Integral Equations and Operator Theory Vol. 13 (1990) 0378-620X/90/060849-0751.50+0.20/0 (c) 1990 Birkh~user Verlag, Basel SMOOTH POINTS OF CERTAIN OPERATOR SPACES Fund Kittaneh and Rahman Younis Let B(H) denote the Banach space of all bounded linear operators on a separable, infinite dimensional Hilbert space H, and let C be the ideal of compact 00 operators on H. It is shown that the unit ball of the Calkin algebra B(H)/C| has no smooth points. We use this to determine the smooth points of the unit ball of B(H). 1. INTRODUCTION Let H be a separable, infinite dimensional complex Hilbert space, and let B(H) denote the Banach space of all bounded linear operators on H. It is well known [5] that the set of extreme points of the unit ball of B(H) coincides with the set of all isometries and coisometries in B(H). In a recent paper [3], Grzaslewicz proved that each extreme point of the unit ball of B(H) is in fact an exposed point. In addition to the concepts of extreme and exposed points of the unit ball of a Banach space, the notion of smooth points plays a central role in the geometry of the unit ball of this space. Recall that a normalized element v of a Banach space V is said to be a smooth point of the unit ball, Ball(V), of V if there exists a unique hyperplane tangent to Ball(V) at v, i.e., there exists a unique functional f 6 V* such that f(v) = HfH = 1. It is easy to see that f is an extreme point of Ball(V*). If x and y are vectors in H, let x | y denote the rank-one operator on H defined by (x | y)z = (z,y)x. Recall that every compact operator T E B(H) has the (Schmidt) representation T = Z aigo i ~ r where {ai} is a i non-increasing sequence of non-negative real numbers converging to zero and {~oi}