MULTI–SMOOTH POINTS OF FINITE ORDER

R. Khalil and A. Saleh

Abstract. A point x of the unit sphere S(X) of the Banach space X is called a multi-smooth point of order n if there exist exactly n-independent continuous linear functionals g_{1}, …, g_{n}, in S( X^{∗}), the unit sphere of the dual of X, such that g_{i}(x) = 1, for 1 ≤ i ≤ n. The object of this paper is to characterize multi-smooth points of some function and operator spaces.

1. Introduction. Let X be a Banach space, X^{∗} be the dual of X, and S(X) the unit sphere of X. The space of bounded linear operators on X will be denoted by L(X). A point x ∈ S(X) is called a smooth point, if there exists a unique bounded linear functional g ∈ S(X^{∗}) such that g(x) = 1. Holub [5], was the first to consider the problem of characterization of smooth points of compact operators on Hilbert spaces. This was generalized by Heinrich [3], to compact operators on arbitrary Banach spaces. Smooth points of S(L(ℓ^{2})) were studied in [9], while those of S(L(ℓ^{p})), 1 ≤ p < ∞, were studied in [1].

Smooth points are associated with the differentiability of the norm at such points. Indeed, x ∈ S(X) is smooth if and only if

lim_{t→0} \frac{∥x + ty∥ − ∥x∥}{t}

exists. We refer to the monograph of Diestel [2], for a detailed relation of smooth points of S(X) and the geometry of X. Smooth points of the unit sphere of function and Banach spaces were discussed in several papers. We refer the reader to [1, 3, 5, 7, 9, 12, 18, 19] for several interesting results on such points.

Now consider ℓ_{2}^{∞} = (R^{2}, ∥∥_{∞}). The unit sphere in such a space is the square with corners (−1, −1), (−1, 1), (1, −1) and (1, 1). We noticed that for each such corner, there exist exactly two independent functionals in S(ℓ_{2}^{1}) which attain their norm at such a corner. It turned out (Section 5 of this paper), this is the case for any finite dimensional normed space. This observation led us to introduce (Definition 1.1) the concept of multi-smooth points (or smooth points of finite order). It is the object of this paper, to characterize the multi-smooth points of order n for many classical Banach spaces and some operator spaces. In particular, we characterize multi-smooth points of order n for the spaces c_{0}, ℓ^{1}, C(K, X), K(ℓ^{p}), the space of compact operators on the Banach space ℓ^{p}, L(ℓ^{p}), 1 < p < ∞, N(ℓ^{2}), the trace class operators, and the Calkin algebra L(ℓ^{2})/K(ℓ^{2}).

2. Smooth Points of Sequence Spaces. In this section we will study multi-smooth points of finite order for c_{0}, and ℓ^{1}. Let us start with

\underline{Definition 1.1}. Let X be a Banach space and S(X) be the unit sphere of X. An element x ∈ S(X) is called a multi-smooth point of order n if there exists nindependent continuous linear functionals g_{1}, …, g_{n}, on X, each of norm one and g_{i}(x) = 1 for all 1 ≤ i ≤ n.

Let c_{0} be the space of bounded sequences that converges to zero. With the sup-norm, c_{0} is a Banach space.

\underline{Theorem 1.1}. Let x ∈ S(c_{0}). The following are equivalent:

(i) x is a smooth point of order n. (ii) x has exactly n-coordinates each of absolute value 1.

\underline{Proof}. (i) → (ii). Let x be a multi-smooth point of order n. Then there exist g_{1}, …, g_{n}, in the unit ball of ℓ^{1}such that < g_{i}, x >= 1 for all 1 ≤ i ≤ n. If x = (a_{i}), then

< g_{i}, x >= ∑_{j=1}^{∞} g_{i}(j)a_{j} = 1.

