A Note on Continuous Restrictions of Linear Maps between Banach Spaces

This note is devoted to the answers to the following questions asked by V. I. Bogachev, B. Kirchheim and W. Schachermayer: 1. Let T : l1 → X be a linear map into the infinite dimensional Banach spaceX. Can one find a closed infinite dimensional subspace Z ⊂ l1 such that T ∣∣ Z is continuous? 2. Let X = c0 or X = lp (1 < p <∞) and let T : X → X be a linear map. Can one find a dense subspace Z of X such that T ∣∣ Z is continuous? This paper continues investigations of [2]. In [2] the following proposition was proved: Let X be a separable Banach space not containing l1 isomorphically. If Y is an arbitrary infinite dimensional Banach space then there exists a linear map T : X → Y such that there does not exist any closed infinite dimensional subspace Z of X such that the restriction of T to Z is continuous. At the end of [2] the following question was proposed: Let T : l1 → X be a linear map into the infinite dimensional Banach space X. Can one find a closed infinite dimensional subspace Z ⊂ l1 such that T ∣∣ Z is continuous? We answer this question in Proposition 1. It was shown in [2] that if X = lp (1 ≤ p < ∞) or X = c0 then for every linear map T : X → X there is an infinite dimensional subspace Z of X such that T ∣∣ Z is continuous. If X = l1 then there is a dense subspace Z of X such that the restriction T ∣∣ Z is continuous. The authors of [2] asked: can one find a dense subspace Z of X such that T ∣∣ Z is continuous in other cases? We answer this question for p 6= 2 in Proposition 4. We use standard Banach space terminology and notation as may be found in [4]. Propositon 1. Let X and Y be infinite dimensional Banach spaces and let X be separable. Then there exists a linear map T : X → Y such that there does not exist any closed infinite dimensional subspace Z of X such that the restriction of T to Z is continuous. Received June 8, 1992. 1980 Mathematics Subject Classification (1991 Revision). Primary 47A05; Secondary 46B20, 46B45. 186 M. I. OSTROVSKII Proof. Let {xα}α∈[0,1] be a Hamel basis of X. Let us denote by eα (α ∈ [0, 1]) the unit vectors of the space l1([0, 1]). Let us introduce a linear map A1 : X → l1([0, 1]) in the following way: we represent x ∈ X as a finite linear combination x = ∑ α aαxα and set A1(x) = ∑ α aαeα. It is well-known that l1([0, 1]) embeds isometrically into l∞. Let A2 : l1([0, 1])→ l∞ be one of the isometric embeddings. Let {yi}i=1 be a minimal sequence in Y . We define the map A3 : l∞ → Y by the equality A3({ai} ∞ i=1) = ∞ ∑