In this section, we first present the existence result for an εapproximate solution for (EP). We also study the continuity of εapproximate solution maps for (PEP).
Theorem 9
Let
K
be a nonempty closed convex subset of a Hadamard space
X
which has the convex hull finite property. For any given
\(\varepsilon>0 \), let
\(f : K \times K \rightarrow\mathbb{R} \)
be a bifunction such that

(i)
for any
\(x\in K \), \(f(x,x) \geq0 \);

(ii)
for every
\(x\in K \), the set
\(\{y\in K : f(x,y) + \varepsilon< 0\} \)
is convex set;

(iii)
for every
\(y\in K \), \(f(\cdot,y) \)
is upper semicontinuous;

(iv)
there exist a compact set
\(L\subseteq X \)
and a point
\(y_{0}\in L\cap K \)
such that
$$ f(x,y_{0}) + \varepsilon< 0, \quad \forall x\in K\backslash L. $$
Then there exists a point
\(x_{0}\in L\cap K \)
such that
$$ f(x_{0},y) + \varepsilon\geq0, \quad\forall y\in K. $$
Proof
Define the mapping \(G : K \rightrightarrows K \) as follows:
$$ G(y) = \bigl\{ x\in K : f(x,y) + \varepsilon\geq0 \bigr\} \quad\mbox{for all } y \in K. $$
We first prove that \(G(y) \) is closed for all \(y\in K \). Let \(\{ x_{m}\} \) be a sequence in \(G(y) \) such that \(x_{m}\rightarrow x' \) for some \(x'\in X \). Then we have \(f(x_{m},y) + \varepsilon\geq0 \) for all m. It follows from (iii) that
$$ f\bigl(x',y\bigr) + \varepsilon\geq\limsup_{m\rightarrow\infty} f(x_{m},y) + \varepsilon\geq0. $$
This implies that \(G(y) \) is closed for all \(y \in K \). By condition (iv) there exists a point \(y_{0} \in K \) for which \(G(y_{0}) \subseteq L \). Since L is compact, we see that \(G(y_{0}) \) is also. We want to prove that \(G(\cdot) \) is a KKM mapping. We prove this by contradiction, suppose that G is not a KKM mapping. Then there exist a finite subset \(A = \{x_{1},x_{2},\ldots,x_{n}\} \) of K and a point \(\hat{x} \in co(A) \) such that \(\hat{x} \notin G(y_{i}) \) for all \(i=1,\ldots,n \). Thus
$$ f(\hat{x},y_{i}) + \varepsilon< 0,\quad \forall i\in\{1,2,\ldots,n\}. $$
That is, for any \(i \in\{1,\ldots,n\} \),
$$ y_{i} \in\bigl\{ y \in K: F (\hat{x}, y) + \varepsilon< 0\bigr\} . $$
Since the set \(\{y \in K: F (\hat{x}, y) + \varepsilon< 0 \} \) is convex, we get
$$ \hat{x} \in co\bigl(\{y_{1},y_{2},\ldots,y_{n}\} \bigr) \subseteq\bigl\{ y\in K : f(\hat {x},y) + \varepsilon< 0 \bigr\} , $$
which is a contradiction to the assumption (i). Then by Lemma 4, there exists a point \(x_{0} \in K \) such that
$$ x_{0}\in\bigcap_{y\in K} G(y). $$
Hence, there exists \(x_{0} \in L \cap K \) such that
$$ f(x_{0},y) + \varepsilon\geq0,\quad \forall y\in K. $$
□
Remark 10

(i)
Our Theorem 9 is closely related to Theorem 3.3 in [13], which is presented in the sense of a minimax inequality.

