Compactness And Subsets Of C(K)

Equicontinuous at a Point

Let

ℱ

be a family of functions of the form

S \to ℝ^{m}

where

S \subseteq ℝ^{n}

, then we say that

ℱ

is equicontinuous at a point

a \in S

if for every

ϵ \in ℝ^{+}

there exists a

δ \in ℝ^{+}

such that for all

x \in S

and

f \in ℱ

‖ x - a ‖ < δ ⟹ ‖ f (x) - f (a) ‖ < ϵ

Equicontinuous Family

We say that

ℱ

is an equicontinuous family of functions of the form

S \to ℝ

diff for every

a \in S

we have that

ℱ

is is equicontinuous at

a

Uniformly Equicontinuous Family

Let

ℱ

be a family of functions of the form

S \to ℝ^{m}

where

S \subseteq ℝ^{n}

, then we say that

ℱ

is uniformly equicontinuous if for every

ϵ \in ℝ^{+}

there exists a

δ \in ℝ^{+}

such that for all

x, y \in S

and

f \in 𝒞

‖ x - y ‖ < δ ⟹ ‖ f (x) - f (y) ‖ < ϵ

Compact Subsets of Continuous Functions with Compact Domain is Equicontinuous

Let

K

be a compact subset of

ℝ^{n}

, a compact subset

ℱ

C (K, ℝ^{m})

is equicontinuous

An Equicontinuous Family of Functions whose Domain is Compact is also Uniformly Equicontinuous

ℱ

is an equicontinuous family of functions of the form

K \to ℝ^{m}

where

K

is compact, then

ℱ

is uniformly equicontinuous

Totally Bounded

We say that a subset

S \subseteq ℝ^{m}

is totally bounded if for any

ϵ \in ℝ^{+}

there exists

a_{1}, \dots, a_{n}

such that

S \subseteq ⋃_{i = 1}^{n} B (a_{i}, ϵ)

A Bounded subset of Rm is Totally Bounded

Suppose that

S \subseteq ℝ^{m}

is bounded, then

S

is totally bounded.

Arzela-Ascoli

Let

K

be a compact subset of

R^{n}

, a subset

ℱ

C (K, R^{m})

is compact iff it is closed, bounded and equicontinuous

Functions Bounded by 1 are Not Compact

Show that

B = {f \in C [0, 1] : {‖ f ‖}_{\infty} \leq 1}

is not compact.

Recall that $B$ is compact iff it is closed bounded and equicontinuous, if we are able to show that $B$ is not equicontinuous, then it wouldn't be compact, we do so by considering the subset of functions $f_{n} (x) = x^{n}$ defined on $[0, 1]$ .

We show that these functions are not equicontinuous, we do this by showing that there exists a point at which the function is not equicontinuous, so let $a = 1$ , $ϵ = \frac{1}{2}$ and let $δ \in ℝ$ we will prove that there exists an $x \in [0, 1]$ and $f_{j}$ such that $| x - 1 | < δ$ and $| f_{j} (x) - f_{j} (1) | \geq \frac{1}{2}$ .

What we will do for $x$ is to simply use $x = 1 - \frac{δ}{2}$ which will satisfy the first inequality, then we will note that for any $α \in (0, 1)$ that the sequence $α^{n} \to 0$ , since $x \in [0, 1]$ then it's also true for our value $x$ and so there is some $j$ such that $x^{j} < \frac{1}{2}$ note that $x^{j} = f_{j} (x)$ and thus we have $| f_{j} (x) - f j (1) | = | x^{j} - 1 | = | 1 - x^{j} | \geq \frac{1}{2}$