Since ∥x∥ = 1 and ∥g_{i}∥ = 1, it follows that |a_{i}| = 1 whenever g_{i}(j) ≠ 0. If |a_{i}| = 1 for more than n-coordinates, say i_{1}, …, i_{m}, with m > n, then < α_{k}δ_{i_{k}}, x >= 1 for 1 ≤ k ≤ m, where (δ_{k}) is the natural basis of ℓ^{1}, and α_{k} ∈ R is chosen such that α_{k}a_{k} = 1. So {α_{1}δ_{i_{1}}, …, α_{m}δ_{i_{m}}} is a set of m independent unit functionals on c_{0}, each attains its norm at x. Since m > n, we get a contradiction to the assumption on x. Similarly, if there exists k-coordinates of x each of absolute value 1, with k > n, then there are only k-independent unit functionals on c_{0} each of which attains its norm at x. Thus, x must have only n-coordinates, each is of absolute value 1.

(ii) → (i). If x is a point in S(c_{0}), with n−coordinates having absolute value 1, say a_{i1}, …, a_{in}, then, as in the proof of the first part, there are n-independent unit elements in ℓ^{1} that attain their norms at x. Further, there is no more and there is no less. So, x is a smooth point of order n.

\underline{Theorem 1.2}. Let x ∈ S(ℓ^{1}). The following are equivalent:

(i) x is a smooth point of order n.

(ii) x has n-coordinates each with value zero.

\underline{Proof}. Let x ∈ S(ℓ^{1}) with x = (x_{i}). Now let x have n-zeros, say x_{1}, …, x_{n}. Define

g_{1} = (0, 0, 0, …, 0, sgn(x_{n+1}), …)
g_{2} = (1, 0, 0, …, 0, sgn(x_{n+1}), …)

⋮
g_{n} = (0, 0, 0, …, 1, sgn(x_{n+1}), …).

Then {g_{1}, g_{2}, …, g_{n}} is a set of n-linearly independent unit functionals on ℓ^{1}, and g_{i}(x) = 1 for all 1 ≤ i ≤ n. Further, if g is any unit functional on ℓ^{1} such that < g, x >= 1 then, g(i) = sgn(x_{i}) for all i ≥ n. And thus, g is in the linear span of g_{1}, …, g_{n}. So, x is a smooth point of order n.

If x does not have n-coordinates each of value zero, then, either there is less than n-independent unit functionals on ℓ^{1} attaining their norm at x, or there is more than n such functionals. In either case x is not multi-smooth point of order n. This ends the proof.

\underline{Remark 1.3}. Since the ℓ^{p} spaces are uniformly convex for all 1 < p < ∞, it follows that S(ℓ^{p}) has no smooth point of order greater than 1.

Let C(K, X) be the space of all continuous functions on the compact set K with values in the Banach space X. Then, C(K, X) is a Banach space under the supnorm ∥f∥_{∞} = sup{∥f(t)∥ : t ∈ K}. It is known [12], that C(K, X) = C(K) ˇ⊗X, the completed injective tensor product of C(K) with X. So extreme points of S(C(K, X))^{∗} are functionals of the form μ ⊗ x^{∗}, with μ extreme in S(C(K))^{∗} and x^{∗} extreme in S(X^{∗}) [15]. We remark here that the dual of C(K) is M(K), the space of regular Borel measures on K.

\underline{Theorem 1.4}. Let f ∈ C(K, X). The following are equivalent:

(i) f ∈ S(C(K, X)) is smooth of order n. (ii) There exist exactly k elements {t_{1}, …, t_{k}} ⊂ K with the following properties

(a) ∥f(t_{i})∥ = 1 for 1 ≤ i ≤ k.

(b) f(t_{i}) is a multi-smooth point of order m_{i}, with m_{1} + ⋯ + m_{k} = n.

