Examples of Uniform Convergence

Uniform Convergence Requires One Error Bound for the Whole Domain

In “Examples of Pointwise Convergence,” the index needed to make the error small could depend on the fixed input. Uniform convergence changes that requirement: once an error tolerance is chosen, a single index must control the error at every point of the domain. Examples are often settled by finding an explicit bound for \(|f_n(x)-f(x)|\) that does not depend on \(x\), and then checking that the bound tends to zero.

The Supremum Criterion for Uniform Convergence gives an equivalent formulation: the suprema of the errors must tend to zero. In the examples below, we will often obtain a direct estimate that works for all inputs. Such an estimate is useful even when the supremum is difficult to calculate exactly. When the supremum can be found, it also describes the largest possible error and can reveal whether that error is attained at a point of the domain.

Geometric Powers Away from the Endpoint

Theorem (Uniform Convergence of Geometric Powers on a Shorter Interval): Let \(0\leq r<1\), and define \(f_n:[0,r]\to\mathbb{R}\) by \(f_n(x)=x^n\) for positive integers \(n\). Then \(f_n\) converges uniformly to the zero function on \([0,r]\).

Proof. If \(r=0\), the domain consists only of \(x=0\), and \(f_n(0)=0\) for every \(n\). The assertion follows. Now suppose \(0<r<1\). For every \(x\in[0,r]\), we have \(0\leq x^n\leq r^n\). Since \(r^n\to0\), this supplies a bound independent of \(x\) that tends to zero.

More explicitly, let \(\varepsilon>0\). Because \(0<r<1\), there is a positive integer \(N\) such that \(r^N<\varepsilon\). For every \(n\geq N\) and every \(x\in[0,r]\),

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

The same \(N\) works for every \(x\) in the interval. Thus \(f_n\) converges uniformly to zero on \([0,r]\). \(\square\)

Worked Example: Powers on \([0,3/4]\)

Consider \(f_n(x)=x^n\) on \([0,3/4]\). The theorem applies with \(r=3/4\), and the estimate is

\sup_{x\in[0,3/4]}|f_n(x)-0| =\sup_{x\in[0,3/4]}x^n =\left(\frac{3}{4}\right)^n.

The equality holds because \(x^n\leq(3/4)^n\) throughout the interval, and equality occurs at \(x=3/4\). Since \((3/4)^n\to0\), the supremum of the errors tends to zero. For a particular tolerance \(\varepsilon>0\), choose \(N\) so that \((3/4)^N<\varepsilon\); then every \(n\geq N\) gives an error less than \(\varepsilon\) at every point of the interval.

The domain matters. In the earlier pointwise example on \([0,1]\), the value at \(x=1\) stays equal to \(1\). There, the supremum error is \(1\) for every \(n\), so convergence is not uniform. Removing the endpoint and restricting further to \([0,3/4]\) produces a bound that shrinks geometrically. A pointwise limit can be the same zero function on both domains below \(1\), while the uniform-convergence conclusion differs because the domains differ.

Uniform Convergence Can Hold on an Unbounded Domain

An unbounded domain does not automatically prevent uniform convergence. What matters is whether the error can be bounded independently of the input. The next family illustrates how a denominator can provide such control even when the domain extends without bound.

Theorem (A Rational Family Converging Uniformly on the Half-Line): For \(n\geq1\), define \(u_n:[0,\infty)\to\mathbb{R}\) by \(u_n(x)=x/(1+nx)\). Then \(u_n\) converges uniformly to zero on \([0,\infty)\).

Proof. Fix \(x\geq0\). The denominator \(1+nx\) is positive. Also, \(1+nx\geq nx\), so for \(x\geq0\),

0\leq u_n(x)=\frac{x}{1+nx}\leq\frac{1}{n}.

For \(x=0\), the left and middle expressions are both zero, and the inequality still holds. The upper bound \(1/n\) tends to zero and is independent of \(x\). Given \(\varepsilon>0\), choose \(N\) such that \(1/N<\varepsilon\). Then for \(n\geq N\) and every \(x\geq0\),

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

This proves uniform convergence on the whole half-line. \(\square\)

Worked Example: Finding the Supremum Error Without Attaining It

For the same functions \(u_n(x)=x/(1+nx)\), the uniform estimate gives \(\sup_{x\geq0}u_n(x)\leq1/n\). To see that this is the exact supremum, take \(x>0\) and compare \(u_n(x)\) with \(1/n\):

\frac{1}{n}-\frac{x}{1+nx} =\frac{1+nx-nx}{n(1+nx)} =\frac{1}{n(1+nx)}>0.

Thus \(u_n(x)<1/n\) for every finite \(x\geq0\). On the other hand,

\frac{1}{n}-u_n(x)=\frac{1}{n(1+nx)}\longrightarrow0 \qquad\text{as }x\to\infty.

Consequently, values of \(u_n(x)\) can be made arbitrarily close to \(1/n\), although no point of the domain gives equality. Hence \(\sup_{x\geq0}u_n(x)=1/n\). The supremum criterion confirms uniform convergence, and the calculation also shows why one must not assume that a supremum is necessarily a maximum. The error bound is sharp even though there is no input where the error equals that bound.

