Notes on Recursive Methods in Economic Dynamics - Stokey & Lucas Theorem 3.1 - C ( X ) is a Banach Space Proof: By Exercise 3.4e, ( C ( X ) , || f − g || ) is a normed vector space (a special case of metric space). We want to show ( C ( X ) , || f − g || ) is complete. So, it is a Banach space. In other words, we want to show that for any Cauchy sequence { f n } in C ( X ), { f n } converges uniformly to f ∈ C ( X ). That means, for any ε > 0, there exists a N ( ε ) ∈ N such that if for all n ≥ N ( ε ), then || f n − f || = sup | ( f n − f )( X ) | ≤ ε and f ∈ C ( X ). Step 1: Find a candidate limit function f Pick x ∈ X and put it to { f n } , { f n ( x ) } now is a sequence of real numbers, not a sequence of functions anymore. | f n ( x ) − f m ( x ) | ≤ sup | ( f n − f m )( X ) | = || f n − f m || The first inequality is by the definition of supremum (the least upper bound). The second equality is by the definition of sup norm (it is also called uniform norm). As | f n ( x ) − f m ( x ) | ≤ || f n − f m || for ∀ n, m ∈ N , { f n ( x ) } is a Cauchy sequence. Given the fact that ( R , | x − y | ) is a complete metric space, any Cauchy sequence from R converges to a point in R As { f n ( x ) } is a sequence from R and is a Cauchy sequence, { f n ( x ) } → f ( x ) ∈ R as n → ∞ . As x is arbitrarily picked from X , other values from X , say y ∈ X also leads to { f n ( y ) } → f ( y ) ∈ R as n → ∞ , f : X → R is well defined, and is our candidate limit function. Step 2: Show that { f n } converges uniformly to f (i.e. || f n − f || → 0 as n → ∞ ) As { f n } is a Cauchy sequence of function by hypothesis, for any ε > 0, there exists a N ( ε ) ∈ N such that if for ∀ n, m ≥ N ( ε ) then || f n − f m || ≤ ε/ 2. Now, pick x ∈ X and for ∀ m ≥ n ≥ N ( ε ), | f n ( x ) − f ( x ) | = | ( f n ( x ) − f m ( x )) + ( f m ( x ) − f ( x )) | ≤ | f n ( x ) − f m ( x ) | + | f m ( x ) − f ( x ) | triangle inequality ≤ || f n − f m || + | f m ( x ) − f ( x ) | see Step 1 ≤ ε/ 2 + | f m ( x ) − f ( x ) | see above From Step 1, we know that { f m ( x ) } → f ( x ) as n → ∞ , that is a pointwise convergence as the convergence depends on x . By the definition of pointwise convergence, for the chosen ε above, there exists a K ( ε, x ) ∈ N such that if for ∀ m ≥ K ( ε, x ) then | f m ( x ) − f ( x ) | ≤ ε/ 2. Moreover, we also require the chosen m ≥ n ≥ N ( ε ). Together, the chosen m should be m ≥ n ≥ N ( ε ) and m ≥ K ( ε, x ). As K ( ε, x ) depends on x , the chosen m may be different for each x ∈ X . For example, if K ( ε, x ) is large for a particular x ∈ X , the chosen m needs to be even larger for that x ∈ X . So, | f n ( x ) − f ( x ) | ≤ ε/ 2 + ε/ 2 = ε As x ∈ X is picked arbitrarily, the above inequality holds for ∀ x ∈ X . To restate, for any ε > 0, there exists N ( ε ) ∈ N such that if for ∀ n ≥ N ( ε ) then | f n ( x ) − f ( x ) | ≤ ε for ∀ x ∈ X . In other words, { f n } converges uniformly to f | f n ( x ) − f ( x ) | ≤ ε for ∀ x ∈ X is the same as | ( f n − f )( X ) | ≤ ε which implies || f n − f || = sup | ( f n − f )( X ) | ≤ ε Step 3: Show that f is continuous and bounded i.e. f ∈ C ( X ) We only prove here that f is continuous on X . We want to show that for any ε > 0 and x, y ∈ X , there exists δ ( ε, y ) > 0 such that if || x − y || R l < δ ( ε, y ) then | f ( x ) − f ( y ) | < ε . Pick ε > 0 and x ∈ X , as we know from Step 2 that || f n − f || ≤ ε for ∀ n ≥ N ( ε ), we can pick a large k such that || f − f k || = || f k − f || ≤ ε/ 3. | f ( x ) − f k ( x ) | ≤ sup | ( f − f k )( X ) | = || f − f k || ≤ ε/ 3 Similar to Step 1 above 1 As our chosen f k ∈ { f n } ⊆ C ( X ), f k is a continuous function on X For the given ε and x , we pick y ∈ X , there exists δ ( ε, y ) > 0 such that if || x − y || R l < δ ( ε, y ) then | f k ( x ) − f k ( y ) | < ε/ 3. | f ( x ) − f ( y ) | = | ( f ( x ) − f k ( x )) + ( f k ( x ) − f k ( y )) + ( f k ( y ) − f ( y )) | ≤ | f ( x ) − f k ( x ) | + | f k ( x ) − f k ( y ) | + | f k ( y ) − f ( y ) | triangle inequality < ε/ 3 + ε/ 3 + ε/ 3 = ε As ε > 0 and y ∈ X is arbitrary, f is continuous on X Q.E.D. Theorem 3.2 - Contraction Mapping Theorem (Part A) Proof: We want to show there exists one and only one v ∈ S such that T v = v ∈ S Step 1: Find a candidate v ∈ S Define { T n } by T 0 x = x and T n x = T ( T n − 1 x ) for n = 1 , 2 , · · · . Pick v 0 ∈ S , create a sequence { v n } ∞ n =0 by v n +1 = T v n . So, v n = T v n − 1 = T ( T v n − 2 ) = T 2 v n − 2 = · · · = T n v n − n = T n v 0 . As T : S → S , { v n } ∞ n =0 ⊆ S ρ ( v 2 , v 1 ) = ρ ( T v 1 , T v 0 ) by v n +1 = T v n ≤ βρ ( v 1 , v 0 ) As T is a contraction mapping with modulus β ∈ (0 , 1) and v 1 , v 0 ∈ S for ∀ n = 1 , 2 , · · · , ρ ( v n +1 , v n ) = ρ ( T v n , T v n − 1 ) ≤ βρ ( v n , v n − 1 ) = βρ ( T v n − 1 , T v n − 2 ) ≤ β 2 ρ ( v n − 1 , v n − 2 ) = · · · ≤ β n ρ ( v n − ( n − 1) , v n − n ) = β n ρ ( v 1 , v 0 ) For any m > n , ρ ( v m , v n ) ≤ ρ ( v m , v m − 1 ) + · · · + ρ ( v n +2 , v n +1 ) + ρ ( v n +1 , v n ) triangle inequality ≤ β m − 1 ρ ( v 1 , v 0 ) + · · · + β n +1 ρ ( v 1 , v 0 ) + β n ρ ( v 1 , v 0 ) = [ β m − 1 + · · · + β n +1 + β n ] ρ ( v 1 , v 0 ) = β n [ β m − n − 1 + · · · + β + 1] ρ ( v 1 , v 0 ) < β n [ · · · + β m − n − 1 + · · · + β + 1] ρ ( v 1 , v 0 ) as β > 0 = β n 1 1 − β ρ ( v 1 , v 0 ) as β ∈ (0 , 1) As β ∈ (0 , 1), β n → 0 as n → ∞ Thus, ρ ( v m , v n ) < β n 1 1 − β ρ ( v 1 , v 0 ) → 0 as n → ∞ So, { v n } ∞ n =0 is a Cauchy sequence. Since ( S, ρ ) is