Some Properties of lp A , X Spaces

and Applied Analysis 3 Theorem 2.2. Let X be a Hausdorff locally convex space, let R be a family of seminorms on X determining its topology, and let A be a set. Then each x ∈ l A,X 1 ≤ p < ∞ is represented by x Σa∈A Ia ◦ x a , 2.1 where Ia : X → l A,X is defined by Ia t b { t, b a, 0, b / a, b ∈ A. 2.2 Proof. We denote byF the family of all finite subsets of the index setA. We write x Σa∈A Ia◦ x a if the net ∑ a∈F Ia ◦ x a : F ∈ F converges to x. Define SF x ∑ a∈F Ia ◦ x a 2.3 for a finite subset F of A. We must prove that the net SF x : F converges to x in l A,X . By the definition of SF x , we have SF x a { x a , a ∈ F, 0, a ∈ A \ F. 2.4 For U ∈ N0 l A,X where N0 l A,X denotes a base of neighborhoods of the origin of l A,X , there exist ε > 0 and r1, r2, . . . , rn ∈ R such that U ⊇ n ⋂ i 1 { z : ∑ a∈A ri ◦ z a p < ε } . 2.5 Since ∑ a∈A r ◦ x a p < ∞ for each r ∈ R, then for i 1 ≤ i ≤ n , we can find Fi ∈ F such that ∑ a∈A\Fi ri ◦ x a p < ε. 2.6 Hence, setting F0 : ⋃n i 1Fi, we have ∑ a∈A ri ◦ x − SF x a p ∑ a∈A\F ri ◦ x a p < ε 2.7 for each F ⊇ F0. This implies x − SF x ∈ U. That is x Σa∈A Ia ◦ x a . Remark 2.3. If X is a normed space and ‖‖p denotes the norm of l A,X , it holds that ‖Ia t ‖p ‖t‖ and ‖Ia‖ 1. 4 Abstract and Applied Analysis Theorem 2.4. Let X be a normed space and let A be a set. Then for each f ∈ l A,X ′ , there exists ψ ∈ l A,X ′ such that f x ∑ a∈A ψ a x a , 2.8 and l A,X ′ l A,X′ , where 1/p 1/q 1 and 1 < p < ∞. Proof. By Theorem 2.2, x ∈ l A,X is represented by x ∑ a∈A Ia x a . 2.9 If f ∈ l A,X ′ , then f x ∑ a∈A f ◦ Ia x a . 2.10 Define ψ : A → X′ by ψ a f ◦ Ia. Next, we prove that ψ ∈ l A,X′ . Let F be an arbitrary finite subset ofA. Since Bishop and Phelps showed that the normattainers are dense in B X,Y for every Banach space X when Y F the symbol F denotes a field that can be either R and C , there exists ξ a in the closed unit ball of X such that ∥ψ a ∥∥ ∣ψ a ξ a ∣∣ 2.11 for each a ∈ F. Let us write ψ a ξ a in the polar form, that is, ψ a ξ a ea ∣ψ a ξ a ∣, 2.12 and define the function x from A to X by x a ∥ψ a ∥q−1e−iθaξ a , if a ∈ F and ψ a ξ a / 0, 0, if a/∈F or ψ a ξ a 0. 2.13 Abstract and Applied Analysis 5 Obviously, x ∈ l A,X . Therefore, for this x, we have ∣f x ∣∣ ∣∣∣∣ ∑ a∈A ψ a x a ∣∣∣∣ ∣∣∣∣ ∑ a∈F ∥ψ a ∥q−1e−iθaeiθa ∣ψ a ξ a ∣∣ ∣∣∣∣ ∑ a∈F ∥ψ a ∥∥q ≤ ∥f∥ ‖x‖ ≤ ∥f∥ ( ∑ a∈F ∥∥ψ a ∥∥q−1 )p )1/p ∥f ∥∥ ( ∑ a∈F ∥ψ a ∥∥q )1/p . 2.14and Applied Analysis 5 Obviously, x ∈ l A,X . Therefore, for this x, we have ∣f x ∣∣ ∣∣∣∣ ∑ a∈A ψ a x a ∣∣∣∣ ∣∣∣∣ ∑ a∈F ∥ψ a ∥q−1e−iθaeiθa ∣ψ a ξ a ∣∣ ∣∣∣∣ ∑ a∈F ∥ψ a ∥∥q ≤ ∥f∥ ‖x‖ ≤ ∥f∥ ( ∑ a∈F ∥∥ψ a ∥∥q−1 )p )1/p ∥f ∥∥ ( ∑ a∈F ∥ψ a ∥∥q )1/p . 2.14 Thus ( ∑ a∈F ∥ψ a ∥∥q )1/q ≤ ∥f∥ < ∞. 2.15 Since F is an arbitrary finite subset of A, we have ∥ψ ∥∥ ( ∑ a∈A ∥ψ a ∥∥q )1/q ≤ ∥f∥ < ∞, 2.16 and so ψ ∈ l A,X′ . Moreover, by Hölder inequality, we have ∣f x ∣∣ ≤ ∑ a∈A ∥ψ a ∥‖x a ‖ ≤ ( ∑ a∈A ∥ψ a ∥∥q 1/q∑ a∈A ‖x a ‖ )1/p ∥ψ ∥‖x‖, 2.17 from which we get ∥f ∥∥ ≤ ∥ψ∥. 2.18 Combining 2.15 and 2.18 yields ‖f‖ ‖ψ‖. Thus we define a linear isometry T : l A,X ′ → l A,X′ with Tf ψ. To prove that T is surjective. Indeed, for ψ ∈ l A,X′ , there exists f defined on l A,X such that f x ∑ a∈A ψ a x a , 2.19 6 Abstract and Applied Analysis that is, Tf ψ. By Mazur-Ulam theorem see 5 , T is a linear isometry from l A,X ′ onto l A,X′ , thus l A,X ′ l ( A,X′ ) . 