\underline{Proof}. (ii) → (i). Let f ∈ S(C(K, X)) and \{t_{1}, …, t_{k}\} ⊂ K satisfy (a) and (b). By (b), f(t_{i}) is a multi-smooth point of order m_{i}. So there exist m_{i} independent functionals x_{i1}^{∗}, … x_{im_{i}}^{∗} in S(X^{∗}), such that < f(t_{i}), x_{ir}^{∗} >= 1 for all 1 ≤ i ≤ k, and 1 ≤ r ≤ m_{i}. Set δ_{i} to denote the Dirac measure at t_{i} for 1 ≤ i ≤ k. Form the following set of linear functionals on C(K, X) : F_{ij} = δ_{i} ⊗ x_{is}^{∗} for 1 ≤ i ≤ k, 1 ≤ s ≤ m_{j}, and 1 ≤ j ≤ k. Since t_{i} ≠ t_{j}, and x_{i1}^{∗}, …, x_{im_{i}}^{∗} are independent, it follows that F_{im_{j}} are independent functionals in S(C(K, X)^{∗}) for which F_{im_{j}}(f) =< f(t_{i}), x_{is}^{∗} >= 1. Hence, there are n-independent functionals in S(C(K, X)^{∗}) and each attains its norm at f. We claim that there are no more independent functionals in S(C(K, X)^{∗}) that attain their norm at f. To see that, let E(f) = {F ∈ S(C(K, X)^{∗}) : F (f) = 1}. By the Alaoglu Theorem and the Krein Milman Theorem, E(f) is the w^{∗}-convex hull of its extreme points. Being extremal [11], extE(f) ⊂ extS(C(K, X)^{∗}) = {δ_{t} ⊗ x^{∗} : x^{∗} ∈ extS(X^{∗})}. For any such extreme functional, one has < f, δ_{t} ⊗ x^{∗} >=< f(t), x^{∗} >= 1, and so ∥f(t)∥ = 1. Consequently, condition (a) forces t to be in E. Further, x^{∗} must be one of the x_{is}^{∗,s}, otherwise condition (b) fails to be true. Hence, f is a multi-smooth point of order n.

(i) → (ii). Let f ∈ S(C(K, X)) be a multi-smooth point of order n. Again, set E(f) = {μ ∈ S(C(K))^{∗} : μ(f) = 1}. One can easily show that E(f) is w^{∗}- compact. Further E(f) is convex. The Krein-Milman Theorem [14], implies that E(f) is the closed convex hull of its extreme points. Looking at f as a linear functional on (C(K, X))^{∗}, we get E(f) an extremal subset of S(C(K, X))^{∗} [11]. Hence [11], ext(E(f)) ⊆ ext(S(C(K, X))^{∗}) = {δ_{t} ⊗ x^{∗} : x^{∗} ∈ extS(X^{∗})}. Since f is a multi-smooth point of order n, then there are exactly n-independent elements in ext(E(f)), say δ_{t_{1}} ⊗ x_{1}^{∗}, …, δ_{t_{n}} ⊗ x_{n}^{∗}. Let δ_{t_{1}}, …, δ_{t_{k}}, k ≤ n, be the independent elements of {δ_{t_{1}}, …, δ_{t_{n}}}. Consequently, {t_{1}, …, t_{k}} and {x_{1}^{∗}, …, x_{n}^{∗}} satisfy (a) and (b). This ends the proof of the theorem.

In case X is the set of real numbers, Theorem 1.4 can be stated as follows.

\underline{Theorem 1.5}. Let f ∈ S(C(K)). The following are equivalent:

(i) f is a multi-smooth point of order n. (ii) There exist exactly {t_{1}, …, t_{n}} in K such that |f(t_{i})| = 1 for 1 ≤ i ≤ n.

\underline{Remark 1.6}. We should remark that for L^{1}, the space of Lebesgue integrable functions (equivalence classes) on any finite measure space, if f ∈ S(L^{1}) is not smooth, then f cannot be a multi-smooth point of any order. This follows from the fact that if f is not smooth, then Z(f) = {t : f(t) = 0} has positive measure. So we can write

Z(f) = ⋃_{i=1}^{∞} D_{i},

with μ(D_{i}) > 0. Then

ϕ_{n}(t) = \{ \begin{matrix}sgnf(t) & if t ∉ Z(f) \\ 1 & if t ∈ D_{n} \\ 0 & if t ∈ Z(f) − D_{n} \end{matrix}

is a sequence of independent functionals in S(L^{∞}) that attains their norm on f.

3. Smooth Points In Operator Spaces. Let X be a Banach space and L(X) be the space of bounded linear operators on X. Let K(X) be the compact operators in L(X). The trace class operators on a Hilbert space H is denoted by N(H) [13]. In this section, we characterize the multi-smooth points of order n of S(K(ℓ^{p})), S(L(ℓ^{p})) and S(N(H)), 1 ≤ p < ∞. Further, we use the idea in [9] to prove that the Calkin Algebra L(H)/K(H) has no multi-smooth points of any order.

