1-COMPLEMENTED SUBSPACES OF SPACES WITH 1-UNCONDITIONAL BASES

BEATA RANDRIANANTOANINA

Abstract. We prove that if X is a complex strictly monotone sequence space with 1-unconditional basis, Y ⊆ X has no bands isometric to ℓ_{2}^{2} and Y is the range of norm-one projection from X, then Y is a closed linear span a family of mutually disjoint vectors in X.

We completely characterize 1-complemented subspaces and norm-one projections in complex spaces ℓ_{p}(ℓ_{q}) for 1 ≤ p, q < ∞.

Finally we give a full description of the subspaces that are spanned by a family of disjointly supported vectors and which are 1-complemented in (real or complex) Orlicz or Lorentz sequence spaces. In particular if an Orlicz or Lorentz space X is not isomorphic to ℓ_{p} for some 1 ≤ p < ∞ then the only subspaces of X which are 1-complemented and disjointly supported are the closed linear spans of block bases with constant coefficients.

1. Introduction

Projections and norm one projections have been studied by many authors. The question about the form of a (norm-one) projection and the structure of its range arises naturally not only in geometry of Banach spaces, but also in approximation theory, spectral theory, ergodic theory; see, e.g., the surveys [ChP, D] for more detailed discussions of applications.

The difficulty in studying 1-complemented subspaces of spaces with 1-unconditional bases arises from the following classical fact due to Lindenstrauss [L] (cf. also [LT, Theorem 3.b.1])

Theorem 1.1. Every space Y with a 1-unconditional basis is 1-complemented in some symmetric space X.

Thus it seems hopeless to give any characterization of 1-complemented subspaces of, even symmetric, spaces with 1-unconditional bases.

The only class of spaces where the full characterization of 1-complemented subspaces was available are the classical spaces ℓ_{p} and c_{0}. Namely it is well known that

1991 Mathematics Subject Classification. 46B20,46B45,41A65. This research was started while the author participated in the Workshop on Linear Analysis

and Probability held at Texas A&M University in College Station, Texas, supported by the NSF Research Group grant MCS DMS-9311902, and completed during the stay at Mathematical Sciences Research Institute in Berkeley, California, supported by the NSF grant DMS-9022140.

every subspace of a Hilbert space is 1-complemented (with the unique orthogonal projection) and in ℓ_{p}, for p ≠ 2, ∞, we have the following result:

Theorem 1.2 ([An], cf. also [LT, Theorem 2.a.4]). Let F ⊂ ℓ_{p}, where 1 ≤ p < ∞, p ≠ 2. Then F is 1-complemented if and only if

(a) F is isometric to ℓ_{p}^{dim F}, or

(b) F is spanned by a family of mutually disjoint vectors.

It is clear that Theorem 1.2(a) cannot be extended to other spaces. Namely Lindberg [Lg] showed a class of Orlicz functions φ (for necessary definitions see Section 2) so that there exists a 1-complemented subspace F in ℓ_{φ} such that F is not even isomorphic to ℓ_{φ}. Altshuler, Casazza and B. L. Lin [ACL] showed a similar example in the class of Lorentz sequence spaces ℓ_{w,p}. However, both of these examples were spanned by a family of mutually disjoint vectors; in fact they were closed linear spans of a block basis with constant coefficients. Also the symmetric space X constructed in Theorem 1.1 was such that Y was isometrically isomorhic to a closed linear span of a block basis with constant coefficients.

It is well known that all such spans are 1-complemented in any symmetric space ([LT, Theorem 3.a.4]), so in fact all of those examples satisfy condition (b) of Theorem 1.2.

In this paper we prove that indeed Theorem 1.2(b) can be extended to a large class of 1-complemented subspaces of complex spaces with 1-unconditional basis.

Namely, if X is a complex, strictly monotone sequence space with a 1-unconditional basis, Y ⊂ X is 1-complemented in X, and Y does not contain a band isometric to ℓ_{2}^{2}, then Y is spanned by a family of disjointly supported vectors (see Corollary 3.2). It is clear that our restrictions on X and Y are necessary (see Remark after Corollary 3.2 and examples in Section 4).

The above-mentioned assumption on Y is satisfied, for example, in all spaces X that do not have a 1-complemented subspace isometric to ℓ_{2}^{2}. We discuss it in greater detail in Section 4.

In Theorem 3.1 we also describe the form of general 1-complemented subspaces of complex strictly monotone spaces.

Our method of proof cannot be extended to real sequence spaces. We use in particular the fact that every 1-complemented subspace of a complex space with 1unconditional basis also has a 1-unconditional basis. The analogous fact is false in real spaces [Le, BFL] (see [R1] for the disscusion in special real spaces).

As a consequence of Theorem 3.1 we obtain a complete characterization of 1complemented subspaces of complex ℓ_{p}(ℓ_{q}), where 1 < p, q < ∞ (Theorems 5.1 and 5.2).

Further we study the subspaces that are spanned by disjointly supported vectors and are 1-complemented in X. Calvert and Fitzpatrick [CF] showed that if all disjointly supported subspaces are 1-complemented in X then X is isometric to ℓ_{p}, for some p, 1 ≤ p < ∞, or to c_{0}.

In Section 6 we completely characterize the disjointly supported subspaces that are 1-complemented in Orlicz and Lorentz sequence spaces (Theorems 6.1 and 6.3). In particular, if a Lorentz or Orlicz space X is not isomorphic to ℓ_{p} for some 1 ≤ p < ∞ then the only disjointly supported subspaces that are 1-complemented are those guaranteed by [LT, Theorem 3.a.4], i.e. spanned by a block basis with constant coefficients. The results of Section 6 are valid for both real and complex spaces.

Acknowledgments . I wish to express my gratitude to Professors W. B. Johnson and A. Koldobsky for valuable suggestions.

2. Preliminaries

In the following we will consider complex Banach spaces X with a normalized 1unconditional basis {e_{i}}_{i∈I}, where ℕ ⊇ I = {1, …, dim X}. Our results are valid in both the finite- and infinite-dimensional case.

If x ∈ X we will write x = (x_{i})_{i∈I} if

x = ∑_{i∈1}^{dim X} x_{i}e_{i} and supp x = {i ∈ ℕ : x_{i} ≠ 0}.

For x ∈ X we will denote by x^{∗} (or sometimes by x^{N}) a norming functional for x, that is, x^{∗} ∈ X^{∗}, ∥x^{∗}∥_{X^{∗}} = 1 and x^{∗}(x) = ∥x∥_{X}.

We say (following [KW], see also [ST]) that an element x ∈ X is hermitian if there exists a hermitian projection P_{x} from X onto span{x}.

Equivalently, x is hermitian if and only if for all y ∈ X, y^{∗} norming for y, and x^{∗} norming for x we have

x^{∗}(y)y^{∗}(x) ∈ ℝ.

The set of all hermitian elements is denoted h(X).

Let {H_{λ} : λ ∈ Λ} be the collection of maximal linear subspaces of h(X). Then {H_{λ} : λ ∈ Λ} are called Hilbert components of X. Kalton and Wood [KW] proved that Hilbert components are well-defined and mutually disjoint.

A Hilbert component H_{λ} is called nontrivial if dim H_{λ} > 1.

For the careful analysis and properties of Hilbert components of various spaces we refer to [KW] and to expository papers [F, R2]. Here we just want to recall some properties which will be used in our arguments.

First, recall that if X has 1-unconditional basis {e_{i}}_{i∈I} then each basis element is hermitian. Moreover Kalton and Wood proved the following:

Theorem 2.1 ([KW, Theorem 6.5]). Let X be a Banach space with a normalized 1-unconditional basis. Then x ∈ X is hermitian in X if and only if

(i) ∥y∥_{X} = ∥y∥_{2} for all y ∈ X with supp y ⊂ supp x, and (ii) for all y, z ∈ X with supp y ∪ supp z ⊂ supp x and for all v ∈ X with supp v ∩

supp x = ∅ if ∥y∥_{X} = ∥z∥_{X} then ∥y + v∥_{X} = ∥z + v∥_{X}.

For our main result we will need the following two facts.

Proposition 2.2 ([KW, Lemma 5.2]). Suppose that x, y are hermitian elements in X. Denote by x^{∗} a norming functional for x.

If x^{∗}(y) ≠ 0 then span{x, y} ⊂ h(X).

Proposition 2.3 ([F, Lemma 4]). Suppose that X has 1-unconditional basis {e_{i}}_{i∈I} and let P : X ⟶ X be norm one projection with range of P equal to Y. Then for all i ∈ I, P e_{i} is a hermitian element in Y.

We will also frequently use the following well-known fact:

Proposition 2.4. Let X be a Banach space with a 1-unconditional basis. Suppose that Y ⊂ X is 1-complemented and a norm-one projection P : X → Y is given by

P (x) = ∑_{i} y_{i}^{∗}(x)y_{i}

where Y = \overline{span}\{y_{i}\} and y_{i}^{∗} is norming for y_{i} for all i.

