Uniform Convergence Preview

One Error Bound for the Whole Domain

A sequence of functions can approach its limit at every input without approaching it at the same rate across the domain. Pointwise convergence records what happens at each fixed input; uniform convergence requires one index to control the approximation error everywhere at once. The uniformly Cauchy criterion from the previous tutorial offers a way to recognize uniform convergence without first knowing the limit. Here we introduce uniform convergence itself and practice reading its quantifiers.

Throughout, \(E\subseteq\mathbb{R}\), \(f_n:E\to\mathbb{R}\) for each positive integer \(n\), and \(f:E\to\mathbb{R}\). The comparison is between \(f_n(x)\) and \(f(x)\), and the central question is whether one index works for every \(x\in E\).

The index \(N\) may depend on \(\varepsilon\), but not on \(x\). By contrast, \(f_n\) converges pointwise to \(f\) on \(E\) if, for each \(x\in E\) and each \(\varepsilon>0\), there exists an \(N\) such that \(n\geq N\) implies \(|f_n(x)-f(x)|<\varepsilon\). In pointwise convergence, \(N\) may depend on both \(\varepsilon\) and \(x\).

Uniform convergence always implies pointwise convergence: for any fixed input \(x\), the index that works over all of \(E\) works at that \(x\). The reverse implication can fail, because different inputs may require arbitrarily late indices. A useful way to make the difference visible is to look for a bound on the error that does not involve \(x\).

Reading the Error Across the Domain

For each index \(n\), the error at \(x\) is \(|f_n(x)-f(x)|\). When \(E\) is nonempty, take the supremum of these errors over \(E\), allowing that supremum to be infinite. This gives one number that measures the largest error, or the least upper bound of the errors, across the domain.

Proof. Suppose first that \(f_n\to f\) uniformly. Let \(\eta>0\). By the definition of uniform convergence, there is an \(N\) such that for every \(n\geq N\) and every \(x\in E\),

Thus \(\eta/2\) is an upper bound for all the errors at index \(n\), so \(0\leq e_n\leq\eta/2<\eta\) whenever \(n\geq N\). This proves that \(e_n\to0\).

Conversely, suppose \(e_n\to0\). Given \(\varepsilon>0\), choose \(N\) such that \(e_n<\varepsilon\) for every \(n\geq N\). For each such \(n\), the definition of supremum gives

The same \(N\) works across \(E\), so \(f_n\to f\) uniformly. \(\square\)

This characterization turns the quantifier condition into a practical estimating task: find a bound for the error that becomes small with \(n\) and does not depend on \(x\). If the supremum is difficult to calculate exactly, any suitable upper bound will do. For example, if \(|f_n(x)-f(x)|\leq b_n\) for all \(x\in E\) and \(b_n\to0\), then the theorem gives uniform convergence.

Worked Examples: Uniform and Pointwise Behavior

Two Consequences of Uniform Control

The finite-domain observation holds for every finite set, not just for the example above. Its proof makes clear why finiteness matters: among finitely many indices chosen at different inputs, one can choose a single largest index.

Proof. If \(E\) is empty, the uniform condition holds vacuously. Otherwise, list its elements as \(x_1,\ldots,x_k\), where \(k\) is a positive integer. Let \(\varepsilon>0\). At each \(x_i\), pointwise convergence supplies a positive integer \(N_i\) such that, for \(n\geq N_i\),

Because there are only finitely many \(N_i\), their maximum \(N=\max\{N_1,\ldots,N_k\}\) exists. If \(n\geq N\), then \(n\geq N_i\) for every \(i\). Consequently the error is less than \(\varepsilon\) at every \(x_i\), hence at every \(x\in E\). This is uniform convergence. \(\square\)

Uniform convergence also preserves pointwise inequalities when both sides have uniform limits. This fact is useful when a sequence of inequalities is easier to verify than a direct description of the limit.

Proof. Fix \(x\in E\). Uniform convergence implies pointwise convergence at this input, so \(f_n(x)\to f(x)\) and \(g_n(x)\to g(x)\). For each \(n\), the hypothesis gives \(f_n(x)\leq g_n(x)\). Suppose, for a contradiction, that \(f(x)>g(x)\). Set \(d=f(x)-g(x)>0\). For sufficiently large \(n\), convergence at \(x\) gives

It follows that

contradicting \(f_n(x)\leq g_n(x)\). Therefore \(f(x)\leq g(x)\). Since \(x\) was arbitrary, the inequality holds throughout \(E\). \(\square\)

How to Choose a Test

For a proposed uniform limit, begin by writing the error \(|f_n(x)-f(x)|\) and seeking an upper bound that tends to zero without depending on \(x\). The supremum characterization packages this strategy into a single numerical sequence of errors. If no such bound is apparent, a moving input \(x_n\) can test whether the errors remain large somewhere in the domain.

A common pitfall is to fix \(x\) too early. Showing that the error tends to zero for each fixed input proves pointwise convergence, but says nothing by itself about inputs that vary with \(n\). The power functions on \([0,1]\) illustrate this distinction: every fixed input below \(1\) eventually gives a small value, while inputs approaching \(1\) retain errors bounded away from zero. On a finite domain, in contrast, only finitely many pointwise thresholds must be combined, which is why pointwise convergence there is automatically uniform.