We start with the following lemma which is a generalization of the result in [19].

\underline{Lemma 2.1}. Let T, T_{i} ∈ S(L(X, Y)), for i = 1, …, k + 1 such that T = T_{1} + ⋯ + T_{k+1}. If ∥T_{i} ± T_{j}∥ ≤ 1 for all i ≠ j, then T cannot be smooth of order r for any r ≤ k.

\underline{Proof}. Since ∥T_{i}∥ = 1, there exists linear functionals F_{i} ∈ S(L(X, Y))^{∗}, such that F_{i}(T_{i}) = 1 for 1 ≤ i ≤ k + 1. For i ≠ j, we have |F_{i}(T_{i} ± T_{j})| = |1 + F_{i}(T_{j})| ≤ ∥F_{i}∥ ∥T_{i} ± T_{j}∥ ≤ 1. So F_{i}(T_{j}) = 0 for all 1 ≤ i ≠ j ≤ k + 1. But this implies that F_{i}(T) = 1 for all 1 ≤ i ≤ k + 1. Further, the functionals F_{1}, …, F_{k+1} are linearly independent. Indeed, if ∑ a_{i}F_{i} = 0, then evaluating at T_{j} we get a_{j} = 0. Thus, T cannot be a multi-smooth point of order r ≤ k.

Now, we prove the following theorem.

\underline{Theorem 2.2}. Let T ∈ S(K(ℓ^{p})), 1 < p < ∞. The following are equivalent:

(i) T is a multi-smooth point of order k. (ii) T attains its norm at exactly k-linearly independent elements, say x_{1}, …, x_{k}.

\underline{Proof}. (i) → (ii). Let T be a multi-smooth point of order k, and E(T) = \{F ∈ S(K(ℓ^{p}))^{∗} : F (T) = 1}. A usual argument, using the Alaoglu Theorem and the Krein Milman Theorem, implies that E(T) is the closed (w^{∗}−) convex hull of its extreme points. Being an extremal set [11], extE(T) ⊂ extS(K(ℓ^{p}))^{∗}. But

(K(ℓ^{p}))^{∗} = ℓ^{p} ˆ⊗ℓ^{q}, \frac{1}{p} + \frac{1}{q} = 1,

where ℓ^{p} ˆ⊗ℓ^{q} is the completed projective tensor product of ℓ^{p} with ℓ^{q} [12]. So, the extreme points of E(T) are of the form x ⊗ y, with ∥x∥ = ∥y∥ = 1. Using the assumption, there are only k independent extreme points x_{1} ⊗ y_{1}, …, x_{k} ⊗ y_{k} in E(T). But then, we have < T, x_{i} ⊗ y_{i} >=< T x_{i}, y_{i} >= 1 for 1 ≤ i ≤ k. Since ∥T ∥ = ∥x_{i}∥ = ∥y_{i}∥ = 1, and every unit vector in ℓ^{p} is a smooth point of S(ℓ^{p}), it follows that < T x_{i}, y_{i} >= ∥T x_{i}∥ = 1 for 1 ≤ i ≤ k. Since the extreme points are independent, it follows that x_{1}, …, x_{k} are independent.

(ii) → (i). Let T attain its norm at k-independent unit vectors, say x_{1}, …, x_{k}. So ∥T x_{i}∥ = 1 for 1 ≤ i ≤ k. Since T x_{i} ∈ ℓ^{p}, there exists a unique y_{i} ∈ S(ℓ^{q}) such that < T x_{i}, y_{i} >= ∥T x_{i}∥ = 1 for 1 ≤ i ≤ k. Hence, x_{1} ⊗ y_{1}, …, x_{k} ⊗ y_{k} are k-independent functionals in S(K(ℓ^{p}))^{∗} for which < T, x_{i} ⊗ y_{i} >= 1, for 1 ≤ i ≤ k. The set E(T) is extremal. Since (K(ℓ^{p}))^{∗} = N(ℓ^{q}) = ℓ^{p} ˆ⊗ℓ^{q}, and ext(ℓ^{p} ˆ⊗ℓ^{q}) = {x ⊗ y : ∥x∥_{p} = ∥y∥_{q} = 1}, then (i) implies that E(T) cannot have more than kindependent extreme points. Thus, {x_{1}⊗y_{1}, …, x_{k} ⊗y_{k}} is a maximal independent set of unit functionals in (K(ℓ^{p}))^{∗} that attain their norm at T, and T is a multismooth point of order k.