(ii)
If \(\varepsilon=0 \), then we have the existence result for an exact solution for (EP) which encloses the result in [12].
By setting \(L = K \) in Theorem 9, the following corollary is immediately obtained.
Corollary 11
Let
K
be a nonempty compact convex subset of a Hadamard space
X
which has the convex hull finite property. For any given
\(\varepsilon>0 \), let
\(f : K \times K \rightarrow\mathbb{R} \)
be a bifunction such that

(i)
for any
\(x\in K \), \(f(x,x) \geq0 \);

(ii)
for every
\(x\in K \), the set
\(\{y\in K : f(x,y) + \varepsilon< 0\} \)
is convex set;

(iii)
for every
\(y\in K \), \(f(\cdot,y) \)
is upper semicontinuous.
Then there exists a point
\(x_{0}\in K \)
such that
$$ f(x_{0},y) + \varepsilon\geq0, \quad \forall y\in K. $$
The following example shows that the εapproximate solution for (EP) depends on ε.
Consider \(\mathbb{R}^{2} \) with the usual Euclidean meter \(d(\cdot,\cdot ) \) and \(\\cdot\\), which are defined by
$$ d(x,y) = \xy\ = \sqrt{(x_{1}y_{1})^{2} + (x_{2}y_{2})^{2}}, $$
where \(x=(x_{1},x_{2}) \) and \(y=(y_{1},y_{2}) \). We define the radial metric \(d_{r} \) by
$$ d_{r}(x,y) = \textstyle\begin{cases} d(x,y), & \mbox{if } y=tx \mbox{ for some } t\in\mathbb{R};\\ d(x,0) + d(y,0), &\mbox{otherwise}. \end{cases} $$
Then \(X:=(\mathbb{R}^{2},d_{r}) \) is an \(\mathbb{R} \)tree with radial meter \(d_{r} \).
Example 12
Let X be an \(\mathbb{R} \)tree with radial meter \(d_{r} \). We put \(K = [[(1,0),(0,1)]] \) and \(f(x,y) = (x_{1}+x_{2})  0.3 \). After calculating we easily get
$$ \widetilde {S}(\varepsilon,\mu) = \textstyle\begin{cases} [[(0,0.3),(0,1)]] \cup[[(0.3,0),(1,0)]], &\mbox{if } 0 \leq \varepsilon < 0.3, \\ [[(1,0),(0,1)]], &\mbox{if } \varepsilon \geq0.3. \end{cases} $$
The following corollary is a sufficient condition for the existence of the parametric εapproximate solution (PEP).
Corollary 13
For any given
\(\varepsilon>0 \), assume that there exists a neighborhood
\(N(\mu_{0}) \)
of
\(\mu_{0} \)
satisfying the following conditions:

(i)
for each
\(\mu\in N(\mu_{0}) \), \(K(\mu) \)
is a nonempty, compact and convex valued;

(ii)
for each
\(\mu\in N(\mu_{0}) \)
and each
\(x\in K(N(\mu_{0})) \), \(f(x,x,\mu) \geq0 \);

(iii)
for each
\(x\in K(N(\mu_{0})) \), the set
\(\{y\in K(N(\mu_{0})) : f(x,y,\mu) + \varepsilon< 0\} \)
is convex set;

(iv)
for each
\(\mu\in N(\mu_{0}) \)
and each
\(y\in K(N(\mu_{0})) \), \(f(\cdot,y,\mu) \)
is upper semicontinuous on
\(K(N(\mu_{0})) \).
Then for each
\(\mu\in N(\mu_{0}) \), \(\widetilde {S}(\varepsilon ,\mu) \)
is nonempty and compact.
Now, we give the sufficient conditions for continuity of an approximate solution S̃ at \((\varepsilon_{0},\mu_{0}) \). Relying on the existence theorem for εapproximation (EP), we assume that the solution of the εapproximation exists for all \((\varepsilon ,\mu)\in \mathbb{R}_{+}\cup\{0\}\times M \).
Theorem 14
Consider the
εapproximate (PEP). We assume that the following conditions hold:
 (C_{1}):

