In this section, we discuss the upper semicontinuity and closedness of the solution mapping \(S_{f}\). Since there have been a number of contributions to existence issues, focusing on stability we always assume that \(S_{f_{1}}(\lambda)\) and \(S_{f}(\lambda)\) are nonempty for all λ in a neighborhood of the considered point λ̄. First of all, we shall establish the upper semicontinuity and closedness of the solution mapping \(S_{f_{1}}\).
Lemma 3.1
For
\((\operatorname{GLQEP}_{\lambda})_{\lambda\in\Lambda}\)
assume that

(i)
E
is usc at
λ̄
and
\(E(\bar{\lambda})\)
is compact;

(ii)
\(K_{2}\)
is lsc in
\(K_{1}(A, \Lambda) \times\{\bar{\lambda}\}\);

(iii)
\(\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})\)
is closed in
\(K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}\)
for
\(i=1,2\).
Then the solution map
\(S_{f_{1}}\)
is both usc and closed at
λ̄.
Proof
We first prove that the solution map \(S_{f_{1}}\) is usc at λ̄. Suppose on the contrary that there exists an open set \(U\supseteq S_{f_{1}}(\bar{\lambda})\) such that for any neighborhood \(N(\bar{\lambda})\) of λ̄, there exists \(\lambda\in N(\bar{\lambda})\) such that \(S_{f_{1}}(\lambda) \nsubseteq U\). In particular, for each \(n\in\Bbb {N}\), there exist sequences \(\{\lambda_{n}\} \subseteq\Lambda\) converging to λ̄ and \(\{ x_{n}\} \subseteq S_{f_{1}}(\lambda_{n}) \subseteq E(\lambda_{n})\) with \(x_{n} \notin U \). By the upper semicontinuity of E and the compactness of \(E(\bar{\lambda})\), one can assume that \(x_{n} \rightarrow x_{0}\), for some \(x_{0} \in E(\bar{\lambda})\). Next, we claim that \(x_{0} \in S_{f_{1}}(\bar{\lambda})\). Again suppose on the contrary that there exists \(y_{0} \in K_{2}(x_{0}, \bar{\lambda})\) such that \(f_{1}(x_{0},y_{0},\bar{\lambda})<0. \) The lower semicontinuity of \(K_{2}\) at \((x_{0}, \bar{\lambda})\), by Lemma 2.5, implies that there exists a sequence \(\{y_{n}\} \) in \(K_{2}(x_{n}, \lambda_{n})\) such that \(y_{n} \rightarrow y_{0}\) as \(n\rightarrow \infty\). For each \(n\in\Bbb {N}\), since \(x_{n} \in S_{f_{1}}(\lambda_{n})\), we have
$$f_{1}(x_{n},y_{n},\lambda_{n})\geq0. $$
It follows from the closedness of \(\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})\) that \(f_{1}(x_{0},y_{0},\bar{\lambda})\geq0\), which leads to a contradiction. Therefore, \(x_{0}\in S_{f_{1}}(\bar{\lambda})\subseteq U\), again a contradiction, since \(x_{n}\notin U\) for all n. Thus, \(S_{f_{1}}\) is usc at λ̄.
Next, we prove that \(S_{f_{1}}\) is closed at λ̄. We suppose on the contrary that \(S_{f_{1}}\) is not closed at λ̄, i.e., there a sequence \(\{\lambda_{n}\}\) converging to λ̄ and \(\{x_{n}\}\subseteq S_{f_{1}}(\lambda_{n})\) with \(x_{n} \rightarrow x_{0}\) but \(x_{0} \notin S_{f_{1}}(\bar{ \lambda})\). The same argument as above ensures that \(x_{0}\in S_{f_{1}}(\bar{ \lambda })\), which gives a contradiction. Therefore, we can conclude that \(S_{f_{1}}\) is closed at λ̄. □
Now, we are in the position to discuss the upper semicontinuity and closedness of the solution mapping \(S_{f}\).
Theorem 3.2
For
\((\operatorname{GLQEP}_{\lambda})_{\lambda\in\Lambda}\)
assume that

