Uniform Convergence

One Index for the Whole Domain

Pointwise convergence studies one input at a time: after fixing $x$, the index needed to make $f_n(x)$ close to its limit may depend on $x$. Uniform convergence strengthens this requirement. For a chosen error tolerance, it asks for one index that works for every input in the domain at once.

This distinction is about the order of the quantifiers. In pointwise convergence, the input is fixed before the index is chosen. In uniform convergence, the index must be chosen before the input is specified. That shared index is what makes uniform convergence useful for controlling an entire family of errors, rather than a single sequence of numerical errors at a time.

The Definition

Definition: Let $E\subseteq\mathbb{R}$ be nonempty, and let $f_n:E\to\mathbb{R}$ for each positive integer $n$. The sequence $(f_n)$ converges uniformly on $E$ to $f:E\to\mathbb{R}$ if, for every $\varepsilon>0$, there is a positive integer $N$ such that $$ n\geq N \quad\Longrightarrow\quad |f_n(x)-f(x)|<\varepsilon \quad\text{for every }x\in E. $$ We write $f_n\to f$ uniformly on $E$.

The index $N$ may depend on $\varepsilon$, but it may not depend on $x$. Once $N$ is selected, the estimate must hold for every point of $E$ and for every $n\geq N$. Uniform convergence implies pointwise convergence: fix any $x\in E$ and apply the uniform estimate at that point. The converse need not hold.

It is useful to define the error at $x$ and index $n$ as $|f_n(x)-f(x)|$. Uniform convergence says that the largest possible error across the domain becomes small, in the sense that all errors are below any chosen tolerance once $n$ is large enough. “Largest possible” should not be taken to mean that some point actually attains the largest error: a supremum need not be attained.

Worked Examples

Worked Example: A Uniform Estimate on a Bounded Interval

For each positive integer $n$, define

f_n(x)=\frac{x^2+1}{n+x^2}, \qquad x\in[-2,2].

We show that $f_n$ converges uniformly on $[-2,2]$ to the zero function. For every $x$ in this interval, $x^2\leq4$, so $x^2+1\leq5$, while $n+x^2\geq n$. Therefore

|f_n(x)-0| =\frac{x^2+1}{n+x^2} \leq\frac{5}{n}.

Given $\varepsilon>0$, choose a positive integer $N$ such that $N>5/\varepsilon$. For every $n\geq N$ and every $x\in[-2,2]$, the estimate gives

|f_n(x)-0|\leq\frac{5}{n}\leq\frac{5}{N}<\varepsilon.

The choice of $N$ depends on $\varepsilon$, not on $x$, so the convergence is uniform.

Worked Example: Pointwise Convergence Without Uniform Convergence

Define $f_n(x)=x/(n+x)$ on $E=[0,\infty)$. Fix $x\in E$. If $x=0$, then $f_n(x)=0$ for every $n$. If $x>0$, then

0\leq f_n(x)=\frac{x}{n+x}\leq\frac{x}{n}\longrightarrow0.

Thus $f_n$ converges pointwise on $E$ to the zero function. To test uniform convergence, let $\varepsilon=1/4$. For every proposed index $N$, take $n=N$ and choose the domain point $x=n$. Substitution gives

|f_n(n)-0| =\frac{n}{n+n} =\frac{1}{2} >\frac{1}{4}.

Consequently, no index $N$ can make the error less than $1/4$ for every $x\in E$ and every $n\geq N$. The convergence is not uniform. The test point changes with the index, which is permitted when checking whether a single index works across the whole domain.

Worked Example: Uniform Convergence on the Whole Real Line

Let $f_n(x)=\sin(nx)/n$ for $x\in\mathbb{R}$. The Sine–Cosine Identity established earlier in this course gives $|\sin y|\leq1$ for every real $y$. Hence, for every $x\in\mathbb{R}$,

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

Given $\varepsilon>0$, choose $N$ such that $N>1/\varepsilon$. Then, for all $n\geq N$ and all real $x$,

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

Thus $f_n$ converges uniformly to zero on $\mathbb{R}$, even though the domain is unbounded. The domain’s size alone does not determine whether convergence is uniform; what matters is whether a single error bound works throughout it.

A Supremum Test for Uniform Convergence

The definition can be expressed using the supremum of the errors. If the errors are unbounded for some $n$, regard their supremum as $+\infty$. This convention lets us state a useful criterion without assuming in advance that every $f_n-f$ is bounded.

Theorem (Supremum Criterion for Uniform Convergence): Let $E$ be nonempty, and let $f_n,f:E\to\mathbb{R}$. Define $$ e_n=\sup_{x\in E}|f_n(x)-f(x)|, $$ allowing $e_n=+\infty$. Then $f_n\to f$ uniformly on $E$ if and only if $e_n\to0$.