Then for any y ∈ Y there exists y^{∗} norming for y and constants K_{i} so that

y^{∗} = ∑_{i} K_{i}y_{i}^{∗}.

Moreover, we have

Proposition 2.5 (Calvert [C]). Let X be a strictly convex reflexive Banach space with strictly convex dual X^{∗}. Let J : X ⟶ X^{∗} be the duality map; ∥Jx∥ = ∥x∥, Jx(x) = ∥x∥^{2}.

Then a closed linear subspace Y of X is 1-complemented in X if and only if J(Y) is a linear subspace of X^{∗}.

Finally we recall a few definitions (see [LT]).

We say that a Banach space X with 1-unconditional basis is strictly monotone if ∥x + y∥ > ∥x∥ for all x, y ≥ 0 with y ≠ 0.

An Orlicz function φ is a convex non-decreasing function φ : [0, ∞) ⟶ [0, ∞] with φ(0) = 0 and φ(1) = 1 or ∞. To any Orlicz function φ we associate the Orlicz space ℓ_{φ} of all sequences of scalars x = (x_{i})_{i} such that

∑_{i=1}^{∞} φ(\frac{|x_{i}|}{ρ}) < ∞ for some ρ > 0,

with the norm

∥x∥_{φ} = inf\{ρ > 0 : ∑_{i=1}^{∞} φ(\frac{|x_{i}|}{ρ}) < 1\}.

Let 1 ≤ p < ∞ and let w = {w_{i}}_{i∈I}, where I = ℕ or I = {1, …, d}, be a nonincreasing sequence such that w_{1} = 1 and w_{i} ≥ 0 for all i. The Banach space of all sequences of scalars x = (x_{i})_{i∈I} for which

∥x∥_{w,p} = sup_{σ∈P(I)}(∑_{i∈I} |x_{σ(i)}|^{p}w_{i})^{\frac{1}{p}} < ∞,

where P(I) is the set of all permutations of I, is denoted ℓ_{w,p} and it is called a Lorentz sequence space (another notation frequently used in the literature is d(w, p)).

Notice that

• ℓ_{φ} is strictly monotone if and only if φ(t) > 0 for all t > 0 and φ(t) < ∞ for all t ≤ 1.

• ℓ_{w,p} is strictly monotone if and only if w_{i} > 0 for all i ∈ I.

• ℓ_{p}(ℓ_{q}) is strictly monotone if and only if p, q ≠ ∞.

For any 1 ≤ p, q ≤ ∞ we denote be ℓ_{p}(ℓ_{q}) the space of sequences of scalars x = (x_{ij})_{i∈I,j∈J} such that

∥x∥_{ℓ_{p}(ℓ_{q})} = ∥(∥(x_{ij})_{j∈J}∥_{ℓ_{q}})_{i∈I} ∥_{ℓ_{p}} < ∞.

We follow standard notations as defined in [LT] and this is also where we refer the reader for all undefined terms.

3. General form of contractive projections

We are now ready to present our main theorem.

Theorem 3.1. Suppose that X is a complex strictly monotone sequence space with 1-unconditional basis {e_{i}} and X ≠ ℓ_{2} and let P be the projection of norm 1 in X. Let {H_{γ} : γ ∈ Γ} be the collection of Hilbert components of Y = P X. Then H′_{γ}s are disjointly supported as elements of X.

Proof. By Proposition 2.3 all {P e_{i}}_{i∈I} are hermitian elements of Y. Let i, j be such that P e_{i} and P e_{j} are not in the same Hilbert component of Y. Assume that there exists k such that k ∈ supp P e_{i} ∩ supp P e_{j}. If P e_{k} ≠ 0, then P^{∗}e_{k}^{∗} is a norming functional for P e_{k} in Y and (P^{∗}e_{k}^{∗}, P e_{i}) = (e_{k}^{∗}, P e_{i}) ≠ 0. Thus, by Proposition 2.2, P e_{k}, P e_{i} are in the same Hilbert component of Y. Similarly P e_{k}, P e_{j} are in the same Hilbert component of Y. But then P e_{i}, P e_{j} are in the same Hilbert component of Y contrary to our assumption.

Thus P e_{k} = 0.

Now suppose that P e_{i} = ∑_{l∈S} α_{l}e_{l}+α_{k}e_{k} for some α_{l}, α_{k} ≠ 0, where S = supp P e_{i}\ {k}.

Since P is a projection and P (e_{k}) = 0 we get

∑_{l∈S} α_{l}e_{l} + α_{k}e_{k} = P e_{i} = P (P e_{i}) = P (∑_{l∈S} α_{l}e_{l}) + α_{k}P (e_{k}) =

= P (∑_{l∈S} α_{l}e_{l}) .

Hence S ≠ ∅ and by strict monotonicity of X

∥P (∑_{l∈S} α_{l}e_{l})∥ = ∥ ∑_{l∈S} α_{l}e_{l} + α_{k}e_{k}∥ > ∥ ∑_{l∈S} α_{l}e_{l}∥

which contradicts the fact that ∥P ∥ = 1.

Thus if P e_{i}, P e_{j} are not in the same Hilbert component of Y then they are disjoint.

Corollary 3.2. Suppose that X is a complex strictly monotone sequence space with 1-unconditional basis {e_{i}} and X ≠ ℓ_{2} and let P be the projection of norm 1 in X.

Suppose that Y = P X ⊂ X has no nontrivial Hilbert components. Then there exist disjointly supported elements {y_{j}}_{j=1}^{m} (m = dim P X ≤ ∞) which span Y = P X. Moreover, for all x ∈ X,

P x = ∑_{j=1}^{m} y_{j}^{∗}(x)y_{j},

where {y_{j}^{∗}}_{j=1}^{m} ⊂ X^{∗} satisfy ∥y_{j}∥ = ∥y_{j}^{∗}∥ = y_{j}^{∗}(y_{j}) = 1 for all j.

Proof. By Proposition 2.3 all {P e_{i}}_{i∈I} are hermitian elements in Y.

By our assumption all Hilbert components of Y are one-dimensional so if P e_{i}, P e_{j} are linearly independent then they belong to different Hilbert components of Y. Therefore, by Theorem 3.1, if P e_{i}, P e_{j} are linearly independent then they are disjoint and Y can be presented as span{P e_{i} : i ∈ I} where I is a collection of such indices i, j that P e_{i}, P e_{j} are mutually disjoint.

Then y_{i} = P e_{i}/∥P e_{i}∥ for all i ∈ I, and for each x ∈ X we have P x = ∑_{i∈I} C_{i}y_{i}, where C_{i} ∈ ℂ are uniquely determined by x. Clearly y_{i}^{∗}(x) =^{def} C_{i}(x) satisfies the conclusion of the theorem.

Notice also that supp y_{i}^{∗} = supp y_{i} for all i ∈ I.

Remarks. 1. Notice that the assumption about X being strictly monotone is important. Indeed, Blatter and Cheney [BCh] (see also [B]) showed examples of 1-complemented hyperplanes in ℓ_{∞}^{3} that are not spanned by disjointly supported vectors.

2. Also the assumption about Y cannot be removed. We discuss it in greater detail in the next section.

3. As mentioned in Introduction, Calvert and Fitzpatrick [CF] showed that if every subspace of the form described in Corollary 3.2 is 1-complemented in X then X is isometric to ℓ_{p}, for some p, 1 ≤ p < ∞, or to c_{0}.

As a consequence of Corollary 3.2 we can express 1-complemented subspaces as an intersection of hyperplanes of special simple form (see [BP] for analogous representation in ℓ_{p}).

Corollary 3.3. Let X and Y be as described in Corollary 3.2. Then Y can be presented as intersection of kernels of functionals f_{j}, such that card(supp f_{j}) ≤ 2 for all j.

4. 1-complemented copies of ℓ_{2}^{2}

In this section we discuss in what situation it is possible that a space X has a 1complemented subspace Y with nontrivial Hilbert components. This clearly reduces to the question of characterizing under what conditions X can have a 1-complemented subspace F that is isometric to ℓ_{2}^{2}. The question that arises here is:

Is it possible that a space X with only 1-dimensional Hilbert components has a 1-complemented subspace isometric to ℓ_{2}^{2}?

One quickly realizes that the answer is yes:

Example 1. Consider the space X = ℓ_{2}(ℓ_{1}) and for x = (x_{ij})_{i,j} ∈ ℓ_{2}(ℓ_{1}) let P x = (x_{i1})_{i}. Then P X is isometric to ℓ_{2}^{2}.

Further there exists an orthogonal projection Q of P X onto a span of any collection of orthogonal vectors from P X = ℓ_{2}.

