In the proof that $x^{2}$ is continuous, the $\delta$ depended on the point $a$: the steeper the graph, the smaller $\delta$ had to be. Uniform continuity asks for one $\delta$ that works everywhere at once. On a closed, bounded interval it comes for free, by the same Bolzano–Weierstrass argument that proves the extreme value theorem, and it is the key to proving that continuous functions can be integrated.
🔑 Key idea
$f$ is uniformly continuous on $S$ when for every $\varepsilon > 0$ there is a $\delta > 0$ such that $|f(x) - f(y)| < \varepsilon$ for all $x, y \in S$ with $|x - y| < \delta$. Theorem: a continuous function on a closed, bounded interval is uniformly continuous. $\frac{1}{x}$ on $(0, 1)$ and $x^{2}$ on $\mathbb{R}$ are continuous but not uniformly continuous.
💡 Why it works
Compare the quantifiers. Continuity: for every $a$ and every $\varepsilon$ there is a $\delta$ (which may depend on $a$). Uniform: for every $\varepsilon$ there is a $\delta$ that works for every $a$. For $\frac{1}{x}$ near $0$, the points $x = \delta$ and $y = \frac{\delta}{2}$ are within $\delta$, yet $\left|\frac{1}{x} - \frac{1}{y}\right| = \frac{1}{\delta}$, which is large: no single $\delta$ handles $\varepsilon = 1$. On $[a, b]$, a failure would produce pairs $x_n, y_n$ with $|x_n - y_n| \to 0$ but $|f(x_n) - f(y_n)| \ge \varepsilon$; Bolzano–Weierstrass gives subsequences converging to a common point $c$, and continuity at $c$ then forces $f(x_{n_k}) - f(y_{n_k}) \to 0$, a contradiction.
Uniform mode for $x^{2}$ on $[0, 3]$: drag the point along the interval. The purple strip shows the best δ at each point; the smallest of them is the one δ that works everywhere.
Worked examples
✏️ Worked example: A line is the easy case
For $f(x) = \frac{3}{2}x + 7$ on $\mathbb{R}$ and $\varepsilon = 0.1$, find the largest uniform $\delta$.
$|f(x) - f(y)| = \frac{3}{2}|x - y|$, the same at every point.
✏️ Worked example: The square on a closed interval
For $f(x) = x^{2}$ on $[0, 2]$ and $\varepsilon = 0.1$, find the largest uniform $\delta$.
For $y = x + t$ with $0 < t < \delta$: $y^{2} - x^{2} = t(2x + t)$, largest when $y = 2$: $t(4 - t)$.
As $t \to \delta$ this approaches $\delta(4 - \delta)$, so we need $\delta(4 - \delta) \le 0.1$, that is, $(2 - \delta)^{2} \ge 3.9$.
Answer: $\delta = 2 - \sqrt{3.9} \approx 0.0252$, set by the steep end of the interval.
✏️ Worked example: Steep but still uniform
For $\sqrt{x}$ on $[0, \infty)$ and $\varepsilon = 0.1$, find the largest uniform $\delta$.
$|\sqrt{x} - \sqrt{y}| \le \sqrt{|x - y|}$, with equality when one point is $0$.
Answer: $\delta = 0.01$: we need $\sqrt{\delta} \le 0.1$.
⚠️ Common mistake
Calling a point-dependent δ uniform. "$\delta = \frac{\varepsilon}{2|a| + 1}$ works for $x^{2}$" is a perfectly good continuity proof at each $a$, but on $\mathbb{R}$ these $\delta$'s shrink to $0$ as $|a|$ grows, so no single one works. Ask: does my δ mention the point? If its smallest value over the set is $0$, the function is not uniformly continuous there.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.