F is bounded : Suppose that F is not bounded, then there exists a sequence {fn}n∈N⊂F such that kfnkK ≥ n, i.e ∃ xn∈ K such that kfn(xn)k ≥ n

Advanced Calculus Handout 4 April 11, 2011 Theorem 4. (Arzela-Ascoli Theorem). Let K be a compact subset of R^p and letF be a subset of the set of continuous functions on K, i.e. F ⊂ {f : K → R^q | f is continuous on K}. Then the following properties are equivalent:

(1) The familyF is bounded and uniformly equicontinuous on K.

(2) Every sequence from F has a subsequence which is uniformly convergent on K to a uniformly continuous function f (which may not belong to F ).

Proof of (2) ⇒ F is bounded : Suppose that F is not bounded, then there exists a sequence {fn}_n∈N⊂F such that kfnkK ≥ n, i.e ∃ xn∈ K such that kfn(xn)k ≥ n. This implies that fncannot have any subsequence that converges uniformly on K, since if f_n had a subsequence, still denoted by f_n, that converges uniformly on K to a uniformly continuous function f (which may not belong to F ), then we have ∞ = lim

n→∞kf_nk_K ≤ lim

n→∞kf_n− f k_K + lim

n→∞kf k_K < ∞ which is a contradiction.

Proof of (2) ⇒ F is uniform equicontinuous: Suppose that F is not uniformly equicontinuous on K, then there exists an ₀ > 0, such that for each n ∈ N, there exist fn ∈ F , and xn, y_n ∈ K satisfying that kxn − ynk < 1

n and kfn(xn) − fn(yn)k ≥ 0. This implies that fn cannot have any subsequence that converges uniformly on K, since if f_n had a subsequence, still denoted by f_n, that converges uniformly on K to a uniformly continuous function f (which may not belong to F ), thus

0 < ₀ ≤ lim

n→∞kf_n(x_n) − f_n(y_n)k = lim

n→∞kf_n(x_n) − f (x_n) + f (x_n) − f (y_n) + f (y_n) − f_n(y_n)k

≤ lim

n→∞kf_n(x_n) − f (x_n)k + lim

n→∞kf (x_n) − f (y_n)k + lim

n→∞kf_n(y_n) − f (y_n)k

≤ lim

n→∞kf_n− f k_K + lim

n→∞kf (x_n) − f (y_n)k + lim

n→∞kf_n− f k_K = 0, which is a contradiction. Note that, in the last equality,

n→∞lim kf_n− f k_K = 0 since f_nconverges to f uniformly on K, and

n→∞lim kf (xn) − f (yn)k = 0 since f is uniformly continuous on K, and kxn− ynk < 1 n.

Proof of (1) ⇒ (2) Step (1): Using diagonal process to extract a subsequence from a given sequence {fn} ⊂F (a bounded family): Let S = K ∩ Q^p = {xi}_i∈N. Note that S, called a dense subset of K, is a countable set, and S = K, i.e. for each x ∈ K, and for each δ > 0, there exists x_i ∈ S such that x_i ∈ B(δ, x) = {z ∈ R^p | kz − xk < δ}.

Suppose that F is bounded and {fn} is any sequence in F , then, since

(∗) {f_n(x₁)} is bounded in R^q ⇒ {f_n} has a subsequence, denoted {f_n¹}, converges at x₁. Next, since

(∗) {f_n¹(x2)} is bounded in R^q ⇒ {f_n¹} has a subsequence, denoted {f_n²}, converges at x2, and x₁ (since {f_n²} is a subsequence of {f_n¹} that converges at x₁).

Inductively, for each k ≥ 2, since (∗)

({f_n^k(x_k+1)} is bounded in R^q ⇒ {f_n^k} has a subsequence, denoted {f_n^k+1}, converges at

x_k+1, and x_j(since {f_n^k+1} is a subsequence of {f_n^j} that converges at x_jfor each j = 1, . . . , k.)

Advanced Calculus Handout 4 (Continued) April 11, 2011

{f_n}

∪

{ f₁¹ f₂¹ f₃¹ · · · f_k¹ · · · }(x₁) → f (x₁)

∪

{ f₁² f₂² f₃² · · · f_k² · · · }(x₂) → f (x₂)

∪

{ f₁³ f₂³ f₃³ · · · f_k³ · · · }(x₃) → f (x₃)

∪

· · ·

∪

{ f₁^k f₂^k f₃^k · · · f_k^k · · · }(x_k) → f (x_k)

By setting g_n = f_nⁿ, we obtain a subsequence of {f_n}. Note that for each k ∈ N, and for each 1 ≤ j ≤ k, gn = f_nⁿ ∈ {f_m^j}_m∈N whenever n ≥ j. This implies that lim

n→∞gn(xj) = f (xj) for each 1 ≤ j ≤ k and for all k ∈ N. Hence, lim

n→∞g_n(x_j) = f (x_j) for each x_j ∈ S.

