The Epsilon Definition of Uniform Convergence

The Epsilon Definition

The previous tutorial compared pointwise and uniform convergence by examining which inputs a chosen index must control. The epsilon definition makes that distinction precise: uniform convergence requires one index to work for every input in the domain. The order of the quantifiers is essential, because changing it changes the kind of convergence being described.

Definition: Let \(E\) be a nonempty set, and let \(f_n:E\to\mathbb{R}\) and \(f:E\to\mathbb{R}\). The sequence \((f_n)\) converges uniformly to \(f\) on \(E\) if, for every \(\varepsilon>0\), there is a positive integer \(N\) such that, whenever \(n\geq N\), the inequality \(|f_n(x)-f(x)|<\varepsilon\) holds for every \(x\in E\).

The index \(N\) may depend on \(\varepsilon\), but it must be chosen before the input \(x\), and it cannot depend on \(x\). In quantified form, the condition is

\text{for every }\varepsilon>0,\ \text{there exists }N\in\mathbb{N}\text{ such that for every }n\geq N\text{ and every }x\in E,\quad |f_n(x)-f(x)|<\varepsilon.

For pointwise convergence, the index may depend on the input: for every \(x\in E\) and every \(\varepsilon>0\), there is an \(N\) such that the error is less than \(\varepsilon\) for all \(n\geq N\). Uniform convergence instead places the choice of \(N\) before the requirement that the estimate hold at every input.

Condition	Choice of index	Required error estimate
Pointwise convergence	May depend on \(x\) and \(\varepsilon\)	For each fixed \(x\)
Uniform convergence	May depend on \(\varepsilon\), not \(x\)	For every \(x\), using the same \(N\)

As established in “Uniform Convergence,” the supremum criterion gives an equivalent way to express uniform convergence when the errors are collected into a single quantity. Here the direct epsilon definition is the central tool: an estimate independent of \(x\) gives a uniform proof, while an error that stays above a fixed positive number somewhere in the domain prevents one.

Reading the Definition in Practice

A proof of uniform convergence usually follows the same sequence. Begin with an arbitrary \(\varepsilon>0\), find an estimate for \(|f_n(x)-f(x)|\) valid for every input, and then choose an index that makes the estimate smaller than \(\varepsilon\). The estimate must remain valid even when \(x\) is large, small, or otherwise varies throughout the domain.

Worked Example: A Uniform Estimate on an Unbounded Domain

For \(n\geq1\), define \(f_n:[0,\infty)\to\mathbb{R}\) by

f_n(x)=\frac{1}{n+x}.

For every \(x\geq0\), \(n+x\geq n>0\), so

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

Let \(\varepsilon>0\). Choose a positive integer \(N>1/\varepsilon\). If \(n\geq N\) and \(x\geq0\), then

|f_n(x)-0|\leq\frac{1}{n}\leq\frac{1}{N}<\varepsilon.

The choice of \(N\) depends only on \(\varepsilon\), and the estimate holds for every \(x\) in the unbounded domain. Thus \(f_n\) converges uniformly to zero on \([0,\infty)\). The domain being unbounded does not by itself prevent uniform convergence; what matters is whether the error can be bounded independently of the input.

Worked Example: A Geometric Error Bound

Fix a real number \(r\) with \(0<r<1\), and define \(g_n:[0,r]\to\mathbb{R}\) by \(g_n(x)=x^n\). We claim that \(g_n\) converges uniformly to zero on \([0,r]\). For every \(x\in[0,r]\),

|g_n(x)-0|=x^n\leq r^n.

Because \(0<r<1\), its powers tend to zero. Given \(\varepsilon>0\), choose a positive integer \(N\) such that \(r^N<\varepsilon\). For \(n\geq N\), the powers decrease, so \(r^n\leq r^N\). Therefore, for every \(x\in[0,r]\),

|g_n(x)-0|=x^n\leq r^n\leq r^N<\varepsilon.

This verifies the quantifiers in the definition. In particular, the same \(N\) works even at the largest input \(x=r\), where the bound \(r^n\) is attained.

When an estimate depends on \(x\), it is not automatically useless, but it must still lead to an index that works throughout the domain. A bound that becomes large near some inputs may fail to provide such an index. The next example shows how the epsilon definition can be contradicted by finding a fixed error that persists at suitable inputs.

Worked Example: A Fixed Error Rules Out Uniform Convergence

For \(n\geq1\), define \(h_n:(0,1]\to\mathbb{R}\) by

h_n(x)=\frac{nx}{1+nx}.

For each fixed \(x>0\), the error from \(1\) is

|h_n(x)-1| =\left|\frac{nx-(1+nx)}{1+nx}\right| =\frac{1}{1+nx}.

Since \(x\) is fixed and positive, \(1/(1+nx)\) tends to zero as \(n\to\infty\). Thus \(h_n\) converges pointwise to the function \(h(x)=1\) on \((0,1]\).

To test uniform convergence, for each \(n\) choose the input \(x_n=1/n\), which belongs to \((0,1]\). Substitution gives

|h_n(x_n)-1| =\frac{1}{1+n(1/n)} =\frac{1}{2}.

For \(\varepsilon=1/3\), the error is therefore not less than \(\varepsilon\) at \(x_n\), no matter how large \(n\) is. Given any proposed index \(N\), take \(n=N\) and \(x=1/N\); the required estimate fails. Hence the convergence is not uniform. The inputs change with the index, which is permitted when testing the uniform requirement that every input be controlled.

The Exact Negation of Uniform Convergence

The definition also gives a systematic way to prove that convergence is not uniform. Negating its quantifiers produces a positive error threshold that cannot be achieved by any proposed index. This is stronger than merely observing that an estimate seems to depend on \(x\): it identifies precisely what a counterexample must establish.

