JOURNAL OF THE CHUNGCHEONG MATHEMATICAL SOCIETY Volume 19, No. 1, March 2006

COMPLETION OF QUASI–NORMED ALGEBRAS AND QUASI–NORMED MODULES

Chun-Gil Park*

Abstract. In this paper, the completion of a quasi-normed algebra and the completion of a quasi-normed module over a are con- structed.

1. Introduction It is well-known that the rational line Q is not complete but can be enlarged to the real line R which is complete. And this completion R of Q is such that Q is dense in R. It is quite important that an arbitrary incomplete normed space can be completed in a similar fashion. Banach spaces play an important role in many branches of and its applications ([2], [4]–[8], [10]). We recall some basic facts concerning quasi-Banach spaces and some pre- liminary results.

Definition 1. ([3], [9]) Let X be a linear space. A quasi- is a real- valued function on X satisfying the following: (1) kxk ≥ 0 for all x ∈ X and kxk = 0 if and only if x = 0. (2) kλxk = |λ| · kxk for all λ ∈ R and all x ∈ X. (3) There is a constant K ≥ 1 such that kx + yk ≤ K(kxk + kyk) for all x, y ∈ X. The pair (X, k · k) is called a quasi-normed space if k · k is a quasi-norm on X. A quasi- is a complete quasi-normed space.

Received by the editors on November 11, 2005. 2000 Mathematics Subject Classifications: Primary 46A16, 46Bxx, 54D35. Key words and phrases: Quasi-normed algebra, Quasi-normed module over a Banach algebra, Completion.

9 10 C. PARK

A quasi-norm k · k is called a p-norm (0 < p ≤ 1) if

kx + ykp ≤ kxkp + kykp for all x, y ∈ X. In this case, a quasi-Banach space is called a p-Banach space.

Definition 2. ([1]) Let (X, k · k) be a quasi-normed space. The quasi- normed space (X, k · k) is called a quasi-normed algebra if X is an algebra and there is a constant C > 0 such that kxyk ≤ Ckxk · kyk for all x, y ∈ X. A quasi-Banach algebra is a complete quasi-normed algebra. If the quasi-norm k·k is a p-norm then the quasi-Banach algebra is called a p-Banach algebra.

Definition 3. Let (A, | · |) be a Banach algebra and X a module over A. A quasi-norm is a real-valued function on X satisfying the following: (1) kxk ≥ 0 for all x ∈ X and kxk = 0 if and only if x = 0. (2) kaxk = |a| · kxk for all a ∈ A and all x ∈ X. (3) There is a constant K ≥ 1 such that kx + yk ≤ K(kxk + kyk) for all x, y ∈ X. The pair (X, k · k) is called a quasi-normed module over A if k · k is a quasi- norm on X. A quasi-Banach module over A is a complete quasi-normed module over A. A quasi-norm k · k is called a p-norm (0 < p ≤ 1) if

kx + ykp ≤ kxkp + kykp for all x, y ∈ X. In this case, a quasi-Banach module over A is called a p-Banach module over A.

In Section 2, the completion of a quasi-normed algebra is constructed. In Section 3, the completion of a quasi-normed module over a Banach algebra is constructed. COMPLETION OF QUASI–NORMED ALGEBRAS 11

2. Completion of quasi-normed algebras In this section, we construct a completion of a quasi-normed algebra.

Definition 4. Let (X, k · kX ) and (Y, k · kY ) be quasi-normed algebras. (1) A mapping L : X → Y is said to be isometric or an isometry if for all x, y ∈ X

kLx − LykY = kx − ykX .

(2) The algebra X is said to be isometric with the algebra Y if there exists a bijective isometry of X onto Y . The algebras X and Y are called isometric algebras.

Theorem 1. Let X = (X, k · kX ) be a quasi-normed algebra. Assume that the quasi-norm k · k is a p-norm. Then there exist a quasi-Banach algebra Xb and an isometry L from X onto a subalgebra Y of Xb which is dense in Xb. The algebra Xb is unique up to isometry.

