Without loss of generality we can assume that assumptions (ii) and (iv) are satisfied for all \(t\in[a,b]\). Moreover, let
$$E:=\bigcup_{i=1}^{2n} E_{i} . $$
Now, observe that by assumption (iii) and Theorem 2.4 of [4] the function h is inductively open. That is, there exists a set \(Y\subseteq X\) such that the function
is open and \(h(Y)=h(X)\). It follows that the multifunction \(T:h(X)\to2^{Y}\) defined by putting, for each \(s\in h(X)\),
$$T(s)=h^{-1}(s)\cap Y $$
is lower semicontinuous in \(h(X)\) with nonempty values. To see this, fix any set \(\Omega_{1}\subseteq Y\), with \(\Omega_{1}\) open in the relative topology of Y. We get
$$\begin{aligned} T^{-}(\Omega_{1})&:=\bigl\{ s\in h(X): T(s)\cap\Omega_{1}\ne \emptyset\bigr\} \\ &=\bigl\{ s\in h(X): h^{-1}(s)\cap Y\cap\Omega_{1}\ne \emptyset\bigr\} \\ &=\bigl\{ s\in h(X): h^{-1}(s)\cap\Omega_{1}\ne\emptyset\bigr\} \\ &=h(\Omega_{1}). \end{aligned}$$
Since the function \(h|_{Y}:Y\to h(X)\) is open, the set \(h(\Omega_{1})\) is open in \(h(X)\). It follows that the set \(T^{-}(\Omega_{1})\) is open in \(h(X)\), hence T is lower semicontinuous in \(h(X)\), as claimed.
Let \(\Psi:[a,b]\times\mathbf{R}^{n}\times \mathbf{R}^{n}\to2^{Y}\) be defined by
$$\Psi(t,x,y):=T\bigl(G(t,x,y)\bigr)=h^{-1}\bigl(G(t,x,y)\bigr)\cap Y $$
(note that Ψ is well defined since \(G(t,x,y)\subseteq h(X)\) for all \((t,x,y)\in[a,b]\times\mathbf{R}^{n}\times\mathbf{R}^{n}\)). We observe the following facts:
-
(a)
the multifunction Ψ is \(\mathcal{L}([a,b])\otimes\mathcal {B}(\mathbf{R} ^{n})\otimes\mathcal{B}(\mathbf{R}^{n})\)-weakly measurable; that is, for each set \(\Omega\subseteq Y\), with Ω open in the relative topology of Y, the set
$$\Psi^{-}(\Omega):=\bigl\{ (t,x,y)\in[a,b]\times\mathbf{R}^{n}\times \mathbf{R}^{n}: \Psi(t,x)\cap\Omega\ne\emptyset\bigr\} $$
belongs to \(\mathcal{L}([a,b])\otimes\mathcal{B}(\mathbf {R}^{n})\otimes\mathcal{B}(\mathbf{R}^{n})\) (this follows from Proposition 13.2.1 of [3], since G is \(\mathcal{L}([a,b])\otimes\mathcal{B}(\mathbf{R}^{n})\otimes \mathcal{B}(\mathbf{R}^{n})\)-measurable and T is lower semicontinuous);
-
(b)
Ψ has nonempty values and for all \(t\in[a,b]\) one has
$$\bigl\{ (x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}:\Psi(t, \cdot, \cdot) \mbox{ is not lower semicontinuous at }(x,y) \bigr\} \subseteq E. $$
Indeed, if \(t\in[a,b]\) and \((\overline{x},\overline{y})\in (\mathbf{R}^{n}\times\mathbf{R}^{n})\setminus E\) are fixed, then the multifunction \(G(t,\cdot,\cdot)\) is lower semicontinuous at \((\overline{x},\overline{y})\), hence (by the lower semicontinuity of T) the multifunction
$$(x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}\to\Psi(t,x,y)=T \bigl(G(t,x,y)\bigr) $$
is lower semicontinuous at \((\overline{x},\overline{y})\), as claimed.
