Lecture 1

1. Periodic functions¶

In the first part of the course we study Fourier series. The Fourier series is a tool to study periodic functions by writing it as an infinite sum of trignomic functions.

Definition 1.1. A function on $\mathbb{R}$ is called periodic provided $f(x+T) = f(x)$ for some $T>0$. We usually tacitly assume $T$ to be the smallest positive period of $f$.

We identify the following three things:

(a) A periodic function on $\mathbb{R}$ with period $2\pi$.

(b) A function $f$ on $[0,2\pi]$ with $f(0) = f(2\pi)$.

(c) A function defined on the unit circle $S^1:=\{(x,y)\in \mathbb{R}^2:x^2+y^2 = 1\}$.

(d) The quotient space $\mathbb{T}:=\mathbb{R}/\mathbb{Z}$. When we say $x\in \mathbb{T}$ we mean $x\in [0,1]$ with 0 and 1 taken as the same point.

From the data given by (a), i.e. a periodic function $f$ with period $2\pi$, we can restrict the function to $[0,2\pi]$ or any interval of length $2\pi$, like $[-\pi,\pi]$ to get data (b). Conversely, if we have a function defined on $[0,T]$ satisfying $f(0) = f(T)$, then we can construct a periodic function on $\mathbb{R}$ by its periodization $f_T(x) = \sum_{k\in \mathbb{Z}}f(x-kT)$ (with modificatilunon at integral multiples of $T$).

The unit circle can be parametrized on $[0,2\pi]$ by $\theta \mapsto e^{i\theta}$. Let $F$ be a function on the circle $S^1$, then $f(\theta):=F(e^{i\theta})$ is a periodic function of period $2\pi$. Similarly, we can also identify $F$ with $g(x) = F(e^{2\pi i x})$, then $g(x)$ is a periodic function with period 1. By definition, $F$ is continuous, differentiable, or integrable... if and only if $f$ the corresponding parametrization $f$ is continuous, differentiable, or integrable.

Let $f$ be a periodic function with period $T$, then $f(Tx)$ is a periodic function with period 1 and $f(2\pi T)$ is period $2\pi$, hence we can restrict our study on period 1 or period $2\pi$ case (use whichever is more convenient). Note that the integral of function on its period is not invarient under scaling: $\int_{0}^1f(Tx)dx = \int_{0}^Tf(x')d(\frac{x'}{T}) = \frac{1}{T}\int_{0}^Tf(x')dx'$, i.e. shrinking period by $T$ also shrinks integral by $T$.

Trignomic functions and trignomic polynomials¶

A real trignomic polynomial is a function of the form $f(t) = \sum_{k = 1}^n (a_k \cos(kt) + b_k\sin(kt))+\frac{a_0}{2}$. In most theoretical calculations it will be easier to use the complex exponentials instead of sines and cosines, thanks to the identity $e^{i\theta} = \cos \theta + i \sin \theta$. Then $\cos t = \frac{e^{it}+e^{-it} }{2}, \sin t = \frac{e^{it} - e^{-it} }{2i}$, and

$f(t) = \sum_{k=1}^n a_k \frac{e^{it}+e^{-it} }{2} + \sum_{k=1}^n b_k \frac{e^{-it} - e^{-it} }{2i} + \frac{a_0}{2}$

$= \sum_{k=1}^n e^{ikt}\frac{a_k-ib_k}{2} + \sum_{k=1}^n e^{-ikt} \frac{a_k+ib_k}{2} + \frac{a_0}{2}$.

Let $c_k = \frac{a_k-ib_k}{2}$, in particular $c_0 = \frac{a_0+0}{2}$, we have $f(t) = \sum_{k=-n}^n c_ke^{ikt}$ with $c_{-k}=\overline{c_k}$. Note that this is because our $f(t)$ is real at the beginning.

Example 1.1. $D_N:=\sum_{n=-N}^{N}e^{int}$ is a trignomic polynomial and $D_N = \frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{1}{2}t)}$.

Proof. Let $\omega = e^{it}$,then $D_N = \sum_{n=-N}^N\omega^n = \sum_{n=0}^N \omega^n + \sum_{n=-N}^{-1}\omega^{n}$. The first term = $\frac{1-\omega^{N+1}}{1-\omega}$, the second term = $\frac{\frac{1}{\omega}(1-\frac{1}{\omega^N})}{1-\frac{1}{\omega}} = \frac{\omega^{-N}-1}{1-\omega}$.

Then sum up we get $D_N = \frac{\omega^{-N}-\omega^{N+1}}{1-\omega} = \frac{\omega^{-(N+1/2)}-\omega^{N+1/2}}{\omega^{-1/2} -\omega^{1/2}} = \frac{2i \sin((N+\frac{1}{2})x)}{2i \sin(\frac{1}{2}x)}$.