Proof. We divide the proof into four steps. b Step I. We construct a quasi-Banach algebra (X, k · kXb ). 0 Let {xn} and {xn} be Cauchy sequences in X. Define {xn} to be equiv- 0 0 alent to {xn}, written {xn} ∼ {xn}, if

0 (2.1) limn→∞kxn − xnkX = 0.

Let Xb be the set of all equivalence classes of Cauchy sequences. We write

{xn} ∈ xb to mean {xn} is a member of xb and a representative of the class xb. We must make Xb into an algebra. To define on Xb the three algebraic operations of an algebra, we consider any x,b yb ∈ Xb and representatives

{xn} ∈ xb and {yn} ∈ yb. We set zn = xn + yn. Then {zn} is Cauchy in X since kzn − zmkX = kxn + yn − (xm + ym)kX ≤ Kkxn − xmkX + Kkyn − ymkX .

We define the sum zb = xb+yb of xb and yb to be the equivalence class for which

{zn} is a representative, i.e., {zn} ∈ zb. This definition is independent of 12 C. PARK the particular choice of Cauchy sequences belonging to xb and yb. In fact, the 0 0 equality (2.1) shows that if {xn} ∼ {xn} and {yn} ∼ {yn}, then {xn +yn} ∼ 0 0 {xn + yn} because

0 0 0 0 kxn + yn − (xn + yn)kX ≤ Kkxn − xnkX + Kkyn − ynkX .

Next, we consider any x,b yb ∈ Xb and representatives {xn} ∈ xb and {yn} ∈ yb. One can easily show that the sequences {xn} and {yn} are bounded. We set zn = xn · yn. Then {zn} is Cauchy in X since kzn − zmkX = kxn · yn − xm · ymkX ≤ Kk(xn − xm)ynkX + Kkxm(yn − ym)kX

≤ KCkxn − xmkX · kynkX + KCkxmkX · kyn − ymkX .

We define zb = xb· yb of xb and yb to be the equivalence class for which {zn} is a representative, i.e., {zn} ∈ zb. This definition is independent of the particular choice of Cauchy sequences belonging to xb and yb. In fact, the equality (2.1) 0 0 0 0 shows that if {xn} ∼ {xn} and {yn} ∼ {yn}, then {xn · yn} ∼ {xn · yn} because

0 0 0 0 0 kxn · yn − xn · yn)kX ≤ Kk(xn − xn)ynkX + Kkxn(yn − yn)kX 0 0 0 ≤ KCkxn − xnkX · kynkX + KCkxnkX · kyn − ynkX .

Similarly, we define the product αxb ∈ Xb of a scalar α and xb to be the equivalence class for which {αxn} is a representative. Again, this defini- tion is independent of the particular choice of a representative of xb. The zero element of Xb is the equivalence class containing all Cauchy sequences which converge to zero. It is not difficult to see that those three algebraic operations have all the properties required by the definition, so that Xb is an algebra. We now set

(2.2) kxb − ybkXb = limn→∞kxn − ynkX , where {xn} ∈ xb and {yn} ∈ yb. We show that this limit exists. We have

p p p p kxn − ynkX ≤ kxn − xmkX + kxm − ymkX + kym − ynkX . COMPLETION OF QUASI–NORMED ALGEBRAS 13

So p p p p kxn − ynkX − kxm − ymkX ≤ kxn − xmkX + kym − ynkX , and a similar inequality with m and n interchanged, i.e.,

p p p p kxm − ymkX − kxn − ynkX ≤ kxn − xmkX + kym − ynkX .

Hence ¯ ¯ ¯ p p ¯ p p (2.3) kxn − ynkX − kxm − ymkX ≤ kxn − xmkX + kym − ynkX .

