2 + 1
4
y2 ± xy

)
, ±

(
x2 + 3

4
y2
)
,

±
[
x2 +

(
c2

4
− 1

)
y2 ± cxy

]
,

±
[
cx2 +

(
c+4

√
1−c

4
− 1

)
y2 ±

(
c + 2

√
1 − c

)
xy

]
(0 ≤ c ≤ 1)

}
.

Theorem 2.3. Let f ∈ P
(
2R2

h( 1
2
)

)∗
with α = f(x2), β = f(y2), γ =

f(xy). Then

∥f∥ = sup
{
|β|,

∣∣α + 1
4
β
∣∣ + |γ|, ∣∣α + 3

4
β
∣∣, ∣∣α + (c2

4
− 1

)
β
∣∣ + c|γ|,∣∣cα + (c+4√1−c

4
− 1

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

Proof. It follows from Theorem 2.2 and the fact that

∥f∥ = sup
{
|f(P)| : P ∈ extB

P
(
2R2

h( 1
2
)

)}.



130 s. g. kim

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

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

Theorem 2.4.

expB
P
(
2R2

h( 1
2
)

) = extB
P
(
2R2

h( 1
2
)

).
Proof. Let (0 ≤ c ≤ 1)

P1(x, y) = y
2 ,

P +2 (x, y) = x
2 + 1

4
y2 + xy ,

P −2 (x, y) = x
2 + 1

4
y2 − xy ,

P3(x, y) = x
2 + 3

4
y2 ,

P +4,c(x, y) = x
2 +

(
c2

4
− 1

)
y2 + cxy ,

P −4,c(x, y) = x
2 +

(
c2

4
− 1

)
y2 − cxy ,

P +5,c(x, y) = cx
2 +

(
c+4

√
1−c

4
− 1

)
y2 + (c + 2

√
1 − c)xy ,

P −5,c(x, y) = cx
2 +

(
c+4

√
1−c

4
− 1

)
y2 − (c + 2

√
1 − c)xy .

Claim 1: P1 = y
2 ∈ expB

P
(
2R2

h( 1
2
)

).
Let f ∈ P

(
2R2

h( 1
2
)

)∗
be such that

α =
1

5
, β = 1 , γ = 0 .

Indeed,

f(P1) = 1 , |f(P ±2 )| =
9

20
, |f(P3)| =

19

20
. (*)

Note that for all 0 ≤ c ≤ 1,

|f(P ±4,c)| =
4

5
−

c2

4
≤

4

5
, (**)

|f(P ±5,c)| = |
√
1 − c +

9c

20
− 1| ≤

11

20
. (***)



exposed polynomials of P
(
2R2

h( 1
2
)

)
131

Hence, by Theorem 2.3, 1 = ∥f∥. We will show that f exposes P1. Let
Q(x, y) = ax2 + by2 + cxy ∈ P

(
2R2

h( 1
2
)

)
such that 1 = ∥Q∥ = f(Q). We will

show that Q = P1. Since P
(
2R2

h( 1
2
)

)
is a finite dimensional Banach space with

dimension 3, by the Krein-Milman Theorem, B
P
(
2R2

h( 1
2
)

) is the closed convex
hull of extB

P
(
2R2

h( 1
2
)

). Then,
Q(x, y) = uP1(x, y) + v

+P +2 (x, y) + v
−P −2 (x, y) + tP3(x, y)

+
∞∑
n=1

λ+n P
+

4,c+n
(x, y) +

∞∑
n=1

λ−n P
−
4,c−n

(x, y)

+

∞∑
m=1

δ+mP
+

5,a+m
(x, y) +

∞∑
m=1

δ−mP
−
5,a−m

(x, y) ,

for some u, v±, t, λ±n , δ
±
m, ∈ R (n, m ∈ N) with 0 ≤ c±n , a±m ≤ 1 and

|u| + |v+| + |v−| + |t| +
∞∑
n=1

|λ+n | +
∞∑
n=1

|λ−n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| = 1 .

We will show that v± = t = λ±n = δ
±
m = 0 for every n, m ∈ N.