complete by hypothesis, { v n } ∞ n =0 ⊆ S converges to a point in S , say v ∈ S . This v is our candidate. Step 2: Show that v ∈ S is a fixed point of T i.e. T v = v ∈ S As T : S → S and v ∈ S , T v ∈ S . So, ρ ( T v, v ) defined. For ∀ n and v 0 ∈ S ρ ( T v, v ) ≤ ρ ( T v, T n v 0 ) + ρ ( T n v 0 , v ) triangle inequality = ρ ( T v, T ( T n − 1 v 0 )) + ρ ( T n v 0 , v ) ≤ βρ ( v, T n − 1 v 0 ) + ρ ( T n v 0 , v ) As T is a contraction mapping with β ∈ (0 , 1); v ∈ S and v n − 1 = T n − 1 v 0 ∈ S = βρ ( v, v n − 1 ) + ρ ( v n , v ) → 0 v n → v as n → ∞ and β ∈ (0 , 1) So, we have ρ ( T v, v ) ≤ 0. As ρ ( T v, v ) is a metric and cannot be negative, ρ ( T v, v ) = 0. In other words, T v = v ∈ S Step 3: Show that this v is unique in S We prove by contradiction. Suppose ˆ v 6 = v and ˆ v ∈ S is another fixed point of T i.e. T ˆ v = ˆ v ∈ S . As ˆ v 6 = v , ρ (ˆ v, v ) := a > 0. We have 0 < a =: ρ (ˆ v, v ) = ρ ( T ˆ v, T v ) = βρ (ˆ v, v ) = βa However, as β ∈ (0 , 1), a 6 = βa . A contradiction. So, ˆ v = v Q.E.D. 2 Theorem 3.2 - Contraction Mapping Theorem (Part B) Proof: for n ≥ 1 ρ ( T n v 0 , v ) = ρ ( T ( T n − 1 v 0 ) , T v ) T v = v as v is a fixed point of T ≤ βρ ( T n − 1 v 0 , v ) As T is a contraction mapping with β ∈ (0 , 1); v n − 1 = T n − 1 v 0 ∈ S and v ∈ S ≤ β [ βρ ( T n − 2 v 0 , v )] = β 2 ρ ( T n − 2 v 0 , v )] · · · ≤ β n ρ ( T n − n v 0 , v ) = β n ρ ( T 0 v 0 , v ) = β n ρ ( v 0 , v ) It also holds for n = 0 as ρ ( T 0 v 0 , v ) = 1 · ρ ( v 0 , v ) = β 0 ρ ( v 0 , v ) Q.E.D. Theorem 3.2 - Corollary 1 Proof: Pick v 0 ∈ S ′ ⊆ S , construct a sequence { v n } ∞ n =0 by v n = T n v 0 for n = 1 , 2 , · · · As T ( S ′ ) ⊆ S ′ by hypothesis, { v n } ∞ n =0 ⊆ S ′ . From the proof of Theorem 3.2 (Part A) Step 1, we know that v n → v ∈ S as n → ∞ . That means, { v n } ∞ n =0 must converge. As S ′ is a closed set by hypothesis and { v n } ∞ n =0 ⊆ S ′ and converges, Theorem 11.1.7 in Bartle implies { v n } ∞ n =0 converges to a point in S ′ . Hence, v ∈ S ′ ⊆ S . If now T ( S ′ ) ⊆ S ′′ , this implies T v ∈ S ′′ as v ∈ S ′ . As v is a fixed point of T , v = T v ∈ S ′′ Q.E.D. Theorem 3.2 - Corollary 2 (N-Stage Contraction Theorem) Proof: By Theorem 3.2, v is a unique fixed point of T N such that T N v = v ∈ S ρ ( T v, v ) = ρ ( T ( T N v ) , T N v ) as T N v = v = ρ ( T N ( T v ) , T N v ) ≤ βρ ( T v, v ) As T N is a contraction mapping with β ∈ (0 , 1); T v ∈ S as T : S → S As ρ ( ., . ) is a metric, ρ ( T v, v ) ≥ 0. suppose ρ ( T v, v ) > 0, this inequality does not hold as β ∈ (0 , 1). So, ρ ( T v, v ) = 0 i.e. T v = v . So, v is also a fixed point of T in addition to T N . Suppose ˆ v 6 = v is also a fixed point of T, T ˆ v = ˆ v T 2 ˆ v = T ( T ˆ v ) = T ˆ v = ˆ v · · · T N ˆ v = ˆ v