Uniform Limits and Continuity

Why the Kind of Convergence Matters

A sequence of functions can converge at every point and still have a limit with very different behavior from its terms. The crucial question is whether the approximation is controlled separately at each point, or uniformly across the whole domain. Uniform convergence gives one error bound that works everywhere; this global control is what makes it useful when studying continuity.

Let $E\subseteq\mathbb{R}$, and let $f_n:E\to\mathbb{R}$ for each positive integer $n$. Pointwise convergence of $(f_n)$ to $f:E\to\mathbb{R}$ means that for each fixed $x\in E$, the real sequence $(f_n(x))$ converges to $f(x)$. The index after which the error is small may depend on $x$. Uniform convergence requires that index to work for every $x$ at once.

Definition: The sequence $(f_n)$ converges uniformly to $f$ on $E$ if, for every $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that whenever $n\geq N$ and $x\in E$, $$ |f_n(x)-f(x)|<\varepsilon. $$ The choice of $N$ may depend on $\varepsilon$, but it must not depend on $x$.

If the difference $f_n-f$ is bounded on $E$, its supremum norm is $\|f_n-f\|_\infty=\sup_{x\in E}|f_n(x)-f(x)|$. In that case, uniform convergence is equivalent to $\|f_n-f\|_\infty\to0$. The definition above is more general: it does not require this supremum to be finite for every early term of the sequence.

Pointwise and Uniform Convergence in Examples

Worked Example: Powers on the Unit Interval

For each positive integer $n$, let $f_n:[0,1]\to\mathbb{R}$ be $f_n(x)=x^n$. If $0\leq x<1$, then $x^n\to0$; at $x=1$, $x^n=1$ for every $n$. Thus the pointwise limit is

f(x)= \begin{cases} 0,&0\leq x<1,\\ 1,&x=1. \end{cases}

This limit is not continuous at $1$. The convergence is not uniform: for any $n$, the error is $x^n$ when $x<1$, and zero at $x=1$. As $x$ approaches $1$ from below, $x^n$ approaches $1$. Therefore

\sup_{x\in[0,1]}|f_n(x)-f(x)|=1

for every $n$. The supremum is not attained, but values of the error can be made arbitrarily close to $1$. Each $f_n$ is continuous, yet the pointwise limit is discontinuous; the failure of uniform convergence is essential to this example.

Worked Example: A Uniform Approximation on the Real Line

Define $f_n:\mathbb{R}\to\mathbb{R}$ by $f_n(x)=x+\frac{\sin x}{n}$, and let $f(x)=x$. Since $|\sin x|\leq1$ for every real $x$,

|f_n(x)-f(x)|=\frac{|\sin x|}{n}\leq\frac1n \qquad(x\in\mathbb{R}).

Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\geq N$, the error is at most $1/n\leq1/N<\varepsilon$, independently of $x$. Thus $f_n\to f$ uniformly on $\mathbb{R}$. Each $f_n$ and the limit $f$ are continuous. Here the whole graph of each $f_n$ stays within a uniformly shrinking vertical distance of the graph of $f$.

Worked Example: Continuous Functions Converging Uniformly to Zero

On $\mathbb{R}$, set $g_n(x)=\frac{x^2}{n(1+x^2)}$. The denominator is positive for every real $x$, so $g_n$ is continuous. Also $0\leq x^2/(1+x^2)\leq1$, since $x^2\leq1+x^2$. Consequently,

|g_n(x)-0|=\frac{x^2}{n(1+x^2)}\leq\frac1n \qquad(x\in\mathbb{R}).

For any $\varepsilon>0$, choose $N$ so that $1/N<\varepsilon$. Then $n\geq N$ implies $|g_n(x)|\leq1/n\leq1/N<\varepsilon$ for every $x$. Hence $g_n$ converges uniformly to the zero function. The estimate works even though the domain is unbounded.

Uniform Limits Are Unique

Uniform convergence is stronger than pointwise convergence, but it does not change what the limit is allowed to be: a sequence cannot converge uniformly to two different functions. The argument uses the same triangle inequality that relates ordinary limits of real sequences.

Theorem: Let $E\subseteq\mathbb{R}$, and suppose $(f_n)$ converges uniformly on $E$ to both $f:E\to\mathbb{R}$ and $g:E\to\mathbb{R}$. Then $f(x)=g(x)$ for every $x\in E$.

