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Definition Let X, Y be metric spaces. Suppose E ⊂ X, f maps E into Y, and p is a limit point of E. We say that lim

x∈E, x→pf (x) = q ∈ Y if

∀  > 0, ∃ δ > 0 such that if x ∈ E and 0 < dX(x, p) < δ then dY(f (x), q) < .

⇐⇒ ∀  > 0, ∃ δ > 0 such that f (Bδ(p) ∩ E \ {p}) ⊂ B(f (p)) ∩ f (E).

Remarks

(a) Let X, Y, E, f, and p be as above. Then

x∈E, x→plim f (x) = q ∈ Y if and only if

n→∞lim f (pn) = q for each sequence {pn} in E such that pn6= p, lim

n→∞pn= p.

Proof

(=⇒) Let {pn} be a sequence in E such that pn6= p ∀ n ∈ N and lim

n→∞pn= p. Thus

∀  > 0, since lim

x→pf (x) = q, ∃ δ > 0 such that if x ∈ E and dX(x, p) < δ =⇒ dY(f (x), q) <  and

with this chosen δ > 0, since lim

n→∞pn= p, ∃N ∈ N such that if n ≥ N =⇒dX(pn, p) < δ.

This implies that

dY(f (pn), q) <  =⇒ lim

n→∞f (pn) = q.

(⇐=) Suppose that lim

n→∞f (pn) 6= q, there exists  > 0 such that for all δ > 0, there exists x = x(δ) such that

dX(x, p) < p and dY(f (x), q) ≥ .

For each n ∈ N, by taking δn = 1

n > 0, there exists pn∈ E such that dX(pn, p) < 1

n and dY(f (pn), q) ≥  ∀ n ∈ N.

This implies that

{pn} ⊂ E is a sequence converging to p such that lim

n→∞f (pn) 6= q.

(b) If f has a limit at p, this limit is unique.

Proof Suppose that lim

x→pf (x) = q ∈ Y and lim

y→pf (y) = r ∈ Y. For each  > 0, there exists a δ > 0 such that

if dX(x, p) < δ =⇒ dY(f (x), q) < 

2 and dY(f (x), r) <  2.

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This implies that

dY(q, r) ≤ dY(q, f (x)) + dY(f (x), r) <  2 + 

2 =  ∀  > 0.

Thus we have

0 ≤ dY(q, r) ≤ lim

→0+

 = 0 =⇒ q = r, i.e. if f has a limit at p, this limit is unique.

Theorem Suppose E ⊂ X, a metric space, p is a limit point of E, f and g are complex functions on E, and

x→plimf (x) = A, lim

x→pg(x) = B for some A, B ∈ C.

Then (a) lim

x→p(f + g)(x) = A + B;

(b) lim

x→p(f g)(x) = AB;

(c) lim

x→p

 f g



(x) = A

B, if B 6= 0.

Remark If f and g map E into Rk, then (a) remains true, and (b) becomes (b0) lim

x→p(f · g)(x) = A · B.

Definition Let X, Y be metric spaces. Suppose E ⊂ X, p ∈ E and f maps E into Y. Then f is said to be continuous at p if

x∈E, x→plim f (x) = f (p).

⇐⇒ ∀  > 0, ∃ δ > 0 such that if x ∈ E and dX(x, p) < δ then dY(f (x), f (p)) < .

⇐⇒ ∀  > 0, ∃ δ > 0 such that f (Bδ(p) ∩ E) ⊂ B(f (p)) ∩ f (E).

We say that f is continuous on E if f is continuous at every point p ∈ E.

Theorems

(a) Let X, Y, Z be metric spaces. Suppose E ⊂ X, f maps E into Y, g maps f (E) into Z, and h is the mapping of E into Z defined by

h(x) = g(f (x)) = g ◦ f (x) ∀ x ∈ E.

If f is continuous at a point p ∈ E and if g is continuous at the point f (p), then h is continuous at p.

Proof Let  > 0 be given. Since g is continuous at f (p), there exists a η > 0 such that g(Bη(f (p)) ∩ f (E)) ⊂ B(g(f (p))) ∩ g(f (E)).

Also since f is continuous at p, there exists a δ > 0 such that f (Bδ(p) ∩ E) ⊂ Bη(f (p)) ∩ f (E).

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Thus we obtain that

g(f (Bδ(p) ∩ E)) ⊂ B(g(f (p))) ∩ g(f (E)) =⇒ h(Bδ(p) ∩ E) ⊂ B(h(p))) ∩ h(E) and this proves that h is continuous at p.

(b) Let X, Y be metric spaces and f : X → Y. Then f is continuous on X if and only if f−1(V )= {x ∈ X | f (x) ∈ V } is open in X for every open set V ⊂ Y.

Proof

(=⇒) Let V be an open subset of Y and p be a point in f−1(V ). Since V is open and f (p) ∈ V, f (p) is an interior point of V and there exists an  > 0 such that

B(f (p)) ⊂ V.

Since f is continuous at p, there exists a δ > 0 such that

f (Bδ(p)) ⊂ B(f (p)) ⊂ V =⇒ Bδ(p) ⊂ f−1(V )

which implies that p is an interior point of f−1(V ). Since p is an arbitrary point of f−1(V ), this shows that f−1(V ) is an open subset of X.

(⇐=) Let  > 0, p be a point in X and V = V(f (p)) ⊂ Y.

Since V is open and p ∈ f−1(V ), f−1(V ) is an open subset in X by the hypothesis and hence there exists a δ > 0 such that

Bδ(p) ⊂ f−1(V ) ⇐⇒ f (Bδ(p)) ⊂ V = B(f (p)) =⇒ f is continuous at p.

