Uniform Cauchy Sequences

Comparing Functions to Each Other

Uniform convergence measures how close each function is to a proposed limit. A related idea asks whether the functions become close to one another, even before a limit has been named. This is the function-sequence version of a Cauchy sequence: after some stage, any two functions in the sequence must differ by little at every point of the domain.

The word “uniform” is essential. For each chosen accuracy, one index must work for every input \(x\), and the same index must work for every pair of later functions. The Cauchy condition does not allow the index to depend on \(x\). This is stronger than requiring the numerical sequence of values at each fixed input to be Cauchy.

Definition: A sequence of functions \(f_n:E\to\mathbb{R}\) is uniformly Cauchy on \(E\) if, for every \(\varepsilon>0\), there is an integer \(N\) such that for all \(m,n\geq N\) and every \(x\in E\), \[ |f_m(x)-f_n(x)|<\varepsilon. \] The index \(N\) may depend on \(\varepsilon\), but not on \(x\), \(m\), or \(n\) once \(m,n\geq N\).

If all the functions belong to \(B(E)\), the same condition can be expressed using the supremum norm: \(\|f_m-f_n\|_\infty<\varepsilon\). In the metric \(d_\infty\) from “The Supremum Norm,” this says that the sequence is Cauchy with respect to that metric. The pointwise definition above is also useful when the functions themselves are unbounded, as long as their pairwise differences satisfy the stated bounds.

Uniform Cauchy Behavior and Pointwise Cauchy Behavior

A uniformly Cauchy sequence is pointwise Cauchy. To see why, fix any \(x\in E\) and any \(\varepsilon>0\). The index supplied by the uniform Cauchy condition works at this particular \(x\), so the real sequence \((f_n(x))\) is Cauchy. Completeness of \(\mathbb{R}\) then gives a pointwise limit at every \(x\). The Uniform Cauchy Criterion established earlier in the course goes further: for real-valued functions, uniform Cauchy behavior is equivalent to uniform convergence to a function. We will use the pairwise estimates here without re-proving that criterion.

Worked Example: A Uniformly Cauchy Sequence on the Half-Line

For \(n\geq1\), define \(f_n:[0,\infty)\to\mathbb{R}\) by \(f_n(x)=1/(n+x)\). Let \(m,n\geq N\). If \(m\leq n\), then for every \(x\geq0\),

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

If \(n\leq m\), the same calculation with the indices exchanged gives the bound \(1/N\). Given \(\varepsilon>0\), choose \(N>1/\varepsilon\). Then \(1/N<\varepsilon\), so \(|f_m(x)-f_n(x)|<\varepsilon\) for every \(m,n\geq N\) and every \(x\geq0\). Thus \((f_n)\) is uniformly Cauchy. The calculation supplies one bound for the entire domain, not merely a bound at each fixed \(x\).

The definition requires control of every sufficiently late pair, not just of consecutive terms. Consecutive differences can sometimes be useful, but a proof using them must also show that the sum of all intervening differences is small. Direct pairwise estimates avoid that extra step.

Worked Example: Pointwise Cauchy but Not Uniformly Cauchy

Define \(p_n:[0,1]\to\mathbb{R}\) by \(p_n(x)=x^n\). At each fixed \(x\in[0,1)\), \(x^n\to0\); at \(x=1\), the values are always \(1\). In particular, for every fixed \(x\), the numerical sequence \((p_n(x))\) converges and is therefore Cauchy.

Nevertheless, the function sequence is not uniformly Cauchy. Let \(N\geq1\) be any proposed Cauchy index, and choose \(m=N\) and \(n=2N\). Set \(x=(1/2)^{1/N}\), which lies in \([0,1]\). Then \(x^N=1/2\) and \(x^{2N}=1/4\), so

|p_N(x)-p_{2N}(x)| =\left|\frac{1}{2}-\frac{1}{4}\right| =\frac{1}{4}.

