[X] (1 ≤ p < ∞) and Grothendieck space

K¨ othe dual of Banach sequence spaces ℓ

[X] (1 ≤ p < ∞) and Grothendieck space

Wu Congxin, Bu Qingying

Abstract. In this paper, we show the representation of K¨othe dual of Banach sequence spacesℓp[X] (1≤p <∞) and give a characterization of that the spacesℓp[X] (1< p <∞) are Grothendieck spaces.

Keywords: vector-valued sequence space; K¨othe dual; GAK-space; Grothendieck space Classification: 46B16

Let X be a Banach space and X^∗ its topological dual, and let B_X denote the closed unit ball ofX. For 1≤p <∞, let

ℓp(X) =n

x= (x_j)∈X^N:kxk_ℓp=X^∞

i=1

kx_ik^p1/p

<∞o , ℓ_p[X] =n

x= (x_j)∈X^N: for each f ∈X^∗, X

i≥1

|f(x_i)|^p <∞o .

And for eachx∈ℓp[X], let

kxk_(ℓp)= supnX

i≥1

|f(x_i)|^p1/p

:f ∈B_X^∗o .

Then (ℓp(X),k · k_ℓp) and (ℓp[X],k · k_(ℓp)) are Banach spaces (see [1], [2], [3]). For x∈X^N, let

x(i≤n) = (x₁, . . . , xn,0,0, . . .), x(i > n) = (0, . . . ,0, x_n+1, x_n+2, . . .).

And let

ℓp[X]r={x∈ℓp[X] : lim

n kx(i > n)k_(ℓp)}= 0.

Ifℓ_p[X]_r=ℓ_p[X], thenℓ_p[X] is said to be a GAK-space [4].

For a vector-valued sequence space S(X) from X, define its K¨othe dual with respect to the dual pair (X, X^∗) (see [4]) as follows:

S(X)^×|_(X,X^∗₎=n

f = (f_j)∈X^∗N: for each x= (x_j)∈S(X), X

i≥1

|f_i(x_i)|<∞o . We denoteS(X)^×|_(X,X^∗₎byS(X)^×simply if the meaning is clear from the context.

∗The authors are supported by The Scientific Fund of China for Ph.D. instructors in Universities.

Lemma 1. For1≤p <∞,(ℓ_p[X]_r)^×=ℓ_p[X]^×.

Proof: It is easy to see thatℓ_p[X]^×⊆(ℓ_p[X]_r)^×. So we only need to prove that (ℓp[X]r)^×⊆ℓp[X]^×.

Forx= (x_j)∈ℓp[X] andt= (t_j)∈c₀, lettx= (t_jx_j). Thenktx(i > n)k_(ℓ

p)≤ kxk_(ℓp)sup_i>n|t_i| implies thattx∈ℓp[X]r. So forf = (f_j)∈(ℓp[X]r)^×, we have

X

i≥1

|f_i(t_ix_i)|<∞.

It follows from the fact thatt∈c₀ was taken arbitrary that X

i≥1

|f_i(x_i)|<∞.

Thus,f ∈ℓp[X]^× and the proof is completed.

Lemma 2. (1) For 1 ≤ p < ∞, ℓ_p[X]^× ⊆ (ℓ_p[X],k · k_(ℓ

p))^∗ and (ℓ_p[X]_r)^× = (ℓp[X]r,k · k_(ℓp))^∗.

(2)Letk·k^∗_(ℓ

p)denote the dual norm ofk·k_(ℓp)on the dual space(ℓp[X],k·k_(ℓp))^∗. Then for eachx∈ℓp[X], we have

kxk_(ℓp)= sup{|hx, fi|:f ∈ℓp[X]^×, kfk^∗_(ℓ

p)≤1}, wherehx, fi=P

i≥1f_i(x_i).

Proof: See Theorem 2.3 in [3].

Lemma 3. Every weak^∗unconditionally Cauchy series inX^∗ is weak uncondition- ally Cauchy series.

Proof: See the proof of p. 49, Corollary 11 in [5].

Lemma 4. For1≤p <∞, ℓ_p[X^∗] =n

f = (f_j)∈X^∗N: for each x∈X, X

i≥1

|f_i(x)|^p<∞o .

Proof: Let

∆ =n

f = (f_j)∈X^∗N: for each x∈X, X

i≥1

|f_i(x)|^p<∞o . By definition, we only need to prove that ∆⊆ℓ_p[X^∗].

