A sequence of functions can converge at every point and still fail to behave like a convergent sequence of functions. The graph may keep a narrow spike that moves, steepens, or concentrates; each fixed input eventually stops seeing the spike, while the functions as whole objects never become uniformly close. The uniform norm is the device that refuses to ignore this global failure. It measures the largest vertical error over the entire domain, so convergence in this norm means that the whole graph has settled down at once.

This is why the uniform norm sits at the meeting point of topology, analysis, and functional analysis. It turns spaces of bounded functions into metric spaces, turns spaces of continuous functions on compact sets into Banach spaces, and gives a precise language for approximation by polynomials, smoothing, and numerical error bounds. The price is that it is an unforgiving norm: a small exceptional region still counts if the function is large there.

[example: Pointwise Convergence Without Uniform Control] Let $f_n: [0,1] \to \mathbb{R}$ be defined by $f_n(x)=x^n$. If $0 \le x < 1$, then $x^n \to 0$, while $f_n(1)=1$ for every $n$. Hence the pointwise limit $f: [0,1] \to \mathbb{R}$ is given by $f(x)=0$ for $0 \le x < 1$ and $f(1)=1$. Fix $n \in \mathbb{N}$. For $0 \le x < 1$, we have $f(x)=0$, so \begin{align*} |f_n(x)-f(x)|=|x^n-0|=x^n. \end{align*} At the endpoint $x=1$, we have $f_n(1)=1$ and $f(1)=1$, so \begin{align*} |f_n(1)-f(1)|=|1-1|=0. \end{align*} Therefore every pointwise error is at most $1$, and so \begin{align*} \sup_{x \in [0,1]} |f_n(x)-f(x)| \le 1. \end{align*} Conversely, if $0<\varepsilon<1$, take $x=(1-\varepsilon)^{1/n}$. Then $0 \le x <1$, and \begin{align*} |f_n(x)-f(x)|=x^n=((1-\varepsilon)^{1/n})^n=1-\varepsilon. \end{align*} Thus the supremum is at least $1-\varepsilon$ for every $0<\varepsilon<1$, so it must be at least $1$. Combining the two inequalities gives \begin{align*} \|f_n-f\|_\infty=\sup_{x \in [0,1]} |f_n(x)-f(x)|=1. \end{align*} The pointwise errors vanish at each fixed $x<1$, but the largest error over the whole interval never becomes small; moreover, the limit function is not continuous at $1$, since $1-1/m \to 1$ while $f(1-1/m)=0$ for every $m$ and $f(1)=1$. [/example]

The example gives the central problem: pointwise convergence is too weak for many analytic operations. If we want continuity to pass to limits, if we want approximation errors to have a single numerical meaning, or if we want a function space to be complete, we need a norm that controls all points simultaneously.

## Definition

Before defining the uniform norm, we need to decide which functions are allowed. If a function is unbounded, the supremum of its absolute value is $\infty$, so the expression cannot be a norm value in $[0,\infty)$. Thus the natural ambient [vector space](/page/Vector%20Space) is $B(E)$, the set of bounded functions $g: E \to \mathbb{R}$, meaning that there exists $M \ge 0$ with $|g(x)| \le M$ for every $x \in E$.

[definition: Uniform Norm] Let $E$ be a set, and let $B(E)$ denote the vector space of bounded functions $g: E \to \mathbb{R}$. The uniform norm on $B(E)$ is the map \begin{align*} \|\cdot\|_\infty: B(E) \to [0,\infty), \qquad \|g\|_\infty = \sup_{x \in E} |g(x)|. \end{align*} [/definition]

The subscript $\infty$ reflects the relation with $L^p$ norms: as $p$ grows, the largest values of a function dominate more strongly. Unlike the $L^\infty$ norm, however, this uniform norm uses the actual supremum over all points, not an essential supremum modulo null sets. If $E = \varnothing$, the usual convention is $\sup \varnothing = 0$, so the unique function on $E$ has uniform norm $0$.

On an arbitrary set, boundedness has to be imposed. In topology and analysis, compactness often supplies boundedness automatically: if $K$ is a compact [topological space](/page/Topological%20Space), then $C(K)$ denotes the vector space of continuous functions $f: K \to \mathbb{R}$, and every such function has finite uniform norm. Some undergraduate notes write this norm as $\|f\|$ when no other norm is present. On this page we write $\|f\|_\infty$ to keep it distinct from Euclidean, $L^p$, Sobolev, and operator norms.

The uniform norm measures the size of one function. To discuss convergence and approximation, we need a distance between two functions. Applying the norm to their difference gives the largest vertical separation between the two graphs.

[definition: Uniform Metric] Let $E$ be a set. The uniform metric on $B(E)$ is the map \begin{align*} d_\infty: B(E) \times B(E) \to [0,\infty), \qquad d_\infty(f,g) = \|f-g\|_\infty. \end{align*} [/definition]

This metric is the geometric form of the norm. A ball in this metric consists of all functions whose graphs lie inside a vertical tube of fixed radius around a given function.

[example: A Uniform Ball Around a Function] Let $E=[0,1]$ and let $f(x)=x^2$. In the uniform metric, the open ball of radius $1/10$ around $f$ consists of exactly those bounded functions whose uniform distance from $f$ is less than $1/10$: \begin{align*} B_\infty(f,1/10)=\{g\in B([0,1]):d_\infty(g,f)<1/10\}. \end{align*} Since $d_\infty(g,f)=\|g-f\|_\infty$ and $(g-f)(x)=g(x)-x^2$, this becomes \begin{align*} B_\infty(f,1/10)=\{g\in B([0,1]):\sup_{x\in[0,1]}|g(x)-x^2|<1/10\}. \end{align*} Thus $g$ belongs to this ball if and only if \begin{align*} |g(x)-x^2|<1/10 \end{align*} for every $x\in[0,1]$. The absolute value inequality is equivalent to \begin{align*} -1/10<g(x)-x^2<1/10. \end{align*} Adding $x^2$ to all three parts gives \begin{align*} x^2-1/10<g(x)<x^2+1/10. \end{align*} So membership in the ball means that the graph of $g$ stays strictly inside the vertical band of radius $1/10$ around the graph of $x^2$ on the entire interval. [/example]

Uniform balls explain why the norm is useful in [approximation theory](/page/Approximation%20Theory): saying that a polynomial is close to a function is not a vague visual statement, but a quantified global error bound.

## Norm Structure

### The Norm Axioms

The uniform norm is useful only if it satisfies the norm axioms. The supremum definition is designed so that the triangle inequality for [real numbers](/page/Real%20Numbers) becomes the triangle inequality for functions after taking the supremum over the domain. This is what lets function estimates be handled by the geometry of normed vector spaces.

[quotetheorem:8257]

The theorem means that $B(E)$ can be studied as a [normed vector space](/page/Normed%20Vector%20Space). In particular, it makes sense to discuss Cauchy sequences of functions, continuous linear maps on $B(E)$, and closed subspaces such as $C(K)$.

### Suprema and Maxima