Subclaim: v± = t = 0.
Assume that v+ ̸= 0. It follows that

1 = f(Q) = uf(P1) + v
+f(P +2 ) + v

−f(P −2 ) + tf(P3) +

∞∑
n=1

λ+n f(P
+
4,cn

)

+

∞∑
n=1

λ−n f(P
−
4,cn

) +

∞∑
m=1

δ+mf(P
+
5,am

) +

∞∑
m=1

δ−mf(P
−
5,am

)

≤ |u| + |v+||f(P +2 )| + |v
−||f(P −2 )| + |t||f(P3)| +

∞∑
n=1

|λ+n ||f(P
+
4,cn

)|

+
∞∑
n=1

|λ−n ||f(P
−
4,cn

)| +
∞∑

m=1

|δ+m||f(P
+
5,am

)| +
∞∑

m=1

|δ−m||f(P
−
5,am

)|

≤ |u| +
9

20
|v+| +

9

20
|v−| +

19

20
|t| +

4

5

∞∑
n=1

|λ+n |

+
4

5

∞∑
n=1

|λ−n | +
11

20

∞∑
m=1

|δ+m| +
11

20

∞∑
m=1

|δ−m| (by (*), (**), (***))



132 s. g. kim

< |u| + |v+| +
9

20
|v−| +

19

20
|t| +

4

5

∞∑
n=1

|λ+n |

+
4

5

∞∑
n=1

|λ−n | +
11

20

∞∑
m=1

|δ+m| +
11

20

∞∑
m=1

|δ−m|

≤ |u| + |v+| + |v−| + |t| +
∞∑
n=1

|λ+n | +
∞∑
n=1

|λ−n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| = 1 ,

which is impossible. Therefore, v+ = 0. Using a similar argument as above,
we have v− = t = 0.

Subclaim: λ±n = δ
±
m = 0 for every n, m ∈ N.

Assume that λ+n0 ̸= 0 for some n0 ∈ N. It follows that

1 = f(Q) = uf(P1) + λ
+
n0
f(P +4,cn0

) +
∑

n∈N,n ̸=n0

λ+n f(P
+

4,c+n
)

+
∞∑
n=1

λ−n f(P
−
4,c−n

) +
∞∑

m=1

δ+mf(P
+

5,a+m
) +

∞∑
m=1

δ−mf(P
−
5,a−m

)

≤ |u| + |λ+n0||f(P
+

4,c+n0
)| +

∑
n∈N,n ̸=n0

|λ+n ||f(P
+

4,c+n
)| +

∞∑
n=1

|λ−n ||f(P
−
4,c−n

)|

+
∞∑

m=1

|δ+m||f(P
+

5,a+m
)| +

∞∑
m=1

|δ−m||f(P
−
5,a−m

)|

< |u| + |λ+n0| +
4

5

∑
n∈N,n̸=n0

|λ+n | +
4

5

∞∑
n=1

|λ−n | +
11

20

∞∑
m=1

|δ+m| +
11

20

∞∑
m=1

|δ−m|

≤ |u| +
∞∑
n=1

|λ+n | +
∞∑
n=1

|λ−n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| = 1 ,

which is impossible. Therefore, λ+n = 0 for every n ∈ N. Using a similar
argument as above, we have λ−n = δ

±
m = 0 for every n, m ∈ N. Therefore,

Q(x, y) = uP1(x, y). Hence u = 1, so Q = P1. Therefore, f exposes P1.
Claim 2: P5,0 = 2xy ∈ expBP

(
2R2

h( 1
2
)

).
Let f ∈ P

(
2R2

h( 1
2
)

)∗
be such that

α = β = 0 , γ =
1

2
.



exposed polynomials of P
(
2R2

h( 1
2
)

)
133

We will show that f exposes P5,0. Indeed, f(P5,0) = 1, f(P1) = 0, f(P
±
2 ) =

±1
2
, f(P3) = 0,

−
1

2
≤ f(P ±4,c) = ±

c

2
≤

1

2
(0 ≤ c ≤ 1) .