(i)
E
is usc at
λ̄
and
\(E(\bar{\lambda})\)
is compact;

(ii)
\(K_{2}\)
is lsc in
\(K_{1}(A, \Lambda) \times\{\bar{\lambda}\}\);

(iii)
\(\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})\)
is closed in
\(K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}\)
for
\(i=1,2\);

(iv)
Z
is lsc in
\(S_{f_{1}}(\bar{\lambda})\times\{\bar{\lambda}\}\).
Then the solution map
\(S_{f}\)
is both usc and closed at
λ̄.
Proof
We first claim that the solution map \(S_{f}\) is usc at λ̄. Suppose there exist an open set \(U\supseteq S_{f}(\bar{\lambda})\), \(\{\lambda_{n}\}\rightarrow\bar{\lambda}\), and \(\{x_{n}\}\subseteq S_{f}(\lambda_{n})\) such that \(x_{n}\notin U\) for all n. By the upper semicontinuity of \(S_{f_{1}}\) at λ̄ and the compactness of \(S_{f_{1}}(\bar{\lambda})\), without loss of generality we can assume that \(x_{n} \rightarrow x_{0}\) as \(n\rightarrow \infty\) for some \(x_{0}\in S_{f_{1}}(\bar{\lambda})\). If \(x_{0}\notin S_{f}(\bar{\lambda})\), there exists \(y_{0}\in Z(x_{0},\bar{\lambda})\) such that \(f_{2}(x_{0},y_{0},\bar{\lambda})<0\). The lower semicontinuity of Z in turn yields \(y_{n}\in Z(x_{n},\lambda_{n})\) tending to \(y_{0}\). Notice that for each \(n\in\Bbb {N}\), \(f_{2}(x_{n},y_{n},\lambda_{n})\geq0\). This together with the closedness of \(\operatorname{lev}_{\geq0}f_{2}(\cdot,\cdot, \bar{\lambda})\) in \(K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}\) implies that \(f_{2}(x_{0},y_{0},\bar{\lambda}) \geq0\), which gives a contradiction. If \(x_{0}\in S_{f}(\bar{\lambda})\subseteq U\), one has another contradiction, since \(x_{n}\notin U\) for all n. Thus, \(S_{f}\) is usc at λ̄.
Now we prove that \(S_{f}\) is closed at λ̄. Suppose on the contrary that there exists a sequence \(\{(\lambda_{n}, x_{n})\} \) converging to \((\bar{\lambda}, x_{0})\) with \(x_{n}\in S_{f}(\lambda_{n})\) but \(x_{0}\notin S_{f}(\bar{\lambda})\). Then \(f_{2}(x_{0},y_{0},\bar{\lambda})<0\) for some \(y_{0}\in Z(x_{0},\bar{\lambda})\). Due to the lower semicontinuity of Z, there is \(y_{n}\in Z(x_{n},\lambda_{n})\) such that \(y_{n}\rightarrow y_{0}\). Since \(x_{n}\in S_{f}(\lambda_{n})\), \(f_{2}(x_{n},y_{n},\lambda_{n})\geq0\). By the closedness of the set \(\operatorname{lev}_{\geq0}f_{2}\), \(f_{2}(x_{0},y_{0},\bar{\lambda})\geq0\), which is impossible since \(f_{2}(x_{0},y_{0},\bar{\lambda})<0\). Therefore, \(S_{f}\) is closed at λ̄. □
Corollary 3.3
For GLQEP, suppose that the conditions (i), (ii), and (iv) given in Theorem
3.2
are satisfied. Further, for each
\(i = 1,2\), assume that
\(f_{i}\)
is upper pseudocontinuous in
\(K_{1}(A, \Lambda) \times K_{2}(A, \Lambda) \times\{\bar{\lambda}\}\). Then the solution map