This $D_N$ is called the Dirichlet kernel and we’ll come back later to study it in detail.

Question: Suppose I tell you that a function $f(t)$, for example, $\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{1}{2}t)}$, is a trignomic polynomial, how do you know its coefficients?

To answer this question we need the following fundamental property of complex exponentials.

Lemma 1.1

The vector space of trignomic polynomials¶

Here is a geometric explanation of the above proposition. Let $V$ be a vector space with inner product $(,)$, let $e_1,\dots,e_m$ be an orthonormal basis of $V$. Then every vector $\mathbf{x}$ in $V$ can be written as $a_1e_1+\cdots+a_ne_n$ for scalers $a_1,\dots,a_n$, and we can calculate $a_k$ by $a_k = (v,e_k)$. In fact, we have $(v,e_k) =(a_1e_1+\cdots+a_ne_n, e_k) = 0+a_k(e_k,e_k) = a_k$.

Now let $V_N$ be trignomic polynomials of degree $N$, i.e, $V_N$ consists of functions of the form $\sum_{n=-N}^N c_n e^{int}, c_n\in \mathbb{C}$. Then $V_N$ is a complex vector space. Define inner product $(,)$ on $V_N$ by $(f,g)\mapsto \frac{1}{2\pi}\int_{0}^{2\pi}f(t)\overline{g(t)} dt$, then $(e^{ikt})_{k=-N}^N$ is an orthonormal basis of $V_N$. So we have $c_n = (f,e_n) = \frac{1}{2\pi}\int_{0}^{2\pi}f(t)\overline{e^{i n t}} dt = \frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{-int}dt$ and $f = \sum_{n=-N}^N c_n e^{int}$.

In the period 1 case the inner product is $\int_0^1 f(x)\overline{g(x)}dx$ and $(e^{2\pi i n x})_{n=-N}^N$ is orthonormal basis.

2. Fourier series¶

Can we write any period function, say, with period $2\pi$, as a sum of complex exponentials? One can soon realize that finite sum will usually not be possible because a trignomic polynomial is always smooth, on the other hand, of course not all periodic functions are smooth. So we cannot avoid infinite sums.

Definition 2.1. Let $f$ be an integrable function on $[0,T]$.

• The n-th Fourier coefficient of $f$ is given by $c_n = \frac{1}{T}\int_{0}^{T}f(t)e^{-(2\pi/T) int}dt$. Sometimes we also use $\hat{f}(n)$ to denote $c_n$.
• $S_N(f):=\sum_{n=-N}^N c_ne^{(2\pi/T)int}$ is the $N$-th Fourier partial sum of $f$.
• The series $S(f):=\sum_{n=-\infty}^{+\infty}c_ne^{(2\pi/T)int}=\lim_{N\to \infty}S_N(f)$ is called the Fourier series of $f$.

The first question is convergence of Fourier series, when and in which sense can we write $f(x) = \sum_{n=-\infty}^{+\infty}c_ne^{inx}$?

Note. In order for the integrals to make sense we need some integrability assumption on $f$. Now you can understand the notion “integrable” as “Riemann integrable”. Later, we’ll introduce a more general notion of Lebesgue integrals, most of the results holds for Lebesgue integrable functions.

The convergence of Fourier series turns out to be delicate.

• If the function is continuously differentiable (i.e. first derivative is continuous), then the Fourier series converges uniformly.
• There exists a continuous function whose Fourier series diverge at some point.
• In the general case, need to consider more general sense of convergence.

Example 2.1 $f(\theta) = \theta$ on $[-\pi,\pi]$. Then the Fourier series of $f$ is given by

when $n\neq 0$ and $c_0 = 0$. So $S(f) = \sum_{n = -\infty}^{+\infty}\frac{(-1)^{n+1}}{in}e^{in\theta} = 2\sum_{n = 1}^{+\infty}(-1)^{n+1}\frac{\sin n\theta}{n}$.

We can use the following Sagemath code to visualize the Fourier partial sums, you can change $N$ and see how it goes. This approach is to directly plot the Fourier partial sum based on our hand calculation.

x = var('x') # Define symbolic variable x
n = var('n') # Define symbolic variable n
N = 20
a(n) = 2*(-1)^(n+1)*(1/n)
Sf=sum(a(n)*sin(n*x),n,1,N) # The Fourier partial sum of f 
plot(Sf,x,-pi,pi)

Example 2.2 $f(x) \equiv 1$ on $[-1/2,1/2]$, find the Fourier series of $f$ on $[-T/2,T/2]$.

