Oscillation and Continuity of a Bounded Function at a Point

# Oscillation and Continuity of a Bounded Function at a Point

Recall from the Oscillation of a Bounded Function at a Point page that if $f$ is a bounded function on $[a, b]$ and $x \in [a, b]$ then the oscillation of $f$ at $x$ is defined to be:

(1)
\begin{align} \quad \omega_f (x) = \lim_{h \to 0} \Omega_f ((x - h, x + h) \cap [a, b]) \end{align}

Earlier, on the Oscillation of a Bounded Function on a Set page, for $T \subseteq [a, b]$ we defined $\Omega_f (T) = \sup \{ f(x) - f(y) : x, y \in T \}$.

We will now look at an extremely important result which can allow us to determine whether a bounded function $f$ is continuous at a point $x \in [a, b]$ based on the value of the oscillation of $f$ at $x$.

 Theorem 1: Let $f$ be a bounded function on $[a, b]$ and let $x \in [a, b]$. Then $f$ is continuous at $c$ if and only if $\omega_f (c) = 0$.
• Proof: $\Rightarrow$ Let $\epsilon > 0$ be given. Suppose that $f$ is continuous at $c$. Then for $\epsilon_1 = \frac{\epsilon}{2}$ there exists a $\delta_1 > 0$ such that if $\mid x - c \mid < \delta_1$, i.e., for $x \in B(c, \delta)$ then:
(2)
\begin{align} \quad \mid f(x) - f(c) \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• We want to show that $\displaystyle{\omega_f (c) = \lim_{h \to 0} \Omega_f ((c - h, c + h) \cap [a, b]) = 0}$. Note that $B(c, h) = (c - h, c + h)$. Let $\delta = \delta_1$. Then if $\mid h \mid < \delta$ we have that $B(c - h, c + h) \subset B(c - \delta, c + \delta)$. Recall that if $T_1, T_2 \subseteq [a, b]$ and $T_1 \subseteq T_2$ then $\Omega_f (T_1) \subseteq \Omega_f(T_2)$. So:
(3)
\begin{align} \quad \sup \{ f(x) - f(y) : x, y \in B(c, h) \cap [a, b] \} \leq \sup \{ f(x) - f(y) : x, y \in B(c, \delta) \cap [a, b] \} \end{align}
• So for all $x, y \in B(c, h) \cap [a, b] \subset B(c, \delta) \cap [a, b]$ we have that $(*)$ holds and:
(4)
\begin{align} \quad f(x) - f(c) \leq \mid f(x) - f(c) \mid < \frac{\epsilon}{2} \quad \mathrm{and} \quad f(y) - f(c) \leq \mid f(y) - f(y) \mid < \frac{\epsilon}{2} \end{align}
• And by the triangle inequality we see that for all $x, y \in B(c, h) \cap [a, b] \subset B(c, \delta) \cap [a, b]$ that:
(5)
\begin{align} \quad f(x) - f(y) \leq \mid f(x) - f(y) \mid \leq \mid f(x) - f(c) \mid + \mid f(y) - f(c) \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon \end{align}
• Therefore $\Omega_f ((c - h, c + h) \cap [a, b]) = \sup \{ f(x) - f(y) : x, y \in (c - h, c + h) \cap [a, b] \} < \epsilon$. So for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $\mid h \mid < \delta$ then $\Omega_f ((c - h, c + h) \cap [a, b]) < \epsilon$ and so:
(6)
\begin{align} \quad \lim_{h \to 0} \Omega_f ((c - h, c + h) \cap [a, b]) = \omega_f (c) = 0 \end{align}
• $\Leftarrow$ Let $\epsilon > 0$ be given. Suppose that $\omega_f(c) = \lim_{h \to 0} \Omega_f((c - h, c + h) \cap [a, b]) = 0$. Then we have that for $\epsilon_1 = \epsilon > 0$ there exists a $\delta_1 > 0$ such that if $\mid h \mid < \delta_1$ then:
(7)
\begin{align} \quad \Omega_f ((c - h, c + h) \cap [a, b]) = \sup \{ f(x) - f(y) : x, y \in (c - h, c + h) \cap [a, b] \} < \epsilon_1 = \epsilon \quad (**) \end{align}
• Let $\delta = \delta_1$. Then for all $x \in B(c, h) = (c - h, c + h) \subset B(c, \delta) = (c - \delta, c + \delta)$ ($\mid h \mid < \delta$) we have that $(**)$ holds and that:
(8)
\begin{align} \quad f(x) - f(c) \leq \mid f(x) - f(c) \mid < \epsilon_1 = \epsilon \end{align}
• So for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in B(c, \delta) = (c - \delta, c + \delta)$ (i.e., $\mid x - c \mid < \delta$) then $\mid f(x) - f(c) \mid < \epsilon$, so $f$ is continuous at $c$. $\blacksquare$