Continuous functions · Selected prerequisite proofs

L7.5.2: Metric continuity and preservation of sequence limits

Proposition 7.5.2.

Let

(X,d_X)

and

(Y,d_Y)

be metric spaces. Then

f \colon X \to Y

is continuous at

c \in X

if and only if for every sequence

\{ x_n \}_{n=1}^\infty

in

X

converging to

c\text{,}

the sequence

\bigl\{ f(x_n) \bigr\}_{n=1}^\infty

converges to

f(c)\text{.}

Proof.

Suppose

f

is continuous at

c\text{.}

Let

\{ x_n \}_{n=1}^\infty

be a sequence in

X

converging to

c\text{.}

Given

\epsilon > 0\text{,}

there is a

\delta > 0

such that

d_X(x,c) < \delta

implies

d_Y\bigl(f(x),f(c)\bigr) < \epsilon\text{.}

So take

M

such that for all

n \geq M\text{,}

we have

d_X(x_n,c) < \delta\text{,}

and then

d_Y\bigl(f(x_n),f(c)\bigr) < \epsilon\text{.}

Hence,

\bigl\{ f(x_n) \bigr\}_{n=1}^\infty

converges to

f(c)\text{.}

On the other hand, suppose

f

is not continuous at

c\text{.}

Then there exists an

\epsilon > 0

such that for every

n \in \N

there exists an

x_n \in X\text{,}

with

d_X(x_n,c) < \nicefrac{1}{n}

such that

d_Y\bigl(f(x_n),f(c)\bigr) \geq \epsilon\text{.}

Then

\{ x_n \}_{n=1}^\infty

converges to

c\text{,}

but

\bigl\{ f(x_n) \bigr\}_{n=1}^\infty

does not converge to

f(c)\text{.}

L7.5.5: Continuous image of a compact set is compact

Lemma 7.5.5.

Let

(X,d_X)

and

(Y,d_Y)

be metric spaces and

f \colon X \to Y

a continuous function. If

K \subset X

is a compact set, then

f(K)

is a compact set.

Proof.

A sequence in

f(K)

can be written as

\bigl\{ f(x_n) \bigr\}_{n=1}^\infty\text{,}

where

\{ x_n \}_{n=1}^\infty

is a sequence in

K\text{.}

The set

K

is compact and therefore there is a subsequence

\{ x_{n_j} \}_{j=1}^\infty

that converges to some

x \in K\text{.}

By continuity,

\begin{equation*} \lim_{j\to\infty} f(x_{n_j}) = f(x) \in f(K) . \end{equation*}

So every sequence in

f(K)

has a subsequence convergent to a point in

f(K)\text{,}

and

f(K)

is compact by Theorem 7.4.11.

L7.5.6: Continuous real functions on nonempty compact spaces attain both extrema

Theorem 7.5.6.

Let

(X,d)

be a nonempty compact metric space and let

f \colon X \to \R

be continuous. Then

f

is bounded and in fact

f

achieves an absolute minimum and an absolute maximum on

X\text{.}

Proof.

As

X

is compact and

f

is continuous,

f(X) \subset \R

is compact. Hence,

f(X)

is closed and bounded. In particular,

\sup f(X) \in f(X)

and

\inf f(X) \in f(X)\text{,}

because both the sup and the inf can be achieved by sequences in

f(X)

and

f(X)

is closed. Therefore, there is some

x \in X

such that

f(x) = \sup f(X)

and some

y \in X

such that

f(y) = \inf f(X)\text{.}

L7.5.11: Continuous maps on compact spaces are uniformly continuous

Theorem 7.5.11.

Let

(X,d_X)

and

(Y,d_Y)

be metric spaces. Suppose

f \colon X \to Y

is continuous and

X

is compact. Then

f

is uniformly continuous.

Proof.

Let

\epsilon > 0

be given. For each

c \in X\text{,}

pick

\delta_c > 0

such that

d_Y\bigl(f(x),f(c)\bigr) < \nicefrac{\epsilon}{2}

whenever

x \in B(c,\delta_c)\text{.}

The balls

B(c,\delta_c)

cover

X\text{,}

and the space

X

is compact. Apply the Lebesgue covering lemma to obtain a

\delta > 0

such that for every

x \in X\text{,}

there is a

c \in X

for which

B(x,\delta) \subset B(c,\delta_c)\text{.}

Suppose

p, q \in X

where

d_X(p,q) < \delta\text{.}

Find a

c \in X

such that

B(p,\delta) \subset B(c,\delta_c)\text{.}

Then

q \in B(c,\delta_c)\text{.}

By the triangle inequality and the definition of

\delta_c\text{,}

\begin{equation*} d_Y\bigl(f(p),f(q)\bigr) \leq d_Y\bigl(f(p),f(c)\bigr) + d_Y\bigl(f(c),f(q)\bigr) < \nicefrac{\epsilon}{2}+ \nicefrac{\epsilon}{2} = \epsilon . \qedhere \end{equation*}

L7.5.12: Continuity of a compact integral in one parameter.

Proposition 7.5.12.

If

f \colon [a,b] \times [c,d] \to \R

is continuous, then

g \colon [c,d] \to \R

defined by

\begin{equation*} g(y) \coloneqq \int_a^b f(x,y) \,dx \qquad \text{is continuous}. \end{equation*}

Proof.

Fix

y \in [c,d]

and let

\epsilon > 0

be given. As

f

is continuous on

[a,b] \times [c,d]\text{,}

which is compact,

f

is uniformly continuous. In particular, there exists a

\delta > 0

such that whenever

z \in [c,d]

and

\sabs{z-y} < \delta\text{,}

we have

\babs{f(x,z)-f(x,y)} < \frac{\epsilon}{b-a}

for all

x \in [a,b]\text{.}

So suppose

\sabs{z-y} < \delta\text{.}

Then

\begin{gathered} \babs{ g(z)- g(y) } = \abs{ \int_a^b f(x,z) \,dx - \int_a^b f(x,y) \,dx } \\ = \abs{ \int_a^b \bigl( f(x,z) - f(x,y) \bigr) \,dx } \leq (b-a) \frac{\epsilon}{b-a} = \epsilon . \qedhere \end{gathered}