The norms of minimal projections onto some subspaces of ℓ_{1}^{(n)}

Tomasz Zaja̧c

IMUJ Preprint 2002/16

We calculate the norms of minimal projections onto some symmetric subspaces of ℓ_{1}^{(n)} (Th. 3.1). This extends Th. 2.6 from [1].

1. Introduction. Let X be a normed space and let V be a linear subspace of X. Denote by P(X, V) the set of all continuous linear projections from X onto V, i.e.,

P(X, V) = {P ∈ L(X, V) : P |_{V} = id_{V} }.

A projection P ∈ P(X, V) is called minimal if

∥P ∥ = λ(V, X) =^{df} inf{∥P ∥ : P ∈ P(X, V)}.

The constant λ(V, X) is called the relative projection constant.

Observe that the problem of finding minimal projections is strictly related to the Hahn-Banach theorem, because we look for an extension of the identity operator id : V → V onto X of minimal norm.

For more information concerning minimal projections (existence, effective formulas, uniqueness or estimates of the norm) the reader is referred to the papers enlisted in references.

The aim of this paper is to determine the norms of minimal projections onto some symmetric subspaces of ℓ_{1}^{(n)} (i.e. ℝ^{n} equipped with the ℓ_{1} norm), (see Definition 2.1 and Theorem 3.1). This extends Theorem 2.6 from [1]. We also give a new example of regular symmetric subspaces. This notion has been introduced in [4].

Now we introduce some notation which will be used later. We denote by S_{V} the unit sphere in a normed space V. The symbol ext(S_{V}) stands for the set of all extremal points of S_{V}. Note that if V is a k-dimensional subspace of ℓ_{1}^{(n)}, then each P ∈ P(ℓ_{1}^{(n)}, V) has the form

AMS Mathematics Subject Classification: 41A35, 41A65, 47B99.

P x = ∑_{i=1}^{k} u^{i}(x)v^{i},

where v^{1}, …, v^{k} is a fixed basis of V and u^{1}, …, u^{k} ∈ ℓ_{∞}^{(n)} satisfy

(1.1)  u^{j}(v^{i}) = ∑_{l=1}^{n} u_{l}^{i}v_{l}^{i} = δ_{ij}.

A point x ∈ X is called a norming point for f ∈ X^{∗} if

x ∈ S_{X} and f(x) = ∥f∥.

Definition 1.1. Let V be a finite dimensional, real Banach space. V is symmetric if there exists a basis v^{1}, …, v^{k} ∈ V such that

∥ ∑_{i=1}^{k} |α_{i}|v^{i}∥ = ∥ ∑_{i=1}^{k} α_{π(i)}v^{i}∥

for any α_{1}, …, α_{k} ∈ ℝ and any permutaton π of indices.

Now let P = ∑_{i=1}^{k} u^{i}(·)v^{i} ∈ P(ℓ_{1}^{(n)}, V). Define

crit(P) = {j ∈ {1, …, n} : ∥P e_{j}∥ = ∥P ∥},

where e_{j} is the j-th unit vector from the canonical basis of ℝ^{n} and for j = 1, …, n

(1.2)  V_{j} = (v_{j}^{1}, …, v_{j}^{k}), U_{j} = (u_{j}^{1}, …, u_{j}^{k}).

Theorem 1.2 ([5], Th. 3, p. 294). Let P = ∑_{i=1}^{k} u^{i}(·)v^{i} ∈ P(ℓ_{1}^{(n)}, V). Then P is minimal if and only if there exists a nonzero k × k matrix M such that for every j ∈ crit(P),

U_{j} = (u_{j}^{1}, …, u_{j}^{k}) = ∥P ∥a^{j},

where a^{j} = (a_{1}^{j}, …, a_{k}^{j}) are the coordinates with respect to v^{1}, …, v^{k} of a norming point for a functional on V associated with MV_{j}, i.e.,

(1.3)  (MV_{j})(x) = ∑_{i=1}^{k}(MV_{j})_{i}x_{i}.

Here x = ∑_{i=1}^{k} x_{i}v^{i}.

Remark 1.3 (see e.g. [17]). If V is a symmetric space, then the identity matrix is one of possible choices for M.