2.20 The proof of this Theorem is finished. Theorem 2.5. LetX be a normed space with an unconditional basis and with a symmetric norm. Then l N, X is also a normed space with an unconditional basis and with a symmetric norm. Moreover, either l N, X is a Hilbert space or each isometry is of type 1.3 . Proof. Suppose that {ek} is an unconditional basis for X with ||ek|| 1. Let eik o, . . . , ek, o, . . . } {{ } ith place . 2.21 By Theorem 2.2, if x i Σ∞ k 1aikek then x ∈ l N, X is represented by x ∑ i∈N k∈N aikeik, 2.22 that is {eik}i∈N,k∈N is a basis for l N, X . Note that x ∑ i∈N,k∈Naikeik is an unconditionally convergent series in l N, X and that {ek} is an unconditional basis for X. Thus {eik}i∈N,k∈N is an unconditional basis for l N, X . by the definition of norm on l N, X and symmetry of norm on X it follows that ∥∥ ∑ aikeik ∥∥ (∑ ‖aikeik‖ )1/p (∑ |aik| )1/p . 2.23 For any permutation of positive integers {pik}, we have ∥∥ ∑ εikaikepik ∥∥ (∑ |aik| )1/p , 2.24 thus l N, X has symmetric norm. By Lemma 2.1, either l N, X is a Hilbert space or each isometry is of type 1.3 . 3. A Simple Proof of an Isometric Extension Result in Hilbert Space Lemma 3.1. Let E and F be normed spaces and let V0 be an isometric operator mapping S1 E into S1 F . If for any λ ∈ R and any x, y ∈ S1 E , ∥V0x − |λ|V0y ∥∥ ≤ ∥x − |λ|y∥, 3.1 then V0 can be isometrically extended to the whole space. Furthermore, when V0 is surjective, V0 can be linearly and isometrically extended to the whole space. Abstract and Applied Analysis 7 Proof. Setand Applied Analysis 7 Proof. Set Vx ⎧ ⎨ ⎩ ‖x‖V0 ( x ‖x‖ ) , if x / 0, 0, if x 0. 3.2 It is easy to see that ‖Vx − Vy‖ ≤ ‖x − y‖ for all x, y ∈ E. In particular, when ‖x‖ ‖y‖ either x or y is zero element, we have ∥Vx − Vy∥ ∥x − y∥. 3.3 Thus, it suffices to prove 3.3 whenever ‖x‖ > ‖y‖ > 0. Suppose, on the contrary, there exist x0, y0 ∈ E such that ‖x0‖ > ‖y0‖ > 0 and ‖Vx0 − Vy0‖ < ‖x0 − y0‖. Define a function on R by φ λ ∥x0 λ ( y0 − x0 ∥. 3.4 The facts that φ λ is a continuous function, φ 1 ‖y0‖ < ‖x0‖ and limλ→ ∞φ λ ∞ assure that there exists λ0 ∈ 1, ∞ such that φ λ0 ‖x0‖ by the intermediate value theorem . Let z0 x0 λ0 y0 − x0 . We see that x0, y0, and z0 lie on a straight line and ‖z0‖ ‖x0‖. Hence ‖z0 − x0‖ ∥z0 − y0 ∥∥ ∥y0 − x0 ∥∥ > ∥Vz0 − Vy0 ∥∥ ∥Vx0 − Vy0 ∥∥ ≥ ‖Vz0 − Vx0‖ ‖z0 − x0‖, 3.5 a contradiction. Thus V0 can be isometrically extended to the whole space, and V is an extension of V0. If V0 is surjective, then the conclusion follows easily from the Mazur-Ulam Theorem. Theorem 3.2. Suppose that E and F are Hilbert spaces and V0 is a surjective isometric operator mapping S1 E onto S1 F . Then V0 can be linearly and isometrically extended to the whole space. Proof. Since V0 is an isometry, we have for all x, y in S1 E that 〈V0 x − V0 ( y ) , V0 x − V0 ( y 〉 〈x − y, x − y〉, 3.6 that is, 2 − 〈V0 x , V0 ( y 〉 − 〈V0 ( y ) , V0 x 〉 2 − 〈x, y〉 − 〈y, x〉, 3.7 and thus we have 〈V0 x , V0 ( y 〉 〈V0 ( y ) , V0 x 〉 〈x, y〉 〈y, x〉. 3.8 8 Abstract and Applied Analysis The last equality gives that 〈V0 x , V0 x 〉 − λ〈V0 x , V0 ( y 〉 − λ〈V0 ( y ) , V0 x 〉 λ〈V0 ( y ) , V0 ( y 〉 1 λ2 − λ〈V0 x , V0 ( y 〉 − λ〈V0 ( y ) , V0 x 〉 1 λ2 − λ〈x, y〉 − λ〈y, x〉 〈x, x〉 − λ〈x, y〉 − λ〈y, x〉 λ2〈y, y〉. 3.9 Thus ∥V0 x − λV0 ( y )∥∥ ∥x − λy∥ 3.10 holds for all λ in R. Now we can apply Lemma 3.1 to obtain the desired result. AcknowledgmentsThe authors of this paper are supported by the NSF of Guangdong Province no. 7300614 . References1 Y. Yılmaz, “Structural properties of some function spaces,” Nonlinear Analysis: Theory, Methods &Applications, vol. 59, no. 6, pp. 959–971, 2004.2 Y. 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