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 A supremum may fail to be attained. This is not a defect in the norm, but it matters when interpreting the word "maximum." The uniform norm is defined with a supremum because arbitrary domains may have missing endpoint values or limiting points that do not belong to the domain. [example: Supremum Not Attained] Let $g:(0,1)\to\mathbb{R}$ be defined by $g(x)=x$. For every $x\in(0,1)$, we have $0<x<1$, so $|g(x)|=|x|=x<1$. Hence $1$ is an upper bound for the set $\{|g(x)|:x\in(0,1)\}$. To see that no smaller upper bound works, fix $M<1$. Choose $n\in\mathbb{N}$ with $n\ge 2$ and $n>1/(1-M)$. Then $0<1/n<1-M$, so \begin{align*} M<1-\frac{1}{n}. \end{align*} Set $x_n=1-1/n$. Since $n\ge 2$, we have $0<x_n<1$, and therefore $x_n\in(0,1)$. At this point, \begin{align*} |g(x_n)|=\left|1-\frac{1}{n}\right|=1-\frac{1}{n}>M. \end{align*} Thus every number $M<1$ fails to bound $|g|$ on $(0,1)$, while $1$ does bound it. Therefore \begin{align*} \|g\|_\infty=\sup_{x\in(0,1)}|g(x)|=1. \end{align*} There is nevertheless no point $x\in(0,1)$ with $|g(x)|=1$, because $|g(x)|=x<1$ for every $x\in(0,1)$. The uniform norm records the least global bound, not necessarily a value attained by the function. [/example] The missing endpoint in the example raises the compactness question: when does the least global bound come from an actual point of the domain? Continuous functions on compact spaces are the basic positive case, and the next theorem is the result that lets analysts replace a supremum by a maximum in compact settings. [quotetheorem:304] For continuous functions on compact domains, the uniform norm is therefore the maximum absolute value. This is the form used in many elementary estimates, but the supremum formula is the more robust definition. ### Algebra Estimates Many analytic arguments multiply functions by coefficients, cutoffs, or nonlinear expressions. The uniform norm is well matched to multiplication because a pointwise product is controlled by the product of the two pointwise bounds. This makes $C(K)$ a normed algebra, not merely a normed vector space. [quotetheorem:8258] Submultiplicativity is the estimate behind many stability arguments: small uniform perturbations of coefficients produce small uniform perturbations of products. [example: Controlling a Nonlinear Error] Let $f,g,h: [0,1] \to \mathbb{R}$ be bounded functions with $\|f-g\|_\infty \le \varepsilon$ and $\|h\|_\infty \le M$. For each $x\in[0,1]$, the product error factors pointwise: \begin{align*} (fh-gh)(x)=f(x)h(x)-g(x)h(x)=(f(x)-g(x))h(x)=((f-g)h)(x). \end{align*} Thus $fh-gh=(f-g)h$, so their uniform norms are equal: \begin{align*} \|fh-gh\|_\infty=\|(f-g)h\|_\infty. \end{align*} For every $x\in[0,1]$, the definition of the uniform norm gives \begin{align*} |f(x)-g(x)|\le \|f-g\|_\infty \le \varepsilon. \end{align*} The same definition gives \begin{align*} |h(x)|\le \|h\|_\infty \le M. \end{align*} Multiplying these two nonnegative inequalities yields \begin{align*} |(f(x)-g(x))h(x)|=|f(x)-g(x)|\,|h(x)|\le \varepsilon M. \end{align*} Taking the supremum over $x\in[0,1]$ gives \begin{align*} \|fh-gh\|_\infty=\sup_{x\in[0,1]} |(f(x)-g(x))h(x)|\le M\varepsilon. \end{align*} So multiplying the approximation error $f-g$ by a coefficient whose uniform size is at most $M$ increases the uniform error by at most the factor $M$. [/example] ## Uniform Convergence ### Norm Convergence of Functions The norm turns convergence of functions into convergence in a [metric space](/page/Metric%20Space). This