A general Pietsch Domination Theorem

A general Pietsch Domination Theorem

J. Math. Anal. Appl. 375 (2011) 371–374 Contents lists available at ScienceDirect Journal of Mathematical Analysis and Applications www.elsevier.com...

111KB Sizes 3 Downloads 128 Views

J. Math. Anal. Appl. 375 (2011) 371–374

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Note

A general Pietsch Domination Theorem Daniel Pellegrino a,∗,1 , Joedson Santos b a b

Departamento de Matemática, Universidade Federal da Paraíba, 58.051-900 João Pessoa, Brazil Departamento de Matemática, Universidade Federal de Sergipe, 49500-000 Itabaiana, Brazil

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 4 June 2010 Available online 18 August 2010 Submitted by Richard M. Aron

In this short communication we show that the unified Pietsch Domination Theorem proved in Botelho et al. (2010) [1] remains true even if we remove two of its hypotheses. © 2010 Elsevier Inc. All rights reserved.

Keywords: Pietsch Domination Theorem Nonlinear absolutely summing operators

1. Introduction Let X , Y and E be (arbitrary) non-void sets, H be a non-void family of mappings from X to Y , G be a Banach space and K be a compact Hausdorff topological space. Let

R: K × E × G −→ [0, ∞)

S : H × E × G −→ [0, ∞)

and

be arbitrary mappings and 0 < p < ∞. A mapping f ∈ H is said to be R S-abstract p-summing if there is a constant C > 0 so that



m 

 1p S( f , x j, b j)

p

  C sup

m 

ϕ∈K

j =1

 1p R (ϕ , x j , b j )

p

,

(1.1)

j =1

for all x1 , . . . , xm ∈ E, b1 , . . . , bm ∈ G and m ∈ N. The main result of [1] proves that under certain assumptions on R and S there is a quite general Pietsch Domination-type Theorem. More precisely R and S must satisfy the three properties below: (1) For each f ∈ H, there is an x0 ∈ E such that

R (ϕ , x0 , b) = S ( f , x0 , b) = 0 for every ϕ ∈ K and b ∈ G. (2) The mapping

R x,b : K −→ [0, ∞)

defined by R x,b (ϕ ) = R (ϕ , x, b)

is continuous for every x ∈ E and b ∈ G.

* 1

Corresponding author. E-mail address: [email protected] (D. Pellegrino). The author is supported by CNPq Grant 301237/2009-3.

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.08.019

©

2010 Elsevier Inc. All rights reserved.

372

D. Pellegrino, J. Santos / J. Math. Anal. Appl. 375 (2011) 371–374

(3) For every

ϕ ∈ K , x ∈ E , 0  η  1, b ∈ G and f ∈ H, the following inequalities hold:

R (ϕ , x, ηb)  η R (ϕ , x, b)

η S ( f , x, b)  S ( f , x, ηb).

and

The Pietsch Domination Theorem from [1] reads as follows: Theorem 1.1. If R and S satisfy (1), (2) and (3) and 0 < p < ∞, then f ∈ H is R S-abstract p-summing if and only if there are a constant C > 0 and a Borel probability measure μ on K such that



 1p R (ϕ , x, b) dμ(ϕ ) p

S ( f , x, b)  C

(1.2)

K

for all x ∈ E and b ∈ G. The aim of this note is to show that the hypotheses (1) and (3) are not necessary. So, Theorem 1.1 is true for arbitrary S (no hypothesis is needed) and the map R just needs to satisfy (2). 2. A recent approach to the Pietsch Domination Theorem In a recent preprint [3] we have extended the Pietsch Domination Theorem from [1] to a more abstract setting, which allows to deal with more general nonlinear mappings in the cartesian product of Banach spaces. In the present note we shall recall the argument used in [3] and a combination of this argument with an interesting argument due to M. Mendel and G. Schechtman (used in [1]) will help us to show that Theorem 1.1 is valid without the hypotheses (1) and (3) on R and S. The first step is to prove Theorem 1.1 without the hypothesis (1). This result is proved in [3] in a more general setting. Since the paper [3] is unpublished and we just need a very particular case, we prefer to sketch the proof for this particular case. The proof of this particular case is essentially Pietsch’s original proof (see [4, p. 232]) on a nonlinear disguise. Theorem 2.1. Suppose that R and S satisfy (2) and (3). If 0 < p < ∞, a map f ∈ H is R S-abstract p-summing if and only if there are a constant C > 0 and a Borel probability measure μ on K such that

1/ p



R (ϕ , x, b) p dμ(ϕ )

S ( f , x, b)  C

(2.1)

K

for all x ∈ E and b ∈ G. Proof. If (2.1) holds it is easy to show that f is R S-abstract p-summing. For the converse, consider the (compact) set P ( K ) of the probability measures in C ( K )∗ (endowed with the weak-star topology). For each (x j )m in E, (b j )m in G and m ∈ N, j =1 j =1 let g := g (x j )m ,(b j )m : P ( K ) → R be defined by j =1

g (ρ ) =

j =1

m  



 S ( f , x j , b j )p − C p

j =1

R (ϕ , x j , b j ) p dρ (ϕ ) K

and F be the set of all such g. From (3), the family F is concave. In fact, if N is a positive integer, g j ∈ F and j = 1, . . . , N are so that α1 + · · · + α N = 1, we have N 

αk gk (ρ ) =

k =1

N 



αk

 p

S ( f , x jk , b jk ) − C

mk  N  



S f , x jk , αk b jk

1 m ,N p (x jk ) j k,k=1 ,(αk k

R (ϕ , x jk , b jk ) dρ (ϕ ) p

K

1 p

p

 − Cp

k =1 j k =1

=g

p

j k =1

k =1



mk  

α j  0,

R



1

ϕ , x jk , αkp b jk

p

dρ (ϕ )

K m ,N

b j ) j k,k=1 k k

(ρ ).

