Equilibrium problems on quasi-weighted graphs

We give a new minimization theorem for equilibrium problems on a quasi-weighted graph. This result generalizes the graphical version of the Ekeland’s variational principle for equilibrium problems on weighted graphs (Alfuraidan and Khamsi in Proc Am Math Soc 9:33–40, 2022).


Preliminaries
We start by recalling the definition of the quasi-metric and some needed terminology on the concept of quasimetric spaces. For more details on these spaces and their relation with EVP, the reader may consult [5]. The pair (T, d) is called a quasi-metric space.
In general, left-limits are not necessarily unique. We denote the set of left-limits by We denote the set of right-limits by (c) A sequence {t n } ⊂ T is left-Cauchy (resp. right-Cauchy) if for each ε > 0 there is an N ∈ N such that d(t n , t m ) < ε (resp. d(t m , t n ) < ε), for any m > n > N .
The following technical lemma will be needed.
is a left-lower semicontinuous function. Now, we present some basic graph theory definitions that will be used in the sequel. For more detail, the reader can consult any graph theory book.
consists of a finite, nonempty set of vertices V and a set of edges E. Each edge e is an ordered pair (t, s) of vertices. (2) If e is an edge of G and s = t, then e is called a loop. If E(G) contains all the loops, then G is a reflexive graph.
A multi-edge is a set of two or more edges having same ends.
In the following definition, we adapt the property (OSC) introduced in [3] to the case of quasi-weighted graphs. d) be a quasi-weighted digraph. We say that G satisfies the property (OSC) if and only if for all G-decreasing sequence {t n } in V (G) and t ∈ l ({t n }), we have

Remark 2.8
In this paper, we study quasi-weighted graphs which make sense in reality. Indeed, we can consider the GPS Manhattan roads as an example of a strongly complete quasi-weighted graph. In order to illustrate this example, consider an oriented square in the plane, see the figure below. The Manhattan quasi-distance is defined in the following way: 1. d(x, y) = 0 if and only if x = y. 2. d(x, y) = length of the path following the orientation going from x to y.
Notice that: (i) d is not symmetric since, for example, where D is the usual Euclidean distance. (ii) If t belong to the path going from x to y, then (t, y) = d(x, y).
(iii) If s does not belong to the path going from x to y, then (x, s) + d(s, y).
Thus d is a quasi-distance. (iv) Consider the following sequence: This example captures the case of any left-Cauchy sequence.

Main results
We are now ready to state our main result.

Theorem 3.1 Let G = (V (G), E(G), d) be a reflexive transitive quasi-weighted digraph such that (V (G), d) is strongly left G-complete. Assume that G satisfies Property (OSC). Let H
Assume that the following conditions hold: .) is lower bounded and G-left lower semicontinuous, ∀s ∈ V (G).
Then, for every ε > 0 and s 0 ∈ V (G), there exists s ∈ V (G) such that (s, s 0 ) ∈ E(G) and Proof If we replace the quasi-metric d(·, ·) by the quasi-metric ε d(·, ·), all the assumptions are satisfied. Therefore and without loss of any generality, we may assume ε = 1. Let s ∈ V (G). Set Since G is reflexive and the property (i), we get s ∈ A(s) thus A(s) =, for any s ∈ V (G). Let t ∈ A(s) and w ∈ A(t). Since G is transitive, we get (w, s) ∈ E(G). for any t ∈ V (G). By construction of {s n }, we know that s n+h ∈ A(s n ), for any n, h ∈ N, which implies Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.