Sequential Proofs of Uniform Continuity

Turning Uniform Continuity into a Sequence Argument

The definition of uniform continuity compares every pair of inputs that are sufficiently close, with one radius working throughout the domain. The Sequential Criterion for Uniform Continuity, established earlier in this course, gives a way to organize the same global question using sequences: it tests what happens to function values along pairs of inputs whose distances tend to zero. We will use that criterion as a tool, not re-prove it.

The sequential viewpoint is especially useful in two situations. To disprove uniform continuity, one can construct pairs that get arbitrarily close while their function values remain separated. To prove it, one can suppose that such pairs exist and use other information about the function to rule them out. This tutorial develops both approaches and uses them to prove two useful results.

A sequential proof does not require the sequences themselves to converge. In many important examples, both input sequences move farther and farther out in the domain; only the distance between the two inputs tends to zero. That is one reason this method detects a failure that pointwise continuity at each fixed input may not reveal.

Constructing a Sequential Witness

A direct way to disprove uniform continuity is to find two input sequences that approach each other while their outputs stay a definite distance apart. The calculations must verify both parts: the input distance tends to zero, and the output distance does not.

When searching for such a witness, it can help to decide first which two output values would make the failure clear, and then solve for inputs that produce those values. The example above chooses inputs whose squares differ by \(\pi\) and whose sine values are exactly opposite. This makes the output calculation exact rather than approximate.

Products of Bounded Uniformly Continuous Functions

Products are not automatically controlled by the uniform continuity of their factors: a difference of products includes both differences between function values and the sizes of those values. A uniform bound on each factor supplies the missing control. The following theorem packages this idea as a sequence argument.

Proof. Since \(f\) and \(g\) are bounded, choose finite constants \(M_f,M_g\geq0\) such that \(|f(x)|\leq M_f\) and \(|g(x)|\leq M_g\) for every \(x\in E\). Let \((x_n)\) and \((y_n)\) be any sequences in \(E\) with \(|x_n-y_n|\to0\). By uniform continuity of \(f\) and \(g\),

For every \(n\), add and subtract \(f(x_n)g(y_n)\) and apply the triangle inequality:

Both terms on the final line tend to zero, so the product difference tends to zero. The Sequential Criterion for Uniform Continuity now implies that \(fg\) is uniformly continuous on \(E\). \(\square\)

The boundedness hypothesis is a genuine part of the theorem, not a technical convenience. Without it, the displayed product estimate contains factors that may be arbitrarily large. Uniform continuity controls the differences of the factors, but it does not by itself bound their values across the domain.

Combining Uniform Continuity Across Separated Pieces

A domain need not be an interval, and a function need not have one formula everywhere. If a domain is a finite union of pieces that stay a positive distance apart, uniform continuity on each piece can be combined into uniform continuity on the whole domain. The separation prevents close inputs from coming from different pieces.

Proof. Use the Sequential Criterion for Uniform Continuity. Let \((x_n)\) and \((y_n)\) be sequences in \(E\) with \(|x_n-y_n|\to0\). We show that \(|f(x_n)-f(y_n)|\to0\). If this were false, there would be an \(\varepsilon_0>0\) and a subsequence of indices \(n_k\) such that

There are only finitely many ordered pairs of pieces. Hence, by passing to a further subsequence, we may assume that \(x_{n_k}\in E_i\) and \(y_{n_k}\in E_j\) for fixed indices \(i,j\). If \(i\ne j\), the separation assumption gives \(|x_{n_k}-y_{n_k}|\geq d_{ij}>0\) for every \(k\), contradicting \(|x_n-y_n|\to0\). Therefore \(i=j\). The restriction of \(f\) to \(E_i\) is uniformly continuous, so \(|x_{n_k}-y_{n_k}|\to0\) implies \(|f(x_{n_k})-f(y_{n_k})|\to0\), contradicting the lower bound \(\varepsilon_0\). Thus the output differences tend to zero for the original sequences. The sequential criterion proves that \(f\) is uniformly continuous on \(E\). \(\square\)

Reading the Sequential Argument Carefully

A useful proof habit is to distinguish what the input sequences are required to do from what they are not required to do. In the Sequential Criterion for Uniform Continuity, the essential input condition is that the pairwise distances tend to zero. The individual sequences need not converge, and they need not be bounded. The example with \(\sin(x^2)\) uses precisely this freedom.

For proofs, the most common structure is to take arbitrary close input sequences and use a quantitative estimate, a bound, or a finite selection argument to force their output differences to zero. The product theorem uses bounds on the factors; the separated-set theorem uses the fact that only finitely many pieces are available and that distinct pieces cannot contain arbitrarily close pairs.

For disproofs, one must provide a single pair of sequences that defeats the required conclusion. It is not enough to show that function values can be large, or that the function varies substantially somewhere. The inputs must get arbitrarily close while the output differences fail to approach zero. Conversely, checking only pairs whose inputs converge to a fixed point may miss failures caused by inputs escaping to infinity.

These arguments illustrate why sequential proofs are more than a change in notation. They make it possible to isolate exactly how uniformity can fail, and they allow global conclusions to follow from sequence estimates that are often easier to organize than a direct search for one common radius.