Thus the uniform Cauchy condition fails for \(\varepsilon=1/4\): no matter how large \(N\) is, there are indices \(m,n\geq N\) and an input \(x\) for which the difference is not less than \(1/4\). The input used to reveal the failure changes with \(N\), which is precisely what pointwise Cauchy behavior does not rule out.

Supremum Norms of Bounded Uniform Cauchy Sequences

When every \(f_n\) is bounded, uniform Cauchy behavior places a common bound on the sizes of all the functions. The argument resembles the boundedness result for uniform limits, but it uses pairwise closeness rather than closeness to a limit. One fixed late function provides the reference point for all later functions.

Theorem (Bounded Supremum Norms for Uniformly Cauchy Sequences): Let \(E\) be nonempty, and suppose \(f_n\in B(E)\) for every \(n\). If \((f_n)\) is uniformly Cauchy on \(E\), then the real sequence \((\|f_n\|_\infty)\) is bounded.

Proof. Apply the uniform Cauchy condition with \(\varepsilon=1\). There is an integer \(N\) such that for every \(n\geq N\) and every \(x\in E\),

$$ |f_n(x)-f_N(x)|<1. $$

The triangle inequality gives \(|f_n(x)|\leq |f_N(x)|+|f_n(x)-f_N(x)|<|f_N(x)|+1\). Taking suprema over \(x\in E\) yields

\|f_n\|_\infty\leq\|f_N\|_\infty+1 \qquad(n\geq N).

This is a finite bound because \(f_N\) is bounded. There are only finitely many earlier indices \(n<N\), and each \(\|f_n\|_\infty\) is finite by hypothesis. The maximum of those finitely many values and \(\|f_N\|_\infty+1\) bounds the entire sequence of norms. \(\square\)

The assumption that each function is bounded matters: it makes each norm finite and lets the fixed reference function give a finite bound. Uniform Cauchy behavior alone does not imply that the functions are bounded.

Worked Example: Uniformly Cauchy Functions Need Not Be Bounded

On \(E=\mathbb{R}\), define \(q_n(x)=x+1/n\). For any \(m,n\geq N\) and \(x\in\mathbb{R}\),

|q_m(x)-q_n(x)| =\left|\frac{1}{m}-\frac{1}{n}\right| \leq\frac{1}{m}+\frac{1}{n} \leq\frac{2}{N}.

Given \(\varepsilon>0\), choose \(N>2/\varepsilon\); the last bound is then less than \(\varepsilon\). Hence \((q_n)\) is uniformly Cauchy. However, each \(q_n\) is unbounded on \(\mathbb{R}\), since \(q_n(x)=x+1/n\) increases without bound as \(x\) increases. The boundedness theorem does not apply, because its hypothesis \(q_n\in B(E)\) fails.

Products of Bounded Uniform Cauchy Sequences

Products require more than pairwise closeness. In the identity for the difference of two products, each difference is multiplied by a function value. If those values have no common bound, a small pairwise difference need not produce a small product difference. For bounded functions, the previous theorem supplies the common bounds needed for a product estimate.

Theorem (Products of Bounded Uniform Cauchy Sequences): Let \(E\) be nonempty. Suppose \(f_n,g_n\in B(E)\) for every \(n\), and suppose both \((f_n)\) and \((g_n)\) are uniformly Cauchy on \(E\). Then \((f_ng_n)\) is uniformly Cauchy on \(E\).

Proof. By the theorem on bounded supremum norms, there are finite constants \(M,K\geq0\) such that \(\|f_n\|_\infty\leq M\) and \(\|g_n\|_\infty\leq K\) for every \(n\). Fix \(\varepsilon>0\), and set

\delta=\frac{\varepsilon}{2(M+K+1)}.

This is positive. Since both sequences are uniformly Cauchy, there is an index \(N\) such that whenever \(m,n\geq N\), for every \(x\in E\),

|f_m(x)-f_n(x)|<\delta \qquad\text{and}\qquad |g_m(x)-g_n(x)|<\delta.

