The Cauchy Criterion for Uniform Convergence

The uniform Cauchy criterion is the engine inside the Weierstrass M-Test: to show $\sum f_k$ converges uniformly, one bounds $|f_k(x)|\le M_k$ with $\sum M_k<\infty$, and the tail estimate $\sum_{k=n+1}^m M_k \to 0$ is exactly what makes the partial sums uniformly Cauchy — precisely the pattern used in Example 1 above. It also underlies why term-by-term integration and differentiation of uniformly convergent series are valid, a cornerstone of real analysis courses covering power series and Fourier series. In functional analysis, the corollary that $B(S,\mathbb{R})$ (and its close relatives $C(S,\mathbb{R})$, the continuous bounded functions, and $\ell^\infty$, bounded sequences) are complete under the sup-norm is the first example of a Banach space, the setting for the entire theory of normed vector spaces used throughout PDE theory, numerical analysis, and quantum mechanics (where state spaces are complete normed or inner-product spaces). The existence-and-uniqueness theorem for ordinary differential equations (Picard–Lindelöf) constructs its solution as the uniform limit of a Cauchy sequence of approximations, verified using exactly this criterion. In numerical analysis, error bounds for approximating a function by a sequence (splines, Fourier partial sums, neural-network approximants under uniform norm) are typically phrased as uniform Cauchy estimates, since one rarely knows the true limiting function in closed form — mirroring Example 1 exactly. In cryptography and complexity theory, uniform approximation bounds for functions (e.g. in constructing pseudorandom generators or in analyzing Boolean function approximation) also implicitly rely on completeness of the relevant function space to guarantee that Cauchy approximation schemes actually converge to a genuine limiting object.

:::quiz[In Example 2, $g_n(x) = x^n(1-x^n)$ on $[0,1]$, what is $\sup_{x\in[0,1]} g_n(x)$?]

• $0$
• $1/4$ ✓
• $1$
• It depends on $n$ and grows without bound

Ask the tutor: How does the uniform Cauchy criterion get used inside the proof of the Weierstrass M-Test, step by step?

Think of each function $f_n$ as a point in a very large space — the space of all bounded functions $S\to\mathbb{R}$ — where the "distance" between two functions $f$ and $g$ is

the largest vertical gap between their graphs. A sequence of functions converging uniformly is just a sequence of points converging in this metric: eventually every $f_n$ lies within a thin horizontal "tube" of radius $\varepsilon$ around the limit function's graph, uniformly across the whole domain. The uniform Cauchy condition asks something more modest and checkable: that the functions eventually get close to each other (all graphs squeezed into a tube of width $\varepsilon$ around one another), not close to some already-known target. The proof that these two notions coincide mirrors exactly why $\mathbb{R}$ is complete: at each fixed point $x$, $(f_n(x))$ is an ordinary Cauchy sequence of real numbers, so it converges by completeness of $\mathbb{R}$; this manufactures a candidate limit function $f(x) = \lim_n f_n(x)$ pointwise. The real work is then showing the convergence is not just pointwise but uniform — that the same $N$ works for every $x$ at once, which follows by letting $m\to\infty$ inside the uniform estimate.

Example 1 (certifying uniform convergence without knowing the limit in closed form). Define, for $x\in[0,2]$,

This partial-sum sequence has no elementary closed form for its limit, yet we can prove uniform convergence on $[0,2]$ via the uniform Cauchy condition, and this is exactly the mechanism underlying the Weierstrass M-Test. For $m>n$,

For $x\in[0,2]$, $|x^k|\le 2^k$ and $|\cos(k!\,x)|\le 1$, so each term is bounded:

Hence, for all $x\in [0,2]$ simultaneously,

This bound is independent of $x$ and of $m$ (for $m>n$), and it tends to $0$ as $n\to\infty$ since $2/3<1$. Given $\varepsilon>0$, choose $N$ so large that $3(2/3)^{N+1}<\varepsilon$ (possible since $(2/3)^{n}\to 0$). Then for all $m>n\ge N$ (and trivially for $m=n$, where the difference is $0$),

So $(f_n)$ is uniformly Cauchy on $[0,2]$. By the theorem, $(f_n)$ converges uniformly on $[0,2]$ to some function $f$ — even though we never wrote $f$ in closed form. This is the general pattern of the Weierstrass M-test: bounding each term by a constant $M_k$ with $\sum M_k <\infty$ makes the tail of the series a uniformly small Cauchy tail.

Example 2 (a sequence that is pointwise Cauchy but fails to be uniformly Cauchy, hence fails to converge uniformly). Define $g_n:[0,1]\to\mathbb{R}$ by

First, pointwise: for each fixed $x\in[0,1]$, $x^n \to 0$ if $0\le x<1$ and $x^n=1$ if $x=1$. So $g_n(x)\to 0$ for $x\in[0,1)$ (since $x^n(1-x^n)\to 0\cdot 1 = 0$) and $g_n(1) = 1\cdot(1-1)=0$. Thus $g_n \to 0$ pointwise on all of $[0,1]$, and in particular $(g_n(x))$ is a Cauchy sequence of real numbers for each fixed $x$ (being convergent). Now test whether $(g_n)$ is uniformly Cauchy, equivalently (by the theorem, contrapositive) whether it converges uniformly to the pointwise limit $g\equiv 0$. Compute $\sup_{x\in[0,1]} g_n(x)$. Let $t = x^n \in [0,1]$ as $x$ ranges over $[0,1]$; then $g_n(x) = t(1-t) = t-t^2$, maximized at $t=1/2$ giving value $1/4$. Since $t=x^n$ does attain the value $1/2$ for $x = (1/2)^{1/n}\in[0,1]$, the supremum $1/4$ is attained:

So $\sup_{x\in[0,1]}|g_n(x) - g_{2n}(x)|$ need not even be checked against a limit — we directly see $\sup_x|g_n(x) - 0| = 1/4$ for every $n$, so $g_n \not\to 0$ uniformly (the sup-distance to the pointwise limit never shrinks below $1/4$). By the contrapositive of the theorem (uniform convergence $\Rightarrow$ uniformly Cauchy), $(g_n)$ cannot be uniformly Cauchy either — let's verify this directly to see the failure concretely. Take $m=2n$: at $x_n = (1/2)^{1/n}$ we have $g_n(x_n) = 1/4$. What is $g_{2n}(x_n)$? Here $x_n^{2n} = (x_n^n)^2 = (1/2)^2 = 1/4$, so $g_{2n}(x_n) = \tfrac14(1-\tfrac14) = \tfrac{3}{16}$. Then

a fixed positive number not depending on $n$. Hence $\sup_{x\in[0,1]}|g_n(x)-g_{2n}(x)| \ge 1/16$ for every $n$, so the uniform Cauchy condition fails (taking $\varepsilon = 1/16$, no $N$ works, since $n$ and $2n$ always violate it). This matches the theorem exactly: pointwise-Cauchy-everywhere is not enough; uniform convergence fails, and correspondingly the uniform Cauchy condition fails.