In Lemma 5.5 we show that if ℓ_{p}(ℓ_{q}) has a 1-complemented subspace F that is isometric to ℓ_{2}^{2} then either q = 2 and F is contained in a Hilbert component of ℓ_{p}(ℓ_{q}) or p = 2 and F is similar to the range of QP in Example 1.

Below we present two more examples: a 1-complemented copy of ℓ_{2}^{2} in a Lorentz space and in a Orlicz space. We do not know a full characterization of spaces X that have 1-complemented copies of ℓ_{2}^{2}, but we suspect that if X is a Lorentz or Orlicz space then X has to be very similar to the examples presented below.

Example 2. Consider Lorentz space ℓ_{w,2} with weight w = (1, 1, w_{3}, w_{4}, …). Then span{e_{1}, e_{2}} ⊂ ℓ_{w,2} is isometric to ℓ_{2}^{2} and clearly it is 1-complemented.

Example 3. Consider 4 dimensional Orlicz space ℓ_{φ} where

φ(t) = \{ \begin{matrix}t^{2} & if 0 ≤ t ≤ a \\ (1 + a)t − a & if a ≤ t ≤ 1 \end{matrix}

and \sqrt{2/3} < a < 1. That is,

∥(x_{1}, …, x_{4})∥_{φ} = inf \{λ : ∑_{i=1}^{4} φ (\frac{|x_{i}|}{λ}) ≤ 1\} .

In fact we have ∑_{i=1}^{4} φ (\frac{|x_{i}|}{∥x∥_{φ}} ) = 1 for all x ∈ ℓ_{φ}. Let F = ker(e_{1}^{∗} + e_{2}^{∗} + e_{3}^{∗}) = span{e_{1} − e_{2}, e_{1} − e_{3}, e_{4}}. then, if x = (x_{1}, x_{2}, x_{3}, x_{4}) ∈ F then ∥x∥_{φ} is a number such that

∑_{i=1}^{3} \frac{|x_{i}|^{2}}{∥x∥_{φ}^{2}} + φ (\frac{|x_{4}|}{∥x∥_{φ}} ) = 1 (4.1)

Indeed, suppose that x ∈ F and denote ∥x∥_{φ} = α. Then

1 = ∑_{i=1}^{4} φ (\frac{|x_{i}|}{α}) ≥ ∑_{i=1}^{4} \frac{|x_{i}|^{2}}{α^{2}} (4.2)

Assume that there is j, 1 ≤ j ≤ 3 such that \frac{|x_{j}|}{α} > a. Then, by (4.2)

∑_{i≠j} \frac{|x_{i}|^{2}}{α^{2}} < 1 − a^{2} .

Hence

|\frac{x_{1} + x_{2} + x_{3}}{α} | ≥ \frac{|x_{j}|}{α} − ∑_{{i=1 \atop i≠j}}^{3} \frac{|x_{i}|}{α} ≥ a − \sqrt{2} \sqrt{1 − a^{2}} > 0

since a > \sqrt{2/3}. But this contradicts the fact that x ∈ F = ker(e_{1}^{∗} + e_{2}^{∗} + e_{3}^{∗}). Thus \frac{|x_{j}|}{α} ≤ a for j = 1, 2, 3, so

1 = ∑_{i=1}^{4} φ (\frac{|x_{i}|}{α}) = ∑_{i=1}^{3} \frac{|x_{i}|^{2}}{α^{2}} + φ (\frac{|x_{4}|}{α}) .

Equation (4.1) and Theorem 2.1 immediately imply that e_{1} − e_{2} and e_{1} − e_{3} are hermitian in F and belong to the same Hilbert component of F, but clearly F is not isometric to ℓ_{2}.

Moreover (4.1) implies that F is 1-complemented in ℓ_{φ}. Indeed, define P : ℓ_{φ} → F by:

\begin{matrix}P (x_{1}, x_{2}, & x_{3}, x_{4}) = \\ (x_{1} − \frac{1}{3}(x_{1} + x_{2} + x_{3}), x_{2} − \frac{1}{3}(x_{1} + x_{2} + x_{3}), x_{3} − \frac{1}{3}(x_{1} + x_{2} + x_{3}), x_{4}) . \end{matrix}

Notice that Q : ℓ_{2}^{3} → ℓ_{2}^{3} defined by

Q(x_{1}, x_{2}, x_{3}) = (x_{1} − \frac{1}{3}(x_{1} + x_{2} + x_{3}), x_{2} − \frac{1}{3}(x_{1} + x_{2} + x_{3}), x_{3} − \frac{1}{3}(x_{1} + x_{2} + x_{3}))

is the norm one projection on this Hilbert space. Let x ∈ ℓ_{φ}. Denote y = P x and ∥P x∥_{φ} = β. Then, by (4.1)

1 = ∑_{i=1}^{3} \frac{|y_{i}|^{2}}{β^{2}} + φ (\frac{|y_{4}|}{β}) = \frac{∥(y_{1}, y_{2}, y_{3})∥_{2}^{2}}{β^{2}} + φ (\frac{|y_{4}|}{β})

= \frac{∥Q(x_{1}, x_{2}, x_{3})∥_{2}^{2}}{β^{2}} + φ (\frac{|y_{4}|}{β}) ≤ \frac{∥(x_{1}, x_{2}, x_{3})∥_{2}^{2}}{β^{2}} + φ (\frac{|x_{4}|}{β})
= \frac{|x_{1}|^{2}}{β^{2}} + \frac{|x_{2}|^{2}}{β^{2}} + \frac{|x_{3}|^{2}}{β^{2}} + φ (\frac{|x_{4}|}{β}) ≤

≤ ∑_{i=1}^{4} φ (\frac{|x_{i}|}{β}) .

Thus ∥x∥_{φ} ≥ β = ∥P x∥_{φ}, i.e. ∥P ∥ ≤ 1.

We finish this section with a lemma characterizing 1-complemented subspaces F in X that are isometric to ℓ_{2}^{2} in terms of norming functionals.

Lemma 4.1. If span{x, y} ⊂ X is 1-complemented in X and span{x, y} is isometric to ℓ_{2}^{2}, then there exist x^{∗} norming for x and y^{∗} norming for y such that, for all a, b ∈ ℂ with |a|^{2} + |b|^{2} = 1, the functional āx^{∗} + b̄y^{∗} is norming for ax + by.

Proof. Let P : X → span{x, y} be a norm-one projection. Then there exist x^{∗} norming for x and y^{∗} norming for y such that

P (u) = x^{∗}(u)x + y^{∗}(u)y for all u ∈ X

and x^{∗}(y) = y^{∗}(x) = 0. Moreover, span{x^{∗}, y^{∗}} ⊂ X^{∗} is isometric to ℓ_{2}^{2}. Thus

(āx^{∗}+b̄y^{∗})(ax+by) = aāx^{∗}(x)+b̄by^{∗}(y) = |a|^{2}+|b|^{2} = ∥āx^{∗}+b̄y^{∗}∥_{x^{∗}}∥ax+by∥_{x}.

5. 1-complemented subspaces of ℓ_{p}(ℓ_{q})

The main results of this section (Theorems 5.1 and 5.2) completely characterize 1-complemented subspaces of the complex space ℓ_{p}(ℓ_{q}), 1 < p, q < ∞, p ≠ q.

To formulate the theorem conveniently we will need some notation. For all x in ℓ_{p}(ℓ_{q}) we will write x = ∑_{i=1}^{n} x_{i}, where n = dim ℓ_{p} ≤ ∞, and x_{i} ∈ ℓ_{q} for all i ≤ n. We will also write

x = ∑_{i=1}^{n} ∑_{j=1}^{m} x_{ij}e_{ij},

where m = dim ℓ_{q} ≤ ∞ and e_{ij} are standard basis elements in ℓ_{p}(ℓ_{q}). We will distinguish two types of support of x:

• the usual support

supp x = {(i, j) ⊂ {1, …, n} × {1, …, m} : x_{ij} ≠ 0},

which we will sometimes call the scalar support of x, and

• the vector support

v-supp x = {i ⊂ {1, …, n} : ℓ_{q} ∋ x_{i} ≠ 0}.

Thus we will have notions of disjointness of elements in ℓ_{p}(ℓ_{q}) in the scalar and vector senses.

Theorem 5.1. Let 1 < p, q < ∞ with p ≠ q, 2. Consider the complex space ℓ_{p}(ℓ_{q}). Then Y ⊂ ℓ_{p}(ℓ_{q}) is 1-complemented if and only if there exist {v^{i}}_{i=1}^{dim Y} ⊂ ℓ_{p}(ℓ_{q}) so that

Y = \overline{span}\{v^{i}\}_{i=1}^{dim Y} ,

and for all i ≠ j ≤ dim Y one of the following conditions holds:

(a) v-supp v^{i} ∩ v-supp v^{j} = ∅, or

(b) v-supp v^{i} = v-supp v^{j}, ∥v_{k}^{i}∥_{q} = ∥v_{k}^{j}∥_{q} for all k ∈ v-supp v^{i}, and (b1) if q ≠ 2 then supp v^{i} ∩ supp v^{j} = ∅, (b2) if q = 2 then v_{k}^{i}, v_{k}^{j} are orthogonal for each k ∈ v-supp v^{i}.

