PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 88(102) (2010), 111–121 DOI: 10.2298/PIM1002111S

2-NORMED ALGEBRAS-I

Neeraj Srivastava, S. Bhattacharya, and S. N. Lal

Communicated by Stevan Pilipović

Abstract. The class of 2-normed real algebras, defined in [33], as shown in [29] is either void or contains only trivial algebras. In this paper, a new definition of real or complex 2-normed algebras and 2-Banach algebras are proposed. Several examples of such algebras are given.

1. Introduction

Following the introductory notion of Banach algebra by Nagumo in 1936 [34], Gelfand et al. put forward a number of fundamental results related to the theory of Banach algebras [21, 22]. This has created an impetus to the study of algebras having manifold applications in the theory of dynamical systems, particle physics etc. Later on important contributions were made by Dunford [12, 13, 14], Yoshida [42], Kaplansky [26, 27], Raubenheimer et al. [35], Burlando [6] and others [2, 4, 5, 10, 11, 36, 37, 38] leading to the development of various Banach algebra techniques which have produced many new results and simplified the theories related to matrices, integral equations and operators.

The spectral radius formula established by Gelfand gives an excellent illustration of interplay between the algebraic structure and the metric structure on the class B(X) of all continuous linear transformations from a complex Banach space X into X (dim X ⩾ 2).

Extensive investigations have been made for the concepts of 2-metric and 2normed linear spaces with different point of views [1, 3, 7, 8, 9, 15, 16, 17, 18, 19, 20, 23, 24, 25, 28, 30, 31, 32, 39, 40, 41].

Considering the concept of 2-normed algebras put forward by Noor Mohammad and Siddiuqe [33], Lal et al. have shown that the class of 2-normed algebras with unity as defined in [33] is either void or contains only trivial algebras [29].

A new definition of 2-normed algebras and an example satisfying this definition is given in Section 5. In a subsequent work we are trying to show that there exist 2-normed algebras (with or without unity) which are not normable and a 2-Banach algebra need not be a 2-Banach space. Spectrum will be defined and some results for spectral radius in case of 2-Banach algebras shall be given establishing that the spectral radius formula relates the algebraic and analytical concepts. We also intend to give the 2-norm version of the Gelfand–Mazur theorem.

2010 Mathematics Subject Classification: 46A, 46H.

2

We recall the definition of 2-normed space [19, 30].

Definition 2.1. Let E be a linear space of dimension greater than one over the field 𝕂, where 𝕂 is the field of real or complex numbers. The real valued function ∥., .∥ on E × E is said to be a 2-norm if it satisfies the following axioms:

(i) ∥x, y∥ = 0, if and only if x and y are linearly dependent in E; (ii) ∥x, y∥ = ∥y, x∥ for all x, y in E; (iii) ∥αx, y∥ = |α| ∥x, y∥ for all α ∈ 𝕂 and for all x, y ∈ E; (iv) ∥x + y, z∥ ⩽ ∥x, z∥ + ∥y, z∥ for all x, y, z ∈ E.

The pair (E, ∥., .∥) is said to be a 2-normed linear space over the field 𝕂.

In [30] several examples of 2-normed complex linear spaces over 𝕂 have been given.

We note here that in a 2-normed linear space (E, ∥., .∥), the 2-norm induces a topology which makes E a locally convex Hausdorff topological vector space. To get this topology, define for each x in E a semi-norm ρ_{x} on E by ρ_{x}(y) = ∥x, y∥, y in E. As dimension E ⩾ 2, for each y (≠ 0) in E there exists an x in E such that x, y are linearly independent and hence, ρ_{x}(y) = ∥x, y∥ ≠ 0. Thus the class {ρ_{x} : x ∈ E} forms a family of semi-norms which separates points in E and the topology formed by this family of semi-norms gives the required topology on E. A typical neighborhood of origin is of the form ⋂_{1=i}^{n}{y ∈ E : ρ_{x_{i}}(y) < ε}, where ε > 0 and {x_{1}, x_{2}, …, x_{n}} is an arbitrary finite subset on E.

We also note here the following elementary but useful proposition [30] :

Proposition 2.1. Let (E, ∥., .∥) be a 2-normed space over 𝕂. Then

(i) ∥x + y, x∥ = ∥x, y∥ for all x, y in E, and

(ii) if for two linearly independent x and y in E, ∥z, x∥ = ∥z, y∥ = 0 for z ∈ E, then z = 0.

3

The following definition of a 2-normed algebra was given [33] in 1987.

Definition 3.1. Let E be a real algebra of dim ⩾ 2 with the 2-norm ∥., .∥. E is said to be a 2-normed algebra if there is some k > 0 such that

∥xy, z∥ ⩽ k∥x, z∥ ∥y, z∥ for all x, y, z ∈ E.

The following theorem [29] establishes that such a class of algebras either does not exist or is trivial.

Theorem 3.1. Let (E, ∥., .∥) be a 2-normed algebra according to Definition 3.1. Then (a) the class of such algebra with unity is void and,

(b) the class of such algebra without unity is trivial i.e., for all x, y ∈ E, xy = 0.

Proof. For a proof see [29]. □

Here we propose the following definition for 2-normed algebras.

Let E be subalgebra of dimension ⩾ 2 of an algebra B, ∥., .∥ be a 2-norm in B and a_{1}, a_{2} ∈ B be linearly independent, non-invertible and be such that for all x, y ∈ E, ∥xy, a_{i}∥ ⩽ ∥x, a_{i}∥ ∥y, a_{i}∥, i = 1, 2. Then E is called a 2-normed algebra with respect to a_{1}, a_{2}.