Remark 1.4. By [6, Th.1] it easy to see that if M is invertible and V_{j} ≠ 0 for j = 1, …, n, then crit(P) = {1, …, n} for any minimal projection.

Let Σ_{k} denote the set of all permutations of {1, …, k} and let ε, σ be the operators from ℝ^{k} onto ℝ^{k} defined by

ε(x) = (ε_{1}x_{1}, …, ε_{k}x_{k}), where ε_{j} ∈ {−1, 1}, j = 1, …, k,
σ(x) = (x_{σ(1)}, …, x_{σ(k)}).

Definition 1.5 ([4], Def. 2.1). Let x ∈ ℝ^{k}. Suppose J : Σ_{k} × {−1, 1}^{k} → {1, …, k!2^{k}} is a fixed bijection such that J(id, (1, …, 1)) = 1. By [[x]] we denote the k × k!2^{k}-matrix with the columns x^{(j)}, j = 1, …, k!2^{k}, where

x^{(j)} = (ε ∘ σ)(x) = (ε_{1}x_{σ(1)}, …, ϵ_{k}x_{σ(k)}).

Here ε ∈ {−1, 1}^{k}, σ ∈ Σ_{k} are chosen so that j = J(σ, ϵ). We will refer to the matrix [[x]] as the block generated by x.

Definition 1.6 ([4], Def. 2.2). Let k, N ∈ ℕ. Put n = Nk!2^{k}. Let x^{1}, …, x^{N} ∈ ℝ^{k}. It is said that a linear subspace V ⊂ ℓ_{1}^{(n)} is generated by (x^{1}, …, x^{N}) if and only if the rows v^{1}, …, v^{k} of the k × Nk!2^{k} matrix [[x^{1}]], …, [[x^{N}]] form a basis of V, where [[x^{i}]] is the block generated by x^{i} for i = 1, …, N. It is easy to check that V is a symmetric space with respect to v^{1}, …, v^{k}. If for i = 1, ..., N x^{i} has at most two coordinates different from zero, then we can treat V as a subspace of l_{1}^{(n)} with n = 4Nk(k − 1).

Definition 1.7. Let k, N ∈ ℕ. Set n = Nk!2^{k}. Suppose x^{1}, …, x^{N} ∈ ℝ \ {0}. Let V be the space generated by x^{1}, …, x^{N}. It is said that V is a regular symmetric subspace if and only if for any j ∈ {1, …, n} the vector

\frac{∑_{i=1}^{k} v_{j}^{i}v^{i}}{∥ ∑_{i=1}^{k} v_{j}^{i}v^{i}∥}

is a norming point for a functional on V associated with V_{j} by (1.3). (By Remark 1.3 we put M = identity matrix).

2. Technical lemmas. We start with

Definition 2.1. Let k ∈ ℕ and k ≥ 2. Put n = 8k(k − 1). Let V ⊂ ℓ_{1}^{(n)} be a linear space generated by (x^{1}, x^{2}), where x^{1}, x^{2} ∈ ℝ^{k}, x^{1} = (1, b, 0, …, 0), x^{2} = (d, c, 0, …, 0), 0 ≤ b, c, d ≤ 1, d ≥ c and d^{2} + c^{2} > 0. Now fix ϵ > 0. Let us extend matrix [[x^{1}]], [[x^{2}]] to the matrix M_{ϵ} by adding k columns of the form ϵe_{i}, where e_{i} is the i-th vector from the canonical basis of ℝ^{k}. By V^{ϵ} we denote a k-dimensional subspace of ℓ_{1}^{(n+k)} spanned by the rows of M_{ϵ}. Observe that V is k-dimensional symmetric subspace of ℓ_{1}^{(n)} and V^{ϵ} is k-dimensional symmetric subspace of ℓ_{1}^{(n+k)}.

Example 2.2. If k = 2 and V is as in Definition 2.1, then

v^{1} = (1, b, −1, −b, −1, −b, 1, b, d, c, −d, −c, −d, −c, d, c),
v^{2} = (b, 1, b, 1, −b, −1, −b, −1, c, d, c, d, −c, −d, −c, −d).

