EXTREME AND SMOOTH POINTS IN LORENTZ AND MARCINKIEWICZ SPACES

HAN JU LEE^{1},ANNA KAMIŃSKA^{2} AND GRZEGORZ LEWICKI^{3}

IMUJ Preprint 2006/01

Abstract. We characterize extreme and smooth points in Lorentz sequence space d(w, 1) and in Marcinkiewicz sequence spaces d_{∗}(w, 1) and d^{∗}(w, 1), which are predual and dual spaces to d(w, 1), respectively. We then apply these characterizations for studying the relationship between the existence and one-complemented subspaces in d(w, 1). We show that a subspace of d(w, 1) is an existence set if and only if it is one-complemented.

Marcinkiewicz and Lorentz spaces play an important role in the theory of Banach spaces. They are key objects for instance in the interpolation theory of linear operators. The origins of the Marcinkiewicz spaces go back to the theorem on weak type operators [23, th. 2.b.15], which was originally due to K. Marcinkiewicz in the 1930-ties. The Lorentz spaces introduced by G.G. Lorentz in 1950, have appeared in a natural way as interpolation spaces between suitable Lebesgue spaces by classical result of Lions and Peetre [23, th. 2.g.18]. This theory has been developed very extensively thereafter and along with these investigations, the theory of Lorentz and Marcinkiewicz spaces, including the studies of their geometric structure, have been evolved independently (e.g. [6, 7, 22, 25]). One can observe that these spaces find also applications in other topics of operator theory. It is worth to mention that Marcinkiewicz spaces d^{∗}(w, 1) have emerged recently many times in the context of norm-attaining linear operators. In the papers [1, 9, 14] it was shown among others, by using the space d^{∗}(w, 1) with specific weight, that the subspace of norm attaining operators is not always dense in the space of all bounded operators, contrary to the Bishop-Phelps theorem for linear functionals. For such types of isometric results the knowledge of geometric properties of the ball is of the utmost importance (see e.g. [9], where the characterization of complex convexity of the Lorentz spaces was the key factor in the proof of the main result).

In this paper we consider the Lorentz and Marcinkiewicz sequence spaces generated by decreasing weight sequences. In the first two sections we shall characterize the smooth and extreme points of the balls in these spaces. In the last section we shall apply these results to study the relationship between the existence and one-complemented subspaces of Lorentz sequence spaces.

2 HAN JU LEE^{1},ANNA KAMIŃSKA^{2} AND GRZEGORZ LEWICKI^{3}

Let’s first agree on basic definitions and notations. Throughout the paper any vector space will be always considered over the field of real numbers ℝ. Given a Banach space (X, ∥ · ∥), by S_{X} and B_{X} we denote the unit sphere and the unit ball of X, respectively. Recall that x ∈ S_{X} is an extreme point of the ball B_{X} whenever x = (x_{1} + x_{2})/2 with x_{i} ∈ S_{X}, i = 1, 2, implies that x = x_{1} = x_{2}. An element x ∈ X is called a smooth point of X if there exists a unique bounded linear functional ϕ ∈ S_{X^{∗}} such that ϕ(x) = ∥x∥. Such functional ϕ is a called a supporting functional of x.

1Department of Mathematics, POSTECH, San 31, Hyoja-dong, Nam-gu, Pohang-shi, Kyungbuk, Republic of Korea

2Department of Mathematical Sciences, The University of Memphis, Memphis, TN 38152, USA

3Department of Mathematics, Jagiellonian University, Reymonta 4, 30-059 Krakow, Poland

A symbol ext C will stand for the set of all extreme points of a convex subset C of X.

Assume that {w(n)} is a decreasing sequence of positive numbers such that lim_{n} w(n) = 0 and ∑_{n=1}^{∞} w(n) = ∞. Let W (n) = ∑_{i=1}^{n} w(i). By card A we denote cardinality of A ⊂ ℕ. For a real sequence x = {x(n)}, by x^{∗} = {x^{∗}(n)} we denote its decreasing rearrangement. Recall that x^{∗}(n) = inf{s > 0 : d_{x}(s) ≤ n}, n ∈ ℕ, where d_{x} is a distribution of x, that is d_{x}(s) = card{k ∈ ℕ : |x(k)| > s}, s ≥ 0. For any x = {x(n)} the support of x is the set supp x = {n ∈ ℕ : x(n) ≠ 0}. We say that two sequences are equimeasurable whenever their distributions coincide. The Lorentz sequence space d(w, 1) is a collection of all real sequences x = {x(n)} such that

∥x∥_{w,1} = ∑_{n=1}^{∞} x^{∗}(n)w(n) < ∞.

It is well known that d(w, 1) is a Banach space under the norm ∥·∥_{w,1}. The Marcinkiewicz sequence space d^{∗}(w, 1) consists of all real sequences x = {x(n)} satisfying

∥x∥_{W} = sup_{n} \frac{∑_{i=1}^{n} x^{∗}(i)}{W (n)} < ∞,

and the subspace d_{∗}(w, 1) of d^{∗}(w, 1) is defined as

d_{∗}(w, 1) = \{x ∈ d^{∗}(w, 1) : lim_{n} \frac{∑_{i=1}^{n} x^{∗}(i)}{W (n)} = 0\} .

Both spaces d^{∗}(w, 1) and d_{∗}(w, 1), equipped with the norm ∥ · ∥_{W}, are Banach spaces and d_{∗}(w, 1) is a closed subspace of d^{∗}(w, 1). It is well known that d_{∗}(w, 1) and d^{∗}(w, 1) are predual and dual spaces of d(w, 1), respectively. Note also that by the assumptions on the weight w, each space d(w, 1), d^{∗}(w, 1) and d_{∗}(w, 1) is contained in the space c_{0}, and thus for any element x in any of these spaces, the distribution function d_{x} is always finite. For more details on the Lorentz and Marcinkiewicz spaces see e.g. ([19, 17, 22]).

1. Smooth points

In this section we characterize smooth points in Lorentz and Marcinkiewicz sequence spaces. We start with some auxiliary lemmas.

Lemma 1.1. Let ϕ = {a(n)} ∈ d(w, 1) be a supporting functional at x ∈ S_{d^{∗}(w,1)}. If there is m ∈ ℕ such that

∑_{i=1}^{m} x^{∗}(i) < W (m),

then a^{∗}(m) = a^{∗}(m + 1).

Proof. Suppose, for a contrary that a^{∗}(m) > a^{∗}(m + 1). Since x is an element of the unit sphere of d^{∗}(w, 1), we have S^{∗}(n) := ∑_{i=1}^{n} x^{∗}(i) ≤ W (n) for all n ∈ ℕ. Thus, in view of S^{∗}(m) < W (m) and by summation by parts, for every l > m,

∑_{i=1}^{l}a^{∗}(i)x^{∗}(i)

= ∑_{i=1}^{m}(a^{∗}(i) − a^{∗}(i + 1))S^{∗}(i) + ∑_{i=m+1}^{l−1}(a^{∗}(i) − a^{∗}(i + 1))S^{∗}(i) + a^{∗}(l)S^{∗}(l)

< ∑_{i=1}^{m}(a^{∗}(i) − a^{∗}(i + 1))W (i) + ∑_{i=m+1}^{l−1}(a^{∗}(i + 1) − a^{∗}(i))W (i) + a^{∗}(l)W (l)

= ∑_{i=1}^{l} a^{∗}(n)w(n) = ∥ϕ∥ .

In view of

lim_{l} ∑_{i=m+1}^{l−1}(a^{∗}(i) − a^{∗}(i + 1))S^{∗}(i) ≤ lim_{l} ∑_{i=m+1}^{l−1}(a^{∗}(i + 1) − a^{∗}(i))W (i)

and

lim_{l} a^{∗}(l)S^{∗}(l) ≤ lim_{l} a^{∗}(l)W (l),

it follows that

(1.1)  ∑_{n=1}^{∞} a^{∗}(n)x^{∗}(n) < ∑_{n=1}^{∞} a^{∗}(n)w(n) = ∥ϕ∥ .

Since W is a supporting functional at x, applying the Hardy-Littlewood inequality, we obtain that

∥ϕ∥ = ϕ(x) = ∑_{n=1}^{∞} a(n)x(n) ≤ ∑_{n=1}^{∞} a^{∗}(n)x^{∗}(n).

This is a contradiction to inequality (1.1) and the proof is done. □

Corollary 1.2. Suppose ϕ = {a(n)} ∈ d(w, 1) is a supporting functional at x ∈ S_{d_{∗}(w,1)}. Then it is finite, i.e., a(n) ≠ 0 for finite numbers of n ∈ ℕ.

Proof. In view of x ∈ d_{∗}(w, 1), there exists N ∈ ℕ such that

N = max \{n : \frac{∑_{i=1}^{n} x^{∗}(i)}{W (n)} = 1\} .

Then ∑_{i=1}^{k} x^{∗}(i) < W (k) for all k > N, and by Lemma 1.1, a^{∗}(N + 1) = a^{∗}(N + 2) = ⋯ = 0, since {a(n)} is an element of c_{0}. □

Proposition 1.3. If x is an element of S_{d^{∗}(w,1)} such that

2 ≤ card \{m : \frac{∑_{i=1}^{m} x^{∗}(i)}{W (m)} = 1\} < ∞,

then there exist two different norm-one supporting functionals in d(w, 1) at x.

Proof. Let

M = max \{m : \frac{∑_{i=1}^{m} x^{∗}(i)}{W (m)} = 1\} ,

and suppose that N < M such that

\frac{∑_{i=1}^{N} x^{∗}(i)}{W (N)} = 1 = \frac{∑_{i=1}^{M} x^{∗}(i)}{W (M)}.

Notice that

x^{∗}(M) = w(M) + ∑_{i=1}^{M−1}(w(i) − x^{∗}(i)) ≥ w(M) ≥ w(M + 1).