K
is continuous at
\(\mu_{0} \)
and
\(K(\mu_{0}) \)
has compact and convex valued;
 (C_{2}):

there exists a neighborhood
\(N(\mu_{0}) \)
of
\(\mu_{0} \)
such that
\(f(\cdot,\cdot,\cdot) \)
is continuous on
\(K(N(\mu_{0}))\times K(N(\mu _{0}))\times\{\mu_{0}\} \);
 (C_{3}):

for each
\(y\in K(\mu_{0}) \), \(f(\cdot,y,\mu_{0}) \)
is a geodesic concave function on
\(K(N(\mu_{0})) \).
Then
\(\widetilde {S}(\cdot,\cdot) \)
is l.s.c. at
\((\varepsilon_{0},\mu_{0}) \).
Proof
We first prove that \(\widetilde {S}(\cdot,\mu_{0}) \) is l.s.c. at \(\varepsilon_{0} \). Suppose not, then there exist an open set \(G\subseteq X \) and a sequence \(\{\varepsilon_{n}\} \) with \(\varepsilon_{n} \rightarrow \varepsilon_{0} \) such that
$$ \widetilde {S}(\varepsilon_{0},\mu_{0}) \cap G \neq \emptyset\quad\mbox{but } \widetilde {S}(\varepsilon_{n},\mu_{0})\cap G= \emptyset \mbox{ for all } n\in\mathbb{N}. $$
(3.1)
Note that \(\widetilde {S}(\varepsilon_{1},\mu_{0})\subseteq \widetilde {S}(\varepsilon_{2},\mu_{0}) \), if \(\varepsilon_{1}\leq\varepsilon_{2} \).
We claim that \(\varepsilon_{0}>\varepsilon_{n} \) for all \(n\in\mathbb{N} \). If not, there exists \(\overline{N}\in\mathbb{N} \) such that \(\varepsilon_{0} \leq\varepsilon_{\overline{N}} \). That is,
$$ \widetilde {S}(\varepsilon_{0},\mu_{0})\subseteq \widetilde {S}( \varepsilon_{\bar{N}},\mu_{0}),\quad \mbox{and so } \emptyset\neq \widetilde {S}(\varepsilon_{0},\mu_{0})\cap G \subseteq \widetilde {S}( \varepsilon_{\bar{N}},\mu_{0})\cap G, $$
which is a contradiction to (3.1). So we have the claim.
Let \(x_{0}\in \widetilde {S}(0,\mu_{0}) \) and \(\bar{x}\in \widetilde {S}(\varepsilon_{0},\mu _{0})\cap G \). We set \(x_{n} = \frac{\varepsilon_{0}  \varepsilon_{n} }{\varepsilon_{0}}\bar{x}\oplus\frac{\varepsilon_{n}}{\varepsilon_{0}}x_{0} \), then \(x_{n}\rightarrow\bar{x} \) as \(n\rightarrow\infty\). Since \(\bar {x}\in G \), there exists \(n_{0}\in\mathbb{N} \) such that
$$ \frac{\varepsilon_{0}  \varepsilon_{n_{0}}}{\varepsilon_{0}}\bar{x}\oplus \frac{\varepsilon_{n_{0}}}{\varepsilon_{0}}x_{0} = x_{n_{0}} \in G. $$
(3.2)
We claim that \(x_{n_{0}}\in \widetilde {S}(\varepsilon_{n_{0}},\mu_{0}) \). Since \(x_{0}\in \widetilde {S}(0,\mu_{0}) \) and \(\bar{x}\in \widetilde {S}(\varepsilon_{0},\mu_{0}) \), we have, for all \(y\in K(\mu_{0}) \),
$$ f(x_{0},y,\mu_{0}) \geq0 \quad\mbox{and}\quad f(\bar{x},y, \mu_{0}) + \varepsilon_{0} \geq0. $$
Hence
