Two Kinds of Convergence
For a sequence of functions, convergence can mean that the values settle down separately at each point, or that one error bound works across the whole domain at once. These conditions are related, but they are not interchangeable. The distinction is encoded in the order of the quantifiers: in pointwise convergence, the stage at which an error becomes small may depend on the point; in uniform convergence, it may not.
Let \(X\) be a set and let \(f_n,f:X\to\mathbb{R}\) be functions for each \(n\in\mathbb{N}\). The topology on \(\mathbb{R}\) is the standard topology. Convergence here concerns the numerical values \(f_n(x)\) and \(f(x)\); no topology on \(X\) is needed for these definitions.
The difference is not whether an error eventually becomes small at each point: both definitions require that. The difference is whether a single index \(N\) can guarantee the error bound at every point. If \(X\) is empty, both definitions hold vacuously for the unique function on \(X\). In the examples below the domains are nonempty.
Uniform Convergence Implies Pointwise Convergence
Proof. If \(X\) is empty, pointwise convergence holds vacuously. Otherwise, let \(x\in X\) and \(\varepsilon>0\) be arbitrary. Uniform convergence gives an \(N\in\mathbb{N}\) such that for every \(n\ge N\) and every \(y\in X\), \(|f_n(y)-f(y)|<\varepsilon\). In particular, this inequality holds when \(y=x\). Since \(x\) and \(\varepsilon\) were arbitrary, the definition of pointwise convergence is satisfied. \(\square\)
This direction follows directly from the quantifiers: a bound valid for every point is valid at any one chosen point. The converse would need a single \(N\) to work across all points, and pointwise convergence does not promise that.
Proof. If \(X=\varnothing\), uniform convergence is vacuous. Suppose \(X\) is finite and nonempty, and fix \(\varepsilon>0\). For each \(x\in X\), pointwise convergence supplies an index \(N_x\) such that \(n\ge N_x\) implies \(|f_n(x)-f(x)|<\varepsilon\). A finite nonempty collection of natural numbers has a maximum, so set \(N=\max_{x\in X}N_x\). If \(n\ge N\), then \(n\ge N_x\) for every \(x\in X\), and hence \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in X\). This is uniform convergence. \(\square\)
Finiteness matters in this argument: it lets us choose one index at least as large as every point-dependent index. On an infinite domain, there need not be a finite maximum of all the required indices.
Uniform Error and the Supremum
When the domain is nonempty, it is often useful to measure the largest error at stage \(n\). Define the extended real number \(E_n=\sup_{x\in X}|f_n(x)-f(x)|\), allowing \(E_n=+\infty\) if the errors are not bounded above.
Proof. Suppose first that the convergence is uniform. Given \(\varepsilon>0\), apply the definition with \(\varepsilon/2\). For all sufficiently large \(n\), every \(x\in X\) satisfies \(|f_n(x)-f(x)|<\varepsilon/2\). Thus \(E_n\le\varepsilon/2<\varepsilon\), so \(E_n\to0\). Conversely, suppose \(E_n\to0\). Given \(\varepsilon>0\), choose \(N\) such that \(n\ge N\) implies \(E_n<\varepsilon\). For every \(x\in X\), \(|f_n(x)-f(x)|\le E_n<\varepsilon\). This is uniform convergence. The use of \(\varepsilon/2\) in the first direction matters: if every error is less than a number, their supremum is at most that number, and need not be strictly less than it. \(\square\)
Worked Examples
Worked Example: Powers on the Unit Interval
Let \(X=[0,1]\) with its usual subspace topology from \(\mathbb{R}\), and define \(f_n(x)=x^n\). Find the pointwise limit and determine whether convergence is uniform.
If \(0\le x<1\), then \(x^n\to0\). For completeness, if \(x=0\) every term is zero. If \(0<x<1\), put \(t=1/x-1>0\), so \(1/x=1+t\). By the binomial theorem, \((1+t)^n\ge1+nt\), since all the remaining terms are nonnegative. Taking reciprocals of positive quantities gives \[ 0\le x^n=(1+t)^{-n}\le\frac{1}{1+nt}\longrightarrow0. \] At \(x=1\), \(f_n(1)=1\) for every \(n\). Therefore the pointwise limit is \[ f(x)= \begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases} \]
To test uniform convergence, compute the error. At \(x=1\), it is zero. For \(0\le x<1\), it is \(x^n\), so it is at most \(1\). Hence \(E_n\le1\). For any \(\eta\) with \(0<\eta<1\), choose \(x=(1-\eta)^{1/n}\), which lies in \([0,1)\). At this point the error is \(x^n=1-\eta\). Since this holds for every such \(\eta\), the supremum is at least \(1\); combined with the upper bound, \(E_n=1\) for every \(n\).
