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 < d_{X}(x, p) < δ then d_{Y}(f (x), q) < .

⇐⇒ ∀ > 0, ∃ δ > 0 such that f (B_{δ}(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 (p_{n}) = q for each sequence {p_{n}} in E such that
pn6= p, lim

n→∞pn= p.

Proof

(=⇒) Let {p_{n}} be a sequence in E such that p_{n}6= p ∀ n ∈ N and lim

n→∞p_{n}= p. Thus

∀ > 0, since lim

x→pf (x) = q, ∃ δ > 0 such that if x ∈ E and d_{X}(x, p) < δ =⇒ d_{Y}(f (x), q) <
and

with this chosen δ > 0, since lim

n→∞p_{n}= p, ∃N ∈ N such that if n ≥ N =⇒d_{X}(p_{n}, p) < δ.

This implies that

d_{Y}(f (p_{n}), q) < =⇒ lim

n→∞f (p_{n}) = 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

d_{X}(x, p) < p and d_{Y}(f (x), q) ≥ .

For each n ∈ N, by taking δn = 1

n > 0, there exists p_{n}∈ E such that
d_{X}(p_{n}, p) < 1

n and d_{Y}(f (p_{n}), q) ≥ ∀ n ∈ N.

This implies that

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

n→∞f (p_{n}) 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 d_{X}(x, p) < δ =⇒ d_{Y}(f (x), q) <

2 and d_{Y}(f (x), r) <
2.

This implies that

d_{Y}(q, r) ≤ d_{Y}(q, f (x)) + d_{Y}(f (x), r) <
2 +

2 = ∀ > 0.

Thus we have

0 ≤ d_{Y}(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 R^{k}, then (a) remains true, and (b) becomes
(b^{0}) 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 (B_{δ}(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(B_{η}(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 (B_{δ}(p) ∩ E) ⊂ B_{η}(f (p)) ∩ f (E).

Thus we obtain that

g(f (B_{δ}(p) ∩ E)) ⊂ B_{}(g(f (p))) ∩ g(f (E)) =⇒ h(B_{δ}(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 (B_{δ}(p)) ⊂ B_{}(f (p)) ⊂ V =⇒ B_{δ}(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

B_{δ}(p) ⊂ f^{−1}(V ) ⇐⇒ f (B_{δ}(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 f_{1}, . . . , f_{k} be real-valued functions on a metric space X, and let f be the mapping of X
into R^{k} defined by

f (x) = (f_{1}(x), . . . , f_{k}(x)) ∀ x ∈ X.

Then f is continuous if and only if each of the functions f_{1}, . . . , f_{k} is continuous.

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

|f_{j}(x) − f_{j}(p)| ≤^{∗} |f (x) − f (p)| =

" _{k}
X

i=1

|f_{i}(x) − f_{i}(p)|^{2}

#^{1/2}

≤^{†}

k

X

i=1

|f_{i}(x) − f_{i}(p)|,

f is continuous at p ⇐⇒ lim

x→p|f (x) − f (p)| = 0

=⇒ lim

x→p|f_{j}(x) − f_{j}(p)| = 0 ∀ 1 ≤ j ≤ k (by ≤^{∗})

⇐⇒ f_{1}, . . . , f_{k} are continuous at p
and

f_{1}, . . . , f_{k} are continuous at p ⇐⇒ lim

x→p|f_{i}(x) − f_{i}(p)| = 0 ∀ 1 ≤ i ≤ k

⇐⇒ lim

x→p k

X

i=1

|f_{i}(x) − f_{i}(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 R^{k}, then f + g and f · g are continuous on X.

Definition A mapping f of a set E into R^{k} 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 {V_{α} | α ∈ I} be an open cover of f (X) = {f (p) | p ∈ X}.

Since

f (X) ⊂ [

α∈I

V_{α} =⇒ X ⊂ f^{−1}([

α∈I

V_{α}) = [

α∈I

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

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

X ⊂

m

[

i=1

f^{−1}(V_{α}_{i}) = f^{−1}(

m

[

i=1

V_{α}_{i}) =⇒ f (X) ⊂

m

[

i=1

V_{α}_{i} =⇒ f (X) is a compact subset of Y.

Remarks

(a) If f is a continuous mapping of a compact metric space X into R^{k}, 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.

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

Hence

f (U ) ∩ f (U^{c}) = ∅ since f is 1 − 1

=⇒ f (U ) ∪ f (U^{c}) = f (U ∪ U^{c}) = f (X) = Y since f is onto

=⇒ f (U ) = Y \ f (U^{c}) is open in Y

=⇒ f^{−1})^{−1}(U ) = f (U )= Y \ f (U^{c}) 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 d_{X}(p, q) < δ, then d_{Y}(f (p), f (q) <

⇐⇒ ∀ > 0, ∃ δ > 0 such that f (B_{δ}(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 (B_{δ(p)}(p)) ⊂ B_{/2}(f (p)).