Since p is an arbitrary point in X, f is continuous on X.

Corollary Let X, Y, f be as above. Then f is continuous if and only if f−1(C)= {x ∈ X | f (x) ∈ C} is closed in X for every closed set C ⊂ Y.

Proof

(=⇒) Let C be a closed subset of Y. Since f : X → Y is continuous and Y \ C is an open subset in Y, the set

f−1(Y \ C) = X \ f−1(C) is open in X ⇐⇒ f−1(C) is closed in X.

(⇐=) For each open subset V of Y, since Y \ V is closed and

f−1(Y \ V ) = X \ f−1(V ) is closed in X ⇐⇒ f−1(V ) is open in X. Hence f : X → Y is continuous on X.

(c) Let f and g be complex continuous functions on a metric space X. Then f + g and f g are continuous on X. Furthermore, if g(x) 6= 0 for all x ∈ E, then f /g is continuous on X.

(d) Let f1, . . . , fk be real-valued functions on a metric space X, and let f be the mapping of X into Rk defined by

f (x) = (f1(x), . . . , fk(x)) ∀ x ∈ X.

Then f is continuous if and only if each of the functions f1, . . . , fk is continuous.

Proof For each 1 ≤ j ≤ k, and for any x, p ∈ X, since

|fj(x) − fj(p)| ≤ |f (x) − f (p)| =

" k X

i=1

|fi(x) − fi(p)|2

#1/2

k

X

i=1

|fi(x) − fi(p)|,

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f is continuous at p ⇐⇒ lim

x→p|f (x) − f (p)| = 0

=⇒ lim

x→p|fj(x) − fj(p)| = 0 ∀ 1 ≤ j ≤ k (by ≤)

⇐⇒ f1, . . . , fk are continuous at p and

f1, . . . , fk are continuous at p ⇐⇒ lim

x→p|fi(x) − fi(p)| = 0 ∀ 1 ≤ i ≤ k

⇐⇒ lim

x→p k

X

i=1

|fi(x) − fi(p)| = 0

=⇒ lim

x→p|f (x) − f (p)| = 0 (by ≤)

⇐⇒ f is continuous at p.

(e) If f and g are continuous mappings of X into Rk, then f + g and f · g are continuous on X.

Definition A mapping f of a set E into Rk is said to be bounded if there is a real number M such that

|f (x)| ≤ M for all x ∈ E.

Theorem Let X be a compact metric space, Y be metric spaces. Suppose f : X → Y is a continuous mapping. Then f (X) is compact.

Proof Let {Vα | α ∈ I} be an open cover of f (X) = {f (p) | p ∈ X}.

Since

f (X) ⊂ [

α∈I

Vα =⇒ X ⊂ f−1([

α∈I

Vα) = [

α∈I

f−1(Vα) =⇒ {f−1(Vα) | α ∈ I} is a cover of X,

and since f is continuous and Vα is an open subset of Y, f−1(Vα) is an open subset of X and {f−1(Vα) | α ∈ I} is an open cover of X. The compactness of X implies that there exist α1, · · · , αm ∈ I such that

X ⊂

m

[

i=1

f−1(Vαi) = f−1(

m

[

i=1

Vαi) =⇒ f (X) ⊂

m

[

i=1

Vαi =⇒ f (X) is a compact subset of Y.

Remarks

(a) If f is a continuous mapping of a compact metric space X into Rk, then f (X) is closed and bounded. Thus, f is bounded.

Proof This follows immediately from the preceding Theorem and the Heine-Borel Theorem (Theorem 2.41 in the text by Rudin).

(b) Suppose f is a continuous real function on a compact metric space X, and M = sup

p∈X

f (p), m = inf

p∈X f (p).

Then there exist points p, q ∈ X such that f (p) = M and f (q) = m.

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Proof Since X is compact and f : X → R is continuous on X, f (X) = {f (p) | p ∈ X} is a bounded and closed subset of R. Thus

sup

p∈X

f (p), inf

p∈Xf (p) exist since f (X) ⊂ R is bounded, and sup

p∈X

f (p), inf

p∈Xf (p) ∈ f (X) since f (X) is closed, hence contains all the limit points,

=⇒ ∃ p, q ∈ X such that f (p) = M, f (q) = m.

(c) Suppose f is a continuous 1 − 1 mapping of a compact metric space X onto a metric space Y. Then the inverse mapping f−1 : Y → X, defined by

f−1(y) = x ∀ y ∈ Y, where x is the only point in X such that f (x) = y.

⇐⇒ f−1(f (x)) = x ∀ x ∈ X.

is a continuous mapping of Y onto X.

Note f : X → Y is 1 − 1 and onto implies that for each subset U of X, we have f−1)−1(U ) = {y ∈ Y | f−1(y) ∈ U } = {y ∈ Y | y ∈ f (U )} = f (U ).

Proof For each open subset U in X, want to show that f−1)−1(U ) is open in Y.

Since Uc = X \ U is closed in the compact space X, Uc is compact in X. Also since f is continuous on X, f (Uc) is a compact subset of Y and hence a closed subset of Y.

Hence

f (U ) ∩ f (Uc) = ∅ since f is 1 − 1

=⇒ f (U ) ∪ f (Uc) = f (U ∪ Uc) = f (X) = Y since f is onto

=⇒ f (U ) = Y \ f (Uc) is open in Y

=⇒ f−1)−1(U ) = f (U )= Y \ f (Uc) is open in Y, i.e. f−1 : Y → X is continuous on Y.

