Pointwise Versus Uniform Convergence

Two Ways to Control the Error

Pointwise and uniform convergence both describe how functions approach a limit, but they ask for different kinds of control. In pointwise convergence, one fixes an input and then asks whether the numerical sequence of function values converges there. Uniform convergence asks for a single index that controls the error at every input in the domain. The difference is the order in which the input and the index are chosen.

Recall the definition of pointwise convergence from the tutorial “Pointwise Convergence.” The sequence \(f_n:E\to\mathbb{R}\) converges pointwise to \(f:E\to\mathbb{R}\) if, for each fixed \(x\in E\), the numerical sequence \(f_n(x)\) converges to \(f(x)\). In epsilon language, for every \(x\in E\) and every \(\varepsilon>0\), there is an index \(N\) such that the error is less than \(\varepsilon\) whenever \(n\geq N\). Here, \(N\) may depend on both \(x\) and \(\varepsilon\).

By contrast, uniform convergence requires that for every \(\varepsilon>0\), there is one \(N\) such that the error is less than \(\varepsilon\) for all \(n\geq N\) and all \(x\in E\). The index may depend on \(\varepsilon\), but it may not depend on \(x\). As established in “Uniform Convergence,” uniform convergence implies pointwise convergence. The converse is not generally true.

This distinction is not about whether the error becomes small at each individual input. Pointwise convergence guarantees that. It is about whether the indices needed at different inputs can be replaced by a single index for the whole domain.

Worked Examples

Pointwise Convergence on a Finite Domain

For a finite domain, only finitely many input-dependent indices need to be coordinated. Taking their maximum produces a single index that works throughout the domain. This argument does not extend directly to an infinite domain: there may be infinitely many required indices with no finite maximum.

Proof. Fix \(\varepsilon>0\). For each \(x\in E\), pointwise convergence gives a positive integer \(N_x\) such that

Because \(E\) is finite and nonempty, the finite collection of indices \(N_x\) has a maximum. Define

If \(n\geq N\), then \(n\geq N_x\) for every \(x\in E\). Therefore \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in E\). This is exactly the uniform convergence condition, so \(f_n\to f\) uniformly on \(E\). \(\square\)

Finiteness matters here. On an infinite domain, pointwise convergence still provides an index for each input, but those indices need not be bounded by one finite number. The first worked example demonstrates this failure: inputs \(x_n=1/n\) continue to exhibit a fixed error even though the error tends to zero at every fixed input.

A Moving-Input Test

A useful way to detect the difference between the two types of convergence is to let the input change as the index changes. Pointwise convergence tests \(f_n(x)\) at each fixed \(x\); uniform convergence must also control errors at inputs chosen differently for different indices. The following criterion makes that idea precise.

Proof. Suppose first that \(f_n\to f\) uniformly on \(E\). Let \((x_n)\) be any sequence of points in \(E\), and fix \(\varepsilon>0\). Uniform convergence supplies an index \(N\) such that, for every \(n\geq N\) and every \(x\in E\),

In particular, this estimate holds at \(x=x_n\), so \(|f_n(x_n)-f(x_n)|<\varepsilon\) for every \(n\geq N\). Thus the errors along the chosen sequence of inputs tend to zero.

Conversely, suppose that for every sequence \((x_n)\) in \(E\), the errors \(|f_n(x_n)-f(x_n)|\) tend to zero. Assume, for contradiction, that \(f_n\) does not converge uniformly to \(f\). By the definition of uniform convergence, there is an \(\varepsilon_0>0\) such that for every positive integer \(N\), there are an index \(n\geq N\) and a point \(x\in E\) with

We can use this failure repeatedly to choose strictly increasing indices \(n_1<n_2<\cdots\) and points \(y_k\in E\) such that

Define a sequence \((x_n)\) in \(E\) by setting \(x_{n_k}=y_k\) for each \(k\), and setting \(x_n=x_0\) at all other indices, where \(x_0\) is any fixed point of the nonempty set \(E\). Along the indices \(n_k\), the errors satisfy

Since \(n_k\to\infty\), these errors cannot tend to zero. This contradicts the assumed property for every sequence of inputs. Therefore \(f_n\to f\) uniformly on \(E\). \(\square\)

This criterion is especially effective for proving that convergence is not uniform: it is enough to find a sequence of inputs along which the errors fail to tend to zero. In the first worked example, the choice \(x_n=1/n\) gives error exactly \(1/2\) at every index. To prove uniform convergence by this criterion, however, one must control the errors for every possible sequence of inputs, which is often less direct than finding a bound independent of \(x\).

What the Comparison Does—and Does Not—Say

Uniform convergence is stronger than pointwise convergence, but the distinction depends on the domain. Restricting functions to a smaller set can change whether convergence is uniform. For example, the functions \(f_n(x)=nx/(1+n^2x^2)\) from the first example do not converge uniformly on \(\mathbb{R}\). On the smaller set \(E=\{0\}\), however, every function is zero, so convergence to zero is uniform there. A uniform estimate on a whole domain also works on each subset, since it remains valid for every point in that subset.

A common mistake is to show that, for each fixed \(x\), the error tends to zero and then treat the index as though it were independent of \(x\). Pointwise convergence permits the needed index to become arbitrarily large as the input changes. When testing uniform convergence, a moving input is therefore not a distraction: it is a direct way to check whether one index truly controls the entire domain.