Notice also that

∑_{i=1}^{M+1} x^{∗}(i) < ∑_{i=1}^{M+1} w(i).

So we have x^{∗}(M + 1) < w(M + 1). Therefore x^{∗}(M) > x^{∗}(M + 1). Hence there is a permutation σ on ℕ such that |x(σ(k))| = x^{∗}(k) for all k = 1, …, M. Now let for y ∈ d^{∗}(w, 1),

ϕ_{1}(y) = \frac{1}{W (N)} ∑_{i=1}^{N} sign(x(σ(i))y(σ(i))

and

ϕ_{2}(y) = \frac{1}{W (M)} ∑_{i=1}^{M} sign(x(σ(i))y(σ(i)).

It is clear that ϕ_{1} ≠ ϕ_{2}, ϕ_{1}(x) = ϕ_{2}(x) = 1 and ∥ϕ_{1}∥ = ∥ϕ_{2}∥ = 1. Thus ϕ_{1} and ϕ_{2} are two different norm-one supporting functionals in d(w, 1) at x. □

Proposition 1.4. Let x be an element of S_{d^{∗}(w,1)}. If

card \{m : \frac{∑_{i=1}^{m} x^{∗}(i)}{W (m)} = 1\} = 1,

then there is a unique norm-one supporting functional ψ in d(w, 1) at x.

Proof. Suppose that

\frac{∑_{i=1}^{m} x^{∗}(i)}{W (m)} = 1

holds for some m ∈ ℕ. Then x^{∗}(m) > x^{∗}(m + 1).

Let ϕ = {a(n)} be a norm-one supporting functional at x, where {a(n)} is an element of d(w, 1). Then by Lemma 1.1,

a^{∗}(1) = a^{∗}(2) = ⋯ = a^{∗}(m),

and

a^{∗}(m + 1) = a^{∗}(m + 2) = ⋯ = 0.

Hence there is an increasing finite sequence j_{1} < j_{2} < ⋯ < j_{m} such that

ϕ(y) = ∑_{k=1}^{m} aλ_{k}y(j_{k}),

where a = a^{∗}(1) and λ_{k} = ±1 for all k = 1, …, m. Since ϕ is a supporting functional at x,

∥ϕ∥ = ∑_{k=1}^{m} aw(k) = ∑_{k=1}^{m} aλ_{k}x(j_{k}) ≤ ∑_{k=1}^{m} ax^{∗}(k) = ∑_{k=1}^{m} aw(k) = 1.

This and the inequality x^{∗}(m) > x^{∗}(m + 1) imply that for k = 1, …, m,

λ_{k}x(j_{k}) = x^{∗}(k) and a = 1/W (n).

Hence |x(j_{k})| = λ_{k}x(j_{k}) and so λ_{k} = sign(x(j_{k})). Thus for y ∈ d^{∗}(w, 1),

ϕ(y) = \frac{1}{W (m)} ∑_{k=1}^{m} sign(x(j_{k}))y(j_{k}).

On the other hand, there is a permutation π on ℕ such that |x(π(k))| = x^{∗}(k) for k = 1, …, m, because x^{∗}(m) > x^{∗}(m + 1). Then the linear functional ψ, defined by

ψ(y) = \frac{1}{W (m)} ∑_{k=1}^{m} sign(x(π(k)))y(π(k)),

is a norm-one supporting functional at x. Since x^{∗}(m) > x^{∗}(m+1) and |x(π(k))| = x^{∗}(k) = |x(j_{k})| for k = 1, …, m, so

{π(k) : k = 1, …, m} = {j_{k} : k = 1, …, m}.

However it implies that ϕ = ψ and completes the proof. □

Theorem 1.5. Let x be an element of S_{d_{∗}(w,1)}. Then x is a smooth point of B_{d_{∗}(w,1)} if and only if

card \{m : \frac{∑_{i=1}^{m} x^{∗}(i)}{W (m)} = 1\} = 1.

Proof. The necessity follows from Proposition 1.4 and sufficiency from Proposition 1.3. □

Since d_{∗}(w, 1) is M-embedded, so (d^{∗}(w, 1))^{∗} = d(w, 1) ⊕_{1} F, where F is the set of singular functionals. If ξ ∈ F then it vanishes on d_{∗}(w, 1) ([15, 17]).

Suppose that ϕ is a supporting functional at x ∈ d^{∗}(w, 1). Then it has a unique representation ϕ = ψ + ξ, where ψ = {a(n)} ∈ d(w, 1) and ξ is a singular linear functional. By M-ideal property we have

∥ϕ∥ = ∥ψ∥ + ∥ξ∥ ≥ ψ(x) + ξ(x) = ϕ(x) = ∥ϕ∥ .

Therefore, both ψ and ξ are supporting functionals at x.

Proposition 1.6. Let x be an element of S_{d^{∗}(w,1)}. Suppose that

\frac{∑_{k=1}^{n} x^{∗}(k)}{W (n)} < 1

holds for all n ∈ ℕ. Then a supporting functional ϕ at x is singular.

Proof. Let ϕ = ψ + ξ be a unique decomposition, where ψ = {a(n)} and ξ is a singular linear functional. Then ψ = {a(n)} is a supporting functional at x, and by Lemma 1.1, a^{∗}(1) = a^{∗}(2) = ⋯ = 0. Thus ψ = 0. □

Proposition 1.7. Let x be an element of S_{d^{∗}(w,1)}. If

lim sup_{n} \frac{∑_{k=1}^{n} x^{∗}(k)}{W (n)} = 1,

then there exist two different norm-one supporting functionals of x.

We need the following lemma.

Lemma 1.8. Let x be an element of S_{d^{∗}(w,1)}. If

lim sup_{n} \frac{∑_{k=1}^{n} x^{∗}(k)}{W (n)} = 1,

then there is a decomposition x = x_{1} + x_{2} such that |x_{1}| ∧ |x_{2}| = 0 and ∥x_{1}∥_{W} = ∥x_{2}∥_{W} = 1.

Proof of Lemma. We shall use mathematical induction. There exists a nonempty finite subset N_{1} of ℕ such that x(i) ≠ 0 for all i ∈ N_{1} and

\frac{∑_{i∈N_{1}} |x(i)|}{W (|N_{1}|)} ≥ (1 − \frac{1}{2}).

Now suppose that there is a disjoint collection {N_{1}, ⋯, N_{m}} of finite subsets of ℕ such that N_{k} is nonempty for all k = 1, …, m and x(i) ≠ 0 for all i ∈ ⋃_{k=1}^{m} N_{k} and for all k = 1, …, m,

\frac{∑_{i∈N_{k}} |x(i)|}{W (|N_{k}|)} ≥ (1 − \frac{1}{2^{k}}).

Taking now y = xχ_{ℕ\⋃_{k=1}^{m} N_{k}}, we have

1 = lim sup_{n} \frac{∑_{i=max\{⋃_{k=1}^{m} N_{k}\}+1}^{n} x^{∗}(i)}{W (n)} ≤ lim sup_{n} \frac{∑_{i=1}^{n} y^{∗}(i)}{W (n)} ≤ 1.

Thus there is a nonempty finite subset N_{m+1} of ℕ \ ⋃_{k=1}^{m} N_{k} such that

\frac{∑_{i∈N_{m+1}} |x(i)|}{W (|N_{m+1}|)} ≥ (1 − \frac{1}{2^{m+1}} ),

which completes the induction process. Set now

G_{1} = ⋃_{k=1}^{∞} N_{2k−1} ∪ (ℕ \ ⋃_{k=1}^{∞} N_{k}), G_{2} = ⋃_{k=1}^{∞} N_{2k},

and x_{1} = xχ_{G_{1}} and x_{2} = xχ_{G_{2}}. It is clear that x = x_{1} + x_{2} and |x_{1}| ∧ |x_{2}| = 0. Moreover,

\frac{∑_{i∈N_{2m−1}} |x(i)|}{W (|N_{2m−1}|)} ≤ \frac{∑_{k=1}^{|N_{2m−1}|} x_{1}^{∗}(k)}{W (|N_{2m−1}|)} ≤ 1,

and

\frac{∑_{i∈N_{2m}} |x(i)|}{W (|N_{2m}|)} ≤ \frac{∑_{k=1}^{|N_{2m}|} x_{2}^{∗}(k)}{W (|N_{2m}|)} ≤ 1

hold for all m. This implies that ∥x_{1}∥_{W} = ∥x_{2}∥_{W} = 1. □

Proof of Proposition. Define a sublinear functional q on d^{∗}(w, 1) as

q(x) = lim sup_{n} \frac{∑_{k=1}^{n} x^{∗}(k)}{W (n)},

for each x ∈ d^{∗}(w, 1). Now, set p(x) = q(x^{+}). Then p is also a sublinear functional satisfying p(x) ≤ ∥x∥_{W} for all x ∈ d^{∗}(w, 1). By the previous lemma, there is a decomposition x = x_{1} + x_{2} such that both |x_{1}| ∧ |x_{2}| = 0 and ∥x_{1}∥_{W} = ∥x_{2}∥_{W} = 1 hold.

Define a linear functional ϕ_{1} so that ϕ_{1}(λ|x_{1}|) = λ on span{|x_{1}|}, where λ ∈ ℝ. Then ϕ_{1}(y) ≤ p(y) on span{|x_{1}|}. Applying now the Hahn-Banach extension theorem, there is an extension ϕ_{1} to d^{∗}(w, 1) such that ϕ_{1}(x) ≤ p(x) holds for all x ∈ d^{∗}(w, 1). Then ϕ_{1} is a positive linear functional with norm one. Indeed, for each x ∈ d^{∗}(w, 1) with x ≥ 0,

−ϕ_{1}(x) = ϕ_{1}(−x) ≤ p(−x) = q((−x)^{+}) = q(0) = 0,

and so ϕ_{1}(x) ≥ 0. Hence

|ϕ_{1}(x)| ≤ ϕ_{1}(|x|) ≤ p(|x|) ≤ ∥x∥_{W} .

