Math2111: Chapter 1: Fourier series. Section 3: Fourier series and pointwise convergence

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.

We consider now the Fourier coefficients of functions $f:[-\pi, \pi] \to \mathbb{R}$ and discuss the convergence behaviour of Fourier series. We will see that the convergence behaviour depends on the smoothness of the function $f$. In the following we explain what we mean by smoothness of the function.

Piecewise continuity

Let $f:[a,b] \to \mathbb{R}$ and let $c \in [a,b]$. We define the one-sided limits

$\displaystyle f(c^{+}) = \lim_{x \to c^{+}} f(x)$

and

$\displaystyle f(c^{-}) = \lim_{x \to c^{-}} f(x).$

Assume that the one-sided limits $f(c^{+})$ and $f(c^{-})$ both exist and are finite.

• If $f(c^{+}) = f(c^{-}) = f(c)$ then we say that $f$ is continuous at the point $c.$
• If $f(c^{+}) = f(c^{-}) \neq f(c)$, or if $f(c^{+}) = f(c^{-})$ but $f(c)$ is not defined, then $f$ has a removable discontinuity at $c.$
• If $f(c^{+}) \neq f(c^{-})$, then $f$ has a jump discontinuity at $c.$

Definition (piecewise continuous)
A function $f:[a,b] \to \mathbb{R}$ is piecewise continuous if it is continuous on $[a,b]$ except at most at a finite number of points where there exist jump discontinuities.

Example
The function

$\displaystyle f(x) = \left\{\begin{array}{ll} x^2 & \mbox{for } 0 \le x < 1, \\ x/2 & \mbox{for } x = 1, \\ 1-x & \mbox{for } 1 < x \le 2 \end{array}\right.$

is piecewise continuous on $[0,2]$, while the function

$\displaystyle f(x) = \left\{ \begin{array}{ll} \mathrm{e}^{-\lceil 1/x \rceil} & \mbox{for } x \neq 0, \\ 0 & \mbox{for } x = 0 \end{array} \right.$

is not piecewise continuous since there exists an infinite number of jump discontinuities at $x = 1/n$ for $n = 2,3,4, \ldots.$ $\Box$

Theorem
If $f$ is piecewise continuous on a closed, bounded interval $[a,b]$, then $\int_a^b f(x) \,\mathrm{d} x$ exists.

We consider several examples of functions some of which are piecewise continuous and some are not.

Example Consider the function

$\displaystyle f(x) = \lfloor 2 \cos x \rfloor, \quad x \in \mathbb{R}.$

Show that $f$ is piecewise continuous on every finite interval. $\Box$

Example Consider the function

$\displaystyle f(x) = \left\{ \begin{array}{ll} 1/x, & \mbox{for } x < 0, \\ 0 & \mbox{for } x = 0. \end{array} \right.$

Show that $f$ is not piecewise continuous on $[0,1]$ since $f(0^{+})$ does not exist. $\Box$

Example Consider the function

$\displaystyle f(x) = \lceil 2 \sin 1/x \rceil, \quad x \in [0,1].$

Show that $f$ is not piecewise continuous on $[0,1]$ since it has infinitely many points of discontinuity on $[0,1]$. $\Box$

Fourier series

We have already seen how to calculate the Fourier coefficients of a trigonometric polynomial of finite degree. In the following we define Fourier series.

Definition (Fourier series)
Let $f: \mathbb{R} \to \mathbb{R}$ be $2 \pi$-periodic, bounded and integrable on $[-\pi,\pi].$ Let the Fourier coefficients be defined by the Euler formulae

$\displaystyle a_k = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos (kx) \, \mathrm{d} x \quad \mbox{for } k \ge 0,$

$\displaystyle b_k = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin (kx) \, \mathrm{d} x \quad \mbox{for } k \ge 1.$

The (formal) series

$\displaystyle Sf(x) := \frac{a_0}{2} + \sum_{k=1}^\infty \left[a_k \cos (kx) + b_k \sin (kx) \right]$

is called the Fourier series of $f.$

In the following we study under which assumptions on the function $f$ the Fourier series $Sf(x)$ converges. In particular, we are interested in pointwise convergence.

Note If the function $f$ is piecewise continuous then it is integrable, but the partial sums

$\displaystyle S_nf(x) := \frac{a_0}{2} + \sum_{k=1}^n \left[a_k \cos (kx) + b_k \sin (kx) \right]$

do not necessarily converge for every point $x \in [-\pi, \pi]$ as $n \rightarrow \infty.$ In fact, there is a continuous function $f$ and an $x \in [-\pi, \pi]$ for which the partial sums $S_n f(x)$ do not converge as $n \rightarrow \infty.$ ($\rhd$ For such a function see for instance E.M. Stein and R. Shakarchi, Princeton lectures in analysis I, Fourier analysis. Princeton University Press, Princeton, 2003. Section 2.2) $\Box$

Hence we need stronger assumptions on the functions to ensure that the Fourier series converges at every point.