The structure of 1-complemented subspaces of ℓ_{2}(ℓ_{q}) can be somewhat more complicated as illustrated in Example 1. We have the following:

Theorem 5.2. Let 1 < q < ∞ with q ≠ 2. Consider the complex space ℓ_{2}(ℓ_{q}) = ℓ_{2}^{n}(ℓ_{q}^{m}), n, m ∈ ℕ ∪ {∞}. Then Y ⊂ ℓ_{2}(ℓ_{q}) is 1-complemented if and only if there exist {v^{i}}_{i=1}^{dim Y} ⊂ ℓ_{p}(ℓ_{q}) so that

Y = \overline{span}\{v^{i}\}_{i=1}^{dim Y} ,

and for all i ≠ j ≤ dim Y one of the conditions (a) or (b) of Theorem 5.1 holds or (c) for each k ∈ v-supp v^{i} ∩ v-supp v^{j} there exists a constant C_{k} ∈ ℂ with

v_{k}^{i} = C_{k}v_{k}^{j}

and for every map σ : {1, …, n} ⟶ {1, …, m} such that (k, σ(k)) ∈ supp v^{i} ∪ supp v^{j} whenever k ∈ v-supp v^{i} ∪ v-supp v^{j}, the vectors

s_{σ}(v^{i}) = (∥v_{k}^{i}∥_{q} \frac{v_{kσ(k)}^{i}}{|v_{kσ(k)}^{i}|})_{k≤n} ∈ ℓ_{2}, and s_{σ}(v^{j}) = (∥v_{k}^{j}∥_{q} \frac{v_{kσ(k)}^{j}}{|v_{kσ(k)}^{j}|})_{k≤n} ∈ ℓ_{2}

(with the convention 0/0 = 0) are orthonormal.

Corollary 5.3. If Y ⊂ ℓ_{p}(ℓ_{q}), 1 < p, q < ∞, p ≠ q, is 1-complemented in ℓ_{p}(ℓ_{q}) if and only if Y is isometric to ∑ ⊕_{p}Y_{i}, where each Y_{i} is isometrically isomorphic to ℓ_{q}^{d_{i}}, d_{i} = dim Y_{i} ∈ ℕ ∪ {∞}.

Proof of Corollary 5.3. The “only if” part follows immediately from Theorems 5.1 and 5.2, and the “if” part when q ≠ 2 is a simple consequence of [Ko, Lemma 6] and Theorems 5.1 and 5.2. When q = 2 the “if” part follows from Lemma 5.5.

For the proof of Theorems 5.1 and 5.2 we use Theorem 3.1 and the following lemmas:

Lemma 5.4. Suppose that x^{1}, x^{2}, ∥x^{1}∥ = ∥x^{2}∥ = 1, are disjointly supported (in the scalar sense). Then span{x^{1}, x^{2}} is 1-complemented in ℓ_{p}(ℓ_{q}), with 1 < p, q < ∞, p ≠ q, if and only if one of the following conditions holds:

(a) v-supp x^{1} ∩ v-supp x^{2} = ∅, in this case F is isometric to ℓ_{p}^{2}, or

(b) v-supp x^{1} = v-supp x^{2} and ∥x_{i}^{1}∥_{q} = ∥x_{i}^{2}∥_{q} for all i ∈ v-supp x^{1}, in this case F is isometric to ℓ_{q}^{2},

Lemma 5.5. Suppose that F ⊂ ℓ_{p}(ℓ_{q}) is isometric to ℓ_{2}^{2} and is 1-complemented in ℓ_{p}(ℓ_{q}), 1 ≤ p, q < ∞, p ≠ q. Then

(a) F is spanned by disjointly supported vectors (in the scalar sense),

(b) p = 2, or

(c) q = 2 and there exists a surjective isometry U of ℓ_{p}(ℓ_{q}) such that UF is spanned by disjointly supported vectors (in the scalar sense).

Remark. It follows immediately from Lemma 5.4 that if Lemma 5.5(a) holds then p = 2 or q = 2.

Lemma 5.6. Let x, y ∈ ℓ_{2}^{n}(ℓ_{q}^{m}), n, m ∈ ℕ ∪ {∞}, 1 ≤ q < ∞, q ≠ 2, ∥x∥ = ∥y∥ = 1. Then F = span{x, y} ⊂ ℓ_{2}(ℓ_{q}) is isometrically isomorphic to ℓ_{2}^{2} and 1-complemented in ℓ_{2}(ℓ_{q}) if and only if for each i ∈ v-supp x ∩ v-supp y there exists a constant C_{i} ∈ ℂ with

x_{i} = C_{i}y_{i}

and for every map σ : {1, …, n} ⟶ {1, …, m} such that (i, σ(i)) ∈ supp x ∪ supp y whenever i ∈ v-supp x ∪ v-supp y, the vectors

s_{σ}(x) = (∥x_{i}∥_{q} \frac{x_{iσ(i)}}{|x_{iσ(i)}|})_{i≤n} ∈ ℓ_{2}, and s_{σ}(y) = (∥y_{i}∥_{q} \frac{y_{iσ(i)}}{|y_{iσ(i)}|})_{i≤n} ∈ ℓ_{2}

(with the convention 0/0 = 0) are orthonormal.

Proof of Theorems 5.1 and 5.2. Since ℓ_{p}(ℓ_{q}) is complex, if Y is 1-complemented in ℓ_{p}(ℓ_{q}), then Y has a 1-unconditional basis {v^{i}}_{i=1}^{dim Y}.

Consider v^{i}, v^{j} for some i ≠ j ≤ dim Y.

If v^{i}, v^{j} belong to different Hilbert components of Y, then by Theorem 3.1, v^{i} and v^{j} are scalarly disjoint and conditions (a), (b) of Theorem 5.1 follow from Lemma 5.4 (since it is clear that span{v^{i}, v^{j}} is 1-complemented in Y and therefore in ℓ_{p}(ℓ_{q})).

If v^{i}, v^{j} are in the same Hilbert component of Y, then span{v^{i}, v^{j}} is isometric to ℓ_{2}^{2} and when p ≠ 2 Lemma 5.5 reduces our considerations to the case of disjointly supported vectors (in the scalar sense), where we apply Lemma 5.4.

When p = 2 we apply Lemma 5.6.

Proof of Lemma 5.4. Notice that when z ∈ ℓ_{p}(ℓ_{q}) then the norming functional z^{∗} for z is given by

z^{∗} = \frac{1}{∥z∥^{p−1}} ∑_{i} ∥z_{i}∥_{q}^{p−q} ∑_{j} |z_{ij}|^{q−1} sgn(\overline{z_{ij}})e_{ij}^{∗}. (5.1)

By Proposition 2.5, span{x^{1}, x^{2}} is 1-complemented in ℓ_{p}(ℓ_{q}) if and only if for all a^{1}, a^{2} ∈ ℂ there exist K_{s}, (= K_{s}(a^{1}, a^{2})), s = 1, 2, in ℂ so that

(a^{1}x^{1} + a^{2}x^{2})^{∗} = K_{1}x^{1∗} + K_{2}x^{2∗}. (5.2)

This is equivalent, by disjointness of x^{1}, x^{2} and (5.1), to the existence of constants K_{s}, (= K_{s}(a^{1}, a^{2})), s = 1, 2, such that for all (i, j) ∈ supp x^{s}:

\begin{aligned}\frac{1}{∥a^{1}x^{1} + a^{2}x^{2}∥^{p−1}} & ∥a^{1}x_{i}^{1} + a^{2}x_{i}^{2}∥_{q}^{p−q}|a^{s}x_{ij}^{s}|^{q−1} sgn(\overline{a^{s}x_{ij}^{s}}) \\ & = K_{s}(a^{1}, a^{2})∥x_{i}^{s}∥_{q}^{p−q}|x_{ij}^{s}|^{q−1} sgn(\overline{x_{ij}^{s}}), \end{aligned}

which is further equivalent to the existence of constants C_{s}, (= C_{s}(a^{1}, a^{2}) = K_{s}(a^{1}, a^{2})· ∥a^{1}x^{1} + a^{2}x^{2}∥^{p−1}), s = 1, 2, such that for all i ∈ v-supp x^{s}:

\frac{∥a^{1}x_{i}^{1} + a^{2}x_{i}^{2}∥_{q}^{p−q}}{∥x_{i}^{s}∥_{q}^{p−q}} = (|a^{s}| + |a^{t}|\frac{∥x_{i}^{t}∥_{q}^{q}}{∥x_{i}^{s}∥_{q}^{q}} )^{\frac{p−q}{q}} = C_{s}(a^{1}, a^{2}),

where t ≠ s and {t, s} = {1, 2}.