For a set A ⊂ X, we set A^{⊥} = {x^{∗} ∈ X^{∗} : x^{∗}(a) = 0 for all a ∈ A}. Now for L(ℓ^{p}), we have the following theorem.

\underline{Theorem 2.3}. Let T ∈ S(L(ℓ^{p})). The following are equivalent:

(i) T is a multi-smooth operator of order k. ii) d(T, K(ℓ^{p})) < 1, and T attains its norm at exactly k-independent unit vectors

in ℓ^{p}.

\underline{Proof}. (ii) → (i). Let x_{1}, …, x_{k} be independent vectors in S(ℓ^{p}) such that ∥T x_{i}∥ = 1 for 1 ≤ i ≤ k. Hence, there exists a unique y_{i} ∈ S(ℓ^{q}) such that < T x_{i}, y_{i} > = ∥T x_{i}∥ = 1 for 1 ≤ i ≤ k, and x_{1} ⊗ y_{1}, …, x_{k} ⊗ y_{k} are k-independent functionals in S(K(ℓ^{p}))^{∗} for which < T, x_{i} ⊗ y_{i} >= 1, for 1 ≤ i ≤ k. We claim that {x_{1} ⊗ y_{1}, …, x_{k} ⊗ y_{k}} is a maximal independent set of unit functionals in (L(ℓ^{p}))^{∗} that attain their norm at T. So let E(T) = {F ∈ (L(ℓ^{p}))^{∗} : F (T) = 1}. But [4], (L(ℓ^{p}))^{∗} = (K(ℓ^{p}))^{∗} ⊕ (K(ℓ^{p}))^{⊥}, where the sum is a direct summand, in the sense it is a direct sum and if F = F_{1} + F_{2}, then ∥F ∥ = ∥F_{1}∥ + ∥F_{2}∥. This implies that extreme points of E(T) are either extreme in (K(ℓ^{p}))^{∗} or extreme points in (K(ℓ^{p}))^{⊥}. If possible let F ∈ extE(T) ∩ (K(ℓ^{p}))^{⊥}. Since d(T, K(ℓ^{p})) < 1, there exists B ∈ K(ℓ^{p}) such that ∥T − B∥ < 1. Then F (T) = F (T − B) ≤ ∥T − B∥ < 1, which contradicts F (T) = 1. Consequently, extE(T) ⊂ (K(ℓ^{p}))^{∗}. But now we can use the same type of argument in Theorem 2.2 to conclude that T is a multi-smooth point of order k.

(ii) → (i). First we show that d(T, K(ℓ^{p})) < 1. If possible let d(T, K(ℓ^{p})) ≥ 1. Since ∥T ∥ = 1 and 0 ∈ K(ℓ^{p}), it follows that d(T, K(ℓ^{p})) = 1. Define the sequence of contractive finite rank projections

P_{n} = ∑_{i=1}^{n} δ_{i} ⊗ δ_{i} ∈ L(ℓ^{p}).

Then ∥I − P_{n}∥ = 1, and T P_{n} is compact for all n. Further, 1 = d(T, K(ℓ^{p})) ≤ ∥T − T P_{n}∥ ≤ ∥T ∥ ∥I − P_{n}∥ = 1, and so ∥T − T P_{n}∥ = 1 for all n. Since P_{n} converges to I strongly, we get:

1 = ∥T − T P_{n}∥ ≤ \underline{lim}_{n} ∥T P_{n} − T P_{m}∥ ≤ \overline{lim}_{n} ∥T P_{n} − T P_{m}∥ ≤ \overline{lim}_{n} ∥P_{n} − P_{m}∥ ≤ 1.