Notice also that ϕ_{1}(|x_{1}|) = 1.

Similarly, we can obtain a positive linear functional ϕ_{2} on d^{∗}(w, 1) satisfying ∥ϕ_{2}∥ = 1 and ϕ_{2}(|x_{2}|) = 1.

Then for each i = 1, 2,

1 = ϕ_{i}(|x_{i}|) ≤ ϕ_{i}(|x_{1}|) + ϕ_{i}(|x_{2}|) = ϕ_{i}(|x|) ≤ 1,

which implies that ϕ_{i}(|x_{j}|) = δ_{i}(j), where δ_{i} is a Dirac-delta function at i = 1, 2, and ϕ_{i}(|x|) = 1 for i = 1, 2. Define an isometry T on d^{∗}(w, 1) as

T y = {sign(x(n))y(n)}_{n=1}^{∞},

for each y ∈ d^{∗}(w, 1). Letting now ψ_{i} = ϕ_{1} ∘ T, i = 1, 2, we obtain two different norm-one supporting functionals at x, which finishes the proof. □

Theorem 1.9. An element x in S_{d^{∗}(w,1)} is a smooth point of B_{d^{∗}(w,1)} if and only if there is m ∈ ℕ such that

\frac{∑_{k=1}^{m} x^{∗}(k)}{W (m)} = 1 > sup_{n≠m} \{\frac{∑_{k=1}^{n} x^{∗}(k)}{W (n)}\} .

Proof. The necessity follows from Propositions 1.3 and 1.7. In order to show the sufficiency, suppose the inequality in the hypothesis is satisfied. Then x^{∗}(m) > x^{∗}(m + 1) and there exists a permutation σ on ℕ such that |x(σ(k))| = x^{∗}(k) for k = 1, …, m. Let ϕ = ψ + ξ be a norm-one supporting functional at x, where ψ ∈ d(w, 1) and ξ is singular. Setting if ξ ≠ 0, then

t = max{σ(k) : k = 1, …, m},

it is clear that ∥xχ_{ℕ\{1,...,t}}∥_{W} < 1. Therefore

∥ξ∥ = ξ(x) = ξ(xχ_{ℕ\{1,...,t}}) < ∥ξ∥,

and so it is a contradiction. Hence ξ = 0. By Proposition 1.4, norm-one supporting functional ψ ∈ d(w, 1) at x is unique, and the proof is done. □

Below we provide a characterization of smooth points in the Lorentz space d(w, 1).

Theorem 1.10. An element x of the unit sphere of d(w, 1) is a smooth point if and only if supp x is infinite and the following condition is satisfied:

(1.2)  Whenever there is k ≥ 1 such that w(k) > w(k + 1), we get x^{∗}(k) > x^{∗}(k + 1).

Proof. We shall show first the necessity. It is easy to see that if supp x is finite, then there are infinitely many supporting functionals at x. Thus assume that supp x is infinite. We shall show that if x^{∗}(k_{0}) = x^{∗}(k_{0} +1) and w(k_{0}) > w(k_{0} +1) for some k_{0} ∈ ℕ, then we can obtain two different supporting functionals at x. It well known that there is a 1 − 1 and onto mapping σ : ℕ → supp x such that x^{∗} = |x ∘ σ|. Choose two sequences y_{1} and y_{2} defined by

y_{1} = \{ \begin{matrix}sign(x(k)) · w(σ^{−1}(k)), & for k ∈ supp x; \\ 0, & otherwise, \end{matrix}

y_{2} = \{ \begin{matrix}y_{1}(k), & k ≠ σ(k_{0}) and k ≠ σ(k_{0} + 1); \\ sign(x(σ(k_{0}))) · \frac{w(k_{0})+w(k_{0}+1)}{2}, & k = σ(k_{0}); \\ sign(x(σ(k_{0} + 1))) · \frac{w(k_{0})+w(k_{0}+1)}{2}, & k = σ(k_{0} + 1). \end{matrix}

Notice that ∥y_{1}∥_{W} = ∥y_{2}∥_{W} = 1. It is also easy to check that y_{1} and y_{2} are two different supporting functionals at x.

Now let x ∈ S_{d(w,1)} satisfy condition (1.2) and let y ∈ S_{d^{∗}(w,1)} be a supporting functional of x. Then

1 = ∑_{k=1}^{∞} x(k)y(k) = ∑_{k=1}^{∞} sign(x(σ(k))) · x^{∗}(k)y(σ(k)),

where x^{∗} = |x ∘ σ|. Taking S(n) = ∑_{k=1}^{n} sign(x(σ(k))) · y(σ(k)) and S′(n) = ∑_{k=1}^{n} y^{∗}(k) we have S(n) ≤ S′(n) ≤ W (n) for every n ∈ ℕ, in view of the Hardy-Littlewood inequality and ∥y∥_{W} = 1. We shall show by induction that for every n ∈ ℕ,

y(σ(n)) = sign(x(σ(n))) · y^{∗}(n) = sign(x(σ(n))) · w(n).

Since lim_{n→∞} x^{∗}(n) = 0, there is m such that m = max{k ≥ 1 : x^{∗}(1) = x^{∗}(k)}. If S(m) < W (m), then by the summation by parts, we get

1 = ∑_{i=1}^{∞} x^{∗}(i)y(σ(i)) sign(x(σ(i))) (1.3)

= ∑_{i=1}^{m}(x^{∗}(i) − x^{∗}(i + 1))S(i) + lim_{l→∞} { ∑_{i=m+1}^{l−1}(x^{∗}(i) − x^{∗}(i + 1))S(i) + x^{∗}(l)S(l)}

< ∑_{i=1}^{m}(x^{∗}(i) − x^{∗}(i + 1))W (i) + lim_{l→∞} { ∑_{i=m+1}^{l−1}(x^{∗}(i) − x^{∗}(i + 1))W (i) + x^{∗}(l)W (l)}

= ∑_{i=1}^{∞} x^{∗}(i)w(i) = 1,

which is a contradiction. So S(m) = W (m). Notice that sup_{k} |y(k)| ≤ y^{∗}(1) ≤ w(1). Since x^{∗}(i) = x^{∗}(j) for every 1 ≤ i, j ≤ m, we also have w(1) = ⋯ = w(m) by the assumption (1.2). This and S(m) ≤ S′(m) ≤ W (m) imply that

S(m) = S′(m) = W (m) = m · w(1),

and sign(x(σ(k))) · y(σ(k)) = y^{∗}(k) = w(k) for k = 1, …, m. Hence

y(σ(1)) = sign(x(σ(1))) · y^{∗}(1) = sign(x(σ(1))w(1).

For the inductive step assume that for every k ≤ n, we have

y(σ(k)) = sign(x(σ(k))) · y^{∗}(k) = sign(x(σ(k))) · w(k).

Let now m = max{k ≥ n + 1 : x^{∗}(n + 1) = x^{∗}(k)}. If S(m) < W (m), then the inequality (1.3) yields a contradiction, and so S(m) = W (m). By the induction hypothesis and by (1.2), we get S(m) − S(n) = W (m) − W (n) = (m − n)w(n + 1) and w(n + 1) ≥ sup_{j≥n+1} |y(σ(j))| = sup_{j≥n+1} y^{∗}(j). Thus for n + 1 ≤ j ≤ m

y(σ(j)) = sign(x(σ(j))) · y^{∗}(j) = sign(x(σ(j))) · w(j).

This completes the induction and the uniqueness of the supporting functional at x has been proved. □

2. Extreme points

A Banach space (X, ∥ · ∥), a collection of real sequences, is said to be a r.i. sequence space if for any x = {x(n)} ∈ X we have ∥x∥ = ∥x^{∗}∥, and for any y = {y(n)} such that |y(n)| ≤ |x(n)| for every n ∈ ℕ, we have that y ∈ X and ∥y∥ ≤ ∥x∥. It is clear that all spaces d(w, 1), d^{∗}(w, 1) and d_{∗}(w, 1) are r.i. sequence spaces.

Proposition 2.1. Let (X, ∥ · ∥) be a r.i. sequence space and x ∈ S_{X} be such that its distribution d_{x} is a finite valued function.

(i) If supp x is finite or equal to ℕ, then x is an extreme point of B_{X} if and only if x^{∗} is an extreme point.

(ii) If x is an extreme point of B_{X}, then x^{∗} is also an extreme point of B_{X}. If in addition X is strictly monotone, then the converse statement is also satisfied.

Proof. Since d_{x}(θ) < ∞ for all θ > 0, lim_{n} x^{∗}(n) = 0 and there exists a 1-1 and onto mapping σ : ℕ → supp x such that for all n ∈ ℕ,

(2.1)  x^{∗}(n) = |x(σ(n))| = λ_{n}x(σ(n)),

where λ_{n} = sign x(σ(n)).

(i). Under the assumptions, σ is a permutation of ℕ, and then the operator

T y(n) = λ_{n}y(σ(n)), y ∈ X,

is an isometry on X such that T x = x^{∗}. We get the conclusion immediately since T preserves extreme points.