Since p ≠ q this is equivalent to the conditions that for all (i, j) ∈ supp x^{s}:

\frac{∥x_{i}^{t}∥_{q}}{∥x_{i}^{s}∥_{q}} = \frac{∥x_{j}^{t}∥_{q}}{∥x_{j}^{s}∥_{q}} .

This means that either v-supp x^{1} ∩ v-supp x^{2} = ∅ or v-supp x^{1} = v-supp x^{2} and ∥x_{i}^{1}∥_{q} = ∥x_{i}^{2}∥_{q} for all i ∈ v-supp x^{1} (since ∥x^{1}∥ = ∥x^{2}∥).

The fact that span{x^{1}, x^{2}} is isometric to ℓ_{p}^{2} or ℓ_{q}^{2}, resp., follows immediately.

Proof of Lemma 5.5.

First we prove that if Part (a) does not hold then p = 2 or q = 2. The proof follows the standard technique of showing that ℓ_{2} can be isometrically embedded in ℓ_{p} only when p is an even integer (see [LV]) and was suggested to me by Alex Koldobsky.

Assume that F = span{x, y}, where ∥x∥ = ∥y∥ = 1 and for all a, b ∈ ℂ

∥ax + by∥ = (|a|^{2} + |b|^{2})^{\frac{1}{2}} .

Then

(|a|^{2} + |b|^{2})^{\frac{p}{2}} = ∥ax + by∥^{p} = ∑_{i=1}^{n}(∑_{j=1}^{m} |ax_{ij} + by_{ij}|^{q})^{\frac{p}{q}} . (5.3)

For each i ≤ n let S_{i} = {j ≤ m : (x_{ij}, y_{ij}) ≠ (0, 0)} = {j ≤ m : (i, j) ∈ supp x ∪ supp y}. We define an equivalence relation R_{i} on the set S_{i} by the condition that (j_{1}, j_{2}) ∈ R_{i} if and only if the pairs (x_{ij_{1}}, y_{ij_{1}}) and (x_{ij_{2}}, y_{ij_{2}}) are proportional.

Let Λ_{i} denote the set of equivalence classes for R_{i} and let J = {i ≤ n : card(Λ_{i}) = 1} and M = {i ≤ n : card(Λ_{i}) > 1}. For each i ∈ J let j_{i} be the representative of the equivalence class of R_{i} (i.e. j_{i} ∈ S_{i}) and for i ∈ M let {j_{α}}_{α∈Λ_{i}} be the set of representatives of each equivalence class. Then there exist positive constants {K_{i}}_{i∈J}, {C_{i,α}}_{i∈M,α∈Λ_{i}} so that (5.3) can be written as

(|a|^{2} + |b|^{2})^{\frac{p}{2}} = ∑_{i∈J} K_{i}|ax_{ij_{i}} + by_{ij_{i}}|^{p} + ∑_{i∈M}(∑_{α∈Λ_{i}} C_{i,α}|ax_{ij_{α}} + by_{ij_{α}}|^{q})^{\frac{p}{q}} , (5.4)

where the pairs (x_{ij_{α}}, y_{ij_{α}})_{α∈Λ_{i}} are mutually linearly independent for each i ∈ M. If there exists i ∈ M and β ∈ Λ_{i} with x_{ij_{β}} · y_{ij_{β}} ≠ 0, then there exist a_{0}, b_{0} ∈ ℂ with

a_{0}x_{ij_{β}} + b_{0}y_{ij_{β}} = 0

and

a_{0}x_{ij_{α}} + b_{0}y_{ij_{α}} ≠ 0 for all α ≠ β, α ∈ Λ_{i}.

Fix b = b_{0} and differentiate (5.4) with respect to a along the real axis at a_{0}. The left-hand side of (5.4) can be differentiated (in this fashion) infinitely many times. But, if q is not an even integer then the ([q] + 1)-st derivative of |ax_{ij_{β}} + b_{0}y_{ij_{β}}|^{q} does not exist at a_{0}.

Hence, if q is not even, x_{ij} · y_{ij} = 0 for all i ∈ M, j ∈ S_{i}. In particular, this implies that card(Λ_{i}) = 2 for all i ∈ M.

Similarly, if there exists i ∈ J with x_{ij_{i}} · y_{ij_{i}} ≠ 0, then there exist a_{0}, b_{0} ∈ ℂ with a_{0}x_{ij_{β}} + b_{0}y_{ij_{β}} = 0, we fix b = b_{0} and differentiate (5.4) with respect to a along the real axis at a_{0}. As before, the left-hand side of (5.4) can be differentiated (in this fashion) infinitely many times. But, if p is not an even integer then the ([p] + 1)-st derivative of |ax_{ij_{i}} + b_{0}y_{ij_{i}}|^{p} does not exist at a_{0}.

Hence, if p is not even, x_{ij} · y_{ij} = 0 for all i ∈ J, j ∈ S_{i}.

Further, by Lemma 4.1, if x^{∗}, y^{∗} denote norming functionals for x, y, respectively, then span{x^{∗}, y^{∗}} is isometric to ℓ_{2}^{2} and is 1-complemented in (ℓ_{p}(ℓ_{q}))^{∗} = ℓ_{p′}(ℓ_{q′}), where 1/p + 1/p′ = 1 = 1/q + 1/q′. By (5.1) relations analogous to R_{i} defined using coefficients of x^{∗} and y^{∗} are identical as the original relations R_{i}. Hence, if q′ is not even, then x_{ij}^{∗} · y_{ij}^{∗} = 0 for all i ∈ M, j ∈ S_{i} and if p′ is not even, then x_{ij}^{∗} · y_{ij}^{∗} = 0 for all i ∈ J, j ∈ S_{i}.

Thus, by (5.1), if q, q′, p, p′ are not all even integers, that is, q ≠ 2 and p ≠ 2 then x and y are (scalarly) disjoint.

We postpone the proof of Part (c), because the proof of Lemma 5.6 is the direct continuation of just presented argument and uses the same notation.

Proof of Lemma 5.6.

The “only if” part: By the first part of the proof of Lemma 5.5, since q ≠ 2, x_{ij} · y_{ij} = 0 for all i ∈ M, j ∈ S_{i}. We will show that M = ∅.

Since span{x, y} = ℓ_{2}^{2} we get, by Lemma 4.1, that for all a, b ∈ ℂ with |a|^{2}+|b|^{2} = 1, the functional āx^{∗} + b̄y^{∗} is norming for ax + by. That is, by (5.1),

∑_{i} ∥ax_{i} + by_{i}∥_{q}^{2−q} ∑_{j} |ax_{ij} + by_{ij}|^{q−1} sgn(\overline{ax_{ij} + by_{ij}})e_{ij}^{∗} =

ā ∑_{i} ∥x_{i}∥_{q}^{2−q} ∑_{j} |x_{ij}|^{q−1} sgn(\overline{x_{ij}})e_{ij}^{∗} + b̄ ∑_{i} ∥y_{i}∥_{q}^{2−q} ∑_{j} |y_{ij}|^{q−1} sgn(\overline{y_{ij}})e_{ij}^{∗}.

Thus if i ∈ M, by disjointness of x_{i}, y_{i}, for each j with (i, j) ∈ supp x we have

∥ax_{i} + by_{i}∥_{q}^{2−q}|ax_{ij}|^{q−1} sgn(\overline{ax_{ij}}) = ā∥x_{i}∥_{q}^{2−q}|x_{ij}|^{q−1} sgn(\overline{x_{ij}}).

Hence, since q ≠ 2,

∥ax_{i} + by_{i}∥_{q} = |a|∥x_{i}∥_{q} = ∥ax_{i}∥_{q}.

Since x_{i} and y_{i} are disjoint and ℓ_{q} is strictly monotone we conclude that y_{i} = 0. But this implies that card(Λ_{i}) = 1, which contradicts the fact that i ∈ M. Thus M = ∅.

Hence J = v-supp x ∪ v-supp y and for each i ∈ v-supp x ∩ v-supp y there exists a constant C_{i} ∈ ℂ with

x_{i} = C_{i}y_{i}.