Now, let \(\overline{\Psi}:[a,b]\times\mathbf{R}^{n}\times\mathbf {R}^{n}\to2^{\mathbf{R}^{n}}\) be defined by
$$\overline{\Psi}(t,x,y):=\overline{\Psi(t,x,y)}. $$
It follows by assumption (iv) and by construction that
$$ \overline{\Psi}(t,x,y)\subseteq \prod_{i=1}^{n} \bigl[\alpha_{i}(t),\beta _{i}(t)\bigr] \quad \mbox{for all } (t,x,y)\in[a,b]\times\mathbf{R}^{n}\times\mathbf{R}^{n}. $$
(4)
By Proposition 2.6 and Theorem 3.5 of [5], the multifunction \(\overline{\Psi}\) is \(\mathcal{L}([a,b])\otimes \mathcal{B}(\mathbf{R} ^{n})\otimes\mathcal{B}(\mathbf{R}^{n})\)-measurable with nonempty values. Moreover, for all \(t\in[a,b]\) one has
$$\bigl\{ (x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}: \overline{\Psi}(t,\cdot ,\cdot)\mbox{ is not l.s.c. at }(x,y) \bigr\} \subseteq E. $$
At this point, let us apply Theorem 2.2 of [1] to the multifunction \(\overline{\Psi}\), choosing \(k=2n\), \(T=[a,b]\), \(X_{i}=\mathbf{R}\) for all \(i=1,\ldots, 2n\), \(S=\mathbf{R}^{n}\) (all the spaces are considered with their Euclidean distance and with the usual Lebesgue measure over their Borel family), and, as above,
$$E:=\bigcup_{i=1}^{2n} E_{i} \subseteq \mathbf{R}^{2n}. $$
We find that there exist \(Q_{1},\ldots,Q_{2n}\subseteq \mathbf{R}\), with \(Q_{i}\in \mathcal{B}(\mathbf{R})\) and \(m_{1}(Q_{i})=0\) for all \(i=1,\ldots,2n\), and a function \(\phi:[a,b]\times\mathbf{R}^{n}\times\mathbf{R}^{n}\to\mathbf {R}^{n}\), such that
- (i)′:
-
\(\phi(t,x,y)\in\overline{\Psi}(t,x,y)\) for all \((t,x,y)\in [a,b]\times\mathbf{R} ^{n}\times \mathbf{R}^{n}\);
- (ii)′:
-
for all \((x,y)\in(\mathbf{R}^{n}\times\mathbf{R}^{n}) \setminus [\bigcup_{i=1}^{2n}(P_{i}^{-1}(Q_{i})\cup E_{i}) ]\), the function \(\phi(\cdot,x,y)\) is \(\mathcal{L}([a,b])\)-measurable;
- (iii)′:
-
for a.e. \(t\in [a,b]\), one has
$$\bigl\{ (x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}: \phi( t, \cdot ,\cdot )\mbox{ is discontinuous at } (x,y) \bigr\} \subseteq \bigcup _{i=1}^{2n} \bigl(E_{i}\cup P_{i}^{-1}(Q_{i})\bigr). $$
Of course, for all \(i=1,\ldots,2n\), one has \(m_{1}(P_{i}(E_{i}\cup P_{i}^{-1}(Q_{i})))=0\).