is stronger than pointwise convergence because the same error tolerance must work for every point in the domain. The definition below records that shared error tolerance directly. [definition: Uniform Convergence] Let $E$ be a set, let $f_n: E \to \mathbb{R}$ be functions, and let $f: E \to \mathbb{R}$ be a function. The sequence $(f_n)_{n=1}^\infty$ converges uniformly to $f$ on $E$ if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n \ge N$ and all $x \in E$, \begin{align*} |f_n(x)-f(x)| < \varepsilon. \end{align*} [/definition] If all $f_n-f$ are bounded, this definition is the same as $\|f_n-f\|_\infty \to 0$. The norm notation compresses the quantifiers without changing their content, and that compression is what allows tools from metric spaces to enter. [quotetheorem:8259] This equivalence is the main reason the uniform norm is the right topology for preserving continuity. Pointwise convergence lets different points require different stages of the sequence; uniform convergence supplies one stage that works everywhere, which is the missing ingredient in continuity arguments. [quotetheorem:258] The theorem says that the uniform norm topology is compatible with continuity. In $C(K)$, limits taken with respect to $\|\cdot\|_\infty$ remain inside $C(K)$. [example: Why Uniformity Preserves Continuity] Let $\varepsilon>0$, and suppose $n$ has been chosen so that $\|f_n-f\|_\infty<\varepsilon/3$. By the definition of the uniform norm, for every $z\in[0,1]$ we have \begin{align*} |f_n(z)-f(z)|\le \|f_n-f\|_\infty<\varepsilon/3. \end{align*} Fix $x\in[0,1]$. Since $f_n$ is continuous at $x$, there exists $\delta>0$ such that whenever $y\in[0,1]$ and $|x-y|<\delta$, we have \begin{align*} |f_n(x)-f_n(y)|<\varepsilon/3. \end{align*} For such a $y$, insert and subtract the two values $f_n(x)$ and $f_n(y)$: \begin{align*} f(x)-f(y)=(f(x)-f_n(x))+(f_n(x)-f_n(y))+(f_n(y)-f(y)). \end{align*} The triangle inequality gives \begin{align*} |f(x)-f(y)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|. \end{align*} The first and third terms are controlled by $\|f_n-f\|_\infty$, while the middle term is controlled by the continuity choice of $\delta$, so \begin{align*} |f(x)-f(y)|<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon. \end{align*} Thus the single index $n$ chosen from the uniform bound controls the approximation error at both $x$ and $y$, which is exactly the extra ingredient needed to transfer continuity from $f_n$ to $f$. [/example] ### Integrals Under Uniform Control Uniform convergence is also strong enough to preserve integrals over finite-measure domains. The pointwise error is bounded by a single number, so the total integral error is bounded by that number times the measure of the domain. This is the standard way a sup-norm estimate becomes an $L^1$ estimate. [quotetheorem:8260] Finite measure is essential here. If the domain has infinite measure, a uniform error that is small at every point may still accumulate over an infinite domain. [example: Uniform Smallness Need Not Give Integrable Smallness on Infinite Measure] Let $E=\mathbb{R}$ with [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^1$, and for each $n\in\mathbb{N}$ let $f_n(x)=1/n$ for every $x\in\mathbb{R}$. Since $n\ge 1$, the value $1/n$ is nonnegative, so for every $x\in\mathbb{R}$, \begin{align*} |f_n(x)|=\left|\frac{1}{n}\right|=\frac{1}{n}. \end{align*} Thus the set of pointwise absolute values is the singleton $\{1/n\}$, and therefore \begin{align*} \|f_n\|_\infty=\sup_{x\in\mathbb{R}}|f_n(x)|=\sup\left\{\frac{1}{n}\right\}=\frac{1}{n}. \end{align*} Since $1/n\to 0$, the sequence converges uniformly to $0$. However, fix $R>0$. On the interval $[-R,R]$ the function $|f_n|$ is the constant $1/n$, so \begin{align*} \int_{-R}^{R}|f_n|\,d\mathcal{L}^1=\int_{-R}^{R}\frac{1}{n}\,d\mathcal{L}^1=\frac{1}{n}\mathcal{L}^1([-R,R])=\frac{2R}{n}. \end{align*} Because $|f_n|\ge 0$ on $\mathbb{R}$, the integral over all of $\mathbb{R}$ is at least the integral over $[-R,R]$: \begin{align*} \int_{\mathbb{R}}|f_n|\,d\mathcal{L}^1\ge \frac{2R}{n}. \end{align*} As $R$ can be chosen arbitrarily large, the numbers $2R/n$ are unbounded above, so \begin{align*} \int_{\mathbb{R}}|f_n|\,d\mathcal{L}^1=\infty. \end{align*} Thus the uniform error can go to zero while the total integral remains infinite, because the domain has infinite measure. [/example] ## Completeness and Banach Spaces ### Uniform Cauchy Sequences A norm is especially powerful when Cauchy sequences converge. Before proving completeness, we need the Cauchy condition in the language of functions. The important feature is that all sufficiently late functions are close with one bound that holds throughout the domain. [definition: Uniformly Cauchy Sequence] Let $E$ be a set. A sequence $(f_n)_{n=1}^\infty$ in $B(E)$ is uniformly Cauchy if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $m,n \ge N$, \begin{align*} \|f_n-f_m\|_\infty < \varepsilon. \end{align*} [/definition] Uniformly Cauchy is the Cauchy condition in the uniform metric. It prevents not just pointwise wandering, but the possibility that the location of the worst error keeps changing with $n$. The next theorem answers the completion question for the largest natural space carrying this norm: every uniformly [Cauchy sequence](/page/Cauchy%20Sequence) of bounded functions has a bounded uniform limit. [quotetheorem:288] Completeness means that uniform limits do not leave the space of bounded functions. For compact domains, the corresponding statement for continuous functions requires one additional ingredient: the limit must remain continuous. The [uniform limit theorem](/theorems/258) supplies exactly that ingredient. [quotetheorem:8261] This theorem is one of the entrances from analysis into functional analysis. It allows the space of continuous functions to be treated as a complete infinite-dimensional geometric object. [example: A Cauchy Sequence of Polynomials with a Non-Polynomial Limit] On $[-1,1]$, define \begin{align*} p_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}. \end{align*} We show that $(p_n)$ is uniformly Cauchy. Let $m>n$ and $x\in[-1,1]$. Then $|x|\le 1$, so \begin{align*} |p_m(x)-p_n(x)|=\left|\sum_{k=n+1}^{m}\frac{x^k}{k!}\right|\le \sum_{k=n+1}^{m}\frac{|x|^k}{k!}\le \sum_{k=n+1}^{m}\frac{1}{k!}. \end{align*} For $k=n+1+j$ with $j\ge 0$, the extra factors after $(n+1)!$ are all at least $2$, hence \begin{align*} k!=(n+1)!(n+2)(n+3)\cdots(n+1+j)\ge (n+1)!2^j. \end{align*} Therefore \begin{align*} \sum_{k=n+1}^{m}\frac{1}{k!}\le \sum_{j=0}^{m-n-1}\frac{1}{(n+1)!2^j}\le \frac{1}{(n+1)!}\sum_{j=0}^{\infty}\frac{1}{2^j}=\frac{2}{(n+1)!}. \end{align*} Taking the supremum over $x\in[-1,1]$ gives \begin{align*} \|p_m-p_n\|_\infty\le \frac{2}{(n+1)!}. \end{align*} Since $2/(n+1)!\to 0$, the sequence $(p_n)$ is Cauchy in the uniform norm. For each $x\in[-1,1]$, the same tail estimate gives \begin{align*} \left|e^x-p_n(x)\right|=\left|\sum_{k=n+1}^{\infty}\frac{x^k}{k!}\right|\le \frac{2}{(n+1)!}, \end{align*} where $e^x=\sum_{k=0}^{\infty}x^k/k!