# Math2111: Chapter 1: Fourier series. Section 5: Convergence of sequences of functions

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

In Section 2 we considered partial sums of Fourier series and we asked what happens when $n \rightarrow \infty$. In this entry we take a more general approach by considering sequences of functions $f_1, f_2, \ldots$ and studying their convergence behaviour.

Let $V$ be a vector space of functions $f:[a,b] \to \mathbb{R}$ equipped with a norm $\|\cdot \|$. We say that a sequence of functions $f_1, f_2, f_3, \ldots \in V$ converges to $f$ in $V$ if $f \in V$ and

$\displaystyle \lim_{n\to \infty} \|f_n - f\| = 0.$

This means that for every $\varepsilon > 0$ there exists an $N > 0$ such that

$\|f_n - f\| < \varepsilon \quad \mbox{for all } n \ge N.$

Exercise Consider the vector space of continuous functions $C([a,b])$ equipped with the maximum norm. Show that the functions

$\displaystyle f_n(x) = \frac{n x^2}{1+n} + \frac{\sin x}{n}, \quad n = 1, 2, \ldots$

converge to

$\displaystyle f(x) = x^2$

as $n \rightarrow \infty$.

Example Consider the vector space of bounded functions $\mathcal{B}([0,1])$ equipped with the supremum norm. Show that the functions

$\displaystyle f_n(x) = x^n, \quad n = 1, 2, \ldots$

converge to

$\displaystyle f(x) = \left\{\begin{array}{rl} 0 & \mbox{if } 0 \le x < 1, \\ & \\ 1 & \mbox{if } x = 1. \end{array} \right.$

as $n \rightarrow \infty$. Show that $f_n$ does not converge to $f$ in the supremum norm however.

Note that the functions $f_n$, $n = 1,2, \ldots$ in the above example are all continuous, but the limit $f$ is not.

Note (Interchanging limits) Limits cannot always be interchanged. For example let

$\displaystyle f_{n}(x) = \frac{1}{1+nx}.$

Then

$\displaystyle \lim_{x \to 0} \lim_{n \to \infty} f_{n}(x) = \lim_{x \to 0} \left(\lim_{n \to \infty} \frac{1}{1+nx} \right) = \lim_{x \to 0} 0 = 0$

and

$\displaystyle \lim_{n \to \infty} \lim_{x \to 0} f_{n}(x) = \lim_{n \to \infty} \left(\lim_{x \to 0} \frac{1}{1+nx} \right) = \lim_{n \to \infty} 1 = 1.$

Hence, in this example,

$\displaystyle \lim_{x \to 0} \lim_{n \to \infty} f_{n}(x) \neq \lim_{n \to \infty} \lim_{x \to 0} f_{n}(x).$

In particular, assume that $f_n:[a,b] \to \mathbb{R}$ is continuous and converges at every point $x \in [a,b]$ to a function $f:[a,b] \to \mathbb{R}$, we do not necessarily have that $f$ is continuous. $\Box$

For the example above we have

$\displaystyle \lim_{x \to 1} \lim_{n \to \infty} f_n(x) \neq \lim_{n \to \infty} \lim_{x \to 1} f_n(x).$

There is, however, a type of convergence which ensures that if the sequence of functions $f_n$ are continuous then the limit $f$ is also continuous. This type of convergence is called uniform convergence.

Uniform convergence

Definition
Let $f_n:[a,b] \to \mathbb{R}$ be a sequence of functions and let $f:[a,b] \to \mathbb{R}.$ We say that $f_n$ converges uniformly to $f$ if for every $\varepsilon > 0$ there is an integer $N$ such that $n \geq N$ implies that

$\displaystyle |f_n(x) - f(x)| < \varepsilon$

for all $x \in [a,b]$.

In other words, let $\|\cdot\|_\infty$ be the supremum norm $\|f\|_\infty = \sup_{a \le x \le b} |f(x)|.$ Then uniform convergence just means that

$\displaystyle \lim_{n \to \infty} \|f_n-f\|_\infty = 0.$

The difference between uniform convergence and pointwise convergence is that for pointwise convergence the choice of $N$ depends on $\varepsilon > 0$ and on $x.$ whereas for uniform convergence $N$ is only allowed to depend on $\varepsilon$ (but not on $x$).

Note that uniform convergence implies pointwise convergence ($\rhd$ can you prove this?), whereas pointwise convergence does not imply uniform convergence.

