A sequence of functions can behave well at every individual point and still behave badly as a family. The pointwise Cauchy condition watches each input $x \in E$ separately; it permits the index at which stability begins to depend on $x$. Uniform Cauchyness forbids that dependence. It asks for a single index after which all functions in the tail are close at every point of the domain at once.

[example: Pointwise Stability Without Uniform Stability] Let $E=(0,1)$ and define $f_n:(0,1)\to \mathbb{R}$ by $f_n(x)=x^n$. Fix $x \in (0,1)$. Since $0<x<1$, the geometric sequence $(x^k)_{k=1}^{\infty}$ converges to $0$, so for every $\varepsilon>0$ there is $K \in \mathbb{N}$ such that $x^k<\varepsilon/2$ whenever $k \ge K$. If $m,n \ge K$, then $x^m\ge 0$ and $x^n\ge 0$, and the triangle inequality gives \begin{align*} |f_n(x)-f_m(x)|=|x^n-x^m|\le |x^n|+|x^m|=x^n+x^m<\varepsilon/2+\varepsilon/2=\varepsilon. \end{align*} Thus $(f_n(x))_{n=1}^{\infty}$ is Cauchy for each fixed $x\in(0,1)$. The index $K$ may depend on the chosen point $x$, and that dependence is exactly what prevents uniform Cauchyness. To see the failure of uniform Cauchyness, take $\varepsilon=1/5$. Let $N\in\mathbb{N}$ be arbitrary, and compare the two tail terms $f_N$ and $f_{2N}$. For $x\in(0,1)$, we have $0<x^N<1$, so multiplying $x^N<1$ by the positive number $x^N$ gives $x^{2N}<x^N$. Hence \begin{align*} |f_N(x)-f_{2N}(x)|=|x^N-x^{2N}|=x^N-x^{2N}. \end{align*} Choose $x=2^{-1/N}$. Since $2^{-1/N}>0$ and $2^{-1/N}<1$, this point lies in $(0,1)$. Also, \begin{align*} x^N=(2^{-1/N})^N=2^{-1}=1/2. \end{align*} Therefore \begin{align*} x^{2N}=(x^N)^2=(1/2)^2=1/4. \end{align*} At this point $x=2^{-1/N}$, the tail difference is \begin{align*} |f_N(x)-f_{2N}(x)|=x^N-x^{2N}=1/2-1/4=1/4>1/5. \end{align*} Thus for every proposed tail index $N$, the indices $N$ and $2N$ are both at least $N$, but there is a point $x\in(0,1)$ where $|f_N(x)-f_{2N}(x)|>1/5$. No single tail index can control all points of $E$ at once, so the sequence is pointwise Cauchy but not uniformly Cauchy on $(0,1)$. [/example]

This example is the basic warning: a limit constructed point by point need not be a limit in the function space one actually wants to use. Uniform Cauchyness is the Cauchy condition appropriate to the sup norm. It is especially useful because it talks only about the sequence itself, not about a candidate limit that may not yet be known.

## Definition

The ordinary Cauchy condition for numbers compares $a_n$ and $a_m$ once $m,n$ are large. For functions, there are many comparisons to make, one at each point of the domain. If the tail index may depend on the point, the family can keep moving its bad behaviour around; the opening example does exactly that. The definition below removes that escape route by requiring one tail index to work simultaneously across the domain.

[definition: Uniformly Cauchy Sequence] Let $E$ be a set, let $(Y,d_Y)$ be a metric space, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n:E \to Y$. The sequence $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$ if for every $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that, for all $m,n \ge N$ and all $x \in E$, \begin{align*} d_Y(f_n(x),f_m(x))<\varepsilon. \end{align*} [/definition]

When $Y=\mathbb{R}$ or $Y=\mathbb{C}$, the metric is usually the absolute value metric, so the condition becomes $|f_n(x)-f_m(x)|<\varepsilon$ for all $x \in E$. The order of the quantifiers is the substance of the definition: $N$ is chosen after $\varepsilon$, but before $m,n$, and before $x$. Throughout the page, a sequence of functions on $E$ means a sequence $(f_n)_{n=1}^{\infty}$ in which every $f_n$ has the same domain $E$ and the same codomain.