So, ˆ v is also a fixed point of T N . As v is the unique fixed point of T N , ˆ v = v . A contradiction. So, v is a unique fixed point of T Q.E.D. Theorem 3.3 - Blackwell’s sufficient conditions for a contraction Proof: for any f, g ∈ B ( X ) such that f ≤ g + || f − g || , T f ≤ T ( g + || f − g || ) by monotonicity ≤ T g + β || f − g || by discounting, β ∈ (0 , 1), g ∈ B ( X ) and || f − g || ≥ 0 ⇐⇒ T f − T g ≤ β || f − g || Similarly, T g ≤ T f + β || f − g || ⇐⇒ − ( T f − T g ) ≤ β || f − g || ⇐⇒ T f − T g ≥ − β || f − g || So, − β || f − g || ≤ T f − T g ≤ β || f − g || ⇐⇒ || T f − T g || ︸︷︷︸ ρ ( T f,T g ) ≤ β || f − g || ︸︷︷︸ ρ ( f,g ) So, T is a contraction mapping with β Q.E.D. 3 Exercise 3.11a Proof: We want to show that the single-valued and u.h.c. correspondence Γ is also l.h.c. We prove by contradiction. Suppose Γ is not l.h.c. (negation of l.h.c.). If y = Γ( x ) = { y } and for any sequence { x n } such that x n → x as n → ∞ , then for any sequence { y n } and for any ε > 0, no matter what N ∈ N you pick, there exists at least one n N ≥ N such that | y n N − y | > ε Moreover, y n = Γ( x n ) for ∀ n ≥ N Pick any subsequence { y n k } ⊆ { y n } . Construct { x n k } ⊆ { x n } such that y n k = Γ( x n k ) for n k ≥ N (for k to be large). As x n → x as n → ∞ , Theorem 3.4.2 in Bartle implies that x n k → x as k → ∞ . As Γ is u.h.c. by hypothesis, there exists a subsequence of { y n k } , say { y n k l } such that y n k l → y = Γ( x ) as l → ∞ . A contradiction, so Γ is l.h.c. As Γ is both l.h.c. and u.h.c., it is continuous. Q.E.D. Theorem 3.4 Proof: As A is closed by hypothesis, Γ( x ) is closed and bounded for ∀ x ∈ X . By Theorem 11.2.5 (Heine-Borel Theorem) in Bartle, Γ( x ) is a compact set for ∀ x ∈ X i.e. Γ is compact-valued. Pick ˆ x ∈ X and any { x n } ⊆ X such that x n → ˆ x as n → ∞ As Γ( x ) 6 = ∅ for ∀ x ∈ X by hypothesis, we construct { y n } such that y n ∈ Γ( x n ) for ∀ n As x n → ˆ x as n → ∞ , there exists a bounded set ˆ X ⊂ X such that { x n } ⊂ ˆ X and ˆ x ∈ ˆ X As Γ( ˆ X ) = { y ∈ Y : y ∈ Γ( x ) , x ∈ ˆ X } and { x n } ⊂ ˆ X and y n ∈ Γ( x n ) for ∀ n , so y n ∈ Γ( ˆ X ) for ∀ n , i.e. { y n } ⊂ Γ( ˆ X ). As ˆ X is bounded, Γ( ˆ X ) is bounded by hypothesis. As { y n } ⊂ Γ( ˆ X ), { y n } is a bounded sequence. Theorem 3.4.8 (Bolzano-Weierstrass Theorem) in Bartle implies that { y n } has a convergent subsequence { y n k } which converges to ˆ y As { x n } converges to ˆ x , Theorem 3.4.2 in Bartle implies that { x n k } also converges to ˆ x ∈ X As y n ∈ Γ( x n ) for ∀ x n ∈ X ⇒ y n k ∈ Γ( x n k ) for ∀ x n k ∈ X , so { ( x n k , y n k ) } ⊂ A := { ( x, y ) ∈ X × Y : y ∈ Γ( x ) , x ∈ X } As y n k → ˆ y and x n k → ˆ x as k → ∞ , { ( x n k , y n k ) } → (ˆ x, ˆ y ) as k → ∞ . As { ( x n k , y n k ) } ⊂ A and A is closed by hypothesis, Theorem 11.1.7 in Bartle implies that (ˆ x, ˆ y ) ∈ A . Thus, ˆ y ∈ Γ(ˆ x ) by the definition of A To restate, Γ is compact-valued, Γ(ˆ x ) 6 = ∅ , and for for any sequence x n → ˆ x as n → ∞ and any sequence { y n } such that y n ∈ Γ( x n ) for ∀ n , there exists a convergent subsequence of { y n } , y n k → ˆ y as k → ∞ such that ˆ y ∈ Γ(ˆ x ). So, Γ is upper hemi-continuous (u.h.c.) at ˆ x . As ˆ x is arbitrary in X , we are done. Q.E.D. Exercise 3.15 Proof: As X is a compact set by hypothesis, Theorem 11.2.5 (Heine-Borel Theorem) in Bartle implies that X is closed and bounded. As X is a bounded set, any { x n } ⊆ X is a bounded sequence. Theorem 3.4.8 (Bolzano-Weierstrass Theorem) in Bartle implies that { x n } has a convergent subsequence, x n k → x as k → ∞ . As X is a closed set and { x n k } ⊆ X and x n k → x as k → ∞ , Theorem 11.1.7 in Bartle implies x ∈ X . For the above { x n k } converging to x ∈ X , we construct { y n k } such that y n k ∈ Γ( x n k ) for ∀ k . As Γ is u.h.c. by hypothesis, there exists a subsequence of { y n k } converging to y ∈ Γ( x ). As y n k ∈ Γ( x n k ) for ∀ k , ( x n k , y n k ) ∈ A for ∀ k . Thus, { ( x n k , y n k ) } ⊆ A . As y ∈ Γ( x ), ( x, y ) ∈ A . As { x n k } converging to x and { y n k } converging to y , ( x n k , y n k ) → ( x, y ) ∈ A as k → ∞ So, { ( x n , y n ) } ⊆ A has a convergent subsequence { ( x n k , y n k ) } ⊆ A converging to ( x, y ) ∈ A . (So, A is a compact set ???) Q.E.D. Theorem 3.5 Proof: Pick ˆ x ∈ X and any sequence { x n } such that x n → ˆ x as n → ∞ . As Γ( x ) 6 = ∅ for ∀ x ∈ X and ˆ x ∈ X , Γ(ˆ x ) 6 = ∅ , so we can pick ˆ y ∈ Γ(ˆ x ). Pick ε > 0 such that the closed set ˆ X = B ε (ˆ x ) ⊆ X . As x n → ˆ x as n → ∞ , there exists N ( ε ) ∈ N such that if for ∀ n ≥ N ( ε ) then x n ∈ ˆ X . Without Loss of Generality (WLOG), we take N = 1, so { x n } ⊂ ˆ X Let D be the boundary of the closed set ˆ X . As { x n } ⊂ ˆ X , x n is a convex combination of an element of D and ˆ x (the center of ˆ X ) for ∀ n . i.e. for ∀ n , there exists α n ∈ (0 , 1) and d n ∈ D such that x n = α n d n + (1 − α n )ˆ x As D is a boundary, D is a bounded set and thus d n is bounded for ∀ n . As x n → ˆ x as n → ∞ , this implies α n → 0 as n → ∞ By hypothesis, there exists a bounded set ˆ Y ⊆ Y such that Γ( x ) ∩ ˆ Y 6 = ∅ for ∀ x ∈ ˆ X . As d n ∈ D ⊂ ˆ X and Γ( x ) ∩ ˆ Y 6 = ∅ for ∀ x ∈ ˆ X , Γ( d n ) ∩ ˆ Y 6 = ∅ for ∀ n , we can then pick ˆ y n ∈ Γ( d n ) ∩ ˆ Y for ∀ n . Define for ∀ n that y n = α n ˆ y n + (1 − α n )ˆ y ˆ y n ∈ Γ( d n ) ∩ ˆ Y ⇒ ˆ y n ∈ Γ( d n ) ⇒ ( d n , ˆ y n ) ∈ A for ∀ n . As ˆ y ∈ Γ(ˆ x ), (ˆ x, ˆ y ) ∈ A . As A is convex by hypothesis, the convex combination