Uniform Continuity¶

Definition¶

A function $f\colon X \to \mathbb{R}$ with $X\subseteq\mathbb{R}$ is said to be uniformly continuous if, for every $\varepsilon >0\text{,}$ there exists a $\delta >0$ such that for all pairs of points $x_1, x_2\in X$ with $|x_1-x_2|<\delta\text{,}$ it holds that $|f(x_1)-f(x_2)|<\varepsilon\text{.}$

The same $\delta$ controls all input pairs of the Function $f$, rather than being selected separately for each point.

Interpretation¶

Uniform continuity means that one rule for "how close is close enough" works everywhere: for each $\varepsilon>0$ there is a single $\delta$ that serves every pair of points of $X$, wherever they lie. Ordinary continuity only asks for a $\delta$ at each point separately, and that $\delta$ may shrink from point to point.

Both rectangles in the figure have the same height, so they correspond to the same $\varepsilon$. Near $x=1$ the graph of $1/x$ is flat enough for a wide rectangle; near $0$ the graph is so steep that only a very narrow one fits. At a point $x_0>0$ the largest $\delta$ that works for $1/x$ is $\delta=\varepsilon x_0^2/(1+\varepsilon x_0)$$\epsilon x_0)$, which tends to $0$ as $x_0\to0$$0$, so no positive $\delta$ serves all points of $(0,1)$.

In the definition both points $x_1$ and $x_2$ are free, and $\delta$ may depend on $\varepsilon$ only. Uniform continuity is therefore a property of $f$ on the whole set $X$, not at a single point: the same function $1/x$ is uniformly continuous on $(1,2)$ and not on $(0,1)$, as in the Examples. A simple way to obtain it is a bound $|f(x_1)-f(x_2)|\le C|x_1-x_2|$ for all $x_1,x_2\in X$$X$, as for $\sin x$; then $\delta=\varepsilon/C$ works.

On a closed bounded interval nothing extra is needed: by the Heine-Cantor theorem, every continuous function on $[a,b]$ is uniformly continuous.

A common slip is to say that $f$ is "uniformly continuous at $x_0$". The phrase has no meaning: uniformity is about all points of $X$ at once.

The definition returns later for functions between more general spaces, with a general distance in place of $|x_1-x_2|$$x_2|$.

Key Results¶

• Every uniformly continuous function is continuous, the converse doesn't necessarily hold.
• $f$ is uniformly continuous on $X$ if and only if given two equivalent sequences $(x_n)_{n=1}^\infty$ and $(y_n)_{n=1}^\infty$ in $X$ (that is, $|x_n-y_n|\to0$ as $n\to\infty$$\infty$), the sequences $(f(x_n))_{n=1}^\infty$ and $(f(y_n))_{n=1}^\infty$ are also equivalent.
• A continuous function $f$ is uniformly continuous on a bounded interval $(a, b)$ if and only if the right limit $\displaystyle f(a^+)$ and the left limit $\displaystyle f(b^-)$ both exist and are finite. In particular, a continuous function that is unbounded on a bounded interval $(a,b)$ is not uniformly continuous there; a bounded function can fail too, for example $\sin(1/x)$ on $(0,1)$, which has no right limit at $0$.
• A differentiable function $f$ is uniformly continuous on $(a,b)$ if $f'$ is bounded. Boundedness of $f'$ is sufficient, not necessary: $\sqrt{x}$ is uniformly continuous on $(0,1)$ although its derivative is unbounded there. A typical failure on an unbounded domain is $f(x)=x^2$ on $\mathbb{R}\text{,}$ whose slope is unbounded.

Examples¶

• The function $f(x)=1/x$ is uniformly continuous on $(1,2)$, but not uniformly continuous on $(0,1)$.
• The function $g(x)=\sin x$ is uniformly continuous on all of $\mathbb{R}$ (hence on any interval), since $|\sin x_1-\sin x_2|\le|x_1-x_2|\text{.}$

Prerequisites¶

• Function

Related Concepts¶

• Continuity

Sources¶

• Math III: Analysis · Lecture 1

Symbols: delta, element of, epsilon, infinity, real numbers, square root, subset

Last updated: 27 October 2026 11:52