uniform continuity কী?

Example 13.20

$x \mapsto x^2$ is continuous on $\R$ but not uniformly continuous: $\abs{(n + \frac 1n)^2 - n^2} = 2 + \frac{1}{n^2} \geq 2$ although the arguments are at distance $\frac 1n \to 0$. On any bounded interval it is Lipschitz, hence uniformly continuous — consistent with Heine’s theorem below.

Example 13.21 (Uniform moduli, explicitly)

On a segment, Heine guarantees a uniform $\delta$; often one can also compute it. For $f(x) = x^2$ on $\intcc{0}{10}$:

so $\delta = \frac{\varepsilon}{20}$ works uniformly (a Lipschitz modulus, linear in $\varepsilon$). For $\sqrt x$ on $\intcc{0}{1}$: $\abs{\sqrt x - \sqrt y} \leq \sqrt{\abs{x - y}}$ (Exercise 13.9), so $\delta = \varepsilon^2$ works — uniform but not linear: near $0$ the square root is steep, and the price appears in the exponent of $\varepsilon$, not in a failure of uniformity. The closing insight: uniform continuity is a spectrum, not a yes/no — the function $\delta(\varepsilon)$, called the modulus, measures how expensive the uniformity is, and Lipschitz is simply its best grade.

Example 13.23 (Bounded, continuous, yet not uniformly so)

The function $f(x) = \sin(x^2)$ is continuous and bounded on $\R$, but not uniformly continuous. Take

yet $f(y_n) - f(x_n) = \sin\bigl(2\pi n + \frac\pi2\bigr) - \sin(2\pi n) = 1 - 0 = 1$ for every $n$: no single $\delta$ can serve $\varepsilon = \frac12$ everywhere. Geometrically, the oscillations of $\sin(x^2)$ accelerate: the graph completes a full wave over shorter and shorter windows, so the horizontal scale a given $\varepsilon$ requires shrinks to zero as $x$ grows. The closing insight: boundedness does not buy uniformity (this example), and unboundedness does not preclude it ($\sqrt x$, Exercise 13.9); what decides is the modulus of oscillation, and Heine’s theorem says compact domains discipline it automatically.