Now, let us apply Theorem 2.1 of [1] with \(f=g=\phi\), taking into account that for a.e. \(t\in[a,b]\) and for all \((x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}\) one has
$$\phi(t,x,y)\in\overline{\Psi}(t,x,y)\subseteq \prod _{i=1}^{n}\bigl[\alpha _{i}(t), \beta_{i}(t)\bigr]. $$
We find that there exist \(u\in W^{2,p}([a,b],\mathbf{R}^{n})\) and a set \(U_{0}\in\mathcal{L}([a,b])\), with \(m_{1}(U_{0})=0\), such that
$$ \textstyle\begin{cases} u^{\prime\prime}(t)=\phi(t,u(t),u^{\prime}(t))& \mbox{for all } t\in [a,b]\setminus U_{0}, \\ u(a)=u(b)=0_{\mathbf{R}^{n}}, \end{cases} $$
and also
$$\bigl(u(t),u^{\prime}(t)\bigr)\notin\bigcup_{i=1}^{2n} \bigl(E_{i}\cup P_{i}^{-1}(Q_{i}) \bigr) \quad \mbox{for all } t\in[a,b]\setminus U_{0}. $$
In particular, by assumption (ii) we get
$$G\bigl(t,u(t),u^{\prime}(t)\bigr)\subseteq F\bigl(t,u(t),u^{\prime}(t) \bigr) \quad \mbox{for all } t\in[a,b]\setminus U_{0}. $$
Consequently, taking into account the continuity of h and the closedness of X, for all \(t\in[a,b]\setminus U_{0}\) we get
$$u^{\prime\prime}(t)=\phi\bigl(t,u(t),u^{\prime}(t)\bigr) \in\overline{\Psi} \bigl(t,u(t),u^{\prime}(t)\bigr) \subseteq h^{-1}\bigl(G \bigl(t,u(t),u^{\prime}(t)\bigr)\bigr). $$
In particular, we get
$$h\bigl(u^{\prime\prime}(t)\bigr)\in G\bigl(t,u(t),u^{\prime}(t)\bigr) \subseteq F\bigl(t,u(t),u^{\prime}(t)\bigr) $$
for all \(t\in[a,b]\setminus U_{0}\) and hence the function u satisfies our conclusion. The proof is now complete.
Remark
We now give two simple examples of application of Theorem 1.1. The first example concerns with the scalar single-valued case, while the second one deals with the vector multivalued case.
Example 2.1
Let \(n=1\) and let \([a,b]\) be any compact interval. Let us consider the problem
$$ \textstyle\begin{cases} \cos(u^{\prime\prime}(t))= F(t,u(t),u^{\prime}(t))& \mbox{for a.e. } t\in[a,b], \\ u(a)=u(b)=0, \end{cases} $$
(5)
where \(F:[a,b]\times\mathbf{R}\times\mathbf{R}\to\mathbf{R}\) is the (single-valued) function defined by putting, for each \((x,y)\in\mathbf{R}^{2}\),
$$F(t,x,y)= \textstyle\begin{cases} 0&\mbox{if }x\in\mathbf{Q}\mbox{ or }y\in\mathbf{Q}, \\ 1&\mbox{otherwise}. \end{cases} $$
Of course, such a function F (which does not depend on t explicitly) is discontinuous at all points
\((x,y)\in\mathbf{R}^{2}\). We observe that Theorem 1.1 can easily be applied by choosing, for any \(p\in [1,+\infty[\),
$$G:[a,b]\times\mathbf{R}\times\mathbf{R}\to\mathbf{R},\qquad G(t,x,y)\equiv1, $$
\(E_{1}=\mathbf{Q}\times\mathbf{R}\), \(E_{2}=\mathbf{R}\times\mathbf {Q}\), \(X=[2\pi,4\pi]\), \(\alpha(t)\equiv2\pi\), \(\beta(t)\equiv4\pi\), \(h(t)=\cos(t)\) (restricted to the interval \([2\pi,4\pi]\)). Indeed, such a function G is \(\mathcal{L}([a,b])\otimes\mathcal {B}(\mathbf{R} )\otimes\mathcal{B}(\mathbf{R})\)-measurable, \(P_{1}(E_{1})=P_{2}(E_{2})=\mathbf{Q}\), and for all \(t\in[a,b]\) one has
$$\begin{aligned}& \bigl\{ (x,y)\in\mathbf{R}^{2}: G(t, \cdot ,\cdot ) \mbox{ is discontinuous at }(x,y) \bigr\} \\& \quad {}\cup \bigl\{ (x,y)\in\mathbf{R}^{2}:G(t,x,y)\ne F(t,x,y) \bigr\} = E_{1}\cup E_{2}. \end{aligned}$$
Moreover, for all \((t,x,y)\in[a,b]\times\mathbf{R}\times\mathbf {R}\) we have
$$G(t,x,y)=1\in h(X)\quad \mbox{and}\quad h^{-1}\bigl(G(t,x,y) \bigr)=h^{-1}(1)\subseteq [2\pi,4\pi]=\bigl[\alpha(t),\beta(t)\bigr]. $$
Finally, observe that assumption (iii) is satisfied since for all \(t\in\operatorname {int}_{\mathbf{R}}(h(X))= \, ]-1,1[ \) the set \(h^{-1}(t)\) contains only two points.
