# Oscillation at Point (Infimum) equals Oscillation at Point (Limit)/Lemma

## Lemma

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 open subset neighborhoods of $x$.

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

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

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

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

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

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

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

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

Then $\map {\omega^L_f} x = \map {\omega_f} x$.

## Proof

We know that $\map {\omega^L_f} x$ and $\map {\omega_f} x$ are real numbers.

We need to prove that $\map {\omega^L_f} x = \map {\omega_f} x$.

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

First, we aim to prove that $\size {\map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x} < \epsilon$ for a small enough $h \in R_{>0}$.

$\map {\omega^L_f} x = \displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$ means by the definition of limit from the right that a strictly positive real number $h_1$ exists such that:

$\size {\map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega^L_f} x)} < \epsilon$

for every $h$ that satisfies: $0 < h < h_1$.

This means in particular that $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$ for every $h$ that satisfies: $0 < h < h_1$.

Let $h'$ be a real number that satisfies: $0 < h' < h_1$.

We observe that $\openint {x - h'} {x + h'} \in N_x$.

Therefore, $\map {\omega_f} {\openint {x - h'} {x + h'} } \in \set {\map {\omega_f} I: I \in N_x}$.

By definition, $\map {\omega_f} x$ is a lower bound for $\set {\map {\omega_f} I: I \in N_x}$.

Accordingly:

$\map {\omega_f} {\openint {x - h'} {x + h'} } \ge \map {\omega_f} x$

The fact that $\map {\omega_f} x \in \R$ implies that:

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

for an $I \in N_x$.

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 number $h_2 \in \R_{>0}$ exists such that $\openint {x - h_2} {x + h_2}$ is a subset of $I$.

Let $h''$ be a real number that satisfies: $0 < h'' < h_2$.

We observe that $\openint {x - h''} {x + h''}$ is a subset of $I$.

We have:

$\map {\omega_f} I \in \R$
$\openint {x - h''} {x + h''}$ is a subset of $I$

Therefore:

$\map {\omega_f} {\openint {x - h''} {x + h''} } \le \map {\omega_f} I$ by Oscillation on Subset

Putting all this together, we get for every $h$ that satisfies: $0 < h < \min \set {h_1, h_2}$:

 $\displaystyle \map {\omega_f} {\openint {x - h} {x + h} }$ $\le$ $\displaystyle \map {\omega_f} I$ $\displaystyle \leadsto \ \$ $\displaystyle \map {\omega_f} x \le \map {\omega_f} {\openint {x - h} {x + h} }$ $\le$ $\displaystyle \map {\omega_f} I$ as $\map {\omega_f} x \le \map {\omega_f} {\openint {x - h} {x + h} }$ is true $\displaystyle \leadsto \ \$ $\displaystyle \map {\omega_f} x \le \map {\omega_f} {\openint {x - h} {x + h} }$ $\le$ $\displaystyle \map {\omega_f} I < \map {\omega_f} x + \epsilon$ as $\map {\omega_f} I < \map {\omega_f} x + \epsilon$ is true $\displaystyle \leadsto \ \$ $\displaystyle \map {\omega_f} x \le \map {\omega_f} {\openint {x - h} {x + h} }$ $<$ $\displaystyle \map {\omega_f} x + \epsilon$ $\displaystyle \leadsto \ \$ $\displaystyle 0 \le \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x$ $<$ $\displaystyle \epsilon$ $\displaystyle \leadsto \ \$ $\displaystyle \size {\map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x}$ $<$ $\displaystyle \epsilon$

Thus, we achieved our first aim.

Next, we get for every $h$ that satisfies: $0 < h < \min \set {h_1, h_2}$:

 $\displaystyle \size {\map {\omega^L_f} x - \map {\omega_f} x}$ $=$ $\displaystyle \size {\map {\omega^L_f} x - \map {\omega_f} {\openint {x - h} {x + h} } + \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x}$ $\displaystyle$ $\le$ $\displaystyle \size {\map {\omega^L_f} x - \map {\omega_f} {\openint {x - h} {x + h} } } + \size {\map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x}$ Triangle Inequality for Real Numbers $\displaystyle$ $<$ $\displaystyle \epsilon + \epsilon$ $\displaystyle$ $=$ $\displaystyle 2 \epsilon$

This holds for every $\epsilon \in \R_{>0}$.

Therefore, $\size {\map {\omega^L_f} x - \map {\omega_f} x} = 0$ as $\size {\map {\omega^L_f} x - \map {\omega_f} x}$ is independent of $\epsilon$.

Accordingly:

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

$\blacksquare$