Let E be a 2-normed algebra with respect to a_{1}, a_{2}. If a sequence {x_{n}} in E satisfying lim_{m,n→∞} ∥x_{n} − x_{m}, a_{i}∥ = 0, i = 1, 2 is such that there exists an x in E for which lim_{n→∞} ∥x_{n} − x, a_{i}∥ = 0, i = 1, 2; then E is a 2-Banach algebra with respect to a_{1}, a_{2}.

Remark. Let E be a 2-normed algebra with respect to a_{1}, a_{2}. Then ∥xa_{i}, a_{i}∥ ⩽ ∥x, a_{i}∥ ∥a_{i}, a_{i}∥ = 0 for i = 1, 2 implying that, ∥xa_{i}, a_{i}∥ = 0 for i = 1, 2 and so xa_{i} and a_{i} are linearly dependent. As a_{1}, a_{2} are linearly independent, a_{i} ≠ 0, i = 1, 2 and so, for all x ∈ E, there exists α_{i}(x) ∈ 𝕂, i = 1, 2 such that xa_{i} = α_{i}(x)a_{i}.

Similarly, for all x ∈ E, there exists β_{i}(x) ∈ 𝕂 such that a_{i}x = β_{i}(x)a_{i}, i = 1, 2.

Now, for i = 1, 2 we have for all x ∈ E,

a_{i}xa_{i} = (a_{i}x)a_{i} = β_{i}(x)a_{i}^{2} = a_{i}(xa_{i}) = a_{i}α_{i}(x)a_{i} = α_{i}(x)a_{i}^{2}

and so, (α_{i}(x) − β_{i}(x))a_{i}^{2} = 0.

Now, if a_{i}^{2} ≠ 0 for any of i = 1, 2, then for this i, α_{i}(x) = β_{i}(x) and we have a_{i}x = xa_{i} = α_{i}(x)a_{i} for all x ∈ E, for this i. If for any of i = 1, 2 a_{i} is invertible, then for all x ∈ E, xa_{i} = α_{i}(x)a_{i} and E will be of dimension 1. But E being a 2-normed space, we have the trivial 2-norm on E, that is, ∥x, y∥ = 0 for all x, y ∈ E. To avoid this and the other trivial case E = {0}, we took a_{1} and a_{2} non-invertible. The linear independence of a_{1} and a_{2} is also needed to get a richer structure for 2-normed algebras, as we will see later on. Also note that α_{i} and β_{i}, i = 1, 2 are homomorphisms. For (xy)a_{i} = x(ya_{i}) = xα_{i}(y)a_{i} = α_{i}(x)α_{i}(y)a_{i} = α_{i}(xy)a_{i} and so, α_{i}(xy) = α_{i}(x) α_{i}(y).

Again, for x, y ∈ E, α, β ∈ 𝕂

(αx + βy)a_{i} = α_{i}(αx + βy)a_{i} = (αα_{i}(x) + βα_{i}(x))a_{i}

and so, α_{i}(αx + βy) = (αα_{i}(x) + βα_{i}(x)).

Thus α_{i}, and similarly β_{i}, are homomorphisms on E, i = 1, 2. We sum up the above discussions in the following theorem.

Theorem 3.2. Let E be a 2-normed algebra with respect to a_{1}, a_{2}. Then there exist four homomorphisms α_{i}, β_{i} on E, i = 1, 2, such that for all x ∈ E, xa_{i} = α_{i}(x)a_{i}, a_{i}x = β_{i}(x)a_{i} and if any of a_{i}, i = 1, 2 be such that a_{i}^{2} ≠ 0, then we have α_{i} = β_{i}, and xa_{i} = a_{i}x = α_{i}(x)a_{i}.

4. Augmentation of unity

Let E be a subalgebra of an algebra B, a_{1}, a_{2} ∈ B, ∥., .∥ be a 2-norm on B and (E, ∥., .∥) be a 2-normed algebra (or a 2-Banach algebra) with respect to a_{1}, a_{2} and E be without unity. Our purpose in this section is to augment unity with E (as in the case of 1-normed algebras).

Consider the space B′ = B × 𝕂, in which for (x, α), (y, β) ∈ B′. Define (x, α) + (y, β) = (x + y, α + β), γ(x, α) = (γx, γα) for γ ∈ 𝕂 and (x, α)(y, β) = (xy + αy + βx, αβ). Then B′ is an algebra with unity e = (0, 1).

Define ∥., .∥′ in B′ by [for (x, α), (y, β) ∈ B′], ∥(x, α), (y, β)∥′ = ∥x, y∥ + ∥βx − αy, a_{1}∥ + ∥βx − αy, a_{2}∥. Then ∥., .∥ is a 2-norm on B′.