$. Hence $\|e^x-p_n\|_\infty\le 2/(n+1)!$, so $p_n\to e^x$ uniformly on $[-1,1]$. The limit is not a polynomial: if $e^x$ were a polynomial of degree $d$, then its $(d+1)$st derivative would be identically zero, but differentiating $e^x$ repeatedly gives $e^x$, which is never zero. Thus a uniformly Cauchy sequence of polynomials can converge uniformly to a continuous non-polynomial function. [/example] ### Closed Uniform Conditions The example shows why completeness of $C(K)$ is larger than completeness of the polynomial subspace. A sequence may converge uniformly to a function outside the smaller algebra that generated it. To build complete subspaces, the condition defining the subspace must be closed under uniform limits. [quotetheorem:287] For instance, spaces of bounded functions satisfying a closed boundary condition, [symmetry condition](/theorems/1360), or linear constraint often become Banach spaces by this theorem. ## Approximation in the Uniform Norm ### Uniform Density Approximation theory asks whether simple functions can approximate complicated ones. The uniform norm makes this question demanding: an approximant must be good everywhere, including near endpoints, peaks, and transition regions. The formal language for this requirement is density in the uniform norm. [definition: Uniformly Dense Subset] Let $E$ be a set, and let $Y \subset B(E)$. A subset $A \subset Y$ is uniformly dense in $Y$ if for every $f \in Y$ and every $\varepsilon > 0$ there exists $a \in A$ such that \begin{align*} \|f-a\|_\infty < \varepsilon. \end{align*} [/definition] Uniform density is the language behind polynomial approximation, Fourier approximation in strong topologies, and numerical schemes with certified worst-case error. On compact intervals, the foundational result is that polynomials are dense in this demanding sense. [quotetheorem:480] The theorem says that polynomials are uniformly dense in $C([a,b])$. It does not say that every [continuous function](/page/Continuous%20Function) is a polynomial; it says that the closure of the polynomial functions in the uniform norm is all of $C([a,b])$. [example: Approximating Absolute Value] The function $f: [-1,1] \to \mathbb{R}$ defined by $f(x)=|x|$ is continuous on $[-1,1]$. Indeed, for $x,y\in[-1,1]$, the [reverse triangle inequality](/theorems/2300) gives \begin{align*} \bigl||x|-|y|\bigr|\le |x-y|. \end{align*} Thus if $|x-y|<\delta$, then $|f(x)-f(y)|<\delta$, so taking $\delta=\varepsilon$ proves continuity. The function is not differentiable at $0$. For $h>0$, \begin{align*} \frac{f(0+h)-f(0)}{h}=\frac{|h|-0}{h}=1. \end{align*} For $h<0$, \begin{align*} \frac{f(0+h)-f(0)}{h}=\frac{|h|-0}{h}=\frac{-h}{h}=-1. \end{align*} The right-hand and left-hand difference quotients have different limits, so $f$ has no derivative at $0$. By *[Weierstrass Approximation Theorem](/theorems/480)*, for every $\varepsilon>0$ there exists a polynomial $p$ such that \begin{align*} \sup_{x\in[-1,1]} \bigl||x|-p(x)\bigr|<\varepsilon. \end{align*} Equivalently, \begin{align*} \||\,\cdot\,|-p\|_\infty<\varepsilon. \end{align*} The approximation is therefore uniform on the whole interval, including at the cusp $x=0$; the target function need not be differentiable. [/example] ### Algebras of Continuous Functions Polynomial approximation is not only about intervals. In higher-dimensional and more structural settings, the key question is whether a chosen algebra of functions has enough information to distinguish points. The [Stone-Weierstrass theorem](/theorems/886) turns that point-separation