Int. J. Contemp. Math. Sciences, Vol. 2, 2007, no. 22, 1077 - 1083
On Banach-Steinhause Theorems in Generalized 2-Normed Spaces M. A¸cıkg¨ oz and M. Menekse University of Gaziantep, Faculty of Science and Arts Department of Mathematics, 27310 Gaziantep, Turkey
[email protected],
[email protected]
Abstract. In this study, we will review Banach-Steinhause theorems of bounded linear operators from a normed space into a generalized 2-normed space. This study is a review on the three works of Zofia Lewandowska. Mathematics Subject Classification: Primary 46A15, Secondary 41A65 Keywords: Generalized 2-normed space, Bounded 2–linear operators, BanachSteinhause theorems 1. Introduction Definition 1. ([1] and [2]) Let X and Y be linear spaces, D be non-empty subset of XxY such that for every x ∈ X, y ∈ Y the sets Dx = {y ∈ Y : (x, y) ∈ D} ,
D y = {x ∈ X : (x, y) ∈ D}
are linear subspaces of the spaces Y and X, respectively. A function ., . : D → [0, ∞) is called a generalized 2-norm on D if it satisfies the following conditions: (N1) x, αy = |α| . x, y = αx, y for all (x, y) ∈ D and every scalar α, (N2) x, y + z = x, y + x, z for all (x, y) , (x, z) ∈ D , (N3) x + y, z = x, z + y, z for all (x, z) , (y, z) ∈ D , Then (D, ., .) is called a 2-normed set. In particular, if D = XxY, then (XxY, ., .) is called a generalized 2-normed space. Moreover if X = Y, then the generalized 2-normed space is denoted by (X, ., .) . Theorem 1. ([1] and [2]) Let (XxY, ., .) be a generalized 2-normed space. (a) A family β of all sets defined by ni=1 {x ∈ X : x, yi < ε}, where n ∈ N, y1 , ..., yn ∈ Y and ε > 0, forms a complete system of neighborhoods of zero for a locally convex topology in Y .
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(b) A family β of all sets defined by ni=1 {y ∈ Y : xi , y < ε}, where n ∈ N, x1 , ..., xn ∈ X and ε > 0, forms a complete system of neighborhoods of zero for a locally convex topology in X. We will denote the above topologies by the symbols τ (X, Y ) and τ (Y, X), respectively. In the case when X = Y , we will denote these topologies by τ1 (X) = τ (X, Y ) and τ2 (X) = τ (Y, X). Definition 2. (i) Let (XxY, ., .) be a generalized 2-normed space and Σ a directed set. A net {xσ }σ∈Σ is convergent to x0 ∈ X in (X, τ (X, Y )) if and only if for all y ∈ Y and each ε > 0 there exists σ0 ∈ Σ such that xσ − x0 , y < ε for all σ ≥ σ0 . Also, a net {yσ }σ∈Σ is convergent to y0 ∈ Y in (Y, τ (Y, X)) if and only if for all x ∈ X and ε > 0 there exists σ0 ∈ Σ such that x, yσ − y0 < ε for all σ ≥ σ0 ([2]) (ii) Let (XxY, ., .) be generalized 2-normed space. A sequence {xn }n≥1 in X is called a Cauchy sequence if for every y ∈ Y and ε > 0 there exists a n ∈ N such that xn − xm , y < ε for all n, m ≥ n0 . (iii) Let (XxY, ., .) be generalized 2-normed space. The space (X, τ (X, Y )) is called sequentially complete if every Cauchy sequence in X is convergent in this space. (iv) Let X be a real normed space and Γ ⊆ Y xY be a 2-normed set, where Y denotes a real linear space. Define F = {(f, g) ∈ L (X, Y )2 : ∀x ∈ X (f (x), g(x)) ∈ Γ & ∃M > 0, ∀x ∈ X f (x) , g (x) ≤ M. x2 } Then, (F, ., .) is a 2-normed set, where the function ., . : F −→ [0, ∞) is defined by f, g = inf M > 0 : ∀x ∈ X f (x) , g (x) ≤ M. x2 . Note that the sets F g = {f ∈ L (X, Y ) : (f , g) ∈ F } & Ff = {g ∈ L (X, Y ) : (f, g ) ∈ F } are linear subspaces of the space L (X, Y ) , for every (f, g) ∈ L (X, Y )2 . Similarly we can define Θ = {(f, g) ∈ L (X, Y )2 : ∀x ∈ X (f (x), g(x)) ∈ Γ & ∃M > 0, ∀x ∈ X f (x) , g (y) ≤ M. x . y} Then, (Θ, ., .) is a 2-normed set, where the function ., . : Θ −→ [0, ∞) is defined by f, g = inf {M > 0 : ∀x ∈ X f (x) , g (y) ≤ M. x . y} .
