One Error Bound for the Whole Domain
When a sequence of functions converges pointwise, its values approach the proposed limit at each fixed input. The rate of approach may depend on that input. Uniform convergence asks for more: after some index, the error must be small at every input, with the same index working throughout the domain. The distinction concerns the order of the quantifiers, not a different kind of distance.
Let \(X\) be a set, let \((Y,d)\) be a metric space, and let \(f_n:X\to Y\) for each \(n\in\mathbb{N}\). Let \(f:X\to Y\) be a proposed limit. No topology on \(X\) is needed to define either kind of convergence.
The order of the quantifiers is important: first choose \(\varepsilon\), then choose \(N\), and then the bound must hold for all \(n\ge N\) and all \(x\in X\). By contrast, pointwise convergence means that for every \(x\in X\) and every \(\varepsilon>0\), there is an index \(N=N(x,\varepsilon)\) such that \(d(f_n(x),f(x))<\varepsilon\) whenever \(n\ge N\). Here the required index may depend on \(x\).
Thus uniform convergence always implies pointwise convergence: the uniform choice of \(N\) is also a valid choice at each fixed \(x\). The reverse implication is not automatic. Also, the inequality in the definition is strict. A bound of \(d(f_n(x),f(x))\le\varepsilon\) does not by itself meet the stated condition for that same \(\varepsilon\), though a stronger estimate may establish it.
Uniform Convergence in a Metric Codomain
Proof. Assume \(f_n\to f\) uniformly, and fix \(x\in X\). For any \(\varepsilon>0\), uniform convergence supplies an \(N\) such that \(d(f_n(y),f(y))<\varepsilon\) for every \(n\ge N\) and every \(y\in X\). In particular, this inequality holds at \(y=x\). Since this works for each \(\varepsilon>0\), the sequence \(f_n(x)\) converges to \(f(x)\). As \(x\) was arbitrary, the convergence is pointwise. If \(X=\varnothing\), the pointwise condition has no points to test and is also satisfied. \(\square\)
The definition does not require a maximum error to exist. It asks directly for an error bound that holds at every point, including when the domain is infinite or unbounded. The following examples apply that criterion without assuming that the domain is compact or that the functions have any continuity property.
Worked Example: Uniform Convergence into the Plane
Take \(X=[0,1]\), with its usual set of real inputs, and take \(Y=\mathbb{R}^2\) with the Euclidean metric. Define \(f_n(t)=(t/n,1/n)\) and \(f(t)=(0,0)\). We verify uniform convergence by estimating the Euclidean distance.
For every \(t\in[0,1]\),
because \(t^2\le1\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>\sqrt{2}/\varepsilon\). For every \(n\ge N\) and every \(t\in[0,1]\), the estimate gives \(d(f_n(t),f(t))\le\sqrt{2}/n\le\sqrt{2}/N<\varepsilon\). The same \(N\) works for all \(t\), so the convergence is uniform.
Check: for the particular tolerance \(\varepsilon=1/2\), the choice \(N=3\) works: \(\sqrt{2}/n\le\sqrt{2}/3<1/2\), since \(2/9<1/4\). This verifies the strict error bound even at the endpoint \(t=1\).
Worked Example: Pointwise but Not Uniform on the Unit Interval
Let \(X=[0,1]\), \(Y=\mathbb{R}\) with its usual metric, and \(f_n(x)=x^n\). The pointwise limit is the function \(f\) given by \(f(x)=0\) for \(0\le x<1\) and \(f(1)=1\). Indeed, at \(x=1\) every term equals \(1\), while for each fixed \(x\) with \(0\le x<1\), the powers \(x^n\) tend to \(0\).
To test uniform convergence to this limit, take \(\varepsilon=1/3\). For each \(n\in\mathbb{N}\), choose the domain point \(x_n=2^{-1/n}\). Since \(0<2^{-1/n}<1\), the proposed limit there is \(f(x_n)=0\). But
For any proposed \(N\), set \(n=N\) and use \(x_N=2^{-1/N}\). At that index and domain point, the error exceeds \(1/3\). Therefore no single \(N\) makes the error less than \(1/3\) at every point for all \(n\ge N\), and the convergence is not uniform.
Check: each selected point is strictly less than \(1\), so the limit value used in the calculation is \(0\), not the value \(f(1)=1\). The domain point is allowed to vary with \(n\); this is precisely what the uniform condition must control.