Thus the sequence is not uniformly convergent: its supremum error does not tend to zero. Equivalently, for every \(n\), the point \(x_n=(3/4)^{1/n}\) lies in \([0,1)\) and gives error \(x_n^n=3/4\). In particular, no index can make the error less than \(1/2\) at every point.
Check: for the selected point \(x_n\), \(f(x_n)=0\) because \(x_n<1\), while \(f_n(x_n)=( (3/4)^{1/n})^n=3/4\). The error is exactly \(3/4\) for every \(n\), as claimed.
Worked Example: A Uniformly Convergent Sequence on the Unit Interval
Again let \(X=[0,1]\) with its usual subspace topology. Define \(g_n(x)=x/n\). Show that \(g_n\) converges uniformly to the zero function \(g(x)=0\).
For every \(x\in[0,1]\), the error is \(|g_n(x)-g(x)|=x/n\). Since \(0\le x\le1\), this is at most \(1/n\), and equality occurs at \(x=1\). Therefore \(E_n=1/n\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). If \(n\ge N\), then for every \(x\in[0,1]\), \[ |g_n(x)-g(x)|=\frac{x}{n}\le\frac{1}{n}\le\frac{1}{N}<\varepsilon. \] The same \(N\) works at every point, so the convergence is uniform. By the theorem, it is also pointwise.
Check: the largest error occurs at \(x=1\), where \(|g_n(1)-g(1)|=|1/n-0|=1/n\). This verifies the computed supremum.
Worked Example: Uniform Convergence on the Real Line
Let \(X=\mathbb{R}\) with its standard topology, and define \(h_n(x)=\sin(nx)/n\). Show that \(h_n\) converges uniformly to zero, even though the domain is unbounded.
For every real \(x\), \(|\sin(nx)|\le1\), so \(|h_n(x)|\le1/n\). This gives \(E_n\le1/n\). The bound is attained at \(x=\pi/(2n)\), because \(n(\pi/(2n))=\pi/2\) and \(\sin(\pi/2)=1\). Thus \(E_n=1/n\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For \(n\ge N\) and every \(x\in\mathbb{R}\), \[ |h_n(x)-0|\le\frac{1}{n}\le\frac{1}{N}<\varepsilon. \] This proves uniform convergence. In particular, it also proves pointwise convergence; at each fixed \(x\), the same error estimate tends to zero.
Check: at \(x=\pi/(2n)\), \(h_n(x)=\sin(\pi/2)/n=1/n\), so the upper bound \(1/n\) is attained and the supremum calculation is exact.
A Common Pitfall: Confusing a Supremum with a Maximum
For the powers example, a tempting but wrong calculation is: “The largest error occurs at \(x=1\), and the error there is zero, so the supremum error is zero.” This confuses a maximum with a supremum. The point \(x=1\) does have zero error, but points strictly below \(1\) can give errors arbitrarily close to \(1\). For example, for any \(0<\eta<1\), the point \(x=(1-\eta)^{1/n}<1\) gives error \(1-\eta\). No point needs to attain the supremum for the supremum to equal \(1\). Therefore \(E_n=1\), not zero, for every \(n\).
The same example also exposes a quantifier error. Pointwise convergence allows the index needed to control the error to depend on \(x\). It does not permit choosing a different point after finding a candidate uniform index and then treating that index as if it controlled every point. Here, at each \(n\), the explicitly chosen \(x_n=(3/4)^{1/n}\) gives error \(3/4\), so the fixed tolerance \(1/2\) is never achieved simultaneously across the interval.
Uniform convergence is useful whenever the aim is a single error guarantee across a domain. Pointwise convergence answers a separate question at each input; uniform convergence supplies one stage after which all inputs satisfy the prescribed tolerance. The examples show that this stronger requirement can hold on an unbounded domain, and can fail on a compact interval: the decisive issue is the error bound, not simply the size of the domain.
Check Your Understanding
Use the definitions and error estimates to justify each answer.
- Write the quantifiers in the definition of pointwise convergence, then compare them with those for uniform convergence.
- For \(f_n(x)=x^n\) on \([0,1]\), what is the pointwise limit at \(x=1\), and why does the error supremum fail to tend to zero?
- For \(g_n(x)=x/n\) on \([0,1]\), compute the supremum error at stage \(n\) and give an index condition that ensures it is less than a specified \(\varepsilon>0\).
- For \(h_n(x)=\sin(nx)/n\) on \(\mathbb{R}\), identify a point where the upper error bound \(1/n\) is attained.
- Why does pointwise convergence on a finite domain imply uniform convergence, and which step in the proof uses finiteness?