Uniformly Cauchy Function Sequences

When Function Values Become Uniformly Close

For a sequence of real numbers, the Cauchy condition says that sufficiently late terms are close to one another. For a sequence of functions, there are two different ways to impose that idea. We could require closeness separately at each input, allowing the stage at which the values become close to depend on the input. Or we could require one stage to work across the entire domain. The second, stronger condition is the uniform Cauchy condition.

This distinction matters because uniform convergence also requires one index to control the approximation error at every point in the domain. The uniform Cauchy condition detects that global behavior without first knowing the function to which the sequence might converge. In this tutorial, we prove that for real-valued functions the two ideas coincide: a sequence is uniformly Cauchy exactly when it converges uniformly to a real-valued function.

Definition: Let \(E\subseteq\mathbb{R}\), and let \(f_n:E\to\mathbb{R}\) for each positive integer \(n\). The sequence \((f_n)\) is uniformly Cauchy on \(E\) if, for every \(\varepsilon>0\), there exists a positive integer \(N\) such that for all \(m,n\geq N\) and every \(x\in E\), \(|f_m(x)-f_n(x)|<\varepsilon\).

The order of the quantifiers is essential. The index \(N\) may depend on \(\varepsilon\), but it cannot depend on \(x\). For comparison, saying that \((f_n(x))\) is Cauchy for every fixed \(x\in E\) allows the required \(N\) to depend on \(x\). That is a pointwise Cauchy condition, and it is weaker.

At each fixed \(x\), uniform Cauchy behavior certainly implies that the real sequence \((f_n(x))\) is Cauchy. Completeness of \(\mathbb{R}\) then supplies a limit value at that \(x\). The key issue is whether the convergence to all those pointwise limit values is uniform. The first theorem answers yes.

The Uniform Cauchy Criterion

Theorem (Uniform Cauchy Criterion): Let \(E\subseteq\mathbb{R}\), and let \(f_n:E\to\mathbb{R}\) for each positive integer \(n\). The sequence \((f_n)\) converges uniformly on \(E\) to a function \(f:E\to\mathbb{R}\) if and only if \((f_n)\) is uniformly Cauchy on \(E\).

Proof. If \(E\) is empty, there is a unique function from \(E\) to \(\mathbb{R}\), and both uniform convergence and the uniformly Cauchy condition hold vacuously. Suppose henceforth that \(E\) is nonempty.

First suppose \((f_n)\) converges uniformly to \(f:E\to\mathbb{R}\). Let \(\varepsilon>0\). By uniform convergence, there is an \(N\) such that for every \(k\geq N\) and every \(x\in E\),

|f_k(x)-f(x)|<\frac{\varepsilon}{2}.

For any \(m,n\geq N\) and \(x\in E\), the triangle inequality gives

|f_m(x)-f_n(x)| \leq |f_m(x)-f(x)|+|f(x)-f_n(x)| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

The same \(N\) works for every \(x\), so \((f_n)\) is uniformly Cauchy on \(E\).

Conversely, suppose \((f_n)\) is uniformly Cauchy on \(E\). Fix any \(x\in E\). Given \(\varepsilon>0\), the uniform Cauchy condition supplies an \(N\) such that \(m,n\geq N\) implies \(|f_m(x)-f_n(x)|<\varepsilon\). Thus the real sequence \((f_n(x))\) is Cauchy. By completeness of \(\mathbb{R}\), it converges to a real number. Define \(f(x)\) to be this limit. Doing this for each \(x\in E\) defines a function \(f:E\to\mathbb{R}\).

It remains to prove that the convergence is uniform. Let \(\varepsilon>0\). Apply the uniformly Cauchy condition with \(\varepsilon/2\). There exists \(N\) such that for all \(m,n\geq N\) and every \(x\in E\),

|f_m(x)-f_n(x)|<\frac{\varepsilon}{2}.

Fix \(n\geq N\) and \(x\in E\). In this inequality, let \(m\) tend to infinity. Since \(f_m(x)\to f(x)\), continuity of the absolute value gives

|f(x)-f_n(x)|\leq\frac{\varepsilon}{2}<\varepsilon.