Theorem (Uniform convergence theorem) If $f_n:[a,b] \to \mathbb{R},$ $n = 1, 2, \ldots$ is a sequence of continuous functions which converge uniformly to a function $f:[a,b]\to \mathbb{R}$, then $f$ is continuous.

A consequence of this is that if $f_n$ is continuous for all $n$ and $f_n \to f$ as $n \to \infty$, then if $f$ has a discontinuity implies that the convergence of $f_n$ to $f$ cannot be uniform.

For Fourier series, this implies the following: Assume that $f$ is periodic with period $L$ and note that the partial sums $S_n f$ are continuous. Then if $f$ is not continuous at some point $c$, then $S_n f$ cannot converge uniformly to $f$ as $n \to \infty$ on any open interval containing $c$.

There is a very convenient test for uniform convergence due to Weierstrass.

Theorem (Weierstrass test)
Suppose $f_1, f_2, \ldots$ is a sequence of functions defined on a set $E$ and suppose

$\displaystyle |f_n(x)|\le M_n \quad \mbox{for all} x \in [a,b] \mbox{ and } n = 1, 2, \ldots.$

Then the series

$\displaystyle \sum_{n=1}^\infty f_n$

converges uniformly on $E$ if $\sum_{n=1}^\infty M_n$ converges.

Example Let $E = [a,b]$ (where $a < b$) and

$\displaystyle f_n(x) = \frac{x^n}{n!}, \quad n = 0, 1, 2, \ldots,$

where $n! = 1 \cdot 2 \cdots n$ and $0! := 1$. We must choose $M_n$, $n = 0, 1, 2, \ldots$ such that

$\displaystyle \left| \frac{x^n}{n!} \right| \le M_n \quad \mbox{for all} x \in [a,b] \mbox{ and } n = 0, 1, 2, \ldots.$

Let $B = \max(|a|,|b|),$ then

$\displaystyle \left| \frac{x^n}{n!} \right| \le \frac{B^n}{n!} \quad \mbox{for all } x \in [a,b] \mbox{ and } n = 0, 1, 2, \ldots,$

that is, we can choose $M_n = \frac{B^n}{n!}.$ Hence it remains to show that the sum

$\displaystyle \sum_{n=0}^\infty M_n = \sum_{n=0}^\infty \frac{B^n}{n!}$

converges. This can be done using the ratio test

$\displaystyle \lim_{n\to \infty} \frac{M_{n+1}}{M_{n}} = \lim_{n \to \infty} \frac{B^{n+1}}{(n+1)!} \frac{n!}{B^n} = \lim_{n\to \infty} \frac{B}{n+1} = 0 < 1.$

Exercise Show that the functions

$\displaystyle f_n(x) = \frac{n x^2}{1+n} + \frac{\sin x}{n}, \quad n = 1, 2, \ldots$

converge uniformly to

$\displaystyle f(x) = x^2$

as $n \rightarrow \infty$.

Mean square convergence of Fourier series

An interesting property of the partial sums $S_N f$ of a Fourier series is that among all trigonometric polynomials of degree $N$, the partial sum $S_N f$ yield the best approximation of $f$ in the $L_2$-norm (or mean square sense). Note that the $L_2$-norm is given by

$\displaystyle \|f\|_{2} = \left(\int_{-\pi}^\pi |f(x)|^2 \, \mathrm{d} x\right)^{1/2}.$

Lemma (Best approximation lemma)
Assume $f$ is integrable and let $S_N f$ denote the $N$th partial sum of the Fourier series, then

$\displaystyle \|f-S_N f\|_2 \le \|f - \frac{c_0}{2} - \sum_{k = 1}^N \left[c_k \cos k x + d_k \sin k x\right] \|_2$

for any real numbers $c_0, c_1, \ldots, c_N$ and $d_1, d_2, \ldots, d_N$.

Exercise Give a geometrical interpretation of this result.

Theorem
Let $f:\mathbb{R} \to \mathbb{R}$ be $2 \pi$ periodic, bounded and integrable on $[-\pi, \pi]$. Then the Fourier series of $f$ converges in the mean square sense, that is,

$\displaystyle \lim_{N\to \infty} \|f-S_Nf\|_2 = 0.$

$\rhd$ You can find a proof of this result on this post.

Further, Parseval’s identity holds

$\displaystyle \|f\|_2^2 = \int_{-\pi}^\pi |f(x)|^2 \, \mathrm{d} x = \frac{\pi}{2} a_0^2 + \pi \sum_{k=1}^\infty \left[a_k^2 + b_k^2 \right].$