Abstract. Let ∞ > p 1 > p 2 > · · · > 1. We construct an easily determined 1-symmetric basic sequence in, which spans a hereditarily 1 subspace without the Schur property. An immediate consequence is the existence of hereditarily 1 subspaces of L 1 without the Schur property.In our notation and terminology we follow mainly [4]- [6]. Recall that an infinite dimensional Banach space X is said to be hereditarily Y if each infinite dimensional subspace X 0 of X contains a further subspace Y 0 ⊆ X 0 which is isomorphic to Y . A Banach space X is said to have the Schur property provided weak convergence of sequences in X implies their norm convergence. It is well known that 1 has the Schur property. As was shown by J. Bourgain and H. P. Rosenthal in [3], there exists a subspace of L 1 having the Schur property, but which does not embed in 1 . The first example of a hereditarily 1 Banach space without the Schur property was constructed by J. Bourgain in [2]. Then P. Azimi and J. N. Hagler in [1] constructed a class of such spaces and investigated their further properties. We construct a class of subspaces of L 1 with the same properties. Actually, we make our construction in the sequence spaceSince X embeds isometrically in L 1 for p 1 ≤ 2 (this can be easily deduced from [6, p. 212]), the construction gives examples of hereditarily 1 subspaces of L 1 without the Schur property. Of course, each of them must be uncomplemented since every complemented subspace of L 1 without the Schur property contains an isomorphic copy of 2 [7].I would like to thank the referee for clarification of some arguments (the proof of Proposition 2 and especially for the idea of considering our example in the setting of the sequence space X, which makes the construction more general, clear and natural) and a number of corrections.