Lipschitz functions and modulus of continuity

Lipschitz functions

A function $f$ is said to be $k$-Lipschitz on $I$ if there exists a constant $k \geq 0$ such that

$$\forall (x, y) \in I^2,\quad \lvert f(x) - f(y) \rvert \leq k \lvert x - y \rvert$$

Such a function is uniformly continuous: for a given $\varepsilon > 0$, the choice $\delta = \dfrac{\varepsilon}{k}$ works everywhere when $k > 0$, and any $\delta$ works if $k = 0$.

In practice, the mean value inequality provides the most convenient criterion: if $f$ is differentiable on $I$ with $\lvert f'(x) \rvert \leq k$ for every $x$, then $f$ is $k$-Lipschitz, hence uniformly continuous.

A hierarchy to remember

$$\text{Lipschitz} \implies \text{uniformly continuous} \implies \text{continuous}$$

Neither implication can be reversed.

The square root function shows it: on $\left[0 \,;\, +\infty\right[$, the inequality $\lvert \sqrt{x} - \sqrt{y} \rvert \leq \sqrt{\lvert x - y \rvert}$ gives uniform continuity with $\delta = \varepsilon^2$. Yet it is not Lipschitz on any neighborhood of $0$, since

$$\frac{\sqrt{x} - \sqrt{0}}{x - 0} = \frac{1}{\sqrt{x}} \xrightarrow[x \to 0^+]{} +\infty$$

The modulus of continuity

The tool that measures all of this at once is the modulus of continuity:

$$\omega_f(t) = \sup \left\{ \lvert f(x) - f(y) \rvert \;:\; (x, y) \in I^2,\; \lvert x - y \rvert \leq t \right\}$$

The function $f$ is uniformly continuous on $I$ if and only if $\omega_f(t) \to 0$ as $t \to 0^+$. It is a reformulation, but it makes error bounds much easier to read.

$f$	$I$	$\omega_f(t)$	Uniformly continuous?
$x \mapsto 3x + 1$	$\mathbb{R}$	$3t$	yes
$x \mapsto \sin x$	$\mathbb{R}$	$\leq t$	yes
$x \mapsto \sqrt{x}$	$\left[0 \,;\, +\infty\right[$	$\sqrt{t}$	yes, not Lipschitz
$x \mapsto x^2$	$\mathbb{R}$	$+\infty$	no
$x \mapsto \dfrac{1}{x}$	$\left]0 \,;\, 1\right]$	$+\infty$	no
$x \mapsto \dfrac{1}{x}$	$\left[1 \,;\, +\infty\right[$	$\leq t$	yes, $1$-Lipschitz

Classic pitfall

Believing that a continuous and bounded function is automatically uniformly continuous. Counterexample: $f(x) = \sin\!\left(\dfrac{1}{x}\right)$ on $\left]0 \,;\, 1\right]$ only takes values in $[-1 \,;\, 1]$, and yet, with

$$x_n = \frac{1}{2\pi n + \tfrac{\pi}{2}} \qquad \text{and} \qquad y_n = \frac{1}{2\pi n}$$

we get $\lvert x_n - y_n \rvert \to 0$ while $\lvert f(x_n) - f(y_n) \rvert = 1$ for every $n$. What is missing here is not the boundedness of $f$, but the closedness of the interval at $0$.