Open Balls of Supremum Metric on Continuous Real Functions on Closed Interval

Theorem
Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.

Let $d: \mathscr C^2 \to \R$ be the supremum metric on $\mathscr C \closedint a b$ defined as:
 * $\ds \forall f, g \in \mathscr C \closedint a b: \map d {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$

where $\sup$ denotes the supremum.

Let $f, g \in \mathscr C \closedint a b$ be such that:
 * $\forall x \in \closedint a b: \map f x < \map g x$

Consider the set $S$, defined as:
 * $S = \set {h \in \mathscr C \closedint a b: \forall x \in \closedint a b: \map f x < \map h x < \map g x}$

Then:
 * $S$ is an open ball of $\struct {\mathscr C \closedint a b, d}$


 * $\exists c \in \R_{>0}: \map g x := \map f x + c$
 * $\exists c \in \R_{>0}: \map g x := \map f x + c$

That is, $f$ and $g$ differ by a constant.

Proof
Recall the definition of open ball:

The open $\epsilon$-ball of $a$ in $M = \struct {A, d}$ is defined as:


 * $\map {B_\epsilon} a := \set {x \in A: \map d {x, a} < \epsilon}$

In this context, the open $\epsilon$-ball of $\phi$ in $\mathscr C \closedint a b$ is defined as:


 * $\ds \map {B_\epsilon} \phi := \set {\rho \in \mathscr C \closedint a b: \sup_{x \mathop \in \closedint a b} \size {\map \rho x - \map \phi x} < \epsilon}$

Necessary Condition
Suppose $g \in \mathscr C \closedint a b$ is defined as:
 * $\forall x \in \closedint a b: \map g x := \map f x + c$

Let $\epsilon = \dfrac c 2$.

Let $\phi \in \mathscr C \closedint a b$ be defined as:
 * $\forall x \in \closedint a b: \map \phi x = \map f x + \epsilon$

Then we have that:

and it is seen by definition of open $\epsilon$-ball of $\phi$:
 * $S = \map {B_\epsilon} \phi$

Sufficient Condition
Let $S$, defined as:
 * $S = \set {h \in \mathscr C \closedint a b: \forall x \in \closedint a b: \map f x < \map h x < \map g x}$

be an open ball in $\mathscr C \closedint a b$.

Thus there exists $\epsilon \in \R_{>0}$ and $\phi \in \mathscr C \closedint a b$ such that:
 * $S = \map {B_\epsilon} \phi$

That is:
 * $\ds \exists \epsilon \in \R_{>0}: \forall \rho \in S: \sup_{x \mathop \in \closedint a b} \size {\map \rho x - \map \phi x} < \epsilon$

it is not the case $f$ and $g$ are such that:
 * $\forall x \in \closedint a b: \map g x - \map f x = c$

for some constant $c \in \R$.

Take $h \in S$ and $r \in \R_{>0}$ such that $\map {B_r} h \subseteq S$.

It is to be shown that $\map {B_r} h \subsetneq S$.

Take $N \in \N$ such that $\dfrac 1 N < r$.

For each $n \ge N$, let $g_n = h + r - \dfrac 1 n $.

Then:
 * $g_n \in \map {B_r} h$

and therefore:
 * $g_n \in S$

which implies that:
 * $\forall x \in \closedint a b: \map {g_n} x < \map g x$

But then:
 * $\ds \forall x \in \closedint a b: \map h x + r = \lim_{n \mathop \to \infty} \map h x + r - \dfrac 1 n = \lim_{n \mathop \to \infty} \map {g_n} x \le \map g x$

By a similar argument:
 * $\forall x \in \closedint a b: \map h x - r \ge \map f x$

and therefore:
 * $\forall x \in \closedint a b: \map g x - \map f x \ge 2 r$

But $g - f$ cannot be the constant function $2 r$.

Hence there is some $x_0 \in \closedint a b$ such that $\map g {x_0} - \map f {x_0} > 2 r$.

So it must be the case that:
 * $\map g {x_0} > \map h {x_0} + r$

or:
 * $\map f {x_0} < \map h {x_0} - r$

or both.

If:
 * $(1): \quad \map g {x_0} > \map h {x_0} + r$

then:
 * $g - h$ is continuous
 * $g - h$ is always greater than or equal to $r$

and $(1)$ holds.

Take $k < \map g {x_0} - \map h {x_0} - r$

and such that $g - k >f$.

Then $g - k \in S$.

But $g - k \notin \map {B_r} h$, because $\map g {x_0} - k - \map h {x_0} > r$.

It follows that $S$ cannot be an open $\epsilon$-ball of $\phi$ in $\mathscr C \closedint a b$.