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2. The result ∞ The main result. Let p ∈ (1, ∞) and consider the reflexive Banach space E = ℓn . Ln=1 1
ℓp Then B(E) is not a Grothendieck space. ∞ Proof. Recall that F = ℓn contains a complemented copy of ℓ . To complete Ln=1 1
ℓ∞ 1 the proof it is enough to embed F as a complemented subspace of B(E). n B n One may identify ℓ1 with a 1-complemented subspace of (ℓ1 ) via the mapping ∗ N ek 7→ ek ⊗ e1 (k 6 n, n ∈ ), ∗ where e1 stands for the coordinate functional associated with e1. Consequently, the space ∞ B n D = (Ln=1 (ℓ1 ))ℓ∞ contains a complemented subspace isomorphic to F . Let ∆: D → B ∞ ∞ (E) be the diagonal embedding, that is, ∆((Tn)n=1) = diag(T1, T2,...) ((Tn)n=1 ∈ D); 1 2 this map is well-defined since the decomposition of E into the subspaces ℓ1,ℓ1,... is un- conditional. It is enough to notice that ∆ has a left-inverse Ξ: B(E) → D given by ∞ B Ξ(Tij)i,j∈N =(Tii)i=1 ((Tij)i,j∈N ∈ (E)), which is bounded. To this end, we shall perform a construction inspired by a trick of Tong (cf. [8, Theorem 2.3] and its proof). With each operator T = (Tij)i,j∈N ∈ B(E) we shall ∞ (n) N associate a sequence S
n=1 of finite-rank perturbations of T such that for each n ∈ we have kS(n)k 6 kT k and the matrix of S(n) agrees with the matrix of the diagonal operator diag(−T11,..., −Tnn, 0, 0,...) at entries (i, j) with i 6 n or j 6 n. This will immediately yield that (n) (n) kΞ(T )k = sup kTnnk = sup k − Snn k 6 sup kS k 6 kT k. n∈N n∈N n∈N Define operators Tk, Tr which have the same columns and rows as T respectively, except the first ones, where we instead set (Tk)i1 = −Ti1 and (Tr)1j = −T1j for i, j ∈ N (these are indeed elements of B(E) as rank-1 perturbations of T ). Certainly, kT k = kTkk = kTrk and the norm of S = (Tk + Tr)/2 does not exceed the norm of T . Arguing similarly, we (n) (n) (1) (n+1) (n) (n) observe that k(Sk +Sr )/2k 6 kT k, where S = S and S =(Sk +Sr )/2(n ∈ N). ∞ (n)  Consequently, S
n=1 is the desired sequence. Remark. The space B(E) shares with the space of operators on a Hilbert space a number of common properties. For instance, since E has a Schauder basis, B(E) can be identified with the bidual of K (E), the space of compact operators on E. Nonetheless, E is plainly not superreflexive and B(E) fails to have weakly sequentially dual for the obvious reason ∗ that ℓ∞ embeds into B(E) . We conjecture that the space of operators on a superreflexive space is Grothendieck (or at least it has weakly sequentially complete dual).

References [1] J. Bourgain, H∞ is a Grothendieck space, Studia Math., 75 (1983). 193–216. [2] J. Diestel, Sequences and Series in Banach Spaces, volume 92 of Graduate Text in Mathematics. Springer Verlag, New York, rst edition, (1984). A REFLEXIVE SPACE WHOSE ALGEBRA OF OPERATORS IS NOT A GROTHENDIECK SPACE 3

[3] J. Diestel and J. J. Uhl Jr., Vector Measures, volume 15 of Math. Surveys. AMS, Providence, RI, (1977). [4] A. Grothendieck, Sur les applications lin´eaires faiblement compactes d’espaces du type C(K), Cana- dian J. Math., 5 (1953), 129–173. [5] W. B. Johnson, A complementary universal conjugate Banach space and its relation to the approxi- mation problem, Israel Journal of Mathematics 13 (3-4) (1972), 301–310. [6] H. Pfitzner, Weak compactness in the dual of a C*-algebra is determined commutatively, Math. Ann. 298, no. 2 (1994), 349–371. [7] D. Soyba¸s, Some Remarks on the Weak Sequential Completeness in the Algebra of Integral Operators, Int. Journal of Math. Analysis, Vol. 4, 2010, no. 4, 167–173. [8] A. Tong, Diagonal submatrices of matrix maps, Pacific J. Math. Volume 32, Number 2 (1970), 551– 559.

Department of Mathematics and Statistics, Fylde College, Lancaster University, Lan- caster LA1 4YF, United Kingdom E-mail address: [email protected]