Although taking the limit may change the strict bound to a weak one, the final bound is still strictly less than \(\varepsilon\). The estimate holds for every \(x\in E\) and every \(n\geq N\), with the same \(N\). Therefore \(f_n\) converges uniformly to \(f\). This proves both directions. \(\square\)

The use of completeness is precisely where the real numbers enter the argument: it guarantees a value \(f(x)\in\mathbb{R}\) at each input. The uniform part of the proof then comes from using the same Cauchy index \(N\) at every input, before taking the pointwise limit.

Worked Applications

Worked Example: A Uniformly Cauchy Sequence with Limit Zero

For each positive integer \(n\), define \(f_n:[0,1]\to\mathbb{R}\) by \(f_n(x)=x^n/n\). Since \(0\leq x\leq1\), we have \(0\leq x^n\leq1\), and hence

|f_n(x)|=\frac{x^n}{n}\leq\frac{1}{n} \qquad(0\leq x\leq1).

For any positive integers \(m,n\) and any \(x\in[0,1]\), it follows that

|f_m(x)-f_n(x)| \leq |f_m(x)|+|f_n(x)| \leq \frac{1}{m}+\frac{1}{n}.

Given \(\varepsilon>0\), choose a positive integer \(N\) such that \(2/N<\varepsilon\). If \(m,n\geq N\), then

|f_m(x)-f_n(x)| \leq\frac{1}{m}+\frac{1}{n} \leq\frac{2}{N}<\varepsilon \qquad(0\leq x\leq1).

Thus \((f_n)\) is uniformly Cauchy. In fact, the first estimate also gives \(|f_n(x)-0|\leq1/n\) everywhere on \([0,1]\), so \(f_n\) converges uniformly to the zero function. The example shows how a simple bound, independent of \(x\), can establish uniform Cauchy behavior directly.

Worked Example: Pointwise Cauchy but Not Uniformly Cauchy

For each positive integer \(n\), define \(g_n:\mathbb{R}\to\mathbb{R}\) by

g_n(x)=\max\{1-n|x|,0\}.

At \(x=0\), \(g_n(0)=1\) for every \(n\). If \(x\ne0\), then \(n|x|\geq1\) whenever \(n\geq1/|x|\), and for such \(n\) we have \(g_n(x)=0\). Thus, at each fixed \(x\), the sequence \(g_n(x)\) is eventually constant. In particular, it is Cauchy at every fixed input.

However, the sequence is not uniformly Cauchy. Let \(N\) be any positive integer, choose \(n\geq N\), and set \(x=1/(2n)\). Direct substitution gives

g_n\left(\frac{1}{2n}\right) =\max\left\{1-n\frac{1}{2n},0\right\} =\frac12,

whereas

g_{2n}\left(\frac{1}{2n}\right) =\max\left\{1-2n\frac{1}{2n},0\right\} =0.

Consequently, \(|g_n(1/(2n))-g_{2n}(1/(2n))|=1/2\), even though both indices are at least \(N\). Taking, for example, \(\varepsilon=1/4\), no choice of \(N\) can meet the uniformly Cauchy condition. The input exposing the difference changes with \(n\); that is exactly the kind of variation the pointwise Cauchy condition does not control.

Worked Example: A Tail Estimate on an Unbounded Domain

For integers \(n\geq2\), define \(h_n:[0,\infty)\to\mathbb{R}\) by

h_n(x)=\sum_{k=2}^{n}\frac{x}{k^2(k+x)}.

For every \(x\geq0\) and \(k\geq2\), the denominator \(k+x\) is positive and \(0\leq x/(k+x)\leq1\). Therefore each summand satisfies

0\leq\frac{x}{k^2(k+x)} =\frac{1}{k^2}\frac{x}{k+x} \leq\frac{1}{k^2}.

Suppose \(m>n\geq2\). The difference \(h_m(x)-h_n(x)\) is the sum of the terms with indices from \(n+1\) through \(m\), so

0\leq h_m(x)-h_n(x) \leq\sum_{k=n+1}^{m}\frac{1}{k^2}.

