Equivalent norms in polynomial spaces.Infinite Analysis Seminar, Celebrating Richard Aron’s work and impact. October, 28th, 2016 Pablo Jim enez Rodr guez Equivalent norms The beginning:

The beginning: the Bohnenblust-Hille inequality.Preparing the ground

Values for k2,q,p and K2,q,p .

Equivalent norms in polynomial spaces.

Pablo Jimenez Rodrıguez

Infinite Analysis Seminar,Celebrating Richard Aron’s work and impact.

October, 28th, 2016

Pablo Jimenez Rodrıguez Equivalent norms



Introductory notation

Definition

α = (α1, α2, . . . , αn) = (N ∪ {0})n is a multiindex.

|α| =∑n

i=1 α, xα = xα11 xα2 · . . . · xαn

n .

P(x) =∑|α|=m aαxα is a homogenous polynomial of

degree m.

P(mKn) ={P : Kn → K :

P is a homogeneous polynomial of degree m over Kn}

If | · | is a norm in Kn and P ∈ P(mKn), we may define‖P‖ = sup{|P(x)| : x ∈ B(Kn,|·|)}





Definition


|α| =∑n

i=1 α, xα = xα11 xα2 · . . . · xαn

n .


degree m.

P(mKn) ={P : Kn → K :







Definition


|α| =∑n

i=1 α, xα = xα11 xα2 · . . . · xαn

n .


degree m.

P(mKn) ={P : Kn → K :







Definition


|α| =∑n

i=1 α, xα = xα11 xα2 · . . . · xαn

n .


degree m.

P(mKn) ={P : Kn → K :







Definition


|α| =∑n

i=1 α, xα = xα11 xα2 · . . . · xαn

n .


degree m.

P(mKn) ={P : Kn → K :







We will be interested in the case where the norm in Kn is chosento be the q−th norm (1 ≤ q ≤ ∞).

Denote ‖P‖q = sup{|P(x)| : ‖x‖q ≤ 1}.Define the p−th norm of P(x) =

∑|α|=m aαxα as

|P|p =

(∑

|α|=m |aα|p)1/p

if 1 ≤ p <∞,max{|aα| : |α| = m} if p =∞.

There exist constants k ,K > 0 so that k‖P‖p ≤ |P|q ≤ K‖P‖p forevery 1 ≤ p, q ≤ ∞.





We will be interested in the case where the norm in Kn is chosento be the q−th norm (1 ≤ q ≤ ∞).Denote ‖P‖q = sup{|P(x)| : ‖x‖q ≤ 1}.

Define the p−th norm of P(x) =∑|α|=m aαxα as

|P|p =

(∑

|α|=m |aα|p)1/p

if 1 ≤ p <∞,max{|aα| : |α| = m} if p =∞.






We will be interested in the case where the norm in Kn is chosento be the q−th norm (1 ≤ q ≤ ∞).Denote ‖P‖q = sup{|P(x)| : ‖x‖q ≤ 1}.Define the p−th norm of P(x) =

∑|α|=m aαxα as

|P|p =

(∑

|α|=m |aα|p)1/p

if 1 ≤ p <∞,max{|aα| : |α| = m} if p =∞.






We will be interested in the case where the norm in Kn is chosento be the q−th norm (1 ≤ q ≤ ∞).Denote ‖P‖q = sup{|P(x)| : ‖x‖q ≤ 1}.Define the p−th norm of P(x) =

∑|α|=m aαxα as

|P|p =

(∑

|α|=m |aα|p)1/p

if 1 ≤ p <∞,max{|aα| : |α| = m} if p =∞.





Theorem (Bohnenblust, Hille, 1931)

There exists a constant Dn,m ≥ 1 such that for every P ∈ P(m`n∞)we have

|P| 2mm+1≤ Dn,m‖P‖∞.

Dn,m can be chosen in a way that it is independent on n.

The value for p = 2mm+1 is optimal

Theorem






|P| 2mm+1≤ Dn,m‖P‖∞.



