The Uniform Convergence Strategy

One Error Bound for the Whole Domain

The Continuity Strategy uses a strict margin to control function values near a point. Uniform convergence asks for a different kind of control: can one bound the error between a function and its limit by the same tolerance at every point in the domain? The key word is uniform. The point at which an estimate is needed does not determine how large the index must be.

This distinction matters in proofs about limits of functions. Pointwise convergence lets the required index depend on the point. Uniform convergence chooses the index first and then controls the error everywhere. That single choice makes it possible to transfer some global properties from approximating functions to their limit.

Definition (Pointwise Convergence): Let \(E\) be a set, and let \(f_n:E\to\mathbb{R}\) and \(f:E\to\mathbb{R}\). The sequence \((f_n)\) converges pointwise to \(f\) on \(E\) if, for every \(x\in E\) and every \(\varepsilon>0\), there is an integer \(N\) such that \(n\geq N\) implies \(|f_n(x)-f(x)|<\varepsilon\).

Definition (Uniform Convergence): The sequence \((f_n)\) converges uniformly to \(f\) on \(E\) if, for every \(\varepsilon>0\), there is an integer \(N\) such that \(n\geq N\) implies \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in E\).

The order of the quantifiers distinguishes the definitions. In pointwise convergence, \(N\) may depend on both \(\varepsilon\) and \(x\). In uniform convergence, \(N\) may depend on \(\varepsilon\), but not on \(x\). Every uniformly convergent sequence converges pointwise, but the reverse implication can fail.

Estimate the Error Before Taking the Limit

A common proof strategy is to find a direct upper bound for \(|f_n(x)-f(x)|\) that does not depend on \(x\) and tends to zero as \(n\) increases. If the bound is smaller than \(\varepsilon\) for all sufficiently large \(n\), then it proves uniform convergence. The estimate need not be sharp; it only needs to control the error everywhere.

Worked Example: A Rational Sequence Converging Uniformly

For \(x\in[0,1]\), define \(f_n(x)=x/(1+nx)\), and let \(f(x)=0\). At \(x=0\), \(f_n(0)=0\). At each \(x>0\), the denominator \(1+nx\) grows without bound, so \(f_n(x)\to0\). Thus the sequence converges pointwise to \(f\).

For uniform convergence, estimate the error at every point. Since \(x\geq0\),

0\leq |f_n(x)-f(x)| =\frac{x}{1+nx} \leq\frac{1}{n}.

The last inequality follows from \(nx\leq1+nx\), after dividing by the positive quantity \(n(1+nx)\). Given \(\varepsilon>0\), choose an integer \(N>1/\varepsilon\). Then, for every \(n\geq N\) and every \(x\in[0,1]\),

|f_n(x)-f(x)|\leq\frac{1}{n}\leq\frac{1}{N}<\varepsilon.

The same index works throughout the interval, so the convergence is uniform.

A useful warning is that pointwise convergence alone does not control the worst error over the domain. The sequence \(f_n(x)=x^n\) on \([0,1]\) converges pointwise to \(f(x)=0\) for \(0\leq x<1\) and \(f(1)=1\). For each positive integer \(n\), take \(x_n=2^{-1/n}\). Then \(0\leq x_n<1\), so \(f(x_n)=0\), while

|f_n(x_n)-f(x_n)|=(x_n)^n=\frac{1}{2}.

Consequently, for every \(n\) there is a point where the error is at least \(1/2\). No index can make the error smaller than, for example, \(1/3\) everywhere. This convergence is not uniform.

Use the Uniform Cauchy Criterion

Sometimes the limit function is not known in advance, or calculating it would be difficult. In that situation, it can be more convenient to compare the functions with one another. The uniform Cauchy criterion says that, for real-valued functions, uniform convergence is equivalent to the errors \(|f_n(x)-f_m(x)|\) becoming uniformly small as both indices grow. Completeness of the real numbers supplies the pointwise limits; the uniform Cauchy estimate then controls convergence to the resulting function.

