What is pointwise convergence? — Mathematics

Definition 10.1 University Mathematics — Year 2 · Chapter 10 — Sequences and Series of Functions

Let $f_n, f \colon X \to \R$ (or $\C$ , or a normed space), $X$ any set. $(f_n)$ converges to $f$ pointwise when $f_n(x) \to f(x)$ for every $x$ ; uniformly when

\norm{f_n - f}_\infty = \sup_{x \in X}\, \abs{f_n(x) - f(x)} \xrightarrow[n \to \infty]{} 0 .

Uniform implies pointwise; on $C(\intcc{a}{b})$ , uniform convergence is exactly convergence in the Banach space $\bigl(C(\intcc{a}{b}), \norm\cdot_\infty\bigr)$ of Chapter 5.

The sequence xn on [0, 1]: the graphs sag toward 0 but all must climb to 1 at x = 1 — the sup distance to the discontinuous pointwise limit never shrinks below a constant. — The sequence $x^n$ on $\intcc{0}{1}$ : the graphs sag toward $0$ but all must climb to $1$ at $x = 1$ — the sup distance to the discontinuous pointwise limit never shrinks below a constant.

Examples

Example 10.2

On $\intcc{0}{1}$ , $f_n(x) = x^n$ converges pointwise to the discontinuous limit $f = \mathbf{1}_{\{1\}}$ ; the convergence is not uniform: $\norm{f_n - f}_\infty \geq f_n\bigl(1 - \tfrac1n\bigr) = (1 - \tfrac1n)^n \to \eu^{-1} \neq 0$ . On $\intcc{0}{a}$ with $a < 1$ it is uniform ( $\sup = a^n \to 0$ ): uniformity is a property of the domain as much as of the sequence.

Example 10.3 (Two limits that refuse to commute)

The whole chapter is about interchanging limits, so here is the smallest possible failure. Let $a_{n,m} = \dfrac{n}{n+m}$ for $n, m \geq 1$ . Then

\lim_{m\to\infty}\Bigl(\lim_{n\to\infty}a_{n,m}\Bigr) = \lim_{m\to\infty} 1 = 1, \qquad \lim_{n\to\infty}\Bigl(\lim_{m\to\infty}a_{n,m}\Bigr) = \lim_{n\to\infty} 0 = 0 :

both iterated limits exist and they differ. Every transfer theorem of this chapter is a licence to commute two limits — $\lim_n$ with $\lim_{x\to a}$ (continuity), with $\int$ (integration), with $\frac{\dd}{\dd x}$ (differentiation) — and uniform convergence is precisely the fee that makes the commutation legal. Closing insight: whenever a “proof” silently swaps two limit operations, this two-line array is the counterexample to hold against it; the sliding bumps of Exercise 10.2 are the same phenomenon wearing an integral sign.

Example 10.5 (Uniformity fails exactly where the limit breaks)

On $\intcc{0}{2}$ , let $f_n(x) = \dfrac{x^n}{1 + x^n}$ . The pointwise limit is a three-piece function:

f(x) = \begin{cases} 0 & 0 \leq x < 1,\\[2pt] \tfrac12 & x = 1,\\[2pt] 1 & 1 < x \leq 2, \end{cases}

discontinuous at $1$ , so by Theorem 10.4 the convergence cannot be uniform on $\intcc{0}{2}$ . On the closed pieces avoiding the threshold it is: for $0 \leq x \leq a < 1$ ,

\sup_{\intcc{0}{a}}\abs{f_n - 0} = \frac{a^n}{1 + a^n} \leq a^n \to 0 ,

and for $1 < b \leq x \leq 2$ ,

\sup_{\intcc{b}{2}}\abs{f_n - 1} = \frac{1}{1 + b^n} \leq b^{-n} \to 0 ,

both suprema computed by monotonicity of $u \mapsto \frac{u}{1+u}$ and of $x \mapsto x^n$ . Closing insight: the failure of uniformity is localized at the discontinuity of the limit — the same geometry as Example 10.2, and the reason the “uniform on every segment inside” discipline recurs all chapter long.

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