Besides, since each R x j ,b j is continuous, it follows that each g ∈ F is also continuous and, moreover, each g ∈ F is convex, since



g λρ1 + (1 − λ)ρ2 = λ g (ρ1 ) + (1 − λ) g (ρ2 ) for all 0  λ  1 and

ρ 1 , ρ2 ∈ P ( K ) .

D. Pellegrino, J. Santos / J. Math. Anal. Appl. 375 (2011) 371–374

We also note that for each g ∈ F there is a measure of K and Weierstrass’ theorem there is a ϕ0 ∈ K so that m 

m 

R (ϕ0 , x j , b j ) p = sup

ϕ ∈ K j =1

j =1

373

μ g ∈ P ( K ) such that g (μ g )  0. In fact, from (2), the compactness

R (ϕ , x j , b j ) p .

Then, considering the Dirac measure exists a μ ∈ P ( K ) so that

μ g = δϕ0 , we deduce g (μ g )  0. So, Ky Fan Lemma (see [4, p. 40]) asserts that there

g (μ)  0 for all g ∈ F and by choosing an arbitrary g with m = 1 the proof is done.

2

3. The main result Note that (1.1) is equivalent to



m 

 1p λ j S( f , x j, b j)

p

 C sup ϕ∈K

j =1

m 

 1p λ j R (ϕ , x j , b j )

p

(3.1)

j =1

for all x1 , . . . , xm ∈ E, b1 , . . . , bm ∈ G, positive real numbers λ j and m ∈ N. The argument is the following: by approximation in (3.1) it is enough to consider rational λ j and thus, by cleaning denominators, integer λ j . Thus, by considering each pair (x j , b j ) repeated λ j times in (1.1) we prove the equivalence. The essence of this argument appears in [2] and is due to M. Mendel and G. Schechtman. Now using (3.1) and invoking Theorem 2.1 we can prove a Pietsch Domination-type Theorem with no hypothesis on S and just supposing that R satisfies (2): Theorem 3.1. Suppose that S is arbitrary and R satisfies (2) and let 0 < p < ∞. A map f ∈ H is R S-abstract p-summing if and only if there are a constant C > 0 and a Borel probability measure μ on K such that



 1p R (ϕ , x, b) dμ(ϕ ) p

S ( f , x, b)  C

(3.2)

K

for all x ∈ E and b ∈ G. Proof. It is clear that if f satisfies (3.2) then f ∈ H is R S-abstract p-summing. Conversely, if f ∈ H is R S-abstract p-summing, then



m 

 1p λ j S ( f , x j , b j )p

  C sup ϕ∈K

j =1

m 

 1p λ j R (ϕ , x j , b j ) p

(3.3)

j =1

for all x1 , . . . , xm ∈ E, b1 , . . . , bm ∈ G, λ1 , . . . , λm ∈ [0, ∞) and m ∈ N. Let

E1 = E × G

and

G1 = K

and define

R : K × E 1 × G 1 −→ [0, ∞)

and

S : H × E 1 × G 1 −→ [0, ∞)

by

R







ϕ , (x, b), λ = |λ| R (ϕ , x, b) and S f , (x, b), λ = |λ| S ( f , x, b).

From (3.3) we conclude that



m 

S f , (x j , b j ), η j

p

 1p

  C sup

j =1

for all x1 , . . . , xm ∈ E, b1 , . . . , bm ∈ G,

ϕ∈K

m 

R

ϕ , (x j , b j ), η j

j =1

η1 , . . . , ηm ∈ K and m ∈ N.

p

 1p

374

D. Pellegrino, J. Santos / J. Math. Anal. Appl. 375 (2011) 371–374

Since R and S satisfy (2) and (3), from Theorem 2.1 we conclude that there is a measure





S f , (x, b), η  C

R



p



μ so that

1 p

ϕ , (x, b), η dμ(ϕ )

K

for all x ∈ E, b ∈ G and

η ∈ K. Taking η = 1 in the above relation we get that



 1p R (ϕ , x, b) dμ(ϕ ) p

S ( f , x, b)  C K

for all x ∈ E and b ∈ G.

2

Acknowledgment The authors thank the referee for helpful suggestions.

References [1] [2] [3] [4]

G. Botelho, D. Pellegrino, P. Rueda, A unified Pietsch domination theorem, J. Math. Anal. Appl. 365 (2010) 269–276. J. Farmer, W.B. Johnson, Lipschitz p-summing operators, Proc. Amer. Math. Soc. 137 (2009) 2989–2995. D. Pellegrino, J. Santos, On summability of nonlinear maps: a new approach, preprint, arXiv:1004.2643v2. A. Pietsch, Operator Ideals, North-Holland, Amsterdam, 1980.