Consequently, by Theorem 1.1, problem (5) has a solution in \(u\in W^{2,p}([a,b],\mathbf{R})\). Such a solution u also satisfies the condition \(u^{\prime\prime}(t)\in[2\pi,4\pi]\) for a.e. \(t\in [a,b]\), hence we get \(u\in W^{2,\infty}([a,b],\mathbf{R})\).
Moreover, observe that Theorem 1.1 can be applied in analogous way also by taking \(X=[2\pi+4k\pi,4\pi+4k\pi]\) and \(\alpha(t)\equiv 2\pi +4k\pi\), \(\beta(t)\equiv4\pi+4k\pi\), with \(k\in\mathbf{N}\). Then, for each \(k\in\mathbf{N}\), we get the existence of a solution \(u_{k}\in W^{2,\infty}([a,b],\mathbf{R})\) such that \(u_{k}^{\prime\prime}(t)\in[2\pi+4k\pi,4\pi+4k\pi]\) for a.e. \(t\in [a,b]\). Hence, problem (5) has infinitely many solutions. Finally, we note that problem (5) does not admit the trivial solution \(u(t)\equiv0\), since \(F(t, x,0)=0\) for all \((t,x)\in[a,b]\times\mathbf{R}\) and \(\cos(0)=1\).
Example 2.2
Let \(n=2\), and let \(\psi:\mathbf{R}^{2}\to\mathbf {R}\) be defined by \(\psi(v,z)=v+z\). In what follows, we denote vectors of \(\mathbf{R}^{2}\) by the notations \(x:=(x_{1},x_{2})\) and \(y:=(y_{1},y_{2})\). Let \(E\subseteq \mathbf{R}^{4}\) be the set of vectors such that at least one component is rational, that is,
$$E:=\bigl\{ (x,y)\in\mathbf{R}^{4} : \mbox{at least one of } x_{1},x_{2},y_{1},y_{2} \mbox{ is rational}\bigr\} . $$
Let \(F:[0,1]\times\mathbf{R}^{2}\times\mathbf{R}^{2}\to2^{\mathbf{R}}\) be defined by
$$F(t,x,y)= \textstyle\begin{cases} {1} &\mbox{if }(x,y)\in E, \\ {[2+t,3+t]}&\mbox{if }(x,y)\notin E\mbox{ and }x_{1}< 1, \\ {[4+t,5+t]}&\mbox{if }(x,y)\notin E\mbox{ and }x_{1}>1. \end{cases} $$
Of course, for any fixed \(t\in[0,1]\), the multifunction \(F(t,\cdot ,\cdot )\)
is not lower semicontinuous at any point \((x,y)\in\mathbf{R}^{4}\). By Theorem 1.1, it is easily seen that for any fixed \(p\in[1,+\infty[\) the problem
$$ \textstyle\begin{cases} \psi(u^{\prime\prime}(t))\in F(t,u(t),u^{\prime}(t))& \mbox{a.e. in } [0,1], \\ u(0)=u(1)=0_{\mathbf{R}^{2}} \end{cases} $$
(6)
has a solution \(u\in W^{2,p}([0,1],\mathbf{R}^{2})\). To this aim, choose \(X=[1,5]\times[1,5]\), \(h:=\psi|_{X}\), \(\alpha(t)\equiv(1,1)\), \(\beta(t)\equiv(5,5)\), \(G:[0,1]\times\mathbf{R}^{2}\times\mathbf{R}^{2}\to2^{\mathbf{R}}\), with
$$G(t,x,y)= \textstyle\begin{cases} {[2+t,3+t]}&\mbox{if }x_{1}\le1, \\ {[4+t,5+t]}&\mbox{if }x_{1}>1 \end{cases} $$
and