Let f ∈ ∆ and t_j ∈ ℓq(1/p+ 1/q = 1). Then P

i≥1|f_i(t_ix)| < ∞ for each x∈X. So the series P

jt_jf_j is weak^∗ unconditionally Cauchy inX^∗ and hence, it is weak unconditionally Cauchy by Lemma 3. That is,P

i≥1|F(t_if_i)|<∞for each F ∈X^∗∗. Since (t_j) is arbitrary inℓq,P

i≥1|F(f_i)|^p <∞and f = (f_j)∈ℓp[X^∗].

The proof is completed.

Lemma 5 (the principle of local reflexivity, [6]). Let X be a normed space and Z^∗∗a finite dimensional subspace ofX^∗∗. For{F_i}ⁿ₁ ⊆Z^∗∗,{f_i}ⁿ₁ ⊆X^∗andε >0, there exists a linear mapT :Z^∗∗→X such thatkTk ≤1 and

|f_i(T F_i)−F_i(f_i)|< ε, i= 1,2, . . . , n.

Proposition 6. ℓp[X^∗∗]^×|_(X^∗∗_,X^∗₎=ℓp[X]^×|_(X,X^∗₎(1≤p <∞).

Proof: It is easy to see thatℓ_p[X]⊆ℓ_p[X^∗∗] implies that ℓp[X^∗∗]^×|_(X^∗∗_,X^∗₎⊆ℓp[X]^×|_(X,X^∗₎ . So we only need to prove that

ℓp[X]^×|_(X,X^∗₎⊆ℓp[X^∗∗]^×|_(X^∗∗_,X^∗₎ .

Letf = (f_j)∈ℓp[X]^×|_(X,X^∗₎ and F = (F_j)∈ℓp[X^∗∗]. For a fixedn∈N, by Lemma 5, there exists a linear mapTn: span{F_i}ⁿ₁ →X such thatkTnk ≤1 and

|F_i(f_i)| ≤ |f_i(T_nF_i)|+ 1/n, i= 1,2, . . . , n.

Now we prove that{(T_nF₁, . . . , T_nF_n,0,0, . . .)}^∞_n=1 is a bounded subset ofℓ_p[X].

By Theorem 1.5 in [2], we have

k(TnF₁, . . . , TnFn,0,0, . . .)k_(ℓ

p)

= supn k

n

X

i=1

s_iT_nF_ik:s= (s_j)∈B_ℓqo

(1/p+ 1/q) = 1

≤supn kTnkk

n

X

i=1

s_iF_ik:s∈B_ℓqo

≤supn k

∞

X

i=1

s_iF_ik:s∈B_ℓqo

=kFk_(ℓ

p).

So{(T_nF₁, . . . , T_nF_n,0,0, . . .)}^∞_n=1 is a bounded subset ofℓ_p[X] and hence, σ(ℓp[X], ℓp[X]^×|_(X,X^∗₎)-bounded. Thus, we have

n

X

i=1

|F_i(f_i)| ≤

n

X

i=1

|f_i(TnF_i)|+ 1≤sup

n≥1

nXⁿ

i=1

|f_i(TnF_i)|o + 1.

Becausen∈Nis arbitrary, it follows that

∞

X

i=1

|F_i(f_i)|<∞.

So we prove thatf = (f_j)∈ℓp[X^∗∗]^×|_(X^∗∗_,X^∗₎ and this completes the proof.

Proposition 7. (ℓ_p[X]^×|_(X,X^∗₎)^×|_(X^∗_,X^∗∗₎=ℓ_p[X^∗∗] (1≤p <∞).

Proof: By Proposition 6, it is easy to see that

ℓp[X^∗∗]⊆(ℓp[X^∗∗]^×|_(X^∗∗_,X^∗₎)^×|_(X^∗_,X^∗∗₎

= (ℓp[X]^×|_(X,X^∗₎)^×|_(X^∗_,X^∗∗₎ . So we only need to prove that

(ℓp[X^∗∗]^×|_(X^∗∗_,X^∗₎)^×|_(X^∗_,X^∗∗₎⊆ℓp[X^∗∗].

LetF = (F_j)∈(ℓp[X^∗∗]^× |_(X^∗∗_,X^∗₎)^× |_(X^∗_,X^∗∗₎. Since f ∈X^∗ andt = (t_j)∈ ℓq (1/p+ 1/q = 1) implies that (t_jf) ∈ℓp[X^∗∗]^× |_(X^∗∗_,X^∗₎, P

i≥1|F_i(t_if)|< ∞.

Thus, P

i≥1|F_i(f)|^p < ∞ and hence, F ∈ ℓ_p[X^∗∗] by Lemma 4. The proof is

completed.

