Pointwise and Uniform Convergence, calculus lesson

Picture the graphs of $x$ , $x^2$ , $x^3$ , $x^4$ on $[0, 1]$ . Each one sags lower than the last, yet every graph still reaches up to height $1$ at the right end. At each single point the heights settle down, but the graphs as a whole never get uniformly close to their limit. That gap between “at each point” and “everywhere at once” decides whether limits of functions keep continuity and whether you may swap a limit with an integral.

Key terms

Sequences and series of functions

Definition A sequence of functions on a set $D \subseteq \mathbb{R}$ is a list $f_1, f_2, f_3, \ldots$ where each $f_n$ is a real function defined on $D$ . A series of functions $\sum_{n=1}^{\infty} f_n$ on $D$ is the sequence of its partial sums

S_N(x) = \sum_{n=1}^{N} f_n(x), \qquad x \in D.

Every convergence question about a series of functions is a question about the sequence $(S_N)$ .

In plain words Instead of a list of numbers you have a list of graphs. Fixing $x$ turns it back into an ordinary sequence of numbers.

Example On $D = [0, 1]$ let $f_n(x) = x^n$ . At $x = \frac{1}{2}$ the values are $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots$ , and at $x = 1$ they are $1, 1, 1, \ldots$ . A power series $\sum c_n x^n$ is a series of functions with $f_n(x) = c_n x^n$ .

Pointwise convergence

Definition The sequence $(f_n)$ converges pointwise on $D$ to a function $f$ if $\lim_{n \to \infty} f_n(x) = f(x)$ for every $x \in D$ . In detail, for every $x \in D$ and every $\varepsilon > 0$ there is an $N$ , which may depend on both $\varepsilon$ and $x$ , such that

n \ge N \implies \lvert f_n(x) - f(x) \rvert < \varepsilon.

A series $\sum f_n$ converges pointwise on $D$ when its partial sums do.

In plain words Check each point on its own. At every $x$ the numbers $f_n(x)$ must settle down, but different points may settle at very different speeds.

Example For $f_n(x) = x^n$ on $[0, 1]$ the pointwise limit is

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

Every $f_n$ is continuous, yet the limit $f$ jumps at $x = 1$ . Pointwise limits can lose continuity.

Uniform convergence

Definition The sequence $(f_n)$ converges uniformly on $D$ to $f$ if for every $\varepsilon > 0$ there is an $N$ , depending only on $\varepsilon$ , such that

n \ge N \implies \lvert f_n(x) - f(x) \rvert < \varepsilon \quad \text{for every } x \in D.

Uniform convergence implies pointwise convergence, but not conversely. If every $f_n$ is continuous on $D$ and $f_n \to f$ uniformly on $D$ , then $f$ is continuous on $D$ .

In plain words One $N$ has to work for every point at once. Past that $N$ , the whole graph of $f_n$ stays inside a band of width $\varepsilon$ around the graph of $f$ .

Example The functions $x^n$ do not converge uniformly on $[0, 1]$ , since their continuous graphs would then have a continuous limit. On the smaller interval $[0, \frac{1}{2}]$ they do converge uniformly to $0$ , because $\lvert x^n - 0 \rvert \le \left(\frac{1}{2}\right)^n$ there, and this bound is below any $\varepsilon > 0$ once $n$ is large.

Uniform error bounds

Definition For functions on $D$ put

d_n = \sup_{x \in D} \lvert f_n(x) - f(x) \rvert.

Then $f_n \to f$ uniformly on $D$ if and only if $d_n \to 0$ . In particular, if there are numbers $e_n \to 0$ with $\lvert f_n(x) - f(x) \rvert \le e_n$ for all $x \in D$ and all $n$ , the convergence is uniform.

In plain words Find the worst error over the whole set for each $n$ . Uniform convergence means that worst error goes to $0$ .

Example On $[0, 1)$ the worst error of $x^n$ against the limit $0$ is $\sup_{0 \le x < 1} x^n = 1$ for every $n$ , so the convergence is not uniform there either. By contrast $f_n(x) = \frac{\sin(nx)}{n}$ satisfies

\lvert f_n(x) - 0 \rvert \le \frac{1}{n} \quad \text{for every real } x,

so $f_n \to 0$ uniformly on $\mathbb{R}$ .

Weierstrass M-test

Definition Let $f_n$ be functions on $D$ . Suppose there are constants $M_n \ge 0$ with

\lvert f_n(x) \rvert \le M_n \quad \text{for all } x \in D \text{ and all } n, \qquad \sum_{n=1}^{\infty} M_n < \infty.

Then $\sum f_n(x)$ converges absolutely for every $x \in D$ , and the series $\sum f_n$ converges uniformly on $D$ .