Note that, for 0 < c ≤ 1,

−1 < f(P ±5,c) = ±
c + 2

√
1 − c

2
< 1 . (†)

Hence, by Theorem 2.3, 1 = ∥f∥. Let

Q(x, y) = uP1(x, y) + v
+P +2 (x, y) + v

−P −2 (x, y) + tP3(x, y)

+

∞∑
n=1

λ+n P
+

4,c+n
(x, y) +

∞∑
n=1

λ−n P
−
4,c−n

(x, y)

+
∞∑

m=1

δ+mP
+

5,a+m
(x, y) +

∞∑
m=1

δ−mP
−
5,a−m

(x, y) ,

for some u, v±, t, λ±n , δ
±
m, ∈ R (n, m ∈ N) with 0 ≤ c±n , a±m ≤ 1 and

|u| + |v+| + |v−| + |t| +
∞∑
n=1

|λ+n | +
∞∑
n=1

|λ−n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| = 1 .

We will show that v± = t = λ±n = δ
±
m = 0 for every n, m ∈ N.

Subclaim: v+ = 0.
Assume that v+ ̸= 0. It follows that

1 = f(Q) = v+f(P +2 ) + v
−f(P −2 ) +

∞∑
n=1

λ+n f(P
+

4,c+n
)

+

∞∑
n=1

λ−n f(P
−
4,c−n

) +

∞∑
m=1

δ+mf(P
+

5,a+m
) +

∞∑
m=1

δ−mf(P
−
5,a−m

)

< |v+| +
1

2
|v−| +

∞∑
n=1

|λ+n ||f(P
+

4,c+n
)| +

∞∑
n=1

|λ−n ||f(P
−
4,c−n

)|

+
∞∑

m=1

|δ+m||f(P
+

5,a+m
)| +

∞∑
m=1

|δ−m||f(P
−
5,a−m

)|

≤ |v+| + |v−| +
∞∑
n=1

|λ+n | +
∞∑
n=1

|λ−n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| ≤ 1 ,



134 s. g. kim

which is impossible. Therefore, v+ = 0. Using a similar argument as
Claim 1, we have v− = λ±n = 0 for every n ∈ N. Hence,

Q(x, y) = uP1(x, y) + tP3(x, y) +

∞∑
m=1

δ+mP
+

5,a+m
(x, y) +

∞∑
m=1

δ−mP
−
5,a−m

(x, y) .

It follows that

1 = f(Q) =
∞∑

m=1

δ+mf(P
+

5,a+m
) +

∞∑
m=1

δ−mf(P
−
5,a−m

)

≤
∞∑

m=1

|δ+m||f(P
+

5,a+m
)| +

∞∑
m=1

|δ−m||f(P
−
5,a−m

)|

≤
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| ≤ 1 ,

which shows that

f(P +
5,a+m

) = f(P −
5,a−m

) =

∞∑
m=1

|δ+m| +
∞∑

m=1

|δ−m| = 1 , u = t = 0 for all m ∈ N .

By (†), P ±
5,a±m

= P5,0 for every m ∈ N and
∑∞

m=1 δ
+
m+

∑∞
m=1 δ

−
m = 1. Therefore,

Q = P5,0. Hence, f exposes P5,0.
Claim 3: P +2 = x

2 + 1
4
y2 + xy ∈ expB

P
(
2R2

h( 1
2
)

).
Let f ∈ P

(
2R2

h( 1
2
)

)∗
be such that

α =
1

2
= β , γ =

3

8
.

We will show that f exposes P2. Indeed, f(P
+
2 ) = 1, f(P

−
2 ) =

1
4
, f(P1) =

1
2
,

f(P ±3 ) =
7
8
. By some calculation, we have

|f(P ±4,c)| ≤
1

2
, |f(P ±5,c)| ≤

57

64
for 0 ≤ c ≤ 1 .

Hence, by Theorem 2.3, 1 = ∥f∥. By similar arguments as Claims 1 and 2, f
exposes P +2 . Obviously, P

−
2 ∈ expBP

(
2R2

h( 1
2
)

).
Claim 4: P +4,0 = x

2 − y2 ∈ expB
P
(
2R2

h( 1
2
)

).



exposed polynomials of P
(
2R2

h( 1
2
)

)
135

Let f ∈ P
(
2R2

h( 1
2
)

)∗
be such that

α =
1

2
= −β , γ = 0 .