Remark Let X, Y and f be as above. For each open set U ⊂ X, it is shown in the proof that f (U ) is open in Y. This implies that f is anopen mapping which maps interior points (of X) to interior points (of Y ) and f−1 is an open mapping from Y onto X. Also note that if E is closed in X, one can show that f (E) is closed in Y and f (∂E) = ∂f (E), f (Int E) = Int f (E).

Definition Let f : X → Y be a mapping of a metric space X into a metric space Y. We say that f is uniformly continuous on X if

∀  > 0, ∃ δ > 0 such that if p, q ∈ X satisfying dX(p, q) < δ, then dY(f (p), f (q) < 

⇐⇒ ∀  > 0, ∃ δ > 0 such that f (Bδ(p)) ⊂ B(f (p)) ∀ p ∈ X.

Theorem Let X be a compact metric space, Y be metric spaces. Suppose f : X → Y is a continuous mapping. Then f is uniformly continuous on X.

Proof Let  > 0 be given.

For each p ∈ X, since f is continuous at p, there exists a δ = δ(p) > 0 such that f (Bδ(p)(p)) ⊂ B/2(f (p)).

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Since

X ⊂ [

p∈X

Bδ(p)/2(p),

the collection {Bδ(p)/2(p) | p ∈ X} is an open cover of the compact space X, and hence there exist p1, · · · , pn∈ X such that

X ⊂

n

[

i=1

Bδ(pi)/2(pi).

Let δ = 1 2 min

1≤i≤nδ(pi). Then for any p, q ∈ X such that dX(p, q) < δ, since p ∈ X ⊂

n

[

i=1

Bδ(pi)/2(pi) =⇒ p ∈ Bδ(pm)/2(pm) for some 1 ≤ m ≤ n

=⇒ dX(q, pm) ≤ d(q, p) + d(p, pm) < δ +δ(pm)

2 ≤ δ(pm)

=⇒ p, q ∈ Bδ(pm)(pm)

=⇒ f (p), f (q) ∈ B(f (pm))

=⇒ dY(f (p), f (q)) ≤ dY(f (p), f (pm)) + dY(f (pm), f (q)) <  2 + 

2 = .

Hence f is uniformly continuous on X.

Examples of continuous but not unformly continuous functions.

(a) Let f (x) = 1

x for x ∈ (0, 1). For each x ∈ (0, 1) and for each fixed 0 < δ < min{x, 1 − x}, since x + δ/2 ∈ (x − δ, x + δ) ∩ (0, 1) and

lim

x→0+|f (x + δ/2) − f (x)| = lim

x→0+

1

x + δ/2− 1 x

= lim

x→0+

δ/2

x(x + δ/2) = ∞, f is not uniformly continuous on the (noncompact) interval (0, 1).

(b) Let f (x) = x2for x ∈ (−∞, ∞). For each fixed δ > 0, since x+δ/2 ∈ (x−δ, x+δ) ⊂ (−∞, ∞) and

x→∞lim |f (x + δ/2) − f (x)| = lim

x→∞

(x + δ/2)2− x2

= lim

x→∞



xδ + δ2 4



= ∞, f is not uniformly continuous on the (noncompact) interval (−∞, ∞).

Theorem Let X, Y be metric spaces and let f : X → Y be a continuous mapping. If E is a connected subset of X, then f (E) is connected.

Proof Assume, on the contrary, that f (E) is disconnected, i.e. there exist nonempty separated subsets A, B of Y such that

f (E) = A ∪ B.

Let (

G = E ∩ f−1(A) H = E ∩ f−1(B)

Since A and B are nonempty separated subsets in Y, G and H are nonempty disjoint subsets of E. Also since f (E) = A ∪ B, we have

E ⊆ f−1(A ∪ B) = f−1(A) ∪ f−1(B) =⇒ E ⊆ G ∪ H ⊆ E =⇒ E = G ∪ H.

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In order to show that E is disconnected, it suffices to show that G and H are separated, i.e.

G ∩ H = ∅ and G ∩ ¯¯ H = ∅.

Since A ⊆ ¯A, we have G ⊆ f−1(A) ⊆ f−1( ¯A).

Also since f is continuous and ¯A is closed in Y, f−1( ¯A) is closed in X and

G ⊆ f¯ −1( ¯A) = f−1( ¯A)

=⇒ G ⊆ f¯ −1( ¯A)

=⇒ f ( ¯G) ⊆ ¯A.

Further since

f (H) = B and ¯A ∩ B = ∅

=⇒ f ( ¯G) ∩ f (H) = ∅

=⇒ G ∩ H = ∅.¯ Similarly, since

B ⊆ ¯B

=⇒ H ⊆ f−1(B) ⊆ f−1( ¯B)

=⇒ H ⊆ f¯ −1( ¯B) = f−1( ¯B)

=⇒ H ⊆ f¯ −1( ¯B)

=⇒ f ( ¯H) ⊆ ¯B

=⇒ f (G) ∩ f ( ¯H) = ∅

=⇒ G ∩ ¯H = ∅.

Thus G and H are separated. This is a contradiction since E is connected.

Corollary Let f : [a, b] → R be continuous. If f (a) < f (b) and if c is a number such that f (a) < c < f (b), then there exists x ∈ (a, b) such that f (x) = c.

Proof Since [a, b] is connected in R (by the Theorem 2.47 in Rudin) and f is continuous on [a, b], f ([a, b]) is connected in R by the preceding Theorem. Hence f ([a, b]) is an interval and if c ∈ (f (a), f (b)) ⊂ f ([a, b]), then there exists x ∈ (a, b) such that f (x) = c.