(ii). Suppose x^{∗} ∈ S_{X} is not an extreme point of B_{X}. Then there exist y, z ∈ S_{X} such that y ≠ z and x^{∗} = (y + z)/2. Hence

x(σ(n)) = \frac{λ_{n}y(n) + λ_{n}z(n)}{2},

and so we get for every n ∈ supp x,

x(n) = \frac{β_{n}y(σ^{−1}(n)) + β_{n}z(σ^{−1}(n))}{2},

where σ^{−1} : supp x → ℕ is 1-1 and onto mapping and β_{n} = sign x(n). Thus setting

\overline{y}(n) = \{ \begin{matrix}β_{n}y(σ^{−1}(n)), & n ∈ supp x; \\ 0, & otherwise, \end{matrix}

\overline{z}(n) = \{ \begin{matrix}β_{n}z(σ^{−1}(n)), & n ∈ supp x; \\ 0, & otherwise, \end{matrix}

we have that \overline{y} and \overline{z} are equimeasurable with y and z, respectively. Hence ∥x∥ = ∥\overline{y}∥ = ∥\overline{z}∥. Moreover, x = (\overline{y}+\overline{z})/2 and \overline{y} ≠ \overline{z}, since there exists m ∈ supp x such that y(σ^{−1}(m)) ≠ z(σ^{−1}(m)) by the assumption that y ≠ z. Thus x is not extreme point of B_{X} as well.

Suppose now that X is strictly monotone, and let x be not extreme point of B_{X}. Then there exist y, z ∈ S_{X} such that y ≠ z, and for all n ∈ ℕ,

x(n) = \frac{y(n) + z(n)}{2}.

It follows that supp y∪supp z ⊂ supp x. Indeed, if there is m ∈ ℕ such that x(m) = 0 and y(m) ≠ 0, then z(m) ≠ 0 and setting ỹ = yχ_{supp x} and z̃ = zχ_{supp x}, we have ∥ỹ∥ < ∥y∥ and ∥z̃∥ < ∥z∥, by strict monotonicity of X. However, x = (ỹ+z̃)/2 and so ∥x∥ ≤ (∥ỹ∥+∥z̃∥)/2 < (∥y∥+∥z∥)/2 = ∥x∥; a contradiction.

By (2.1), for all n ∈ ℕ,

x^{∗}(n) = \frac{λ_{n}y(σ(n)) + λ_{n}z(σ(n))}{2}.

Since supports of y and z are included in supp x, |y ∘ σ| and |z ∘ σ| are equimeasurable with y and z, respectively. It is also clear that they are different. Thus taking y_{0}(n) = λ_{n}y(σ(n)) and z_{0}(n) = λ_{n}z(σ(n)) we have that ∥y_{0}∥ = ∥z_{0}∥ = 1, y_{0} ≠ z_{0}, and x^{∗} = (y_{0} + z_{0})/2. Thus x^{∗} is not an extreme point, which completes the proof.

□

Theorem 2.2. An element x ∈ S_{d^{∗}(w,1)} is an extreme point of B_{d^{∗}(w,1)} if and only if x^{∗} = w.

Proof. Recall that whenever x ∈ d^{∗}(w, 1) then d_{x}(θ) < ∞ for every θ > 0. Assume that x^{∗} ≠ w, where ∥x∥_{W} = 1. In view of Proposition 2.1 (ii) it is enough to show that if x^{∗} is an extreme point, then x^{∗} = w. We shall prove it by use of induction. Suppose, on the contrary, that x^{∗}(1) < w(1). We have three possible cases.

Case (1). Suppose first that x^{∗}(1) > x^{∗}(2) > x^{∗}(3) holds. Then choose an ϵ > 0 such that x^{∗}(1) + ϵ < w(1)

x^{∗}(1) + ϵ > x^{∗}(2) − ϵ > x^{∗}(3), and
x^{∗}(1) − ϵ > x^{∗}(2) + ϵ > x^{∗}(3).

Then by setting

y = (x^{∗}(1) + ϵ, x^{∗}(2) − ϵ, x^{∗}(3), …) and
z = (x^{∗}(1) − ϵ, x^{∗}(2) + ϵ, x^{∗}(3), …),

it is easy to see that ∥y∥_{W} , ∥z∥_{W} ≤ 1 and x^{∗} = \frac{y+z}{2} hold. Hence it is a contradiction to the fact that x^{∗} is an extreme point.

Case (2). Suppose that x^{∗}(1) = x^{∗}(2) = ⋯ = x^{∗}(m) > x^{∗}(m + 1) for some m ≥ 2. Then for every 1 ≤ k < m,

kx^{∗}(k) < w(1) + ⋯ + w(k).

Indeed, if kx^{∗}(1) = w(1) + ⋯ + w(k) for some 1 ≤ k < m, then w(k) < x^{∗}(1). Hence

mx^{∗}(1) = w(1) + ⋯ + w(k) + (m − k)x^{∗}(1) > w(1) + ⋯ + w(m),

which is a contradiction to the fact that ∥x∥_{W} = 1.

Now choose ϵ > 0 such that (m − 1)x^{∗}(1) + ϵ < w(1) + ⋯ + w(m − 1) and x^{∗}(1) − ϵ > x^{∗}(m + 1) hold. Setting

y = (x^{∗}(1) + ϵ, x^{∗}(1), …, x^{∗}(1) − ϵ, x^{∗}(m + 1), …)
z = (x^{∗}(1) − ϵ, x^{∗}(1), …, x^{∗}(1) + ϵ, x^{∗}(m + 1), …)

Then it is easy to see that ∥y∥_{W} = ∥z∥_{W} ≤ 1 and x^{∗} = \frac{y+z}{2} . This is also a contradiction.

Case (3). Suppose that x^{∗}(1) > x^{∗}(2) = x^{∗}(3) = ⋯ = x^{∗}(m) > x^{∗}(m + 1) for some m ≥ 3. Then it is easy to see that for 1 ≤ k < m,

x^{∗}(1) + (k − 1)x^{∗}(2) < w(1) + ⋯ + w(k).

Indeed, if x^{∗}(1) + (k − 1)x^{∗}(2) = w(1) + ⋯ + w(k) for some 2 ≤ k < m, then (k − 1)x^{∗}(2) ≥ w(1) − x^{∗}(1) + (k − 1)w(k) > (k − 1)w(k). That is, x^{∗}(2) > w(k). Hence

x^{∗}(1) + (m − 1)x^{∗}(2) = w(1) + ⋯ + w(k) + (m − k)x^{∗}(2)
> w(1) + ⋯ + w(k) + w(k + 1) + ⋯ + w(m).

It is impossible since x^{∗} has norm one.

Then choose ϵ > 0 such that x^{∗}(1) > x^{∗}(2) + ϵ, x^{∗}(2) − ϵ > x^{∗}(m + 1) and

x^{∗}(1) + (m − 2)x^{∗}(2) + ϵ < w(1) + ⋯ + w(m − 1).

Letting

y = (x^{∗}(1), x^{∗}(2) + ϵ, …, x^{∗}(2) − ϵ, x^{∗}(m + 1), …), and
z = (x^{∗}(1), x^{∗}(2) − ϵ, …, x^{∗}(2) + ϵ, x^{∗}(m + 1), …),

Then ∥y∥_{W} = ∥z∥_{W} ≤ 1 and x^{∗} = \frac{y+z}{2} , which is impossible since x^{∗} is an extreme point.

Therefore, we show that x^{∗}(1) = w(1). For the use of induction, suppose that x^{∗}(k) = w(k) for 1 ≤ k ≤ n. If x^{∗}(n + 1) < w(n + 1), then exactly same argument for cases (1), (2) and (3), shows that it is a contradiction. Hence, x^{∗}(n + 1) = w(n + 1). Therefore if x^{∗} is an extreme point, then x^{∗} = w.

Now we show that x is an extreme point of the unit ball of d^{∗}(w, 1) if x^{∗} = w. Let A = supp x and β : A → A be a 1-1 and onto mapping such that the sequence {|x(β(n))|} is decreasing on A that is |x(β(j))| ≤ |x(β(i))| whenever i < j. Consider γ : ℕ → ℕ defined as γ(n) = β(n) for n ∈ A and γ(n) = n for n ∉ A. Then

(T y)(n) = sign x(γ(n))y(γ(n))

is a linear isometry on d^{∗}(w, 1) such that

(T x)(n) = |x(γ(n))|.

Hence x is an extreme point whenever |x ∘ γ| is an extreme point. In view of that we can assume that x ∈ S_{d^{∗}(w,1)} is non-negative, decreasing whenever restricted to its support A, and x^{∗} = w. Now let y, z ∈ S_{d^{∗}(w,1)} be such that for all n ∈ ℕ,

x(n) = \frac{y(n) + z(n)}{2}.

In view of the assumptions on x, letting A = {n_{1}, n_{2}, … }, where n_{1} < n_{2} < …, we have

x(n_{k}) = w(k), k ∈ ℕ.

We shall show first that

y(n_{k}) = z(n_{k}) = x(n_{k}), k ∈ ℕ.

Let further

ỹ = yχ_{A} and z̃ = zχ_{A}.

We shall apply mathematical induction. Let n = n_{1}. Assuming that |y(n_{1})|/w(1) > 1 we get a contradiction since

1 ≥ \frac{ỹ^{∗}(1)}{w(1)} ≥ \frac{|y(n_{1})|}{w(1)} > 1.

Thus |y(n_{1})| ≤ w(1) and similarly |z(n_{1})| ≤ w(1). Then y(n_{1}) = z(n_{1}) = w(1) = x(n_{1}). Notice also that we have ỹ^{∗}(1) = y(n_{1}) = z̃^{∗}(1) = z(n_{1}). Assume now that

ỹ^{∗}(i) = y(n_{i}) = w(i) = z(n_{i}) = z̃^{∗}(i)

for all i = 1, …, m and some m > 1. Then

\frac{∑_{i=1}^{m+1} ỹ^{∗}(i)}{W (m + 1)} = \frac{∑_{i=1}^{m} w(i) + ỹ^{∗}(m + 1)}{W (m + 1)} ≤ 1,

and so ỹ^{∗}(m + 1) ≤ w(m + 1). Hence for all i ≥ m + 1,

ỹ^{∗}(i) ≤ w(m + 1),

and since there exists j ≥ m + 1 such that ỹ^{∗}(j) = y(n_{m+1}), we have

|y(n_{m+1})| ≤ w(m + 1).