$$ \frac{\varepsilon_{0}  \varepsilon_{n_{0}}}{\varepsilon_{0}}f(x_{0},y,\mu_{0}) \geq0 \quad\mbox{and} \quad \frac{\varepsilon_{n_{0}}}{\varepsilon_{0}}f(\bar {x},y,\mu_{0}) + \frac{\varepsilon_{n_{0}}}{\varepsilon_{0}} \varepsilon_{0} \geq0. $$
It follows from (iii) that
$$ f \biggl(\frac{\varepsilon_{0}  \varepsilon_{n_{0}}}{\varepsilon_{0}}\bar {x}\oplus\frac{\varepsilon_{n_{0}}}{\varepsilon_{0}}x_{0},y, \mu_{0} \biggr) \geq\frac{\varepsilon_{0}  \varepsilon_{n_{0}}}{\varepsilon_{0}}f(x_{0},y,\mu _{0}) + \frac{\varepsilon_{n_{0}}}{\varepsilon_{0}}f(\bar{x},y,\mu_{0}) + \varepsilon_{n_{0}} \geq0. $$
Therefore, \(x_{n_{0}}\in \widetilde {S}(\varepsilon_{n_{0}},\mu_{0}) \), which leads to a contradiction to (3.1). We can conclude that \(\widetilde {S}(\cdot,\mu_{0}) \) is l.s.c. at \(\varepsilon_{0} \).
Next, we prove that \(\widetilde {S}(\varepsilon ,\cdot) \) is l.s.c. at \(\mu_{0} \) for all \(\varepsilon> 0 \). By the assumption of nonemptiness of the exact solution, we set the following strict εapproximate solution maps:
$$ \widetilde {S}_{S}(\varepsilon,\mu) : = \bigl\{ x\in K(\mu_{0}) : f(x,y,\mu_{0}) + \varepsilon> 0, \forall x\in K(\mu_{0}) \bigr\} . $$
It is clear that, for any \(\mu\in N(\mu_{0}) \),
$$ \emptyset\neq \widetilde {S}(0,\mu) \subseteq \widetilde {S}_{S}(\varepsilon, \mu)\subseteq \widetilde {S}(\varepsilon,\mu). $$
(3.3)
We show that \(\widetilde {S}_{S}(\varepsilon,\cdot) \) is l.s.c. at \(\mu_{0} \) for all \(\varepsilon >0 \). If not, there are \(\bar{x}\in \widetilde {S}_{S}(\varepsilon,\mu_{0}) \) and a sequence \(\mu_{n} \rightarrow\mu_{0} \), but for all \(x_{n}\in \widetilde {S}_{S}(\varepsilon,\mu_{n}) \), \(x_{n} \nrightarrow\bar{x} \). Since \(K(\mu_{0}) \) is l.s.c. at \(\mu_{0} \), there exists \(\bar{x}_{n} \in K(\mu_{n}) \) with \(\bar{x}_{n} \rightarrow\bar{x} \). By the above contradiction assumption, there must be a subsequence \(\{\bar{x}_{n_{k}}\} \) of \(\{ \bar{x}_{n}\} \) such that \(\bar{x}_{n_{k}}\notin \widetilde {S}_{S}(\varepsilon,\mu_{n}) \), i.e., there exists \(y_{n_{k}}\in K(\mu_{n_{k}}) \) such that
$$ f(\bar{x}_{n_{k}},y_{n_{k}},\mu_{n_{k}}) + \varepsilon\leq0. $$
(3.4)
Since \(K(\cdot) \) is u.s.c. at \(\mu_{0} \) and \(K(\mu_{0}) \) is compact valued, there exists \(\bar{y}\in K(\mu_{0}) \) such that \(y_{n_{k_{j}}}\rightarrow\bar{y} \). From (3.4) and continuity of \(f(\cdot,\cdot,\cdot) \) on \(K(N(\mu_{0})) \times K(N(\mu_{0})) \times\{ \mu_{0}\} \), we get
$$ f(\bar{x},\bar{y},\mu_{0}) + \varepsilon\leq0, $$
which leads to a contradiction. Hence \(\widetilde {S}_{S}(\varepsilon,\cdot ) \) is l.s.c. at \(\mu_{0} \).
