Normal Convergence¶

Definition¶

Let $(f_n)_{n=1}^\infty$ be a sequence of bounded functions on $X$ . We say the series $\displaystyle \sum_{n=1}^\infty f_n(x)$ is normally convergent if the series of uniform norms of the terms of the series converges, i.e., $\displaystyle \sum_{n=1}^\infty \|f_n\|_\infty<+\infty.$

Interpretation¶

Normal convergence asks that the worst-case sizes $\|f_n\|_\infty$ of the terms form a convergent numerical series. Instead of studying $\displaystyle \sum_n f_n(x)$ at every $x$ , one attaches a single number to each term, its uniform norm; if these numbers are summable, the series of functions is controlled everywhere at once.

Values 1 over n squared of the uniform norms of the terms, decreasing to 0, and their partial sums increasing towards pi squared over 6, drawn as a horizontal line.

In the figure, $\|f_n\|_\infty=1/n^2$ $1/ n^{2}$ . The bars are the norms, decreasing to $0$ , and the partial sums of the norms increase towards $\pi^2/6\approx1.645$ $1.645$ . Because the series of norms converges, its tail bounds the error everywhere: after $N$ terms, the partial sum of the function series differs from its limit by at most $\displaystyle \sum_{n>N}1/n^2$ at every $x$ , less than $0.1$ for $N=10$ $10$ .

$\displaystyle \sum_n\|f_n\|_\infty=\sum_n 1/n^2$ converges, so the partial sums stay within the shrinking band around the limit.

The condition involves only numbers, so it is checked with the tests for numerical series, such as comparison with a $p$ -series. Normal convergence is exactly the hypothesis of the Weierstrass $M$ -test, which turns it into uniform convergence; at each fixed $x$ it also gives absolute convergence of $\displaystyle \sum_n f_n(x)$ , since $|f_n(x)|\le\|f_n\|_\infty$ $∥ f_{n} ∥_{\infty}$ . The name refers to the norm.

The implications in the Key Results go one way only. The Examples give a series that converges absolutely and uniformly but not normally. A uniformly convergent series also need not converge absolutely: the constant functions $f_n(x)=(-1)^n/n$ form a uniformly convergent alternating series, while $\displaystyle \sum_n\|f_n\|_\infty$ is the harmonic series. Normal convergence is a convenient sufficient condition, not a description of all uniformly convergent series.

Key Results¶

Normal convergence implies uniform absolute convergence (the series $\displaystyle \sum_n|f_n(x)|$ converges uniformly), which in turn implies uniform convergence, which in turn implies pointwise convergence, i.e.,

\begin{aligned} \sum_{n=1}^\infty \|f_n\|_\infty<+\infty &\implies \sum_{n=1}^\infty |f_n(x)|\rightrightarrows g\\ &\implies \sum_{n=1}^\infty f_n(x)\rightrightarrows f\\ &\implies \sum_{n=1}^\infty f_n(x)\to f. \end{aligned}

Here $g$ is the sum of the absolute series; it differs from $f$ unless every $f_n\ge0\text{.}$

Examples¶

Uniform absolute convergence does not imply normal convergence. On $X=\mathbb{N}$ $N$ , consider

f_n(x)=\begin{cases} 1/n, &x=n,\\[2pt] 0, &x\neq n. \end{cases}

The absolute series converges uniformly: its tail after the first $N$ terms has supremum $1/(N+1)$ $1)$ , which tends to zero. However, $\|f_n\|_\infty=1/n$ $1/ n$ , so the series of uniform norms is the harmonic series and diverges.

Prerequisites¶

Weierstrass $M$ -test

Sources¶

Math III: Analysis · Lecture 2

Aliases: normally convergent Symbols: infinity, natural numbers, pi, sum

Last updated: 20 October 2026 03:02