Analogously we can show that

|z(n_{m+1})| ≤ w(m + 1),

and so

ỹ^{∗}(m + 1) = y(n_{m+1}) = z(n_{m+1}) = z̃^{∗}(m + 1) = w(m + 1) = x(n_{m+1}).

This completes the induction. It remains to show that y(i) = z(i) = 0 for all i ∉ A. Notice that ỹ^{∗} ≤ y^{∗} and we have for every n ≥ 1,

1 ≥ \frac{∑_{i=1}^{n} y^{∗}(i)}{W (n)} ≥ \frac{∑_{i=1}^{n} ỹ^{∗}(i)}{W (n)} = 1.

Hence ỹ^{∗} = y^{∗} and in view of lim_{n} y^{∗}(n) = 0 we have y(i) = 0. Similarly z(i) = 0 for every i ∉ A. □

Remark 2.3. The extreme points for the unit ball of finite dimensional Marcinkiewicz sequence spaces are characterized in [8].

Lemma 2.4. The Lorentz space d(w, 1) is strictly monotone.

Proof. If x < y, x, y ∈ d(w, 1), then x^{∗} ≤ y^{∗} and there exists m ∈ ℕ such that x^{∗}(m) < y^{∗}(m). This results from the simple fact that d_{y} is finite. It follows that ∥x∥_{w,1} < ∥y∥_{w,1}. □

The next result follows immediately from Proposition 2.1 and Lemma 2.4.

Corollary 2.5. An element x ∈ S_{d(w,1)} is extreme point of the ball B_{d(w,1)} if and only if x^{∗} is an extreme point of B_{d(w,1)}.

Theorem 2.6. An element x ∈ S_{d(w,1)} is extreme point of the ball B_{d(w,1)} if and only if there exists n_{0} ∈ ℕ such that x^{∗}(i) = 1/W (n_{0}) for i = 1, ..., n_{0}, x^{∗}(i) = 0 for i > n_{0} and w(1) > w(n_{0}), provided n_{0} > 1.

Proof. In view of Corollary 2.5 we assume that x = x^{∗}. Suppose first that x ∈ S_{d(w,1)} is an extreme point of B_{d(w,1)} and let

n_{0} = sup{n ∈ ℕ : x(n) = x(1)}.

Since d(w, 1) ⊂ c_{0}, it is clear that n_{0} ∈ ℕ. We shall show that x(n_{0} + 1) = 0. Let for a contrary x(n_{0} + 1) > 0 and set

n_{1} = max{n ∈ ℕ : x(n) = x(n_{0} + 1)}.

Setting d = min{x(1) − x(n_{0} + 1), x(n_{0} + 1) − x(n_{1} + 1)}, we have d > 0. Fix b > 0 such that

b(1 + \frac{W (n_{0})}{W (n_{1}) − W (n_{0})}) < d.

Define

y =(x(1) − b, …, x(n_{0}) − b, x(n_{0} + 1) + \frac{bW (n_{0})}{W (n_{1}) − W (n_{0})}, …,

x(n_{1}) + \frac{bW (n_{0})}{W (n_{1}) − W (n_{0})}, x(n_{1} + 1), x(n_{1} + 2) …)

and

z =(x(1) + b, …, x(n_{0}) + b, x(n_{0} + 1) − \frac{bW (n_{0})}{W (n_{1}) − W (n_{0})}, …,

x(n_{1}) − \frac{bW (n_{0})}{W (n_{1}) − W (n_{0})}, x(n_{1} + 1), x(n_{1} + 2) …).

Note that y ≠ z and x = (y + z)/2. By the choice of b and d, y = y^{∗} and z = z^{∗}. Hence

∥y∥_{w,1} = ∑_{j=1}^{∞} y(j)w(j) = ∑_{j=1}^{∞} x(j)w(j) − bW (n_{0}) + \frac{bW (n_{0})}{W (n_{1}) − W (n_{0})} ∑_{j=n_{o}+1}^{n_{1}} w(j)

= ∑_{j=1}^{∞} x(j)w(j) = 1.

Analogously, we can show that ∥z∥_{w,1} = 1. It contradicts the assumption that x is an extreme point and consequently x(n_{0} + 1) = 0, as required. If n_{0} > 1 and w(1) = w(n_{0}), define for 0 < b < x(1),

y_{b} = (x(1) + b, x(2), …, x(n_{0} − 1), x(n_{0}) − b, x(n_{0} + 1), …)

and

z_{b} = (x(1) − b, x(2), …, x(n_{0} − 1), x(n_{0}) + b, x(n_{0} + 1), …).

It is easy to see that ∥y_{b}∥_{w,1} = ∥z_{b}∥_{w,1} = 1, z_{b} ≠ y_{b} and x = (y_{b} + z_{b})/2. So, x is not an extreme point, which is a contradiction. Thus we showed as required that if n_{0} > 1, then w(1) > w(n_{0}),.

Now assume x ∈ d(w, 1), and n ∈ ℕ are such that x(i) = 1/W (n) for i = 1, ..., n, x(i) = 0 for i > n and w(1) > w(n) if n > 1. We shall show that x is an extreme point of B_{d(w,1)}. If n = 1 this is an easy consequence of Lemma 2.4. Suppose that n > 1. Let x = (y + z)/2 with ∥y∥_{w,1} = ∥z∥_{w,1} = 1 and y ≠ z. By Lemma 2.4, y(i) = z(i) = 0 for i > n. Indeed, if y(i) ≠ 0 for some i > n, then z(i) = −y(i). Defining y^{1} = (y(1), ...y(n), 0, ...), z^{1} = (z(1), ...z(n), 0, ...), we have that x = (y^{1} + z^{1})/2. But by strict monotonicity, ∥z^{1}∥_{w,1} < ∥z∥_{w,1} = 1 and ∥y^{1}∥_{w,1} < ∥y∥_{w,1} = 1 and so ∥x∥_{w,1} < 1; a contradiction. Define

I_{1} = {i = 1, …, n : y(i) > 1/W (n)},
I_{2} = {i = 1, …, n : y(i) = 1/W (n)},

and

I_{3} = {i = 1, …, n : y(i) < 1/W (n)}.

By the strict monotonicity, we have y, z ≥ 0. Otherwise, we can choose ỹ, z̃ such that |ỹ| < |y|, |z̃| < |z| and x = \frac{ỹ+z̃}{2}. Hence ∥ỹ∥_{w,1} < ∥y∥_{w,1} < 1 and ∥z̃∥_{w,1} < ∥z∥_{w,1} < 1, which is a contradiction to the fact that x = \frac{y+z}{2} .

Let for i = 1, 2, 3, k_{i} = card I_{i}. Since d(w, 1) is strictly monotone, y ≠ x and ∥y∥_{w,1} = 1, so k_{1} > 0 and k_{3} > 0. Without loss of generality, permuting coordinates of y and z, if necessary, we can assume that y^{∗} = y. Since ∥y∥_{w,1} = 1,

y(1)w(1) + ⋯ + y(n)w(n) = 1.

By ∥z∥_{w,1} = 1 and x = (y + z)/2 we also have

y(1)w(n) + ⋯ + y(n)w(1) = 1.

Moreover, by the assumption w(1) > w(n), and by Hardy-Littlewood inequality, we have

1 = y(1)w(1) + ⋯ + y(n)w(n)
> y(1)w(n) + y(n)w(1) + (y(2)w(2) + … + y(n − 1)w(n − 1))
≥ y(1)w(n) + y(n)w(1) + (y(2)w(n − 1) + ⋯ + y(n − 1)w(2))
= y(1)w(n) + ⋯ + y(n)w(1) = 1,

which is a contradiction. The proof is complete. □

3. Applications

In this section we shall study the relationship between the existence and one-complemented subspaces of Lorentz space d(w, 1), applying the characterization of smooth points in d(w, 1) (Theorem 1.10) and extreme points in its dual d^{∗}(w, 1) (Theorem 2.2).

Let X be a Banach space and let C ⊂ X be a non-empty set. A continuous surjective mapping P : X → C is called a projection onto C, whenever P |_{C} = Id, that is P^{2} = P.

Given a subspace V of a Banach space X, by P (X, V) we denote the set of all linear, bounded projections from X onto V. Recall that a closed subspace V of a Banach space X is called one-complemented if there exists a norm one projection P ∈ P (X, V). Setting for each x ∈ X,

M_{C}(x) = {z ∈ X : ∥z − c∥ ≤ ∥x − c∥ for any c ∈ C},

it is clear that x ∈ M_{C}(x) for every x ∈ X and M_{C}(c) = {c} for every c ∈ C. Letting Min C be a subset of X consisting an element x such that M_{C}(x) = {x}, we say that C ⊂ X is optimal if Min C = C. Observe that for any C ⊂ X, C ⊂ Min C.

This notion has been introduced by Beauzamy and Maurey in [3], where basic properties concerning optimal sets can be found.

A set C ⊂ X is called an existence set of best coapproximation (existence set for brevity), if for any x ∈ X, R_{C}(x) ≠ ∅, where

(3.1)  R_{C}(x) = {d ∈ C : ∥d − c∥ ≤ ∥x − c∥ for any c ∈ C}.

It is clear that any existence set is an optimal set. The converse, in general, is not true. However, by [3, Prop. 2], if X is one-complemented in X^{∗∗} and strictly convex, then any optimal subset of X is an existence set in X, which, in particular, holds true for strictly convex spaces X, such that X = Z^{∗} for some Banach space Z.

Existence and optimal sets have been studied by many authors from different points of view, mainly in the context of approximation theory and functional analysis (see e.g. [2, 3, 4, 5, 10, 12, 13, 18, 21, 26]). There is also a large literature concerning one-complemented subspaces (see e.g. a survey paper [24] and a recent paper [16]).