Worked Example: Pointwise but Not Uniform on the Real Line
Let \(X=\mathbb{R}\), \(Y=\mathbb{R}\) with the usual metric, and \(f_n(x)=x/n\). For a fixed \(x\in\mathbb{R}\), \(|f_n(x)|=|x|/n\to0\), so the pointwise limit is \(f(x)=0\).
However, use \(\varepsilon=1/2\). For any \(n\in\mathbb{N}\), choose \(x_n=n\), which is a real number in the domain. Then
Given any proposed index \(N\), take \(n=N\) and \(x_N=N\). The error is \(1\), so the required bound fails. Hence this convergence is pointwise but not uniform. The calculation uses a moving input; pointwise convergence only tests each fixed input as \(n\) increases.
Finite Domains and Uniformity
A finite domain provides a situation in which pointwise convergence does force uniform convergence. The reason is that only finitely many point-dependent indices need to be accommodated.
Proof. If \(X=\varnothing\), then for every \(\varepsilon>0\) the uniform condition holds vacuously, for example with \(N=1\). Now suppose \(X\ne\varnothing\), and fix \(\varepsilon>0\). For each \(x\in X\), pointwise convergence gives an index \(N_x\in\mathbb{N}\) such that \(d(f_n(x),f(x))<\varepsilon\) whenever \(n\ge N_x\). Since \(X\) is finite and nonempty, the finite set of indices \(\{N_x:x\in X\}\) has a maximum; call it \(N\). If \(n\ge N\), then \(n\ge N_x\) for every \(x\in X\), so \(d(f_n(x),f(x))<\varepsilon\) for every \(x\). This is uniform convergence. \(\square\)
Worked Example: Combining Pointwise Bounds on a Finite Domain
Let \(X=\{a,b\}\) and \(Y=\mathbb{R}\) with its usual metric. Define \(f_n(a)=1/n\), \(f_n(b)=1/\sqrt{n}\), and \(f(a)=f(b)=0\). At \(a\), \(1/n\to0\); at \(b\), \(1/\sqrt{n}\to0\). We can also produce one index for both points.
For any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) such that \(N>\max\{1/\varepsilon,1/\varepsilon^2\}\). If \(n\ge N\), then \(1/n\le1/N<\varepsilon\), and \(1/\sqrt{n}\le1/\sqrt{N}<\varepsilon\). Thus both errors are less than \(\varepsilon\) for every \(n\ge N\), proving uniform convergence.
Check: with \(\varepsilon=1/4\), \(N=17\) meets the conditions because \(17>4\) and \(17>16\). At every \(n\ge17\), the errors satisfy \(1/n\le1/17<1/4\) and \(1/\sqrt n\le1/\sqrt{17}<1/4\); the second strict inequality follows from \(17>16\).
A Common Quantifier Mistake
Common pitfall. In the example \(f_n(x)=x^n\) on \([0,1]\), it is wrong to argue that convergence is uniform because, for each fixed \(x<1\), the powers \(x^n\) eventually become small. That argument fixes \(x\) first, so its index may depend on \(x\); it establishes only pointwise convergence. The explicit calculation with \(x_n=2^{-1/n}\) shows the error is \(1/2\) at the \(n\)-th term and the \(n\)-dependent input \(x_n\). Since \(1/2>1/3\) for every \(n\), there is no common index for the tolerance \(1/3\).
Uniform convergence is useful whenever a conclusion requires control over a whole family of inputs rather than a separate conclusion at each fixed input. Its defining test is always the same: given one positive error tolerance, find one index after which every domain point satisfies the strict distance bound. Pointwise convergence supplies such indices separately; uniform convergence requires a shared one.
Check Your Understanding
For each question, identify whether the index may depend on the domain point, and justify your conclusion using the definition.
- State the quantifiers in the definition of uniform convergence and compare them with those for pointwise convergence.
- For \(f_n(x)=1/n\) on \(\mathbb{R}\), with proposed limit \(f(x)=0\), verify uniform convergence directly.
- For \(f_n(x)=x^n\) on \([0,1]\), what point \(x_n\) gives an error of \(1/2\) from the pointwise limit at index \(n\)?
- Why does pointwise convergence on a finite domain imply uniform convergence?
- Does pointwise convergence on an infinite domain always fail to be uniform? Give an example or a counterexample to this claim.