Worked Example: A Reciprocal Family with an Attained Maximum Error

Define \(v_n:[0,\infty)\to\mathbb{R}\) by \(v_n(x)=1/(n+x)\). For fixed \(x\geq0\), the denominator \(n+x\) tends to infinity, so \(v_n(x)\to0\). To check uniform convergence, observe that \(n+x\geq n\), and therefore

0<|v_n(x)-0|=\frac{1}{n+x}\leq\frac{1}{n} \qquad(x\geq0).

This bound tends to zero independently of \(x\), so \(v_n\to0\) uniformly. In this case the supremum error is attained at \(x=0\):

\sup_{x\geq0}|v_n(x)| =\frac{1}{n+0} =\frac{1}{n}.

The two rational examples both converge uniformly on an unbounded domain, but their sharp error bounds behave differently at the domain’s edge: the supremum for \(v_n\) occurs at \(0\), while that for \(u_n\) is approached only as \(x\) grows without bound. In both cases, a direct bound independent of \(x\) is enough to prove uniform convergence.

Derivative Bounds on Bounded Intervals

Another useful technique is to control changes in a function by a bound on its derivative. The sine function \(S\) defined earlier in the course satisfies \(S(0)=0\) and \(S'(t)=C(t)\), with \(|C(t)|\leq1\). The Mean Value Theorem therefore gives \(|S(t)|\leq|t|\) for every real \(t\): if \(t\neq0\), apply the theorem between \(0\) and \(t\); the case \(t=0\) is immediate. This estimate turns a bound on the input into a bound on the function value.

Worked Example: Sine at a Shrinking Scale

Fix a finite number \(A\geq0\), and define \(w_n:[-A,A]\to\mathbb{R}\) by \(w_n(x)=S(x/n)\). For every \(x\in[-A,A]\), the derivative estimate just established gives

|w_n(x)-0| =\left|S\left(\frac{x}{n}\right)\right| \leq\left|\frac{x}{n}\right| \leq\frac{A}{n}.

If \(A>0\), then \(A/n\to0\), so the errors tend to zero uniformly on \([-A,A]\). If \(A=0\), the domain contains only zero and every error is zero, so uniform convergence holds in that case as well. Thus \(w_n\to0\) uniformly on every fixed bounded interval.

The restriction to a bounded interval is essential for this estimate: it uses \(|x|\leq A\). On all of \(\mathbb{R}\), this particular bound gives \(|S(x/n)|\leq|x|/n\), and the right side has no finite bound independent of \(x\). This does not by itself prove that uniform convergence fails on \(\mathbb{R}\); it shows only that this estimate cannot establish it there. A proof must control the error over the actual domain, not just at each fixed input.

How to Read a Uniform Error Estimate

The examples use several forms of the same strategy. First identify the proposed limit. Then express the error, simplify it, and seek a bound that is valid for every input in the domain. Finally, verify that the bound tends to zero as \(n\) increases. The required bound may be exact, as in the two supremum calculations, or merely sufficient, as in the sine example.

State the candidate limit.
Use pointwise calculations, when needed, to identify the function \(f\) against which the errors \(|f_n(x)-f(x)|\) will be measured.

Seek a bound across the domain.
Estimate the error by a quantity depending on \(n\) but not on \(x\). Check that the estimate includes boundary points and every other part of the stated domain.

Show the bound vanishes.
Given an arbitrary positive tolerance, choose one index that makes the bound small. That index must work for every input in the domain.

Check whether the domain affects the estimate.
A restriction such as \(|x|\leq A\) may be exactly what makes a bound uniform. Do not transfer a conclusion to a larger domain without a new argument.

These examples also clarify the difference between pointwise and uniform reasoning. A pointwise proof can hold \(x\) fixed and use a bound involving that particular \(x\). A uniform proof must prevent the error from becoming large at any point of the domain as \(n\) varies. The Moving-Input Criterion for Uniform Convergence provides another way to detect failure: if suitable inputs \(x_n\) keep the errors away from zero, uniform convergence cannot hold. For positive examples, however, an explicit domain-wide estimate is often the most direct proof.

Takeaway: To prove uniform convergence, bound \(|f_n(x)-f(x)|\) by an expression tending to zero that works for every \(x\) in the domain. The domain is part of the claim: bounded restrictions, endpoints, and unbounded regions can change which estimates are available.

Check Your Understanding

Use the estimates and examples above to answer the following questions.

Why does \(x^n\) converge uniformly to zero on \([0,3/4]\), while the same sequence does not converge uniformly to zero on \([0,1]\)?
For \(u_n(x)=x/(1+nx)\), why is \(1/n\) the supremum error even though no finite \(x\geq0\) attains it?
Where does the supremum error for \(v_n(x)=1/(n+x)\) occur, and what is its value?
Which estimate proves uniform convergence of \(S(x/n)\) on \([-A,A]\), and where is boundedness of the domain used?
What is the difference between a supremum and a maximum in the rational-family examples?