In plain words If each term is trapped under a number that does not depend on $x$ , and those numbers form a convergent series, the series of functions converges uniformly.

Example Consider the series

\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2}

on $\mathbb{R}$ and take $M_n = \frac{1}{n^2}$ . Since $\sum \frac{1}{n^2}$ converges, the series converges uniformly on all of $\mathbb{R}$ . Its terms are continuous, so its sum is a continuous function.

Exchanging limits and integrals

Definition Let each $f_n$ be continuous on $[a, b]$ and let $f_n \to f$ uniformly on $[a, b]$ . Then $f$ is continuous on $[a, b]$ and

\lim_{n \to \infty} \int_a^b f_n(x) \, dx = \int_a^b f(x) \, dx.

For series, if each $f_n$ is continuous on $[a, b]$ and $\sum f_n$ converges uniformly on $[a, b]$ , then

\int_a^b \sum_{n=1}^{\infty} f_n(x) \, dx = \sum_{n=1}^{\infty} \int_a^b f_n(x) \, dx.

Pointwise convergence alone is not enough.

In plain words With uniform convergence the area under $f_n$ approaches the area under $f$ . Without it, area can hide in a thin spike that slides away.

Example On $[0, 1]$ let $f_n(x) = n x e^{-n x^2}$ . For each fixed $x > 0$ the exponential wins and $f_n(x) \to 0$ , and $f_n(0) = 0$ , so the pointwise limit is $f = 0$ . But

\int_0^1 n x e^{-n x^2} \, dx = \frac{1 - e^{-n}}{2} \to \frac{1}{2} \neq 0 = \int_0^1 f(x) \, dx.

The convergence is not uniform, since $f_n\left(\frac{1}{\sqrt{n}}\right) = \sqrt{n} \, e^{-1}$ grows without bound.

Derivatives need more hypotheses

Definition Uniform convergence of $f_n$ does not imply convergence of $f_n'$ . The standard result is this. Let each $f_n$ be continuously differentiable on $[a, b]$ , let $(f_n(x_0))$ converge for at least one $x_0 \in [a, b]$ , and let $(f_n')$ converge uniformly on $[a, b]$ . Then $(f_n)$ converges uniformly on $[a, b]$ to a differentiable function $f$ , and

f'(x) = \lim_{n \to \infty} f_n'(x) \quad \text{for all } x \in [a, b].

For series the condition is that the series of derivatives $\sum f_n'$ converges uniformly.

In plain words A graph can be uniformly close to a flat line while wiggling steeply. To differentiate the limit you need control of the derivatives themselves.

Example The functions $f_n(x) = \frac{\sin(nx)}{n}$ converge uniformly to $0$ on $\mathbb{R}$ . Their derivatives $f_n'(x) = \cos(nx)$ give $f_n'(0) = 1$ for every $n$ , not $0$ , and at $x = \pi$ they give $(-1)^n$ , which does not converge at all.

Check yourself

Show that $f_n(x) = \frac{x}{1 + nx}$ converges uniformly to $0$ on $[0, \infty)$ .

Show the answer

For $x > 0$ we have

\frac{x}{1 + nx} < \frac{x}{nx} = \frac{1}{n},

and $f_n(0) = 0$ . So $\lvert f_n(x) \rvert \le \frac{1}{n}$ for all $x \ge 0$ , a bound that goes to $0$ and does not depend on $x$ .

Does $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ converge uniformly on $[-1, 1]$ ?

Show the answer

Yes. For $\lvert x \rvert \le 1$ we have

\left\lvert \frac{x^n}{n^2} \right\rvert \le \frac{1}{n^2},

and $\sum \frac{1}{n^2}$ converges, so the M-test applies.

Let $f_n(x) = \frac{x^n}{n}$ on $[0, 1]$ . Does $f_n$ converge uniformly? Do the derivatives converge uniformly?

Show the answer

Yes for the functions. On $[0, 1]$ we have $\lvert f_n(x) \rvert \le \frac{1}{n}$ , so $f_n \to 0$ uniformly. No for the derivatives. They are $f_n'(x) = x^{n-1}$ , which converge pointwise to a limit that is $0$ for $x < 1$ and $1$ at $x = 1$ . That limit is discontinuous while each $f_n'$ is continuous, so the convergence is not uniform. Indeed $\lim f_n'(1) = 1$ , while the limit function $0$ has derivative $0$ at $x = 1$ .

Pointwise and Uniform Convergence

Key terms

Sequences and series of functions

Pointwise convergence

Uniform convergence

Uniform error bounds

Weierstrass M-test

Exchanging limits and integrals

Derivatives need more hypotheses

Check yourself

How this lesson connects

Builds on

Used later in