Proof of (1) ⇒ (2) Step (2): Using the uniform equicontinuity of F to show that the subsequence g_n (of f_n) converges uniformly to f on K: In particular, we shall show that the sequence g_n (which is a subsequence of {f_n} ⊂F ) satisfies the Cauchy criterion for uniform con- vergence on K to a uniformly continuous function f (which may not belong to F ), i.e.

for each  > 0 and for each x ∈ K, there exists L ∈ N such that if m, n ≥ L, then kgn(x)−g_m(x)k < .

For each  > 0, since F is uniformly equicontinuous on K, there exists a δ = δ() > 0 such that if x, y ∈ K satisfying that kx − yk < δ, then kh(x) − h(y)k <  for all h ∈ F . This implies that for each each x ∈ K, since S = K ⊂ ∪^∞_i=1B(δ, x_i), and each g_n ∈F , there exists an xi ∈ S such that (∗) kx − x_ik < δ =⇒ kg_n(x) − g_n(x_i)k < 

3 for all n ∈ N.

Also, for the same  > 0, since lim

n→∞g_n(x_i) = f (x_i), there exists L ∈ N such that (†) for any m, n ≥ L, we have kg_m(x_i) − g_n(x_i)k < 

3.

Therefore, combining (∗) and (†), we have for each x ∈ K and for any m, n ≥ L, we have kg_n(x) − g_m(x)k ≤ kg_n(x) − g_n(x_i)k + kg_n(x_i) − g_m(x_i)k + kg_m(x_i) − g_m(x)k<

3 +  3 + 

3 = .

This implies that for any m, n ≥ L, kg_n − g_mk_K < , and, g_n converges uniformly on K to a uniformly continuous function f (which may not belong to F ).

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Advanced Calculus Handout 4 (Continued) April 11, 2011 Study guide of Chapter 3:

Outlines of (3.1): Apply the Implicit Function Theorem to study the following question:

Q: Let W be an open subset of R^m× Rⁿ, F : W ⊂ R^m× Rⁿ → Rⁿ be a (vector-valued) function of class C¹, and (a, b) be a point in W ⊂ R^m× Rⁿ satisfying F (a, b) = 0. When is there

(1) a function f (x), defined in some open set in Rⁿ containing a, and (2) an open set U ⊂ W ⊂ R^m+n containing (a, b),

such that for (x, y) ∈ U,

F (x, y) = 0 ⇐⇒ y = f (x)?

Read e.g. Theorem 3.1, Theorem 3.9 in the book, and Exercises 1 − 3, 5 − 7, 9 in (3.1)

Outlines of (3.2), (3.3), (3.5): There are three common ways of locally representing smooth k−dimensional manifolds (e.g. curves, or surfaces if k = 1 or 2.) in Rⁿ :

Type (1) : as the graph of a function, y = f (x), where f is of class C¹, x ∈ B, and B is a connected open set in R^k;

Type (2) : as the locus of an equation F (x, y) = 0, where F is of class C¹;

Type (3) : parametrically, as the range of a C¹ function g : U ⊂ R^m → Rⁿ whose differential Dg has rank k everywhere in a connected open set U.

Note that if a manifold can be represented locally in Type (1), then it can be represented in both Types (2) and (3) by setting F (x, y) = y − f (x), and g(x) = (x, f (x)) for x ∈ B ⊂ R^k, respectively. It is natural to ask the following converse, i.e.

Q: When can a manifold given locally in Types (2) or (3) be represented in Type (1)?

Read e.g. Theorem 3.11, Theorem 3.15, Theorem 3.21 in the book, and Exercises 1 − 3 in (3.2), and 1, 2 in (3.3).

Outlines of (3.4), (3.5): A question concerned here is:

Q: When can a mapping (or a transformation) f : Rⁿ → Rⁿ of class C¹ have a local inverse? How smooth is the inverse (if it exists)?

Read e.g. the following The Inverse Function Theorem in the book.

Theorem 3.18: Let U and V be open sets in Rⁿ, a ∈ U, and b = f (a). Suppose that f : U → V is a mapping of class C¹ and the Fr´echet derivative Df (a) is invertible (i.e. the Jacobian det Df (a) is nonzero). Then ∃ neighborhoods M ⊂ U and N ⊂ V of a and b, respectively, so that f is a one-to-one map from M onto N, and the inverse map f⁻¹ from N to M is also C¹. Moreover, if y = f (x) ∈ N, D(f⁻¹)(y) = [Df (x)]⁻¹.

Note that if f is C¹ with nonzero det Df (a) at an interior point a, then b = f (a) is also an interior point of f (U ), i.e. The Inverse Function Theorem also tells us that when an interior point in the domain is mapped to an interior point in the range.

Also, read e.g. Theorem 3.20, Theorem 3.22 in the book, and Exercises 1, 2, 5 in (3.4) and 1 in (3.5).

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