Analogously, if ϵ > 0, k = 2 and V^{ϵ} is as in Definition 2.1, then

v^{1} = (1, b, −1, −b, −1, −b, 1, b, d, c, −d, −c, −d, −c, d, c, ϵ, 0),
v^{2} = (b, 1, b, 1, −b, −1, −b, −1, c, d, c, d, −c, −d, −c, −d, 0, ϵ).

Remark 2.3. Let V and V^{ϵ} be as in Definition 2.1. It is easy to check that

∥v^{1}∥ = 4(k − 1)(1 + b + c + d),
∥v^{1}∥^{ϵ} = ∥v^{1}∥ + ϵ,
∥v^{1} + bv^{2}∥ = 4(k − 2)b^{2} + 4(2k − 3)b + 4(k − 1)+
4(c + d)(kb − 2b + k − 1) − 2c + 2bd + 2|c − bd|,
∥v^{1} + bv^{2}∥^{ϵ} = ∥v^{1} + bv^{2}∥ + (1 + b)ϵ,
∥dv^{1} + cv^{2}∥ = 4(k − 2)c^{2} + 4(2k − 3)cd + 4(c + d)(kb − 2b + k − 1)+
4kd^{2} − 4d^{2} + 2bd − 2c + 2|c − bd|,
∥dv^{1} + cv^{2}∥^{ϵ} = ∥dv^{1} + cv^{2}∥ + (d + c)ϵ.

Here and subsequently ∥ · ∥ (resp. ∥ · ∥^{ϵ}) stands for the norm in V (resp. V^{ϵ}) induced from ℓ_{1}^{(n)} (resp. ℓ_{1}^{(n+k)}).

The proofs of the following two lemmas appeared in [1] and [2].

Lemma 2.4. Let V = Span[v^{1}, …, v^{k}] be a k-dimensional (k ≥ 2) subspace of ℓ_{1}^{(n)}. Then x = ∑_{i=1}^{k} x_{i}v^{i} ∈ ext(S_{V}) if and only if the matrix W consisting of all vectors V_{j} (see (1.2)) orthogonal to x has rank k − 1 and ∥x∥ = 1. We understand that V_{j} is orthogonal to x if

V_{j}(x) = ∑_{i=1}^{k}(V_{j})_{i}x_{i} = ∑_{i=1}^{k} v_{j}^{i}x_{i} = 0.

Corollary 2.5. Let V = Span[v^{1}, v^{2}] be a 2-dimensional subspace of ℓ_{1}^{(n)}. Then x = x_{1}v^{1} + x_{2}v^{2} ∈ ext(S_{V}) if and only if there exists V_{j} ≠ 0 which is orthogonal to x and ∥x∥ = 1.

Lemma 2.6. Let ∥ · ∥ be a norm on ℝ^{k} satisfying

∥(x_{1}, …, x_{k})∥ = ∥(x_{1}, …, x_{k−1}, −x_{k}∥.

Let u = (u_{1}, …, u_{k−1}, 0) ∈ (ℝ^{k}, ∥ · ∥)^{∗}. If x = (x_{1}, …, x_{k}) is a norming point for u then so is (x_{1}, …, x_{k−1}, 0).

The next lemma is related to Lemma 2.5 from [1].

Lemma 2.7. Let V^{ϵ} be the space given by Definition 2.1. If x = ∑_{j=1}^{k} x_{j}v^{j} ∈ ext(S_{V^{ϵ}}) is a norming point for V_{j} (j = 1, …, n + k), then (x_{1}, …, x_{k}) = cV_{l} for some l ∈ {1, …, n + k} and c ∈ ℝ chosen so that ∥x∥^{ϵ} = 1.

Proof. If k = 2, the result follows from Corollary 2.5. So suppose that k ≥ 3. Let x ∈ ext(S_{V^{ϵ}}) be a norming point for V_{j}. Because of the symmetry, without loss of generality we can assume that V_{j} = (d, c, 0, …, 0), V_{j} = (1, b, 0, …, 0) or V_{j} = (1, 0, …, 0). First we consider the case V_{j} = (d, c, 0, …, 0). Observe that

(2.1)  ∥x∥^{ϵ} = ∥x∥ + ϵ ∑_{i=1}^{k} |x_{i}|.

