Examples of Nonuniform Convergence

How Nonuniform Convergence Appears

In “Examples of Uniform Convergence,” domain-wide error bounds showed how one index can control the error at every input. Nonuniform convergence is the opposite situation: the sequence may converge at each fixed input, yet its errors cannot be made small everywhere with a single index. Often the obstruction is concentrated near a boundary, moves across the domain as the index changes, or occurs farther and farther out on an unbounded domain.

A useful way to detect such behavior is to look for inputs \(x_n\) that depend on \(n\). Pointwise convergence tests each fixed \(x\); it does not prevent a sequence of changing inputs from keeping the error large. The Moving-Input Criterion for Uniform Convergence, established earlier in the course, formalizes this test. We will use direct calculations to identify such inputs and to find exact supremum errors where possible.

A Boundary Layer That Prevents Uniform Convergence

Consider functions that are zero at the left endpoint but rise rapidly toward one just to its right. As the sequence index grows, the transition occurs closer to that endpoint. Every fixed positive input eventually lies beyond the transition, but there is no single index that controls the error uniformly all the way down to zero.

Theorem (A Rational Boundary Layer Is Not Uniformly Convergent): For each positive integer \(n\), define \(f_n:[0,1]\to\mathbb{R}\) by \(f_n(x)=x/(x+1/n)\). Then \(f_n\) converges pointwise to the function \(f\) given by \(f(0)=0\) and \(f(x)=1\) for \(0<x\leq1\), but the convergence is not uniform on \([0,1]\).

Proof. At \(x=0\), we have \(f_n(0)=0\) for every \(n\), so the limit there is zero. Fix \(x\) with \(0<x\leq1\). Then

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

This proves the stated pointwise limit. For \(x>0\), the error is

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

At \(x=0\), the error is zero. For every \(x>0\), the displayed error is less than one, but it approaches one as \(x\) decreases to zero. Thus its supremum over \([0,1]\) is one for every \(n\). In particular, the supremum errors do not tend to zero, so the Supremum Criterion for Uniform Convergence shows that convergence is not uniform.

One can also see the obstruction with inputs depending on \(n\). Set \(x_n=1/n\), which belongs to \([0,1]\). Then

f_n(x_n)=\frac{1/n}{1/n+1/n}=\frac12, \qquad f(x_n)=1, \qquad |f_n(x_n)-f(x_n)|=\frac12.

The errors along these inputs remain fixed at one half. This gives a direct witness to failure of uniform convergence. \(\square\)

Worked Example: The Error Supremum Is Not a Maximum

For the functions in the theorem, the error is zero at \(x=0\), while at every \(x>0\) it is \(1/(1+nx)\). The error never equals one: at positive \(x\), its denominator is strictly greater than one, and at zero the error is zero. Nevertheless, values of the error get arbitrarily close to one. Indeed, for any fixed \(n\), as \(x\) approaches zero through positive values,

\frac{1}{1+nx}\longrightarrow1.

Therefore the supremum error equals one, even though no point of the domain attains it. This distinction matters: uniform convergence is governed by the supremum, not by whether an input realizes the largest error. The boundary point itself has zero error, but points arbitrarily close to it prevent the supremum error from becoming small.

Removing the Problem Region Can Restore Uniform Convergence

The same sequence behaves differently if the domain stays a positive distance from zero. This illustrates why a convergence claim must always name its domain. The pointwise limit remains one on the restricted interval, and there the error has a bound that is independent of the input.

Theorem (Uniform Convergence Away from the Boundary Layer): Fix \(0<\delta\leq1\). The functions \(f_n(x)=x/(x+1/n)\) converge uniformly to the constant function \(1\) on \([\delta,1]\).

Proof. For every \(x\in[\delta,1]\),

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

The right side does not depend on \(x\), and it tends to zero as \(n\) tends to infinity. Explicitly, given \(\varepsilon>0\), choose a positive integer \(N\) such that \(1/(1+N\delta)<\varepsilon\). For every \(n\geq N\) and every \(x\in[\delta,1]\), we have

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

The same \(N\) works throughout the interval, which proves uniform convergence there. \(\square\)

Worked Example: A Uniform Estimate on \([1/4,1]\)

Set \(\delta=1/4\). On \([1/4,1]\), the error in approximating one satisfies

|f_n(x)-1| =\frac{1}{1+nx} \leq\frac{1}{1+n/4} =\frac{4}{n+4}.

For example, when \(n=36\), this bound is \(4/40=1/10\); when \(n\) increases, the bound tends to zero. More generally, for a tolerance \(\varepsilon>0\), any integer \(N\) satisfying \(4/(N+4)<\varepsilon\) works for every \(n\geq N\) and every \(x\in[1/4,1]\). On the full interval \([0,1]\), no such estimate can tend to zero: inputs arbitrarily close to zero have errors close to one. Restricting away from zero removes precisely that region.

A Moving Peak Can Defeat Uniform Convergence

Nonuniformity need not come from a discontinuous pointwise limit. An error can instead form a peak whose location changes with \(n\). At every fixed positive input the peak eventually moves away, but its height remains unchanged. The following example has a continuous zero limit and an exactly computable peak.