It is obvious that any one-complemented subspace is an existence set. However the converse, in general, is not true. By a deep result of Lindenstrauss [21] there exist a Banach space X and a subspace V of X, with codim V = 2, such that:

(a) V is one-complemented in any Y, where Y ⊃ V is a hyperplane in X i.e Y = f^{−1}({0}) for some f ∈ X^{∗} \ {0}.

(b) V is not one-complemented in X.

This fact together with the simple observation stated as Lemma 3.1 below, gives an example of a subspace being an existence set which is not one-complemented.

Lemma 3.1. Let X be a Banach space and let V ⊂ X, V ≠ {0} be a linear subspace. Then V is an existence set in X if and only if for any x ∈ X \ V, there exists P_{x} ∈ P (Z_{x}, V) with ∥P_{x}∥ = 1. Here Z_{x} = V ⊕ [x], where [x] denotes the linear span generated by x.

Proof. Assume that for any x ∈ X \ V there exists P_{x} ∈ P (Z_{x}, V), ∥P_{x}∥ = 1. Fix z ∈ Z_{x} and v ∈ V. Note that

∥P_{x}z − v∥ = ∥P_{x}(z − v)∥ ≤ ∥z − v∥.

Hence P_{x}z ∈ R_{V} (z) and so V is an existence set in X. Now assume that V is an existence set in X and fix x ∈ X \ V. Take any d ∈ R_{V} (x). Since any z ∈ Z_{x} can be uniquely expressed as z = αx + v for some v ∈ V and α ∈ ℝ, we can define P_{x} : Z_{x} → V by

P_{x}z = αd + v.

It is easy to see that P_{x} ∈ P (Z_{x}, V). To show that ∥P_{x}∥ = 1, fix y = αx + v ∈ Z_{x}, with α ≠ 0. Since d ∈ R_{V} (x),

∥P_{x}y∥ = ∥αd + v∥ = |α|∥d + v/α∥ ≤ |α∥x + v/α∥ = ∥αx + v∥ = ∥y∥,

which completes the proof. □

In [3] the following result has been proved.

Theorem 3.2. ([3], Prop. 5). Let V ≠ {0} be a linear subspace of a smooth, reflexive and strictly convex Banach space X. If V is an optimal set then V is one-complemented in X. If X is a smooth Banach space, then any subspace of X which is an existence set is one-complemented. Moreover, in both cases a norm-one projection from X onto V is uniquely determined.

We shall show here that the above result can be true in spaces that are not smooth. We will prove that any subspace of d(w, 1) which is an existence set must be one-complemented, which cannot be deduced from Theorem 3.2 because by Theorem 1.10, d(w, 1) is not a smooth space. Just recently [20], a similar result has been proved for spaces c_{0}, c, ℓ_{1} and a large class of Musielak-Orlicz sequence spaces equipped with the Luxemburg norm. These facts provide a partial answer to the question stated in [3], p. 125 concerning generalization of Theorem 3.2 to non-smooth case.

One of the main tools in our investigations, stated below, has been recently proved in [20].

Theorem 3.3. Let X be a Banach space and let V ⊂ X, be a linear subspace. Assume that V is an existence set and V ≠ {0}. Put

G_{V} = {v ∈ V \ {0} : there exists a unique f ∈ S_{X^{∗}} : f(v) = ∥v∥}.

Assume that the norm closure of G_{V} in X is equal to V. Then there exists a unique projection P ∈ P (X, V) such that ∥P ∥ = 1. Consequently, V is one-complemented in X.

For further reference we state the next well-known result.

Lemma 3.4. Let X, Y be two Banach spaces, V ⊂ X be a linear subspace and let T : X → Y be a linear isometry. Then V is an existence set in X if and only if T (V) is an existence set in T (X). Also V is one-complemented in X if and only if T (V) is one complemented in T (X).

For n ∈ ℕ and a decreasing sequence of positive numbers {w(1), ..., w(n)} define a finite dimensional Lorentz space

d^{n}(w, 1) = (ℝ^{n}, ∥ · ∥_{w,1}),

where

∥x∥_{w,1} = ∑_{j=1}^{n} x^{∗}(j)w(j).

Before we state the main result we shall prove several auxiliary lemmas.

Lemma 3.5. Let {C_{j}}_{j∈ℕ} be a family of finite, nonempty subsets of ℕ such that C_{i} ∩ C_{j} = ∅ for i ≠ j. Define for j ∈ ℕ

X_{C_{j}} = {x ∈ d(w, 1) : x(i) = x(k) for any i, k ∈ C_{j}}.

Let

X_{C} = ⋂_{j=1}^{∞} X_{C_{j}}.

Then X_{C} is one-complemented in d(w, 1). The same result applies to d^{n}(w, 1). In this case we consider a finite family of nonempty, pairwise disjoint subsets of {1, ..., n}.

Proof. Let for j ∈ ℕ, C_{j} = {i_{1}, ...i_{k_{j}}}, where k_{j} = card C_{j}.

Set for x ∈ d(w, 1), j ∈ ℕ, P_{j}x = (z(1), …, z(n), …), where z(i) = (∑_{l∈C_{j}} x(l))/k_{j} if i ∈ C_{j}, and z(i) = x(i) in the opposite case. It is clear that P_{j} ∈ P (d(w, 1), X_{C_{j}}). We also have that ∥P_{j}∥ = 1. Indeed, since for any permutation σ : ℕ → ℕ, the mapping T_{σ} : d(w, 1) → d(w, 1) given by T_{σ}x = x ∘ σ is a linear, surjective isometry of d(w, 1). So by Lemma 3.4, we can assume that C_{j} = {1, ..., k_{j}}. Let x ∈ S_{d(w,1)}, and set for l = 2, ..., k_{j}

x^{l} = (x(l), x(l + 1), …, x(k_{j}), x(1), …, x(l − 1), x(k_{j} + 1), …).

Then x + ∑_{l=2}^{k_{j}} x^{l} = k_{j}(P_{j}x), and

∥P_{j}x∥_{w,1} = ∥(x + ∑_{l=2}^{k_{j}} x^{l})/k_{j}∥_{w,1} ≤ (∥x∥_{w,1} + ∑_{l=2}^{k_{j}} ∥x^{l}∥_{w,1})/k_{j} = 1,

since ∥x^{l}∥_{w,1} = ∥x∥_{w,1} = 1 for l = 2, ..., k_{j}. Thus ∥P_{j}∥ = 1. Now define for j ∈ ℕ

X_{j} = ⋂_{m=1}^{j} X_{C_{m}},

and

Q_{j} = P_{1} ∘ P_{2} ∘ ⋯ ∘ P_{j}.

Since C_{i} ∩ C_{k} = ∅, for i ≠ k, so Q_{j} ∈ P (d(w, 1), X_{j}). By the above reasoning, ∥Q_{j}∥ = 1.

Now, fix x ∈ d(w, 1). Define Qx = ((Qx)(1), ...(Qx)(n), ...), where (Qx)(i) = x(i) if i ∉ ⋃_{j∈ℕ} C_{j} and (Qx)(i) = ∑_{l∈C_{j}} x(l))/k_{j} if i ∈ C_{j}. Since C_{i} ∩ C_{j} = ∅ for i ≠ j,

(Qx)(i) = lim_{j}(Q_{j}x)(i),

for any i ∈ ℕ. Now we show that Qx ∈ d(w, 1) for any x ∈ d(w, 1). Indeed, for any x ∈ B_{d(w,1)} and any j ∈ ℕ, we have ∥Q_{j}x∥ ≤ 1 since ∥Q_{j}∥ = 1. In view of d(w, 1) = (d_{∗}(w, 1))^{∗}, and the fact that d_{∗}(w, 1) is separable, the weak* topology on B_{d(w,1)} is metrizable. Thus by the Banach - Alaoglu theorem, there exists a subsequence {j_{k}} and Rx ∈ B_{d(w,1)}, with Q_{j_{k}}x → Rx weakly^{∗} in d(w, 1). In particular for any i ∈ ℕ we have

(Rx)(i) = lim_{k}(Q_{j_{k}}x)(i).

This shows that Qx = Rx, and consequently Qx ∈ d(w, 1). Note also that Qx ∈ X_{C} and for any x ∈ X_{C}, Qx = x. Since Qx = Rx, Qx ∈ B_{d(w,1)}, for any x ∈ B_{d(w,1)}. Thus Q is a linear projection from d(w, 1) onto X_{C} with ∥Q∥ = 1, which completes the proof. The case of d^{n}(w, 1) can be proved in the similar way. □

The next lemma is well-known but for the sake of completeness we include its proof here.

Lemma 3.6. Let X be a Banach space and let x ∈ X. Define

D(x) = {f ∈ B_{X^{∗}} : f(x) = ∥x∥}.

Then

∅ ≠ ext D(x) ⊂ ext B_{X^{∗}}.

Proof. If x = 0, then D(x) = B_{X^{∗}} which shows our claim. So assume x ≠ 0. By the Hahn-Banach Theorem, D(x) ≠ ∅. Note that D(x) is a convex, weakly^{∗} closed subset of B_{X}. By the Banach-Alaoglu and the Krein-Milman Theorems, ext D(x) ≠ ∅. We show that ext D(x) ⊂ ext B_{X^{∗}}. Let f ∈ ext D(x). Assume f = (f_{1} + f_{2})/2 and f_{1}, f_{2} ∈ S_{X^{∗}}. Hence

∥x∥ = f(x) = (f_{1}(x) + f_{2}(x))/2.

