Uniform Continuity
Summarize
Introduction
Ordinary continuity describes the local behaviour of a function, where small changes in the input near each point produce small changes in the output. This property can depend on the specific point, and in many contexts a stronger form of continuity is required to control oscillations consistently across the entire domain. Uniform continuity addresses this need by requiring a single tolerance for input variations to apply everywhere in the set.
This stability property leads to several consequences:
- A uniformly continuous function on a dense subset extends continuously to the closure of that subset.
- Uniformly continuous functions map Cauchy sequences to Cauchy sequences.
- Uncontrolled oscillations are prevented whenever the domain is compact.
Oscillation is the variation in the output values of a function over a given region, that is, how much the function rises and falls within a small portion of the domain. For example, the function $\sin(1/x)$ near the origin exhibits infinitely many oscillations within any small interval, which makes its behaviour difficult to control.
Definition
Consider a function $f : A \subset \mathbb{R} \to \mathbb{R}$. The function is uniformly continuous on $A$ if, for every $\varepsilon > 0$, there exists a $\delta > 0$ such that for all $x, y \in A$ the following implication holds:
$$ |x - y| < \delta \to |f(x) - f(y)| < \varepsilon $$
The essential feature of this definition is that $\delta$ depends solely on $\varepsilon$, and not on the particular points $x$ and $y$. The same $\delta$ must be valid for every pair of points in the domain.
The graph shows the global nature of the condition. The same $\delta$ controls the function's variation throughout the domain, so whenever $|x - y| < \delta$, it follows that $|f(x) - f(y)| < \varepsilon$ for all $x, y \in A$.
The distinction between continuity and uniform continuity is clarified by comparing the standard definition of continuity at a point $x_0$. A function is continuous at $x_0$ if, for every $\varepsilon > 0$, there exists a $\delta > 0$, which may depend on $x_0$, such that:
$$ |x - x_0| < \delta \to |f(x) - f(x_0)| < \varepsilon $$
The distinction lies in the dependence of $\delta$:
- For ordinary continuity, $\delta$ may vary with each point in the domain.
- For uniform continuity, a single value of $\delta$ is valid for all points in the set.
Heine–Cantor theorem
The Heine–Cantor theorem establishes the conditions under which continuity implies uniform continuity. On compact sets, the local property of continuity suffices to guarantee a single global control parameter $\delta$ applicable to the entire domain.
Let $K \subset \mathbb{R}$ be a compact set. If a function $f : K \to \mathbb{R}$ is continuous on $K$, then it is uniformly continuous on $K$. Since compactness in $\mathbb{R}$ is equivalent to being closed and bounded, the theorem admits the following equivalent formulation: if $f : [a,b] \to \mathbb{R}$ is continuous on a closed and bounded interval $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.
The theorem identifies compactness as the structural property of the domain that prevents local continuity from degenerating into purely pointwise behaviour and guarantees instead a global form of regularity.
Sequential characterization
Uniform continuity can be equivalently formulated in terms of sequences. A function $f : A \to \mathbb{R}$ is uniformly continuous on $A$ if and only if, for every pair of sequences $(x_n)$ and $(y_n)$ in $A$ that satisfy $|x_n - y_n| \to 0$, we also have:
$$ |f(x_n) - f(y_n)| \to 0 $$
This formulation shows that uniformly continuous functions preserve infinitesimal closeness between sequences, which is also why they map Cauchy sequences to Cauchy sequences.
This criterion is particularly useful for establishing that a function is not uniformly continuous, as shown in Example 1.
Example 1
Consider the function $f(x) = x^2$ on $\mathbb{R}$. This function is continuous everywhere, since it is a polynomial, but it is not uniformly continuous on $\mathbb{R}$. The reason is that the function grows ever faster as $|x|$ increases: for points far from the origin, even a small variation in $x$ produces a large variation in $f(x)$, and no single $\delta$ can control this behaviour across the whole real line.
The failure of uniform continuity on $\mathbb{R}$ can be made precise through the sequential characterization. Consider the sequences $x_n = n + \frac{1}{n}$ and $y_n = n$. Their difference tends to zero:
$$|x_n - y_n| = \frac{1}{n} \to 0$$
The difference between the images of these sequences, however, does not:
$$|f(x_n) - f(y_n)| = \left(n + \frac{1}{n}\right)^2 - n^2 = 2 + \frac{1}{n^2} \to 2$$
Since $|f(x_n) - f(y_n)|$ does not tend to zero, the function fails the sequential criterion, so it is not uniformly continuous on $\mathbb{R}$.
By contrast, when the function is restricted to a closed and bounded interval $[a, b]$, it becomes uniformly continuous. The interval is compact, so the Heine–Cantor theorem applies and the growth of the function is bounded over the finite domain.
Example 2
Consider the function $f(x) = 1/x$ on the open interval $(0, 1)$. The interval is bounded but not closed, so it is not compact and the Heine–Cantor theorem does not apply. The function is continuous at every point of $(0, 1)$, yet it is not uniformly continuous there. As $x$ approaches $0$, the function grows without bound, and its variation over a fixed small displacement becomes arbitrarily large near the left endpoint.
The sequential characterization makes the failure precise. Consider the sequences $x_n = \frac{1}{n}$ and $y_n = \frac{1}{n+1}$, both lying in $(0, 1)$. Their difference tends to zero:
$$|x_n - y_n| = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} \to 0$$
The difference between the images, however, equals one for every $n$:
$$|f(x_n) - f(y_n)| = |n - (n+1)| = 1$$
Since $|f(x_n) - f(y_n)|$ does not tend to zero, the function is not uniformly continuous on $(0, 1)$. The contrast with Example 1 isolates two distinct mechanisms: there the domain was unbounded, while here the domain is bounded and uniform continuity fails because the function itself grows without bound near the boundary.