Theorem






|P| 2mm+1≤ Dn,m‖P‖∞.



Theorem






|P| 2mm+1≤ Dn,m‖P‖∞.


The value for p = 2mm+1 is optimal.

Theorem






|P| 2mm+1≤ Dn,m‖P‖∞.



Theorem






|P| 2mm+1≤ DR,m‖P‖∞.



Theorem






|P| 2mm+1≤ DR,m‖P‖∞.



Theorem

The real Bohnenblust-Hille constant is hypercontractive.

lim supm D1/mR,m = 2






|P| 2mm+1≤ DR,m‖P‖∞.



Theorem

The real Bohnenblust-Hille constant is hypercontractive.

lim supm D1/mR,m = 2




The Hardy-Littlewood constant.

There are constants Hm,n,p and Lm,n,p so that

|P| pp−m≤ Hm,n,p‖P‖p for m < p ≤ 2m,

|P| 2mpmp+p−2m

≤ Lm,n,p‖P‖p for 2m ≤ p ≤ ∞.

The choice of Hm,n,p and Lm,n,p can be made independent of n.The exponents p

m−p for m < p ≤ 2m and 2mpmp+p−2m for

2m ≤ p ≤ ∞ are optimal.

Definition

The smallest constant we can fit in the inequalities above are theHardy-Littlewood constants.

DR,m = Lm,∞







|P| 2mpmp+p−2m

≤ Lm,n,p‖P‖p for 2m ≤ p ≤ ∞.




Definition


DR,m = Lm,∞







|P| 2mpmp+p−2m

≤ Lm,n,p‖P‖p for 2m ≤ p ≤ ∞.




Definition


DR,m = Lm,∞






|P| pp−m≤ Hm,p‖P‖p for m < p ≤ 2m,

|P| 2mpmp+p−2m

≤ Lm,p‖P‖p for 2m ≤ p ≤ ∞.




Definition


DR,m = Lm,∞







|P| 2mpmp+p−2m

≤ Lm,p‖P‖p for 2m ≤ p ≤ ∞.




Definition


DR,m = Lm,∞







|P| 2mpmp+p−2m

≤ Lm,p‖P‖p for 2m ≤ p ≤ ∞.




Definition


DR,m = Lm,∞







|P| 2mpmp+p−2m

≤ Lm,p‖P‖p for 2m ≤ p ≤ ∞.




Definition


DR,m = Lm,∞




If we work within polynomials of a fixed degree, we define

DR,m(n) := max{ |P| 2m

m+1

‖P‖∞

},

Hm,p(n) := max{ |P| p

p−m

‖P‖p

}, if m ≤ p ≤ 2m,

Lm,p(n) := max{ |P| 2mp

mp+p−2m

‖P‖p

}, if 2m ≤ p ≤ ∞,

for every P ∈ P(mRn).






m+1

‖P‖∞

},


p−m

‖P‖p

}, if m ≤ p ≤ 2m,


mp+p−2m

‖P‖p

}, if 2m ≤ p ≤ ∞,







m+1

‖P‖∞

},


p−m

‖P‖p

}, if m ≤ p ≤ 2m,


mp+p−2m

‖P‖p

}, if 2m ≤ p ≤ ∞,







m+1

‖P‖∞

},


p−m

‖P‖p

}, if m ≤ p ≤ 2m,


mp+p−2m

‖P‖p

}, if 2m ≤ p ≤ ∞,







m+1

‖P‖∞

},


p−m

‖P‖p

}, if m ≤ p ≤ 2m,


mp+p−2m

‖P‖p

}, if 2m ≤ p ≤ ∞,





Generalizing those ideas.

Definition

Let 1 ≤ p, q ≤ ∞. We define

k ′m,n,p,q := max{‖P‖p : P ∈ B(P(mRn),|·|q)

},

Km,n,q,p = max{|P|q : P ∈ B(P(mRn),‖·‖p)

}

Hm,p(n) = Km,n, pp−m

,p, if m < p ≤ 2m,

Lm,p(n) = Km,n, 2mpmp+p−2m

,p if 2m ≤ p ≤ ∞.Hm,p ≥ supn Km,n, pp−m

,p for m < p ≤ 2m,

Lm,p ≥ supn Km,n, 2mpmp+p−2m

,p for 2m ≤ p ≤ ∞.