For \(k\geq2\), we have \(k^2\geq k(k-1)\), and hence \(1/k^2\leq1/[k(k-1)]\). Also,

\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.

The resulting sum telescopes, giving

|h_m(x)-h_n(x)| \leq\sum_{k=n+1}^{m}\frac{1}{k^2} \leq\sum_{k=n+1}^{m}\left(\frac{1}{k-1}-\frac{1}{k}\right) =\frac{1}{n}-\frac{1}{m} <\frac{1}{n}.

If \(n>m\), interchange the indices in the absolute difference; if \(m=n\), the difference is zero. Given \(\varepsilon>0\), choose \(N\geq2\) such that \(1/N<\varepsilon\). Then for all \(m,n\geq N\) and \(x\geq0\), the difference is less than or equal to \(1/\min\{m,n\}\leq1/N<\varepsilon\). Thus \((h_n)\) is uniformly Cauchy on the unbounded domain \([0,\infty)\), and the Uniform Cauchy Criterion guarantees a uniform limit there.

A Consequence for Uniform Continuity

The criterion is useful not only for identifying a limit. Once uniform convergence has been obtained from the uniformly Cauchy condition, results about uniform limits can be applied. Earlier in this course, the theorem “Uniform Limits Preserve Continuity” established that a uniform limit of continuous functions is continuous. For uniform continuity, one can make a related argument directly.

Theorem: Let \(E\subseteq\mathbb{R}\), and let \(f_n:E\to\mathbb{R}\) be uniformly continuous for every \(n\). If \((f_n)\) is uniformly Cauchy on \(E\), then its uniform limit \(f:E\to\mathbb{R}\) is uniformly continuous on \(E\).

Proof. By the Uniform Cauchy Criterion, there is a function \(f:E\to\mathbb{R}\) such that \(f_n\) converges uniformly to \(f\). Let \(\varepsilon>0\). Uniform convergence gives an index \(n\) such that for every \(x\in E\),

|f(x)-f_n(x)|<\frac{\varepsilon}{3}.

The function \(f_n\) is uniformly continuous on \(E\). Therefore there exists a \(\delta>0\) such that for \(x,y\in E\), \(|x-y|<\delta\) implies

|f_n(x)-f_n(y)|<\frac{\varepsilon}{3}.

For any such \(x,y\), the triangle inequality yields

\begin{aligned} |f(x)-f(y)| &\leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &=\varepsilon. \end{aligned}

This \(\delta\) works for all \(x,y\in E\), so \(f\) is uniformly continuous on \(E\). No common \(\delta\) for the whole sequence \((f_n)\) was assumed: after choosing one sufficiently accurate \(f_n\), its own uniform continuity supplies the needed \(\delta\). \(\square\)

What to Keep in View

The Uniform Cauchy Criterion translates a question about an unknown limit into a question about pairs of known functions. To prove uniform convergence, it can be easier to estimate \(|f_m(x)-f_n(x)|\) than to identify the limit and estimate its distance from \(f_n(x)\). A bound on the tail that is independent of \(x\) is especially useful.

The main pitfall is allowing the index \(N\) to depend on \(x\). That proves only that the values form a Cauchy sequence at each separate input. The moving-spike example shows why this does not suffice: at every fixed point the sequence eventually settles down, while an input that changes with the indices continues to reveal a fixed-size difference. For uniform Cauchy behavior, one \(N\) must control every input at once.

Check Your Understanding

Use the definition, estimates, and proof in this tutorial to answer the following questions.

Which quantifier in the definition of uniform Cauchy behavior distinguishes it from being Cauchy at each fixed input?
Where is completeness of \(\mathbb{R}\) used in the proof of the Uniform Cauchy Criterion?
Why does the reverse direction of the criterion use \(\varepsilon/2\) in the Cauchy estimate?
For \(g_n(x)=\max\{1-n|x|,0\}\), what input gives a difference of \(1/2\) between \(g_n\) and \(g_{2n}\)?
In the tail estimate for \(h_n\), why is the bound independent of \(x\)?
Does the uniform-continuity consequence require one distance \(\delta\) to work for every \(f_n\)?