The projection methods in countably normed spaces
 Nashat Faried^{1}Email author and
 Hany A ElSharkawy^{1}
https://doi.org/10.1186/s1366001405400
© Faried and ElSharkawy; licensee Springer 2015
Received: 9 April 2014
Accepted: 20 December 2014
Published: 5 February 2015
Abstract
It is well known that a normed space is uniformly convex (smooth) if and only if its dual space is uniformly smooth (convex). We extend the notions of uniform convexity (smoothness) from normed spaces to countably normed spaces ‘in which there is a countable number of compatible norms’. We get some fundamental links between Lindenstrauss duality formulas. A duality property between uniform convexity and uniform smoothness of countably normed spaces is also given. Moreover, based on the compatibility of those norms, it is interesting to show that ‘from any point in a real uniformly convex complete countably normed space, the nearest point to a nonempty convex closed subset of the space is the same for all norms’, which is helpful in further studies for fixed points.
Keywords
1 Introduction
Definition 1.1
(Uniformly convex space [1–3])
A normed linear space E is called uniformly convex if for any \(\epsilon\in(0,2]\) there exists a \(\delta=\delta(\epsilon)> 0\) such that if \(x,y \in E\) with \(\x\=1\), \(\y\=1\) and \(\xy\ \geq\epsilon\) then \(\\frac{1}{2} (x+y)\ \leq1\delta\).
Definition 1.2
Definition 1.3
(Uniformly smooth space [1–3])
Definition 1.4
Let K be a nonempty convex subset of a real normed linear space E. For strict contraction selfmappings of K into itself, with a fixed point in K, a well known iterative method ‘the celebrated Picard method’ has successfully been employed to approximate such fixed points. If, however, the domain of a mapping is a proper subset of E (and this is the case in several applications), and it maps K into E, this iteration method may not be well defined. In this situation, for Hilbert spaces and uniformly convex uniformly smooth Banach spaces, this problem has been overcome by the introduction of the metric projection in the recursion formulas (see, for example, [4–6]).
Definition 1.5
(Metric projection [1, 4, 5, 7])
It is our purpose in this paper to extend the notion of uniform convexity and uniform smoothness of Banach spaces to countably normed spaces. Moreover, by extending some theorems to the case of uniformly convex uniformly smooth countably normed spaces, we prove the existence and uniqueness of nearest points in these spaces. Our theorems generalize some results of [3, 8, 9].
2 Preliminaries
Definition 2.1
(Countably normed space [10, 11])
Remark 2.2
([11])
Proposition 2.3
([10])
Let E be a countably normed space. Then E is complete if and only if \(E = \bigcap_{n = 1}^{\infty} E_{n}\).
Each Banach space \(E_{n}\) has a dual, which is a Banach space and denoted by \(E_{n}^{*}\).
Proposition 2.4
([10])
Remark 2.5
A countably normed space is metrizable and its metric d can be defined by \(d(x, y) = \sum_{i=1}^{\infty} \frac{1}{2^{i}} \frac{\ x  y\_{i}}{1 + \x  y\_{i}}\).
Example 2.6
An example of a countably normed space is the space of entire functions that are analytic in the unit disc \(z < 1\) with the topology of the uniform convergence on any closed subset of the disc and with the collection of norms \(\x(z)\_{n} = \max_{z\leq1  \frac{1}{n}} x(z)\).
Example 2.7
For \(1 < p < \infty\), the space \(\ell^{p+0} := \bigcap_{q > p} \ell^{q}\) is a countably normed space. In fact, one can easily see that \(\ell^{p+0} = \bigcap_{n} \ell^{p_{n}}\) for any choice of a monotonic decreasing sequence \(\{p_{n}\}\) converging to p. Using Proposition 2.3 and the fact that \(\ell^{p_{n}}\) is Banach for every n, it is clear now that the countably normed space \(\ell^{p+0}\) is complete.
3 Main results
In this section, we give new definitions and prove our main theorems.
Definition 3.1
A countably normed space E is said to be uniformly convex if \((E_{i}, \\cdot\_{i})\) is uniformly convex for all i, i.e., if for each i, \(\forall \epsilon> 0\), \(\exists \delta _{i}(\epsilon) >0 \) such that if \(x, y \in E_{i}\) with \(\x\_{i} = 1 = \y\_{i}\) and \(\x  y\_{i} \ge\epsilon\), then \(1  \\frac{x + y}{2}\_{i} > \delta_{i}\).