Hence, lim_{n} ∥T P_{n} − T P_{m}∥ = 1. Thus, there exists an increasing subsequence of Q_{r} = P_{n_{r+1}} − P_{n_{r}} such that ∥T Q_{r}∥ > 1 − \frac{1}{1+r} for all r ∈ N. Clearly

∑_{r=0}^{∞} Q_{r}x = x

for all x ∈ ℓ^{p}. Consider the following n + 1 operators

J_{k}x = ∑_{r=0}^{∞} Q_{rn+(r+k−1)}x.

Note that J_{1} + ⋯ + J_{n+1} = I. Set T_{i} = T R_{i} for 1 ≤ i ≤ n + 1. Then

∑_{i=1}^{n+1} T_{i} = T.

Further, ∥T_{i}∥ = 1 for 1 ≤ i ≤ n + 1. Indeed, ∥T_{i}∥ = ∥T R_{i}∥ ≤ 1. On the other hand

∥T_{i}∥ ≥ ∥T_{i}Q_{rn+r+i−1}∥

= ∥T R_{i}Q_{rn+r+i−1}∥

= ∥T Q_{rn+r+i−1}∥ , (since R_{i}Q_{rn+r+i−1} = Q_{rn+r+i−1})

> 1 − \frac{1}{1 + rn + r + i − 1} = 1 − \frac{1}{rn + r + i}.

Since this is true for all r ∈ N, we get ∥T_{i}∥ = 1. Since ∥J_{i} ± J_{j}∥ ≤ 1, it follows that ∥T_{i} ± T_{j}∥ ≤ 1. Lemma 2.1 now implies that d(T, K(ℓ^{p})) < 1. But now, one can use a similar argument as in (ii) → (i) of Theorem 2.2 to get the result. This ends the proof.

We turn now to operators on Hilbert spaces.

4. Smooth Points In L(H). For a Hilbert space H, let L(H) be the space of bounded linear operators on H and N(H) be the space of trace class operators [17]. The quotient algebra J = L(H)/K(H) is called the Calkin algebra. In [13], it was shown that the unit ball of the Calkin algebra has no smooth points. In this section we show that the unit ball of the Calkin algebra has no multi-smooth points of finite order. Multi-smooth points of finite order of S(N(H)) are also characterized in this section.

\underline{Theorem 3.1}. Let T ∈ S(N(H)). The following are equivalent:

(i) T is a multi-smooth point of order n = \binom{2m}{r}. (ii) T has a Schmidt representation

T = ∑ λ_{i}e_{i} ⊗ θ_{i}

with

∑ λ_{i} = 1,

where either (e_{i}) needs r-elements to form an orthonormal basis and (θ_{i}) needs m-elements to form an orthonormal basis, or visa-versa.

\underline{Proof}. (i) → (ii). Let E(T) = \{F ∈ S(L(H)) : F (T) = 1\}. Then E(T) is a w^{∗}-compact set in L(H), and hence, the closed convex hull of its extreme points.

Since the extreme points of S(L(H)) are the isometries and the co-isometries [4], then (i) implies that E(T) has n = \binom{2m}{r} independent isometries or co-isometries, say {F_{1}, …, F_{n}}. Now if F_{i}(T) = F_{j}(T) for some i and j, then

∑ λ_{k} < F_{i}(e_{k}), θ_{k} >= ∑ λ_{k} < F_{j}(e_{k}), θ_{k} > .

Consequently, F_{i}(e_{k}) = F_{j}(e_{k}) and F_{i}^{∗}(θ_{k}) = F_{j}^{∗}(θ_{k}) for all k. So, if (e_{k}) and (θ_{k}) are both complete then F_{i} = F_{j}, contradicting (i). So both (e_{k}) and (θ_{k}) are not complete. Let [u_{1}, u_{2}, …] denote the closed linear span of {u_{1}, u_{2}, … } in H. Consider the orthogonal decomposition of the Hilbert space H into orthogonal subspaces:

\begin{matrix}H = [e_{1}, e_{2}, …] ⊕ M \\ H = [θ_{1}, θ_{2}, …] ⊕ W. \end{matrix}

If the dimension of M is infinite, then any linear isometric operator A defined on [e_{1}, e_{2}, …] has an infinite number of independent extensions. Similarly, assume W is infinite dimensional. Consequently, T is not multi-smooth of finite order. Thus, M and W are of finite dimension, say

M = [g_{1}, …, g_{r}] and W = [h_{1}, …, h_{m}].