These examples clarify the relationship between continuity and uniform continuity. In general:
- Continuity does not imply uniform continuity. A function may be continuous at every point of its domain without admitting a single global $\delta$ that works for all pairs of points.
- Uniform continuity is the stronger condition. If a function is uniformly continuous on $A$, then it is necessarily continuous at every point of $A$.
Uniform continuity is a property of the function together with its domain, not of the formula alone. The map $x \mapsto x^2$ is uniformly continuous on every bounded interval $[a, b]$ and fails to be so on $\mathbb{R}$, even though the rule defining it is the same.
Lipschitz continuity
A particular form of uniform continuity is Lipschitz continuity. A function $f : A \subset \mathbb{R} \to \mathbb{R}$ is Lipschitz continuous on $A$ if there exists a constant $L > 0$, called a Lipschitz constant, such that for all $x, y \in A$:
$$ |f(x) - f(y)| \le L |x - y| $$
This condition maps nearby points to nearby values and establishes a global linear bound on the function's oscillation, since the variation of the output is proportional to the variation of the input. The same constant $L$ must work everywhere on the domain. The implication for uniform continuity follows directly: given any $\varepsilon > 0$, selecting $\delta = \varepsilon / L$ guarantees that:
$$ |x - y| < \delta \to |f(x) - f(y)| \le L |x - y| < L \delta = \varepsilon $$
Therefore, every Lipschitz continuous function is uniformly continuous.
The converse does not hold in general. A function may be uniformly continuous without satisfying any global linear bound of this form, so Lipschitz continuity is a strictly stronger condition, as Example 3 shows.
A useful criterion links Lipschitz continuity to differentiability. If a function $f$ is differentiable on an interval and its derivative is bounded, that is, there exists $M > 0$ such that for all $x$ in the interval:
$$ |f'(x)| \le M $$
then, by the mean value theorem, $f$ is Lipschitz continuous with constant $L = M$. For any two points $x, y$ in the interval, there exists $c$ between them such that:
$$ f(x) - f(y) = f'(c)(x - y) $$
Taking absolute values yields $|f(x) - f(y)| \leq M|x - y|$, which is the Lipschitz condition with constant $L = M$.
Geometrically, Lipschitz continuity restricts the steepness of the graph. The slopes of all secant lines are uniformly bounded in magnitude by $L$, so the function cannot oscillate more rapidly than a fixed linear rate.
Example 3
The function $f(x) = \sqrt{x}$ on $[0, 1]$ is uniformly continuous but not Lipschitz continuous. The interval is compact and the function is continuous, so the Heine–Cantor theorem guarantees uniform continuity on $[0, 1]$. The function is not Lipschitz continuous because its rate of change is unbounded near the origin.
Suppose a Lipschitz constant $L$ existed, so that $|\sqrt{x} - \sqrt{y}| \le L|x - y|$ for all $x, y \in [0, 1]$. Choosing $y = 0$ reduces this to $\sqrt{x} \le Lx$ for every $x \in (0, 1]$. Since $x > 0$, we divide both sides by $x$, which gives:
$$\frac{1}{\sqrt{x}} \le L$$
The left side grows without bound as $x \to 0^+$, so no finite $L$ can satisfy the inequality for every $x$. The function is therefore uniformly continuous without being Lipschitz continuous, which confirms that Lipschitz continuity is strictly stronger than uniform continuity.
Example 4
The Heine–Cantor theorem guarantees uniform continuity whenever the domain is compact. Compactness is sufficient but not necessary, however, and uniform continuity can also hold on domains that are not compact. This example illustrates that situation.
Consider the function $f(x) = \sin(x)$ defined on the entire real line $\mathbb{R}$. The domain is unbounded and not compact, so uniform continuity is not guaranteed a priori and must be verified directly.
Given $\varepsilon > 0$ and arbitrary $x, y \in \mathbb{R}$, the goal is to find a single $\delta > 0$, depending only on $\varepsilon$, such that $|x - y| < \delta$ implies $|\sin(x) - \sin(y)| < \varepsilon$. Applying the sum-to-product identity gives:
$$|\sin(x) - \sin(y)| = 2\left|\cos\!\left(\frac{x+y}{2}\right)\sin\!\left(\frac{x-y}{2}\right)\right|$$
Since $|\cos(\cdot)| \leq 1$ everywhere, this simplifies to:
$$|\sin(x) - \sin(y)| \leq 2\left|\sin\!\left(\frac{x-y}{2}\right)\right|$$
Applying the standard inequality $|\sin(t)| \leq |t|$, valid for all $t \in \mathbb{R}$, gives:
$$|\sin(x) - \sin(y)| \leq 2 \cdot \frac{|x - y|}{2} = |x - y|$$
This is the Lipschitz condition with constant $L = 1$, holding for every pair of points in $\mathbb{R}$ without restriction. Selecting $\delta = \varepsilon$, whenever $|x - y| < \delta$ we obtain:
$$|\sin(x) - \sin(y)| \leq |x - y| < \delta = \varepsilon$$
Here $\delta$ depends solely on $\varepsilon$, so $f$ is uniformly continuous on $\mathbb{R}$.
The underlying reason is geometric: the sine function oscillates indefinitely, but its oscillations remain bounded in amplitude and its slope never exceeds one in absolute value, since $|f'(x)| = |\cos(x)| \leq 1$ everywhere. This global bound on the rate of change prevents the uncontrolled growth observed in Example 1 and ensures that a uniform $\delta$ is attainable across the entire real line.