Now (5.3), (5.4) and definition of J imply that for every map σ : {1, …, n} ⟶ {1, …, m} such that (i, σ(i)) ∈ supp x ∪ supp y whenever i ∈ v-supp x ∪ v-supp y we have (with the convention 0/0 = 0):

|a|^{2} + |b|^{2} = ∥ax + by∥^{2} = ∑_{i∈J}(∑_{j∈S_{i}} |ax_{ij} + by_{ij}|^{q})^{\frac{2}{q}}

= ∑_{i∈J} (∑_{j∈S_{i}} (\frac{|x_{ij}|}{|x_{iσ(i)}|}|ax_{iσ(i)} + by_{iσ(i)}|)^{q})^{\frac{2}{q}}

= ∑_{i∈J} (∑_{j∈S_{i}} \frac{|x_{ij}|^{q}}{|x_{iσ(i)}|^{q}} )^{\frac{2}{q}} |ax_{iσ(i)} + by_{iσ(i)}|^{2}

= ∑_{i∈J} \frac{∥x_{i}∥_{q}^{2}}{|x_{iσ(i)}|^{2}} |ax_{iσ(i)} + by_{iσ(i)}|^{2}

= ∑_{i∈J} |a\frac{∥x_{i}∥_{q}}{|x_{iσ(i)}|}x_{iσ(i)} + b\frac{∥y_{i}∥_{q}}{|y_{iσ(i)}|}y_{iσ(i)}|^{2} .

Thus the vectors

s_{σ}(x) = (∥x_{i}∥_{q} \frac{x_{iσ(i)}}{|x_{iσ(i)}|})_{i≤n} ∈ ℓ_{2}, and s_{σ}(y) = (∥y_{i}∥_{q} \frac{y_{iσ(i)}}{|y_{iσ(i)}|})_{i≤n} ∈ ℓ_{2}

(with the convention 0/0 = 0) are orthonormal.

The “if” part: It is clear from above calculations that if x, y are of the form described in Lemma 5.6 then span{x, y} is isometrically isomorphic to ℓ_{2}^{2}.

Further, by (5.1) we have for all a, b ∈ ℂ:

(ax + by)^{∗} = \frac{1}{(|a|^{2} + |b|^{2})^{\frac{1}{2}}} ∑_{i} ∥ax_{i} + by_{i}∥_{q}^{2−q} ∑_{j} |ax_{ij} + by_{ij}|^{q−1} sgn(\overline{ax_{ij} + by_{ij}})e_{ij}^{∗}

= \frac{1}{(|a|^{2} + |b|^{2})^{\frac{1}{2}}} ( ∑_{i∈v-supp x\v-supp y} ∥ax_{i}∥_{q}^{2−q} ∑_{j} |ax_{ij}|^{q−1} sgn(\overline{ax_{ij}})e_{ij}^{∗}

+ ∑_{i∈v-supp x∩v-supp y} ∥ax_{i} + bC_{i}x_{i}∥_{q}^{2−q} ∑_{j} |ax_{ij} + bC_{i}x_{ij}|^{q−1} sgn(\overline{ax_{ij} + bC_{i}x_{ij}})e_{ij}^{∗}

+ ∑_{i∈v-supp y\v-supp x} ∥by_{i}∥_{q}^{2−q} ∑_{j} |by_{ij}|^{q−1} sgn(\overline{by_{ij}})e_{ij}^{∗})

= \frac{1}{(|a|^{2} + |b|^{2})^{\frac{1}{2}}} (ā ∑_{i∈v-supp x\v-supp y} ∥x_{i}∥_{q}^{2−q} ∑_{j} |x_{ij}|^{q−1} sgn(\overline{x_{ij}})e_{ij}^{∗}

+ \overline{a + bC_{i}} ∑_{i∈v-supp x∩v-supp y} ∥x_{i}∥_{q}^{2−q} ∑_{j} |x_{ij}|^{q−1} sgn(\overline{x_{ij}})e_{ij}^{∗}

+ b̄ ∑_{i∈v-supp y\v-supp x} ∥y_{i}∥_{q}^{2−q} ∑_{j} |y_{ij}|^{q−1} sgn(\overline{y_{ij}})e_{ij}^{∗})

= \frac{1}{(|a|^{2} + |b|^{2})^{\frac{1}{2}}} (āx^{∗} + b̄y^{∗}).

Thus, by Lemma 2.5, span{x, y} is 1-complemented.

Proof of Lemma 5.5(c). Assume that q = 2. First notice that, if z ∈ ℓ_{p}(ℓ_{2}) and ∥z∥ = 1, then the norming functional z^{∗} for z is given by

z^{∗} = ∑_{i=1}^{n} ∥z_{i}∥_{2}^{p−2} · ∑_{j=1}^{m} \overline{z_{ij}}e_{ij} (5.5)

Let F = span{x, y}, where ∥x∥ = ∥y∥ = 1 and ∥ax + by∥ = (|a|^{2} + |b|^{2})^{1/2} for all a, b ∈ ℂ. Let U_{i} : ℓ_{m}^{2} → ℓ_{m}^{2} be a surjective isometry such that, for all i ≤ n,

U_{i}x_{i} = ∥x_{i}∥_{2}e_{1}.

Define an isometry U : ℓ_{p}(ℓ_{2}) → ℓ_{p}(ℓ_{2}) by

U((z_{i})_{i=1}^{n}) = (U_{i}z_{i})_{i=1}^{n}.

Then UF = span{Ux, Uy} is isometric to ℓ_{2}^{2} and is 1-complemented in ℓ_{p}(ℓ_{2}). Thus, by Lemma 4.1, for all a, b ∈ ℂ with |a|^{2} + |b|^{2} = 1, we have

(aUx + bUy)^{∗} = ā(Ux)^{∗} + b̄(Uy)^{∗}.

Hence, by (5.5) and by the choice of U, we get for all i ≤ n and all j ≥ 2:

\begin{aligned}∥a(Ux)_{i} + b(Uy)_{i}∥_{2}^{p−2} & · (ā · 0 + b̄ · \overline{(Uy)_{ij}}) = \\ & ā · ∥(Ux)_{i}∥_{2}^{p−2} · 0 + b̄ · ∥(Uy)_{i}∥_{2}^{p−2}\overline{(Uy)_{ij}}. \end{aligned} (5.6)

Now for each i with ∥(Ux)_{i}∥_{2} ≠ 0 we consider two cases: either

1. there exists j ≥ 2 with (Uy)_{ij} ≠ 0, or

2. (Uy)_{i} = ((Uy)_{i1}, 0, 0, …, 0). In case (1) we have

∥a(Ux)_{i} + b(Uy)_{i}∥_{2}^{p−2} = ∥(Uy)i∥_{2}^{p−2}

whenever |a|^{2} + |b|^{2} = 1. In particular, since p ≠ 2 and ∥(Ux)_{i}∥_{2} = ∥(Uy)_{i}∥_{2}, we get

∥a\frac{(Ux)_{i}}{∥(Ux)_{i}∥_{2}} + b\frac{(Uy)_{i}}{∥(Uy)_{i}∥_{2}} ∥_{2} = 1.

Thus (Ux)_{i}/∥(Ux)_{i}∥_{2} and (Uy)_{i}/∥(Uy)_{i}∥_{2} form an orthonormal basis for ℓ_{2}. Since (Ux)_{i}/∥(Ux)_{i}∥ = (1, 0, 0, …), the vectors (Ux)_{i} and (Uy)_{i} are disjoint.

In case (2), i.e. when (Uy)_{i} = ((Uy)_{i1}, 0, 0, …, 0), then by (5.5) and by the form of U we get for all a, b with |a|^{2} + |b|^{2} = 1:

\begin{aligned}|a(Ux)_{i1} + b(Uy)_{i1}|^{p−2} & · (ā(Ux)_{i1} + b̄ \overline{(Uy)_{i1}}) = \\ & ā(Ux)_{i1}^{p−2}(Ux)_{i1} + b̄|(Uy)_{i1}|^{p−2}\overline{(Uy)_{i1}}. \end{aligned} (5.7)

Let a ∈ ℝ_{+} and b = ce^{iθ}, where c ∈ ℝ_{+}, a^{2}+c^{2} = 1 and θ is such that e^{−iθ} ·\overline{(Uy)_{i1}} = |(Uy)_{i1}|. Set α = (Ux)_{i1} > 0, β = |(Uy)_{i1}| ≥ 0. Then (5.7) becomes

(aα + cβ)^{p−1} = aα^{p−1} + cβ^{p−1}, (5.8)

and this equation holds for all a, c ≥ 0 with a^{2} + c^{2} = 1.

For any fixed u, v ≥ 0 define a function f_{u,v} : [0, 1] → ℝ by

f_{u,v}(a) = au + \sqrt{1 − a^{2}}v.

It is not difficult to check that f_{u,v} attains its maximum on [0, 1] at the point a_{0} = a_{0}(u, v) = u(u^{2} + v^{2})^{−1/2} and the maximum value of f_{u,v} is equal to M(u, v) = (u^{2} + v^{2})^{1/2}.