Applying Lemma 2.6 k − 2 times we see that y = x_{1}v^{1} + x_{2}v^{2} is also a norming point for V_{j}. In particular ∥y∥^{ϵ} = 1. Since y = \frac{x+z}{2} where z = y − ∑_{j=3}^{k} x_{j}v^{j} and ∥x∥ = ∥z∥ we have ∥y∥ ≤ ∥x∥. By (2.1) we get x_{i} = 0 for i = 3, …, k, which shows that x = y. Observe that (x_{1}, …, x_{k}) is orthogonal to e_{i} for i = 3, …, k. Since x ∈ ext(S_{V^{ϵ}}), by Lemma 2.4, the result follows. The same argument applies to the case V_{j} = (1, b, 0, …, 0).

The case V_{j} = (1, 0, …, 0) may be handled in much the same way (Lemma 2.6 should be applied k − 1 times).

Lemma 2.8. Let k, N ∈ ℕ, k ≥ 2. Put n = Nk!2^{k}. Let x^{1}, …, x^{N} ∈ ℝ^{k} be such that x^{i} and x^{j} are linearly independent for all i ≠ j. Let V ⊂ ℓ_{1}^{(n)} be generated by (x^{1}, …, x^{N}). Assume that

(2.2)  \frac{∥x^{1}∥_{e}}{∥ ∑_{i=1}^{k} x_{i}^{1}v^{i}∥} ≤ ⋯ ≤ \frac{∥x^{N}∥_{e}}{∥ ∑_{i=1}^{k} x_{i}^{N}v^{i}∥},

here ∥ · ∥_{e} stands for the Euclidean norm in ℝ^{k}.

Then none of the points of the form \frac{∑_{i=1}^{k} x_{i}^{l}v^{i}}{∥ ∑_{i=1}^{k} x_{i}^{l}v^{i}∥}, where l < p, is norming for each functional on V associated with the first column of [[x^{p}]], p = 1, …, N (see Definition 1.5, (1.2) and (1.3)).

Proof. Fix p, l ∈ {1, …, N} such that l < p. Using classical Schwarz inequality and (2.2) we have

\frac{x_{1}^{p}x_{1}^{l} + ⋯ + x_{k}^{p}x_{k}^{l}}{∥ ∑_{i=1}^{k} x_{i}^{l}v^{i}∥} < \frac{∥x^{p}∥_{e}∥x^{l}∥_{e}}{∥ ∑_{i=1}^{k} x_{i}^{l}v^{i}∥} ≤ \frac{(x_{1}^{p})^{2} + ⋯ + (x_{k}^{p})^{2}}{∥ ∑_{i=1}^{k} x_{i}^{p}v^{i}∥} .

The next lemma follows from Theorem 3.1.6 from [12].

Lemma 2.9. Let V and V^{ϵ} be as in Definition 2.1. Then

(a) λ(V, ℓ_{1}^{(n)}) = lim_{ϵ→0} λ(V^{ϵ}, ℓ_{1}^{(n+k)}),

(b) λ(V, ℓ_{1}^{(n)}) = lim_{l→∞} λ(V_{l}, l_{1}^{(n)}),

where V_{l} ⊂ l_{1}^{(n)} is generated by (1, b_{l}, 0, …, 0), (d_{l}, c_{l}, 0, …, 0) and b_{l} → b, c_{l} → c and d_{l} → d (l → ∞).

3. The main result. Theorem 3.1. Let V be a k-dimensional (k ≥ 2) subspace of ℓ_{1}^{(n)} given by Definition 2.1. Consider the following cases:

(A)  \begin{matrix}\frac{1 + b^{2}}{∥v^{1} + bv^{2}∥} ≥ \frac{d + bc}{∥dv^{1} + cv^{2}∥}, \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥} ≥ \frac{1}{∥v^{1}∥}, \\ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥} ≥ \frac{d + bc}{∥v^{1} + bv^{2}∥}, \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥} ≥ \frac{d}{∥v^{1}∥}, \end{matrix}

\begin{matrix}\frac{1}{∥v^{1}∥} ≥ \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥}, \frac{1}{∥v^{1}∥} ≥ \frac{d + bc}{∥dv^{1} + cv^{2}∥}, \\ \frac{d}{∥v^{1}∥} ≥ \frac{d + bc}{∥v^{1} + bv^{2}∥}, \frac{d}{∥v^{1}∥} ≥ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥}, \end{matrix} (B)

\begin{matrix}\frac{1 + b^{2}}{∥v^{1} + bv^{2}∥} ≥ \frac{1}{∥v^{1}∥}, \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥} ≥ \frac{d + bc}{∥dv^{1} + cv^{2}∥}, \\ \frac{d + bc}{∥v^{1} + bv^{2}∥} ≥ \frac{d}{∥v^{1}∥}, \frac{d + bc}{∥v^{1} + bv^{2}∥} ≥ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥}, \end{matrix} (C)

\begin{matrix}\frac{d + bc}{∥dv^{1} + cv^{2}∥} ≥ \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥}, \frac{d + bc}{∥dv^{1} + cv^{2}∥} ≥ \frac{1}{∥v^{1}∥}, \\ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥} ≥ \frac{d + bc}{∥v^{1} + bv^{2}∥}, \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥} ≥ \frac{d}{∥v^{1}∥}, \end{matrix} (D)

\begin{matrix}\frac{1}{∥v^{1}∥} ≥ \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥}, \frac{1}{∥v^{1}∥} ≥ \frac{d + bc}{∥dv^{1} + cv^{2}∥}, \\ \frac{d + bc}{∥v^{1} + bv^{2}∥} ≥ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥}, \frac{d + bc}{∥v^{1} + bv^{2}∥} ≥ \frac{d}{∥v^{1}∥}, \end{matrix} (E)

\begin{matrix}\frac{1}{∥v^{1}∥} ≥ \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥}, \frac{1}{∥v^{1}∥} ≥ \frac{d + bc}{∥dv^{1} + cv^{2}∥}, \\ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥} ≥ \frac{d}{∥v^{1}∥}, \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥} ≥ \frac{d + bc}{∥v^{1} + bv^{2}∥}, \end{matrix} (F)

\begin{matrix}\frac{d + bc}{∥dv^{1} + cv^{2}∥} ≥ \frac{1}{∥v^{1}∥}, \frac{d + bc}{∥dv^{1} + cv^{2}∥} ≥ \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥}, \\ \frac{d}{∥v^{1}∥} ≥ \frac{d + bc}{∥v^{1} + bv^{2}∥}, \frac{d}{∥v^{1}∥} ≥ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥}, \end{matrix} (G)

\begin{matrix}\frac{1 + b^{2}}{∥v^{1} + bv^{2}∥} ≥ \frac{1}{∥v^{1}∥}, \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥} ≥ \frac{d + bc}{∥dv^{1} + cv^{2}∥}, \\ \frac{d}{∥v^{1}∥} ≥ \frac{d + bc}{∥v^{1} + bv^{2}∥}, \frac{d}{∥v^{1}∥} ≥ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥}. \end{matrix} (H)

If A is satisfied, then V is a regular symmetric subspace and

λ(V, ℓ_{1}^{(n)}) = \frac{1}{\frac{4(k−1)(1+b^{2})}{∥v^{1}+bv^{2}∥} + \frac{4(k−1)(d^{2}+c^{2})}{∥dv^{1}+cv^{2}∥}} .

If is B satisfied, then

λ(V, ℓ_{1}^{(n)}) = \frac{∥v^{1}∥}{4(k − 1)(1 + d)}.

If is C satisfied, then

λ(V, ℓ_{1}^{(n)}) = \frac{∥v^{1} + bv^{2}∥}{4(k − 1)(1 + b^{2} + bc + d)}.

If is D satisfied, then

λ(V, ℓ_{1}^{(n)}) = \frac{∥dv^{1} + cv^{2}∥}{4(k − 1)(c^{2} + d^{2} + bc + d)}.

If is E satisfied, then

λ(V, ℓ_{1}^{(n)}) = \frac{1}{\frac{4(k−1)}{∥v^{1}∥} + \frac{4(k−1)(bc+d)}{∥v^{1}+bv^{2}∥}} .