Definition



},


}Hm,p(n) = Km,n, p

p−m,p, if m < p ≤ 2m,



,p for m < p ≤ 2m,


,p for 2m ≤ p ≤ ∞.





Definition



},


}Hm,p(n) = Km,n, p

p−m,p, if m < p ≤ 2m,


,p if 2m ≤ p ≤ ∞.

Hm,p ≥ supn Km,n, pp−m

,p for m < p ≤ 2m,


,p for 2m ≤ p ≤ ∞.





Definition



},


}Hm,p(n) = Km,n, p

p−m,p, if m < p ≤ 2m,



,p for m < p ≤ 2m,


,p for 2m ≤ p ≤ ∞.





Definition


k ′m,n,q,p := max{‖P‖p : P ∈ B(P(mRn),|·|q)

},


}

We will be working on R2.





Definition


k ′m,n,q,p := max{‖P‖p : P ∈ B(P(mRn),|·|q)

},


}






Definition


k ′2,q,p := max{‖P‖p : P ∈ B(P(mRn),|·|q)

},

K2,q,p = max{|P|q : P ∈ B(P(mRn),‖·‖p)

}


In the present talk, I will give someexact values and bounds for k ′2,p,q, K2,q,p for certain values of p

and q. Remark that, if k2,q,p = 1k ′2,q,p

, then k2,p,q and K2,p,q are the

best possible constants for which the inequalities

k2,p,q‖P‖p ≤ |P|q ≤ K2,q,p‖P‖phold for every P ∈ P(mR2).





Definition



},


}

We will be working on R2. In the present talk, I will give someexact values and bounds for k ′2,p,q, K2,q,p for certain values of p

and q.

Remark that, if k2,q,p = 1k ′2,q,p








Definition



},


}

We will be working on R2. In the present talk, I will give someexact values and bounds for k ′2,p,q, K2,q,p for certain values of p

and q. Remark that, if k2,q,p = 1k ′2,q,p










We will make use of a corollary of the Krein-Milmann theorem,according to which the maximum of a continuous convex functionover a convex set is attained over the set of extreme points.

ext(B|·|q) =

{±ek : 1 ≤ k ≤ m + 1 if q = 1,

{∑m+1

k=1 εkek : εk = ±1 if q =∞,S|·|q if 1 < q <∞.}




We will make use of a corollary of the Krein-Milmann theorem,according to which the maximum of a continuous convex functionover a convex set is attained over the set of extreme points.

ext(B|·|q) =

{±ek : 1 ≤ k ≤ m + 1 if q = 1,

{∑m+1

k=1 εkek : εk = ±1 if q =∞,S|·|q if 1 < q <∞.}




Theorem (Richard Aron and M. Klimek)

Let us denote ‖(a, b, c)‖R := supx∈[−1,1] |ax2 + bx + c|. Then,

‖(a, b, c)‖R =

∣∣∣b24a − c

∣∣∣ if |b| < 2|a| and

ca + 1 < 1

2(| b2a | − 1)2,

|a + c |+ |b| otherwise.

If α < β are real numbers and‖(a, b, c)‖[α,β] := supx∈[α,β] |ax2 + bx + c |, then

‖(a, b, c)‖[α,β] =

∥∥∥∥∥((

α− β2

)2

a,α2 − β2

2a +

α− β2

b,

(α + β

2

)2

a +α + β

2b + c

)∥∥∥∥∥R

.




Theorem (Richard Aron and M. Klimek)

Let us denote ‖(a, b, c)‖R := supx∈[−1,1] |ax2 + bx + c|. Then,

‖(a, b, c)‖R =

∣∣∣b24a − c

∣∣∣ if |b| < 2|a| and

ca + 1 < 1

2(| b2a | − 1)2,

|a + c |+ |b| otherwise.