Note that for (x, α), (y, β) ∈ B′, ∥(x, α), (y, β)∥′ = 0 iff ∥x, y∥ = 0 and ∥βx − αy, a_{i}∥ = 0, i = 1, 2 iff x and y are linearly dependent and βx − αy are also linearly dependent to a_{1}, a_{2}. But as a_{1}, a_{2} are linearly independent, βx = αy. If (x, α), (y, β) are linearly dependent in B′, then for (y, β) = (0, 0), ∥(x, α), (y, β)∥ = 0 and if (y, β) ≠ 0, we have for some γ ∈ 𝕂 such that (x, α) = γ(y, β) that is, x = γy and α = γβ and so, x, y are linearly dependent giving αy = (γβ)y = β(γy) = βx and so ∥(x, α), (y, β)∥′ = 0. Conversely, if ∥(x, α), (y, β)∥ = 0, we have x, y linearly dependent and βx = αy. Now, if y = 0 either x = 0 or β = 0. If β = 0, then (y, β) = (0, 0) and (x, α) and (y, β) are linearly dependent. If β ≠ 0 but x = 0, for α = 0, (x, α) and (y, β) are linearly dependent and for α ≠ 0 for some γ ∈ 𝕂, α = γβ and we have, (x, α) = γ(y, β) and (x, α) and (y, β) are linearly dependent. If y ≠ 0, then as x, y are linearly dependent there exists γ ∈ 𝕂 such that x = γy. We assume that γ ≠ 0, that is, x ≠ 0 as if x = 0, the augment above will imply that (x, α), (y, β) are linearly dependent. x = γy and βx = αy imply that β(γy) = αy and as y ≠ 0, α = γβ and so (x, α) = γ(y, β) that is (x, α) and (y, β) are linearly dependent.

It is easy to show that ∥., .∥′ satisfies the other 2-norm axioms and consequently (B′, ∥., .∥′) is a 2-normed space.

We claim that E′ = E × 𝕂 is a 2-normed algebra with respect to b_{1} = (ka_{1}, 0), b_{2} = (ka_{2}, 0) ∈ B′ with 2-norm ∥., .∥′, for some suitable k > 0 if (E′, ∥., .∥) is a 2-normed algebra. To see this, let (x, α), (y, β) ∈ E′. Then

∥(x, α)(y, β), b_{i}∥′ = ∥(xy + αy + βx, αβ), (ka_{i}, 0)∥′
= ∥xy + αy + βx, ka_{i}∥ + ∥0(xy + αy + βx) − αβka_{i}, a_{1}∥
+ ∥0(xy + αy + βx) − αβka_{i}, a_{2}∥
⩽ k∥x, a_{i}∥ ∥y, a_{i}∥ + k|α| ∥y, a_{i}∥ + k|β| ∥x, a_{i}∥
+ k|αβ| [∥a_{i}, a_{1}∥ + ∥a_{i}, a_{2}∥]

and

∥(x, α), b_{i}∥′ = ∥(x, α), (ka_{i}, 0)∥′
= k∥x, a_{i}∥ + ∥0x − αka_{i}, a_{1}∥ + ∥0x − αka_{i}, a_{2}∥
= k∥x, a_{i}∥ + k|α| [∥a_{i}, a_{1}∥ + ∥a_{i}, a_{2}∥],
∥(y, β), b_{i}∥′ = k∥y, a_{1}∥ + |β|k [∥a_{i}, a_{1}∥ + ∥a_{i}, a_{2}∥].

For i = 1,

∥(x, α), b_{i}∥′ ∥(y, β), b_{i}∥′ = [k∥x, a_{1}∥ + |α|k∥a_{1}, a_{2}∥] [k∥y, a_{1}∥ + |β|k∥a_{1}, a_{2}∥]
= k^{2}∥x, a_{1}∥ ∥y, a_{1}∥ + k^{2}|α| ∥a_{1}, a_{2}∥ ∥y, a_{1}∥
+ k^{2}|β| ∥a_{1}, a_{2}∥ ∥x, a_{1}∥ + k^{2}|αβ| ∥a_{1}, a_{2}∥^{2}.

Now, if ∥a_{1}, a_{2}∥ ⩾ 1, we take k = 1 and if ∥a_{1}, a_{2}∥ < 1, take k = \frac{1}{∥a_{1},a_{2}∥} (∥a_{1}, a_{2}∥ ≠ 0 as a_{1}, a_{2} are linearly independent) and we see that ∥(x, α)(y, β), b_{1}∥′ ⩽ ∥(x, α), b_{1}∥′ ∥(y, β), b_{1}∥′.

Similarly, ∥(x, α)(y, β), b_{2}∥′ ⩽ ∥(x, α), b_{2}∥′ ∥(y, β), b_{2}∥′ and we see that (E′, ∥., .∥′) is a 2-normed algebra with unity e = (0, 1) with respect to b_{i}, i = 1, 2.

That (E′, ∥., .∥′) is a 2-Banach algebra with respect to b_{i}, i = 1, 2 if (E′, ∥., .∥) is a 2-Banach algebra as is also easy to see.

Let {(x_{n}, α_{n})} be a sequence in E′ such that lim_{n,m} ∥(x_{n}, α_{n})−(x_{m}, α_{m}), b_{i}∥′ = 0 for i = 1, 2. Then lim_{n,m}{k∥x_{n} −x_{m}, a_{i}∥+k|α_{n} −α_{m}| [∥a_{i}, a_{1}∥+∥a_{i}, a_{2}∥]} = 0 for i = 1, 2; which implies that lim_{n,m} ∥x_{n}−x_{m}, a_{i}∥ = 0, i = 1, 2 and lim_{n,m} |α_{n}−α_{m}| = 0. But then there exists an x ∈ E such that lim_{n} ∥x_{n} − x, a_{i}∥ = 0, i = 1, 2 (as (E, ∥., .∥) is a 2-Banach algebra with respect to a_{i}, i = 1, 2) and as {α_{n}} is Cauchy, there exists α ∈ 𝕂 such that lim_{n} |α_{n} − α| = 0 and hence lim_{n}{k∥x_{n} − x, a_{i}∥ + k|α_{n} − α| [∥a_{i}, a_{1}∥ + ∥a_{i}, a_{2}∥]} = 0 and so lim_{n} ∥(x_{n}, α_{n}) − (x, α), b_{i}∥′ = 0. This shows that (E′, ∥., .∥′) is a 2-Banach algebra with respect to b_{i}, i = 1, 2.

