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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 {fn } in C(X), {fn } 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 ||fn − f || = sup|(fn − f )(X)| ≤ ε and f ∈ C(X). Step 1: Find a candidate limit function f Pick x ∈ X and put it to {fn }, {fn (x)} now is a sequence of real numbers, not a sequence of functions anymore. |fn (x) − fm (x)| ≤ sup|(fn − fm )(X)| = ||fn − fm || 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 |fn (x) − fm (x)| ≤ ||fn − fm || for ∀n, m ∈ N, {fn (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 {fn (x)} is a sequence from R and is a Cauchy sequence, {fn (x)} → f (x) ∈ R as n → ∞. As x is arbitrarily picked from X, other values from X, say y ∈ X also leads to {fn (y)} → f (y) ∈ R as n → ∞, f : X → R is well defined, and is our candidate limit function. Step 2: Show that {fn } converges uniformly to f (i.e. ||fn − f || → 0 as n → ∞) As {fn } 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 ||fn − fm || ≤ ε/2. Now, pick x ∈ X and for ∀m ≥ n ≥ N (ε), |fn (x) − f (x)| = |(fn (x) − fm (x)) + (fm (x) − f (x))| ≤ |fn (x) − fm (x)| + |fm (x) − f (x)| triangle inequality ≤ ||fn − fm || + |fm (x) − f (x)| see Step 1 ≤ ε/2 + |fm (x) − f (x)| see above From Step 1, we know that {fm (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 |fm (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, |fn (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 |fn (x) − f (x)| ≤ ε for ∀x ∈ X. In other words, {fn } converges uniformly to f . |fn (x) − f (x)| ≤ ε for ∀x ∈ X is the same as |(fn − f )(X)| ≤ ε which implies ||fn − f || = sup|(fn − 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||Rl < δ(ε, y) then |f (x) − f (y)| < ε. Pick ε > 0 and x ∈ X, as we know from Step 2 that ||fn − f || ≤ ε for ∀n ≥ N (ε), we can pick a large k such that ||f − fk || = ||fk − f || ≤ ε/3. |f (x) − fk (x)| ≤ sup|(f − fk )(X)| = ||f − fk || ≤ ε/3 Similar to Step 1 above 1 As our chosen fk ∈ {fn } ⊆ C(X), fk 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||Rl < δ(ε, y) then |fk (x) − fk (y)| < ε/3. |f (x) − f (y)| = |(f (x) − fk (x)) + (fk (x) − fk (y)) + (fk (y) − f (y))| ≤ |f (x) − fk (x)| + |fk (x) − fk (y)| + |fk (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 v0 ∈ S, create a sequence {vn }∞ n=0 by vn+1 = T vn . So, vn = T vn−1 = T (T vn−2 ) = T 2 vn−2 = · · · = T n vn−n = T n v0 . As T : S → S, {vn }∞ n=0 ⊆ S. ρ(v2 , v1 ) = ρ(T v1 , T v0 ) by vn+1 = T vn ≤ βρ(v1 , v0 ) As T is a contraction mapping with modulus β ∈ (0, 1) and v1 , v0 ∈ S for ∀n = 1, 2, · · · , ρ(vn+1 , vn ) = ρ(T vn , T vn−1 ) ≤ βρ(vn , vn−1 ) = βρ(T vn−1 , T vn−2 ) ≤ β 2 ρ(vn−1 , vn−2 ) = · · · ≤ β n ρ(vn−(n−1) , vn−n ) = β n ρ(v1 , v0 ) For any m > n, ρ(vm , vn ) ≤ ρ(vm , vm−1 ) + · · · + ρ(vn+2 , vn+1 ) + ρ(vn+1 , vn ) triangle inequality m−1 n+1 n ≤β ρ(v1 , v0 ) + · · · + β ρ(v1 , v0 ) + β ρ(v1 , v0 ) m−1 n+1 n = [β + ··· + β + β ]ρ(v1 , v0 ) n m−n−1 = β [β + · · · + β + 1]ρ(v1 , v0 ) n m−n−1 < β [· · · + β + · · · + β + 1]ρ(v1 , v0 ) as β > 0 1 = βn ρ(v1 , v0 ) as β ∈ (0, 1) 1−β 1 As β ∈ (0, 1), β n → 0 as n → ∞. Thus, ρ(vm , vn ) < β n 1−β ρ(v1 , v0 ) → 0 as n → ∞. So, {vn }∞n=0 is a Cauchy sequence. ∞ Since (S, ρ) is complete by hypothesis, {vn }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 v0 ∈ S ρ(T v, v) ≤ ρ(T v, T n v0 ) + ρ(T n v0 , v) triangle inequality n−1 n = ρ(T v, T (T v0 )) + ρ(T v0 , v) n−1 ≤ βρ(v, T v0 ) + ρ(T n v0 , v) As T is a contraction mapping with β ∈ (0, 1); v ∈ S and vn−1 = T n−1 v0 ∈ S = βρ(v, vn−1 ) + ρ(vn , v) →0 vn → 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 v̂ 6= v and v̂ ∈ S is another fixed point of T i.e. T v̂ = v̂ ∈ S. As v̂ 6= v, ρ(v̂, v) := a > 0. We have 0 < a =: ρ(v̂, v) = ρ(T v̂, T v) = βρ(v̂, v) = βa However, as β ∈ (0, 1), a 6= βa. A contradiction. So, v̂ = v. Q.E.D. 2 Theorem 3.2 - Contraction Mapping Theorem (Part B) Proof: for n ≥ 1 ρ(T n v0 , v) = ρ(T (T n−1 v0 ), T v) T v = v as v is a fixed point of T n−1 ≤ βρ(T v0 , v) As T is a contraction mapping with β ∈ (0, 1); vn−1 = T n−1 v0 ∈ S and v ∈ S ≤ β[βρ(T n−2 v0 , v)] = β 2 ρ(T n−2 v0 , v)] ··· ≤ β n ρ(T n−n v0 , v) = β n ρ(T 0 v0 , v) = β n ρ(v0 , v) It also holds for n = 0 as ρ(T 0 v0 , v) = 1 · ρ(v0 , v) = β 0 ρ(v0 , v) Q.E.D. Theorem 3.2 - Corollary 1 Proof: Pick v0 ∈ S 0 ⊆ S, construct a sequence {vn }∞ n 0 0 n=0 by vn = T v0 for n = 1, 2, · · · . As T (S ) ⊆ S by hypothesis, ∞ {vn }n=0 ⊆ S . From the proof of Theorem 3.2 (Part A) Step 1, we know that vn → v ∈ S as n → ∞. That means, {vn }∞ 0 n=0 must converge. As S 0 is a closed set by hypothesis and {vn }∞ 0 n=0 ⊆ S and converges, Theorem 11.1.7 in Bartle implies {vn }∞ 0 0 0 00 00 0 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 00 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 v̂ 6= v is also a fixed point of T, T v̂ = v̂ 2 T v̂ = T (T v̂) = T v̂ = v̂ ··· N T v̂ = v̂ So, v̂ is also a fixed point of T N . As v is the unique fixed point of T N , v̂ = 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|| ≤ β ||f − g|| So, T is a contraction mapping with β. Q.E.D. | {z } | {z } ρ(T f,T g) ρ(f,g) 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 {xn } such that xn → x as n → ∞, then for any sequence {yn } and for any ε > 0, no matter what N ∈ N you pick, there exists at least one nN ≥ N such that |ynN − y| > ε. Moreover, yn = Γ(xn ) for ∀n ≥ N . Pick any subsequence {ynk } ⊆ {yn }. Construct {xnk } ⊆ {xn } such that ynk = Γ(xnk ) for nk ≥ N (for k to be large). As xn → x as n → ∞, Theorem 3.4.2 in Bartle implies that xnk → x as k → ∞. As Γ is u.h.c. by hypothesis, there exists a subsequence of {ynk }, say {ynk l } such that ynk 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 x̂ ∈ X