Maximum theorem

Let θ ∈ Θ {\displaystyle \theta \in \Theta } , and suppose G {\displaystyle G} is an open set containing ( A ∩ B ) ( θ ) {\displaystyle (A\cap B)(\theta )} . If A ( θ ) ⊆ G {\displaystyle A(\theta )\subseteq G} , then the result follows immediately. Otherwise, observe that for each x ∈ A ( θ ) ∖ G {\displaystyle x\in A(\theta )\setminus G} we have x ∉ B ( θ ) {\displaystyle x\notin B(\theta )} , and since B {\displaystyle B} is closed there is a neighborhood U x × V x {\displaystyle U_{x}\times V_{x}} of ( θ , x ) {\displaystyle (\theta ,x)} in which x ′ ∉ B ( θ ′ ) {\displaystyle x'\notin B(\theta ')} whenever ( θ ′ , x ′ ) ∈ U x × V x {\displaystyle (\theta ',x')\in U_{x}\times V_{x}} . The collection of sets { G } ∪ { V x : x ∈ A ( θ ) ∖ G } {\displaystyle \{G\}\cup \{V_{x}:x\in A(\theta )\setminus G\}} forms an open cover of the compact set A ( θ ) {\displaystyle A(\theta )} , which allows us to extract a finite subcover G , V x 1 , … , V x n {\displaystyle G,V_{x_{1}},\dots ,V_{x_{n}}} . By upper hemicontinuity, there is a neighborhood U θ {\displaystyle U_{\theta }} of θ {\displaystyle \theta } such that A ( U θ ) ⊆ G ∪ V x 1 ∪ ⋯ ∪ V x n {\displaystyle A(U_{\theta })\subseteq G\cup V_{x_{1}}\cup \dots \cup V_{x_{n}}} . Then whenever θ ′ ∈ U θ ∩ U x 1 ∩ ⋯ ∩ U x n {\displaystyle \theta '\in U_{\theta }\cap U_{x_{1}}\cap \dots \cap U_{x_{n}}} , we have A ( θ ′ ) ⊆ G ∪ V x 1 ∪ ⋯ ∪ V x n {\displaystyle A(\theta ')\subseteq G\cup V_{x_{1}}\cup \dots \cup V_{x_{n}}} , and so ( A ∩ B ) ( θ ′ ) ⊆ G {\displaystyle (A\cap B)(\theta ')\subseteq G} . This completes the proof. ◻ {\displaystyle \square }

Fix θ ∈ Θ {\displaystyle \theta \in \Theta } , and let ε > 0 {\displaystyle \varepsilon >0} be arbitrary. For each x ∈ C ( θ ) {\displaystyle x\in C(\theta )} , there exists a neighborhood U x × V x {\displaystyle U_{x}\times V_{x}} of ( θ , x ) {\displaystyle (\theta ,x)} such that whenever ( θ ′ , x ′ ) ∈ U x × V x {\displaystyle (\theta ',x')\in U_{x}\times V_{x}} , we have f ( x ′ , θ ′ ) < f ( x , θ ) + ε {\displaystyle f(x',\theta ')<f(x,\theta )+\varepsilon } . The set of neighborhoods { V x : x ∈ C ( θ ) } {\displaystyle \{V_{x}:x\in C(\theta )\}} covers C ( θ ) {\displaystyle C(\theta )} , which is compact, so V x 1 , … , V x n {\displaystyle V_{x_{1}},\dots ,V_{x_{n}}} suffice. Furthermore, since C {\displaystyle C} is upper hemicontinuous, there exists a neighborhood U ′ {\displaystyle U'} of θ {\displaystyle \theta } such that whenever θ ′ ∈ U ′ {\displaystyle \theta '\in U'} it follows that C ( θ ′ ) ⊆ ⋃ k = 1 n V x k {\displaystyle C(\theta ')\subseteq \bigcup _{k=1}^{n}V_{x_{k}}} . Let U = U ′ ∩ U x 1 ∩ ⋯ ∩ U x n {\displaystyle U=U'\cap U_{x_{1}}\cap \dots \cap U_{x_{n}}} . Then for all θ ′ ∈ U {\displaystyle \theta '\in U} , we have f ( x ′ , θ ′ ) < f ( x k , θ ) + ε {\displaystyle f(x',\theta ')<f(x_{k},\theta )+\varepsilon } for each x ′ ∈ C ( θ ′ ) {\displaystyle x'\in C(\theta ')} , as x ′ ∈ V x k {\displaystyle x'\in V_{x_{k}}} for some k {\displaystyle k} . It follows that

f ∗ ( θ ′ ) = sup x ′ ∈ C ( θ ′ ) f ( x ′ , θ ′ ) < max k = 1 , … , n f ( x k , θ ) + ε ≤ f ∗ ( θ ) + ε , {\displaystyle f^{*}(\theta ')=\sup _{x'\in C(\theta ')}f(x',\theta ')<\max _{k=1,\dots ,n}f(x_{k},\theta )+\varepsilon \leq f^{*}(\theta )+\varepsilon ,}

which was desired. ◻ {\displaystyle \square }

Fix θ ∈ Θ {\displaystyle \theta \in \Theta } , and let ε > 0 {\displaystyle \varepsilon >0} be arbitrary. By definition of f ∗ {\displaystyle f^{*}} , there exists x ∈ C ( θ ) {\displaystyle x\in C(\theta )} such that f ∗ ( θ ) < f ( x , θ ) + ε 2 {\displaystyle f^{*}(\theta )<f(x,\theta )+{\frac {\varepsilon }{2}}} . Now, since f {\displaystyle f} is lower semicontinuous, there exists a neighborhood U 1 × V {\displaystyle U_{1}\times V} of ( θ , x ) {\displaystyle (\theta ,x)} such that whenever ( θ ′ , x ′ ) ∈ U 1 × V {\displaystyle (\theta ',x')\in U_{1}\times V} we have f ( x , θ ) < f ( x ′ , θ ′ ) + ε 2 {\displaystyle f(x,\theta )<f(x',\theta ')+{\frac {\varepsilon }{2}}} . Observe that C ( θ ) ∩ V ≠ ∅ {\displaystyle C(\theta )\cap V\neq \emptyset } (in particular, x ∈ C ( θ ) ∩ V {\displaystyle x\in C(\theta )\cap V} ). Therefore, since C {\displaystyle C} is lower hemicontinuous, there exists a neighborhood U 2 {\displaystyle U_{2}} such that whenever θ ′ ∈ U 2 {\displaystyle \theta '\in U_{2}} there exists x ′ ∈ C ( θ ′ ) ∩ V {\displaystyle x'\in C(\theta ')\cap V} . Let U = U 1 ∩ U 2 {\displaystyle U=U_{1}\cap U_{2}} . Then whenever θ ′ ∈ U {\displaystyle \theta '\in U} there exists x ′ ∈ C ( θ ′ ) ∩ V {\displaystyle x'\in C(\theta ')\cap V} , which implies

f ∗ ( θ ) < f ( x , θ ) + ε 2 < f ( x ′ , θ ′ ) + ε ≤ f ∗ ( θ ′ ) + ε {\displaystyle f^{*}(\theta )<f(x,\theta )+{\frac {\varepsilon }{2}}<f(x',\theta ')+\varepsilon \leq f^{*}(\theta ')+\varepsilon }

which was desired. ◻ {\displaystyle \square }