5. Examples

In this section we give examples of 2-normed algebras and 2-Banach algebras. It is well known that a finite dimensional algebra over 𝕂 is an 1-normed (and hence 1-Banach) algebra for a suitably defined 1-norm. Our first example shows that a finite dimensional algebra with dim ⩾ 2 also becomes a 2-normed (and 2-Banach) algebra with respect to a_{1}, a_{2} (suitably chosen) with suitably defined 2-norm.

Example 5.1. Let E be a finite dimensional (dim E = n ⩾ 2) algebra over 𝕂 and let {e_{1}, …, e_{n}} be a basis for E. Let a, b be two symbols (not in E) and define B = {x + αa + βb : x ∈ E, α, β ∈ 𝕂}, with the agreement that x + αa + βb = 0, if and only if x = 0, α = β = 0, for x ∈ E, α, β ∈ 𝕂.

For y_{i} = x_{i} + α_{i}a + β_{i}b ∈ B, i = 1, 2, α ∈ 𝕂, define y_{1} + y_{2} = (x_{1} + x_{2}) + (α_{1} + α_{2})a + (β_{1} + β_{2})b and αy_{1} = αx_{1} + (αα_{1})a + (αβ_{1})b, y_{1}y_{2} = x_{1}x_{2} + α_{1}α_{2}a + β_{1}β_{2}b. Then B is an algebra over 𝕂 and if E has unit e, then ẽ = e + a + b is the unit of B.

For x = ∑_{i=1}^{n} α_{i}e_{i} + s_{1}a + s_{2}b, y = ∑_{i=1}^{n} β_{i}e_{i} + t_{1}a + t_{2}b ∈ B define ∥x, y∥ by

∥x, y∥^{2} = (∑_{i=1}^{n} |α_{i}|^{2}+|s_{1}|^{2}+|s_{2}|^{2})(∑_{i=1}^{n} |β_{i}|^{2}+|t_{1}|^{2}+|t_{2}|^{2})−|∑_{i=1}^{n} α_{i}\overline{β_{i}}+s_{1}\overline{t_{1}}+s_{2}\overline{t_{2}}|^{2}

Then ∥., .∥ defines a 2-norm in B (See [30]).

On E, define ∥.∥_{1} by ∥x∥_{1} = ∥x, a∥ for x ∈ E. Note that ∥.∥_{1} is a 1-norm on E. Let ∥.∥ be a 1-norm on E so that (E, ∥.∥) is a 1-normed algebra. Then E being finite dimensional, both 1-norms, ∥.∥_{1} and ∥.∥ on E are equivalent and hence there exist k_{1}, k_{2} > 0 such that for every x ∈ E, ∥x∥_{1} ⩽ k_{1}∥x∥ and ∥x∥ ⩽ k_{2}∥x∥_{1}.

For x, y ∈ E, we have

∥xy, a∥ = ∥xy∥_{1} ⩽ k_{1}∥xy∥ ⩽ k_{1}∥x∥ ∥y∥ ⩽ k_{1}k_{2}∥x∥_{1}k_{2}∥y∥_{1} = k_{1}k_{2}^{2}∥x, a∥ ∥y, a∥

and so for a_{1} = k_{1}k_{2}^{2}a, we have for all x, y ∈ E, ∥xy, a_{1}∥ ⩽ ∥x, a_{1}∥ ∥y, a_{1}∥.

Similarly, for suitably chosen k_{3} > 0, we have for a_{2} = k_{3}b, ∥xy, a_{2}∥ ⩽ ∥x, a_{2}∥ ∥y, a_{2}∥ for all x, y ∈ E and (E, ∥., .∥) becomes a 2-normed algebra over 𝕂 with respect to a_{1}, a_{2}.

To see that (E, ∥., .∥) is a 2-Banach algebra with respect to a_{1}, a_{2}, let {x_{n}} be a sequence in E so that lim_{n,m→∞} ∥x_{n}−x_{m}, a_{i}∥ = 0, i = 1, 2. Then lim_{n,m} ∥x_{n}−x_{m}∥_{1} = 0, and E being finite dimensional (E, ∥.∥_{1}) is a Banach space and hence there is an x ∈ E, such that lim_{n} ∥x_{n} − x∥_{1} = 0 or equivalently lim_{n} ∥x_{n} − x, a_{1}∥ = 0. Now if we define for x ∈ E, ∥x∥_{2} = ∥x, a_{2}∥, then (E, ∥.∥_{2}) also becomes a 1-normed space and the norms ∥.∥_{1} and ∥.∥_{2} on E being equivalent lim_{n} ∥x_{n} − x∥_{2} = 0 or equivalently lim_{n} ∥x_{n} − x, a_{2}∥ = 0, and the proof that (E, ∥., .∥) is a 2-Banach algebra with respect to a_{1}, a_{2} is complete.

Before we proceed further, we note the following: for x, y ∈ E, we say x ⊥_{B} y (that is x is orthogonal to y in Birkhoff’s sense) if ∥x∥ ⩽ ∥x + αy∥ for all α ∈ 𝕂.