If α < β are real numbers and‖(a, b, c)‖[α,β] := supx∈[α,β] |ax2 + bx + c |, then

‖(a, b, c)‖[α,β] =

∥∥∥∥∥((

α− β2

)2

a,α2 − β2

2a +

α− β2

b,

(α + β

2

)2

a +α + β

2b + c

)∥∥∥∥∥R

.




Theorem

‖ax2 + by2 + cxy‖1 = max

{∥∥∥∥(a + b + c

4,a− b

2,a + b − c

4

)∥∥∥∥R,∥∥∥∥(a + b − c

4,b − a

2,a + b + c

4

)∥∥∥∥R

}

‖ax2 + by2 + cxy‖∞ = max{‖(a, c , b)‖R, ‖(b, c, a)‖R.




Theorem

‖ax2 + by2 + cxy‖1 = max

{∥∥∥∥(a + b + c

4,a− b

2,a + b − c

4

)∥∥∥∥R,∥∥∥∥(a + b − c

4,b − a

2,a + b + c

4

)∥∥∥∥R

}

‖ax2 + by2 + cxy‖∞ = max{‖(a, c , b)‖R, ‖(b, c, a)‖R.




Theorem (Y.S. Choi, S.G. Kim, H. Ki)

The extreme polynomials of B(P(2R2),‖·‖1) are of the form

P(x , y) = ±x2 ± y2 ± 2xy ,

P(x , y) = ±√

4|t|−t22 (x2 − y2) + txy , where |t| ∈ (2, 4].

Theorem (Y.S. Choi, S.G. Kim)

The extreme polynomials of B(P(2R2),‖·‖∞) are of the form

P(x , y) = ±x2,

P(x , y) = ±y2,

±(tx2 − ty2 ± 2

√t(1− t)xy

), where t ∈ [12 , 1].






P(x , y) = ±x2 ± y2 ± 2xy ,

P(x , y) = ±√

4|t|−t22 (x2 − y2) + txy , where |t| ∈ (2, 4].



P(x , y) = ±x2,

P(x , y) = ±y2,

±(tx2 − ty2 ± 2

√t(1− t)xy

), where t ∈ [12 , 1].






P(x , y) = ±x2 ± y2 ± 2xy ,

P(x , y) = ±√

4|t|−t22 (x2 − y2) + txy , where |t| ∈ (2, 4].



P(x , y) = ±x2,

P(x , y) = ±y2,

±(tx2 − ty2 ± 2

√t(1− t)xy

), where t ∈ [12 , 1].






P(x , y) = ±x2 ± y2 ± 2xy ,

P(x , y) = ±√

4|t|−t22 (x2 − y2) + txy , where |t| ∈ (2, 4].



P(x , y) = ±x2,

P(x , y) = ±y2,

±(tx2 − ty2 ± 2

√t(1− t)xy

), where t ∈ [12 , 1].






P(x , y) = ±x2 ± y2 ± 2xy ,

P(x , y) = ±√

4|t|−t22 (x2 − y2) + txy , where |t| ∈ (2, 4].



P(x , y) = ±x2,

P(x , y) = ±y2,

±(tx2 − ty2 ± 2

√t(1− t)xy

), where t ∈ [12 , 1].






P(x , y) = ±x2 ± y2 ± 2xy ,

P(x , y) = ±√

4|t|−t22 (x2 − y2) + txy , where |t| ∈ (2, 4].



P(x , y) = ±x2,

P(x , y) = ±y2,

±(tx2 − ty2 ± 2

√t(1− t)xy

), where t ∈ [12 , 1].




Theorem (B. Grecu)

Let p > 2. The extreme points of the unit ball of P(2`2p) are

P(x , y) = ±(x2 − y2),

P(x , y) = ax2 + cy2, ac ≥ 0, |a|p

p−2 + |c|p

p−2 = 1,

P(x , y) = ±(αp−βp

α2+β2 (x2 − y2) + 2αβ αp−2+βp−2

α2+β2 xy), α, β ≥

0, αp + βp = 1.

