The Timpson Lyceum · Real Analysis (2020)

Lecture 11 (Real Analysis 2020)

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 $1 < x < 3$):

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

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 $\delta = \min\{1,\ \varepsilon/C\}$.