Theorem (Failure Criterion for Uniform Convergence): Let \(E\) be nonempty and let \(f_n,f:E\to\mathbb{R}\). The sequence \(f_n\) does not converge uniformly to \(f\) on \(E\) if and only if there is an \(\varepsilon_0>0\) such that for every positive integer \(N\), there exist \(n\geq N\) and \(x\in E\) satisfying \(|f_n(x)-f(x)|\geq\varepsilon_0\).

Proof. Uniform convergence means that for every \(\varepsilon>0\), some positive integer \(N\) makes \(|f_n(x)-f(x)|<\varepsilon\) hold for every \(n\geq N\) and every \(x\in E\). Negating “for every \(\varepsilon>0\), there exists \(N\)” gives “there exists \(\varepsilon_0>0\), such that for every \(N\),” the required estimate does not hold for all \(n\geq N\) and all \(x\in E\). For each such \(N\), failure of that universal estimate means there is at least one \(n\geq N\) and at least one \(x\in E\) for which the strict inequality fails. Failure of \(|f_n(x)-f(x)|<\varepsilon_0\) is exactly \(|f_n(x)-f(x)|\geq\varepsilon_0\). This proves the stated condition. Reversing these steps shows that the condition prevents the epsilon definition from holding. \(\square\)

The distinction between \(>\) and \(\geq\) here comes from negating a strict estimate: the negation of “less than \(\varepsilon_0\)” is “greater than or equal to \(\varepsilon_0\).” In the preceding example, \(\varepsilon_0=1/3\) works, since for every \(N\) one can take \(n=N\) and \(x=1/N\), giving an error of \(1/2\).

Stability Under Small Perturbations

The epsilon definition also shows that a uniformly convergent sequence is not substantially changed by errors that are uniformly small across the domain. This is useful when a complicated sequence can be compared with a simpler one. The comparison error must itself be controlled for all inputs, rather than merely tending to zero at each fixed input.

Theorem (Stability Under Uniformly Vanishing Perturbations): Let \(E\) be nonempty, and suppose \(f_n:E\to\mathbb{R}\) converges uniformly to \(f:E\to\mathbb{R}\). Suppose also that \(g_n:E\to\mathbb{R}\) and that there are nonnegative numbers \(a_n\) with \(a_n\to0\) such that \(|g_n(x)-f_n(x)|\leq a_n\) for every \(n\) and every \(x\in E\). Then \(g_n\) converges uniformly to \(f\) on \(E\).

Proof. Let \(\varepsilon>0\). Since \(f_n\to f\) uniformly, there is a positive integer \(N_1\) such that, whenever \(n\geq N_1\) and \(x\in E\),

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

Since \(a_n\to0\), there is a positive integer \(N_2\) such that \(a_n<\varepsilon/2\) whenever \(n\geq N_2\). Set \(N=\max\{N_1,N_2\}\). For every \(n\geq N\) and every \(x\in E\), the triangle inequality and the assumed perturbation bound give

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

This is the epsilon definition of uniform convergence for \(g_n\) to \(f\), so the conclusion follows. \(\square\)

Both parts of the hypothesis have a uniform role: the sequence \(f_n\) is uniformly close to \(f\), and \(g_n\) is uniformly close to \(f_n\). Pointwise smallness of the perturbation alone would not justify the estimate with one index for every input.

Worked Example: Comparing Two Sequences

On \([0,\infty)\), let

f_n(x)=\frac{1}{n+x}, \qquad g_n(x)=\frac{1}{n+x}+\frac{x}{n^2(1+x)}.

The first sequence converges uniformly to zero by the earlier estimate. The difference between the sequences satisfies, for every \(x\geq0\),

|g_n(x)-f_n(x)| =\frac{x}{n^2(1+x)} \leq\frac{1}{n^2},

because \(x\leq1+x\). The bounds \(a_n=1/n^2\) are nonnegative and tend to zero. The stability theorem therefore implies that \(g_n\) also converges uniformly to zero. Directly, for every \(x\geq0\),

|g_n(x)| \leq \frac{1}{n}+\frac{1}{n^2}.

Given \(\varepsilon>0\), choose \(N\) so large that \(1/N+1/N^2<\varepsilon\). For \(n\geq N\), both terms on the right are no larger than their values at \(N\), so \(|g_n(x)|\leq1/N+1/N^2<\varepsilon\) for every \(x\geq0\). This confirms the uniform conclusion with an explicit bound.

Common Pitfalls

A frequent error is to choose an index after fixing an input and then conclude that the convergence is uniform. That argument establishes only pointwise convergence unless the resulting index can be replaced by one independent of the input. Another mistake is to use a bound that is valid at each \(x\) but whose dependence on \(x\) prevents a single choice of \(N\). Always return to the quantifiers: after choosing \(\varepsilon\), the index must work for every \(x\) at once.

Takeaway: Uniform convergence means that for each positive error tolerance, one index controls all inputs and all later terms. A domain-independent error estimate proves uniform convergence; a fixed positive error that persists at selected inputs disproves it.

Check Your Understanding

Use the epsilon definition and the results in this tutorial to answer the following questions.

In the definition of uniform convergence, which choices may depend on \(\varepsilon\), and which may not depend on \(x\)?
For \(f_n(x)=1/(n+x)\) on \([0,\infty)\), what estimate gives a uniform choice of index?
Why does an error equal to \(1/2\) at inputs \(x_n=1/n\) rule out uniform convergence to \(1\) in the third worked example?
State the failure criterion for uniform convergence, including the inequality that replaces the strict epsilon estimate.
Why must the perturbation bound in the stability theorem hold for every input, not only for each fixed input separately?

The Epsilon Definition of Uniform Convergence

What You'll Learn