Lipschitz Continuity, H¨older Continuity and Fixed Point Theorem

Definition A map f : U ⊂ Rm → Rn is called a Lipschitz continuous map on U if there exists a constant C ≥ 0 such that

kf (x) − f (y)k ≤ Ckx − yk for all x, y ∈ U.

If one can choose a (Lipschitz) constant C < 1 such that the above Lipschitz condition hold on U, then f is called a contraction map.

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Remarks

(a) If f is Lipschitz on an open subset U of Rm, then f is uniformly continuous on U since for each  > 0 there is a δ() = 

1 + C > 0 such that

if x, y ∈ U and if kx − yk < δ then kf (x) − f (y)k ≤ C kx − yk < C δ < .

(b) Let f (x) = |x| for x ∈ R. Since |f (x) − f (y)| = ||x| − |y|| ≤ |x − y|, f is Lipschitz with Lipschitz constant C = 1. Note that f is not differentiable at x = 0.

(c) Let 0 < α < 1, and let f be defined by

f (x) = xα for x ∈ [0, ∞).

For x, y ∈ [0, ∞), assuming for example 0 ≤ y < x and taking into account that α − 1 ≤ 0, we get

|f (x) − f (y)| = xα− yα = Z x

y

αtα−1dt

≤ Z x

y

α(t − y)α−1dt since 0 ≤ t − y ≤ t and α − 1 ≤ 0 =⇒ tα−1 ≤ (t − y)α−1

= (x − y)α

= |x − y|α.

Thus f satisfies the H¨older condition

|f (x) − f (y)| ≤ |x − y|α ∀ x, y ∈ [0, ∞)

and it is called a H¨older continuous function with the exponent α in [0, ∞).

(d) Let E be a bounded subset of Rm, f : E → Rn be a function defined on E and let 0 < α < β < 1. Then f is called a H¨older continuous function with the exponent αin E if there exists Cα ≥ 0 such that

kf (x) − f (y)k ≤ Cαkx − ykα for all x, y ∈ E.

Note that if

f is Lipschitz continuous in E

i.e. ∃ C ≥ 0 such that kf (x) − f (y)k ≤ C kx − yk for all x, y ∈ E

=⇒ f is H¨older continuous with the exponent 0 < β < 1 in E

i.e. ∃ Cβ ≥ 0 such that kf (x) − f (y)k ≤ Cβkx − ykβ for all x, y ∈ E

=⇒ f is H¨older continuous with the exponent 0 < α < 1 in E

i.e. ∃ Cα ≥ 0 such that kf (x) − f (y)k ≤ Cαkx − ykα for all x, y ∈ E

=⇒ f is uniformly continuous in E

=⇒ f is continuous in E.

Also note that the function f (x) = x for x ∈ R is Lipschitz continuous in R, but it is not H¨older continuous with exponent 0 < α < 1.

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Definition Let U be a subset of Rm, and f be a map that maps U into U, i.e. f : U → U. A point p ∈ U is said to be a fixed point of f if f (p) = p.

Fixed Point Theorem for Contractions Let f : Rm → Rm be a contraction map. Then f has a unique fixed point.

Proof Let x1 be arbitrary point in Rm and set x2 = f (x1); inductiveley, set xn+1= f (xn), n ∈ N.

Observe that

kx3− x2k = kf (x2) − f (x1)k ≤ Ckx2− x1k and, inductively, that

kxn+1− xnk = kf (xn) − f (xn−1)k ≤ Ckxn− xn−1k ≤ Cn−1kx2− x1k.

If m ≥ n, then

kxm− xnk ≤ kxm− xm−1k + kxm−1− xm−2k + · · · + kxn+1− xnk

≤ {Cm−2 + Cm−3+ · · · + Cn−1}kx2− x1k

≤ Cn−1

1 − Ckx2− x1k → 0 as n → ∞ since 0 < C < 1.

Therefore, {xn} is a Cauchy sequence in the complete space Rm. Let

p = lim

n→∞xn. Then p is a fixed point of f since

p = lim

n→∞xn+1= lim

n→∞f (xn) = f ( lim

n→∞xn) = f (p).

In fact, there is only one fixed point for f. Suppose p, q are two distinct fixed points of f, then kp − qk = kf (p) − f (q)k ≤ Ckp − qk =⇒ p = q.

Uniform Convergence

Definition Let X be a metric space and let

C(X) = {f : X → C | f is a complex-valued continuous, bounded function defined on X}.

For each f ∈ C(X), define the supremum norm k kX : C(X) → R by kf kX = sup

x∈X

|f (x)| = sup {|f (x)| | x ∈ X}.

Remarks

(a) Since f is bounded, kf kX < ∞ for all f ∈ C(X).

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(b) kf kX ≥ 0 and kf kX = 0 if and only if f (x) = 0 ∀x ∈ X.

(c) kcf kX = |c| kf kX ∀c ∈ C.

(d) kf + gkX ≤ kf kX + kgkX ∀f, g ∈ C(X).

(e) Let the function d : C(X) × C(X) → R be defined by

d(f, g) = kf − gkX ∀ f, g ∈ C(X).

Then C(X), k kX is a metric space.

Definition Let E be a subset of a metric space X. For each n ∈ N, let f, fn : E → C be complex-valued functions defined on E.

We say that fn converges to f pointwise on E if

n→∞lim fn(x) = f (x) for each x ∈ E, and fn converges uniformly on E to f if

∀  > 0, ∃N = N () ∈ N such that if n ≥ N, then |fn(x) − f (x)| ≤  for all x ∈ E.