Since

X ⊂ [

p∈X

B_{δ(p)/2}(p),

the collection {B_{δ(p)/2}(p) | p ∈ X} is an open cover of the compact space X, and hence there
exist p_{1}, · · · , p_{n}∈ X such that

X ⊂

n

[

i=1

B_{δ(p}_{i}_{)/2}(p_{i}).

Let δ = 1 2 min

1≤i≤nδ(p_{i}). Then for any p, q ∈ X such that d_{X}(p, q) < δ, since
p ∈ X ⊂

n

[

i=1

B_{δ(p}_{i}_{)/2}(p_{i}) =⇒ p ∈ B_{δ(p}_{m}_{)/2}(p_{m}) for some 1 ≤ m ≤ n

=⇒ d_{X}(q, p_{m}) ≤ d(q, p) + d(p, p_{m}) < δ +δ(p_{m})

2 ≤ δ(p_{m})

=⇒ p, q ∈ B_{δ(p}_{m}_{)}(p_{m})

=⇒ f (p), f (q) ∈ B_{}(f (p_{m}))

=⇒ d_{Y}(f (p), f (q)) ≤ d_{Y}(f (p), f (p_{m})) + d_{Y}(f (p_{m}), 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) = x^{2}for 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}− x^{2}

= lim

x→∞

xδ + δ^{2}
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.

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 ⊂ R^{m} → R^{n} 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.

Remarks

(a) If f is Lipschitz on an open subset U of R^{m}, 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) = x^{α} 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)| = x^{α}− y^{α} =
Z x

y

αt^{α−1}dt

≤ Z x

y

α(t − y)^{α−1}dt since 0 ≤ t − y ≤ t and α − 1 ≤ 0 =⇒ t^{α−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 R^{m}, f : E → R^{n} 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 C_{α} ≥ 0 such that

kf (x) − f (y)k ≤ Cαkx − yk^{α} 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. ∃ C_{β} ≥ 0 such that kf (x) − f (y)k ≤ C_{β}kx − yk^{β} 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 − yk^{α} 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.

Definition Let U be a subset of R^{m}, 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 : R^{m} → R^{m} be a contraction map. Then f
has a unique fixed point.

Proof Let x_{1} be arbitrary point in R^{m} and set x_{2} = f (x_{1}); inductiveley, set
x_{n+1}= f (x_{n}), n ∈ N.

Observe that

kx_{3}− x_{2}k = kf (x_{2}) − f (x_{1})k ≤ Ckx_{2}− x_{1}k
and, inductively, that

kx_{n+1}− x_{n}k = kf (x_{n}) − f (x_{n−1})k ≤ Ckx_{n}− x_{n−1}k ≤ C^{n−1}kx_{2}− x_{1}k.

If m ≥ n, then

kx_{m}− x_{n}k ≤ kx_{m}− x_{m−1}k + kx_{m−1}− x_{m−2}k + · · · + kx_{n+1}− x_{n}k

≤ {C^{m−2} + C^{m−3}+ · · · + C^{n−1}}kx_{2}− x_{1}k

≤ C^{n−1}

1 − Ckx_{2}− x_{1}k → 0 as n → ∞ since 0 < C < 1.

Therefore, {x_{n}} is a Cauchy sequence in the complete space R^{m}.
Let

p = lim

n→∞x_{n}.
Then p is a fixed point of f since

p = lim

n→∞x_{n+1}= lim

n→∞f (x_{n}) = f ( lim

n→∞x_{n}) = 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 k_{X} = sup

x∈X

|f (x)| = sup {|f (x)| | x ∈ X}.

Remarks

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

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

(c) kcf k_{X} = |c| kf k_{X} ∀c ∈ C.

(d) kf + gk_{X} ≤ kf k_{X} + kgk_{X} ∀f, g ∈ C(X).

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

d(f, g) = kf − gk_{X} ∀ f, g ∈ C(X).

Then C(X), k k_{X} is a metric space.

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

We say that f_{n} converges to f pointwise on E if

n→∞lim f_{n}(x) = f (x) for each x ∈ E,
and f_{n} 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 {f_{n}} and f be complex-valued
continuous, bounded function defined on E. Then f_{n} converges uniformly on E to f if and
only if

n→∞lim kf_{n}− f k_{E} = lim

n→∞sup

x∈E

|f_{n}(x) − f (x)| = 0

⇐⇒ ∀ > 0, ∃ N = N () ∈ N such that if m, n ≥ N then kfm− f_{n}k_{E} ≤ .