We will show that f exposes P4,0. Indeed,

f(P +4,0) = 1 , |f(P1)| =
1

2
, |f(P ±2 )| =

3

8
, |f(P3)| =

1

8
.

Note that

|f(P ±4,c)| = 1 −
c2

8
< 1 for 0 < c ≤ 1 .

Note that, for 0 ≤ c ≤ 1,

|f(P ±5,c)| =
3c + 4 − 4

√
1 − c

8
≤

7

8
.

Hence, by Theorem 2.3, 1 = ∥f∥. By similar arguments as Claims 1 and 2, f
exposes P +4,0.

Claim 5: P3 = x
2 + 3

4
y2 ∈ expB

P
(
2R2

h( 1
2
)

).
Let f ∈ P

(
2R2

h( 1
2
)

)∗
be such that

α =
5

8
, β =

1

2
, γ = 0 .

We will show that f exposes P3. Indeed,

f(P3) = 1 , |f(P1)| =
1

2
, |f(P ±2 )| =

3

4
.

Note that

|f(P ±4,c)| ≤
1

4
, |f(P ±5,c)| ≤

1

3
for 0 ≤ c ≤ 1 .

Hence, by Theorem 2.3, 1 = ∥f∥. By similar arguments as Claims 1 and 2, f
exposes P3.

Claim 6: P +5,1 = x
2 − 3

4
y2 + xy ∈ expB

P
(
2R2

h( 1
2
)

).
Let f ∈ P

(
2R2

h( 1
2
)

)∗
be such that

α =
11

16
, β = −

1

4
, γ =

1

8
.



136 s. g. kim

We will show that f exposes P +5,1. Indeed,

f(P +5,1) = 1 , |f(P1)| =
1

4
, |f(P ±2 )| ≤

3

4
, |f(P3)| =

1

2
.

Note that

3

4
≤ f(P ±4,c) < 1 , −

1

4
≤ f(P ±5,c) < 1 for 0 ≤ c < 1 .

Hence, by Theorem 2.3, 1 = ∥f∥. By similar arguments as Claims 1 and 2, f
exposes P +5,1. Obviously, P

−
5,1 ∈ expBP

(
2R2

h( 1
2
)

).
Claim 7: P +4,c = x

2 + (c
2

4
− 1)y2 + cxy ∈ expB

P
(
2R2

h( 1
2
)

) for 0 < c < 1.
Let f ∈ P

(
2R2

h( 1
2
)

)∗
be such that

α =
3

4
−

c2

16
, β = −

1

4
, γ =

c

8
.

Indeed,

f(P +4,c) = 1 ,
3

4
≤ f(P −4,c) = 1 −

c2

4
< 1 , |f(P1)| =

1

4
,

1

2
≤ f(P ±2 ) ≤

3

4
,

1

2
≤ f(P3) <

9

16
.

(*)

Note that for every t ∈ [0, 1] with t ̸= c,

f(P +4,t) = −
1

16
t2 +

c

8
t +

(
1 −

c2

16

)
and

f(P −4,t) = −
1

16
t2 −

c

8
t +

(
1 −

c2

16

)
.

Hence, we have, for every t ∈ [0, 1] with t ̸= c,

1 < min

{
1 −

c2

16
, 1 −

(1 − c)2

16

}
≤ f(P +4,t) < 1 (**)

and

−1 < 1 −
(1 + c)2

16
≤ f(P −4,t) ≤ 1 −

c2

16
< 1 .



exposed polynomials of P
(
2R2

h( 1
2
)

)
137

Note that, for every t ∈ [0, 1],

f(P +5,t) =

(
−c2 + 2c + 11

16

)
t +

(
c − 1
4

)√
1 − t +

1

4

and

f(P −5,t) =

(
−c2 − 2c + 11

16

)
t +

(
c + 1

4

)√
1 − t +

1

4
.

Hence, we have that, for every t ∈ [0, 1],

−1 <
c

4
≤ f(P +5,t) ≤

−c2 + 2c + 15
16

< 1 (***)

and

−1 <
c + 2

4
≤ f(P −5,t) ≤

−c2 − 2c + 15
16

< 1 .