If \(\inf\delta_{i} > 0\), then one may call the space E equiuniformly convex.
Definition 3.2
Proposition 3.3
A countably normed linear space E is uniformly convex if and only if for each i we have \(\delta_{E_{i}} (\epsilon) > 0 \) for all \(\epsilon\in(0,2]\).
Proof
Assume that \((E_{i}, \\cdot\_{i})\) is uniformly convex for all i. Then, for each i, given \(\epsilon> 0\) there exists \(\delta_{i} > 0\) such that \(\delta_{i} \le1  \ \frac{x + y}{2} \_{i}\) for every x and y in \(E_{i}\) such that \(\x\_{i} = 1 = \y\_{i}\) and \(\x  y\_{i} \ge\epsilon\). Therefore \(\delta_{E_{i}} (\epsilon) \ge\delta_{i} > 0\) for all i.
Conversely, assume that for each i, \(\delta_{E_{i}} (\epsilon) > 0 \) for all \(\epsilon \in(0,2]\). Fix \(\epsilon\in(0,2]\) and take x, y in \(E_{i}\) with \(\ x\_{i} = 1 = \y\_{i}\) and \(\x  y\_{i} \ge\epsilon\), then \(0 < \delta_{E_{i}} (\epsilon) \le1  \ \frac{x + y}{2} \_{i}\) and therefore \(\ \frac{x + y}{2} \_{i} \le1  \delta_{i}\) with \(\delta_{i} = \delta_{E_{i}} (\epsilon)\) which does not depend on x or y. Then \((E_{i}, \\cdot\_{i})\) is uniformly convex for all i and hence the countably normed space E is uniformly convex. □
In Proposition 3.3, we showed the conditions of equivalence to uniform convexity of countably normed spaces, and now in Theorem 3.4, we show the conditions of equivalence to uniform smoothness of countably normed spaces.
Theorem 3.4
Proof
Assume that \((E_{i}, \\cdot\_{i})\) is uniformly smooth for each i and if \(\epsilon> 0\), then there exists \(\delta_{i} > 0\) such that \(\frac{\x + y\_{i} + \x  y\_{i}}{2}  1 < \frac{\epsilon}{2} \y\ _{i}\) for every x, y in \(E_{i}\) with \(\x\_{i} = 1\) and \(\y\_{i} = \delta _{i}\). This implies that for each i, we have \(\rho_{E_{i}} (t) < \frac {\epsilon}{2} t\) for every \(t < \delta_{i}\).
Conversely, for each i, given \(\epsilon> 0\) suppose that there exists \(\delta_{i} > 0\) such that \(\frac{\rho_{E_{i}} (t)}{t} < \frac {\epsilon}{2}\) for every \(t < \delta_{i}\). Let x, y be in \(E_{i}\) such that \(\x\_{i} = 1\) and \(\y\_{i} = \delta_{i}\). Then with \(t = \y\_{i}\) we have \(\x+y\_{i} +\xy\_{i} < 2 + \epsilon\y\_{i}\). Then \((E_{i}, \\cdot\_{i})\) is uniformly smooth for all i and hence the countably normed space E is uniformly smooth. □
Now, we prove one of the fundamental links between the Lindenstrauss duality formulas.
Proposition 3.5
Proof
The following two theorems give or determine some duality property concerning uniform convexity and uniform smoothness of countably normed spaces.
