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

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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:

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

$\blacksquare$