property into uniform density. [quotetheorem:886] The separation hypothesis says that the algebra has enough functions to distinguish points of the space. The conclusion says that algebraic generation plus point separation is enough to approximate arbitrary continuous functions in the strongest common topology on $C(K)$. [example: Trigonometric Approximation] Let $K=\{(\cos t,\sin t):0\le t\le 2\pi\}$ be the unit circle, and identify a continuous periodic function on $[0,2\pi]$ with the continuous function it induces on $K$. Let $A$ be the algebra generated by the constant functions and the coordinate functions \begin{align*} c(\cos t,\sin t)=\cos t. \end{align*} \begin{align*} s(\cos t,\sin t)=\sin t. \end{align*} Elements of $A$ are finite sums of products of constants, $c$, and $s$, so they are exactly finite real linear combinations of functions of the form $(\cos t)^j(\sin t)^k$. We verify the point-separation hypothesis. If $p=(\cos u,\sin u)$ and $q=(\cos v,\sin v)$ are distinct points of $K$, then \begin{align*} (\cos u,\sin u)\ne(\cos v,\sin v). \end{align*} Therefore either $\cos u\ne\cos v$ or $\sin u\ne\sin v$. In the first case, $c(p)\ne c(q)$; in the second case, $s(p)\ne s(q)$. Since $c,s\in A$, the algebra $A$ separates points of $K$ and contains the constants. By *Stone-Weierstrass Theorem*, for every continuous function $F:K\to\mathbb{R}$ and every $\varepsilon>0$, there exists $a\in A$ such that \begin{align*} \sup_{z\in K}|F(z)-a(z)|<\varepsilon. \end{align*} Writing $z=(\cos t,\sin t)$, this says \begin{align*} \sup_{0\le t\le 2\pi}|F(\cos t,\sin t)-a(\cos t,\sin t)|<\varepsilon. \end{align*} Thus every continuous periodic function can be uniformly approximated by finite algebraic combinations of $\sin t$ and $\cos t$, and these are trigonometric polynomials after rewriting products of powers of $\sin t$ and $\cos t$ as finite sums of sines and cosines of integer multiples of $t$. [/example] Uniform approximation is stronger than approximation in an integral norm. A function may be small in $L^1$ while having a large narrow spike; the uniform norm detects the spike. [example: Small Area with Large Uniform Error] For $n \in \mathbb{N}$, define $f_n: [0,1] \to \mathbb{R}$ by $f_n(x)=1$ for $0 \le x \le 1/n$ and $f_n(x)=0$ for $1/n < x \le 1$. Since $1/n \le 1$, the interval $[0,1/n]$ lies inside $[0,1]$. For every $x\in[0,1]$, the value $f_n(x)$ is either $0$ or $1$, so $|f_n(x)|=f_n(x)$. On $[0,1/n]$ we have $|f_n(x)|=1$, and on $(1/n,1]$ we have $|f_n(x)|=0$. Therefore \begin{align*} \int_0^1 |f_n|\,d\mathcal{L}^1=\int_{[0,1/n]}1\,d\mathcal{L}^1+\int_{(1/n,1]}0\,d\mathcal{L}^1=\mathcal{L}^1([0,1/n])+0=\frac{1}{n}. \end{align*} Thus $\|f_n\|_{L^1([0,1])}=1/n$, and since $1/n\to 0$, the sequence converges to $0$ in $L^1$. The uniform norm behaves differently. Because $f_n(x)$ is always either $0$ or $1$, we have \begin{align*} |f_n(x)|\le 1 \end{align*} for every $x\in[0,1]$, so $\|f_n\|_\infty\le 1$. At the point $x=0$, however, \begin{align*} |f_n(0)|=|1|=1. \end{align*} Hence the supremum is at least $1$, and combining the two inequalities gives \begin{align*} \|f_n\|_\infty=\sup_{x\in[0,1]}|f_n(x)|=1. \end{align*} The area under the spike is $1/n$ and tends to $0$, but the spike still reaches height $1$ for every $n$, so the functions vanish in $L^1$ but not in the uniform norm. [/example] ## Comparison with Other Norms ### Integral Norms The uniform norm gives pointwise control everywhere on the domain. Integral norms average size, and essential supremum norms ignore