Next, we claim that
$$ \widetilde {S}(\varepsilon,\mu_{0}) \subseteq cl \bigl( \widetilde {S}_{S}(\varepsilon,\mu_{0}) \bigr). $$
(3.5)
For any \(x_{0}\in \widetilde {S}(\varepsilon,\mu_{0}) \) and \(x_{1}\in \widetilde {S}_{S}(\varepsilon,\mu_{0}) \), we put \(x_{t} = (1t)x_{0} \oplus tx_{1} \), \(t\in (0,1) \). Then \(x_{t} \rightarrow x_{0} \) as \(t\rightarrow+0 \). Since \(f(\cdot,y,\mu_{0}) \) is geodesic concave, we have
$$\begin{aligned} f(x_{t},y,\mu_{0}) + \varepsilon \geq& (1t)f(x_{0},y, \mu_{0}) + tf(x_{1},y,\mu _{0}) + \varepsilon \\ =& (1t) \bigl(f(x_{0},y,\mu_{0}) + \varepsilon \bigr) + t \bigl(f(x_{1},y,\mu _{0}) + \varepsilon \bigr) \\ \geq& 0, \quad\forall y\in K(\mu_{0}). \end{aligned}$$
Hence, \(x_{t}\in \widetilde {S}_{S}(\varepsilon,\mu_{0}) \). This implies that \(x_{0}\in cl (\widetilde {S}_{S}(\varepsilon,\mu_{0}) ) \). So, we have the claim. Hence, for all \(\varepsilon> 0 \), \(\widetilde {S}_{S}(\varepsilon,\cdot) \) is l.s.c. at \(\mu_{0} \). Thus, for any \(\mu_{m} \rightarrow\mu_{0} \)
$$ cl \bigl( \widetilde {S}_{S}(\varepsilon,\mu_{0}) \bigr) \subseteq \liminf \widetilde {S}_{S}(\varepsilon,\mu_{m}). $$
Combining the last inequality with (3.3) and (3.5), we have
$$ \widetilde {S}(\varepsilon,\mu_{0}) \subseteq cl \bigl( \widetilde {S}_{S}( \varepsilon,\mu_{0}) \bigr) \subseteq\liminf \widetilde {S}_{S}( \varepsilon,\mu_{m}) \subseteq\liminf \widetilde {S}(\varepsilon, \mu_{m}). $$
Therefore \(\widetilde {S}(\varepsilon,\cdot) \) is l.s.c. at \(\mu_{0} \). We can conclude from Step 1 and 2 that \(\widetilde {S}(\cdot,\cdot) \) is l.s.c. at \((\varepsilon _{0},\mu_{0}) \). □
Theorem 15
Assume that the conditions (C_{1})(C_{2}) hold. Then
\(\widetilde {S}(\cdot ,\cdot)\)
is u.s.c. at
\((\varepsilon_{0},\mu_{0}) \).
Proof
We prove by contradiction, suppose that \(\widetilde {S}(\cdot,\cdot)\) is not u.s.c. at \((\varepsilon_{0},\mu_{0}) \). Then there exist an open neighborhood U of \(\widetilde {S}(\varepsilon_{0},\mu_{0}) \) and a sequence \(\{(\varepsilon_{n},\mu_{n})\} \) converging to \((\varepsilon_{0},\mu_{0}) \) such that
$$ \widetilde {S}(\varepsilon_{n},\mu_{n}) \nsubseteq U, \quad \forall n\in\mathbb{N}. $$
Then there exists some \(x_{n}\in \widetilde {S}(\varepsilon_{n},\mu_{n}) \) such that
$$ x_{n} \notin U, \quad \forall n\in\mathbb{N}. $$
(3.6)
Since \(x_{n} \in \widetilde {S}(\varepsilon_{n},\mu_{n}) \), we have \(x_{n} \in K(\mu_{n}) \). By the assumption, \(K(\cdot) \) is u.s.c. being compact valued at \(\mu_{0} \), then we see that there exists a subsequence \(\{x_{n_{k}}\} \) such that \(x_{n_{k}} \rightarrow x^{*} \).