Hence, by Theorem 2.3, 1 = ∥f∥. We will show that f exposes P +4,c. Let
Q(x, y) = ax2 + by2 + cxy ∈ P

(
2R2

h( 1
2
)

)
such that 1 = ∥Q∥ = f(Q). We will

show that Q = P +4,c. By the Krein-Milman Theorem,

Q(x, y) = uP1(x, y) + v
+P +2 (x, y) + v

−P −2 (x, y) + tP3(x, y)

+

∞∑
n=1

λ+n P
+

4,c+n
(x, y) +

∞∑
n=1

λ−n P
−
4,c−n

(x, y)

+
∞∑

m=1

δ+mP
+

5,a+m
(x, y) +

∞∑
m=1

δ−mP
−
5,a−m

(x, y) ,

for some u, v±, t, λ±n , δ
±
m, ∈ R (n, m ∈ N) with 0 ≤ c±n , a±m ≤ 1 and

|u| + |v+| + |v−| + |t| +
∞∑
n=1

|λ+n | +
∞∑
n=1

|λ−n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| = 1 .

We will show that u = v± = t = λ−n = δ
±
m = 0 for every n, m ∈ N. Assume



138 s. g. kim

that δ+m0 ̸= 0 for some m0 ∈ N. It follows that

1 = f(Q) = uf(P1) + v
+f(P +2 ) + v

−f(P −2 ) + tf(P3) +

∞∑
n=1

λ+n f(P
+

4,c+n
)

+

∞∑
n=1

λ−n f(P
−
4,c−n

) +

∞∑
m=1

δ+mf(P
+

5,a+m
) +

∞∑
m=1

δ−mf(P
−
5,a−m

)

<
1

4
|u| +

3

4
|v+| +

3

4
|v−| +

9

16
|t| +

∞∑
n=1

|λ+n |

+

∞∑
n=1

|λ−n | + |δ
+
m0

| +
∑

m̸=m0

|δ+m| +
∞∑

m=1

|δ−m| (by (*), (**), (***)) ≤ 1 ,

which is impossible. Therefore, δ+m = 0 for every m ∈ N. Using a similar
argument as above, we have u = v± = t = λ−n = 0. Therefore,

Q(x, y) =
∞∑
n=1

λ+n P
+

4,c+n
(x, y) .

We will show that if c+n0 ̸= c for some n0 ∈ N, then λ
+
n0

= 0. Assume that
λ+n0 ̸= 0. It follows that

1 = f(Q) = λ+n0f(P
+

4,c+n0
) +

∑
n̸=n0

λ+n f(P
+

4,c+n
)

< |λ+n0| +
∑
n ̸=n0

|λ+n | = 1 ,

which is impossible. Therefore, λ+n = 0 for every n ∈ N. Therefore,

Q(x, y) =

( ∑
c+n =c

λ+n

)
P +4,c(x, y) = P

+
4,c(x, y) .

Therefore, f exposes P +4,c. Obviously, P
−
4,c ∈ expBP

(
2R2

h( 1
2
)

) for 0 < c ≤ 1.
Claim 8: P +5,c = cx

2 +

(
c+4

√
1−c

4
−1

)
y2 +(c+2

√
1 − c)xy ∈ expB

P
(
2R2

h( 1
2
)

)
for 0 < c < 1.



exposed polynomials of P
(
2R2

h( 1
2
)

)
139

Let f ∈ P
(
2R2

h( 1
2
)

)∗
be such that

α =
1

2

(
1 −

c + 4
√
1 − c

4

)
, β = −

c

2
, γ =

c + 2
√
1 − c

4
.

Note that

0 ≤ α <
3

8
, −

1

2
< β ≤ 0 ,

1

4
< γ ≤

1

2
.

We will show that f exposes P +5,c. Indeed,

f(P +5,c) = 1 , |f(P1)| <
1

2
, 0 < f(P +2 ) <

1

2
,

− 1 < f(P −2 ) < −
1

8
, −

1

8
≤ f(P3) < 0 .

(*)

Note that for every t ∈ [0, 1],

f(P +4,t) = −
c

8
t2 +

(
c + 2

√
1 − c

4

)
t +

1

2
+

3c

8
−

√
1 − c
2

and

f(P −4,t) = −
c

8
t2 −

(
c + 2

√
1 − c

4

)
t +

1

2
+

3c

8
−

√
1 − c
2

.