Theorem 8. For1≤p <∞,ℓp

∨

⊗X, the injective tensor product ofℓp andX, is isometrically isomorphic to the space(ℓp[X]r,k · k_(ℓp)).

Proof: For each u= Pn

i=1t⁽ⁱ⁾⊗x_i ∈ ℓ_p⊗X (t⁽ⁱ⁾ ∈ ℓ_p, x_i ∈ X), define x_u = (Pn

i=1t⁽ⁱ⁾₁ x_i,Pn

i=1t⁽ⁱ⁾₂ x_i, . . .). Then kxuk_(ℓp)= supn

|X

k≥1

s_kf(

n

X

i=1

t⁽ⁱ⁾_k x_i)|:f ∈B_X^∗, s∈B_ℓqo

= supn

|

n

X

i=1

f(x_i)ht⁽ⁱ⁾, si|:f ∈B_X^∗, s∈B_ℓqo

=λ(u) (see [7, p. 223]) (1/p+ 1/q= 1).

LetM = sup_1≤i≤nkx_ik. It follows from the above equality that

kxu (j > k)k_(ℓp)= supn

|

n

X

i=1

f(x_i)ht⁽ⁱ⁾, s(j > k)i|:f ∈B_X^∗, s∈B_ℓqo

≤MsupnXⁿ

i=1

|ht⁽ⁱ⁾, s(j > k)i|:s∈B_ℓqo .

SinceB_ℓq is weak^∗ compact, Theorem 6.11 in [8] implies that limk kxu (j > k)k_(ℓ

p)= 0.

So,xu ∈ℓp[X]rand we can define a map ϕ:ℓp⊗X →ℓp[X]rbyϕ(u) =xu. It is easy to see thatϕis a linear isometrically isomorphic map fromℓ_p⊗X to ℓ_p[X]_r. Next, we only need to prove thatϕis surjective.

Forx= (x₁, . . . , x_n,0,0, . . .), if we let u=Pn

i=1e_i⊗x_i (wheree_i=

(0, . . . ,0,1⁽ⁱ⁾,0,0, . . .)), then x= ϕ(u). Notice that limnx(j ≤ n) = xfor each x∈ℓp[X]r. Soϕis surjective and the proof is completed.

For two Banach spacesX andY, letB^∧(X, Y),I(X, Y) andN(X, Y) denote the class of integral bilinear functionals onX×Y, the class of integral operators from X toY and the class of nuclear operators fromX toY respectively (see p. 232 and p. 170 in [7]).

Theorem 9. Let1≤p <∞and1/p+ 1/q= 1. Then f = (f_j)∈ℓp[X]^×|_(X,X^∗₎ if and only if there exist anr= (rj)∈ℓ1 a bounded sequence{s⁽ⁿ⁾}^∞_n=1 of ℓq and a bounded sequence{h_n}^∞_n=1 ofX^∗ such that

f_i=X

n≥1

r_ns⁽ⁿ⁾_i h_n, i= 1,2, . . . .

Proof: Necessity. Let f = (f_j) ∈ ℓ_p[X]^×. By Lemma 1 and Lemma 2, f ∈ (ℓ_p[X]_r,k · k_(ℓ

p))^∗. So Theorem 8 implies that there is anψ^∗ ∈ (ℓ_p⊗^∨ X)^∗ corre- sponding tof. By Definition 6 in [7, p. 232], there is anψ∈ B^∧(ℓp, X) corresponding to ψ^∗. Furthermore, by Corollary 12 in [7, p. 237], there exists a T_ψ ∈ I(ℓp, X^∗) corresponding toψ. Since Corollary 10 in [7, p. 235] and Theorem 6 in [7, p.248]

guarantee that I(ℓ_p, X^∗) = N(ℓ_p, X^∗), there are an r = (r_j) ∈ ℓ₁, a bounded sequence{s⁽ⁿ⁾}^∞_n=1 ofℓ_q and a bounded sequence{h_n}^∞_n=1 ofX^∗ such that

T_ψ(t) =X

n≥1

r_nht, s⁽ⁿ⁾ih_n, for t∈ℓ_p.

Now for each i≥1 and each x∈X, by the above corresponding relations, we have

T_ψ(e_i)(x) =ψ(e_i, x) =ψ^∗(e_i⊗x) =hϕ(e_i⊗x), fi=f_i(x).

Thus

f_i=T_ψ(e_i) =X

n≥1

rns⁽ⁿ⁾_i hn, i= 1,2, . . . .