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Similarly the sets Θg = {f ∈ L (X, Y ) : (f , g) ∈ Θ} & Θf = {g ∈ L (X, Y ) : (f, g ) ∈ Θ} are linear subspaces of the space L (X, Y ) , for every (f, g) L (X, Y )2 . 2. Some Results Theorem 2. Let (X, .) be a Banach space, (Y, ., .) be a generalized 2normed space and {fn }n≥1 a sequence of elements from Θg for some g ∈ L (X, Y ) . Then the following conditions are equivalent: (a) The sequence of 2-norms {fn , g}n≥1 is bounded. (b) ∃M > 0 ∀x, y ∈ X, x ≤ 1, y ≤ 1, ∀n ≥ 1, fn (x) , g (y) ≤ M. (c) The following conditions are true: (i) ∀x ∈ X, ∃Mx > 0, ∀y ∈ X , y ≤ 1, ∀n ≥ 1, fn (x) , g (y) ≤ Mx . (ii) ∀y ∈ X, ∃My > 0, ∀x ∈ X , x ≤ 1, ∀n ≥ 1, fn (x) , g (y) ≤ My . Proof. At first, let us suppose that the sequence of 2-norms {fn , g}n≥1 is bounded. From this, it follows that there exists a positive number M such that fn , g ≤ M for each n ∈ N. Thus for x, y ∈ X, x ≤ 1, y ≤ 1 and n ∈ N, we obtain fn (x) , g (y) ≤ fn , g . x . y ≤ M . Now, let the condition (b) be satisfied. We fix x ∈ X\ {0} . Then for each y ∈ X, y ≤ 1 and n ∈ N we obtain the inequalities: x x fn (x) , g (y) = fn . x , g (y) = x . fn , g (y) ≤ M. x x x If we choose Mx = M. x , then we have the condition (i). Moreover, for x = 0 the condition (i) is satisfied for every positive number Mx .In similar way, taking My = M. y for each y ∈ X \ {0} and any positive number for y = 0, we obtain (ii). Conversely, let (i) and (ii) be satisfied. In XxX, let us define a norm by the formula, (x, y)∗ = x + y for each (x, y) ∈ XxX. It is easy to verfy that (XxX, .∗ ) is a Banach space. Put Anm = {(x, y) ∈ XxX : fn (x) , g (y) ≤ m} and Bm =
∞
Anm
n=1
for m, n ∈ N. We shall show that the sets Bm are closed in (XxX, . ∗ ) for each m ∈ N. At first, we shall show that sets Anm are closed in this space. Let m, n ∈ N and let {(xk , yk ) : k ∈ N} ⊂ Anm be a sequence converging to (x , y ) ∈ XxX. Then fn (xk ) , g (yk ) ≤ m and (xk , yk ) − (x , y )∗ → 0, k → ∞. The last condition is equivalent to the following: xk − x → 0 and yk − y → 0, which implies the convergence of the sequence {xk : k ∈ N} , {yk : k ∈ N} .