Proof. Fix $\varepsilon>0$. Uniform convergence to $f$ gives an index $N_1$ such that $n\geq N_1$ implies $|f_n(x)-f(x)|<\varepsilon/2$ for every $x\in E$. Uniform convergence to $g$ gives $N_2$ with the corresponding bound $|f_n(x)-g(x)|<\varepsilon/2$. Choose $n$ at least as large as both indices. For every $x\in E$, the triangle inequality gives

|f(x)-g(x)| \leq |f(x)-f_n(x)|+|f_n(x)-g(x)| <\varepsilon.

This holds for every $\varepsilon>0$, so $|f(x)-g(x)|=0$ for each $x\in E$. Therefore $f=g$ on $E$. $\square$

What Uniform Convergence Says About Boundedness

Uniform convergence controls the difference between a function and its approximants. It does not, by itself, say that any of them is bounded. Boundedness transfers when a sufficiently late approximant is bounded, or when all approximants are bounded. It also transfers in the other direction: if the limit is bounded, then sufficiently late approximants are bounded.

Theorem: Suppose $f_n:E\to\mathbb{R}$ converges uniformly to $f:E\to\mathbb{R}$. If every $f_n$ is bounded on $E$, then $f$ is bounded on $E$. Conversely, if $f$ is bounded, then every sufficiently late $f_n$ is bounded.

Proof. For the first claim, uniform convergence supplies an index $N$ such that $|f_N(x)-f(x)|<1$ for every $x\in E$. Because $f_N$ is bounded, there is $M\geq0$ such that $|f_N(x)|\leq M$ on $E$. Thus, for every $x\in E$,

|f(x)|\leq|f(x)-f_N(x)|+|f_N(x)|<1+M.

So $f$ is bounded. For the converse, choose $M\geq0$ with $|f(x)|\leq M$ on $E$. Uniform convergence gives $N$ such that $n\geq N$ implies $|f_n(x)-f(x)|<1$ for every $x\in E$. Hence

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

Each such $f_n$ is bounded. $\square$

Worked Example: Uniform Convergence Need Not Give Bounded Functions

On $E=\mathbb{R}$, define $f_n(x)=x+1/n$ and $f(x)=x$. For every $x\in\mathbb{R}$,

|f_n(x)-f(x)|=\frac1n, \qquad \sup_{x\in\mathbb{R}}|f_n(x)-f(x)|=\frac1n.

Since $1/n\to0$, the convergence is uniform. Nevertheless, every $f_n$ is unbounded: as $x$ ranges over $\mathbb{R}$, $x+1/n$ does too. The limit $f(x)=x$ is also unbounded. Thus uniform convergence alone does not imply that the functions, or their limit, are bounded.

It is also important not to use the boundedness theorem with an arbitrary early term in place of a sufficiently late one. For example, let $f_1(x)=0$ and $f_n(x)=x$ for $n\geq2$, on $\mathbb{R}$. This sequence converges uniformly to $f(x)=x$, because every term from $n=2$ onward equals the limit. The first term is bounded, but the limit is unbounded. The proof of the theorem uses an index at which the function is both bounded and uniformly close to the limit; boundedness of an unrelated early term is not enough.

How Uniform Approximation Connects to Continuity

The defining distinction can now be read as a distinction in control. At each fixed $x$, pointwise convergence eventually makes $f_n(x)$ close to $f(x)$, but the required index may grow as $x$ changes. Uniform convergence rules out that dependence: one index controls the error over all of $E$.

This matters for continuity because continuity compares function values at nearby inputs. When a continuous approximant is uniformly close to a limit, the limit's change between two nearby points can be estimated using three changes: from the limit to the approximant at the first point, within the approximant, and back to the limit at the second point. The first and last changes are controlled by uniform convergence; the middle one is controlled by continuity of the approximant. The next tutorial develops this argument into a complete proof that uniform limits of continuous functions are continuous.

The example $f_n(x)=x^n$ on $[0,1]$ shows why the uniform hypothesis matters: pointwise convergence alone allowed continuous functions to approach a discontinuous limit. The examples with errors bounded by $1/n$, by contrast, demonstrate the kind of global control used in the continuity argument. The essential task in applications is therefore to find an error estimate that is independent of the input $x$, not merely to check convergence at each fixed input.

Check Your Understanding

Use the definitions, examples, and results above 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 $x^n$ converge pointwise but not uniformly to the stated limit on $[0,1]$?
What additional boundedness condition on the approximants allows boundedness to pass to a uniform limit?
Why does the bounded first term in the example $f_1=0$, $f_n(x)=x$ for $n\geq2$, not imply that the limit is bounded?
Describe the three changes used to compare the values of a limit function at two nearby inputs with values of a continuous approximant.

What You'll Learn