Sufficiency. Let M = sup_n≥1ks⁽ⁿ⁾kq and N = sup_n≥1khnk. Then, for each x= (x_j)∈ℓp[X], we have

X

i≥1

|s⁽ⁿ⁾_i h_n(x_i)| ≤M Nkxk_(ℓ

p), for n≥1.

And so

X

i≥1

|f_i(x_i)| ≤X

n≥1

|rn|X

i≥1

|s⁽ⁿ⁾_i hn(x_i)|<∞.

Therefore,f ∈ℓ_p[X]^× and the proof is completed.

Theorem 10. For1< p <∞,(ℓ_p[X]^×,k · k^∗_(ℓ

p))is a GAK-space.

Proof: Letf = (f_j)∈ℓp[X]^×. Then by Theorem 9, there exist anr= (r_j)∈ℓ₁, a bounded sequence{s⁽ⁿ⁾}^∞₁ ofℓqand a bounded sequence{hn}^∞₁ ofX^∗such that

f_i=X

n≥1

r_ns⁽ⁿ⁾_i h_n, i= 1,2, . . . .

Without loss of generality, we can assume thatks⁽ⁿ⁾kq≤1 andkhnk ≤1 forn≥1.

Thus, forx∈ℓp[X] withkxk_(ℓp)≤1, we have X

i≥1

|s⁽ⁿ⁾_i hn(x_i)| ≤ kxk_(ℓp)≤1 for n≥1.

So

nX

i≥1

|s⁽ⁿ⁾_i hn(x_i)|

n≥1:kxk_(ℓp)≤1o

⊆B_ℓ∞.

Letε >0. ThenB_ℓ∞is weak^∗compact implies that there exists ann₀ ∈Nsuch that

X

n>n⁰

|rn|X

i≥1

|s⁽ⁿ⁾_i hn(x_i)|< ε/2, x∈ℓp[X], kxk_(ℓp)≤1.

SinceB_ℓp is weakly compact set and

n(h_n(x_i))_i≥1:x∈ℓ_p[X], kxk_(ℓ

p)≤1, n≥1o

⊆B_ℓp, there is ak₀ ∈Nsuch that for eachk > k₀,

X

i>k

|s⁽ⁿ⁾_i hn(x_i)|< ε/2krk₁ forx∈ℓ_p[X] with kxk_(ℓ

p)≤1 andn= 1,2, . . . , n₀. Thus, for eachx∈ℓ_p[X] with kxk_(ℓ

p)≤1 and eachk > k₀, we have X

i>k

|f_i(x_i)| ≤

n⁰

X

n=1

|r_n|X

i>k

|s⁽ⁿ⁾_i h_n(x_i)|+ X

n>n0

|r_n|X

i>k

|s⁽ⁿ⁾_i h_n(x_i)|

≤X^∞

n=1

|rn|

ε/2krk₁+ X

n>n⁰

|rn|X

i≥1

|s⁽ⁿ⁾_i hn(x_i)|< ε.

So fork > k₀,

kf (j > k)k^∗_(ℓ

p)= supn

|hx, f (j > k)i|:x∈ℓ_p[X], kxk_(ℓ

p)≤1o

= supn

|X

i>k

f_i(x_i)|:kxk_(ℓ

p)≤1o

< ε.

Therefore, lim_kkf (j > k)k^∗_(ℓ

p)= 0 andf ∈(ℓ_p[X],k · k^∗_(ℓ

p))_r.

For 1< p <∞, by Theorem 10 and [4, Proposition 4.9], we have (∗) (ℓ_p[X]^×|_(X,X^∗₎)^×|_(X^∗_,X^∗∗₎= (ℓ_p[X]^×|_(X,X^∗₎,k · k^∗_(ℓ

p))^∗. Now, if we letk·k^∗∗_(ℓ

p)denote the dual norm ofk·k^∗_(ℓ

p)on the dual space (ℓp[X]^×|_(X,X^∗₎, k·k^∗_(ℓ

p))^∗, then by Proposition 7 and Lemma 2, the normk·k_(ℓp)on the spaceℓp[X^∗∗] is equal to the normk · k^∗∗_(ℓ

p).

Similarly as the proof of Theorem 3.6 in [3], we have the following two proposi- tions.

Proposition 11. Letf⁽ⁿ⁾∈ℓp[X]^×(1≤p <∞). Then that σ(ℓ_p[X]^×|_(X,X^∗₎,(ℓ_p[X]^×|_(X,X^∗₎)^×|_(X^∗_,X^∗∗₎)−lim

n f⁽ⁿ⁾= 0 is equivalent to

(a) σ(X^∗, X^∗∗)−limnf_i⁽ⁿ⁾= 0fori≥1; and (b) sup_n≥1kf⁽ⁿ⁾k^∗_(ℓ

p)<∞

if and only if((ℓp[X]^×|_(X,X^∗₎)^×|_(X^∗_,X^∗∗₎,k · k^∗∗_(ℓ

p))is a GAK-space.