[example: Shrinking Linear Functions Are Uniformly Cauchy] Let $E=[0,1]$ and define $f_n:[0,1]\to \mathbb{R}$ by $f_n(x)=x/n$. We show that for every $\varepsilon>0$ there is one tail index $N$ that controls $|f_n(x)-f_m(x)|$ for all $x\in[0,1]$ at once. Fix $\varepsilon>0$. Choose $N\in\mathbb{N}$ such that $N>2/\varepsilon$. Let $m,n\ge N$ and let $x\in[0,1]$. Then \begin{align*} |f_n(x)-f_m(x)|=\left|\frac{x}{n}-\frac{x}{m}\right|. \end{align*} Since $\frac{x}{n}=x\cdot\frac{1}{n}$ and $\frac{x}{m}=x\cdot\frac{1}{m}$, distributivity gives \begin{align*} \frac{x}{n}-\frac{x}{m}=x\left(\frac{1}{n}-\frac{1}{m}\right). \end{align*} Therefore, using $|ab|=|a||b|$, \begin{align*} \left|\frac{x}{n}-\frac{x}{m}\right|=\left|x\left(\frac{1}{n}-\frac{1}{m}\right)\right|=|x|\left|\frac{1}{n}-\frac{1}{m}\right|. \end{align*} Because $x\in[0,1]$, we have $|x|=x\le 1$, so \begin{align*} |x|\left|\frac{1}{n}-\frac{1}{m}\right|\le \left|\frac{1}{n}-\frac{1}{m}\right|. \end{align*} By the triangle inequality for the absolute value, \begin{align*} \left|\frac{1}{n}-\frac{1}{m}\right|=\left|\frac{1}{n}+\left(-\frac{1}{m}\right)\right|\le \left|\frac{1}{n}\right|+\left|-\frac{1}{m}\right|. \end{align*} Since $m,n\in\mathbb{N}$, both $1/n$ and $1/m$ are positive, and hence \begin{align*} \left|\frac{1}{n}\right|+\left|-\frac{1}{m}\right|=\frac{1}{n}+\frac{1}{m}. \end{align*} Because $m,n\ge N$, we have $1/n\le 1/N$ and $1/m\le 1/N$. Combining the estimates, \begin{align*} |f_n(x)-f_m(x)|\le \frac{1}{n}+\frac{1}{m}\le \frac{1}{N}+\frac{1}{N}=\frac{2}{N}. \end{align*} The choice $N>2/\varepsilon$ implies $2/N<\varepsilon$, so \begin{align*} |f_n(x)-f_m(x)|<\varepsilon. \end{align*} The index $N$ was chosen using only $\varepsilon$, not $x$, $m$, or $n$ beyond the tail condition $m,n\ge N$. Hence $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $[0,1]$. [/example]

The computation illustrates the main technique: estimate the difference by a bound independent of $x$. The moment the bound still depends on $x$, the argument has not yet proved uniform Cauchyness.

## Quantifiers and Sup Norm Geometry

### Uniform Quantifiers

The definition looks small, but most errors with uniform Cauchyness are quantifier errors. Pointwise Cauchyness permits $N$ to depend on $x$. Uniform Cauchyness does not. This difference is invisible if one tests only finitely many points, but it becomes decisive on infinite domains.

### Sup Norm Viewpoint

A sequence of functions may also be viewed as a sequence in a metric space of functions. This viewpoint explains why the definition is not a new kind of convergence but the usual Cauchy condition applied to a stronger metric.

A compact way to read uniform Cauchyness is to measure the worst possible distance between two functions over the whole domain. This gives a metric on any class of functions for which that worst-case distance is always finite, and it gives an extended-valued comparison on the larger class of all functions.

[definition: Uniform Distance] Let $E$ be a nonempty set, let $(Y,d_Y)$ be a metric space, and let $\mathcal{F} \subset Y^E$ be a set of functions such that \begin{align*} \sup_{x \in E} d_Y(f(x),g(x))<\infty \end{align*} for all $f,g \in \mathcal{F}$. The uniform distance on $\mathcal{F}$ is the map $d_\infty:\mathcal{F}\times \mathcal{F}\to [0,\infty)$ defined by \begin{align*} d_\infty(f,g)=\sup_{x \in E} d_Y(f(x),g(x)). \end{align*} [/definition]

With this notation, uniform Cauchyness says exactly that $d_\infty(f_n,f_m) \to 0$ as $m,n \to \infty$, provided the sequence lies in a function class where the relevant uniform distances are finite. The original quantifier definition remains the safer primary definition because it does not first require choosing such a finite-distance function class.

[definition: Sup Norm] Let $E$ be a nonempty set and let $B(E,\mathbb{R})$ be the [vector space](/page/Vector%20Space) of bounded functions $f:E \to \mathbb{R}$. The sup norm is the map $\|\cdot\|_\infty:B(E,\mathbb{R})\to [0,\infty)$ defined by \begin{align*} \|f\|_\infty=\sup_{x \in E}|f(x)|. \end{align*} [/definition]

For real-valued bounded functions, uniform distance is just the sup norm of the difference: $d_\infty(f,g)=\|f-g\|_\infty$. This observation is useful only if it is made precise: it tells us when a quantified condition over points can be replaced by a single metric condition in a function space. The next theorem gives that translation, so later completeness arguments can use ordinary [Cauchy sequence](/page/Cauchy%20Sequence) language without losing the uniform quantifier.

[quotetheorem:9530]