Since {xn} and {yn} are Cauchy, we can make the right side as small as we please. This implies that the limit in (2.2) exists since R is complete. We must show that the limit in (2.2) is independent of the particular 0 0 choice of representatives. If {xn} ∼ {xn} and {yn} ∼ {yn}, then by (2.1) ¯ ¯ ¯ p 0 0 p ¯ 0 p 0 p kxn − ynkX − kxn − ynkX ≤ kxn − xnkX + kyn − ynkX , which tends to zero as n → ∞. This implies the assertion

0 0 lim kxn − ynkX = lim kxn − ynkX . n→∞ n→∞

b Now we prove that k · kXb in (2.2) is a quasi-norm on X.

Obviously, k · kXb satisfies Definition 1 (1) and (2). Furthermore, since k · kX is a quasi-norm on X, there is a constant K ≥ 1 such that

kx + ykX ≤ K(kxkX + kykX ) for all x, y ∈ X. Thus

kxb + ybkXb ≤ K(kxbkXb + kybkXb )

b b for all x,b yb ∈ X. So k · kXb is a quasi-norm on X.

Note that the inequality (2.2) implies that the quasi-norm k · kXb is equiv- alent to a p-norm. Step II. We construct an isometry L : X → Y ⊂ Xb. 14 C. PARK

With each b ∈ X we associate the class b ∈ Xb which contains the constant Cauchy sequence (b, b ··· ). This defines a mapping L : X → Y onto the subalgebra Y = L(X) ⊂ Xb. The mapping L is given by b 7→ b = Lb, where (b, b, ··· ) ∈ b. We see that L is an isometry since (2.2) becomes simply

b kb − bckXb = kb − ckX .

Here bc is the class of {yn} where yn = c for all n ∈ N. Any isometry is injective, and L : X → Y is surjective since L(X) = Y . Hence Y and X are isometric. From the definition it follows that on Y the operations of algebra induced from Xb agree with those induced from X by means of L.

We show that Y is dense in Xb. We consider any xb ∈ Xb. Let {xn} ∈ xb. For every ² > 0 there is an N ∈ N such that ² kx − x k < n N X 2 for all n > N. Let (xN , xN , ··· ) ∈ xcN . Then xcN ∈ Y . By (2.2), ² kxb − xcN k b = lim kxn − xN kX ≤ < ². X n→∞ 2 This shows that every ²-neighborhood of the arbitrary xb ∈ Xb contains an element of Y . Hence Y is dense in Xb. Step III. We prove the completeness of Xb.

Let {xcn} be any Cauchy sequence in Xb. Since Y is dense in Xb, for every xcn there is a zcn ∈ Y such that 1 (2.4) kxc − zck < . n n Xb n Hence by Definition 1 (3),

kzcm − zcnkXb ≤ Kkzcm − xcmkXb + Kkxcm − xcnkXb + Kkxcm − zcnkXb K K < + Kkxc − xck + m m n Xb n and this is less than any given ² > 0 for sufficiently large m and n because

{xcn} is a Cauchy sequence. Hence {zcn} is a Cauchy sequence. Since L : COMPLETION OF QUASI–NORMED ALGEBRAS 15

−1 X → Y is isometric and zcn ∈ Y , the sequence {zn}, where zn = L (zcn), is a Cauchy sequence in X. Let xb ∈ Xb be the class to which {zn} belongs.

We show that xb is the limit of {xcn}. By Definition 1 (3) and (2.4),

kxcn − xbkXb ≤ Kkxcn − zcnkXb + Kkzcn − xbkXb K (2.5) < + Kkzc − xbk . n n Xb

Since {zn} ∈ xb and zcm ∈ Y , so that (zm, zm, zm, ··· ) ∈ zcm, the inequality (2.5) becomes

K kxcn − xbk b < + K lim kzn − zmkX X n m→∞ and the right side is smaller than any given ² > 0 for sufficiently large n.