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As a consequence these sequences are bounded. There exists K > 0 such that the inequalities xk ≤ K, yk ≤ K are true for each k ∈ N. Using these results we get fn (x ) , g (y ) ≤ M + K. fn , g . xk − x + K. fn , g . yk − y + fn , g . xk − x . yk − y . Letting k → ∞, we obtain fn (x ) , g (y ) ≤ M, which means that (x , y ) ∈ Anm .Therefore the sets Anm are closed for each n, m ∈ N, and hence the sets Bm are closed in (XxX, .∗ ) . ∞ Now we shall show that the equality XxX = Bm is true. Let m=1 x x, y ∈ X, x = 0.Then x = 1. By virtue (ii) there exists My > 0 such that x , gy ≤ My for each n ∈ N. Thus fn (x) , g (y) ≤ My . x for each fn x n ∈ N.If x = 0, then x ≤ 1 and fn (x) , g (y) = 0, g (y) = 0 = My . 0 . as a consequence, for every x, y ∈ X the sequence {fn (x) , g (y) : n ∈ N} is bounded.From this it follows that for any point (x, y) ∈ XxX there exists ∞ m ∈ N such that fn (x) , g (y) ≤ M for every m ∈ n,i.e. (x, y) ∈ Bm .
Thus XxX =
∞ m=1
m=1
Bm .By the well known Baire theorem there exists a set Bm0
with non-empty interior. Therefore Bm0 contains some closed ball with the center (x0 , y0 ) and radius r.Denote it by K ((x0 , y0 ) , r) . Thus for each n ∈ N and (x, y) ∈ K ((x0 , y0) , r) , we have fn (x) , g (y) ≤ m0 . Let us take x, y ∈ X such that x ≤ 2r and y ≤ 2r . Then (x, y)∗ = x + y ≤ r and (x, y)∗ = (x + x0 , y + y0 ) − (x0 , y0)∗ ≤ r. Therefore fn (x + x0 ) , g (y + y0 ) ≤ m0 . In particular, fn (x0 ) , g (y0 ) ≤ m0 .Thus, fn (x) , g (y) ≤ fn (x + x0 ) , g (y + y0 ) + fn (x + x0 ) , g (y0 ) + fn (x0 ) , g (y + y0 ) + fn (x0 ) , g (y0 ) ≤ 2m0 + fn (x0 ) + fn (x0 ) , g (y0 ) + fn (x0 ) , g (y) + g (y0 ) ≤ 4m0 + fn (x) , g (y0 ) + fn (x0 ) , g (y) . So we have shown that the inequalities x ≤ condition
r 2
and y ≤
r 2
imply the
fn (x) , g (y) ≤ 4m0 + fn (x) , g (y0 ) + fn (x0 ) , g (y) . Now, let x, y ∈ X, x ≤ 1 and y ≤ 1. Because r2 x ≤ 2r and r2 y ≤ 2r , then r r r
r fn x , g y ≤ 4m0 + fn x , g (y0 ) + fn (x0 ) , g y . 2 2 2 2
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As a sequence, we obtain 16m0 2 + (fn (x) , g (y0 ) + fn (x) , g (y)) for each n ∈ N. r2 r Applying (i) we have that there exists Mx0 > 0 such that for every y ∈ X, y ≤ 1 and n ∈ N the inequality fn (x0 ) , g (y) ≤ Mx0 is true. However the assumption (ii) implies there exists My0 > 0 such that for every x ∈ X, x ≤ 1 and n ∈ N the inequality fn (x) , g (y0 ) ≤ My0 is satisfied. So fn (x) , g (y) ≤
16m0 2 + (My0 + Mx0 ) r2 r for each n ∈ N and x, y ∈ X such that x ≤ 1, y ≤ 1.Therefore fn (x) , g (y) ≤
