Real Analysis: Continuity

Uniform Continuity

Uniform Continuity

A function $f : (X, d_X) \rightarrow (Y, d_Y)$ is uniformly continuous on $X$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

$$d_Y(f(x), f(y)) < \varepsilon$$

for all $x, y \in X$ where $d_X(x, y) < \delta.$

While continuity is a property of a function at a single point in its domain, uniform continuity is a property of a function regarding its entire domain. Thus, continuity is a local property of a function, whereas uniform continuity is a global property. A notable consequence of this distinction is that while uniform continuity implies continuity by definition, the reverse is not true. As shown in the problems below, continuity does not imply uniform continuity, even for some of the simplest functions.

Compactness And Uniform Continuity

An important theorem connects compactness, continuity, and uniform continuity: If $f$ is a continuous function of a compact metric space, then it is uniformly continuous.

This connection between compactness and uniform continuity is critical for proving key theorems about integrals, such as the fact that continuity implies integrability as well as the fundamental theorem of calculus.

Problems

  1. Give an example of a continuous function that is not uniformly continuous.

    Proof by contradiction. Let $f(x) = x^2.$ Because $f$ is a monomial, it is continuous. Assume $f$ is uniformly continuous along $\mathbb{R}.$ Then for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $|x^2 - y^2| < \varepsilon$ whenever $|x - y| < \delta.$ Pick $x = c\delta$ and $y = (c + \frac{1}{2})\delta.$ Then

    $$|x - y| = \dfrac{1}{2}\delta < \delta.$$

    The uniform continuity of $f$ implies that

    $$|x^2 - y^2| = (c + \dfrac{1}{2})^2\delta^2 - c^2\delta^2 = (c + \dfrac{1}{4})\delta^2 < \varepsilon.$$

    By the Archimedean property, there exists a $c \in \mathbb{R}$ such that

    $$(c + \dfrac{1}{4})\delta^2 > \varepsilon.$$

    However, this contradicts the uniform continuity of $f.$ Therefore $f$ was not uniformly continuous after all.

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  2. Let $f : (X, d_X) \rightarrow (Y, d_Y)$ be continuous, and let $(X, d_X)$ be compact. Show that $f$ is uniformly continuous.

    Let $f : (X, d_X) \rightarrow (Y, d_Y)$ be continuous, and let $(X, d_X)$ be compact. Because $f$ is continuous, for each $x \in X$ there is a positive number $S(x)$ such that

    $$d_Y(f(x), f(y)) < \frac{\varepsilon}{2} \text{ whenever } d_X(x, y) < S(x), \text{ and }y \in X.$$

    Let $T(x) = \{ y : d_X(x, y) < \frac{1}{2}T(x) \}.$ Because $x \in T(x),$ it follows that $\mathbf{T} = \{ T(x) : x \in X \}$ is an open cover of $X.$ Because $X$ is compact, there is a finite open subcover of $\mathbf{T},$ call it $V = \{T(x_1), \ldots, T(x_n)\}.$ Pick $\delta = \min\{ S(x_1), \ldots, S(x_n) \}.$ It follows that $\delta > 0.$

    Let $x, y \in X$ such that $d_X(x, y) < \delta.$ Since $V$ is a finite cover of $X,$ it follows that there is some $m \in \{1, \ldots, n\}$ such that $x \in T(x_m),$ and so $d_X(x, x_m) < \frac{1}{2}S(x_m).$ By the triangle inequality, we see that

    $$d_X(y, x_1) \leq d_X(y, x) + d_X(x, x_m) < \delta + \dfrac{1}{2}S(x_m) \leq S(x_m).$$

    Likewise, we can see that

    $$d_Y(f(x), f(y)) \leq d_Y(f(x), f(x_m)) + d_Y(f(x_m), f(y)) < \varepsilon.$$

    Therefore $f$ is uniformly continuous on $X.$

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