Lemma 5.1. There exists, for an 0 ≠ x ∈ E, a ϕ ∈ E^{∗} such that ∥ϕ∥ = 1 and ϕ(x) = ∥x∥.

Lemma 5.2. [25] Let x, y ∈ E, x ≠ 0. Then x ⊥_{B} y if and only if there exists a ϕ ∈ E^{∗} such that ∥ϕ∥ = 1, ϕ(x) = ∥x∥ and ϕ(y) = 0.

Lemma 5.3. For x, y ∈ E, both nonzero, there exists an α ∈ 𝕂 such that x ⊥_{B} (αx + y).

Proof. From Lemma 5.1, there exists a ϕ ∈ E^{∗} such that ∥ϕ∥ = 1 and ϕ(x) = ∥x∥. Now, let α = −ϕ(y)/ϕ(x). Then ϕ(αx + y) = 0 and the lemma follows by Lemma 5.2. □

Lemma 5.4. [28] Let (E, ∥.∥) be a 1-normed linear space over 𝕂. For x, y ∈ E, define ∥x, y∥ by

∥x, y∥ = sup_{ϕ,Ψ∈E^{∗}∥ϕ∥=∥Ψ∥=1} |ϕ(x)Ψ(y) − ϕ(y)Ψ(x)|.

Then (E, ∥., .∥) is a 2-normed linear space over 𝕂.

The above lemma leads us to the following: let (E, ∥., .∥) be a 1-normed linear space over 𝕂. The 2-norm defined above is called the induced 2-norm.

We also note the following:

Lemma 5.5. [28] Let (E, ∥.∥) be a 1-normed linear space over 𝕂 and let (E, ∥., .∥) be the corresponding induced 2-normed linear space. Then for all x, y ∈ E, ∥x, y∥ ⩽ 2∥x∥ ∥y∥. If further x ⊥_{B} y or y ⊥_{B} x, then ∥x∥ ∥y∥ ⩽ ∥x, y∥.

Let E be an algebra over 𝕂, ϕ : E → 𝕂 be such that ϕ is linear and for all x, y ∈ E, ϕ(xy) = ϕ(x)ϕ(y), we say ϕ is a 𝕂-homomorphism on E.

Before we give our next example as the following theorem, we note that from Theorem 3.2 it is obvious that not all 1-normed algebras (or 1-Banach algebras) can be made into a 2-normed algebra (or 2-Banach algebra) respectively as the existence of 𝕂-homomorphism on a 1-normed algebra is not guaranteed though Theorem 3.2 states that 𝕂-homomorphism on a 2-normed algebra (or 2-Banach algebra) (E, ∥., .∥) with respect to a_{1}, a_{2} is guaranteed.

Theorem 5.1. Let (E, ∥.∥) (dim E ⩾ 2) be a 1-normed algebra (or 1-Banach algebra) over 𝕂 with or without unity, over which a nontrivial 𝕂-homomorphism θ exists. Then there exist an algebra A over 𝕂 of which E is a subalgebra and a 2-norm ∥., .∥ on A, a_{1}, a_{2} ∈ A such that (E, ∥., .∥) becomes a 2-normed algebra (or a 2-Banach algebra respectively) with respect to a_{1}, a_{2}.

Proof. We prove the theorem for the case of 1-normed algebras. In the course of the proof, we will see that the case of 1-Banach algebras is also proved.

Let (E, ∥ . ∥) be a 1-normed algebra over 𝕂. If E is without unity, we augment the unity in the usual way and extend θ over the extended 1-normed algebra E ×𝕂. For (x, α), (y, β) ∈ E × 𝕂, we define (x, α) + (y, β) = (x + y, α + β), (x, α)(y, β) = xy + αy + β and ∥(x, α)∥ = ∥x∥ + |α|.

Unit of E × 𝕂 is given by e = (0, 1) and θ(x, α) = θ(x) + α over E × 𝕂 becomes a 𝕂-homomorphism. We can now without any loss of generality assume that 1normed algebra (E, ∥.∥) is with unity. Let Ẽ be the completion of E. Now, let a ∈ E be such that ∥a∥ < 1 and |θ(a)| ⩾ 1. Then θ(e − α^{−1}a) = 0 where α = θ(a). Write b = e − α^{−1}a. Then θ(b) = 0 and as ∥α^{−1}a∥ < 1, b is invertible in Ẽ and hence is invertible in E itself. But θ being a nontrivial 𝕂-homomorphism, θ(b) ≠ 0. Hence |θ(a)| < 1 and we conclude that θ is continuous over E and hence θ can be extended continuously to θ̃ over Ẽ and θ̃ is a nontrivial 𝕂-homomorphism over Ẽ with ∥θ̃∥ = ∥θ∥ = 1. Therefore (Ẽ, ∥.∥) is a 1-Banach algebra, with unity e and θ̃ is a nontrivial 𝕂-homomorphism over Ẽ. Then Ẽ is isometrically isomorphic to a closed subalgebra M of B(X), the 1-Banach algebra of all bounded linear operators on a 1-Banach-space (X, ∥.∥_{x}) over 𝕂. Now, let a, b be two symbols (not in X) and consider the space X_{1} = {x + αa + βb : x ∈ X, α, β ∈ 𝕂}, with the agreement that x + αa + βb = 0 if and only if x = 0, α = β = 0. For, y_{i} = x_{i} + α_{i}a + β_{i}b ∈ X_{1}, i = 1, 2 define y_{1} + y_{2} = (x_{1} + x_{2}) + (α_{1} + α_{2})a + (β_{1} + β_{2})b ∈ X_{1} and αy_{1} = (αx_{1}) + (αα_{1})a + (αβ_{1})b ∈ X_{1}, for α ∈ 𝕂.

