Lecture 8 - fouriernew

Summery of Fourier series part¶

In the Fourier series part we mainly studied the convergence of Fourier series, here is a breif summery of points covered in the lectures and the book. It is strongly recommended to read this wikipedia page for a complete summery of convergence of Fourier series:

Note

The Fourier series of an integrable function converges in $L^2$ to the function.
There are two approaches to study the pointwise convergence of Fourier series.
- Global approach: find condition such that the Fourier series converges absolutely, this is usually growth restrictions of Fourier coefficients.
- Local approach: directly estimate the pointwise difference of Fourier partial sum and the value of function without making global estimate on the Fourier integrals.
In particular:
- If $f$ is Lipschitz continuous on the circle, then its Fourier series converges uniformly. If $f$ is Lipschitz continuous at $x_0$ , then its Fourier series at $x_0$ converges to $f(x_0)$ (see theorem below).
- If $f$ is piecewise $\mathscr{C}^1$ with jump discontinunity, then $S_N(f)(x)\to \frac{f(x^+)+ f(x^-)}{2}$ .

In the global approach we have the following results:

Regularity of f	Decay of Fourier coefficients	Reason
Integrable	$\to 0$	Riemann-Lebesgue lemma
$\mathscr{C}^k$	$o(\frac{1}{n^k})$	Derivative theorem + Riemann-Lebegue Lemma
bounded monotonic	$O(\frac{1}{n})$	Page 93 exercise 17
Lipschitz	$O(\frac{1}{n})$	Page 91 Exercise 15

Note that when $f$ is Lipschitz or $\mathscr{C}^1$ , the decay condition it self does not imply absolute convergence of Fourier series. However, the Fourier series does converges absolutely (Page 92, Exercise 16).

We have also the local convergence criteria:

Theorem¶

Let $f$ be integrable on the circle. If $f$ is differentiable at $x_0$ , then $S_N(f)(x_0)\to f(x_0)$ .

Proof. Let $F(t)$ be function on the circle defined by

F(t) = \begin{cases} \frac{f(x_0+t) - f(x_0)}{t},\; t> 0\\ f'(x_0), \; t=0 \end{cases}

(1)

Then $F(t)$ is $L^1$ . To see this, since $\lim_{t\to 0}F(t) = f'(x_0)$ , $F$ is bounded near 0, and $F$ is difference of two $L^1$ functions when $t$ is away from 0.

Then

\begin{split} |S_N(f)(x_0) - f(x_0)| &= |f*D_N(x_0) - f(x_0)| \\ &\leq \frac{1}{2\pi}|\int_{-\pi}^\pi tF(t)D_N(t)dt | \\ & = \frac{1}{2\pi}|\int_{-\pi}^\pi F(t)\frac{t}{\sin(\frac{1}{2}t)} \sin ((N+\frac{1}{2})t)dt| \end{split}

(2)

Note that $\frac{t}{\sin(\frac{1}{2}t)}$ is continuous, $\sin((N+\frac{1}{2})t) = \sin(Nt) \cos(\frac{1}{2}t)+ \cos(Nt) \sin(\frac{1}{2}t)$ . $F(t)$ is integrable, the right hand side is integrable of the form “integrable $\cdot \sin(Nt)$ ” and “integrable $\cdot \cos(Nt)$ ”, which goes to zero as a consequence of Riemann Lebesuge lemma.

Note that we only need $F$ to be bounded near 0, so it sufficies to assume $f$ is merely Lipschitz continuous at $x_0$ , i.e. $|f(x_0+t)-f(x_0)|\leq C|t|$ for small enough $t$ .

Example of continuous but everywhere non-differentiable function¶

See Page 113 of Stein-Shakarchi.

The Fourier transform¶

The Fourier series for periodic functions

Roughly speaking, the Fourier coefficient associates a function on the circle a sequence of numbers $(\hat{f}(n))_{n\in \mathbb{Z}}$ such that

The $L^2$ norm is perserved, $\|f\|_{L^2(\mathbb{T})} = \|(\hat{f}(n))\|_{l^2} = \sum_{n \in \mathbb{Z}}|\hat{f}(n)|^2$ .
The convolution theorem: $f* g \mapsto (\hat{f}(n)\cdot \hat{g}(n))_{n\in \mathbb{Z}}$ .
The derivative theorem: $f'\mapsto (\frac{2\pi}{T} i n \hat{f}(n))_{n\in \mathbb{Z}}$ .
Given the data of Fourier coefficients $\hat{f}(n)$ , we can recover back the function by taking Fourier series $\sum_{n\in \mathbb{Z}}\hat{f}(n)e^{\frac{2\pi}{T}inx}$ .