changes on null sets. On finite-measure spaces, however, a uniform estimate immediately gives every finite $L^p$ estimate. [quotetheorem:8262] The reverse estimate is false without additional hypotheses. Large values on small sets can make the uniform norm large while keeping every integral norm small. [example: No Reverse Bound from $L^p$ to Uniform Norm] Let $1 \le p < \infty$, and define $f_n:[0,1]\to\mathbb{R}$ by \begin{align*} f_n(x)=n^{1/p}\text{ for }0\le x\le 1/n,\qquad f_n(x)=0\text{ for }1/n<x\le 1. \end{align*} For every $x\in[0,1]$, the value of $|f_n(x)|^p$ is $n$ on $[0,1/n]$ and $0$ on $(1/n,1]$, because \begin{align*} |n^{1/p}|^p=n. \end{align*} Therefore \begin{align*} \int_0^1 |f_n(x)|^p\,dx=\int_0^{1/n} n\,dx+\int_{1/n}^1 0\,dx. \end{align*} The two integrals are \begin{align*} \int_0^{1/n} n\,dx=n\left(\frac{1}{n}-0\right)=1. \end{align*} and \begin{align*} \int_{1/n}^1 0\,dx=0. \end{align*} Thus \begin{align*} \int_0^1 |f_n(x)|^p\,dx=1. \end{align*} Taking the $p$th root gives \begin{align*} \|f_n\|_{L^p([0,1])}=\left(\int_0^1 |f_n(x)|^p\,dx\right)^{1/p}=1^{1/p}=1. \end{align*} For the uniform norm, every value of $|f_n(x)|$ is either $n^{1/p}$ or $0$, so \begin{align*} |f_n(x)|\le n^{1/p} \end{align*} for every $x\in[0,1]$. Hence $\|f_n\|_\infty\le n^{1/p}$. At $x=0$, however, \begin{align*} |f_n(0)|=n^{1/p}. \end{align*} So $\|f_n\|_\infty\ge n^{1/p}$, and the two inequalities give \begin{align*} \|f_n\|_\infty=n^{1/p}. \end{align*} If a constant $C$ satisfied $\|f\|_\infty\le C\|f\|_{L^p([0,1])}$ for every bounded function $f$ on $[0,1]$, then applying it to $f_n$ would give \begin{align*} n^{1/p}=\|f_n\|_\infty\le C\|f_n\|_{L^p([0,1])}=C. \end{align*} But $n^{1/p}\to\infty$ as $n\to\infty$, so no fixed $C$ can satisfy these inequalities for all $n$. Therefore the $L^p$ norm cannot uniformly control the pointwise supremum norm on all bounded functions on $[0,1]$. [/example] ### Essential Supremum The previous example shows that integral norms can miss narrow spikes. Measure theory introduces another supremum-type norm that deliberately ignores sets of measure zero. This is indispensable for $L^\infty$ spaces, where functions are identified up to almost-everywhere equality, but it is different from the pointwise uniform norm. [definition: Essential Supremum Norm] Let $(E,\mathcal{E},\mu)$ be a [measure space](/page/Measure%20Space), and let $L^\infty(E,\mu)$ be the vector space of essentially bounded [measurable functions](/page/Measurable%20Functions) $f: E \to \mathbb{R}$ modulo equality $\mu$-a.e. The essential supremum norm is the map \begin{align*} \|\cdot\|_{L^\infty(E)}: L^\infty(E,\mu) \to [0,\infty), \qquad \|f\|_{L^\infty(E)} = \operatorname{ess\,sup}_{x \in E} |f(x)|. \end{align*} [/definition] It is not the same object as the uniform norm on pointwise-defined bounded functions. [example: Uniform Norm Versus Essential Supremum] Let $f: [0,1]\to\mathbb{R}$ be defined by $f(0)=1$ and $f(x)=0$ for $0<x\le 1$. For every $x\in[0,1]$, the value of $|f(x)|$ is either $1$ or $0$, so \begin{align*} |f(x)|\le 1. \end{align*} Hence \begin{align*} \sup_{x\in[0,1]}|f(x)|\le 1. \end{align*} At the point $x=0$, we have \begin{align*} |f(0)|=|1|=1. \end{align*} Therefore the supremum is at least $1$, and combining the two inequalities gives \begin{align*} \|f\|_\infty=\sup_{x\in[0,1]}|f(x)|=1. \end{align*} For the essential supremum with respect to Lebesgue measure, observe that \begin{align*} \{x\in[0,1]: |f(x)|>0\}=\{0\}. \end{align*} The singleton $\{0\}$ has Lebesgue measure $0$, so $|f(x)|=0$ for Lebesgue-almost every $x\in[0,1]$. Thus $0$ is an almost-everywhere upper bound for $|f|$, which gives \begin{align*} \|f\|_{L^\infty([0,1])}\le 0. \end{align*} Since an essential supremum norm is always nonnegative, \begin{align*} 0\le \|f\|_{L^\infty([0,1])}. \end{align*} Hence \begin{align*} \|f\|_{L^\infty([0,1])}=0. \end{align*} The uniform norm records the point value $f(0)=1$, while the essential supremum ignores that value because it occurs only on a Lebesgue-null set. [/example] The example is discontinuous, so it leaves a natural question: can continuity prevent a function from hiding all of its largest value on a null set? On compact intervals the answer is yes, because any large value of a continuous function persists on an interval of positive measure. [quotetheorem:8263] The interval hypothesis gives enough measure around every point of maximum. It is the elementary model for a broader principle: continuity turns a large point value into largeness on a neighbourhood, so a measure with enough support cannot ignore the supremum. ## Operators and Functionals ### Point Evaluation The uniform norm lets pointwise operations become continuous maps between normed spaces. Point evaluation is the simplest example. It is not continuous in many integral norms, but it is continuous in the uniform norm because the norm controls every point value. [definition: Evaluation Functional] Let $E$ be a set and let $x_0 \in E$. The evaluation functional at $x_0$ is the [linear map](/page/Linear%20Map) \begin{align*} \operatorname{ev}_{x_0}: B(E) \to \mathbb{R}, \qquad \operatorname{ev}_{x_0}(f)=f(x_0). \end{align*} [/definition] The definition gives a linear map, but functional analysis asks whether that map is continuous. Since $|f(x_0)|$ is controlled by the worst value of $|f|$ on all of $E$, the uniform norm should make point evaluation bounded. The next theorem computes its exact operator norm. [quotetheorem:8265] This theorem is one way to remember the meaning of the uniform norm: it is strong enough to make every point value a stable observable. [example: Evaluation Is Not Controlled by $L^1$] On the concrete class of bounded measurable representatives on $[0,1]$, define $f_n:[0,1]\to\mathbb{R}$ by $f_n(x)=n$ for $0\le x\le 1/n$ and $f_n(x)=0$ for $1/n<x\le 1$. Since $n\in\mathbb{N}$, we have $1/n\in(0,1]$, so the interval $[0,1/n]$ is contained in $[0,1]$. At the point $0$, the first branch applies, and therefore \begin{align*} f_n(0)=n. \end{align*} For every $x\in[0,1]$, the value $f_n(x)$ is either $n$ or $0$, so $|f_n(x)|=f_n(x)$. Hence \begin{align*} \|f_n\|_{L^1([0,1])}=\int_0^1 |f_n(x)|\,dx=\int_0^{1/n} n\,dx+\int_{1/n}^{1}0\,dx. \end{align*} The first integral is \begin{align*} \int_0^{1/n} n\,dx=n\left(\frac{1}{n}-0\right)=1, \end{align*} and the second integral is \begin{align*} \int_{1/n}^{1}0\,dx=0. \end{align*} Therefore \begin{align*} \|f_n\|_{L^1([0,1])}=1+0=1. \end{align*} Suppose there were a constant $C\ge 0$ such that every bounded measurable representative $f$ on $[0,1]$ satisfied \begin{align*} |f(0)|\le C\|f\|_{L^1([0,1])}. \end{align*} Applying this estimate to $f_n$ gives \begin{align*} n=|f_n(0)|\le C\|f_n\|_{L^1([0,1])}=C. \end{align*} This would force $n\le C$ for every $n\in\mathbb{N}$, which is impossible because the natural numbers are unbounded. Thus no constant $C$ can control the representative value $|f(0)|$ by the $L^1$ norm on this concrete function class. This is not a bounded functional on the quotient space $L^1([0,1])$: changing a