Hence the arbitrary Cauchy sequence {xcn} in Xb has the limit xb ∈ Xb, and Xb is complete. Step IV. We show the uniqueness of Xb up to isometry. e If (X, k · kXe ) is another complete metric algebra with a subalgebra Z dense in Xe and isometric with X, then for any x,e ye ∈ Xe we have sequences

{xfn}, {yfn} in Z such that xfn → xe and yfn → ye. So

kxe − yek e = lim kxfn − yfnk e X n→∞ X follows from ¯ ¯ ¯kxe − yekp − kxf − yfkp ¯ ≤ kxe − xfkp + kye − yfkp → 0. Xe n n Xe n Xe n Xe

Here the inequality is similar to (2.3). Since Z is isometric with Y ⊂ Xb and Y = Xb, the norms on Xe and Xb must be the same. Hence Xe and Xb are isometric. ¤

3. Completion of quasi-normed modules over a Banach algebra In this section, we construct a completion of a quasi-normed module over a Banach algebra. 16 C. PARK

Definition 5. Let (X, k·kX ) and (Y, k·kY ) be quasi-normed modules over a Banach algebra A. (1) A mapping L : X → Y is said to be isometric or an isometry if for all x, y ∈ X

kLx − LykY = kx − ykX .

(2) The A-module X is said to be isometric with the A-module Y if there exists a bijective isometry of X onto Y . The A-modules X and Y are called isometric A-modules.

Theorem 2. Let X = (X, k · kX ) be a quasi-normed module over a Banach algebra A. Assume that the quasi-norm k · k is a p-norm. Then there exist a quasi-Banach module Xb over A and an isometry L from X onto a submodule Y of Xb which is dense in Xb. The quasi-Banach A-module Xb is unique up to isometry.

b Proof. We construct a quasi-Banach A-module (X, k · kXb ). 0 Let {xn} and {xn} be Cauchy sequences in X. Define {xn} to be equiv- 0 0 alent to {xn}, written {xn} ∼ {xn}, if

0 (3.1) limn→∞kxn − xnkX = 0.

Let Xb be the set of all equivalence classes of Cauchy sequences. We write

{xn} ∈ xb to mean {xn} is a member of xb and a representative of the class xb. We must make Xb into an A-module. To define on Xb the two algebraic operations of a module, we consider any x,b yb ∈ Xb and representatives {xn} ∈ xb and {yn} ∈ yb. We set zn = xn + yn. Then {zn} is Cauchy in X since kzn − zmkX = kxn + yn − (xm + ym)kX ≤ Kkxn − xmkX + Kkyn − ymkX .

We define the sum zb = xb+yb of xb and yb to be the equivalence class for which

{zn} is a representative, i.e., {zn} ∈ zb. This definition is independent of the particular choice of Cauchy sequences belonging to xb and yb. COMPLETION OF QUASI–NORMED ALGEBRAS 17

Similarly, we define the product axb ∈ Xb of a ∈ A and xb to be the equivalence class for which {axn} is a representative. Again, this definition is independent of the particular choice of a representative of xb. The zero element of Xb is the equivalence class containing all Cauchy sequences which converge to zero. It is not difficult to see that those two algebraic operations have all the properties required by the definition, so that Xb is an A-module. We now set

(3.2) kxb − ybkXb = limn→∞kxn − ynkX , where {xn} ∈ xb and {yn} ∈ yb. By the same method as in the proof of b Theorem 1, one can show that k · kXb in (3.2) is a quasi-norm on X. Next, we construct an isometry L : X → Y ⊂ Xb. With each b ∈ X we associate the class b ∈ Xb which contains the constant Cauchy sequence (b, b ··· ). This defines a mapping L : X → Y onto the submodule Y = L(X) ⊂ Xb. The mapping L is given by b 7→ b = Lb, where (b, b, ··· ) ∈ b. We see that L is an isometry since (3.2) becomes simply

b kb − bckXb = kb − ckX .

Here bc is the class of {yn} where yn = c for all n ∈ N. Any isometry is injective, and L : X → Y is surjective since L(X) = Y . Hence Y and X are isometric. From the definition it follows that on Y the operations of module induced from Xb agree with those induced from X by means of L. The rest of the proof is similar to the proof of Theorem 1. ¤

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* Department of Mathematics Chungnam National University Daejeon 305-764, Korea E-mail: [email protected]