Proof

(=⇒) Suppose {f_{n}} 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 |f_{n}(x) − f_{m}(x)| ≤ |f_{n}(x) − f (x)| + |f (x) − f_{m}(x)| ≤

=⇒ if n, m ≥ N then kf_{n}− f_{m}k ≤ .

(⇐=) 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 |f_{n}(x) − f_{m}(x)| ≤ kf_{n}− f_{m}k ≤ (∗)

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

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

Letting m → ∞ in (∗), we get

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

m→∞|f_{n}(x) − f_{m}(x)| ≤ .

Hence f_{n} converges uniformly on E to f.

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

Proof For each > 0, since f_{n} converges uniformly on E to f, there is an integer N such
that

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

if q ∈ E and d_{X}(p, q) < δ then |f_{N}(p) − f_{N}(q)| <
3.
Hence for each p ∈ E and for each q ∈ E satisfying d_{X}(p, q) < δ, we have

|f (p) − f (q)| ≤ |f (p) − f_{N}(p)| + |f_{N}(p) − f_{N}(q)| + |f_{N}(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 k_{E} 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 f_{n}in C(E) converges to f ∈ C(E) since one can show that f_{n}converges uniformly
to f on E.

Proof If {f_{n}} be a Cauchy sequence in C(E), k k_{E}, 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 f_{n}: [a, b] → C be a sequence of continuously differentiable functions defined
on I = [a, b]. If f_{n}(x_{0}) converges for some x_{0} ∈ [a, b] and if f_{n}^{0} : [a, b] → C converges uni-
formly on [a, b], then f_{n} 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

[f_{m}(x) − f_{n}(x)] − [f_{m}(x_{0}) − f_{n}(x_{0})] = [f_{m}^{0} (y) − f_{n}^{0}(y)](x − x_{0})

=⇒ kf_{m}− f_{n}k_{I} ≤ |f_{m}(x_{0}) − f_{n}(x_{0})| + (b − a)kf_{m}^{0} − f_{n}^{0}k_{I}.

This implies that f_{n} 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 f^{n} : R → R be defined by f^{n}(x) = x^{2}+ nx

n for each x ∈ R. Show that
f_{n} 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 f_{n} converges uniformly to f ≡ 0 on R.

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

f_{n}(x) = x^{n} for each x ∈ [0, 1],
and let f defined by

f (x) =

(0 if 0 ≤ x < 1 1 if x = 1.

Show that f_{n} 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

(∗) f_{m}(x) = lim

n→∞(cos m!πx)^{2n} =

(1 if m! x ∈ Z, 0 if m! x /∈ Z.

Then f_{m} 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→∞f_{m}(x) = 1 for all x ∈ Q.

On the other hand, if x 6∈ Q, since m! x /∈ Z for all m ∈ N, we have
f_{m}(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 f_{m}(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

|f_{m}(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 f_{n}(x) = 1

nx. For what values of f does

n→∞lim f_{n}(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 + n^{2}x^{2}. Show that lim

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

(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 k_{R}≥ lim

x6=1, x→1|f_{n}(x) − f (x)|

=⇒ lim sup

n→∞

kf_{n}− f k_{R}≥ lim

n→∞ lim

x6=1, x→1 |(cos πx)^{2n}| = 1 6= 0.

(d) Let f_{n} be defined on the interval I = [0, 1] by the formula

f_{n}(x) =

1 − nx if 0 ≤ x ≤ 1 n

0 if 1

n < x ≤ 1 Show that lim

n→∞f_{n}(x) exists on I. Is this convergence uniform on I?

Note for each n ∈ N, kfn− f k_{I} ≥ |f_{n}(1/2n)| = 1/2 =⇒ lim sup

n→∞

kf_{n}− f k_{I} 6= 0.

(e) Let f_{n} be defined on the interval I = [0, 1] by the formula

f_{n}(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 k_{I} ≥ |f_{n}(1/2n) − f (1/2n)| ≥ 1/2 − 1/2n

=⇒ lim sup

n→∞

kf_{n}− f k_{I} 6= 0.

Definitions Let E be a subset of a metric space X and let
f_{n} : E → C

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

(a) {f_{n}} is pointwise bounded on E if {f_{n}(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 kf_{n}k = sup{|f_{n}(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 R^{p} and letF be a collection of func-
tions which are continuous on K and have values in R^{q}. The following properties are equivalent:

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

(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 {f_{n}}_{n∈N}⊂F such that kfnk_{K} ≥ n, i.e. ∃ x_{n} ∈ K such that kf_{n}(x_{n})k ≥ n.