\(S_{f}\)
is both usc and closed at
λ̄.
Proof
It is suffice to derive the condition (iii) that given in Theorem 3.2. For \(i = 1,2\), suppose \(\{ (x_{n},y_{n},\lambda _{n})\} \) is any sequence in \(\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})\) such that \((x_{n},y_{n},\lambda_{n})\rightarrow (\bar{x},\bar{y}, \bar{\lambda})\) as \(n\rightarrow \infty\). Assume that on the contrary that \((\bar{x},\bar{y}, \bar{\lambda})\notin \operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})\), which implies that \(f_{i}(\bar{x},\bar{y}, \bar{\lambda})<0 = f_{i}(\bar{x},\bar{x}, \bar{\lambda})\). The upper pseudocontinuity of \(f_{i}\) at \((\bar{x},\bar{y}, \bar{\lambda})\) implies that
$$0=f_{i}(\bar{x},\bar{x}, \bar{\lambda}) > \limsup_{n\rightarrow \infty} f_{i}(x_{n},y_{n},\lambda_{n})\geq0, $$
which gives a contradiction. Hence, we can conclude that \((\bar{x},\bar{y}, \bar{\lambda}) \in \operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})\). Now, the closedness of \(\operatorname{lev}_{\geq0}f_{i}(\cdot,\cdot, \bar{\lambda})\) is proved. Applying Theorem 3.2, we get the desired result. □
The following examples show that all assumptions imposed in Theorem 3.2 are very essential and cannot be relaxed.
Example 3.4
(The upper semicontinuity and compactness in (i) are crucial)
Let \(A = X = \mathbb {R}\), \(\Lambda= [0,1]\), \(\bar{\lambda}= 0\). Define the mappings \(K_{1}\), \(K_{2}\), and f by
$$K_{1}(x, \lambda) = (\lambda, 1 + \lambda] \quad\mbox{and}\quad K_{2}(x, \lambda) = (0, 1] $$
and
$$f(x,y,\lambda) = \bigl(x(xy)^{2}(1+ \lambda), 2^{\lambda+ xy}x(x y) \bigr). $$
Then we have
$$E(0) = (0, 1] \quad\mbox{and}\quad E(\lambda) = (\lambda, 1+ \lambda], \quad \forall\lambda\in(0,1]. $$
Hence, \(E(0)\) is not compact and E is not usc. Indeed, we choose an open set \(U := (0, 3/2) \supseteq E(0) = (0, 1] \). We observe that, for any \(\varepsilon> 0\), we can choose \(\lambda'= \varepsilon/2 \in N(0, \varepsilon)\) such that
$$E\bigl(\lambda'\bigr) = ( \varepsilon/2, \varepsilon/2 ] \nsubseteq U. $$
Clearly, the conditions (ii) and (iii) are all satisfied. Easy calculations yield
$$S_{f_{1}}(\lambda)= \textstyle\begin{cases} (0,1], & \mbox{if } \lambda=0, \\ (\lambda, 1 + \lambda], & \mbox{if } \lambda\neq0, \end{cases} $$
and \(Z(x,\lambda)=\{x\}\). Hence, assumption (iv) is satisfied. Direct computations give
$$S_{f}(\lambda) = \textstyle\begin{cases} (0,1], & \mbox{if } \lambda=0, \\ (\lambda, 1 + \lambda], & \mbox{if } \lambda\neq0. \end{cases} $$
It is evident that \(S_{f}\) is neither usc nor closed at \(\bar{\lambda}= 0\). This is caused by the fact that E is neither upper semicontinuous nor compactvalued at \(\bar{\lambda}=0\).