Since equation (5.4) can be written as

(f_{α,β}(a))^{p−1} = f_{α^{p−1},β^{p−1}(a)},

we have

a_{0}(α, β) = a_{0}(α^{p−1}, β^{p−1}),
M(α, β)^{p−1} = M(α^{p−1}, β^{p−1}).

Thus

(\frac{α}{\sqrt{α^{2} + β^{2}}} )^{p−1} = \frac{α}{\sqrt{α^{2} + β^{2}}} .

Since p ≠ 2, we conclude that either α = 0 or α(α^{2} + β^{2})^{−1/2} = 1, i.e., β = 0. But α = (Ux)_{i1} = ∥(Ux)_{i}∥_{2} and β = |(Uy)_{i1}| = ∥(Uy)_{i}∥_{2}. Thus (Ux)_{i} = 0 or (Uy)_{i} = 0. Hence Ux and Uy are disjoint.

Remark. Lemmas 5.4-5.6 are all valid (with the presented proofs) both in the complex and real case. We suspect that Theorems 5.1 and 5.2, too, are true in the real case, but our method of proof does not work then.

6. 1-complemented disjointly supported subspaces of Orlicz and Lorentz spaces

In this section we fully characterize subspaces of (real or complex) Orlicz and Lorentz spaces that are spanned by disjointly supported elements and 1-complemented.

In particular, it follows from Theorems 6.1 and 6.3 that in “most” Orlicz and Lorentz spaces the only 1-complemented disjointly supported subspaces are those spanned by a block basis with constant coefficients (of some permutation of the original basis).

Theorem 6.1. Let ℓ_{φ} be a (real or complex) Orlicz space and let x, y ∈ ℓ_{φ}, be disjoint elements such that ∥x∥_{φ} = ∥y∥_{φ} = 1 and span{x, y} is 1-complemented in ℓ_{φ}.

Then one of three possibilities holds:

(1) card(supp x) < ∞ and |x_{i}| = |x_{j}| for all i, j ∈ supp x; or

(2) there exists p, 1 ≤ p ≤ ∞, such that φ(t) = Ct^{p} for all t ≤ ∥x∥_{∞}; or

(3) there exists p, 1 ≤ p ≤ ∞, and constants C_{1}, C_{2}, γ ≥ 0 such that C_{2}t^{p} ≤ φ(t) ≤ C_{1}t^{p} for all t ≤ ∥x∥_{∞} and such that, for all j ∈ supp x,

|x_{j}| = γ^{k(j)} · ∥x∥_{∞}

for some k(j) ∈ ℤ.

For the proof of the theorem we will need the following (well-known?) lemma, whose proof is outlined in [Z]. For the convenience of the reader we provide its proof below.

Lemma 6.2. Let a > 0 and suppose that φ : [0, a] → ℝ is an increasing differentiable function with φ(0) = 0. Suppose that there exist a < α, β < 1 so that, for all u ≤ a,

φ(αu) = βφ(u). (6.1)

Then there exist p > 0 and C_{1}, C_{2} > 0 such that, for all u ≤ a,

C_{2}u^{p} ≤ φ(u) ≤ C_{1}u^{p}.

Moreover, if φ(u) ≢ C · u^{p}, there exists γ > 0 such that (6.1) is satisfied (with the corresponding β) if and only if α = γ^{k} for some k ∈ ℤ.

Proof of Theorem 6.1. Let z ∈ ℓ_{φ}. By [GH] the norming functional z^{∗} of z is given by:

z_{i}^{∗} = \frac{1}{C} sgn(\overline{z_{i}})φ′(\frac{|z_{i}|}{∥z∥_{φ}} ),

where C is a constant depending on z. By Proposition 2.4, for all b ∈ ℂ there exist constants K_{1}, K_{2} such that

(x + by)^{∗} = K_{1}x^{∗} + K_{2}y^{∗}.

Since x and y are disjoint, there exists a constant K = K(b) so that for all i ∈ supp x

sgn(\overline{x_{i}})φ′ (\frac{|x_{i}|}{∥x + by∥_{φ}} ) = K · sgn(\overline{x_{i}})φ′(|x_{i}|).

Now for all , 0 < t < 1 there exists b ∈ ℂ so that ∥x + by∥_{φ} = t^{−1}. Thus, for all 0 < t ≤ 1, there exists C_{t} > 0 so that for all i ∈ supp x

φ′(|x_{i}| · t) = C_{t}φ′(|x_{i}|).

Hence for all i, j ∈ supp x and for all t ≤ 1

\frac{φ′(|x_{i}| · t)}{φ′(|x_{i}|)} = \frac{φ′(|x_{j}| · t)}{φ′(|x_{j}|)}.

Set

β = \frac{φ′(|x_{i}|)}{φ′(|x_{j}|)}, u = |x_{j}| · t, α = \frac{|x_{i}|}{|x_{j}|}.

In this notation we have

φ′(αu) = βφ′(u)

for all u such that 0 ≤ u ≤ |x_{j}|. Thus

φ(αu) = βαφ(u)

for all u such that 0 ≤ u ≤ |x_{j}|.

Let j ∈ supp x be such that |x_{j}| = ∥x∥_{∞}. If there exists i ∈ supp x with |x_{i}| ≠ |x_{j}| = ∥x∥_{∞} then, by Lemma 6.2, condition (2) or (3) holds.

Proof of Lemma 6.2. Let p = log_{α}(β). Let m_{0} ∈ ℤ be the smallest integer with α^{m_{0}} ≤ a.

If α^{m_{0}} < u ≤ a we have

φ(u) ≤ φ(a) ≤ \frac{φ(a)}{α^{m_{0}p}} · u^{p}

and

φ(u) ≥ φ(α^{m_{0}}) ≥ \frac{φ(α^{m_{0}})}{a^{p}} · u^{p}.

If α^{m+1} < u ≤ α^{m} for some m ≤ m_{0} we have

φ(u)≤φ(α^{m})=βφ(α^{m−1})=⋯=β^{m−m_{0}}φ(α^{m_{0}})= \frac{φ(α^{m_{0}})}{β^{m_{0}+1}}(α^{p})^{m+1} ≤ \frac{φ(α^{m_{0}})}{β^{m_{0}+1}} · u^{p}

and

φ(u)≥φ(α^{m+1})=βφ(α^{m})=⋯=β^{m+1−m_{0}}φ(α^{m_{0}})= \frac{φ(α^{m_{0}})}{β^{m_{0}−1}} · (φ^{p})^{m}≥ \frac{φ(α^{m_{0}})}{β^{m_{0}−1}}u^{p}.

Set C_{1} = max{φ(a)/α^{m_{0}p}, φ(α^{m_{0}})/β^{m+1}} and C_{2} = min{φ(α^{m_{0}})/a^{p}, φ(α^{m_{0}})/β^{m_{0}−1}}. Then

C_{2}u^{p} ≤ φ(u) ≤ C_{1}u^{p}

for all u with 0 ≤ u ≤ a.

Further define a function h_{φ} : (−∞, ln a] → ℝ by

h_{φ}(t) = \frac{d}{ds}(ln(φ(e^{s})))|_{s=t}.

Then, by (6.1), h_{φ}(t + ln(α)) = h_{φ}(t) for all t ≤ ln a. Thus, since α ≠ 1, either

• h_{φ} is constant, that is, there exists a constant K so that φ(u) = K · u^{p} for all u ≤ a, or

• h_{φ} is periodic, that is, there exists w, with minimal |w|, so that h_{φ}(t+w) = h_{φ}(t)

for all t ≤ ln a. Thus there exists γ > 0 (namely γ = e^{w}) and k ∈ ℤ such that α = γ^{k}.

Our next theorem describes disjointly supported 1-complemented subspaces of (real or complex) Lorentz sequence spaces.

Theorem 6.3. Let ℓ_{w,p}, with 1 < p < ∞, be a real or complex Lorentz sequence space. Suppose that {x_{i}}_{i∈I} are mutually disjoint elements of ℓ_{w,p} such that card(I) ≥ 2 and F = \overline{span}\{x_{i}\}_{i∈I} is 1-complemented in ℓ_{w,p}. Suppose, moreover, that w_{ν} ≠ 0 for all ν ≤ Σ =^{def} ∑_{i∈I} card(supp x_{i}) (≤ ∞).

Then

(a) w_{ν} = 1 for all ν ≤ Σ, or

(b) |x_{il}| = |x_{ik}| for all i ∈ I and all k, l ∈ supp x_{i}.

Proof. With each element z ∈ ℓ_{w,p} we associate a decreasing sequence of positive numbers (z̃_{i})_{i=1}^{l(z)} and “level sets” A_{i}(z) defined inductively as follows:

\begin{aligned}z̃_{1} & = ∥z∥_{∞}, \\ z̃_{2} & = max\{|z_{j}| : j ∈ ℕ \ A_{1}(z)\}, \end{aligned}  \begin{aligned}A_{1}(z) & = \{j ∈ ℕ : |z_{j}| = z̃_{1}\}, \\ A_{2}(z) & = \{j ∈ ℕ : |z_{j}| = z̃_{2}\}, \end{aligned}

and so on. Note that l(z) is the largest integer such that z̃_{l(z)} > 0 and supp z = ⋃_{i=1}^{l(z)} A_{i}(z).

