Uniform Convergence¶

Definition¶

Let $(f_n)_{n=1}^\infty$ be a sequence of functions defined on $X \subseteq \mathbb{R}$ and $f\colon X \rightarrow \mathbb{R}$ be another function. We say that $(f_n)_{n=1}^\infty$ converges uniformly to $f$ on $X$ and write $f_n\rightrightarrows f\text{,}$ if for every $\varepsilon > 0\text{,}$ there exists a positive integer $N$ such that for all $n > N\text{,}$ the inequality $|f_n(x) - f(x) | < \varepsilon$ holds for all $x \in X\text{.}$

The objects converging are Function values on a common domain. A single index $N$ must control the error at every input, which distinguishes uniform convergence from separate numerical limits.

Interpretation¶

Uniform convergence gives one error bound for the whole domain. A single index $N$ , chosen for a given $\varepsilon$ , must work for every $x$ at once. The errors at different points need not decrease at identical rates; they must eventually all be smaller than the same tolerance.

A function f with a shaded tube of vertical half-width epsilon around its graph; the graph of f1 leaves the tube, while the graph of f n for large n stays inside it everywhere.

Uniform convergence $f_n\to f$ means that for every $\varepsilon>0$ $0$ , the graph of $f_n$ lies in the $\varepsilon$ -tube around $f$ for all sufficiently large $n$ . Thus $|f_n(x)-f(x)|<\varepsilon$ holds for all $x$ at once, equivalently $\sup_x|f_n(x)-f(x)|\to0$ $0$ .

The difference with pointwise convergence lies only in the order of the quantifiers: pointwise, $N$ may depend on both $\varepsilon$ and $x$ ; uniformly, $N$ depends on $\varepsilon$ alone. The functions $x^n$ on $[0,1]$ show why this matters. They converge pointwise to the function that is $0$ on $[0,1)$ and $1$ at $x=1$ $1$ , but $\sup_{0\le x<1}x^n=1$ for every $n$ , so the convergence is not uniform. The limit is discontinuous although every $x^n$ is continuous.

Uniform convergence is exactly the control needed to pass continuity and integrals to the limit, as the Key Results state. The quantity $\sup_x|f_n(x)-f(x)|$ later gets its own name, the uniform norm, and the same idea is used to show that series of functions, such as power series, converge uniformly.

Key Results¶

Uniform convergence implies pointwise convergence.
Uniform convergence preserves boundedness and continuity (uniform limit theorem). On a closed bounded interval $[a,b]$ it preserves Riemann integrability and allows $\displaystyle\lim_{n\to\infty}\int_a^b f_n=\int_a^b f\text{.}$ It does not by itself preserve differentiability. If the $f_n$ are differentiable on an open interval and $f_n'$ also converges uniformly there, then $f$ is differentiable and $f'=\lim_n f_n'$ $lim_{n} f_{n}^{'}$ . This additional condition is sufficient, not necessary: $f_n(x)=\sin(nx)/n$ converges uniformly to the differentiable zero function on $\mathbb{R}$ , while $f_n'(\pi)=(-1)^n$ does not converge.

Examples¶

The sequence $f_n(x) = x/n$ converges uniformly to $f(x)=0$ on every bounded interval $[-R,R]$ (since $|x/n|\le R/n$ $R / n$ ), but only pointwise on $\mathbb{R}\text{.}$
The functions $f_n(x) = \dfrac{n}{nx + 1}$ on $(0,1]$ converge pointwise, but not uniformly, to $f(x)=1/x\text{,}$ because $|f_n(x)-1/x|=\dfrac{1}{x(nx+1)}$ is unbounded near $0$ for every $n\text{.}$ (On $[0,1]$ there is no pointwise limit at all, since $f_n(0)=n\to\infty\text{.}$ $\infty .$ )

Prerequisites¶

Function

Sources¶

Math III: Analysis · Lecture 1

Aliases: converges uniformly, uniformly convergent Symbols: element of, epsilon, infinity, integral, pi, real numbers, subset

Last updated: 27 October 2026 11:52