Theorem 3.6
Proof
 ‘⟹’:

Assume that \((E_{i_{0}}^{*}, \\cdot\_{i_{0}})\) is not uniformly convex for some \(i_{0}\). Therefore, \(\delta _{E_{i_{0}}^{*}}(\epsilon_{0}) = 0\) for some \(\epsilon_{0} \in(0, 2]\). Using Proposition 3.5, we get for any \(\tau > 0\),which shows that E is not uniformly smooth.$$0 < \frac{\epsilon_{0}}{2} \leq\frac{\rho_{E_{i_{0}}}(\tau)}{\tau},\quad \mbox{hence } \lim _{\tau} \frac{\rho_{E_{i_{0}}}(\tau)}{\tau} \neq0, $$
 ‘⟸’:

Assume that E is not uniformly smooth, thenthis means that there exists \(\epsilon> 0\) such that for every \(\delta > 0\) we can find \(t_{\delta}\) with \(0 < t_{\delta} < \delta\) and \(\rho_{E_{i_{0}}}(t_{\delta}) \geq t_{\delta}\epsilon\). Consequently, one can choose a sequence \((\tau_{n})\) such that \(0 < \tau_{n} < 1\), \(\tau_{n} \rightarrow 0\), and \(\rho _{E_{i_{0}}}(\tau_{n}) \geq\epsilon \tau_{n} >\frac{\epsilon}{2} \tau_{n}\). Using Proposition 3.5, for every n there exists \(\epsilon_{n} \in(0, 2]\) such that$$\exists i_{0} \mbox{:} \quad \lim_{t\to0^{+}} \frac{\rho_{E_{i_{0}}(t)}}{t} \neq0, $$which implies$$\frac{\epsilon}{2} \tau_{n} \leq\frac{\tau_{n} \epsilon_{n}}{2}  \delta _{E_{i_{0}}^{*}}(\epsilon_{n}), $$in particular \(\epsilon< \epsilon_{n}\) and \(\delta_{E_{i_{0}}^{*}}(\epsilon _{n})\rightarrow 0\). Recalling the fact that \(\delta_{E^{*}}\) is a nondecreasing function we get \(\delta_{E_{i_{0}}^{*}}(\epsilon) \leq\delta _{E_{i_{0}}^{*}}(\epsilon_{n})\rightarrow 0\). Therefore \(E^{*}\) is not uniformly convex.$$0 < \delta_{E_{i_{0}}^{*}}(\epsilon_{n}) \leq\frac{\tau_{n}}{2} ( \epsilon_{n}  \epsilon), $$
The proof of the following theorem is easy.
Theorem 3.7
The following example is a direct application on the previous theorems.
Example 3.8
(Uniformly convex and uniformly smooth countably normed space)
It is well known that the space \(\ell^{2}=\{(x_{n}):x_{n} \in\mathbb{R}, \forall n\in\mathbb{N},\sum_{n=1}^{\infty} x_{n}^{2} <\infty\}\) with the norm \(\(x_{n})\_{2}=\sqrt{\sum_{n=1}^{\infty} x_{n}^{2}}\) is uniformly convex normed space. Besides, it is uniformly smooth, because \((\ell ^{2})^{*} =\ell^{2}\).
On \(\ell^{2}\), we define a countable number of seminorms by \(p_{i, i+1}((x_{n}))=\sqrt{x_{i}^{2}+x_{i+1}^{2}}\), where \(i=1,3,5,\ldots\) . Also defining a countable number of compatible norms on \(\ell^{2}\) by \(\ (x_{n})\_{i}=p_{i, i+1}((x_{n}))+\(x_{n})\_{2}\). Then \((\ell^{2}, \{\\cdot\_{i}, i=1,3,5,\ldots\})\) is a countably normed space.
Now, it is clear that \((\ell^{2}, \{\\cdot\_{i}, i=1,3,5,\ldots\})\) is uniformly smooth (convex) countably normed space, as its completion \(\ell^{2}_{i}=(\ell^{2}, \\cdot\_{i})\) is uniformly smooth (convex) complete normed space for all i.
Proposition 3.9
Proof
The first direction is trivial.
Conversely, assume that \(\x_{n}  x_{m}\_{i} \rightarrow 0\), ∀i. Then, for each i, \(\forall \epsilon> 0\), \(\exists n_{i}\): \(n, m > n_{i}\) implies \(\x_{n}  x_{m}\_{i} < \frac{\epsilon}{2}\).
In the following theorem, we establish one of the most important and interesting geometric property in uniformly convex countably normed spaces, the metric projection point \(\bar{x}\) ‘the solution of the minimization problem’ is well defined, that is to say, we have ‘existence and uniqueness’ for all norms.
Theorem 3.10
Claim
Let E be a real uniformly convex Banach space. If \(\{ x_{n}\}\) is a sequence in E: (a) \(\lim_{n\to\infty} \x_{n}\ = 1\) and (b) \(\lim_{n, m\to\infty} \x_{n} + x_{m}\ = 2\). Then \(\{x_{n}\}\) is a Cauchy sequence in \((E, \\cdot\)\).
Proof of the claim
Proof of Theorem 3.10
Since \(x\notin K\) and K is closed for each i, we have \(d_{i} := \inf_{\xi\in K}\x \xi\_{i} > 0\).
Declarations
Acknowledgements
The authors would like to thank the editor and the referees for their valuable comments.
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Authors’ Affiliations
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