### 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}∗ δ_{0}= δ_{0}∗ δ_{k}:= δ_{k},

δ_{n}∗ δ_{n}:= 1

p^{n−1}_{0} (p0− 1)δ_{0}+

n−1

X

k=1

p^{k−1}_{0} δ_{k}+p_{0}− 2
p_{0}− 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_{n}dµ(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,

(p_{0}− 1)p^{k−1}_{0} , 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)p^{n−1}_{0} < ∞, (2.1)

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

|x_{n}| (p_{0}− 1)p^{n−1}_{0} ^{1}_{q}

< 1.

But for each n ∈ N,

(p_{0}− 1)p^{n−1}_{0} ^{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}(p_{0}− 1)p^{n−1}_{0} < |x_{n}|^{q}(p_{0}− 1)p^{n−1}_{0} .
By (2.1) we getP∞

n=1|x_{n}|^{r}(p_{0}− 1)p^{n−1}_{0} < ∞, 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

(p^{n−1}_{0} (p0−1)n^{an})^{1}^{r} if n ≥ 1,
then x ∈ L^{r}(Z+) − L^{q}(Z+).

Proof. From

k x k^{r}_{r}=

∞

X

n=0

|x_{n}|^{r}m({n})

= 1 +

∞

X

n=1

1

p^{n−1}_{0} (p_{0}− 1)n^{a}^{n}(p_{0}− 1)p^{n−1}_{0}

= 1 +

∞

X

n=1

1
n^{a}^{n} < ∞

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 ∈ (_{a}^{1}

t0, 1), and there is N_{0} ∈ N such that _{a}^{a}_{t0}^{n} < s < 1 for all
n ≥ N0. Hence,

1
n^{s} < 1

n

an at0

for all n ≥ N_{0}. We have
kx k

r at0r at0 =

∞

X

n=0

|x_{n}|

r

at0 m({n})

= 1 +

∞

X

n=1

1

p^{n−1}_{0} (p_{0}− 1)n^{a}^{n}^{1}_{r}

r at0

(p_{0}− 1)p^{n−1}_{0}

= 1 +

∞

X

n=1

(p0− 1)p^{n−1}_{0} ^{1−}_{at0}^{1} 1
n

an at0

.

Since P∞ n=1 1

n^{s} = ∞ and

(p_{0}− 1)p^{n−1}_{0} ^{1−}

1 at0 ≥ 1, we have

∞

X

n=N0

1
n^{s} ≤

∞

X

n=N0

1 n

an t0

≤ 1 +

∞

X

n=1

(p_{0}− 1)p^{n−1}_{0} ^{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}− x_{0}|^{q}+

∞

X

i=1

|x_{n,i}− x_{i}|^{q}(p_{0}− 1)p^{i−1}_{0} < 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_{q}for 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)p^{n−1}_{0} ^{1}_{q}

= kχnk_{q}≤ M kχ_{n}k_{r}= M (p0− 1)p^{n−1}_{0} ^{1}_{r}
,
and so,

(p_{0}− 1)p^{n−1}_{0} ^{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^{q}^{n}(Z+), (2.2)

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

n=1L^{q}^{n}(Z+)) =S∞

n=1L^{q}^{n}(Z+). So, applying Theorem
2 (put Z_{n}:= (L^{q}^{n}(Z+), k · k_{q}_{n}) 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.

References

[1] R. M. Aron, V. I. Gurariy, J. B. Seoane-Sep´ulveda: Lineability and spaceability of sets of functions on R, Proc. Amer. Math. Soc., 133, n. 3 (2005), 795–803.

[2] W. R. Bloom, H. Heyer: Harmonic analysis of probability measures on hypergroups, De GruYte, Berlin, 1995.

[3] G. Botelho, D. Diniz, V. V. F´avaro, D. Pellegrino: Spaceability in Banach and quasi-Banach sequence spaces, Linear Algebra Appl., 434, n. 5 (2011), 1255–1260.

[4] G. Botelho, V. V. F´avaro, D. Pellegrino, J. B. Seoane-Sep´ulveda:

Lp[0, 1]\S

q>pLq[0, 1] is spaceable for every p > 0, Linear Algebra Appl., 436 (2012), 2963–2965.

[5] C. F. Dunkl, D. E. Ramirez: A family of countably compact P^{∗}-hypergroups, Trans.

Amer. Math. Soc., 202 (1975), 339–356.

[6] V. P. Fonf, V. I. Gurariy, M. I. Kadets: An infinite dimensional subspace of C[0, 1]

consisting of nowhere differentiable functions, C. R. Acad. Bulgare Sci., 52 (1999), 13–16.

[7] V. I. Gurariy, L. Quarta: On lineability of sets of continuous functions, J. Math. Anal.

Appl., 294 (2004), 62–72.

[8] H. Hahn: Uber folgen linearer operationen, Monatsh. Math. Phys., 32 (1922), 3–88.

[9] D. Kitson, R. M. Timoney: Operator ranges and spaceability, J. Math. Anal. Appl., 378 (2011), 680–686.