Proposition 12. Letf⁽ⁿ⁾∈(ℓ_p[X]_r)^∗ (1≤p <∞). Then σ((ℓp[X]r)^∗, ℓp[X]r)−limf⁽ⁿ⁾= 0

if and only ifσ(X^∗, X)−lim_nf_i⁽ⁿ⁾= 0fori≥1 andsup_n≥1kf⁽ⁿ⁾k^∗_(ℓ

p)<∞.

We say a Banach spaceXto be a Grothendieck space if every weak^∗null sequence onX^∗ is weak null sequence (see [7, p. 179]). Leonard [1] has proved thatℓp(X) (1< p <∞) is a Grothendieck space if and only ifXis a Grothendieck space. Now we have

Theorem 13. For1< p <∞. The Banach space (ℓ_p[X]_r,k · k_(ℓ

p))is a Grothen- dieck space if and only if

(i) X is a Grothendieck space; and (ii) (ℓp[X^∗∗],k · k_(ℓ

p))is a GAK-space.

Proof: Sufficiency. By (ii), (ℓ_p[X],k · k_(ℓ

p)) is a GAK-space, i.e. ℓ_p[X]_r=ℓ_p[X].

Letf⁽ⁿ⁾∈(ℓp[X],k · k_(ℓp))^∗ such that σ(ℓp[X]^∗, ℓp[X])−lim

n f⁽ⁿ⁾= 0.

By Proposition 12, we have

σ(X^∗, X)−lim

n f_i⁽ⁿ⁾= 0, i= 1,2, . . . and

sup

n≥1

kf⁽ⁿ⁾k^∗_(ℓ

p)<∞.

By (i), we have

σ(X^∗, X^∗∗)−lim

n f_i⁽ⁿ⁾= 0, i= 1,2, . . . . By (ii) and Propositions 2, 6, 7, the space ((ℓp[X]^∗)^×|_(X^∗_,X^∗∗₎,k · k_(ℓ

p)) is a GAK- space. So Proposition 11 guarantees that

σ(ℓp[X]^∗,(ℓp[X]^∗)^×)−lim

n f⁽ⁿ⁾= 0.

It follows from (∗) that

σ(ℓ_p[X]^∗, ℓ_p[X]^∗∗)−lim

n f⁽ⁿ⁾= 0.

and completes the sufficiency.

Necessity. To prove (i), letfn∈X^∗ (n≥1) such that σ(X^∗, X)−lim

n f_n= 0.

Letf⁽ⁿ⁾= (f_n,0,0, . . .) forn≥1. Thenf⁽ⁿ⁾∈(ℓ_p[X]_r)^∗ and σ((ℓ_p[X]_r)^∗, ℓ_p[X]_r)−lim

n f⁽ⁿ⁾= 0.

So

σ((ℓp[X]r)^∗,(ℓp[X]r)^∗∗)−lim

n f⁽ⁿ⁾= 0 and hence,σ(X^∗, X^∗∗)−limnfn= 0. (i) follows.

For (ii), letf⁽ⁿ⁾∈ℓ_p[X]^×|_(X,X^∗₎ such that σ(X^∗, X^∗∗)−lim

n f_i⁽ⁿ⁾= 0, i= 1,2, . . . and

sup

n≥1

kf⁽ⁿ⁾k^∗_(ℓ

p)<∞.

By Lemmas 1, 2 and Proposition 12, we have σ((ℓp[X]r)^∗, ℓp[X]r)−lim

n f⁽ⁿ⁾= 0.

And hence,

σ((ℓp[X]r)^∗,(ℓp[X]r)^∗∗)−lim

n f⁽ⁿ⁾= 0.

It follows from (∗) that

σ(ℓp[X]^×|_(X,X^∗₎,(ℓp[X]^×|_(X,X^∗₎)^×|_(X^∗_,X^∗∗₎)−lim

n f⁽ⁿ⁾= 0.

So Propositions 6, 7, 11 imply that (ℓ_p[X^∗∗],k·k_(ℓ

p)) is a GAK-space and (ii) follows.

The proof is completed.

Corollary 14. If ℓ_p[X]_r (1 < p < ∞) is a Grothendieck space, then ℓ_p[X] is a GAK-space.

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Department of Mathematics, Harbin Institute of Technology, Harbin, 150006 China (Received September 23, 1992)