of ( d n , ˆ y n ) and (ˆ x, ˆ y ), i.e. ( x n , y n ) is in A for ∀ n { ( x n , y n ) } ⊂ A ⇒ y n ∈ Γ( x n ) for ∀ n As ˆ y n ∈ Γ( d n ) ∩ ˆ Y ⇒ ˆ y n ∈ ˆ Y for ∀ n and ˆ Y is a bounded set, ˆ y n is bounded for ∀ n . Since α n → 0 as n → ∞ , y n → ˆ y ∈ Γ(ˆ x ) as n → ∞ Thus, Γ is lower hemi-continuous (l.h.c.) at ˆ x . ˆ x is the center of ˆ X , if ˆ x is at the boundary of X , ˆ X * X . So, ˆ x can only be in the interior of X . As ˆ x is arbitrary in the interior of X , we are done. Q.E.D. 4 Theorem 3.6 (Theorem of Maximum) Proof: Pick x ∈ X As Γ( x ) 6 = ∅ and is compact, and f ( x, y ) is continuous by hypothesis, Theorem 11.3.5.2 (Weierstrass Theorem / Max. Min. Theorem) in Bartle implies that h ( x ) := max y ∈ Γ( x ) f ( x, y ) exists. So, G ( x ) := { y ∈ Γ( x ) : f ( x, y ) = h ( x ) } 6 = ∅ . As Γ( x ) is compact by hypothesis. Theorem 11.2.5 (Heine-Borel Theorem) in Bartle implies that Γ( x ) is closed and bounded. As Γ( x ) is bounded and G ( x ) ⊆ Γ( x ), G ( x ) is bounded. Construct a sequence { y n } such that { y n } ⊆ G ( x ) and y n → y as n → ∞ As { y n } ⊆ G ( x ) ⊆ Γ( x ) and Γ( x ) is closed, Theorem 11.1.7 in Bartle implies that y ∈ Γ( x ). As { y n } ⊆ G ( x ), f ( x, y n ) = h ( x ) for ∀ n . As f ( x, y ) is continuous, h ( x ) = lim n →∞ h ( x ) = lim n →∞ f ( x, y n ) = f ( x, lim n →∞ ( y n )) = f ( x, y ) So, the limit point y ∈ G ( x ). Thus, any convergent sequence from G ( x ) will converge to a point in G ( x ). Theorem 11.1.7 in Bartle implies that G ( x ) is closed. As G ( x ) is bounded and closed, Theorem 11.2.5 (Heine-Borel Theorem) in Bartle implies G ( x ) is compact for each x ∈ X Pick x ∈ X , choose { x n } be any sequence in X such that x n → x as n → ∞ . Construct { y n } such that y n ∈ G ( x n ) ⊆ Γ( x n ) for ∀ n . As Γ is continuous by hypothesis, it is both l.h.c. and u.h.c. As y n ∈ Γ( x n ) for ∀ n , Γ is compact-valued by hypothesis and Γ is u.h.c., there exists a subsequence of { y n } , y n k → y ∈ Γ( x ) as k → ∞ . Pick z ∈ Γ( x ). As Γ is l.h.c., there exists N ∈ N and a sequence { z n k } such that if for ∀ n k ≥ N then z n k → z ∈ Γ( x ) and z n k ∈ Γ( x n k ). As { y n k } ⊆ { y n } and y n ∈ G ( x n ) for ∀ n , so y n k ∈ G ( x n k ) for ∀ k i.e. y n k is a maximizer in Γ( x n k ) for ∀ k . So, f ( x n k , y n k ) ≥ f ( x n k , z n k ) for ∀ k . As f is continuous by hypothesis, f ( lim k →∞ ( x n k ) , lim k →∞ ( y n k )) = lim k →∞ f ( x n k , y n k ) ≥ lim k →∞ f ( x n k , z n k ) = f ( lim k →∞ ( x n k ) , lim k →∞ ( z n k )) As x n → x as n → ∞ , Theorem 3.4.2 in Bartle implies that lim k →∞ ( x n k ) = x . So, f ( x, y ) ≥ f ( x, z ) As z is arbitrary in Γ( x ), the above inequality implies that y is a maximizer in Γ( x ), i.e. y ∈ G ( x ). So, G is u.h.c. at x ∈ X As x is arbitrary in X , we are done. Q.E.D. Lemma 3.7 Proof: As Γ( x ) is compact for ∀ x ∈ X and f ( x, y ) is continuous in y , Theorem 11.3.5.2 (Weierstrass Theorem / Max. Min. Theorem) in Bartle implies that g ( x ) := arg max y ∈ Γ( x ) f ( x, y ) exists and g is well defined. As Γ( x ) is convex for ∀ x ∈ X and f ( x, y ) is strictly concave in y , Theorem 7.14 in Sundaram implies that FOC is a sufficient condition for a unique global maximum. So, there is only one y ∈ Γ( x ) such that f ( x, y ) is globally maximized. So, g is a single-valued correspondence (i.e. a function), not a multi-valued correspondence. Theorem 3.6 (Theorem of Maximum) implies that g is upper hemi-continuous (u.h.c.). As g is a single-valued correspondence (i.e. a function) and u.h.c., Exercise 3.11a implies g is a continuous function. As X is compact, Exercise 3.15 implies that A is compact. For each ε > 0, we define A ε := { ( x, y ) ∈ A : || g ( x ) − y || ≥ ε } ⊆ A If A ε = ∅ for each ε > 0, then || g ( x ) − y || < ε for ∀ ( x, y ) ∈ A i.e. for ∀ y ∈ Γ( x ). This means that y is very close to g ( x ) for ∀ y ∈ Γ( x ) as ε can be arbitrarily small. This implies that Γ( x ) = { g ( x ) } , a singleton. The result is trivial. If A ε 6 = ∅ for at least one ε > 0 which is small enough, say ˆ ε A ε 6 = ∅ for ∀ ε such that 0 < ε ≤ ˆ ε , define δ ( ε ) := min ( x,y ) ∈ A ε | f ( x, g ( x )) − f ( x, y ) | (the chosen y ∈ Γ( x ) such that || g ( x ) − y || ≥ ε must not be equal to the unique global maximizer g ( x ), but is the y ∈ Γ( x ) such that || g ( x ) − y || ≥ ε and is the closest to the global maximizer g ( x )). As f is continuous on A by hypothesis, f is continuous on A ε ⊆ A | | is also a continuous function. As A ε is compact and | f ( x, g ( x )) − f ( x, y ) | is continuous on A ε , Theorem 11.3.5.2 (Weierstrass Theorem / Max. Min. Theorem) in Bartle implies that δ ( ε ) exists and δ is well- defined. If ( x, g ( x )) ∈ A ε , 0 = || g ( x ) − g ( x ) || ≥ ε > 0. A contradiction. So, ( x, g ( x )) / ∈ A ε for ∀ x ∈ X , this implies that | f ( x, g ( x )) − f ( x, y ) | 6 = 0 for ∀ ( x, y ) ∈ A ε As | | ≥ 0, | f ( x, g ( x )) − f ( x, y ) | > 0 for ∀ ( x, y ) ∈ A ε This implies that δ ( ε ) := min ( x,y ) ∈ A ε | f ( x, g ( x )) − f ( x, y ) | > 0. Then, ( x, y ) ∈ A ( i.e. y ∈ Γ( x )) s.t. || g ( x ) − y || ≥ ε implies | f ( x, g ( x )) − f ( x, y ) | ≥ min ( x,y ) ∈ A ε | f ( x, g ( x )) − f ( x, y ) | := δ ( ε ) > 0 By logic, if A then B implies if not B then not A, So, the above can be restated as ( x, y ) ∈ A ( i.e. y ∈ Γ( x )) s.t. | f ( x, g ( x )) − f ( x, y ) | < δ ( ε ) implies || g ( x ) − y || < ε Q.E.D. 5

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