Worked Example: A Peak That Moves Toward Zero

Define \(g_n:[0,1]\to\mathbb{R}\) by

g_n(x)=\frac{nx}{1+n^2x^2}.

At \(x=0\), \(g_n(0)=0\). If \(x>0\) is fixed, then

0\leq g_n(x) =\frac{nx}{1+n^2x^2} \leq\frac{nx}{n^2x^2} =\frac{1}{nx} \longrightarrow0.

Hence \(g_n\) converges pointwise to zero on \([0,1]\). To find its largest error, put \(y=nx\), so \(y\geq0\). The inequality \((y-1)^2\geq0\) gives \(2y\leq1+y^2\), and therefore

0\leq g_n(x)=\frac{y}{1+y^2}\leq\frac12.

Equality holds when \(y=1\), that is, at \(x=1/n\), which belongs to \([0,1]\) for every positive integer \(n\). Thus \(\sup_{x\in[0,1]}|g_n(x)|=1/2\) for every \(n\), so the sequence does not converge uniformly to zero. The peak has fixed height one half, while its location \(1/n\) moves toward zero.

This example emphasizes a useful distinction. For the rational boundary-layer sequence, the supremum error is approached near the endpoint but not attained. For \(g_n\), the supremum error is attained at a point that itself changes with \(n\). In either case, fixed-input limits alone do not reveal the obstruction; the error must be examined across the whole domain.

An Unbounded Domain Can Also Hide Large Errors

An unbounded domain creates another possible failure: inputs far from the origin can have large errors even as each fixed input behaves well. This contrasts with examples where denominator estimates gave uniform convergence on the half-line. Here the changing inputs escape farther and farther out.

Worked Example: Errors Escaping to Infinity

For \(n\geq1\), define \(h_n:[0,\infty)\to\mathbb{R}\) by \(h_n(x)=n/(n+x)\). For each fixed \(x\geq0\),

h_n(x)=\frac{1}{1+x/n}\longrightarrow1.

Thus the pointwise limit is the constant function \(h(x)=1\). The error at a finite input is

|h_n(x)-1| =1-\frac{n}{n+x} =\frac{x}{n+x}.

For every \(x\geq0\), this error is less than one, but it approaches one as \(x\) tends to infinity. Hence its supremum on \([0,\infty)\) is one for every \(n\), and convergence is not uniform. A moving-input calculation makes the failure explicit: with \(x_n=n^2\),

|h_n(x_n)-1| =\frac{n^2}{n+n^2} =\frac{n}{n+1} \longrightarrow1.

Each fixed \(x\) gives an error tending to zero, while inputs chosen farther out keep the error close to one. The pointwise statement and the uniform statement test different things.

Reading a Nonuniformity Argument

A proof of nonuniform convergence should identify why no single index controls the error everywhere. An exact supremum calculation is one route. Another is to find inputs \(x_n\) for which the errors fail to tend to zero; the Moving-Input Criterion for Uniform Convergence then rules out uniform convergence. In either approach, the proposed pointwise limit must first be identified correctly.

A common pitfall is to treat a weak estimate as proof of nonuniformity. For example, an estimate that does not tend to zero does not show that the actual errors stay large; it may simply be too crude. The examples above establish failure by calculating the errors themselves, their suprema, or specific moving-input errors. Conversely, showing that an error is small at every fixed input is only pointwise reasoning. Uniform convergence requires control over all inputs at once.

Find the pointwise limit.
Check each fixed input, including endpoints and any points where a formula changes behavior.

Write the actual error.
Simplify \(|f_n(x)-f(x)|\) before drawing a conclusion about uniform convergence.

Search across the domain.
Look for a boundary region, a moving peak, or inputs escaping to infinity where the error remains large.

Verify the obstruction.
Calculate the supremum error or exhibit changing inputs with errors bounded away from zero. A failed upper estimate alone is not enough.

The examples also show that nonuniformity can depend strongly on the domain. The boundary-layer sequence fails to converge uniformly on \([0,1]\), but it converges uniformly after the interval is restricted to \([\delta,1]\) for any \(\delta>0\). On an unbounded domain, a sequence can converge pointwise while errors persist at increasingly distant inputs. Always include the domain in the claim and in the error analysis.

Takeaway: To prove nonuniform convergence, find an obstruction that survives as the index grows: a supremum error that does not vanish or a sequence of inputs whose errors stay away from zero. Pointwise convergence checks fixed inputs; uniform convergence must control the entire domain.

Check Your Understanding

Use the error calculations and domain comparisons above to answer the following questions.

What is the pointwise limit of \(f_n(x)=x/(x+1/n)\) at \(x=0\) and at \(x>0\)?
Why is the supremum error for this boundary-layer sequence equal to one even though it is not attained?
For \(g_n(x)=nx/(1+n^2x^2)\), where is the peak, and what is its height?
Why does restricting the rational boundary-layer sequence to \([\delta,1]\), with \(\delta>0\), give uniform convergence?
For \(h_n(x)=n/(n+x)\), which inputs show that errors remain large on the half-line?