We assume that r ≤ m (if r ≥ m then we go to the adjoint operator of A). Since A(g_{i}) = h_{i}, and A(g_{i}) = −h_{i} gives two independent extensions of A, A + g_{1} ⊗ h_{1} and A − g_{1} ⊗ h_{1} are independent. So there are \binom{2m}{r} isometries (or co-isometries) attaining their norm at T, and T is multi-smooth of order n = \binom{2m}{r}.

For (ii) → (i), the proof is clear.

\underline{Problem}. Characterization of smooth points of N(ℓ^{p}) = ℓ^{q} ˆ⊗ℓ^{p}, the nuclear operators on ℓ^{p}, 1 < p ≠ 2 < ∞, is an open problem [7]. This is due to the fact that characterization of extreme points of S(L(ℓ^{p})), 1 < p ≠ 2 < ∞, is still an open problem [8].

\underline{Theorem 3.2}. The unit ball of J = L(H)/C_{∞}(H) has no multi-smooth points of finite order.

\underline{Proof}. For T ∈ L(H), we let ∥T ∥_{e} denote the norm of T in J. Then there exists an orthonormal sequence (x_{n}) in H such that lim_{n→∞} ∥T x_{n}∥ = 1 = ∥T ∥_{e} [12]. Let A be any infinite subset of ℕ, the set of natural numbers. Let M be the closure of span{x_{i} : i ∈ A}, and M̂ be the closure of span{x_{i} : i ∈ ℕ\A}. Set P to denote the orthogonal projection onto M, B = T P, and S = T (I − P). So B + S = T. Using the argument in [13] and the Hahn-Banach Theorem, we get two independent unit functionals in (L(H))^{∗}, say f_{1} and f_{2}, such that f_{1}(B) = 1, f_{1}(D) = 0 for all D ∈ span{C_{∞}, S} and f_{2}(S) = 1, f_{2}(D) = 0 for all D ∈ span{C_{∞}, B}. Consequently f_{1}(B + S) = f_{1}(B) = 1, and f_{2}(B + S) = f_{2}(S) = 1. This implies that for each infinite subset of ℕ, we have two unit independent functionals attaining their norm at T. Since different infinite subsets of ℕ give different independent pairs of unit functionals on L(H) attaining their norm at T, it follows that T cannot be a multi-smooth point of finite order. This ends the proof.

5. Smooth Points Related To Extreme Points. There is a very nice relation between multi-smooth points of order n and extreme points. Such a relation is given in the following theorem.

\underline{Theorem 4.1}. Let X be a finite dimensional Banach space and x ∈ S(X). If x is a smooth point of order n = dim(X), then x is an extreme point of the unit ball of X.

\underline{Proof}. Let x be a smooth point of order n of S(X). Thus, there exist n independent unit functionals on X, say f_{1}, …, f_{n} such that f_{i}(x) = 1 for all 1 ≤ i ≤ n. If possible assume x is not an extreme point. So there exists a y ≠ z in S(X) such that x = \frac{1}{2}(y + z). It follows that f_{i}(y) = f_{i}(z) = 1 for all 1 ≤ i ≤ n. Thus, y − z is a non zero element in the kernel of n-independent linear functionals in a space X with dimension n. This cannot be true since ∩ ker(f_{i}) = {0}. Thus, x is an extreme point.

\underline{Acknowledgment}. The authors would like to thank the referee for his sound comments that helped the authors in the present form of the paper.

\underline{References}

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Mathematics Subject Classification (2000): 46B20, 46B99

R. Khalil Department of Mathematics University of Jordan Amman, Jordan email: roshdi@ju.edu.jo

A. Saleh Department of Mathematics University of Jordan Amman, Jordan