Proof of the Uniform Cauchy Criterion

What the Criterion Must Establish

The Uniform Cauchy Criterion, stated earlier in this course, says that a sequence of real-valued functions on a nonempty set is uniformly convergent if and only if it is uniformly Cauchy. This tutorial proves both directions. The central issue in the converse is existence: pairwise closeness in late tails must produce an actual function that the sequence approaches.

For real numbers, the ordinary Cauchy criterion gives that existence because the real numbers are complete: every Cauchy sequence of real numbers converges to a real limit. For functions, we apply that fact separately at each input to construct a candidate limit. The uniform part then comes from using the same tail index for every input.

The order of these requirements matters. First choose an accuracy \(\varepsilon\); then choose one tail index \(N\); after that, the estimate must work for every pair of indices in the tail and every input in the domain. If the required index changes with \(x\), the condition is only pointwise Cauchy behavior, not uniform Cauchy behavior.

Uniform Convergence Implies Uniform Cauchy Behavior

Suppose \(f_n\) converges uniformly to a function \(f\). Then every sufficiently late function is uniformly close to \(f\). Two late functions are close to one another by the triangle inequality, since each is close to the same \(f\).

Proof. Let \(\varepsilon>0\). Uniform convergence gives an integer \(N\) such that for every \(n\geq N\) and every \(x\in E\),

If \(m,n\geq N\), then for every \(x\in E\), the triangle inequality gives

The same \(N\) works for every \(x\), so the sequence is uniformly Cauchy. \(\square\)

This direction uses no completeness assumption: it only uses the existence of a uniform limit and the triangle inequality. The converse is where completeness of \(\mathbb{R}\) becomes essential.

Constructing the Limit from the Uniform Cauchy Condition

Now suppose \((f_n)\) is uniformly Cauchy on \(E\). Fix any \(x\in E\). The uniform Cauchy condition implies that the real sequence \(f_n(x)\) is Cauchy: for each positive accuracy, the same tail index works at this particular input. Completeness of \(\mathbb{R}\) therefore gives a real limit. Define \(f(x)\) to be that limit.

This construction defines a function \(f:E\to\mathbb{R}\), because for each \(x\) the limit exists and is unique. To prove uniform convergence, we must do more than know that each \(f_n(x)\) tends to \(f(x)\). We must use one index that controls the error simultaneously at every input.

Proof. The forward direction is the proposition just proved. For the converse, assume \((f_n)\) is uniformly Cauchy on \(E\).

Fix \(x\in E\). Given \(\varepsilon>0\), choose \(N\) from the uniform Cauchy condition. For \(m,n\geq N\), we then have

Thus \((f_n(x))\) is a Cauchy sequence of real numbers. By completeness of \(\mathbb{R}\), it converges. Define

It remains to prove uniform convergence to \(f\). Let \(\varepsilon>0\). Since \((f_n)\) is uniformly Cauchy, there is one \(N\) such that for every \(m,n\geq N\) and every \(x\in E\),

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

The index \(N\) was chosen independently of \(x\), and the last estimate holds for every \(n\geq N\) and every \(x\in E\). This is uniform convergence of \(f_n\) to \(f\). \(\square\)

The limit step may change the strict pairwise inequality into a non-strict inequality: a sequence of numbers all less than \(\varepsilon/2\) can converge to \(\varepsilon/2\). That causes no difficulty, because \(\varepsilon/2\) is still strictly less than the requested accuracy \(\varepsilon\).

Worked Examples Using the Criterion

A Consequence: Bounded Functions Form a Complete Space

The criterion also proves a useful completeness result for the supremum norm. Recall that \(B(E)\) is the set of bounded real-valued functions on \(E\), with \(\|u\|_\infty=\sup_{x\in E}|u(x)|\). A Cauchy sequence in this norm is exactly a uniformly Cauchy sequence, because \[ \|f_m-f_n\|_\infty<\varepsilon \] means that \(|f_m(x)-f_n(x)|<\varepsilon\) for every \(x\in E\).

Proof. Let \((f_n)\) be Cauchy in \(B(E)\). It is uniformly Cauchy, so the Uniform Cauchy Criterion gives a function \(f:E\to\mathbb{R}\) such that \(f_n\to f\) uniformly. We must check that \(f\) is bounded; uniform convergence alone does not make this automatic without an additional argument.

Choose an index \(N\) such that \[ |f_n(x)-f_N(x)|<1 \] for every \(n\geq N\) and every \(x\in E\). Fix \(x\) and let \(n\) tend to infinity. Since \(f_n(x)\to f(x)\), continuity of absolute value gives

Therefore, for every \(x\in E\),

The right-hand side is finite because \(f_N\in B(E)\). Hence \(f\in B(E)\). Uniform convergence is convergence in the supremum norm, so \((f_n)\) converges in \(B(E)\), as required. \(\square\)

The boundedness check is an important final step: the pointwise limit produced by the criterion is initially just a real-valued function. To conclude convergence within \(B(E)\), we must show that this limit also belongs to \(B(E)\).

Reading the Proof as a Quantifier Argument

The key distinction between the two halves of the criterion is where the limit comes from. If uniform convergence is already known, the triangle inequality compares two terms through that limit. If uniform Cauchy behavior is known instead, completeness of \(\mathbb{R}\) first produces a limit at each point, and the uniform tail estimate then controls the error to it.

A common pitfall is to prove only that \(f_n(x)\) is Cauchy for each fixed \(x\). That establishes pointwise convergence, but says nothing by itself about one index working over the whole domain. In the converse proof, the same index \(N\) controls all inputs because it was chosen from the uniform Cauchy condition before the input \(x\) was fixed. This is the step that upgrades pointwise convergence to uniform convergence.