$c_n = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}1\cdot e^{-(2\pi/T)inx}dx = -\frac{1}{2\pi in}(e^{-\frac{n\pi}{T}i}-e^{-\frac{n\pi}{T}i}) = \frac{\sin(\frac{n\pi}{T})}{n\pi}$ for $n\neq 0$ and $c_0 = \frac{1}{T}$.

In the case $T = 2\pi$, the Fourier series is $\sum_{n=-\infty}^{+\infty}\frac{\sin(n/2)}{n\pi}e^{inx} = \frac{1}{2\pi}+\sum_{n=1}^n \frac{2\sin(n/2)}{n\pi}\cos(nx)$.

Actually Sagemath has a built in method to calculate the Fourier series for us, as shown in the following code:

x = var('x') # Declare variable x
f = piecewise([[(-pi,-1/2),0],[(-1/2,1/2),1],[(1/2,pi),0] ] ) # Define f as a piecewise function
FS5=f.fourier_series_partial_sum(5,pi) # For a piecewise function, there is a method called "fourier_series_partial_sum", we calculate 5 terms
print(FS5) # print the result
P1 = f.plot() # plot the graph of f
P2 = plot(FS5,x,-pi,pi,linestyle="--") # plot the graph of fourier partial sum
(P1+P2).show(title="5 terms") # print the graph
FS100=f.fourier_series_partial_sum(50,pi)
P3 = plot(FS100,x,-pi,pi).show(title="100 terms") # What happens if we calculate 100 terms?

From the picture we can see that even though we approximate $f$ hard enough, the approximation seems to be OK for points far from the jump. But near the jump we there is still big error remaining. This is called the Gibbs phenomenon.

3. Pointwise convergence (simple case)¶

First observation as a consequence of Weierstrass $M$-test:

If $\sum_{n=-\infty}^{+\infty }|c_n|<\infty$, then $S(f)$ converges uniformly.

In this case we call the Fourier series of $f$ converge absolutely. For a postive series, it converges when the coefficients decays sufficiently fast, and from calculus we know that $O(\frac{1}{n})$ is not enough and $O(\frac{1}{n^2})$ is enough. In fact, we can show using Fourier series of the quadratic function that $\sum_{n=1}^{+\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ (in exercise).

Proposition 3.1 (The derivative theorem) Let $f$ be continuously differentiable on $\mathbb{T}$ (this means $f$ is continuously differentiable on $[0,1]$ with $f(0) = f(1)$), then $\widehat{{f'}}(n) = 2\pi in\hat{f}(n)$.

Proof. This is just integration by part. $\widehat{f'}( n) = \int_{0}^{1}f'(x)e^{-2\pi i n x}dx = \int_{0}^1e^{-2\pi in x}df = f(x)e^{-2\pi i n x}|_{0}^1 - \int_{0}^1 f(x)(-2\pi i n)e^{-2\pi i n x} dx = 2\pi i n \int_{0}^1 f(x)e^{-2\pi i nx }dx = 2\pi i n \hat{f}(n)$.

Remark. In the period $T$ case $\widehat{f}(n) = \frac{2\pi}{T} i n\hat{f}(n)$.

We call a function on $S^1$ integrable if $\int_{0}^{2\pi}|f(x)|dx<\infty$. We use $L^1(S^1)$ to denote integrable functions on $S^1$. For $f\in L^1(S^1)$, the quantity $\int_{0}^{2\pi}|f(x)|dx$ is called the $L^1$-norm of $f$. We have the following handy uniform bound by the $L^1$-norm for the Fourier coefficients:

Lemma. $|\hat{f}(n)|\leq \int_{0}^1|f(x)|dx$. If we denote $\int_{0}^1|f(x)|dx$ by $\|f\|_{L^1}$ then $|\widehat{f}(n)|\leq \|f\|_{L^1}$.

Proof. $|\hat{f}(n)| = |\int_{0}^1f(x)e^{-2\pi i n x}dx|\leq \int_{0}^1|f(x)|dx$.

Corollary. If $f$ has continuous second derivative then $\widehat{f}(n) =O(\frac{1}{n^2})$, hence the Fourier series of $f$ converges uniformly.

Proof. Apply the derivative theorem twice we get $\widehat{f''}(n) = -4\pi^2n^2\hat{f}(n)$. By the $L^1$ estimate we have $|\hat{f''}(n)| \leq \int_{0}^1|f''(x)|dx = \|f''\|_{L^1}$. So $|\hat{f}(n)|\leq \frac{1}{4\pi}\|f''\| \cdot \frac{1}{n^2} = \text{constant}\cdot \frac{1}{n^2}$.