Theorem (Uniform Cauchy Criterion): Let \(E\) be a set, and let \(f_n:E\to\mathbb{R}\). The sequence \((f_n)\) converges uniformly on \(E\) to a real-valued function if and only if, for every \(\varepsilon>0\), there is an integer \(N\) such that \(n,m\geq N\) implies \(|f_n(x)-f_m(x)|<\varepsilon\) for every \(x\in E\).

Proof. First suppose \(f_n\to f\) uniformly. Given \(\varepsilon>0\), choose \(N\) such that \(n\geq N\) implies \(|f_n(x)-f(x)|<\varepsilon/2\) for every \(x\in E\). For \(n,m\geq N\), the triangle inequality gives, for every \(x\in E\),

|f_n(x)-f_m(x)| \leq |f_n(x)-f(x)|+|f(x)-f_m(x)| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

This proves the uniform Cauchy condition.

Conversely, suppose the uniform Cauchy condition holds. Fix \(x\in E\). For every \(\varepsilon>0\), the condition gives an \(N\) such that \(n,m\geq N\) implies \(|f_n(x)-f_m(x)|<\varepsilon\). Thus \((f_n(x))\) is a Cauchy sequence of real numbers. By completeness of \(\mathbb{R}\), it converges. Define \(f(x)=\lim_{m\to\infty}f_m(x)\) for each \(x\in E\).

We now prove uniform convergence to \(f\). Given \(\varepsilon>0\), apply the assumed condition with \(\varepsilon/2\) in place of \(\varepsilon\). It supplies an \(N\) such that, whenever \(n,m\geq N\),

|f_n(x)-f_m(x)|<\frac{\varepsilon}{2} \qquad\text{for every }x\in E.

Fix \(n\geq N\) and \(x\in E\), and let \(m\to\infty\). Since \(f_m(x)\to f(x)\), the absolute-value limit theorem gives

|f_n(x)-f(x)|\leq\frac{\varepsilon}{2}<\varepsilon.

The choice of \(N\) did not depend on \(x\), so this proves uniform convergence. \(\square\)

Worked Example: Uniform Convergence of Geometric Partial Sums

Fix \(r\) with \(0<r<1\), and define \(S_N(x)=\sum_{k=0}^{N}x^k\) on \([0,r]\). To use the uniform Cauchy criterion, let \(M>N\). The finite geometric-sum identity gives

|S_M(x)-S_N(x)| =\sum_{k=N+1}^{M}x^k =x^{N+1}\frac{1-x^{M-N}}{1-x}.

For \(x\in[0,r]\), every term in the sum is nonnegative, \(x^{N+1}\leq r^{N+1}\), and \(1-x^{M-N}\leq1\). Also \(1-x\geq1-r>0\). Therefore,

|S_M(x)-S_N(x)| \leq\frac{r^{N+1}}{1-r}.

If \(N\geq M\), interchanging the two indices gives the same bound with the smaller index in the exponent. Since \(r^{N+1}/(1-r)\to0\), the partial sums satisfy the uniform Cauchy condition. The theorem gives a uniform limit. The finite geometric-sum formula also identifies it: for \(x\in[0,r]\),

S_N(x)=\frac{1-x^{N+1}}{1-x} \longrightarrow \frac{1}{1-x}.

The bound above is uniform in \(x\), so the convergence to \(1/(1-x)\) is uniform on the whole interval.

Transfer Uniform Continuity Through a Uniform Limit

The Uniform Limit of Continuous Functions theorem from the Approximation Strategy says that a uniform limit of continuous functions is continuous. Uniform convergence also allows a stronger conclusion when the approximating functions are uniformly continuous: their uniform limit is uniformly continuous. This is useful when continuity at each point is not enough, and one needs a single distance tolerance that works throughout the domain.

Theorem (Uniform Limits Preserve Uniform Continuity): Let \(E\) be a subset of a metric space, and let \(f_n:E\to\mathbb{R}\) be uniformly continuous for every \(n\). If \(f_n\) converges uniformly on \(E\) to \(f:E\to\mathbb{R}\), then \(f\) is uniformly continuous on \(E\).