Theorem (G.A. Munoz-Fernandez, D. Pellegrino, J.B.Seoane-Sepulveda, A. Weber)

The extreme points of the unit ball of P(2`22) are

±(ax2 − ay2 + 2

√1− a2xy

), a ∈ [−1, 1],

±(x2 + y2).




Theorem (B. Grecu)


P(x , y) = ±(x2 − y2),

P(x , y) = ax2 + cy2, ac ≥ 0, |a|p

p−2 + |c|p

p−2 = 1,

P(x , y) = ±(αp−βp

α2+β2 (x2 − y2) + 2αβ αp−2+βp−2

α2+β2 xy), α, β ≥

0, αp + βp = 1.



±(ax2 − ay2 + 2

√1− a2xy

), a ∈ [−1, 1],

±(x2 + y2).




Theorem (B. Grecu)


P(x , y) = ±(x2 − y2),

P(x , y) = ax2 + cy2, ac ≥ 0, |a|p

p−2 + |c|p

p−2 = 1,

P(x , y) = ±(αp−βp

α2+β2 (x2 − y2) + 2αβ αp−2+βp−2

α2+β2 xy), α, β ≥

0, αp + βp = 1.



±(ax2 − ay2 + 2

√1− a2xy

), a ∈ [−1, 1],

±(x2 + y2).




Theorem (B. Grecu)


P(x , y) = ±(x2 − y2),

P(x , y) = ax2 + cy2, ac ≥ 0, |a|p

p−2 + |c|p

p−2 = 1,

P(x , y) = ±(αp−βp

α2+β2 (x2 − y2) + 2αβ αp−2+βp−2

α2+β2 xy), α, β ≥

0, αp + βp = 1.



±(ax2 − ay2 + 2

√1− a2xy

), a ∈ [−1, 1],

±(x2 + y2).




Theorem (B. Grecu)


P(x , y) = ±(x2 − y2),

P(x , y) = ax2 + cy2, ac ≥ 0, |a|p

p−2 + |c|p

p−2 = 1,

P(x , y) = ±(αp−βp

α2+β2 (x2 − y2) + 2αβ αp−2+βp−2

α2+β2 xy), α, β ≥

0, αp + βp = 1.



±(ax2 − ay2 + 2

√1− a2xy

), a ∈ [−1, 1],

±(x2 + y2).




Theorem (B. Grecu)


P(x , y) = ±(x2 − y2),

P(x , y) = ax2 + cy2, ac ≥ 0, |a|p

p−2 + |c|p

p−2 = 1,

P(x , y) = ±(αp−βp

α2+β2 (x2 − y2) + 2αβ αp−2+βp−2

α2+β2 xy), α, β ≥

0, αp + βp = 1.



±(ax2 − ay2 + 2

√1− a2xy

), a ∈ [−1, 1],

±(x2 + y2).




Theorem (From the results proved by Aron and Klimek.)

For p, q ∈ {1,∞} we have

k2,q,p =

1 if q = p = 1,

1 if q = 1, p =∞,1 if q =∞, p = 1,13 if q = p =∞.

Extremal polynomials are given:

p1,1(x , y) = ±x2, ±y2,p1,∞(x , y) = ±x2, ±y2,±xy ,p∞,1(x , y) = ±x2 ± y2 ± xy ,

p∞,∞(x , y) = ±(x2 + y2 ± xy).




Theorem (From the results proved by Aron and Klimek.)

For p, q ∈ {1,∞} we have

k2,q,p =

1 if q = p = 1,

1 if q = 1, p =∞,1 if q =∞, p = 1,13 if q = p =∞.


p1,1(x , y) = ±x2, ±y2,p1,∞(x , y) = ±x2, ±y2,±xy ,p∞,1(x , y) = ±x2 ± y2 ± xy ,

p∞,∞(x , y) = ±(x2 + y2 ± xy).