$$\begin{aligned}& E_{1}:=\mathbf{Q}\times\mathbf{R}\times\mathbf{R}\times\mathbf {R}, \qquad E_{2}:=\mathbf{R}\times\mathbf{Q} \times\mathbf{R}\times \mathbf{R}, \\& E_{3}:=\mathbf{R}\times\mathbf{R}\times\mathbf{Q} \times \mathbf{R}, \qquad E_{4}:=\mathbf{R}\times \mathbf{R}\times \mathbf{R}\times\mathbf{Q}. \end{aligned}$$
Of course, such a multifunction G is \(\mathcal{L}([0,1])\otimes\mathcal{B}(\mathbf{R}^{2})\otimes \mathcal{B}(\mathbf{R}^{2})\)-measurable with nonempty closed values. Moreover, if one fix any \(t\in[0,1]\), then the multifunction \(G(t,\cdot,\cdot)\) is lower semicontinuous at each point \((x,y)\in\mathbf{R}^{4}\), with \(x_{1}\ne1\). Consequently, for all \(t\in[0,1]\) we have
$$\begin{aligned}& \bigl\{ (x,y)\in\mathbf{R}^{2}\times\mathbf{R}^{2}: G(t, \cdot ,\cdot )\mbox{ is not lower semicontinuous at }(x,y) \bigr\} \\& \quad {}\cup \bigl\{ (x,y)\in\mathbf{R}^{2}\times\mathbf {R}^{2}:G(t,x,y)\nsubseteq F(t,x,y) \bigr\} = \bigcup _{i=1}^{4} E_{i}=E. \end{aligned}$$
Clearly, for all \(i=1,2,3,4\), one has \(P_{i}(E_{i})=\mathbf{Q}\). Now, observe that for all \((t,x,y)\in[0,1]\times\mathbf{R}^{2}\times \mathbf{R}^{2}\) we have
$$G(t,x,y)\subseteq [2,6]\subseteq h(X)=[2,10] $$
and
$$h^{-1}\bigl(G(t,x,y)\bigr)\subseteq X=[1,5]\times[1,5]=\bigl[ \alpha_{1}(t),\beta_{1}(t)\bigr] \times\bigl[ \alpha_{2}(t),\beta_{2}(t)\bigr]. $$
Finally, let us show that for all \(s\in\operatorname{int}_{\mathbf{R}}(h(X))=\, ]2,10[\), we have \(\operatorname{int}_{X} (h^{-1}(s))=\emptyset\) (though this fact is quite intuitive - since h is never locally constant - we shall provide an explicit proof). To this aim, fix \(s\in\operatorname{int}_{\mathbf{R}}(h(X))\), and let \((v_{0},z_{0})\in h^{-1}(s)\). Therefore, \((v_{0},z_{0})\in X\) and \(h(v_{0},z_{0})=v_{0}+z_{0}=s\). Let Ω be an open set in X such that \((v_{0},z_{0})\in\Omega\). Of course, one can find \(r>0\) such that
$$\bigl([v_{0}-r,v_{0}+r]\times[z_{0}-r,z_{0}+r] \bigr)\cap X\subseteq \Omega. $$
Let \(v_{1}\) be any point in \([v_{0}-r,v_{0}+r]\cap[1,5]\), with \(v_{1}\ne v_{0}\). We have
$$(v_{1},z_{0})\in\Omega $$
and
$$h(v_{1},z_{0})=v_{1}+z_{0}\ne v_{0}+z_{0}=s. $$
Hence, the set \(h^{-1}(s)\) has empty interior in X, as claimed. Thus, all the assumptions of Theorem 1.1 are satisfied. Consequently, problem (6) has at least a solution \(u\in W^{2,p}([0,1],\mathbf{R}^{2})\).
As a matter of fact, since \(u^{\prime\prime}(t)\in X\) for a.e. \(t\in[0,1]\), we get \(u\in W^{2,\infty}([0,1],\mathbf{R}^{2})\). As before, we note that problem (6) does not admit the trivial solution \(u(t)\equiv0_{\mathbf{R}^{2}}\).