# Oscillation of a Bounded Function at a Point

Recall from the Oscillation of a Bounded Function on a Set page that if $f$ is a bounded function on $[a, b]$ and $T \subseteq [a, b]$ then the oscillation of $f$ on $T$ is denoted $\Omega_f (T)$ and is defined as:

(1)We will now define what the oscillation of $f$ at a point $x \in [a, b]$ is.

Definition: Let $f$ be a bounded function on $[a, b]$ and let $x \in [a, b]$. The Oscillation of $f$ at $x$ is denoted $\omega_f (x)$ and is defined as $\displaystyle{\omega_f (x) = \lim_{h \to 0} \Omega_f ((x - h, x + h) \cap [a, b])}$. |

We should note that for a bounded function $f$ on $[a, b]$ and $x \in [a, b]$ that $\omega_f(x)$ always exists. To show this, consider the function:

(2)Consider two sets $T_1, T_2 \subseteq [a, b]$ and suppose that $T_1 \subseteq T_2$. Then:

(3)Now take $\lim_{h \to 0} g(h)$. For $h$ sufficiently small, we will have that $(x - h, x + h) \subset [a, b]$, and so from above, $g$ is a decreasing function. However, $\Omega_f (T) \geq 0$ for all $T \subseteq [a, b]$ and so $\omega_f (x) = \lim_{h \to 0} g(h)$ always exists.

To visualize the oscillation of $f$ at $x$, consider a closed interval $[a, b]$ with $x \in [a, b]$. Then for every $h > 0$, we have that $(x - h, x + h)$ is an open interval centered at $h$. We only care about points contained in $[a, b]$, so then $(x - h, x + h) \cap [a, b]$ is the set of points that are in the open interval $(x - h, x + h)$ and are also in $[a, b]$. As $h \to 0$, the open interval $(x - h, x + h)$ shrinks, and furthermore, $\Omega_f ((x - h, x + h) \cap [a, b])$ has less points to formulate a supremum from. That limit is equal to the oscillation of $f$ at $x$.

For a rather simple example, consider the function $f(x) = x$ on the interval $[0, 1]$. For $x = 0.5$, consider the oscillation of $f$ at $x$:

(4)Notice that for $h < \frac{1}{2}$ we have that $\left ( \frac{1}{2} - h, \frac{1}{2} + h \right ) \subset [0, 1]$ and so as $h \to 0$, $\left ( \frac{1}{2} - h, \frac{1}{2} + h \right ) \cap [0, 1] \to \left \{ \frac{1}{2} \right \}$, so:

(5)