Proof. Suppose first that $f_n\to f$ uniformly. Let $\eta>0$. By the definition, there is an $N$ such that, whenever $n\geq N$, every $x\in E$ satisfies

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

Taking the supremum over $x$ gives $e_n\leq\eta/2<\eta$ for every $n\geq N$. Thus $e_n\to0$, and the same estimate also shows that $e_n$ is finite for all sufficiently large $n$.

Conversely, suppose $e_n\to0$. Given $\varepsilon>0$, choose $N$ such that $e_n<\varepsilon$ for every $n\geq N$. For every $x\in E$, the definition of supremum implies

|f_n(x)-f(x)|\leq e_n<\varepsilon \qquad(n\geq N).

The same $N$ works for every $x$, so $f_n\to f$ uniformly. $\square$

When every error function is bounded, this criterion is often written by saying that the supremum of the error tends to zero. It does not require an input where the supremum is attained. The estimates in the examples work directly for every input, so they establish bounds on the supremum whether or not any particular point realizes it.

Consequences of Uniform Error Control

Uniform error estimates can be combined with other bounds that hold across the domain. The following result shows why multiplication by a bounded function is compatible with uniform convergence.

Theorem (Multiplication by a Bounded Function): Suppose $f_n\to f$ uniformly on $E$, and $h:E\to\mathbb{R}$ is bounded. Then $hf_n\to hf$ uniformly on $E$.

Proof. Since $h$ is bounded, there is a constant $M\geq0$ such that $|h(x)|\leq M$ for every $x\in E$. If $M=0$, then $h(x)=0$ everywhere, and $hf_n=hf=0$, so the claim holds. Now suppose $M>0$. Given $\varepsilon>0$, uniform convergence of $f_n$ to $f$ gives an index $N$ such that for all $n\geq N$ and all $x\in E$,

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

For those $n$ and $x$, it follows that

|h(x)f_n(x)-h(x)f(x)| =|h(x)|\,|f_n(x)-f(x)| \leq M|f_n(x)-f(x)| <\varepsilon.

The index works throughout $E$, so $hf_n\to hf$ uniformly. $\square$

Worked Example: Applying a Bounded Multiplier

For $f_n(x)=\sin(nx)/n$ on $\mathbb{R}$, the previous example showed that $f_n\to0$ uniformly. Let $h(x)=\cos x$. The Sine–Cosine Identity also gives $|\cos x|\leq1$ for every real $x$. Therefore

|h(x)f_n(x)-0| =|\cos x|\frac{|\sin(nx)|}{n} \leq\frac{1}{n} \qquad(x\in\mathbb{R}).

Given $\varepsilon>0$, any positive integer $N>1/\varepsilon$ makes this error less than $\varepsilon$ for all $n\geq N$ and all real $x$. Thus $\cos(x)\sin(nx)/n$ converges uniformly to zero on $\mathbb{R}$.

A related consequence is that if $f$ is bounded and $f_n\to f$ uniformly, then the functions $f_n$ are bounded by one common constant for all sufficiently large $n$. Indeed, choose $K\geq0$ such that $|f(x)|\leq K$ for all $x$. Uniform convergence with tolerance $1$ provides $N$ such that

|f_n(x)|\leq |f(x)|+|f_n(x)-f(x)|<K+1 \qquad(n\geq N,\ x\in E).

This conclusion concerns the sequence from some index onward; the finitely many earlier functions need not be bounded.

Why the Shared Index Matters

Uniform convergence supplies a single error estimate across the domain. That is why it can be used in results that require controlling the functions at many points simultaneously. For example, the theorem on uniform limits of continuous functions established earlier in this course says that a uniform limit of continuous functions is continuous. The shared index is essential: pointwise convergence alone does not provide the uniform control used in that result.

A common pitfall is to prove that the error tends to zero for each fixed $x$ and then conclude that convergence is uniform. The example $x/(n+x)$ shows why that is insufficient: its pointwise error tends to zero, but the input $x=n$ keeps the error equal to $1/2$. Conversely, an unbounded domain does not automatically prevent uniform convergence; the estimate $1/n$ for $\sin(nx)/n$ works for every real input.

Takeaway: Uniform convergence means that, for each tolerance, one index controls the error at every point of the domain. A reliable method is to seek a bound for $|f_n(x)-f(x)|$ that does not depend on $x$, then show that the bound tends to zero.

Check Your Understanding

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

In the definition of uniform convergence, which quantities may the index $N$ depend on, and which may it not depend on?
Why does the estimate $|f_n(x)|\leq 5/n$ for every $x\in[-2,2]$ prove uniform convergence to zero?
For $f_n(x)=x/(n+x)$ on $[0,\infty)$, what input demonstrates that the convergence to zero is not uniform?
How does the supremum criterion express uniform convergence in terms of the errors?
Why does multiplying a uniformly convergent sequence by a bounded function preserve uniform convergence?

What You'll Learn