## Overview This lecture develops two refinements of the idea of continuity. First is **uniform continuity**, a stronger property that controls how fast a function can grow over an interval, captured by a sequential definition. Second is the **epsilon-delta criterion**, the classical definition of continuity at a point that you will most often see in a college calculus course. We prove that every uniformly continuous function is continuous, show by counterexample that $x^2$ and $1/x$ fail to be uniformly continuous, and establish the major theorem that a continuous function on a closed bounded interval is automatically uniformly continuous. We then prove the epsilon-delta criterion is logically equivalent to the sequential definition of continuity. ## Key Concepts **Intuition for uniform continuity.** A uniformly continuous function is one whose growth over any interval can be bounded. For any interval length you choose, there is a single bound the function never exceeds on an interval of that length, no matter where the interval sits. - $\sin x$ is uniformly continuous: its values stay in $[-1,1]$, so it never changes by more than $2$ on any interval. - Any straight line is uniformly continuous: on an interval of length $h$ it always grows by the same fixed amount. - $f(x) = x^2$ is **not** uniformly continuous: over successive unit intervals it grows by $3, 5, 7, 9, \dots$ — the growth on a fixed interval length is unbounded as you move right, so no single bound works. **Definition (uniform continuity, sequential form).** A function $f : D \to \mathbb{R}$ is **uniformly continuous** provided that whenever $(u_n)$ and $(v_n)$ are sequences in $D$ with $$\lim_{n\to\infty}(u_n - v_n) = 0,$$ it follows that $$\lim_{n\to\infty}\big(f(u_n) - f(v_n)\big) = 0.$$ **Relation to continuity.** Uniform continuity implies continuity, but not conversely. (Recall ordinary continuity: $\lim x_n = x_0 \implies \lim f(x_n) = f(x_0)$.) **Closed bounded intervals.** The recurring reason these intervals matter is **sequential compactness**: every sequence in a closed bounded interval $[a,b]$ has a subsequence converging to a point of $[a,b]$. Whenever a proof concerns continuity of a function on $[a,b]$, sequential compactness is almost always the tool used. **Key theorem.** If $f : [a,b] \to \mathbb{R}$ is continuous on a **closed bounded interval**, then $f$ is **uniformly continuous**. The proof argues by contradiction: if $f(u_n)-f(v_n) \not\to 0$, then (passing to a subsequence) there is an $\varepsilon > 0$ with $|f(u_{n_k}) - f(v_{n_k})| \ge \varepsilon$ for all $k$. Sequential compactness extracts subsequences $u_{n_k}, v_{n_k} \to x_0 \in [a,b]$ (they share the limit because $u_n - v_n \to 0$). Continuity at $x_0$ forces $f(u_{n_k}) - f(v_{n_k}) \to f(x_0) - f(x_0) = 0$, contradicting the $\ge \varepsilon$ bound. **The epsilon-delta criterion.** A function $f : D \to \mathbb{R}$ satisfies the **epsilon-delta criterion at $x_0$** provided that $$\forall \varepsilon > 0 \ \exists \delta > 0 \ \text{ such that } \ |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon.$$ Geometrically: for **every** open interval (of half-width $\varepsilon$) around $f(x_0)$, you can find an open interval (of half-width $\delta$) around $x_0$ that maps entirely inside it. A function with a jump fails this — near the jump, any $\delta$-interval maps some points outside the small $\varepsilon$-interval. **Equivalence theorem.** $f$ is continuous at $x_0$ **if and only if** $f$ satisfies the epsilon-delta criterion at $x_0$. So the two notions are interchangeable as definitions of continuity. **Negating the criterion** (used in the equivalence proof): $f$ fails the criterion at $x_0$ exactly when $$\exists \varepsilon_0 > 0 \ \text{ such that } \ \forall \delta > 0 \ \exists x \ \text{ with } \ |x - x_0| < \delta \ \text{ and } \ |f(x) - f(x_0)| \ge \varepsilon_0.$$ Taking $\delta = \tfrac{1}{n}$ produces a sequence $x_n \to x_0$ whose images stay $\ge \varepsilon_0$ away from $f(x_0)$ — contradicting continuity. ## Worked Example **Show $f(x) = x^3$ satisfies the epsilon-delta criterion at $x_0 = 2$.** We must produce, for any $\varepsilon > 0$, a $\delta > 0$ with $|x - 2| < \delta \implies |x^3 - 8| < \varepsilon$. Factor to expose the controllable term $|x-2|$, then bound the cofactor by restricting $\delta \le 1$ (which forces $1 < x < 3$): $$ \begin{aligned} |x^3 - 8| &= |x - 2|\,\big|x^2 + 2x + 4\big| \\ &< |x - 2|\,\big(3^2 + 2\cdot 3 + 4\big) \qquad (\text{since } 1 < x < 3) \\ &= 19\,|x - 2| \\ &< 19\,\delta, \qquad \text{so choose } \delta = \min\!\left\{1,\ \tfrac{\varepsilon}{19}\right\} \\ &\le 19 \cdot \tfrac{\varepsilon}{19} = \varepsilon. \end{aligned} $$ Taking $\delta = \min\{1,\ \varepsilon/19\}$ makes $|x-2| < \delta \implies |x^3 - 8| < \varepsilon$, as required. **Counterexample: $f(x) = x^2$ on $\mathbb{R}$ is not uniformly continuous.** Take $u_n = n + \tfrac1n$ and $v_n = n$. Then $u_n - v_n = \tfrac1n \to 0$, yet $$ \begin{aligned} f(u_n) - f(v_n) &= \left(n + \tfrac1n\right)^2 - n^2 \\ &= n^2 + 2 + \tfrac{1}{n^2} - n^2 \\ &= 2 + \tfrac{1}{n^2} \ \longrightarrow\ 2 \ \neq\ 0. \end{aligned} $$ The image differences do not vanish, so $x^2$ is not uniformly continuous. (Similarly, $1/x$ on $(0,2]$ fails using $u_n = \tfrac{1}{n+1},\ v_n = \tfrac1n$, whose reciprocals differ by a constant.) ## Summary - **Uniform continuity** strengthens continuity by bounding a function's growth on any interval; formally, $u_n - v_n \to 0 \implies f(u_n) - f(v_n) \to 0$ for all sequences in the domain. - Uniform continuity $\implies$ continuity, but the converse fails in general. - $x^2$ and $1/x$ (on a non-closed domain) are continuous but **not** uniformly continuous — choose sequences whose inputs close in while their outputs stay a fixed distance apart. - On a **closed bounded interval**, continuity upgrades to uniform continuity, proved via sequential compactness. - The **epsilon-delta criterion** ($\forall\varepsilon\,\exists\delta:\ |x-x_0|<\delta \implies |f(x)-f(x_0)|<\varepsilon$) is logically **equivalent** to the sequential definition of continuity. - Proving the criterion at a point: factor out $|x - x_0|$, restrict $\delta \le 1$ to bound the remaining factor by a constant $C$, then set $\delta = \min\{1,\ \varepsilon/C\}$.

Lecture 11 (Real Analysis 2020)

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✦ Hall of Aces

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