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E-GOVERNMENT IN ZIMBABWE: AN OVERVIEW OF PROGRESS MADE AND CHALLENGES AHEAD

Ab Hamid Ganie1*, Mobin Ahmad2 and Neyaz Ahmad Sheikh3
  1. Department of Mathematics SSM College of Engineering Technology, Pattan, Jammu and Kashmir
  2. Department of Mathematics Faculty of Science Jazan University, Saudi Arabia profmobin@yahoo.com
  3. Department of Mathematics National Institute of Technology, Srinagar, Jammu and Kashmir neyaznit@yahoo.co.in
Corresponding author: Ab Hamid Ganie, E-mail: ashamidg@rediffmail.com
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Abstract

This paper surveyed the level of preparedness by the Zimbabwe government in using information and communication technologies to enhance the range and quality of services provided to the citizen; and determined the extent of, and continuous improvement efforts of Zimbabwe leaders towards the attainment of connected government. Data for the study was a secondary data adapted from the Internet. The UN global e-government readiness ranking and the models of e-government were downloaded, adapted and used as a benchmark for measuring the web readiness of the Zimbabwe governments in 2005, 2008, 2010 and 2012. It was discovered that the Zimbabwean government has demonstrated its willingness to apply information and communication technologies (ICTs) in their public administration, but is at the emerging stages. A serious impediment which characterized e-government readiness in Zimbabwe is low level of human capital and knowledge economy. The implications include poor provision of government services and underutilization of ICTs facilities in Zimbabwe which might result in the widening of ‘access divide’ between the rich and the poor.

INTRODUCTION

A sequence space is defined to be a linear space of real or complex sequences. Throughout the paper N, R and C denotes the set of non-negative integers, the set of real numbers and the set of complex numbers respectively. Letω denote the space of all sequences (real or complex). Let l and c be Banach spaces of bounded and convergent sequences image supremum norm image . Let T denote the shift operator onω , that is, image image and so on. A Banach limit L is defined on l∞ as a non-negative linear functional such that L is invariant i.e., L(Sx)=L(x) and L(e)=1, e=(1,1,1,…) [1].
Lorentz, called a sequence {xn} almost convergent if all Banach limits of x, L(x), are same and this unique Banach limit is called F-limit of x [1]. In his paper, Lorentz proved the following criterian for almost convergent sequences.
A sequence image l∞ ∈ is almost convergent with F-limit L(x) if and only if
image
where, image uniformly in n≥0.
We denote the set of almost convergent sequences by f.
Several authors including Duran [2], Ganie et al. [3-7], King [8], Lorentz [1] and many others have studied almost convergent sequences. Maddox [9,10] has defined x to be strongly almost convergent to a number α if
image
By [f] we denote the space of all strongly almost convergent sequences. It is easy to see that image
The concept of paranorm is related to linear matric spaces. It is a generalization of that of absolute value. Let X be a linear space. A function P:x→R is called a paranorm, if [11,12].
image
image
image
image (triangle inequality)
image is a sequence of scalars with λn→λ (n→∞) and (xn) is a sequence of vectors with image then image image image ),(continuity of multiplication of vectors).
A paranorm p for which p(x)=0 implies x=0 is called total. It is well known that the metric of any linear metric space is given by some total paranorm [10].
The following inequality will be used throughout this paper. Let p=(pk) be a sequence of positive real numbers with image and let image For image . We have that (Equation 1) [9,11].
image (1)
Nanda defined the following [13,14]:
image
image
image
The difference sequence spaces,
image
where X= ∞ l , C and C0, were studied by Kizmaz [15].
It was further generalized by Ganie et al. [5], Et and Colak [16], Sengonul [17] and many others.
Further, it was Tripathy et al. [18] generalized the above notions and unified these as follows:
image
Where
image
and
image
Recently, M. Et [19] defined the following:
image
image
image
Following Maddox [20]and Ruckle [21], a modulus function g is a function from [0,∞) to [0,∞) such that
(i) g(x)=0 if and only if x=0,
(ii)image
(iii) g is increasing,
(iv) g if continuous from right at x=0.
Maddox [10] introduced and studied the following sets:
image
image
of sequences that are strongly almost convergent to zero and strongly almost convergent.
Let p=(pk) be a sequence of positive real numbers with image and H=max(1, M).