If is F satisfied, then

λ(V, ℓ_{1}^{(n)}) = \frac{1}{\frac{4(k−1)}{∥v^{1}∥} + \frac{4(k−1)(d^{2}+c^{2})}{∥dv^{1}+cv^{2}∥}} .

If is G satisfied, then

λ(V, ℓ_{1}^{(n)}) = \frac{1}{\frac{4(k−1)(d+bc)}{∥dv^{1}+cv^{2}∥} + \frac{4d(k−1)}{∥v^{1}∥}} .

If is H satisfied, then

λ(V, ℓ_{1}^{(n)}) = \frac{1}{\frac{4(k−1)(1+b^{2})}{∥v^{1}+bv^{2}∥} + \frac{4d(k−1)}{∥v^{1}∥}} .

Proof. Let

Z_{1} = {(b, c, d) : b ≠ 1},
Z_{2} = {(b, c, d) : c ≠ bd},
Z_{3} = {(b, c, d) : c ≠ d},
Z_{4} = {(b, c, d) : b ≠ 0},
Z_{5} = {(b, c, d) : c ≠ 0}.

Put

Z = ⋂_{j=1}^{5} Z_{j}.

Since all the cases can be shown in a similar way, we will only give the proof of the case (A). We divide our proof into two parts. Part I. Assume that (b, c, d) ∈ Z. Hence there exists a sequence ϵ_{l} such that ϵ_{l} > 0, ϵ_{l} → 0 (l → ∞) and exactly one of the cases (A) – (H) for ∥ · ∥^{ϵ_{l}} instead of ∥ · ∥ is satisfied with strict inequalities. This means that for any functional V_{j} there exists exactly one norming point in V^{ϵ_{l}}. Fix l ∈ ℕ. Without loss of generality we can assume that

\begin{matrix}\frac{1 + b^{2}}{∥v^{1} + bv^{2}∥^{ϵ_{l}}} > \frac{d + bc}{∥dv^{1} + cv^{2}∥^{ϵ_{l}}} , \frac{1 + b^{2}}{∥v^{1} + bv^{2}∥^{ϵ_{l}}} > \frac{1}{∥v^{1}∥^{ϵ_{l}}} , \\ \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥^{ϵ_{l}}} > \frac{d + bc}{∥v^{1} + bv^{2}∥^{ϵ_{l}}} , \frac{d^{2} + c^{2}}{∥dv^{1} + cv^{2}∥^{ϵ_{l}}} > \frac{d}{∥v^{1}∥^{ϵ_{l}}} . \end{matrix}

By the symmetry of V^{ϵ_{l}} it is necessary to find norming points for (1, 0, …, 0), (1, b, 0, …, 0) and (d, c, 0, …, 0). By Lemma 2.7, \frac{v^{1}+bv^{2}}{∥v^{1}+bv^{2}∥^{ϵ_{l}}} is the only norming point for (1, b, …, 0),

\frac{dv^{1}+cv^{2}}{∥dv^{1}+cv^{2}∥^{ϵ_{l}}} is the only norming point for (d, c, 0, …, 0) and \frac{v^{1}}{∥v^{1}∥^{ϵ_{l}}} is the only norming point for (1, 0, …, 0). Let P_{ϵ_{l}} denotes a minimal projection onto V^{ϵ_{l}} (since V^{ϵ_{l}} is finite dimensional such a projection exists, see e.g. Corollary 3.1.2 in [12], by the way this P_{ϵ_{l}} is the only minimal projection, which follows from Theorem 1.2). By Theorem 1.2, Remark 1.4 and (1.1) applied to i = j = 1, we get

∑_{s=1}^{n+k} u_{s}^{1}v_{s}^{1} = ∑_{s=1}^{n+k} ∥P_{ϵ_{l}}∥a_{s}^{1}v_{s}^{1} = 1.

Hence

∥P_{ϵ_{l}}∥ = \frac{1}{\frac{4(k−1)(1+b^{2})}{∥v^{1}+bv^{2}∥^{ϵ_{l}}} + \frac{4(k−1)(d^{2}+c^{2})}{∥dv^{1}+cv^{2}∥^{ϵ_{l}}} + \frac{ϵ_{l}}{∥v^{1}∥^{ϵ_{l}}}} .