Proof. Let \(\varepsilon>0\). By uniform convergence, choose an index \(n\) such that

|f_n(x)-f(x)|<\frac{\varepsilon}{3} \qquad\text{for every }x\in E.

Keep this index fixed. Since \(f_n\) is uniformly continuous, there is a \(\delta>0\) such that \(x,y\in E\) and \(d(x,y)<\delta\) imply \(|f_n(x)-f_n(y)|<\varepsilon/3\). For such \(x,y\), the triangle inequality yields

\begin{aligned} |f(x)-f(y)| &\leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\varepsilon. \end{aligned}

Thus the same \(\delta\) works for every pair \(x,y\in E\). This is precisely uniform continuity of \(f\). \(\square\)

Worked Example: Uniform Approximation of the Absolute-Value Function

On \(\mathbb{R}\), define \(f_n(x)=\sqrt{x^2+1/n}\) and \(f(x)=|x|\). First, \(f_n\) is uniformly continuous. Indeed, it is the Euclidean norm of the vector \((x,1/\sqrt{n})\), and the reverse triangle inequality gives

|f_n(x)-f_n(y)| \leq \left|(x,1/\sqrt{n})-(y,1/\sqrt{n})\right| =|x-y|.

So each \(f_n\) is Lipschitz with constant \(1\). Next, \(f_n(x)\geq|x|\), and \((|x|+1/\sqrt{n})^2\geq x^2+1/n\). Taking nonnegative square roots shows \(f_n(x)\leq|x|+1/\sqrt{n}\). Consequently, for every \(x\in\mathbb{R}\),

0\leq |f_n(x)-f(x)|\leq\frac{1}{\sqrt{n}}.

The bound tends to zero independently of \(x\), so \(f_n\to f\) uniformly on \(\mathbb{R}\). The theorem therefore implies that \(f(x)=|x|\) is uniformly continuous. In this case the conclusion can also be checked directly, since \(||x|-|y||\leq|x-y|\); the example illustrates how a uniform approximation can transfer the property.

Keep the Quantifiers and Hypotheses in View

Uniform convergence is not a claim that the functions are close at every index, nor does it require a formula for the largest error. It requires that, for each chosen tolerance, one index controls the error at every point thereafter. When proving it, write down the quantifiers and make sure the index is independent of the point.

The distinction from pointwise convergence is especially important when applying the Uniform Limit of Continuous Functions theorem: its hypothesis is uniform convergence, not merely pointwise convergence. The example \(x^n\) on \([0,1]\) shows why that hypothesis cannot be dropped. Similarly, the theorem on uniform limits of uniformly continuous functions needs both uniform convergence and uniform continuity of the approximating functions. Check each hypothesis before transferring a property to the limit.

Identify the target.
Write the error \(|f_n(x)-f(x)|\), or compare \(|f_n(x)-f_m(x)|\) if using the uniform Cauchy criterion.

Find a bound independent of the point.
Estimate the error by a quantity depending only on the index and known to become small.

Choose the index from the tolerance.
Make the index large enough for the bound to be below the requested tolerance, with no dependence on the domain point.

Transfer a property only under its hypotheses.
For continuity, use the Uniform Limit of Continuous Functions theorem; for uniform continuity, use the proved theorem above.

The central habit is to look for one estimate that holds everywhere. When such an estimate is unavailable, the uniform Cauchy criterion offers another route: control all sufficiently late pairs of functions uniformly, then use completeness to construct the limit.

Check Your Understanding

Use the definitions and proof strategies from this tutorial to answer the following questions.

In the definition of uniform convergence, which variables may the index \(N\) depend on, and which may it not depend on?
Why does the choice \(x_n=2^{-1/n}\) show that \(x^n\) does not converge uniformly to its pointwise limit on \([0,1]\)?
In the converse direction of the Uniform Cauchy Criterion, where is completeness of \(\mathbb{R}\) used?
Why does the proof that uniform limits preserve uniform continuity use \(\varepsilon/3\) for each of three errors?
What hypothesis must be added to pointwise convergence of continuous functions to apply the Uniform Limit of Continuous Functions theorem?