Uniform vs Pointwise Convergence: A Case Study

April 9, 2026 · Luciano Muratore

It is well-known that a sequence of functions can converge to a limit in more than one sense. The question is: how strong is that convergence?

This is not a trivial distinction. Pointwise convergence asks whether, for each fixed point, the sequence eventually gets close to the limit. Uniform convergence asks for something stronger: can we find a single threshold that works simultaneously for all points?

To make this concrete, let us work through a specific example.

The Sequence

For each $n \in \mathbb{N}$, define $f_n : [0, 1] \to \mathbb{R}$ by

$f_n(x) = \begin{cases} 1, & x \in \left[\frac{1}{n}, \frac{2}{n}\right], \\ 0, & \text{otherwise.} \end{cases}$

Each $f_n$ is bounded and its support is precisely the interval $\left[\frac{1}{n}, \frac{2}{n}\right]$, a window that slides toward $0$ as $n \to \infty$.

Pointwise Convergence

Recall the definition: a sequence $(f_n)$ converges pointwise to $f$ on $A$ if

$\forall x \in A : \lim_{n \to \infty} f_n(x) = f(x).$

The limit is taken with respect to $n$ for each fixed $x$ — the point is frozen, and we ask whether the sequence eventually settles there.

For our sequence, fix $x \in [0,1]$.

If $x = 0$, then $x \notin \left[\frac{1}{n}, \frac{2}{n}\right]$ for any $n$, so $f_n(0) = 0$ for all $n$.

If $x > 0$, choose $N \in \mathbb{N}$ such that $N > \frac{2}{x}$. Then for all $n \geq N$,

$\frac{2}{n} < x \implies x \notin \left[\frac{1}{n}, \frac{2}{n}\right] \implies f_n(x) = 0.$

So $\lim_{n \to \infty} f_n(x) = 0$ for every $x \in [0, 1]$, and we conclude $f_n \xrightarrow{\text{pointwise}} 0$ on $[0,1]$.

Failure of Uniform Convergence

Uniform convergence imposes a global bound: $(f_n)$ converges uniformly to $f$ on $A$ if

$\forall \varepsilon > 0,\ \exists N \in \mathbb{N} : \forall n \geq N,\ \forall x \in A,\ |f_n(x) - f(x)| < \varepsilon,$

or equivalently, $\sup_{x \in A} |f_n(x) - f(x)| \to 0$.

With $f \equiv 0$, let us compute this supremum. Since $f_n(x) = 1$ for every $x \in \left[\frac{1}{n}, \frac{2}{n}\right]$, we have

$\sup_{x \in [0,1]} |f_n(x) - 0| = 1 \quad \forall n.$

This supremum never goes to $0$, so $f_n \xrightarrow{\text{uniformly}} \hspace{-2.6em}\setminus\hspace{1.6em} 0$ on $[0, 1]$.

Why Compactness Restores Uniform Convergence

The failure above traces back to a single culprit: the point $0$. The supports $\left[\frac{1}{n}, \frac{2}{n}\right]$ accumulate at $0$ as $n \to \infty$, so no single $N$ can push the bump away from points near $0$.

Now, what if we restrict to a compact subset $K \subset (0, 1]$? Since $K$ is closed and bounded and $0 \notin K$, there exists $a > 0$ such that $K \subset [a, 1]$. In other words, $K$ stays bounded away from the problematic accumulation point:

$\text{dist}(K, 0) = \inf_{x \in K} x = a > 0.$

This is the key. Since $\text{supp}(f_n) = \left[\frac{1}{n}, \frac{2}{n}\right]$, choose $N > \frac{2}{a}$. Then for all $n \geq N$,

$\text{supp}(f_n) \cap K = \emptyset \implies f_n(x) = 0 \quad \forall x \in K.$

Therefore $\sup_{x \in K} |f_n(x) - 0| = 0$, and $f_n \xrightarrow{\text{uniformly}} 0$ on $K$.

I want to make a pause here. The compactness of $K$ is not the direct reason for uniform convergence — it is the positive distance from $0$ that does the work. Compactness in $(0,1]$ guarantees that such a distance exists. If we had taken a non-compact set like $(0, 1]$ itself, the infimum of $x$ over the set would be $0$, and the argument would collapse.

The moral: uniform convergence is a global property, and it can fail precisely where the limit function is approached in a non-uniform manner. Compactness, when it excludes accumulation points of the supports, is the geometric condition that restores uniformity.