Since ∥f_{1}∥ = ∥f_{2}∥ = 1, f_{1}(x) = f_{2}(x) = ∥x∥, which gives f_{1}, f_{2} ∈ D(x). Since f ∈ ext D(x), f_{1} = f_{2}, as required. □

Lemma 3.7. Let v ∈ d(w, 1) \ {0} be such that v = v^{∗} and card(supp v) = ∞. Let

D_{1} = {k ∈ ℕ : v(1) = v(k)} and n_{1} = max D_{1}.

For i ≥ 2, let

D_{i} = {k ∈ ℕ : v(n_{i−1} + 1) = v(k)} and n_{i−1} = max D_{i−1}.

Set

E(v) = {f ∈ ext B_{d^{∗}(w,1)} : f(v) = ∥v∥_{w,1}}.

Then f ∈ E(v) if and only if f = w ∘σ, where σ : ℕ → ℕ is a permutation such that for any k ∈ D_{i} and l ∈ D_{i+1} we have w(σ(k)) ≥ w(σ(l)) and

∑_{k∈D_{i}} w(σ(k)) = ∑_{k∈D_{i}} w(k)

for any i ∈ ℕ.

Proof. Assume f ∈ E(v). Since v = v^{∗} and card(supp(v)) = ∞, by Theorem 2.2, f = w ∘ σ for some permutation σ : ℕ → ℕ. Assume on the contrary, that there exist i ∈ ℕ, k ∈ D_{i} and l ∈ D_{i+1}, with w(σ(k)) < w(σ(l)). Define σ_{1} : ℕ → ℕ by σ_{1}(l) = σ(k), σ_{1}(k) = σ(l) and σ_{1}(n) = σ(n) for n ∉ {k, l}. Since v(k) > v(l), and w(σ(k)) < w(σ(l)),

∥v∥_{w,1} = ∑_{j=1}^{∞} v(j)w(σ(j)) = f(v) < ∑_{j=1}^{∞} v(j)w(σ_{1}(j)),

which is a contradiction. Now, applying the induction argument, we show that

∑_{k∈D_{i}} w(σ(k)) = ∑_{k∈D_{i}} w(k)

for any i ∈ ℕ. Let

Z_{1} = {j ∈ ℕ : w(j) = w(1)},

m_{1} = max Z_{1} and for i ≥ 2

Z_{i} = {j ∈ ℕ : w(j) = w(m_{i−1} + 1)},

where m_{i} = max Z_{i}. If D_{1} ⊂ Z_{1}, then σ(D_{1}) ⊂ Z_{1}, since f(v) = ∥v∥_{w,1}. Consequently, since card(σ(D_{1})) = n_{1},

∑_{i∈D_{1}} w(i) = n_{1}w(1) = ∑_{i∈D_{1}} w(σ(i)),

as required. If m_{1} < n_{1}, there is i_{0} ≥ 2 such that

D_{1} = Z_{1} ∪ ⋯ ∪ Z_{i_{0}−1} ∪ (Z_{i_{0}} ∩ D_{1}).

Since f(v) = ∥v∥_{w,1}, Z_{j} ⊂ σ(D_{1}) for any j ≤ i_{0} − 1 and

σ(D_{1}) ⊂ Z_{1} ∪ ⋯ ∪ Z_{i_{0}}.

Hence

σ(D_{1}) = Z_{1} ∪ ⋯ Z_{i_{0}−1} ∪ (Z_{i_{0}} ∩ σ(D_{1}))

and

∑_{k∈D_{1}} w(σ(k)) = ∑_{k∈D_{1}} w(k).

Now assume that

∑_{k∈D_{i}} w(σ(k)) = ∑_{k∈D_{i}} w(k)

for i ≤ u − 1. We will show that

∑_{k∈D_{u}} w(σ(k)) = ∑_{k∈D_{u}} w(k).

First we show that for any j ∈ D_{u},

(3.2)  w(n_{u−1}) ≥ w(σ(j)) ≥ w(n_{u} + 1),

where for k ∈ ℕ n_{k} = max D_{k}. By the induction hypothesis, there exists l ∈ ⋃_{i=1}^{u−1} D_{i} with w(σ(l)) = w(n_{u−1}). Hence by the first part of the proof,

w(n_{u−1}) = w(σ(l)) ≥ w(σ(j)).

If there exists j ∈ D_{u} with w(σ(j)) < w(n_{u} + 1), then ⋃_{i=1}^{u} D_{i} \ σ(⋃_{i=1}^{u} D_{i}) ≠ ∅. Since σ is a permutation there exists l > n_{u} ≥ j with w(σ(l)) ≥ w(n_{u}) > w(σ(j)). Since f(v) = ∥v∥_{w,1}, this leads to a contradiction. Now J_{u} = {j ∈ ℕ : Z_{j} ⊂ D_{u}}. We show that for any j ∈ J_{u}, Z_{j} ⊂ σ(D_{u}). Fix j ∈ J_{u}. Since w(n_{u−1}) > w(σ(k)) for any k ∈ σ^{−1}(Z_{j}), and f(v) = ∥v∥_{w,1} Z_{j} ⊂ ⋃_{i=u}^{∞} σ(D_{i}). If Z_{j} \ σ(D_{u}) ≠ ∅, there exists l > n_{u} with w(σ(l)) ≥ w(n_{u}) and k ≤ n_{u} with w(σ(k)) < w(n_{u}); a contradiction with the first part of the proof. Now let set

B_{u} = {Z_{j} : Z_{j} ∩ D_{u−1} ≠ ∅ and Z_{j} ∩ D_{u} ≠ ∅, }
F_{u} = {Z_{j} : Z_{j} ∩ D_{u} ≠ ∅ and Z_{j} ∩ D_{u+1} ≠ ∅, }

Since w(n) is decreasing B_{u} (F_{u} resp.)= ∅ or B_{u} = {Z_{i_{0}}} (F_{u} = {Z_{j_{0}}} resp.). If D_{u} ⊂ Z_{i_{0}} or D_{u} ⊂ Z_{j_{0}}, then J_{u} = ∅. Since f(v) = ∥v∥_{w,1}, σ(D_{u}) ⊂ Z_{i_{0}} or σ(D_{u}) ⊂ Z_{j_{0}}. By (3.2) w(σ(k)) = w(n_{u−1} + 1) for any k ∈ σ^{−1}(Z_{i_{0}}) or for any k ∈ σ^{−1}(Z_{j_{0}}). Hence

∑_{k∈D_{u}} w(k) = card(D_{u})w(n_{u−1} + 1) = ∑_{k∈D_{u}} w(σ(k)).

If B_{u} = {Z_{i_{0}}} and F_{u} = {Z_{j_{0}}} then by the induction hypothesis card(σ(Z_{i_{0}})∩D_{u}) = card(Z_{i_{0}}∩D_{u}). If J_{u} ≠ ∅, then for any j ∈ J_{u} card(Z_{j}) = card(Z_{j} ∩ D_{u}) = card(σ(D_{u}) ∩ Z_{j}), since Z_{j} ⊂ σ(D_{u}). Since card(D_{u}) = card(σ(D_{u})), also card(Z_{j_{0}} ∩ D_{u}) = card(σ(Z_{j_{0}}) ∩ D_{u}). Note that

∑_{i∈D_{u}} w(i) = card(Z_{i_{0}} ∩ D_{u})w(n_{u−1} + 1) + ∑_{j∈J_{u}} card(Z_{j})w(m_{j})

+ card(Z_{j_{0}} ∩ D_{u})w(n_{u})

= card(σ(Z_{i_{0}}) ∩ D_{u})w(n_{u−1} + 1) + ∑_{j∈J_{u}} card(σ(D_{u}) ∩ Z_{j})w(m_{j})

+ card(σ(Z_{j_{0}}) ∩ D_{u})w(n_{u}) = ∑_{i∈D_{u}} w(σ(i)).

If (B_{u} = ∅ or F_{u} = ∅) and J_{u} ≠ ∅, then reasoning as above we get that

∑_{i∈D_{u}} w(i) = ∑_{i∈D_{u}} w(σ(i)),

which shows our claim. In order to prove the converse, note that

∥v∥_{w,1} = ∑_{n=1}^{∞} w(n)v(n) = ∑_{i∈ℕ} v(n_{i}) ∑_{j∈D_{i}} w(j)

= ∑_{i∈ℕ} v(n_{i}) ∑_{j∈D_{i}} w(σ(j)) = ∑_{n=1}^{∞} v(n)w(σ(n)).

Thus the proof is complete. □

Lemma 3.8. Let V ⊂ d(w, 1) be a linear subspace. Set

G_{V} = {v ∈ V \ {0} : there exists a unique f ∈ S_{V^{∗}} : f(v) = ∥v∥_{w,1}}

and for any k ∈ ℕ, let

C_{k} = {j ∈ ℕ : x(j) = x(k) for any x ∈ V }.

Let D_{i} and Z_{j} be such as in Lemma 3.7. If v ∈ G_{V} is such that v = v^{∗} and card(supp v) = ∞, and i ∈ ℕ is such that D_{i} \ Z_{j} ≠ ∅ for any j ∈ ℕ, then D_{i} = C_{k} for some k ∈ ℕ.

Proof. Since D_{i} \ Z_{j} ≠ ∅ for any j ∈ ℕ, there exists k ∈ D_{i}, k + 1 ∈ D_{i} with w(k) > w(k + 1). We show that C_{k} = D_{i}. Indeed, the inclusion C_{k} ⊂ D_{i} is obvious by definition of C_{k} and D_{i}. In order to show the opposite inclusion, assume on the contrary that there exists l ∈ D_{i} \ C_{k}. If l ≥ k + 1, define for x ∈ d(w, 1)

h_{1}(x) = ∑_{m=1}^{∞} x(m)w(m),

h_{2}(x) = ∑_{m≠l,k}^{∞} x(m)w(m) + x(k)w(l) + x(l)w(k).

