A spaceability result in the context of hypergroups

A spaceability result in the context of hypergroups

Seyyed Mohammad Tabatabaie

Department of Mathematics, University of Qom, Qom, Iran sm.tabatabaie@qom.ac.ir

Bentol Hoda Sadathoseyni

Department of Mathematics, University of Qom, Qom, Iran sb.sadathosseini@stu.qom.ac.ir

Received: 14.8.2017; accepted: 4.12.2017.

Abstract. In this paper, by an elementary constractive technique, it is shown that L^r(Z+) − S

q<rL^q(Z⁺) is non-empty, where Z⁺is the dual of a compact countable hypergroup introduced by Dunkl and Ramirez. Also, we prove that for each r > 1, L^r(Z+)−S

q<rL^q(Z+) is spaceable.

Keywords:locally compact hypergroup, spaceability, L^p–space

MSC 2000 classification:primary 46A45, secondary 43A62, 46B45

1 Introduction and Notations

Suppose that X is a topological vector space. A subset S ⊆ X is called spaceable if S ∪ {0} is large enough to contain a closed infinite dimensional subspace of X. This concept was first introduced in [6] and then the term spaceable was used in [1]. There have been several further works on this notion (for example see [4], [7] and [9]). It is well-known that for each 0 < p, there are sequences in lp not belonging to S

q<plq. Usually, this fact is proved by non-constructive techniques, althought a constructive proof, depending on the Principle of Uniform Boundedness can be found in [8]. Actually, in [3] it is proved that for every p > 0, the set lp−S

q<plq is even spaceable.

Let Z+ be the set of non-negative integers equipped with discrete topology, M (Z+) be the space of all Radon measures on Z+, and p0 be a fixed prime number. For any k ∈ Z+ and distinct non-zero m, n ∈ Z+ we put

δ_k∗ δ₀= δ₀∗ δ_k:= δ_k,

δ_n∗ δ_n:= 1

pⁿ⁻¹₀ (p0− 1)δ₀+

n−1

X

k=1

p^k−1₀ δ_k+p₀− 2 p₀− 1δ_n, http://siba-ese.unisalento.it/ c 2018 Universit`a del Salento

and δ_m ∗ δ_n := δ_max{m,n}. Then, by [5], Z+ equipped with the convolution

∗ : M (Z+) × M (Z+) → M (Z+) defined by µ ∗ ν :=

Z

Z+

Z

Z+

δ_m∗ δ_ndµ(m)dν(n), (µ, ν ∈ M (Z+),

and the identity mapping on Z+as involution, is a commutative discrete hyper- group. Also, the measure m defined by

m({k}) :=







1, if k = 0,

(p₀− 1)p^k−1₀ , if k ≥ 1,

is a Haar measure for Z+. For more details about locally compact hypergroups see [2].

In the sequel, we consider Z+ with above structure, and for each s > 0 we denote L^s(Z+) := L^s(Z+, m).

In this paper, by an elementary technique, we give an algorithm that gen- erates lots of sequences in L^p(Z+) but not in L^q(Z+) for every q < p. This argument can be regarded as a variant of the technique that was used to prove the Banach-Steinhaus Theorem in [8]. Also, we prove that for each r > 1, the set L^r(Z+) −S

q<rL^q(Z+) is spaceable in L^r(Z+).

2 Main Results

This is well-known that:

Lemma 1. There is a sequence (an)^∞_n=1 of real numbers such that an > 1 for all n, limn→∞an= 1, and P∞

n=1 1

n^an converges.

Lemma 2. If 0 < q < r < ∞, then L^q(Z+) ⊆ L^r(Z+).

Proof. Let f := (xn)^∞_n=0∈ L^q(Z+). Since kf k^q_q= |x0|^q+

∞

X

n=1

|x_n|^q(p0− 1)pⁿ⁻¹₀ < ∞, (2.1)

there is a positive number N such that for each n ≥ N ,

|x_n| (p₀− 1)pⁿ⁻¹₀
¹_q

< 1.

But for each n ∈ N,

(p₀− 1)pⁿ⁻¹₀
¹_q

≥ 1.

Hence, for all n ≥ N , |x_n| < 1, and then |x_n|^r < |x_n|^q. Therefore, for each n ≥ N we have

|x_n|^r(p₀− 1)pⁿ⁻¹₀ < |x_n|^q(p₀− 1)pⁿ⁻¹₀ . By (2.1) we getP∞

n=1|x_n|^r(p₀− 1)pⁿ⁻¹₀ < ∞, i.e. f ∈ L^r(Z+). ^QED

Lemma 3. Let q < r, p0 be a prime number, and (an)^∞_n=1 be a sequence as in Lemma 1. If the sequence x = (x_n)^∞_n=0 is defined by

x_n:=







1 if n = 0,

1

(pⁿ⁻¹₀ (p0−1)n^an)^1r if n ≥ 1, then x ∈ L^r(Z+) − L^q(Z+).

Proof. From

k x k^r_r=

∞

X

n=0

|x_n|^rm({n})

= 1 +

∞

X

n=1

1

pⁿ⁻¹₀ (p₀− 1)n^an(p₀− 1)pⁿ⁻¹₀

= 1 +

∞

X

n=1

1 n^an < ∞

we conclude that x ∈ L^r(Z+). Let 0 < q < r. Since _a^r

n < r for every n and limn→∞ r

an = r, there is some t0∈ N such that q < _a^r_t0. Since

n→∞lim a_n at0

= 1 at0

< 1,

we can choose s ∈ (_a¹

t0, 1), and there is N₀ ∈ N such that _a^a_t0ⁿ < s < 1 for all n ≥ N0. Hence,

1 n^s < 1

n

an at0

for all n ≥ N₀. We have kx k

r at0r at0 =

∞

X

n=0

|x_n|

r

at0 m({n})

= 1 +

∞

X

n=1





1

pⁿ⁻¹₀ (p₀− 1)n^an
1_r





r at0

(p₀− 1)pⁿ⁻¹₀

= 1 +

∞

X

n=1

(p0− 1)pⁿ⁻¹₀
¹⁻_at0¹ 1 n

an at0

.

