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 − yRl < 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 − yRl < 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 (NStage 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 + βf − g by discounting, β ∈ (0, 1), g ∈ B(X) and f − g ≥ 0 ⇐⇒ T f − T g ≤ βf − g Similarly, T g ≤ T f + βf − g ⇐⇒ −(T f − T g) ≤ βf − g ⇐⇒ T f − T g ≥ −βf − g So, −βf − g ≤ T f − T g ≤ βf − 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 singlevalued 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 (HeineBorel Theorem) in Bartle, Γ(x) is a compact set for ∀x ∈ X i.e. Γ is compactvalued. 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 (BolzanoWeierstrass 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 compactvalued, 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 hemicontinuous (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 (HeineBorel 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 (BolzanoWeierstrass 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 hemicontinuous (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 (HeineBorel 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 (HeineBorel 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 compactvalued 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 singlevalued correspondence (i.e. a function), not a multivalued correspondence. Theorem 3.6 (Theorem of Maximum) implies that g is upper hemicontinuous (u.h.c.). As g is a singlevalued 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
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