Taking the limit on both sides as l → ∞ and applying Lemma 2.9(a), we complete the proof of Part I. Part II. Suppose (b, c, d) ∉ Z. Observe that for any nontrivial polynomial p of three variables the set Z_{p} = {z ∈ ℝ^{3} : p(z) = 0} has empty interior in ℝ^{3}. Hence there exists a sequence (b_{l}, c_{l}, d_{l}) ∈ Z such that b_{l} → b, c_{l} → c, d_{l} → d, (l → ∞). By Part I and Lemma 2.9(b) we get the result.

Example 3.2. According to Lemma 2.8, for every choice of b, c, d (as in Definition 2.1) at least one of the cases (A)-(H) holds. Note that, by Lemma 2.7, V is a regular symmetric subspace if and only if (A) is satisfied. Moreover, (A) is satisfied e.g. when c = d = 1, b ∈ (0, \frac{1}{3}) for k = 2, b ∈ (\frac{1}{2}, \frac{\sqrt{41}−5}{2}) for k = 3, b ∈ (\frac{k−2}{k−1}, \frac{2k−5}{2k−4}) for k > 3. (B) is satisfied e.g. when d = 1, c = \frac{1}{4}, b ∈ (0, \frac{1}{4}) for k ≥ 3 (for k = 2 the case (B) never happens).

(C) is satisfied e.g. when d = 1, c = \frac{3}{100}, b = \frac{1}{100} for k = 2; d = 1, c = 1− \frac{1}{k}, b = 1− \frac{1}{k} − \frac{1}{k^{2}} for k ≥ 3. (D) is satisfied e.g. when d = c = 1, b ∈ (\frac{1}{3}, 1) for k = 2, b ∈ (\frac{4k−9}{4k−8}, 1) for k ≥ 3. (E) is satisfied e.g. when d = 1, c = 1 − \frac{1}{k}, b = \frac{1}{k^{2}} for k ≥ 3 (for k = 2 the case (E) never happens). (F) is satisfied e.g. when c = d = 1, b ∈ (0, \frac{k−2}{k−1}) for k ≥ 3 (for k = 2 the case (F) never happens). (G) is satisfied e.g. when d = 1, b = 1 − \frac{1}{k}, c = \frac{1}{k^{2}} for k ≥ 3 (for k = 2 the case (G) never happens). (H) is satisfied e.g. when b = d = 1, c ∈ (0, \frac{k−2}{k−1}) for k ≥ 3 (for k = 2 the case (H) never happens).

Remark 3.3. If k = 2, then the proof of Theorem 3.1 may be simplified. We may then apply Corollary 2.5, by which we do not have to use the space V^{ϵ} in the proof.

Example 3.4. Let k = 2, d = 1, b < c (0 ≤ b, c ≥ 1). Applying Remark 3.3 it is easily seen that it is sufficient to consider only the cases A,C and D, which take the simple form

(A)  1 ≥ b(c + 2), c ≥ \sqrt{2} − 1 and λ(V, ℓ_{1}^{(16)}) = \frac{1}{\frac{b^{2}+1}{b+c+2} + \frac{c^{2}+1}{2(c+1)}} ,

(C)  \sqrt{2} − 1 ≥ c and λ(V, ℓ_{1}^{(16)}) = \frac{b + c + 2}{b^{2} + bc + 2},
(D)  b(c + 2) ≥ 1 and λ(V, ℓ_{1}^{(16)}) = \frac{2(c + 1)}{c^{2} + bc + 2}.

Remark 3.5. Similar method can be applied for any finite mumber of blocks [[x^{1}]], …, [[x^{p}]], where x^{j} = (a_{j}, b_{j}, 0, …, 0) for a_{j}, b_{j} ∈ ℝ, j = 1, …, p. However, it is necessary to consider much more cases then in Theorem 3.1.

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Author’s address: Tomasz Zaja̧c Institute of Mathematics University of Agriculture Al. Mickiewicza 24/28 30-059 Kraków Poland

E-mail: rmzajac@cyf-kr.edu.pl