Remarks

(a) (Cauchy Criterion for Uniform Convergence) Let {fn} and f be complex-valued continuous, bounded function defined on E. Then fn converges uniformly on E to f if and only if

n→∞lim kfn− f kE = lim

n→∞sup

x∈E

|fn(x) − f (x)| = 0

⇐⇒ ∀  > 0, ∃ N = N () ∈ N such that if m, n ≥ N then kfm− fnkE ≤ .

Proof

(=⇒) Suppose {fn} converges uniformly on E, and let f be the limit function. Then there is an integer N such that

if n ≥ N and x ∈ E then |fn(x) − f (x)| ≤  2

=⇒ if n, m ≥ N and x ∈ E then |fn(x) − fm(x)| ≤ |fn(x) − f (x)| + |f (x) − fm(x)| ≤ 

=⇒ if n, m ≥ N then kfn− fmk ≤ .

(⇐=) Suppose for each  > 0 there is an integer N such that if n, m ≥ N then kfn− fmk ≤ 

=⇒ if n, m ≥ N and x ∈ E then |fn(x) − fm(x)| ≤ kfn− fmk ≤  (∗)

=⇒ ∀ x ∈ E, {fn(x)} is a Cauchy sequence in C

=⇒ ∀ x ∈ E, {fn(x)} converges to a limit f (x) ∈ C.

Letting m → ∞ in (∗), we get

if n ≥ N and x ∈ E then |fn(x) − f (x)| = lim

m→∞|fn(x) − fm(x)| ≤ .

Hence fn converges uniformly on E to f.

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(b) Theorem Let E be a subset of a metric space X and let fn : E → C be a sequence of continuous complex-valued functions defined on E. If fn converges uniformly to f : E → C on E then f is continuous.

Proof For each  > 0, since fn converges uniformly on E to f, there is an integer N such that

if n ≥ N and if x ∈ E then |fn(x) − f (x)| <  3. For each p ∈ E, since fN is continuous on E, there exists a δ > 0 such that

if q ∈ E and dX(p, q) < δ then |fN(p) − fN(q)| <  3. Hence for each p ∈ E and for each q ∈ E satisfying dX(p, q) < δ, we have

|f (p) − f (q)| ≤ |f (p) − fN(p)| + |fN(p) − fN(q)| + |fN(q) − f (q)| < .

This implies that f is continuous on E.

(c) Theorem Let E be a subset of a metric space X and let C(E), k kE denote the space of bounded complex-valued continuous functions defined on E equipped with the supremum norm. Then the metric space C(E), k kE is a complete metric space, i.e. every Cauchy sequence fnin C(E) converges to f ∈ C(E) since one can show that fnconverges uniformly to f on E.

Proof If {fn} be a Cauchy sequence in C(E), k kE, then for each x ∈ E, since {fn(x)} is a Cauchy sequence in (the complete space) C, {fn(x)} converges to a limit f (x) ∈ C. Also since {fn} is a Cauchy sequence of continuous functions in C(E), k kE, the convergence is uniform on E and hence the limit function f is continuous on E.

(d) Theorem Let fn: [a, b] → C be a sequence of continuously differentiable functions defined on I = [a, b]. If fn(x0) converges for some x0 ∈ [a, b] and if fn0 : [a, b] → C converges uni- formly on [a, b], then fn converges uniformly on [a, b] to a function f.

Proof For any x ∈ I = [a, b] and for any m, n ∈ N, there exists a y depending on m, n such that

[fm(x) − fn(x)] − [fm(x0) − fn(x0)] = [fm0 (y) − fn0(y)](x − x0)

=⇒ kfm− fnkI ≤ |fm(x0) − fn(x0)| + (b − a)kfm0 − fn0kI.

This implies that fn converges uniformly on I to a function f and f is continuous on I.

Examples

(a) For each n ∈ N, let fn : R → R be defined by fn(x) = x

n for each x ∈ R. Show that fn

converges to f ≡ 0 pointwise (not uniformly) on R.

(b) For each n ∈ N, let fn : R → R be defined by fn(x) = x2+ nx

n for each x ∈ R. Show that fn converges to f (x) = x pointwise (not uniformly) on R.

(c) For each n ∈ N, let fn : R → R be defined by fn(x) = 1

nsin(nx + n) for each x ∈ R. Show that fn converges uniformly to f ≡ 0 on R.

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(d) For each n ∈ N, let fn : [0, 1] → R be defined by

fn(x) = xn for each x ∈ [0, 1], and let f defined by

f (x) =

(0 if 0 ≤ x < 1 1 if x = 1.

Show that fn converges to pointwise (not uniformly) on [0, 1]. Note that the continuity, the integral or the derivative is not preserved by pointwise convergence.

(e) For each m ∈ N, let

(∗) fm(x) = lim

n→∞(cos m!πx)2n =

(1 if m! x ∈ Z, 0 if m! x /∈ Z.

Then fm is continuous almost everywhere in R.

Now if x ∈ Q, since x = p

q for some p, q ∈ Z, we have m! x ∈ Z for all m ≥ q =⇒ lim

m→∞fm(x) = 1 for all x ∈ Q.

On the other hand, if x 6∈ Q, since m! x /∈ Z for all m ∈ N, we have fm(x) = 0 for all x /∈ Q.

Let f be defined by

(∗∗) f (x) =

(1 if x ∈ Q, 0 if x 6∈ Q.

Then f is discontinuous everywhere in R and note that

m→∞lim fm(x) = f (x) =

(1 if x ∈ Q 0 if x 6∈ Q and for each m ∈ N, by comparing (∗) and (∗∗), we have

sup

x∈Q

|fm(x) − f (x)| = 1.