X_{1} is a linear space over 𝕂. In X_{1} define ∥.∥_{x_{1}} by ∥x + αa + βb∥_{x_{1}} = ∥x∥_{x} + |α|+|β| for x+αa+βb ∈ X_{1}. ∥.∥_{x_{1}} defines a 1-norm in X_{1} with ∥a∥_{x_{1}} = ∥b∥_{x_{1}} = 1, and ∥x∥_{x_{1}} = ∥x∥_{x} for x ∈ X.

Let {x_{n} + α_{n}a + β_{n}b} be a Cauchy sequence in X_{1}. Then {x_{n}} is Cauchy in X and {α_{n}}, {β_{n}} are also Cauchy in 𝕂 and so {x_{n}}, {α_{n}}, {β_{n}} converge to x ∈ X and α, β ∈ 𝕂 respectively and {x_{n} + α_{n}a + β_{n}b} to x + αa + βb in X_{1} and so, (X_{1}, ∥x∥_{x_{1}}) is a 1-Banach space.

For T ∈ M, let T̃ be defined on X_{1} by

T̃(x + αa + βb) = T (x) + θ̃_{1}(T)[αa + βb] for all x + αa + βb in X_{1},

where θ̃_{1} is the 𝕂-homomorphism over M corresponding to θ̃ over Ẽ. For x+αa+βb ∈ X_{1}, T ∈ M, we have

∥T̃(x + αa + βb)∥_{x_{1}} = ∥T (x)∥_{x} + |θ̃_{1}(T)| (|α| + |β|)
⩽ ∥T ∥_{x}∥x∥_{x} + ∥T ∥_{x}(|α| + |β|)
= ∥T ∥_{x}∥(x + αa + βb)∥_{x_{1}} for all x + αa + βb ∈ X_{1}

and so, T̃ ∈ B(X_{1}) and as ∥T̃∥_{x_{1}} ⩽ ∥T ∥_{x} we have ∥T̃∥_{x_{1}} = ∥T ∥_{x}.

Let M̃ = {T̃ ∈ B(X_{1}) : T̃ as defined above for T ∈ M}. Then M̃ is a 1-Banach algebra with unity, a closed subalgebra of B(X_{1}).

Let T_{a}, T_{b} ∈ B(X_{1}) be defined as T_{a}(x+αa+βb) = αa and T_{b}(x+αa+βb) = βb, for x+αa+βb ∈ X_{1}. Then, ∥T_{a}∥_{x_{1}} = ∥T_{b}∥_{x_{1}} = 1. For T̃ ∈ M̃, and x+αa+βb ∈ X_{1}, we have

T̃T_{a}(x + αa + βb) = T̃(αa) = θ̃_{1}(T)αa = θ̃_{1}(T)T_{a}(x + αa + βb) = T_{a}T̃(x + αa + βb)

and so T̃T_{a} = T_{a}T̃ = θ̃_{1}(T)T_{a}. Similarly T̃T_{b} = T_{b}T̃ = θ̃_{1}(T)T_{b} and T_{a}^{2} = T_{a}, T_{b}^{2} = T_{b}.

Now, if we identify B(X_{1}) with the 1-Banach algebra B of which Ẽ is a closed 1-normed subalgebra, (writing ∥.∥_{B} for the norm in B) and T_{a}, T_{b} be identified with a, b respectively in B, then we have, for each x ∈ Ẽ, xa = ax = θ̃(x)a and xb = bx = θ̃(x)b, and a^{2} = a, b^{2} = b, ∥a∥_{B} = ∥b∥_{B} = 1.

Let us equip B with the induced 2-norm. We claim that (Ẽ, ∥., .∥), ∥., .∥ is the induced 2-norm on B, is a 2-Banach algebra with respect to 2a, 2b. To prove the claim, we first prove that (Ẽ, ∥., .∥) is a 2-normed algebra with respect to 2a, 2b. We begin by showing that for every x, y ∈ Ẽ, ∥xy, a∥ ⩽ 2∥x, a∥ ∥y, a∥.

Let x, y ∈ Ẽ, x ≠ 0 ≠ y. Then by Lemma 5.3 there exist α_{1}, α_{2} ∈ 𝕂 such that a ⊥_{B} (α_{1}a + x) and a ⊥_{B} (α_{2}a + y) and let x_{1} = α_{1}a + x, x_{2} = α_{2}a + y, then x_{1}x_{2} = α_{1}α_{2}a^{2} +α_{1}ay +α_{2}xa+xy = α_{1}α_{2}a+α_{1}θ̃(y)a+α_{2}θ̃(x)a+xy and we have, by Proposition 2.1, ∥x_{1}x_{2}, a∥ = ∥xy, a∥. Now

∥xy, a∥ = ∥x_{1}x_{2}, a∥
⩽ 2∥x_{1}x_{2}∥_{B}∥a∥_{B}  (by Lemma 5.5)
⩽ 2∥x_{1}∥_{B}∥x_{2}∥_{B}.1  (as (B, ∥.∥_{B}) is a 1-normed algebra)
⩽ 2∥x_{1}, a∥ ∥x_{2}, a∥  (by Lemma 5.5 as a ⊥_{B} x_{1}, a ⊥_{B} x_{2})
= 2∥x, a∥ ∥y, a∥  (again by Proposition 2.1).