Note that h_{1}(v) = h_{2}(v) = ∥v∥_{w,1}. Since l ∉ C_{k} and k ∈ C_{k}, there exists z ∈ V such that z(k) ≠ z(l). We have

h_{1}(z) − h_{2}(z) = z(k)(w(k) − w(l)) + z(l)(w(l) − w(k)) = (w(k) − w(l))(z(k) − z(l)).

It follows that h_{1}(z) ≠ h_{2}(z) since w(k) > w(k + 1) ≥ w(l) and z(k) ≠ z(l). Thus h_{1} ≠ h_{2} on V and so v ∉ G_{V}; a contradiction. If l < k, consider g_{1}, g_{2} ∈ d^{∗}(w, 1) = (d(w, 1))^{∗} defined by:

g_{1}(x) = ∑_{m=1}^{∞} x(m)w(m),

g_{2}(x) = ∑_{m≠i,k}^{∞} x(m)w(m) + x(l)w(k + 1) + x(k + 1)w(l).

Note that g_{1}(v) = g_{2}(v) = ∥v∥_{w,1}. Since l ∉ C_{k} and by the above proof k + 1 ∈ C_{k}, there exists y ∈ V such that y(l) ≠ y(k + 1). By the following equality

g_{1}(y) − g_{2}(y) = y(l)(w(l) − w(k + 1)) + y(k + 1)(w(k + 1) − w(l))
= (w(l) − w(k + 1))(y(l) − y(k + 1)),

and in view of w(l) ≥ w(k) > w(k + 1) and y(l) ≠ y(k + 1), we have that g_{1}(y) ≠ g_{2}(y). Thus v ∉ G_{V}; a contradiction. Thus the sets D_{i} and C_{k} coincide. □

Now we are able to state the main result of this section.

Theorem 3.9. Let V ⊂ d(w, 1), V ≠ {0} be a linear subspace. If V is an existence set then V is one-complemented.

Proof. Let

supp V = ⋃_{v∈V} supp v.

First we assume that card (supp V) = ∞. For any i ∈ ℕ define

C_{i,1} = {j ∈ ℕ, j ≠ i : x(i) = x(j) for any x ∈ V },
C_{i,2} = {j ∈ ℕ, j ≠ i : x(i) = −x(j) for any x ∈ V },

and

C_{i} = {i} ∪ C_{i,1} ∪ C_{i,2}.

Note that for any i ≠ j, C_{i} = C_{j} or C_{i} ∩ C_{j} = ∅. Since d(w, 1) ⊂ c_{0} and card (supp V) = ∞, C_{i} is a finite, nonempty set for any i ∈ ℕ. Set i_{1} = 1, i_{2} = min{ℕ \ C_{i_{1}}} and i_{n} = min{ℕ \ ⋃_{j=1}^{n−1} C_{i_{j}}}. Note that ℕ = ⋃_{j=1}^{∞} C_{i_{j}} and C_{i_{j}} ∩ C_{i_{k}} = ∅ for j ≠ k. Since for any permutation σ : ℕ → ℕ and {ϵ(n)} with ϵ(n) = ±1, the mapping T x = {ϵ(n)x(σ(n))} is a linear, surjective isometry of d(w, 1), in view of Lemma 3.4, we can assume without loss of generality that C_{i_{j},2} = ∅ for any j ∈ ℕ. For simplicity we shall further denote the sets {C_{i_{j}}} by {C_{i}}. Let X_{C} be the space considered in Lemma 3.5, generated by the sets C_{i} defined above. By Lemma 3.5, X_{C} is one-complemented in d(w, 1). By the construction of the sets C_{i}, and Lemma 3.4, we can assume that V ⊂ X_{C} for modified sets C_{i}. Thus in order to show that V is one-complemented in d(w, 1), it is enough to demonstrate that V is one-complemented in X_{C}. We will apply Theorem 3.3. Let

G_{V} = {v ∈ V \ {0} : there exists a unique f ∈ S_{V^{∗}} : f(v) = ∥v∥_{w,1}},

and

G_{V,C} = {v ∈ V \ {0} : there exists a unique f ∈ S_{(X_{C})^{∗}} : f(v) = ∥v∥_{w,1}}.

We shall show that G_{V} = G_{V,C}. Note that by the Hahn-Banach Theorem, G_{V,C} ⊂ G_{V} . To prove the converse, assume that v ∈ G_{V} . We need to show that v is a smooth point in X_{C}.

Note also that card (supp v) = ∞. Indeed, if we assume that supp v = {1, …, n}, then in view of card (supp V) = ∞, there exist j > n and y ∈ V with y(j) ≠ 0. Defining for x ∈ d(w, 1)

f_{1}(x) = ∑_{m=1}^{∞} x(m)w(m)

and

f_{2}(x) = f_{1}(x) − 2x(j)w(j),

we have f_{1}(v) = f_{2}(v) = ∥v∥_{w,1} and |f_{i}(x)| ≤ ∥x∥_{w,1}, i = 1, 2, by the Hardy inequality. Thus ∥f_{1}|_{V} ∥ = ∥f_{2}|_{V} ∥ = 1. Since also f_{1}(y) ≠ f_{2}(y), so v ∉ G_{V}; a contradiction. Thus supp v is infinite. Let

E(v, C) = {f ∈ ext B_{(X_{C})^{∗}} : f(v) = ∥v∥_{w,1}}.

By Lemma 3.6 applied to v and X_{C}, E(v, C) ≠ ∅. We shall show that card E(v, C) = 1. Recall that

E(v) = {f ∈ ext B_{d^{∗}(w,1)} : f(v) = ∥v∥_{w,1}}.

We have the following inclusion

E(v, C) ⊂ E(v)|_{X_{C}} = {h|_{X_{C}} : h ∈ E(v)}.

Indeed, let g ∈ S_{V^{∗}} such that g(v) = ∥v∥_{w,1}. Since v ∈ G_{V} , g is uniquely determined, g ∈ ext B_{V^{∗}} and thus for any f ∈ E(v, C), f|_{V} = g. Hence

E(v, C) = {f ∈ ext B_{(X_{C})^{∗}} : f|_{V} = g}

and

E(v) = {h ∈ ext B_{d^{∗}(w,1)} : h|_{V} = g}.

If f ∈ E(v, C) then the set of all norm preserving extensions of f to d(w, 1) is denoted by

G(f) = {h ∈ B_{d^{∗}(w,1)} : h|_{X_{C}} = f}.

Since G(f) is non-empty and weakly^{∗} compact, by the Krein-Milman Theorem ext G(f) ≠ ∅. It is clear that for any h ∈ ext G(f), h|_{X_{C}} = f and h|_{V} = g, which shows the required inclusion.

Now we claim that for any h ∈ E(v), and x ∈ X_{C}

h(x) = ∑_{n=1}^{∞} w(n)x(n).

In fact, by Lemma 3.7, h = w ∘ σ, where the permutation σ is such that for any i ∈ ℕ

∑_{j∈D_{i}} w(j) = ∑_{j∈D_{i}} w(σ(j))

and D_{i} are such as in Lemma 3.7. Therefore, it is enough to demonstrate that for any i ∈ ℕ and any x ∈ X_{C}

∑_{j∈D_{i}} x(j)w(j) = ∑_{j∈D_{i}} x(j)w(σ(j)).

Fix i ∈ ℕ. If D_{i} ⊂ Z_{j} for some j ∈ J_{i} (see Lemma 3.7) then for any k ∈ D_{i}

w(k) = w(m_{j}) = w(σ(k)).

Hence

∑_{j∈D_{i}} x(j)w(j) = w(m_{j}) ∑_{j∈D_{i}} x(j) = ∑_{j∈D_{i}} x(j)w(σ(j)).

If Z_{j} \ D_{i} ≠ ∅ for any j ∈ ℕ, then by Lemma 3.8, D_{i} = C_{k} for some k ∈ ℕ, and in view of Lemma 3.7 we get

∑_{j∈D_{i}} x(j)w(j) = ∑_{j∈C_{k}} x(j)w(j) = x(n_{i}) ∑_{j∈D_{i}} w(i)

= x(n_{i}) ∑_{j∈D_{i}} w(σ(i)) = ∑_{j∈D_{i}} x(j)w(σ(j)),

which shows our claim. Thus E(v)|_{X_{C}} consists of exactly one element and consequently card E(v, C) = 1, since E(v, C) ⊂ E(v)|_{X_{C}} and E(v, C) is nonempty.

By Lemma 3.6, v is a smooth point in X_{C} and consequently v ∈ G_{V,C}. Thus G_{V} = G_{V,C}. Since V is an existence set in d(w, 1) and V ⊂ X_{C} ⊂ d(w, 1), V is an existence set in X_{C}. Moreover, by separability of d(w, 1) and by the Mazur Theorem [11, Theorem 4.12], that the collection of smooth points in a separable Banach space X is dense in X, G_{V} is dense in X_{C}. Applying now Theorem 3.3 to V and X_{C}, there exists a norm-one projection P ∈ P (X_{C}, V). In view of Lemma 3.5 we can also find a norm-one projection Q ∈ P (d(w, 1), X_{C}). Hence R = P ∘ Q is a norm-one projection from d(w, 1) onto V. The proof is complete in the case when supp V is infinite.

If supp V is a finite set, by Lemma 3.4, we can assume that supp V = {1, ..., n} for some n ∈ ℕ. In this case we can consider V as a subspace of d^{n}(w, 1). Since V is an existence set in d(w, 1), V is also an existence set in d^{n}(w, 1). Reasoning as above we can show that V is one-complemented in d^{n}(w, 1). Since the norm in d(w, 1) is monotone, the mapping

Qx = (x(1), ...x(n), 0, ...)

is a norm-one projection from d(w, 1) onto d^{n}(w, 1). Hence V is one-complemented in d(w, 1), as required. □

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