Example 3.5
(The lower semicontinuity of \(K_{2}\) in \(K_{1}(A, \Lambda) \times\{\bar{\lambda}\}\) is essential)
Let \(A = X = \mathbb {R}\), \(\Lambda= [0,1]\), \(\bar{\lambda}= 0\). Define the mappings \(K_{1}\), \(K_{2}\), and f by
$$K_{1}(x, \lambda) = K_{2}(x, \lambda) = \textstyle\begin{cases} [1, 0], & \mbox{if } \lambda= 0, \\ {[}\frac{1}{2}, 0], & \mbox{if } \lambda\neq0, \end{cases} $$
and
$$f(x,y,\lambda) = \bigl( (1 +\lambda) (y  x), 2^{\lambda y}(y x) \bigr). $$
Then we have
$$E(\lambda) = \textstyle\begin{cases} [1, 0], & \mbox{if } \lambda= 0, \\ {[}\frac{1}{2}, 0], & \mbox{if } \lambda\neq0, \end{cases} $$
which shows that E is usc at 0 and \(E(0)\) is compact, that is, (i) is satisfied. Clearly, the condition (iii) in Theorem 3.2 is satisfied. Furthermore, easy calculations yield
$$S_{f_{1}}(\lambda) = \textstyle\begin{cases} \{1\}, & \mbox{if } \lambda=0, \\ \{\frac{1}{2}\} , & \mbox{if } \lambda\neq0, \end{cases} $$
and \(Z(x, \lambda) = \{x\}\), which is lsc in \(S_{f_{1}}(\bar{\lambda})\times\{\bar{\lambda}\}\). Direct calculation gives
$$S_{f}(\lambda) = \textstyle\begin{cases} \{1\}, & \mbox{if } \lambda=0, \\ \{\frac{1}{2}\} , & \mbox{if } \lambda\neq0. \end{cases} $$
It is evident that \(S_{f}\) is neither usc nor closed at \(\bar{\lambda}= 0\). This is caused by the fact that \(K_{2}\) is not lsc at \(\bar{\lambda}=0\). Indeed, we observe that \((0,0) \in K_{1}(A, \Lambda) \times\{0\}\) and \(\{ (\frac{1}{n}, \frac{1}{n} )\} \rightarrow (0,0)\). We choose \(y := 1 \in K_{2}(0,0) = [1, 0]\) such that there is not any sequence \(\{y_{n}\}\) in \(K_{2}(\frac{1}{n}, \frac{1}{n}) = [\frac{1}{2}, 0]\) converging to y.
Example 3.6
(The lower semicontinuity of Z in \(S_{f_{1}}(\bar{\lambda})\times\{\bar{\lambda}\}\) are crucial)
Let \(A = X = \mathbb {R}\), \(\Lambda= [0,1]\), \(\bar{\lambda}= 0\). Define the mappings \(K_{1}\), \(K_{2}\), and f by
$$K_{1}(x, \lambda) = K_{2}(x, \lambda) = [0,1] $$
and
$$f(x,y,\lambda) = \bigl(\lambda(x  y), 2^{\lambda y}(y x) \bigr). $$
Hence Conditions (i), (ii), and (iii) clearly hold. By direct calculations, we can get
$$\begin{aligned}& S_{f_{1}}(\lambda) = \textstyle\begin{cases} [0,1], & \mbox{if } \lambda= 0, \\ \{1\}, & \mbox{if } \lambda\neq0, \end{cases}\displaystyle \\& Z(x, \lambda) = \textstyle\begin{cases} [0,1], & \mbox{if } \lambda= 0, \\ \{x\}, & \mbox{if } \lambda\neq0. \end{cases}\displaystyle \end{aligned}$$
Hence Z is not lsc in \([0,1]\times\{ 0 \}\). Further
$$S_{f}(\lambda) = \textstyle\begin{cases} \{0\}, & \mbox{if } \lambda= 0, \\ \{1\}, & \mbox{if } \lambda\neq0. \end{cases} $$
It is evident that \(S_{f}\) is neither usc nor closed at \(\bar{\lambda}= 0\).