This implies that the convergence is not uniform on R.

Examples

(a) For each n ∈ N, let fn be defined for x > 0 by fn(x) = 1

nx. For what values of f does

n→∞lim fn(x) exist? Is the convergence uniform on the entire (0, ∞)? Is the convergence uniform for x ≥ 1?

(b) For each n ∈ N, let fn be defined on R by fn(x) = nx

1 + n2x2. Show that lim

n→∞fn(x) exists for all x ∈ R. Is this convergence uniform on R?

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(c) Show that lim

n→∞(cos πx)2n exists for all x ∈ R. Is the limiting function contiuous? Is this convergence uniform?

Note for each n ∈ N, kfn− f kR≥ lim

x6=1, x→1|fn(x) − f (x)|

=⇒ lim sup

n→∞

kfn− f kR≥ lim

n→∞ lim

x6=1, x→1 |(cos πx)2n| = 1 6= 0.

(d) Let fn be defined on the interval I = [0, 1] by the formula

fn(x) =





1 − nx if 0 ≤ x ≤ 1 n

0 if 1

n < x ≤ 1 Show that lim

n→∞fn(x) exists on I. Is this convergence uniform on I?

Note for each n ∈ N, kfn− f kI ≥ |fn(1/2n)| = 1/2 =⇒ lim sup

n→∞

kfn− f kI 6= 0.

(e) Let fn be defined on the interval I = [0, 1] by the formula

fn(x) =





nx if 0 ≤ x ≤ 1

n(1 − x) n

n − 1 if 1

n < x ≤ 1 Show that lim

n→∞fn(x) exists on I. Is this convergence uniform on I? Is the convergence uniform on [c, 1] for c > 0?

Note for each n ∈ N, kfn− f kI ≥ |fn(1/2n) − f (1/2n)| ≥ 1/2 − 1/2n

=⇒ lim sup

n→∞

kfn− f kI 6= 0.

Definitions Let E be a subset of a metric space X and let fn : E → C

be a sequence of complex-valued functions defined on E. We say that

(a) {fn} is pointwise bounded on E if {fn(x) | n ∈ N} is bounded for each x ∈ E, i.e.

∃ a function φ : E → R such that |fn(x)| ≤ φ(x) ∀ x ∈ E and ∀ n ∈ N.

(b) {fn} is bounded (or uniformly bounded) on E if

∃ a constant M such that kfnk = sup{|fn(x)| | x ∈ E} ≤ M ∀ n ∈ N.

Definition Let E be a subset of a metric space X and letF be a set of functions defined on E.

Then F is said to be uniformly equicontinuous on E if

∀  > 0, ∃ δ = δ() > 0 such that if x, y ∈ E and kx−yk < δ, then kf (x)−f (y)k <  ∀ f ∈ F .

Theorem (Arzel`a-Ascoli) Let K be a compact subset of Rp and letF be a collection of func- tions which are continuous on K and have values in Rq. The following properties are equivalent:

(a) The family F is bounded and uniformly equicontinuous on K.

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(b) Every sequence from F has a subsequence which is uniformly convergent on K.

Proof

(b) ⇒ (a) Claim: F is bounded on K.

Proof Suppose that F is not bounded, then

∃ a sequence {fn}n∈N⊂F such that kfnkK ≥ n, i.e. ∃ xn ∈ K such that kfn(xn)k ≥ n.

This implies that fn cannot have any subsequence that converges uniformly on K.

Otherwise if fn had a subsequence, still denoted by fn, that converges uniformly on K to a uniformly continuous function f, then

∞ = lim

n→∞kfnkK ≤ lim

n→∞kfn− f kK+ lim

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

Hence, F is bounded on K.

Claim: F is uniformly equicontinuous on K.

Proof Suppose that F is not uniformly equicontinuous on K, then

∃ 0 > 0 such that ∀ n ∈ N, ∃ fn ∈F and xn, yn ∈ K such 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.

Otherwise if fn had a subsequence, still denoted by fn, that converges uniformly on K to a uniformly continuous function f, then

0 < 0 ≤ lim

n→∞kfn(xn) − fn(yn)k

= lim

n→∞kfn(xn) − f (xn) + f (xn) − f (yn) + f (yn) − fn(yn)k

≤ lim

n→∞kfn(xn) − f (xn)k + lim

n→∞kf (xn) − f (yn)k + lim

n→∞kfn(yn) − f (yn)k

≤ lim

n→∞kfn− f kK+ lim

n→∞kf (xn) − f (yn)k + lim

n→∞kfn− f kK

= 0 which is a contradiction.

Note that in the last equality

n→∞lim kfn− f kK = 0 since fnconverges 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. Hence, F is uniformly equicontinuous on K.

(a) ⇒ (b) Use the diagonal process to find a convergent subsequence.

Let S = K ∩ Qp = {xi}i∈N.

Then S is a countable dense subset of K such that S = K, i.e.

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∀ x ∈ K and ∀ δ > 0, ∃ xi ∈ S such that x ∈ Bδ(xi) = {z ∈ Rp | kz − xik < δ}.

Let {fn} be a sequence inF . Then

{fn(x1)} is bounded in Rq ⇒ {fn} has a subsequence {fn1} converges at x1. {fn1(x2)} is bounded in Rq ⇒ {fn1} has a subsequence {fn2} converges at x2, x1.

· · · ·

{fnk−1(xk)} is bounded in Rq ⇒ {fnk−1} has a subsequence {fnk} converges at xk, xk−1, . . . , x1.

· · · · i.e.