representative at the single point $0$, a Lebesgue-null set, can change $f(0)$ without changing its $L^1$ equivalence class. [/example] ### Nonlinear Maps The uniform norm also controls composition when the outer function is Lipschitz. This is a common way nonlinear expressions inherit convergence, since a scalar Lipschitz bound can be applied at every point and then promoted to a function-space estimate. [quotetheorem:8267] The theorem turns scalar Lipschitz estimates into function-space estimates. It is often the first step in proving stability of nonlinear equations in a sup-norm setting. [example: Squaring on a Uniformly Bounded Set] Let $M\ge 0$, and let $f,g:E\to[-M,M]$ be bounded functions. For any $s,t\in[-M,M]$, \begin{align*} |s^2-t^2|=|(s-t)(s+t)|=|s-t|\,|s+t|. \end{align*} Since $|s|\le M$ and $|t|\le M$, the triangle inequality gives \begin{align*} |s+t|\le |s|+|t|\le M+M=2M. \end{align*} Therefore \begin{align*} |s^2-t^2|\le 2M|s-t|. \end{align*} Applying this with $s=f(x)$ and $t=g(x)$, for each $x\in E$ we get \begin{align*} |f(x)^2-g(x)^2|\le 2M|f(x)-g(x)|. \end{align*} Also $|f(x)-g(x)|\le \|f-g\|_\infty$ by the definition of the uniform norm, so \begin{align*} |f(x)^2-g(x)^2|\le 2M\|f-g\|_\infty. \end{align*} Taking the supremum over $x\in E$ gives \begin{align*} \|f^2-g^2\|_\infty=\sup_{x\in E}|f(x)^2-g(x)^2|\le 2M\|f-g\|_\infty. \end{align*} Thus on the set of functions whose values lie in $[-M,M]$, the squaring map is Lipschitz with constant $2M$, and hence is uniformly continuous in the uniform norm. [/example] ## Beyond and Connected Topics The uniform norm is the starting point for [Banach Space](/page/Banach%20Space) methods. The completeness of $C(K)$ under $\|\cdot\|_\infty$ makes continuous functions a central example in functional analysis, and it provides a concrete model for abstract theorems about bounded linear maps, dual spaces, and compact operators. It is also the natural norm for [Continuity](/page/Continuity) and [Uniform Convergence](/page/Uniform%20Convergence). The theorem that uniform limits of continuous functions are continuous is one of the basic bridges between point-set topology and analysis. Approximation theory uses the uniform norm to give strong error guarantees. The Weierstrass and Stone-Weierstrass theorems say that algebraically simple functions can approximate continuous functions with a controlled worst-case error. The relation with [Lebesgue Integration](/page/Integral), $L^p$ spaces, and [Sobolev Space](/page/Sobolev%20Space) is more delicate. Integral norms measure average size, Sobolev norms measure weak derivatives, and the uniform norm measures pointwise size. Embedding theorems are precisely the results that convert derivative or integrability information into uniform control. In complex analysis, the uniform norm on compact sets is the topology behind locally uniform convergence of holomorphic functions. This leads to powerful stability theorems: limits of holomorphic functions remain holomorphic when convergence is uniform on compact subsets. ## References Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Androma, [Banach Space](/page/Banach%20Space). Androma, [Continuity](/page/Continuity). Androma, [Uniform Convergence](/page/Uniform%20Convergence). Androma, [Integral](/page/Integral). Androma, [Sobolev Space](/page/Sobolev%20Space). Walter Rudin, *Principles of Mathematical Analysis* (1976). John B. Conway, *A Course in Functional Analysis* (1990). Erwin Kreyszig, *Introductory Functional Analysis with Applications* (1978).