Some inclusions on the real line:
- Differentiable with a bounded derivative \(\subset\) Lipschitz continuous \(\subset\) absolutely continuous \(\subset\) uniformly continuous \(\subset\) continuous
Proofs:
- Mean Value Theorem,
- Triangle inequality,
- Definition of absolute continuity specialized to one interval,
- Definition of uniform continuity
There is a function discontinuous precisely on \({\mathbf{Q}}\).
\(f(x) = \frac 1 n\) if \(x = r_n \in {\mathbf{Q}}\) is an enumeration of the rationals, and zero otherwise. The limit at every point is 0.
There do not exist functions that are discontinuous precisely on \({\mathbf{R}}\setminus {\mathbf{Q}}\).
\(D_f\) is always an \(F_\sigma\) set, which follows by considering the oscillation \(\omega_f\). Use that \(\omega_f(x) = 0 \iff f\) is continuous at \(x\), and \(D_f = \cup_n A_{\frac 1 n}\) where \(A_\varepsilon = \left\{{\omega_f \geq \varepsilon}\right\}\) is closed.
An alternative characterization of uniform continuity: \begin{align*} \left\|\tau_{y} f-f\right\|_{u} \rightarrow 0 \text { as } y \rightarrow 0 \end{align*}
If \(f\) is Lipschitz on \(X\), then \(f\) is uniformly continuous on \(X\).
Supposing that \begin{align*} {\left\lVert {f(x) - f(y)} \right\rVert} \leq C {\left\lVert {x-y} \right\rVert} ,\end{align*} for a fixed \({\varepsilon}\) take \(\delta({\varepsilon}) \coloneqq{\varepsilon}/C\), then \begin{align*} {\left\lVert {f(x) - f(y)} \right\rVert} &\leq C {\left\lVert {x-y} \right\rVert} \\ &\leq C \delta \\ &= C \qty{{\varepsilon}/C} \\ &= {\varepsilon} .\end{align*}
Every continuous function \(f:X\to Y\) where \(X\) is a compact metric space is uniformly continuous. As a result, if \(f:U\to {\mathbf{R}}\) is continuous, then \(f\) is uniformly continuous on any \(K \subseteq U\) compact.
Fix \({\varepsilon}>0\), we’ll find a \(\delta\) that works for all \(x\in X\) uniformly. For every \(x\in X\), pick a \(\delta_x\) neighborhood satisfying the conditions for (assumed) continuity. Take an open cover by \(\delta_x/2\) balls, extract a finite subcover, take \(\delta\) the minimal radius.
If \(\mathcal F \subset C(X)\) is a family of continuous functions on \(X\), then \(\mathcal F\) equicontinuous at \(x\) iff
\begin{align*} \forall \varepsilon > 0 ~~\exists U \ni x \text{ such that } y\in U \implies {\left\lvert {f(y) - f(x)} \right\rvert} < \varepsilon \quad \forall f\in \mathcal{F} .\end{align*}