Hence, we have for every t ∈ [0, 1],

− 1 <
1

2
+

3c

8
−

√
1 − c
2

≤ f(P +4,t) ≤
c + 1

2
< 1 , (**)

− 1 <
1

2
−
√
1 − c ≤ f(P −4,t) ≤

1

2
+

3c

8
−

√
1 − c
2

< 1 .

Note that for every t ∈ [0, 1] with t ̸= c,

f(P +5,t) =
1

2
t +

√
1 − c

√
1 − t +

c

2

and

f(P −5,t) =

(
1 − c −

√
1 − c

2

)
t − (c +

√
1 − c)

√
1 − t +

c

2
.

Hence, we have for every t ∈ [0, 1] with t ̸= c,

− 1 < min
{
c

2
+
√
1 − c,

c + 1

2

}
≤ f(P +5,t) < 1 , (***)

− 1 < −
(
c

2
+

√
1 − c

)
≤ f(P −5,t) ≤

1

2
−
√
1 − c < 1 .



140 s. g. kim

Hence, by Theorem 2.3, 1 = ∥f∥. Let Q(x, y) = ax2 + by2 + cxy in
P
(
2R2

h( 1
2
)

)
such that 1 = ∥Q∥ = f(Q). By the Krein-Milman Theorem,

Q(x, y) = uP1(x, y) + v
+P +2 (x, y) + v

−P −2 (x, y) + tP3(x, y)

+

∞∑
n=1

λ+n P
+

4,c+n
(x, y) +

∞∑
n=1

λ−n P
−
4,c−n

(x, y)

+
∞∑

m=1

δ+mP
+

5,a+m
(x, y) +

∞∑
m=1

δ−mP
−
5,a−m

(x, y) ,

for some u, v±, t, λ±n , δ
±
m, ∈ R (n, m ∈ N) with 0 ≤ c±n , a±m ≤ 1 and

|u| + |v+| + |v−| + |t| +
∞∑
n=1

|λ+n | +
∞∑
n=1

|λ−n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m| = 1 .

We will show that u = v± = t = λ±n = δ
−
m = 0 for every n, m ∈ N. Assume

that λn0 ̸= 0 for some n0 ∈ N. It follows that

1 = f(Q) = uf(P1) + v
+f(P +2 ) + v

−f(P −2 ) + tf(P3) +
∞∑
n=1

λ+n f(P
+

4,c+n
)

+

∞∑
n=1

λ−n f(P
−
4,c−n

) +

∞∑
m=1

δ+mf(P
+

5,a+m
) +

∞∑
m=1

δ−mf(P
−
5,a−m

)

<
1

2
|u| +

1

2
|v+| +

1

2
|v−| +

1

2
|t| + |λ+n0| +

∑
n̸=n0

|λ+n | +
∞∑

m=1

|δ+m| +
∞∑

m=1

|δ−m|

≤ 1 (by (*), (**), (***)) ,

which is impossible. Therefore, λ+n = 0 for every n ∈ N. Using a similar
argument as above, we have u = v± = t = λ−n = δ

−
m = 0 for every n, m ∈ N.

Therefore,

Q(x, y) =

∞∑
m=1

δ+mP
+

5,a+m
(x, y) .

We will show that if a+m0 ̸= c for some m0 ∈ N, then δ
+
m0

= 0. Assume that



exposed polynomials of P
(
2R2

h( 1
2
)

)
141

δ+m0 ̸= 0. It follows that

1 = f(Q) = δ+m0f(P
+

5,a+m0
) +

∑
m ̸=m0

δ+mf(P
+
5,am

)

< |δ+m0| +
∑

m ̸=m0

|δ+m| = 1

which is impossible. Therefore, δ+m0 = 0. Therefore,

Q(x, y) =

( ∑
am=a

δ+m

)
P +5,c(x, y) = P

+
5,c(x, y) .

Therefore, f exposes P +5,c. Obviously, P
−
5,c ∈ expBP

(
2R2

h( 1
2
)

) for 0 < c < 1.
Therefore, we complete the proof.

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