This implies that f_{n} 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→∞kf_{n}k_{K} ≤ lim

n→∞kf_{n}− f k_{K}+ lim

n→∞kf k_{K} < ∞ 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, y_{n} ∈ K
such that

kx_{n}− y_{n}k < 1

n and kf_{n}(xn) − fn(yn)k ≥ 0.

This implies that f_{n} cannot have any subsequence that converges uniformly on K.

Otherwise if f_{n} had a subsequence, still denoted by f_{n}, that converges uniformly on K to a
uniformly continuous function f, then

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_{n}converges to f uniformly on K,
and

n→∞lim kf (x_{n}) − f (y_{n})k = 0 since f is uniformly continuous on K, and kx_{n}− y_{n}k < 1
n.
Hence, F is uniformly equicontinuous on K.

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

Let S = K ∩ Q^{p} = {x_{i}}_{i∈N}.

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

∀ x ∈ K and ∀ δ > 0, ∃ x_{i} ∈ S such that x ∈ B_{δ}(x_{i}) = {z ∈ R^{p} | kz − x_{i}k < δ}.

Let {f_{n}} be a sequence inF . Then

{f_{n}(x_{1})} is bounded in R^{q} ⇒ {f_{n}} has a subsequence {f_{n}^{1}} converges at x_{1}.
{f_{n}^{1}(x_{2})} is bounded in R^{q} ⇒ {f_{n}^{1}} has a subsequence {f_{n}^{2}} converges at x_{2}, x_{1}.

· · · ·

{f_{n}^{k−1}(x_{k})} is bounded in R^{q} ⇒ {f_{n}^{k−1}} has a subsequence {f_{n}^{k}} converges at x_{k}, x_{k−1}, . . . , x_{1}.

· · · · i.e.

{ f_{1} f_{2} · · · f_{k} · · · }
S

{ f_{1}^{1} f_{2}^{1} · · · f_{k}^{1} · · · }(x_{1}) → f (x_{1})
S

{ f_{1}^{2} f_{2}^{2} · · · f_{k}^{2} · · · }(x_{i}) → f (x_{i}) for 1 ≤ i ≤ 2
S

{ f_{1}^{3} f_{2}^{3} · · · f_{k}^{3} · · · }(x_{i}) → f (x_{i}) for 1 ≤ i ≤ 3
S

... S

{ f_{1}^{k} f_{2}^{k} · · · f_{k}^{k} · · · }(x_{i}) → f (x_{i}) for 1 ≤ i ≤ k
S

...

By setting g_{n}= f_{n}^{n}, we obtain a subsequence of {f_{n}}.

For each j ∈ N, since

g_{n} = f_{n}^{n}∈ {f_{m}^{j}}_{m∈N} ∀ n ≥ j =⇒{g_{n}| n ≥ j} ⊂ {f_{m}^{j}} and lim

n→∞g_{n}(x_{j}) = lim

m→∞f_{m}^{j}(x_{j}) = f (x_{j}) ∀ j ∈ N.
Hence,

n→∞lim gn(xj) = f (xj) ∀ xj ∈ S.

Claim: The subsequence g_{n} (of f_{n}) 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, x_{i}) and F is uniformly equicon-
tinuous on K,

(∗) ∃ x_{i} ∈ S such that kx − x_{i}k < δ =⇒ kg_{n}(x) − g_{n}(x_{i})k <

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

n→∞g_{n}(x_{i}) = f (x_{i}), there exists L ∈ N such that
(†) ∃ L ∈ N such that if m, n ≥ L =⇒ kgm(x_{i}) − g_{n}(x_{i})k <

3.

Thus for each x ∈ K and for each m, n ≥ L, combining (∗) and (†), 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, we have kg_{n}− g_{m}k_{K} < ,
i.e. g_{n} 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) f_{n}(x) = x + n for x ∈ [0, 1]

(b) fn(x) = x^{n} for x ∈ [0, 1]

(c) f_{n}(x) = 1

1 + (x − n)^{2} for x ∈ [0, ∞)

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

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

| f_{n}(x) − f_{n}(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 {f_{n}} is uniformly equicontinuous on K and, since {f_{n}} is bounded by the hypothesis, {f_{n}}
has a uniformly convergent subsequence by the Arzel`a-Ascoli Theorem.

Example Let the functions f_{n} : [a, b] → R be uniformly bounded continuous functions. Set
F_{n}(x) =

Z x a

f_{n}(t) dt for x ∈ [a, b].