We will show that \(x^{*} \in \widetilde {S}(\varepsilon_{0},\mu_{0}) \). Suppose to the contrary that \(x^{*} \notin \widetilde {S}(\varepsilon_{0},\mu_{0}) \). Then there exists \(y^{*}\in A(\mu_{0}) \) such that
$$ f\bigl(x^{*},y^{*},\mu_{0}\bigr) + \varepsilon< 0. $$
(3.7)
Since \(K(\cdot) \) is l.s.c. at \(\mu_{0} \) and \(y^{*} \in K(\mu_{0}) \), \(\varepsilon_{n} \rightarrow\varepsilon _{0} \) and \(\mu_{n} \rightarrow\mu_{0} \), we see that there exists \(y_{n} \in \widetilde {S}(\varepsilon_{n},\mu_{n}) \) such that \(y_{n} \rightarrow y^{*} \). It follows from \(y_{n} \in K(\mu_{n}) \) that
$$ f(x_{n},y_{n},\mu_{n}) + \varepsilon\geq0, \quad \forall n\in\mathbb{N}. $$
Condition (C_{2}) gives \(f(x^{*},y^{*},\mu_{0}) + \varepsilon\geq0 \). This is a contradiction with (3.7). Hence, we have \(x^{*} \in \widetilde {S}(\varepsilon_{0},\mu_{0})\subseteq U \). Since \(x_{n} \rightarrow x^{*} \) and U is an open set, there exists some \(n_{0}\in\mathbb{N} \) such that \(x_{n}\in U \) for all \(n\geq n_{0} \), which is a contradiction with (3.6). Therefore, \(\widetilde {S}(\cdot ,\cdot) \) is u.s.c. at \((\varepsilon_{0},\mu_{0}) \). □
Theorem 16
Consider the
εapproximate (PEP). We assume that the conditions (C_{1})(C_{3}) hold. Then
\(\widetilde {S}(\cdot,\cdot) \)
is continuous at
\((\varepsilon_{0},\mu_{0}) \).
The following example illustrates that Theorem 16 cannot apply with exact solution maps to (PEP).
Example 17
Let X be an \(\mathbb{R} \)tree with radial meter \(d_{r} \). We put \(M=[0,1] \) and \(K(\mu) = [[(1,0),(0,1)]] \) and \(f(x,y,\mu) = \mu (x_{1}+x_{2}) \). Then the assumptions of Theorem 9 are satisfied. Direct computations give us the approximate solution set, for positive \(\varepsilon >0 \) and \(\mu\in M \),
$$ \widetilde {S}(\varepsilon ,\mu) = \textstyle\begin{cases} [[(1,0),(0,1)]], &\mbox{if } \mu=0, \\ [ [ (\min\{ \frac{\varepsilon }{\mu},1 \},0 ), (0,\min\{ \frac{\varepsilon }{\mu},1 \} ) ] ], & \mbox{if } \mu\neq0. \end{cases} $$
We see that \(\widetilde {S}(\cdot,\cdot) \) is not l.s.c. at \((0,0) \). Indeed, for \((\varepsilon _{n},\mu_{n})= (\frac{1}{2n},\frac{1}{n} )\rightarrow (0,0) \) and \((1,0) \in \widetilde {S}(0,0) \), it is clear that there is no sequence in \(\widetilde {S}(\frac{1}{2n},\frac{1}{n} ) = [ [ (\frac{1}{2},0 ), (0,\frac{1}{2} ) ] ] \) which converges to \((1,0) \). Hence, \(\widetilde {S}(\cdot,\cdot ) \) is not l.s.c. at \((0,0) \).