The Fourier transform can be seen as a generalization of this correspondence to non-periodic functions on $\mathbb{R}$ . The idea is that a non-periodic function can be viewed as a “periodic function with period = ∞”.

Motivation¶

Let’s consider the step function

f(x) = \begin{cases} 1, x\in [-1/2,1/2]\\ 0, \text{ else}\end{cases}

(3)

which is a non-periodic function on $\mathbb{R}$ . For any $T>1$ , we can view $f$ as a function defined on $[\frac{T}{2},\frac{T}{2}]$ , and make it a $T$ -periodic function $f_T(x):=\sum_{k\in \mathbb{Z}}f(x - kT)$ . Observe that, when $T\to \infty$ , $f_T(x)\to f(x)$ . So to use Fourier analysis to study $f$ , we can look at behavior of Fourier coefficients of $f_T$ as $T\to \infty$ .

By direct computation we get

\widehat{f_T}(n) = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} f_T(t)e^{-\frac{2\pi}{T}int}dt = \frac{1}{T}\int_{-\frac{1}{2}}^{\frac{1}{2} }1\cdot e^{-\frac{2\pi}{T}int}dt =\begin{cases} \frac{\sin(\frac{\pi n}{T})}{\pi n}, n\neq 0\\ \frac{1}{T}, n = 0 \end{cases} .

(4)

We can see that when $T$ becomes larger and larger, the Fourier coefficients takes value on a function, more and more slowly with higher and higher “definition”. If you take, for example, $|n|\leq 100$ , then when $T$ is small, it samples the function $\frac{1}{T}\frac{\sin \pi s}{\pi s}$ on largher interval but more rough; When $T$ is large, the first 100 coefficients only give you information about the function $\frac{1}{T}\frac{\sin \pi s}{\pi s}$ on a small interval, but with greater detail. Also, when $T$ is big, the $\frac{1}{T}$ makes the Fourier coefficients small. Let’s get rid of the shrinking by scaling the function back by $T$ . Then we obtained a function $g(s)$ such that $T\widehat{f_T}(n) = g(\frac{n}{T})$ , in this case we know that $g(s) = \frac{\sin \pi s}{\pi s}$ .

To get the function $g$ in a more intrinsic way, when $T$ is large, $\frac{n}{T}$ samples through the real line. Let $s = \frac{n}{T}$ so that $n = sT$ , then

g(s) = T\widehat{f}(Ts) = T\cdot \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-\frac{2\pi}{T} i n t} dt = \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-2\pi i s t}dt.

(5)

When $T\to \infty$ , the left hand side gives a real function, and the right hand side is, by definition, the improper integral $\int_{-\infty}^{+\infty}f(t)e^{-2\pi i s t}dt$ .

The $g$ we get is the Fourier transform of $f$ . Of course, the argument needs some assumption to work but it gives us a motivation of Fourier transform.

Definition, inversion and first examples¶

Let $f$ be an $L^1$ function on $\mathbb{R}$ . The Fourier transform of $f$ is given by

\mathcal F f(s) = \int_{-\infty}^{+\infty} f(t)e^{-2\pi i t s} dt.

(6)

Note that to be $L^1$ or bounded support is too restrictive for a satisfactory theory of Fourier transforms, we just add this assumption for simplicity. The Fourier transform can be defined on much wider objects, as we shall see later.

The inversion integral¶

The Fourier coefficients can recover the function under certain regularity assumptions or growth conditions (they are the same). One expects same behavior for Fourier transforms. Suppose we know the Fourier transform of $f$ is $F(s)$ , can we recover $f$ ? For simplicity let’s assume $f$ is bounded support (we also call it time limited) so that we can obtain its periodization $f_T$ without convergence issues. Then we can try to get $f_T$ by Fourier series and then recover $f$ by letting $T\to \infty$ . Using Fourier series expansion on $f_T$ and equation $T\widehat{f_T}(n) = F(\frac{n}{T})$ we get

\begin{split} f_T(x) &= \sum_{n= -\infty}^{+\infty} \widehat{f_T}(n)e^{\frac{2\pi }{T}i nx}\\ & = \sum_{n = -\infty}^{+\infty} \frac{1}{T} F(\frac{n}{T})e^{2\pi i \frac{n}{T}x} \end{split}

(7)

This is an Riemann sum of, which goes to $\int_{-\infty}^{+\infty}F(s)e^{2\pi i x s} ds$ , when $T\to \infty$ , and the left hand side goes to $f(x)$ . The argument suggests that we can get $f$ back by taking integral

\int_{-\infty}^{+\infty}F(s)e^{2\pi i s x}.

(8)

This is the inverse Fourier transform. We need assumption to make this work, which we shall examine later.