Next we consider the continuous case.

Proposition 3.2. Let $f$ be an integrable function on the circle. If $\hat{f}(n) = 0$ for all $n\in\mathbb{Z}$, then $f(x)$ vanishes at continuous points, i.e. for every $x\in[0,2\pi]$ such that $f$ is continuous at $x$, $f(x) = 0$.

Note. The Fourier coefficients vanishes means $f$ is orthogonal to all trignomic polynomials.

Proof.

Method 1. Assume $f(x_0)\neq 0$ for some continuous point $x_0$. We can moreover assome $f(x_0)>0$, then since $f$ is continuous at $x_0$, it must be $\geq \epsilon$ near $x_0$. Then the idea is to construct a sequence of trignomial polynomials $p_k$ such that $p_k$ goes to $+\infty$ near $x_0$ and the negative part of $p_k$ is controlled, so that $\int_{0}^{2\pi}f(t)p_k(t)dt \to \infty$, which contradicts to the assumption that $f$ is orthogonal to $p_k$. Precisely, fix $\epsilon > 0$ and choose small $\delta > 0$ such that $f(x)\geq \epsilon$ on $(x_0-\delta,x_0+\delta)$. Let $p(x)=\cos(x-x_0)+\alpha$ for small $\alpha > 0$. Let $p_k(x) = p(x)^k$. We choose α so small that $|p(x)|<1$ so $p_k(x)\to 0$ when $x\notin (x_0-\delta,x_0+\delta)$. Then $0=\int_{0}^{2\pi}f(x)p_k(x)dx = \int_{x_0-\delta}^{x_0+\delta}f(x)p_k(x)dx + \int_{x\notin(x_0-\delta,x_0+\delta)}f(x)p_k(x)dx$. The first term go to $+\infty$ when $k\to \infty$, the second term go to 0 when $k\to \infty$, contradiction.

Note: the second limit is a consequence of the dominated convergence theorem. We want to have $\lim_{k\to \infty}\int_{x\notin(x_0 - \delta,x_0 + \delta)}f(x)p_k(x)dx = \int_{x \notin (x_0-\delta,x_0+\delta)}\lim_{k\to \infty}f(x)p_k(x)dx = \int_{x\notin (x_0-\delta,x_0+\delta)}f(x)\cdot 0 dx = 0$. To change the limit and integral we need the condition that $|f(x)p_k(x)|\leq \text{some } L^1 \text{ function}$ on $(x_0-\delta,x_0+\delta)$. Since $|p_k(x)|\leq 2$ and $f(x)\in L^1(S^1)$, we can take $2|f|$.

If you want a more elementary proof see the book Theorem 2.1.

The second method involves Lemma 5.1 and Theorem 5.2 in Chapter 2 of Stein-Shakarchi, that we’ll do later.

Method 2. Let $F_N(x)=\frac{1}{N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$ which is a trignomic polynomial called Fejer kernel. Assume $f$ is continuous at 0, the Fejer theorem (later) implies $\frac{1}{2\pi}\int_{0}^{2\pi}f(x)F_N(x)dx\to f(0)$ when $N\to \infty$. Since $f$ is orthogonal to trignomic polynomials, this shows $f(0)=0$. If $x_0\neq 0$, let $g(x) = f(x+x_0)$, apply the result to $g(x)$ shows $g(0) = 0 = f(x_0)$.

Corollary. Let $f$ be an integrable function on the circle such that the Fourier series of $f$ converges absolutely. If $f$ is continuous at $x_0$, then the Fourier series of $f$ converges to $f$ at $x_0$, i.e. $\lim_{N\to \infty}S_N(f)(x_0) = f(x_0)$.

Proof. Since the Fourier series of $f$ converges absolutely which implies $S(f)$ converges uniformly, hence $S(f)$ is a continuous function. Note that the Fourier coefficients of $S(f)$ is precisely $\hat{f}(n)$, so $S(f)$ and $f$ has same Fourier coefficients. Then $S(f)$ and $f$ coincide on continuous points by Proposition 3.2.

Appendix: Plot of Direchlet kernel and Fejer kernel¶

N=20
x=var('x')
DN=sin((N+1/2)*x)/sin(1/2*x)
FN=(1/N)*sin(N*x/2)^2/(sin(x/2)^2)
P1=plot(DN)
P1.show(title="Direchlet kernel")
P2=plot(FN)
P2.show(title="Fejer kernel")

We can also plot an approximate process for different values, for example the Fejer kernel

for N in [20,40,60,80]:
    FN=(1/N)*sin(N*x/2)^2/(sin(x/2)^2)
    P2=P2+plot(FN)
P2.show()