Since P∞ n=1 1

n^s = ∞ and

(p₀− 1)pⁿ⁻¹₀
¹⁻

1 at0 ≥ 1, we have

∞

X

n=N0

1 n^s ≤

∞

X

n=N0

1 n

an t0

≤ 1 +

∞

X

n=1

(p₀− 1)pⁿ⁻¹₀
¹⁻

1 at0 1

n

an at0

≤ k x k

r at0r at0

,

and so, x /∈ L

r

at0(Z+). Therefore, x /∈ L^q(Z+) because q < _a^r

t0 and L^q(Z+) ⊆ L

r

at0(Z+). ^QED

Theorem 1. If 1 ≤ q < r < ∞, then L^q(Z+) is not closed in L^r(Z+).

Proof. First Proof: Let C_c(Z+) be the space of all functions from Z+ into C with finite support. We have Cc(Z+) ⊆ L^q(Z+) ⊆ L^r(Z+), and Cc(Z+) is dense in L^q(Z+) and in L^r(Z+). So, if L^q(Z+) is closed in L^r(Z+), then we have L^q(Z+) = L^r(Z+), a contradiction.

Second Proof: Let T : L^q(Z+) −→ L^r(Z+) be the inclusion mapping. We claim that T is continuous. For this, suppose that (f_n)^∞_n=0 is a sequence in L^q(Z+) with fn := (xn,i)^∞_i=0, and fn → f in L^q(Z+), where f := (xi)^∞_i=0 ∈ L^q(Z+). For each 0 < ε < 1 there is N ∈ N such that for all n ≥ N ,

kf_n− f k^q_q= |x_n,0− x₀|^q+

∞

X

i=1

|x_n,i− x_i|^q(p₀− 1)pⁱ⁻¹₀ < ε < 1.

Hence, |x_n,i− x_i|^q< 1 for all n ≥ N and i ∈ Z+. Since q < r, we get

|x_n,i− x_i|^r≤ |x_n,i− x_i|^q,

where n ≥ N and i ∈ Z+, and so kf_n−f k_r ≤ kf_n−f k_qfor all n ≥ N . Therefore, f_n→ f with respect to the norm k · k_r, and T is continuous.

Now, we show that k · kq and k · kr are not equivalent. In contrast, suppose that there exists M > 0 such that for each f ∈ L^q(Z+), kf k_q≤ M kf k_r. For any n ∈ N, let χn be the characteristic function of the set {n}. Then for all n ∈ N, we have

(p0− 1)pⁿ⁻¹₀
¹_q

= kχnk_q≤ M kχ_nk_r= M (p0− 1)pⁿ⁻¹₀
¹_r , and so,

(p₀− 1)pⁿ⁻¹₀
¹_q−¹_r

≤ M,

which is a contradiction, since ¹_q−¹_r > 0. Therefore, the norms k · kq and k · kr

are not equivalent. Finally, suppose that L^q(Z+) is closed in L^r(Z+). Since T is continuous, the induced mapping

T : Le ^q(Z+) −→ T (L^q(Z+)) = (L^q(Z+), k · k_r), T (f ) := T (f ) = f,e is a bijective continuous linear operator between Banach spaces. So, by the Open Mapping Theorem, eT is an isomorphism, and in particular the norms k · kq and k · k_r are equivalent on L^q(Z+), a contradiction. Therefore, L^q(Z+) is not closed

in L^r(Z+). ^QED

By [9, Theorem 2.4], we can conclude:

Corollary 1. If 0 < q < r < ∞, then L^r(Z+) − L^q(Z+) is spaceable.

Here, we recall the following result (see [9, Theorem 3.3]).

Theorem 2. Suppose that X is a Frechet space, and for each n = 1, 2, . . ., Zn is a Banach space. Let for each n ∈ N, Tn : Zn −→ X be bounded lin- ear operators. If Y := span(S∞

n=1T_n(Z_n)) is not closed in X, then X − Y is spaceable.

Theorem 3. If r > 1, then

L^r(Z+) −[

q<r

L^q(Z+)

is spaceable.

Proof. The sequence x ∈ L^r(Z+) constructed in Lemma 3 does not belong to L^q(Z+) for all q < r, i.e. L^r(Z+) −S

q<rL^q(Z+) is non-empty.

Let q < r. Since Cc(Z+) is dense in L^r(Z+), and C_c(Z+) ⊆ L^q(Z+) ⊆ L^r(Z+),

L^q(Z+) is dense in L^r(Z+), and so S

q<rL^q(Z+) is not closed in L^r(Z+). By Lemma 2, the setS

q<rL^q(Z+) is actually a countable union:

[

q<r

L^q(Z+) =

∞

[

n=1

L^qn(Z+), (2.2)

where q1< q2 < · · · < r, and moreover, the sequence (qn)nconverges to r. Easily, one can see that span (S∞

n=1L^qn(Z+)) =S∞

n=1L^qn(Z+). So, applying Theorem 2 (put Z_n:= (L^qn(Z+), k · k_qn) and T_n: Z_n,→ L^r(Z+)), L^r(Z+) −S

q<rL^q(Z+)

is spaceable. ^QED

Acknowledgements. We would like to thank the referee of this paper for a careful reading and helpful comments. Also, we are specially grateful to Professor G. Botelho for his effective comments and suggestions on this paper.

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