Oscillation at Point (Infimum) equals Oscillation at Point (Limit)

Theorem
Let $f: D \to \R$ be a real function where $D \subseteq \R$.

Let $x$ be a point in $D$.

Let $N_x$ be the set of neighborhoods of $x$.

Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$ as defined by:


 * $\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$

where $\map {\omega_f} I$ is the oscillation of $f$ on a real set $I$:


 * $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

Let $\map {\omega^L_f} x$ be the oscillation of $f$ at $x$ as defined by:
 * $\map {\omega^L_f} x = \ds \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$

Then:


 * $\map {\omega_f} x \in \R$ $\map {\omega^L_f} x \in \R$

and, if $\map {\omega_f} x$ and $\map {\omega^L_f} x$ exist as real numbers:


 * $\map {\omega_f} x = \map {\omega^L_f} x$

Necessary Condition
Let $\map {\omega_f} x \in \R$.

We need to prove:


 * $\map {\omega^L_f} x \in \R$
 * $\map {\omega^L_f} x = \map {\omega_f} x$

where:
 * $\map {\omega^L_f} x = \ds \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$


 * $\map {\omega_f} {\openint {x - h} {x + h} } = \sup \set {\size {\map f y - \map f z}: y, z \in \openint {x - h} {x + h} \cap D}$


 * $\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$


 * $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

Let $\epsilon \in \R_{>0}$.

Then an $I \in N_x$ exists such that:


 * $\map {\omega_f} I - \map {\omega_f} x < \epsilon$ by Infimum of Set of Oscillations on Set is Arbitrarily Close

Let $I$ be such an element of $N_x$.

We observe in particular that $\map {\omega_f} I \in \R$.

A neighborhood in $N_x$ contains an open subset that contains the point $x$.

So, $I$ contains such an open subset as $I \in N_x$.

Therefore, a $\delta \in \R_{>0}$ exists such that $\openint {x - \delta} {x + \delta}$ is a subset of $I$.

Let $h$ be a real number that satisfies: $0 < h < \delta$.

We observe that $\openint {x - h} {x + h} \subset I$.

Also, $\openint {x - h} {x + h} \in N_x$.

We have:


 * $I \in N_x$


 * $\openint {x - h} {x + h} \in N_x$


 * $\openint {x - h} {x + h} \subset I$


 * $\map {\omega_f} I \in \R$

from which follows by Oscillation on Subset:


 * $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$


 * $\map {\omega_f} {\openint {x - h} {x + h} } \le \map {\omega_f} I$

We have that:
 * $\map {\omega_f} {\openint {x - h} {x + h} } \in \set {\map {\omega_f} {I'}: I' \in N_x}$

as $\openint {x - h} {x + h} \in N_x$.

Also, $\map {\omega_f} x$ is a lower bound for $\set {\map {\omega_f} {I'}: I' \in N_x}$ by the definition of $\map {\omega_f} x$.

Therefore:
 * $\map {\omega_f} {\openint {x - h} {x + h} } \ge \map {\omega_f} x$

We find:

which means that $\ds \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$ exists and equals $\map {\omega_f} x$ by the definition of limit.

In other words, $\map {\omega^L_f} x \in \R$ and $\map {\omega^L_f} x = \map {\omega_f} x$.

Sufficient Condition
Let $\map {\omega^L_f} x \in \R$.

We need to prove:


 * $\map {\omega_f} x \in \R$


 * $\map {\omega_f} x = \map {\omega^L_f} x$

where:


 * $\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$


 * $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$


 * $\map {\omega^L_f} x = \ds \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$

We have:
 * $\ds \lim_{h \mathop \to 0^+} \map{\omega_f} {\openint {x - h} {x + h} } \in \R$ as $\map {\omega^L_f} x \in \R$

Therefore, $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$ for a small enough $h$ in $\R_{>0}$ by the definition of limit.

Let $h$ be such a real number.

We observe that $\openint {x - h} {x + h}$ is a neighborhood in $N_x$.

We have:


 * $\openint {x - h} {x + h} \in N_x$


 * $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$

Accordingly:


 * $\map {\omega_f} x \in \R$ by Infimum of Set of Oscillations on Set

$\map {\omega_f} x = \map {\omega^L_f} x$ follows by Lemma.

This finishes the proof of the theorem.