For k2,q,1 and k2,q,∞ (q ∈ (1,∞)), we used numericalapproximation.

k2,q,1 = 1 for every q ∈ (1,∞), and ±e1, ±e2 are extremal.k2,q,∞ = 31/q−1 for every q ∈ (1,∞), and ±(a,−a,−a) areextremal.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1/3

0.4

0,5

0,6

0,7

0,8

0,9

1

k2,q,∞

31q−1




For k2,q,1 and k2,q,∞ (q ∈ (1,∞)), we used numericalapproximation.k2,q,1 = 1 for every q ∈ (1,∞), and ±e1, ±e2 are extremal.

k2,q,∞ = 31/q−1 for every q ∈ (1,∞), and ±(a,−a,−a) areextremal.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1/3

0.4

0,5

0,6

0,7

0,8

0,9

1

k2,q,∞

31q−1




For k2,q,1 and k2,q,∞ (q ∈ (1,∞)), we used numericalapproximation.k2,q,1 = 1 for every q ∈ (1,∞), and ±e1, ±e2 are extremal.k2,q,∞ = 31/q−1 for every q ∈ (1,∞), and ±(a,−a,−a) areextremal.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1/3

0.4

0,5

0,6

0,7

0,8

0,9

1

k2,q,∞

31q−1




For k2,q,1 and k2,q,∞ (q ∈ (1,∞)), we used numericalapproximation.k2,q,1 = 1 for every q ∈ (1,∞), and ±e1, ±e2 are extremal.k2,q,∞ = 31/q−1 for every q ∈ (1,∞), and ±(a,−a,−a) areextremal.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1/3

0.4

0,5

0,6

0,7

0,8

0,9

1

k2,q,∞

31q−1







Theorem

Let p ∈ (1,∞). Then,

k2,q,p =

{1 if q = 1,22/p

3 if q =∞ and p ≥ 43 .


p1,p(x , y) = ±x2,±y2,p∞,p(x , y) = ±(x2 + y2 + xy).




Theorem


k2,q,p =

{1 if q = 1,22/p

3 if q =∞ and p ≥ 43 .


p1,p(x , y) = ±x2,±y2,p∞,p(x , y) = ±(x2 + y2 + xy).




Theorem


k2,q,p =

{1 if q = 1,22/p

3 if q =∞ and p ≥ 43 .


p1,p(x , y) = ±x2,±y2,p∞,p(x , y) = ±(x2 + y2 + xy).







Theorem (From the results proved by Choi, Kim and Ki)

For q, p ∈ {1,∞},

K2,q,p =

2 + 2

√2 if q = p = 1,

1 +√

2 if q = 1, p =∞,4 if q =∞, p = 1,

1 if q = p =∞.





For q, p ∈ {1,∞},

K2,q,p =

2 + 2

√2 if q = p = 1,

1 +√

2 if q = 1, p =∞,4 if q =∞, p = 1,

1 if q = p =∞.






P1,1(x , y) = ±√

2

2(x2 − y2) + (2 +

√2)xy ,

P1,∞(x , y) = ±

(2 +√

2

4x2 − 2 +

√2

4y2 ±

√2

2xy

),

P∞,1(x , y) = ±4xy ,

P∞,∞(x , y) = ±x2,±y2,±(

1

2x2 − 1

2y2 ± xy

).








Define

fq,1(t) =(

21−q(4t − t2)q/2 + tq)1/q

, t ∈ [2, 4],

fq,∞(t) =(

2tq + 2q(t − t2)q/2)1/q

, t ∈ [2, 4].

For every q ∈ [1,∞),

K2,q,1 = max{fq,1(t) : t ∈ [2, 4]},

K2,q,∞ = max{fq,∞(t) : t ∈ [1

2, 1]}.




K2,q,1 = 4 and K2,q,∞ = 21/q for every q ≥ 2.


Pq,1(x , y) = ±4xy ,

Pq,∞(x , y) = ±(x2 − y2), for q ≥ 2.