For i ≤ l(z) introduce also

s_{0}(z) = 0, s_{i}(z) = ∑_{j=1}^{i} card(A_{i}(z)),

L_{i}(z) = {s_{i−1}(z) + 1, …, s_{i}(z)} ⊂ ℕ,

and let δ_{i} : A_{i}(z) → L_{i}(z) be a bijection.

Finally, for any set A ⊂ ℕ denote by P(A) the set of all permutations of A.

In this notation we can easily describe norming functionals z^{N} for z. Namely, for each j with 1 ≤ j ≤ l(z), there exists a family of coefficients {λ_{σ}}_{σ∈P(L_{j}(z))} such that λ_{σ} ≥ 0, ∑_{σ∈P(L_{j})} λ_{σ} = 1 and

(|z_{k}^{N}|)_{k∈A_{j}(z)} = (\frac{z̃_{j}}{∥z∥})^{p−1} ∑_{σ∈P(L_{j}(z))} λ_{σ}(w_{σ(δ_{j}(k))})_{k∈A_{j}(z)}.

In particular, we can compute the ℓ_{1}-norm of z^{N} restricted to a level set A_{j}(z)

∥(|z_{k}^{N}|)_{k∈A_{j}(z)}∥_{ℓ_{1}} = (\frac{z̃_{j}}{∥z∥})^{p−1} ∑_{n∈L_{j}(z)} w_{n}. (6.2)

Notice that the left hand side of (6.2) does not depend on the choice of the norming functional z^{N} for z.

Now assume that l(x_{1}) > 1. We will show that w_{ν} = 1 for all natural numbers ν ≤ Σ, where Σ =^{def} ∑_{i∈I} card(supp x_{i}) as defined above.

If ν ≤ Σ, there exists n ≤ l_{1}(x_{1}) such that

ν − s_{n}(x_{1}) ≤ Σ − card(supp x_{1}).

Further, there exist μ ∈ ℕ and {j_{i}}_{i=2}^{μ} ⊂ ℕ such that j_{i} ≤ l(x_{i}) for i = 2, …, μ and

∑_{i=2}^{μ} s_{j_{i}}(x_{i}) =: M ≥ ν − s_{N}(x_{1}).

Choose {a_{i}}_{i=1}^{μ} ⊂ ℝ_{+} such that, for all i with 2 ≤ i ≤ μ,

a_{1}(x̃_{1})_{1} > a_{i}(x̃_{i})_{j_{i}} > a_{1}(x̃_{1})_{2}

and a_{1}(x̃_{1})_{n} > a_{i}(x̃_{i})_{j} for all j > j_{i}.

To shorten the notation, set x = ∑_{i=1}^{μ} a_{1}x_{1}. Then

a_{1}(x̃_{1})_{1} = x̃_{1} and A_{1}(x) = A_{1}(x_{1}). (6.3)

Moreover, there exists k ∈ ℕ with 2 < k ≤ 1 + ∑_{i=2}^{μ} j_{i} such that, for all α satisfying 2 ≤ α ≤ n,

a_{1}(x̃_{1})_{α} = x̃_{k+(α−2)}, A_{k+(α−2)}(x) = A_{α}(x_{1}), (6.4)

and

s_{k−1}(x) = s_{1}(x_{1}) + M.

Thus

L_{1}(x) = L_{1}(x_{1}) = {1, …, s_{1}(x_{1})}, (6.5)

and, for all α with 2 ≤ α ≤ n,

L_{k+(α−2)}(x) = M + L_{α}(x_{1}) = {M + s_{α−1}(x_{1}) + 1, …, M + s_{α}(x_{1})}. (6.6)

22 BEATA RANDRIANANTOANINA

By Proposition 2.4 there exist norming functionals x^{N} for x and x_{i}^{N} for x_{i}, and constants K_{i}, where i = 1, …, μ, such that

x^{N} = ∑_{i=1}^{μ} K_{i}x_{i}^{N}.

Thus, by (6.2) and (6.3),

(\frac{x̃_{1}}{∥x∥})^{p−1} ∑_{j∈L_{1}(x)} w_{j} = K_{1} · (\frac{a_{1}(x̃_{1})_{1}}{∥x_{1}∥} )^{p−1} ∑_{j∈L_{1}(x_{1})} w_{j}.

Hence, by (6.5),

K_{1} · (\frac{∥x∥}{∥x_{1}∥})^{p−1} = 1. (6.7)

Moreover, by (6.2) and (6.4), for all α with 2 ≤ α ≤ n we get:

(\frac{x̃_{k+(α−2)}}{∥x∥} )^{p−1} ∑_{j∈L_{k+(α−2)}(x)} w_{j} = K_{1} · (\frac{a_{1}(x̃_{1})_{α}}{∥x_{1}∥} )^{p−1} ∑_{j∈L_{α}(x_{1})} w_{j}.

Hence, by (6.7) and (6.6),

∑_{j=s_{α−1}(x_{1})+1}^{s_{α}(x_{1})} w_{j+M} = ∑_{j=s_{α−1}(x_{1})+1}^{s_{α}(x_{1})} w_{j}.

Since {w_{j}} is a decreasing sequence of numbers we immediately conclude that, for all α with 2 ≤ α ≤ n,

w_{s_{α−1}(x_{1})+1} = w_{s_{α}(x_{1})+M}.

Since M ≥ 1 we get w_{s_{1}(x_{1})+1} = w_{s_{2}(x_{1})+M} = w_{s_{2}(x_{1})+1} = w_{s_{3}(x_{1})+M} = ⋯ = w_{s_{n}(x_{1})+M}. (6.8)

Finally, choose {b_{i}}_{i=1}^{μ} ⊂ ℝ_{+} in such a way that, for all i with 2 ≤ i ≤ μ,

b_{i}(x̃_{i})_{j_{i}} > b_{1}(x̃_{1})_{1}.

Now set y = ∑_{i=1}^{μ} b_{i}x_{i}. Then there exists t ∈ ℕ, with 1 ≤ t ≤ 1 + ∑_{i=2}^{μ} j_{i}, such that for all α with 1 ≤ α ≤ n we have b_{1}(x̃_{1})_{α} = ỹ_{t+(α−1)} ; A_{α}(x_{1}) = A_{t+(α−1)}(y). (6.9) Similarly, as before,

s_{t+(α−1)}(y) = s_{α}(x_{1}) + M

and L_{t+(α−1)}(y) = M + L_{α}(x_{1}) = {M + s_{α−1}(x_{1}) + 1, …, M + s_{α}(x_{1})}. (6.10)

Again, by Proposition 2.4 there exist norming functionals y^{N} for y and x_{i}^{N} for x_{i}, and constants K′_{i}, where i = 1, …, μ, such that

y^{N} = ∑_{i=1}^{μ} K′_{i}x_{i}^{N}.

Thus, by (6.2) and (6.9) we get, for all α with 1 ≤ α ≤ n,

(\frac{ỹ_{t+(α−1)}}{∥y∥} )^{p−1} · ∑_{j∈L_{t+(α−1)}(y)} w_{j} = K′_{1}(\frac{(x̃_{1})_{α}}{∥x∥} )^{p−1} · ∑_{j∈L_{α}(x_{1})} w_{j}.

Hence, by (6.10),

(\frac{1}{∥y∥})^{p−1} · ∑_{j=s_{α−1}(x_{1})+1}^{s_{α}(x_{1})} w_{j+M} = K′_{1} · (\frac{1}{∥x∥})^{p−1} ∑_{j=s_{α−1}(x_{1})+1}^{s_{α}(x_{1})} w_{j}. (6.11)

If α = 2, by (6.8) we conclude that

(\frac{1}{∥y∥})^{p−1} = K′_{1} · (\frac{1}{∥x∥})^{p−1} .

Thus, when α = 1, (6.11) becomes

∑_{j=1}^{s_{1}(x_{1})} w_{j+M} = ∑_{j=1}^{s_{1}(x_{1})} w_{j}.

Thus

w_{1} = w_{s_{1}(x_{1})+M}

and since M ≥ 1, by (6.8) we get

w_{1} = w_{s_{n}(x_{1})+M}.

Since ν ≤ s_{n}(x_{1}) + M we conclude that

1 = w_{1} = w_{ν},

which ends the proof.

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Department of Mathematics, The University of Texas at Austin, Austin, TX 78712 Current address: Mathematical Sciences Research Institute, 1000 Centennial Drive, Berkeley,

CA 94720

  E-mail address: brandri@math.utexas.edu