If either of x or y in Ẽ is zero, then ∥xy, a∥ = 0 = ∥x, a∥ ∥y, a∥, and we have, for every x, y ∈ Ẽ, ∥xy, a∥ ⩽ 2∥x, a∥ ∥y, a∥. This proves that ∥xy, 2a∥ ⩽ ∥x, 2a∥ ∥y, 2a∥, and similarly it can be shown that ∥xy, 2b∥ ⩽ ∥x, 2b∥ ∥y, 2b∥.

Now to complete the proof of the claim, which proves the theorem also, it remains to be shown that (Ẽ, ∥., .∥) is a 2-Banach algebra with respect to 2a, 2b. To prove this let {x_{n}} be a sequence in Ẽ satisfying lim_{m,n→∞} ∥x_{m} − x_{n}, 2a∥ = lim_{m,n→∞} ∥x_{m}−x_{n}, 2b∥ = 0. By Lemma 5.1, there exists a ϕ ∈ B^{∗} such that ∥ϕ∥ = 1 and ϕ(a) = ∥a∥_{B} = 1. We define a sequence {x^{n}} in B by, x^{n} = x_{n} − ϕ(x_{n})a.

Then ϕ(x^{n}) = ϕ(x_{n}) − ϕ(x_{n})ϕ(a) = 0, for each n ∈ ℕ, and so, for each m, n ∈ ℕ, ϕ(x^{m} − x^{n}) = 0 and so a ⊥_{B} (x^{m} − x^{n}) using Lemma 5.2.

Now, by Lemma 5.5, and Proposition 2.1, we have for each m, n ∈ ℕ,

∥x^{m} − x^{n}∥_{B} = ∥x^{m} − x^{n}∥_{B}∥a∥_{B} ⩽ ∥x_{m} − x_{n}, a∥ → 0 as m, n → ∞

which implies that {x^{n}} is a Cauchy sequence in (B, ∥.∥_{B}). □

Now we consider the following cases:

Case 1. Let the sequence {ϕ(x_{n})} be bounded. There is a subsequence of {ϕ(x_{n})} which is convergent in 𝕂, to say s. Then the corresponding subsequence of {x^{n}} in B converges to some x_{0} − sa in B. But {x^{n}} being Cauchy in B, it converges to x_{0} − sa in B, and as the subsequence of {ϕ(x_{n})} converges to s, it follows that {x_{n}} converges to x_{0} in B. Since {x_{n}} is in Ẽ and Ẽ closed in B, we conclude that {x_{n}} converges to x_{0} in Ẽ in ∥.∥_{B} norm. Using Lemma 5.5, we now have ∥x_{n} − x_{0}, 2a∥ ⩽ 2∥x_{n} − x_{0}∥_{B}∥2a∥_{B} → 0 as n → ∞ and we have an x_{0} ∈ Ẽ such that lim_{n→∞} ∥x_{n} − x_{0}, 2a∥ = 0.

Similarly it can be shown that lim_{n→∞} ∥x_{n} − x_{0}, 2b∥ = 0, and our claim is proved in this case.

Case 2. Let the sequence {ϕ(x_{n})} be unbounded and for infinitely many n’s say for n_{1} < n_{2} < …, ϕ(x_{n_{i}}) = 0. In this case x^{n_{i}} = x_{n_{i}} and as we have observed in the Case 1, {x^{n_{i}}} = {x_{n_{i}}} is a Cauchy sequence in B and hence converges to some x_{0} ∈ B. But Ẽ is closed in B and {x_{n_{i}}} is in Ẽ and hence {x_{n_{i}}} converges to x_{0} in Ẽ. Again as observed above, {x_{n}} is Cauchy in B. Hence, {x_{n}} converges to x_{0} ∈ Ẽ. Then we conclude that, as in the Case 1, lim_{n→∞} ∥x_{n} − x_{0}, 2a∥ = lim_{n→∞} ∥x_{n} − x_{0}, 2b∥ = 0 and the claim is proved in this case also.

Case 3. Let the sequence {ϕ(x_{n})} be unbounded and ϕ(x_{n}) = 0 for finitely many n’s only. We shed x_{n}’s for which ϕ(x_{n}) = 0 and writing the remaining sequence by {x_{n}} again, we have {x_{n}} in Ẽ such that ϕ(x_{n}) ≠ 0 for all n ∈ ℕ.

Now, as {ϕ(x_{n})} is unbounded, {1/ϕ(x_{n})} is bounded. Let s_{n} = 1/ϕ(x_{n}), y^{n} = s_{n}x_{n} − a, for all n ∈ ℕ. Then ϕ(y^{n}) = s_{n}ϕ(x_{n}) − ϕ(a) = 1 − 1 = 0 and so, for all m, n ∈ ℕ, ϕ(y^{m} −y^{n}) = 0, and we have by Lemmas 5.2, 5.5 and Proposition 2.1,

∥y^{m} − y^{n}∥_{B} = ∥a∥_{B}∥y^{m} − y^{n}∥_{B} ⩽ ∥y^{m} − y^{n}, a∥
⩽ ∥s_{n}x_{n} − s_{n}x_{m}, a∥ + ∥(s_{n} − s_{m})x_{m}, a∥
⩽ |s_{n}| ∥x_{n} − x_{m}, a∥ + |s_{n} − s_{m}| ∥x_{m}, a∥

Note that lim_{m,n→∞} ∥x_{m} − x_{n}, a∥ = 0 implies that {∥x_{n}, a∥} is bounded, for there exists ℕ such that ∥x_{m} − x_{n}, a∥ < 1, for all m, n ⩾ N and this implies that ∥x_{m}−x_{N}, a∥ < 1, for all m ⩾ N. Consequently ∥x_{m}, a∥−∥x_{N}, a∥ ⩽ ∥x_{m}−x_{N}, a∥ < 1 for all m ⩾ N and we get ∥x_{m}, a∥ < 1 + ∥x_{N}, a∥ for all m ⩾ N and this proves that ∥x_{m}, a∥ is bounded.