K2,q,1 = 4 and K2,q,∞ = 21/q for every q ≥ 2.Extremal polynomials are given:

Pq,1(x , y) = ±4xy ,

Pq,∞(x , y) = ±(x2 − y2), for q ≥ 2.








Pq,1(x , y) = ±4xy ,

Pq,∞(x , y) = ±(x2 − y2), for q ≥ 2.

For q = 43 , the maximum of fq,1(t) is attained at






Pq,1(x , y) = ±4xy ,

Pq,∞(x , y) = ±(x2 − y2), for q ≥ 2.







Pq,1(x , y) = ±4xy ,

Pq,∞(x , y) = ±(x2 − y2), for q ≥ 2.







Pq,1(x , y) = ±4xy ,

Pq,∞(x , y) = ±(x2 − y2), for q ≥ 2.

For q = 43 , the maxima of fq,∞(t) is attained at






Pq,1(x , y) = ±4xy ,

Pq,∞(x , y) = ±(x2 − y2), for q ≥ 2.









Theorem

If q > 1, then

K2,q,2 =

2 if q ≥ 2,

2

(1+2

1q−2

)1/q

(1+2

2(q−1)q−2

)1/q if 1 < q < 2.


Pq,2(x , y) = ±(x2 − y2) q ≥ 2,

Pq,2(x , y) = ±(a0x

2 − a0y2 + 2

√1− a20xy

), 1 < q < 2,

where a0 =

(1 + 2

2(q−1)q−2

)−1/2.




Theorem

If q > 1, then

K2,q,2 =

2 if q ≥ 2,

2

(1+2

1q−2

)1/q

(1+2

2(q−1)q−2

)1/q if 1 < q < 2.


Pq,2(x , y) = ±(x2 − y2) q ≥ 2,

Pq,2(x , y) = ±(a0x

2 − a0y2 + 2

√1− a20xy

), 1 < q < 2,

where a0 =

(1 + 2

2(q−1)q−2

)−1/2.




Theorem

If q > 1, then

K2,q,2 =

2 if q ≥ 2,

2

(1+2

1q−2

)1/q

(1+2

2(q−1)q−2

)1/q if 1 < q < 2.


Pq,2(x , y) = ±(x2 − y2) q ≥ 2,

Pq,2(x , y) = ±(a0x

2 − a0y2 + 2

√1− a20xy

), 1 < q < 2,

where a0 =

(1 + 2

2(q−1)q−2

)−1/2.







Theorem

Let q, p > 2. Then,

K2,q,p = 2max{ 1q, 2p}.


Pq,p(x , y) = ±22/pxy , q ≥ p

2,

Pq,p(x , y) = ±(x2 − y2), q 2. Then,

K2,q,p = 2max{ 1q, 2p}.


Pq,p(x , y) = ±22/pxy , q ≥ p

2,

Pq,p(x , y) = ±(x2 − y2), q 2. Then,

K2,q,p = 2max{ 1q, 2p}.


Pq,p(x , y) = ±22/pxy , q ≥ p

2,

Pq,p(x , y) = ±(x2 − y2), q 2. Then,

K2,q,p = 2max{ 1q, 2p}.


Pq,p(x , y) = ±22/pxy , q ≥ p

2,

Pq,p(x , y) = ±(x2 − y2), q 2. Then,

K2,q,p = 2max{ 1q, 2p}.


Pq,p(x , y) = ±22/pxy , q ≥ p

2,

Pq,p(x , y) = ±(x2 − y2), q 2. Then,

K2,∞,p = 22pmax{1

q,}.


Pq,p(x , y) = ±22/pxy , q ≥ p

2,

Pq,p(x , y) = ±(x2 − y2), q 2. Then,

K2,∞,p = 22p .


Pq,p(x , y)= ±22/pxy , q ≥ p

2,

Pq,p(x , y) = ±(x2 − y2), q 2. Then,

K2,∞,p = 22p .


P∞,p(x , y)= ±22/pxy , ∞ ≥ p

2,

Pq,p(x , y)= ±(x2 − y2), q <p

2.








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