fn , g = sup {fn (x) , g (y) : x, y ∈ X, x ≤ 1, y ≤ 1} 16m0 + 2r (Mx0 + My0 ) ≤ r2 for each n ∈ N. So the sequence {fn , g : n ∈ N} is bounded and the proof is complete. Theorem 3. ([3]) Let (X, .) be a Banach space, (Y, ., .) a generalized 2-normed space and g a linear operator from X to Y.If {fn : n ∈ N} ⊂ θg is pointwise convergent to f ∈ L (X, Y ) and satisfies one of the conditions (a) , (b) , (c) from Theorem 2.1 then f ∈ θg . Theorem 4. Let A be a linearly dense set in a Banach space (X, .) , (Y, ., .) be a generalized 2-normed space such that (Y, τ1 (Y )) is a Hausdoff sequentially complete space. Let g be a linear operator from X into Y and {fn }n≥1 ⊆ Θg . Then the following conditions are equivalent: (a) The sequence {fn }n≥1 is pointwise convergent to f ∈ L (X, Y ) and the conditions (i), (ii) from Theorem 2 are satisfied. (b) The sequence {fn }n≥1 is pointwise convergent to f ∈ Θg on the set A and the sequence of 2-norms {fn , g}n≥1 is bounded. Proof. If the sequence {fn (x) : n ∈ N} is convergent to f (x) ∈ Y for each x ∈ X, then it convergent also for x ∈ A ⊂ X. Moreover this follows from theorem 2.1 and theorem 3-the sequence {fn , g : n ∈ N} is bounded and f ∈ θg . Now, we will suppose that the sequence {fn : n ∈ N} is pointwise convergent to f ∈ θg on the set A and the conditions (i) , (ii) hold. Let X0 be the vector subspace of the Banach space X generated by A. So X0 is normed space. Let x, y ∈ X0 .Then x = a1 x1 + ... + ak xk , y = b1 y1 + ... + bt yt , where ai , bj ∈ R, xi , yj ∈ A, i = 1, 2, 3, ..., k; j = 1, 2, 3, ..., t and k, t ∈ N. Thus, it follows from assumptions on fn , f, g that fn (x) − f (x) , g (y) = a1 (fn (x1 ) − f (x1 )) + ... + ak (fn (xk ) − f (xk )) , b1 g (y1 ) + b2 g (y2 ) + ... + bt g (yt ) .
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Using properties of 2-norms, we get
fn (x) − f (x) , g (y) ≤
t k
|ai bj | . fn (xi ) − f (xi ) , g (yj ) .
i=1 j=1
Because lim fn (xi ) − f (xi ) , g (yj ) = 0 for ach xi , yj ∈ A, n→∞
then lim fn (x) − f (x) , g (y) = 0.i.e. the sequence {fn : n ∈ N} is conn→∞
vergent to f on X0 .Let fn , g ≤ M for every n ∈ N. Let us take a number ε > 0, x ∈ X and y ∈ X such that y = 0. Since X0 is a dense set in X, we can ε choose x0 ∈ X0 , x0 = 0 such that x − x0 < 6.M.y . Moreover, there exists ε y0 ∈ X0 with the property y − y0 < 6.M.x0 . The sequence {fn (x0 ) : n ∈ N} is convergent in (Y, τ1 (Y )) , so it is a Cauchy sequence in this space. Therefore, there exists a number n0 such that fn (x0 ) − fm (x0 ) , g (y0 ) < 3ε for each n, m ≥ n0 . As a consequence, we obtain, fn (x) − fm (x) , g (y) ≤
fn (x) − fn (x0 ) , g (y) + fn (x0 ) − fm (x0 ) , g (y0 ) + fm (x0 ) − fm (x) , g (y)
≤
fn , g . x − x0 . y + fn (x0 ) − fm (x0 ) , g (y − y0 ) + g (y0 )
≤
+ fm , g . x − x0 . y 2M. x − x0 . y + fn (x0 ) − fm (x0 ) , g (y − y0 )
0, ∀y ∈ X , y ≤ 1, ∀n ≥ 1, fn (x) , gn (y) ≤ Mx . (ii) ∀y ∈ X, ∃My > 0, ∀x ∈ X , x ≤ 1, ∀n ≥ 1, fn (x) , gn (y) ≤ My . References [1] Z. Lewandowska, Linear operators on generalized 2-normed spaces, Bull. Math. Soc. Sci. Math. Roumanie (N.S.) 42(90), no.4 (1999), 353-368. [2] Z. Lewandowska, Z., On 2-normed sets, Glas.Math.Ser.III, 38(58), no.1, (2003) 99-110. [3] Z. Lewandowska, Z., (2003), Banach Steinhaus theorems for bounded linear operators with values in a generalized 2-normed space, Glas. Mat. Ser.III, 38(58), no.2, 329-340.
Received: March 23, 2007