and any {xn } ⊆ X such that xn → x̂ as n → ∞. As Γ(x) 6= ∅ for ∀x ∈ X by hypothesis, we construct {yn } such that yn ∈ Γ(xn ) for ∀n. As xn → x̂ as n → ∞, there exists a bounded set X̂ ⊂ X such that {xn } ⊂ X̂ and x̂ ∈ X̂. As Γ(X̂) = {y ∈ Y : y ∈ Γ(x), x ∈ X̂} and {xn } ⊂ X̂ and yn ∈ Γ(xn ) for ∀n, so yn ∈ Γ(X̂) for ∀n, i.e. {yn } ⊂ Γ(X̂). As X̂ is bounded, Γ(X̂) is bounded by hypothesis. As {yn } ⊂ Γ(X̂), {yn } is a bounded sequence. Theorem 3.4.8 (Bolzano-Weierstrass Theorem) in Bartle implies that {yn } has a convergent subsequence {ynk } which converges to ŷ. As {xn } converges to x̂, Theorem 3.4.2 in Bartle implies that {xnk } also converges to x̂ ∈ X. As yn ∈ Γ(xn ) for ∀xn ∈ X ⇒ ynk ∈ Γ(xnk ) for ∀xnk ∈ X, so {(xnk , ynk )} ⊂ A := {(x, y) ∈ X × Y : y ∈ Γ(x), x ∈ X}. As ynk → ŷ and xnk → x̂ as k → ∞, {(xnk , ynk )} → (x̂, ŷ) as k → ∞. As {(xnk , ynk )} ⊂ A and A is closed by hypothesis, Theorem 11.1.7 in Bartle implies that (x̂, ŷ) ∈ A. Thus, ŷ ∈ Γ(x̂) by the definition of A. To restate, Γ is compact-valued, Γ(x̂) 6= ∅, and for for any sequence xn → x̂ as n → ∞ and any sequence {yn } such that yn ∈ Γ(xn ) for ∀n, there exists a convergent subsequence of {yn }, ynk → ŷ as k → ∞ such that ŷ ∈ Γ(x̂). So, Γ is upper hemi-continuous (u.h.c.) at x̂. As x̂ 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 {xn } ⊆ X is a bounded sequence. Theorem 3.4.8 (Bolzano-Weierstrass Theorem) in Bartle implies that {xn } has a convergent subsequence, xnk → x as k → ∞. As X is a closed set and {xnk } ⊆ X and xnk → x as k → ∞, Theorem 11.1.7 in Bartle implies x ∈ X. For the above {xnk } converging to x ∈ X, we construct {ynk } such that ynk ∈ Γ(xnk ) for ∀k. As Γ is u.h.c. by hypothesis, there exists a subsequence of {ynk } converging to y ∈ Γ(x). As ynk ∈ Γ(xnk ) for ∀k, (xnk , ynk ) ∈ A for ∀k. Thus, {(xnk , ynk )} ⊆ A. As y ∈ Γ(x), (x, y) ∈ A. As {xnk } converging to x and {ynk } converging to y, (xnk , ynk ) → (x, y) ∈ A as k → ∞. So, {(xn , yn )} ⊆ A has a convergent subsequence {(xnk , ynk )} ⊆ A converging to (x, y) ∈ A. (So, A is a compact set ???) Q.E.D. Theorem 3.5 Proof: Pick x̂ ∈ X and any sequence {xn } such that xn → x̂ as n → ∞. As Γ(x) 6= ∅ for ∀x ∈ X and x̂ ∈ X, Γ(x̂) 6= ∅, so we can pick ŷ ∈ Γ(x̂). Pick ε > 0 such that the closed set X̂ = Bε (x̂) ⊆ X. As xn → x̂ as n → ∞, there exists N (ε) ∈ N such that if for ∀n ≥ N (ε) then xn ∈ X̂. Without Loss of Generality (WLOG), we take N = 1, so {xn } ⊂ X̂. Let D be the boundary of the closed set X̂. As {xn } ⊂ X̂, xn is a convex combination of an element of D and x̂ (the center of X̂) for ∀n. i.e. for ∀n, there exists αn ∈ (0, 1) and dn ∈ D such that xn = αn dn + (1 − αn )x̂ As D is a boundary, D is a bounded set and thus dn is bounded for ∀n. As xn → x̂ as n → ∞, this implies αn → 0 as n → ∞. By hypothesis, there exists a bounded set Ŷ ⊆ Y such that Γ(x) ∩ Ŷ 6= ∅ for ∀x ∈ X̂. As dn ∈ D ⊂ X̂ and Γ(x) ∩ Ŷ 6= ∅ for ∀x ∈ X̂, Γ(dn ) ∩ Ŷ 6= ∅ for ∀n, we can then pick ŷn ∈ Γ(dn ) ∩ Ŷ for ∀n. Define for ∀n that yn = αn ŷn + (1 − αn )ŷ ŷn ∈ Γ(dn ) ∩ Ŷ ⇒ ŷn ∈ Γ(dn ) ⇒ (dn , ŷn ) ∈ A for ∀n. As ŷ ∈ Γ(x̂), (x̂, ŷ) ∈ A. As A is convex by hypothesis, the convex combination of (dn , ŷn ) and (x̂, ŷ), i.e. (xn , yn ) is in A for ∀n. {(xn , yn )} ⊂ A ⇒ yn ∈ Γ(xn ) for ∀n. As ŷn ∈ Γ(dn )∩ Ŷ ⇒ ŷn ∈ Ŷ for ∀n and Ŷ is a bounded set, ŷn is bounded for ∀n. Since αn → 0 as n → ∞, yn → ŷ ∈ Γ(x̂) as n → ∞. Thus, Γ is lower hemi-continuous (l.h.c.) at x̂. x̂ is the center of X̂, if x̂ is at the boundary of X, X̂ * X. So, x̂ can only be in the interior of X. As x̂ 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) := maxy∈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 {yn } such that {yn } ⊆ G(x) and yn → y as n → ∞. As {yn } ⊆ G(x) ⊆ Γ(x) and Γ(x) is closed, Theorem 11.1.7 in Bartle implies that y ∈ Γ(x). As {yn } ⊆ G(x), f (x, yn ) = h(x) for ∀n. As f (x, y) is continuous, h(x) = limn→∞ h(x) = limn→∞ f (x, yn ) = f (x, limn→∞ (yn )) = 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 {xn } be any sequence in X such that xn → x as n → ∞. Construct {yn } such that yn ∈ G(xn ) ⊆ Γ(xn ) for ∀n. As Γ is continuous by hypothesis, it is both l.h.c. and u.h.c. As yn ∈ Γ(xn ) for ∀n, Γ is compact-valued by hypothesis and Γ is u.h.c., there exists a subsequence of {yn }, ynk → y ∈ Γ(x) as k → ∞. Pick z ∈ Γ(x). As Γ is l.h.c., there exists N ∈ N and a sequence {znk } such that if for ∀nk ≥ N then znk → z ∈ Γ(x) and znk ∈ Γ(xnk ). As {ynk } ⊆ {yn } and yn ∈ G(xn ) for ∀n, so ynk ∈ G(xnk ) for ∀k i.e. ynk is a maximizer in Γ(xnk ) for ∀k. So, f (xnk , ynk ) ≥ f (xnk , znk ) for ∀k. As f is continuous by hypothesis, f (limk→∞ (xnk ), limk→∞ (ynk )) = limk→∞ f (xnk , ynk ) ≥ limk→∞ f (xnk , znk ) = f (limk→∞ (xnk ), limk→∞ (znk )) As xn → x as n → ∞, Theorem 3.4.2 in Bartle implies that limk→∞ (xnk ) = 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 maxy∈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 Aε := {(x, y) ∈ A : ||g(x) − y|| ≥ ε} ⊆ A If Aε = ∅ 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 Aε 6= ∅ for at least one ε > 0 which is small enough, say ε̂. Aε 6= ∅ for ∀ε such that 0 < ε ≤ ε̂, define δ(ε) := min(x,y)∈Aε |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 Aε ⊆ A. |.| is also a continuous function. As Aε is compact and |f (x, g(x)) − f (x, y)| is continuous on Aε , 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)) ∈ Aε , 0 = ||g(x) − g(x)|| ≥ ε > 0. A contradiction. So, (x, g(x)) ∈ / Aε for ∀x ∈ X, this implies that |f (x, g(x)) − f (x, y)| 6= 0 for ∀(x, y) ∈ Aε . As |.| ≥ 0, |f (x, g(x)) − f (x, y)| > 0 for ∀(x, y) ∈ Aε . This implies that δ(ε) := min(x,y)∈Aε |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)∈Aε |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