{ f1 f2 · · · fk · · · } S

{ f11 f21 · · · fk1 · · · }(x1) → f (x1) S

{ f12 f22 · · · fk2 · · · }(xi) → f (xi) for 1 ≤ i ≤ 2 S

{ f13 f23 · · · fk3 · · · }(xi) → f (xi) for 1 ≤ i ≤ 3 S

... S

{ f1k f2k · · · fkk · · · }(xi) → f (xi) for 1 ≤ i ≤ k S

...

By setting gn= fnn, we obtain a subsequence of {fn}.

For each j ∈ N, since

gn = fnn∈ {fmj}m∈N ∀ n ≥ j =⇒{gn| n ≥ j} ⊂ {fmj} and lim

n→∞gn(xj) = lim

m→∞fmj(xj) = f (xj) ∀ j ∈ N. Hence,

n→∞lim gn(xj) = f (xj) ∀ xj ∈ S.

Claim: The subsequence gn (of fn) converges uniformly to f on K.

Proof 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 < 

3 ∀ h ∈F . This implies that for each each x ∈ K, since S = K ⊂

[

i=1

B(δ, xi) and F is uniformly equicon- tinuous on K,

(∗) ∃ xi ∈ S such that kx − xik < δ =⇒ kgn(x) − gn(xi)k < 

3 ∀ n ∈ N. With the given  > 0, since lim

n→∞gn(xi) = f (xi), there exists L ∈ N such that (†) ∃ L ∈ N such that if m, n ≥ L =⇒ kgm(xi) − gn(xi)k < 

3.

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Thus for each x ∈ K and for each m, n ≥ L, combining (∗) and (†), we have kgn(x) − gm(x)k ≤ kgn(x) − gn(xi)k + kgn(xi) − gm(xi)k + kgm(xi) − gm(x)k<

3 +  3 + 

3 = .

This implies that

for any m, n ≥ L, we have kgn− gmkK < , i.e. gn converges uniformly on K to a uniformly continuous function f .

Examples Justify that the Arzel`a-Ascoli Theorem is not applicable in the following sequence of functions.

(a) fn(x) = x + n for x ∈ [0, 1]

(b) fn(x) = xn for x ∈ [0, 1]

(c) fn(x) = 1

1 + (x − n)2 for x ∈ [0, ∞)

Examples Let fn : [0, 1] → R be continuous and be such that |fn(x)| ≤ 100 for every n and for all x ∈ [0, 1] and the derivatives fn0(x) exist and are uniformly bounded on (0, 1). Prove that fn has a uniformly convergent subsequence.

Proof Let M be a constant such that |fn0(x)| ≤ M for all x ∈ (0, 1). By the mean value theorem, we get

| fn(x) − fn(y) | ≤ M |x − y| ∀x, y ∈ [0, 1] and ∀ n ∈ N.

For each  > 0, let δ = 

1 + M. This inequality implies that

if x, y ∈ [0, 1] satisfying |x − y| < δ then |f (x) − f (y)| ≤ M |x − y| < M δ = M  1 + M ≤ .

Hence {fn} is uniformly equicontinuous on K and, since {fn} is bounded by the hypothesis, {fn} has a uniformly convergent subsequence by the Arzel`a-Ascoli Theorem.

Example Let the functions fn : [a, b] → R be uniformly bounded continuous functions. Set Fn(x) =

Z x a

fn(t) dt for x ∈ [a, b].

Prove that Fn has a uniformly convergent subsequence.

Proof Since kFnk ≤ kfnk(b − a), Fn is uniformly bounded. Also, since |Fn0(x)| ≤ kfnk, Fn is equicontinuous by the preceding result. Hence, Fn has a uniformly convergent subsequence by Arzel`a-Ascoli Theorem.

Example Let f : [a, b] → R be differentiable satisfying

0 < m ≤ f0(x) ≤ M for every x ∈ [a, b]

and let

f (a) < 0 < f (b).

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Given x1 ∈ [a, b], define the sequence {xn} by xn+1 = xn− 1

Mf (xn) for each n ∈ N.

Prove that this sequence is well-defined and converges to the unique root ¯x of the equation f (¯x) = 0 in [a, b].

Hint Consider φ : [a, b] → R defined by φ(x) = x − f (x)

M . Note that the φ map [a, b] into [a, b] = dom(f ).

Proof For any x ∈ [a, b], since 0 < f0(t) ≤ M for all t ∈ [a, b], we have f (x) − f (a) =

Z x a

f0(t)dt ≤ M Z x

a

dt = M (x − a) which implies that

a ≤ x − f (x)

M +f (a)

M ≤ x −f (x) M ,

where we have used the assumption that f (a) < 0 in the last inequality.

Similarly,

f (b) − f (x) = Z b

x

f0(t)dt ≤ M Z b

x

dt = M (b − x) which implies that

b ≥ f (b)

M + x − f (x)

M ≥ x −f (x) M ,

where we have used the assumption that f (b) > 0 in the last inequality.

Hence, the function

a ≤ φ(x) = x − f (x) M ≤ b maps [a, b] into [a, b], and the sequence

xn+1= xn− 1

Mf (xn) = φ(xn), for n ∈ N, is well defined if x1 ∈ [a, b].

For any x, y ∈ [a, b], since

|φ(x) − φ(y)| = |x − y −f0(c)

M (x − y)|, for some c lying between x and y.

Hence,

|φ(x) − φ(y)| ≤ (1 − m

M)|x − y|

and φ is Lipschitz with Lipschitz constant 0 ≤ 1 − m M < 1.