Piecewise differentiable

Consider a function $f:[a,b] \to \mathbb{R}$ and a point $c \in [a, b]$. We write

$\displaystyle D^{+}f(c) = \lim_{h \rightarrow 0^{+}} \frac{f(c+h) - f(c^{+})}{h}$

if this one-sided limit exists and analogously we write

$\displaystyle \begin{array}{lcl} D^{-}f(c) & = & \lim_{h \rightarrow 0^{-}} \frac{f(c+h) - f(c^{-})}{h} \\ && \\ & = & D^{-}f(c) = \lim_{h \rightarrow 0^{+}} \frac{f(c^{-}) - f(c-h)}{h}. \end{array}$

A function $f$ is differentiable at $c$ if and only if $f(c^{+}) = f(c) = f(c^{-})$ and $D^{+}f(c) = D^{-}f(c).$

Note $D^{+}f(c)$ is not necessarily the same as $\displaystyle f^\prime(x^{+}) = \lim_{x \to c^{+}} f^\prime(x).$ For instance, let

$\displaystyle f(x) = x^2 \sin \frac{1}{x}.$

Then $f(0^{+}) = 0$ and

$\displaystyle D^{+}f(0) = \lim_{h \to 0^{+}} \frac{h^2 \sin (1/h) - 0}{h} = 0,$

but

$f^\prime(x)= 2x \sin \frac{1}{x} - \cos \frac{1}{x}$

has no limit as $x \to 0^{+}$. ($\rhd$ See also here.) $\Box$

Definition
A function $f:[a,b] \to \mathbb{R}$ is piecewise differentiable if the one-sided limits $D^{\pm}f$ exist everywhere and $f$ is differentiable on $[a,b]$ except at (at most) a finite number of points.

Lemma
If $f:[a,b] \to \mathbb{R}$ is piecewise differentiable, then $f$ is piecewise continuous.

Example
The function $f(x) = \sqrt{|x|}$ is continuous on $[-1,1]$ but is not piecewise differentiable because $D^{+}f(0)$ and $D^{-}f(0)$ do not exist. $\Box$

Example
The function

$\displaystyle f(x) = \left\{\begin{array}{ll} 1-x & \mbox{for } x < 0, \\ x & \mbox{for } x \ge 0, \end{array} \right.$

is piecewise differentiable on $[-1,1]$. $\Box$

Pointwise convergence

The following result gives conditions on the function $f$ under which the Fourier series $Sf(x)$ converges and under which $S_nf(x) \to f(x)$ as $n \rightarrow \infty.$

Theorem
Let $c \in \mathbb{R}$ and suppose that the function $f: \mathbb{R} \to \mathbb{R}$ has the following properties:

1. $f$ is $2 \pi$-periodic,
2. $f$ is piecewise continuous on $[-\pi,\pi],$
3. $D^{+}f$ and $D^{-}f$ exist.

If $f$ is continuous at $c$, then

$\displaystyle Sf(c) = f(c),$

whereas if $\small f$ has a jump discontinuity at $c$, then

$\displaystyle Sf(c) = \frac{1}{2} \left[f(c^{+}) + f(c^{-}) \right].$

What is remarkable about this result is that the convergence of the Fourier series at a point $x$ depends only on the behaviour of the function $f$ near $x.$ This is not so obvious, since the Fourier coefficients are defined by integrating the function multiplied by $\cos (kx)$ or $\sin (kx)$ over the interval $[-\pi, \pi].$

$\rhd$ For further results on pointwise convergence of Fourier series see here and here.

4 responses to “Math2111: Chapter 1: Fourier series. Section 3: Fourier series and pointwise convergence”

1. DummyStudent

so removable discontinuities have no effect on the piecewise continuity of the graph on some interval??

• Hi, good question. First notice that removable discontinuities are different from jump discontinuities. A discontinuity at a point $c$ is removable if the left-hand and right-hand limits coincide, whereas it is a jump discontinuity if the left-hand and right-hand limits exist (and are finite) but are different. The definition of piecewise continuity assumes that the function does NOT have any removable discontinuities (i.e. it assumes that all removable discontinuities have been removed already). If a function has a removable discontinuity, then it is NOT piecewise continuous.

Note that, in view of the pointwise convergence theorem of Fourier series, Fourier series do not have removable discontinuities. That is why it is convenient to assume that piecewise continuous functions do not have removable discontinuities, since the Fourier series converges to the function which has all its removable discontinuities removed already.

2. Thien Nguyen

How do we prove pointwise convergence? Reading around it says that pointiwse convergence is prooved using an episilon-delta (i.e. for (x-a) < delta, f(x) – f(a) < epsilon), pointwise convergence is proven by finding an epsilon that is a function of both delta and x.

However, I can't seem to find any worked examples, or myself work out how to apply it to a case like tute question 5 in the tutorial for week 8.

Furthermore, if we are to ask to prove 'pointwise continuity' and on a given domain, do we have to prove pointiwse convergence across all points within the domain? If so it seems as if we have proven uniform convergence on the domain…

For example, take a function f_n(x) on [0,1] and another function g(x). If we are asked to prove that f_n(x) converges pointwise to g(x), then do we prove the pointwise convergence for all points in the interval [0,1]? I'm just a bit hazy on the procedure we should follow to prove pointwise convergence…

3. Hi Thien, as you said, you prove pointwise convergence by proving convergence at each point.

For instance, to prove that a sequence of functions $f_n:[0,1]\to\mathbb{R}$ converges pointwise to ${}f:[0,1]\to\mathbb{R}$ you have to show that for all $x\in[0,1]$ you have $\lim_{n\to\infty} f_n(x)=f(x)$.

For example for $f_n(x)=x^n$ we have $f_n(1)=1$ for all $n$ and hence $\lim_{n\to\infty} f_n(1)=1$. If $0 \le x <1$, then

$\displaystyle \lim_{n\to\infty} f_n(x)=\lim_{n\to\infty}x^n=0.$

Uniform convergence is different, here you need to prove that $\lim_{n\to\infty} \sup_{0\le x\le 1} |f_n(x)-f(x)|=0.$ In the example above, we have for each ${}n$ that $\sup_{0\le x\le 1} |f_n(x)-f(x)|=\sup_{0\le x\textless 1} x^n=1,$ hence $\lim_{n\to\infty} \sup_{0\le x\le 1} |f_n(x)-f(x)|=\lim_{n\to\infty} 1= 1.$