For such \(m,n,x\), add and subtract \(f_m(x)g_n(x)\), then use the triangle inequality and the uniform bounds:

\begin{aligned} |f_m(x)g_m(x)-f_n(x)g_n(x)| &\leq |f_m(x)|\,|g_m(x)-g_n(x)| +|g_n(x)|\,|f_m(x)-f_n(x)|\\ &\leq (M+K)\delta\\ &=\frac{\varepsilon(M+K)}{2(M+K+1)} &<\varepsilon. \end{aligned}

The last inequality holds also when \(M+K=0\), because then its left-hand side is \(0\), which is less than \(\varepsilon\). The bound is independent of \(x\), so the product sequence is uniformly Cauchy. \(\square\)

Worked Example: Applying the Product Theorem

On \([0,1]\), let \(f_n(x)=1+x/n\) and \(g_n(x)=2-x/n\). For \(m,n\geq N\),

|f_m(x)-f_n(x)| =x\left|\frac{1}{m}-\frac{1}{n}\right| \leq\frac{1}{N}, \qquad |g_m(x)-g_n(x)| =x\left|\frac{1}{m}-\frac{1}{n}\right| \leq\frac{1}{N}.

Thus both sequences are uniformly Cauchy. Also, \(1\leq f_n(x)\leq2\) and \(1\leq g_n(x)\leq2\) for every \(x\in[0,1]\) and \(n\geq1\). The product theorem therefore shows that \((f_ng_n)\) is uniformly Cauchy. Directly, the products are

f_n(x)g_n(x) =\left(1+\frac{x}{n}\right)\left(2-\frac{x}{n}\right) =2+\frac{x}{n}-\frac{x^2}{n^2}.

For \(m,n\geq N\), subtracting these expressions and using \(0\leq x\leq1\) gives \[ |f_m(x)g_m(x)-f_n(x)g_n(x)| \leq \left|\frac1m-\frac1n\right| +\left|\frac1{m^2}-\frac1{n^2}\right| \leq\frac{1}{N}+\frac{1}{N^2}. \] This bound tends to zero as \(N\) increases and confirms the uniform Cauchy behavior.

Why Product Bounds Cannot Be Omitted

The boundedness condition in the product theorem is not merely a technical convenience. On \(\mathbb{R}\), take \(f_n=g_n=q_n\), where \(q_n(x)=x+1/n\). As shown above, \((q_n)\) is uniformly Cauchy. But its product sequence is

q_n(x)^2=x^2+\frac{2x}{n}+\frac{1}{n^2}.

For any positive integer \(N\), compare indices \(N\) and \(2N\). Their product difference is

q_N(x)^2-q_{2N}(x)^2 =\frac{x}{N}+\frac{3}{4N^2}.

For each fixed \(N\), this expression is unbounded as \(x\) ranges over \(\mathbb{R}\). In particular, it cannot be less than a fixed positive \(\varepsilon\) for every \(x\). Thus the product sequence is not uniformly Cauchy. Small uniform differences between unbounded functions do not control products unless there is also a uniform bound on their sizes.

When working with a uniform Cauchy sequence, first identify what the definition actually controls: pairwise differences, uniformly over the domain. It implies pointwise Cauchy behavior, and for bounded functions it also yields a common bound on supremum norms. That bound is what makes product estimates possible. The Uniform Cauchy Criterion from earlier in the course then connects the pairwise condition to uniform convergence, but the pairwise estimates remain useful in their own right.

Takeaway: Uniform Cauchy behavior requires one index to control every late pair of functions at every input. For bounded function sequences, it also bounds all supremum norms and is stable under products of two such sequences.

Check Your Understanding

Use the uniform Cauchy definition and the results proved here to answer the following questions.

In the uniform Cauchy definition, which choices may the index \(N\) depend on, and which may it not depend on?
Why is every uniformly Cauchy sequence pointwise Cauchy?
For \(p_n(x)=x^n\) on \([0,1]\), what choice of \(x\) shows that the sequence is not uniformly Cauchy?
Where does boundedness enter the proof that products of bounded uniformly Cauchy sequences are uniformly Cauchy?
Why does \(q_n(x)=x+1/n\) show that the product conclusion can fail without a common bound on function values?

What You'll Learn