Hence, φ is a contraction and has a unique fixed point ¯x ∈ [a, b].

Since

|xn+1− ¯x| = |φ(xn) − φ(¯x)| ≤ (1 − m

M)|xn− ¯x| ≤ (1 − m

M)n|x1− ¯x| ≤ (1 − m

M)n(b − a), and

¯

x = lim

n→∞xn+1 = lim

n→∞xn− lim

n→∞

f (xn)

M = ¯x − f (¯x) M . Hence, f (¯x) = 0.

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Example For each k ∈ N, let fk: R → R be a differentiable function satisfying

|fk0(x)| ≤ 1 for all x ∈ R and fk(0) = 0.

Prove that {fk} has a subsequence converges piontwise on Q.

Proof For each x ∈ R, since |fk(x)| = |fk(x) − fk(0)| ≤ |fk0(ck) (x − 0)| ≤ |x|, where ck lies between x and 0, the set {fk(x)}k=1 is bounded.

Let Q = {x1, x2, · · · }. Since {fk(x1)} is bounded, we can extract a convergent subsequence, denoted {fk1(x1)}, out of {fk(x1)}. By the boundedness of {fk1(x2)}, we can extract a con- vergent subsequence, denoted by {fk2(x2)}, out of {fk1(x2)}. Inductively, by the boundedness of {fkj(xj+1)}, we can extract a convergent subsequence {fkj+1(xj+1)}, out of {fkj(xj+1)} for all j ∈ N.

For each j ∈ N, since fjj ∈ {fk}, {fjj} is a convergent subsequence of {fk} that converges at each xi ∈ Q.

Example LetF be a bounded and uniformly equicontinuous collection of functions on D ⊂ Rp to R and let f be defined on D → R by

f(x) = sup{f (x) | f ∈F , x ∈ D}

Show that f is continuous on D to R. Show that the conclusion may fail if F is not uniformly equicontinuous.

Stone-Weierstrass Theorem Let f : [a, b] → C be a continuous complexed-valued function.

Then there exists a sequence of polynomials Pn such that

n→∞lim Pn(x) = f (x) uniformly on [a, b].

If f is a real-valued function, then Pn may be taken as real-valued.

Proof We may assume, without loss of generality, that [a, b] = [0, 1]

by considering the composition of

f : [a, b] → C and x = a + t(b − a) : [0, 1] → [a, b].

We may also assume that

f (0) = f (1) = 0, otherwise we consider the function g defined by

g(x) = f (x) − f (0) − x[f (1) − f (0)] for x ∈ [0, 1]

to obtain that

g(0) = g(1) = 0.

Note that if g can be obtained as the limit of a uniformly convergent sequence of polynomials, it is clear that the same is true for f, since f (x) − g(x) = f (0) + x[f (1) − f (0)] is a polynomial.

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Extend f from [0, 1] to R by letting

f (x) =

(f (x) x ∈ [0, 1]

0 x ∈ R \ [0, 1].

Then f : R → C is uniformly continuous on R.

For each n ∈ N, let Qn(x) = cn(1 − x2)n, where cn is chosen such that Z 1

−1

Qn(x) dx = 1 i.e. cn Z 1

−1

(1 − x2)n = 1.

Since

Z 1

−1

(1 − x2)ndx = 2 Z 1

0

(1 − x2)ndx

≥ 2 Z 1/

n 0

(1 − x2)ndx

≥ 2 Z 1/

n 0

(1 − n x2) dx

≥ 4

3√ n

> 1

√n. Hence we have

cn <√

n ∀ n ∈ N.

For any δ > 0, this implies that Qn(x) = cn(1 − x2)n ≤√

n (1 − δ2)n ∀ δ ≤ |x| ≤ 1 =⇒ Qn → 0 uniformly in δ ≤ |x| ≤ 1.

Now set

Pn(x) = Z 1

−1

f (x + t) Qn(t) dt for each x ∈ [0, 1].

Since f (x) =

(f (x) x ∈ [0, 1]

0 x ∈ R \ [0, 1]. =⇒ f (x + t) = 0 if x + t /∈ [0, 1] (i.e. t /∈ [−x, 1 − x]), we have

Pn(x) =

Z 1−x

−x

f (x + t) Qn(t) dt

= Z 1

0

f (s) Qn(s − t) ds by setting s = x + t, ds = dt

= Z 1

0

f (s) cn1 − (s − x)2n

ds

= a polynomial in x.

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This shows that {Pn} is a sequence of polynomials.

Given  > 0, we choose δ > 0 such that

if |y − x| < δ =⇒ |f (y) − f (x)| <  2. Let M = sup

x∈[0,1]

|f (x)|. Since

Z 1

−1

Qn(t) dt = 1 and Qn(t) ≥ 0, we see that for 0 ≤ x ≤ 1,

|Pn(x) − f (x)| =

Z 1

−1

f (x + t) Qn(t) dt − f (x) Z 1

−1

Qn(t) dt

=

Z 1

−1

f (x + t) − f (x) Qn(t) dt

≤ Z 1

−1

|f (x + t) − f (x)| Qn(t) dt

≤ 2M Z −δ

−1

Qn(t) dt +  2

Z δ

−δ

Qn(t) dt + 2M Z 1

δ

Qn(t) dt

≤ 4M√

n 1 − δ2n

+  2

<  if n is sufficiently large and this proves that Pn converges to f uniformly on [0, 1].

Corollary For each interval [−a, a] ⊂ R, there exists a sequence of real-valued polynomials {Pn} defined on [−a, a] such that

( Pn(0) = 0

n→∞lim Pn(x) = |x| uniformly on [−a, a].

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