Again {s_{n}} being bounded, there is a subsequence of s_{n}, say {s_{n_{i}}}, converging to some s ∈ 𝕂. Then, as lim_{m,n→∞} ∥x_{m} − x_{n}, a∥ = 0, the above inequality shows that the sequence {y^{n_{i}}} is a Cauchy sequence in B and hence, converges in (B, ∥.∥_{B}). But as for all m, n ∈ ℕ, ∥s_{m}x_{m} − s_{n}x_{n}∥_{B} = ∥y^{m} − y^{n}∥_{B} and {y^{n_{i}}} is Cauchy in B, {s_{n_{i}}x_{n_{i}}} converges in B and as {s_{n_{i}}x_{n_{i}}} is in Ẽ and Ẽ is closed, {s_{n_{i}}x_{n_{i}}} converges to say sy in Ẽ.

Now, for x ∈ Ẽ, define ∥x∥_{1} = max{∥x, a∥, ∥x, b∥}. Observe that for x ∈ Ẽ, ∥x∥_{1} = 0 if and only if x is linearly dependent on both a and b, that is if and only if x = 0 as both a, b ∉ Ẽ. Thus ∥.∥_{1} defines a 1-norm in Ẽ.

Let ( ˜Ẽ, ∥.∥_{1}) be the completion of (Ẽ, ∥.∥). Since for j = 1, 2, a_{1} = 2a, a_{2} = 2b, we have lim_{m,n→∞} ∥x_{m}−x_{n}, a_{j}∥ = 0, and therefore, lim_{m,n→∞} ∥x_{m}−x_{n}∥_{1} = 0, that is, {x_{n}} is Cauchy in ( ˜Ẽ, ∥.∥_{1}). Hence, there is an x_{0} ∈ ˜ Ẽ such that lim_{n→∞} ∥x_{n} − x_{0}∥_{1} = 0, and as {s_{n_{i}}} converges to s in 𝕂, lim_{n→∞} ∥s_{n_{i}}x_{n_{i}} − sx_{0}∥_{1} = 0. But by Lemma 5.5, we have for j = 1, 2 ∥s_{n_{i}}x_{n_{i}} − sy, a_{j}∥ ⩽ 2∥s_{n_{i}}x_{n_{i}} − sy∥_{B}∥a_{j}∥_{B} → 0 as n_{i} → ∞. As {s_{n_{i}}x_{x_{i}}} converges to sy in B, it follows that lim_{n_{i}→∞} ∥s_{n_{i}}x_{n_{i}} −sy∥_{1} = 0, which implies that x_{0} = y ∈ Ẽ, and we conclude that, there exists an x_{0} ∈ Ẽ, s ∈ K, such that for j = 1, 2 lim_{n_{i}→∞} ∥s_{n_{i}}x_{n_{i}} − sx_{0}, a_{j}∥ = 0.

Now observe that ∥s_{n_{i}}x_{n_{i}} − sx_{0}, a_{j}∥ ⩾ |s_{n_{i}}| ∥x_{n_{i}} − x_{0}, a_{j}∥ − |s_{n_{i}} − s| ∥x_{0}, a_{j}∥ and therefore as n_{i} → ∞ we have |s| lim_{n_{i}→∞} ∥x_{n_{i}} − x_{0}, a_{j}∥ ⩽ 0 which implies that lim_{n_{i}→∞} ∥x_{n_{i}} − x_{0}, a_{j}∥ = 0 for j = 1, 2. Now lim_{m,n} ∥x_{m} − x_{n}, a_{j}∥ = 0 and lim_{n_{i}→∞} ∥x_{n_{i}} − x_{0}, a_{j}∥ = 0 imply that for ε > 0 there exists N such that ∥x_{n+p} − x_{n}, a_{j}∥ < ε for all n ⩾ N, p ⩾ 0, and ∥x_{n_{i}} − x_{0}, a_{j}∥ < ε for all n_{i} ⩾ N.

For all n > N, let n_{i} > N and then ∥x_{n}−x_{0}, a_{j}∥ ⩽ ∥x_{n}−x_{n_{i}}, a_{j}∥+∥x_{n_{i}}−x_{0}, a_{j}∥ < ε + ε, and so, lim_{n→∞} ∥x_{n} − x_{0}, a_{j}∥ = 0 for j = 1, 2. This establishes our claim and the theorem is completely established.

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(Srivastava and Lal) National Academy for Scientific Development Education and Research “Shashi Villa”, Mahamanapuri Colony B.H.U. Post Office, Varanasi 221 005, India (Received 22 11 2008)

    neeraj bhuin@yahoo.co.in
    snlmath@gmail.com

Department of Mathematics, Faculty of Science Banaras Hindu University, Varanasi 221 005, India

    sbhatta@bhu.ac.in