MAIN RESULTS:

In the present paper, we define the spaces image and image as follows:
image
image
image
Where (pk) is any bounded sequence of positive real numbers.
Theorem 1: Let (pk) be any bounded sequence and g be any modulus function. Then image and image are linear space over the set of complex numbers.
Proof: We shall prove the result forimage and the others follows on similar lines. Let image Now for image , we can find positive numbers Aα,Bᵝ such that image image Since f is subadditive and image is linear
image
image
image
image
As n→∞, uniformly in m. This proves that image is linear and the result follows.
Theorem 2: Let g be any modulus function. Then
image
Proof: We shall prove the result for image and the second shall be proved on similar lines. Let image
Now, by definition of g, we have
image
image
Thus, for any number L, there exists a positive integer KL such that image we have
image
image
Since, image , we have x= ∈ image ), and the proof of the result follows.
Theorem 3: image is a paranormed space with
image
Proof: From Theorem 2, for each image image exists. Also, it is trivial that image and image for x=0. Since, h(0)=0, we have image for x=0. Since, image therefore, by Minkowski’s inequality and by definition of g for each n that
image
image
image
which shows that image is sub-additive. Further, let α be any complex number. Therefore, we have by definition of g, we have
image
where, Sα is an integer such that α<Sα. Now, let α→0 for any fixed x with image By definition of g for image we have for image that (Equation 2)
image (2)
As g is continuous, we have, for 1≤n≤N and by choosing α so small that (Equation 3)
image (3)
Consequently, (2) and (3) gives that image as α→0.
Theorem 4: Let X be any of the spaces [f,g], [f,g]0 and [f,g]. Then,image is strict. In general, image all j=1,2,…,r-1 and the inclusion is strict.
Proof: We give the proof for the space [f, g]∞ and others can be proved similarly. So, let image Then, we have image
Since, g is increasing function, we have
image
image
image
Thus, image Continuing in this way, we shall get image for j=1,2,…,r-1. The inclusion is strict. For this, we consider x=(kr) and is in image but does not belong to imagefor f(x)=x and n=1. ( if x=(kr), then image and imagefor all image
Theorem 5: image
Proof: The proof is obvious from Theorem 4 above.
Theorem 6: Let g, g1 and g2 be any modulus functions. Then,
(i) image
(ii) image
Proof: Let ε be given small positive number and choose δ with 0< δ<1 such that g(t)< ε for 0<t≤ δ. We put image and consider
image
where the first summation is over image and second summation is over yk+m> δ. As g is continuous, we have (Equation 4)
image (4)
and for yk+m> δ, we use the fact that
image
Now, by definition of g, we have for yk+m> δ that
image
Thus (Equation 5),
image (5)
Consequently, we see from (4) and (5) that image
To prove (ii), we have from (1) that
image
Let image Consequently, by adding above inequality form k=1 to k=n, we have image and the result follows.
Theorem 7: Let g, g1 and g2 be any modulus functions. Then,
image
image
image
image
Proof: The follows as a routine verification as of the Theorem 6.
Theorem 8: The spaces image and image are not solid in general.
Proof: To show that the spaces image and image not solid in general, we consider the following example.
Let pk=1 for all k and g(x)=x with r=1=n. Then, image but image when αk=(-1)k for all image Hence is result follows.
From above Theorem, we have the following corollary.
Corollary 9: The spaces image and image are not perfect.
Theorem 10: The spaces image and image are not symmetric in general.
Proof : To show that the spaces image and image are not perfect in general, to show this, let us consider pk=1 for all k and g(x)=x with n=1. Then, image Let the re-arrangement of (xk) be (yk) where (yk) is defined as follows,
image
Then, image and this proves the result.

References