Prove that F_{n} has a uniformly convergent subsequence.

Proof Since kF_{n}k ≤ kf_{n}k(b − a), F_{n} is uniformly bounded. Also, since |F_{n}^{0}(x)| ≤ kf_{n}k, F_{n} is
equicontinuous by the preceding result. Hence, F_{n} has a uniformly convergent subsequence by
Arzel`a-Ascoli Theorem.

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

0 < m ≤ f^{0}(x) ≤ M for every x ∈ [a, b]

and let

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

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

Mf (x_{n}) 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 < f^{0}(t) ≤ M for all t ∈ [a, b], we have
f (x) − f (a) =

Z x a

f^{0}(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

f^{0}(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

x_{n+1}= x_{n}− 1

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

For any x, y ∈ [a, b], since

|φ(x) − φ(y)| = |x − y −f^{0}(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

|x_{n+1}− ¯x| = |φ(x_{n}) − φ(¯x)| ≤ (1 − m

M)|x_{n}− ¯x| ≤ (1 − m

M)^{n}|x_{1}− ¯x| ≤ (1 − m

M)^{n}(b − a),
and

¯

x = lim

n→∞x_{n+1} = lim

n→∞x_{n}− lim

n→∞

f (xn)

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

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

|f_{k}^{0}(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 |f^{k}(x)| = |fk(x) − fk(0)| ≤ |f_{k}^{0}(ck) (x − 0)| ≤ |x|, where ck lies
between x and 0, the set {f_{k}(x)}^{∞}_{k=1} is bounded.

Let Q = {x1, x_{2}, · · · }. Since {f_{k}(x_{1})} is bounded, we can extract a convergent subsequence,
denoted {f_{k}^{1}(x_{1})}, out of {f_{k}(x_{1})}. By the boundedness of {f_{k}^{1}(x_{2})}, we can extract a con-
vergent subsequence, denoted by {f_{k}^{2}(x_{2})}, out of {f_{k}^{1}(x_{2})}. Inductively, by the boundedness
of {f_{k}^{j}(x_{j+1})}, we can extract a convergent subsequence {f_{k}^{j+1}(x_{j+1})}, out of {f_{k}^{j}(x_{j+1})} for all
j ∈ N.

For each j ∈ N, since fj^{j} ∈ {f_{k}}, {f_{j}^{j}} is a convergent subsequence of {f_{k}} that converges at each
x_{i} ∈ Q.

Example LetF be a bounded and uniformly equicontinuous collection of functions on D ⊂ R^{p}
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 P_{n} such that

n→∞lim P_{n}(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.

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) = c_{n}(1 − x^{2})^{n}, where c_{n} is chosen such that
Z 1

−1

Q_{n}(x) dx = 1 i.e. c_{n}
Z 1

−1

(1 − x^{2})^{n} = 1.

Since

Z 1

−1

(1 − x^{2})^{n}dx = 2
Z 1

0

(1 − x^{2})^{n}dx

≥ 2 Z 1/√

n 0

(1 − x^{2})^{n}dx

≥ 2 Z 1/√

n 0

(1 − n x^{2}) dx

≥ 4

3√ n

> 1

√n. Hence we have

c_{n} <√

n ∀ n ∈ N.

For any δ > 0, this implies that
Q_{n}(x) = c_{n}(1 − x^{2})^{n} ≤√

n (1 − δ^{2})^{n} ∀ δ ≤ |x| ≤ 1 =⇒ Q_{n} → 0 uniformly in δ ≤ |x| ≤ 1.

Now set

P_{n}(x) =
Z 1

−1

f (x + t) Q_{n}(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) Q_{n}(s − t) ds by setting s = x + t, ds = dt

= Z 1

0

f (s) c_{n}1 − (s − x)^{2}n

ds

= a polynomial in x.

This shows that {P_{n}} 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

Q_{n}(t) dt = 1 and Q_{n}(t) ≥ 0,
we see that for 0 ≤ x ≤ 1,

|P_{n}(x) − f (x)| =

Z 1

−1

f (x + t) Q_{n}(t) dt − f (x)
Z 1

−1

Q_{n}(t) dt

=

Z 1

−1

f (x + t) − f (x) Q_{n}(t) dt

≤ Z 1

−1

|f (x + t) − f (x)| Q_{n}(t) dt

≤ 2M Z −δ

−1

Q_{n}(t) dt +
2

Z δ

−δ

Q_{n}(t) dt + 2M
Z 1

δ

Q_{n}(t) dt

≤ 4M√

n 1 − δ^{2}n

+ 2

< if n is sufficiently large
and this proves that P_{n} 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].