Advanced Analysis

Fourier Series

Video. Fourier Series Introduction

Some Topics Covered Fourier coefficients, Riemann Lebesgue Lemma via density of step functions
Definition
A complex Fourier series of period \(1\) is a formal sum
\[ \sum_{n=-\infty}^\infty a_n e^{2 \pi i n x}\]
where the coefficients \(\{a_n\}_{n=-\infty}^\infty\) is a doubly-infinite sequence of complex numbers (and \(i\) is the imaginary unit: \(i^2 = -1\)).
Note
We can integrate complex-valued functions using the machinery we've already developed for real functions. If \(f\) is a complex-valued function on \([a,b]\), we say it's Riemann integrable when its real and imaginary parts are. In other words, we write \(f = f_R + i f_I\) for unique real functions \(f_R\) and \(f_I\) and define
\[ \int_a^b f(t) dt = \int_a^b f_R(t) + i \int_a^b f_I(t) dt \]
whenever the right-hand side is defined.
Fact
Given two integers \(n\) and \(m\),
\[ \int_0^1 e^{2 \pi i nx} e^{-2 \pi i m x} dx = \begin{cases} 1 & n = m \\ 0 & n \neq m \end{cases}. \]
Proof
If \(n=m\), then \(e^{2 \pi i nx} e^{-2 \pi i mx} = e^{2 \pi i (n-m)x} = e^{0} = 1\) (and the integral of \(1\) on \([0,1]\) is simply \(1\)). When \(n \neq m\), we still have that \(e^{2 \pi i nx} e^{-2 \pi i mx} = e^{2 \pi i (n-m)x}\), but now we can integrate the exponential and use the fact that \(e^{2 \pi i k} = 1\) for each \(k \in {\mathbb Z}\) to compute the integral.
Using this fact and formally integrating a complex Fourier series term-by-term, if one has the identity
\[ f(x) = \sum_{n=-\infty}^\infty a_n e^{2 \pi i n x}\]
for a suitably nice function \(f\) and some coefficients \(\{a_n\}_{n=-\infty}^\infty\), then it should be the case that
\[ \int_0^1 f(x) e^{-2 \pi i nx} dx = a_n. \]
(This would be rigorously justifiable, for example, if it were known that the Fourier series converged uniformly.) This observation is called Fourier's trick.
Definition
Given a Riemann-integrable complex-valued function \(f\) on \([0,1]\), we define the \(n\)-th Fourier coefficient \(\widehat{f}(n)\) as
\[ \widehat{f}(n) := \int_0^1 f(x) e^{-2 \pi i nx} dx \]
for each \(n \in \mathbb{Z}\).
The fundamental question of Fourier series is to establish when (and in what sense) we have
\[ f(x) = \sum_{n=-\infty}^\infty \widehat{f}(n) e^{2 \pi i nx}. \]
Before we can say anything at all about the convergence of the series, it would be necessary to establish that the Fourier coefficients tend to zero as \(n \rightarrow \pm \infty\) (since otherwise the sum couldn't possibly be convergent).
Theorem (Riemann-Lebesgue Lemma)
Suppose that \(f\) is a complex-valued Riemann-integrable function on \([0,1]\). Then \(|\widehat{f}(n)| \rightarrow 0\) as \(n \rightarrow \pm \infty\). Moreover, for each \(n \in \mathbb{Z}\),
\[ |\widehat{f}(n)| \leq \int_0^1 |f(t)|dt. \]
We will prove the Riemann-Lebesgue Lemma in a series of steps.

Step 0: The bound for coefficients This is just a consequence of the triangle inequality for integrals:
\[{}|\widehat{f}(n)| \leq \int_0^1 |e^{-2 \pi i n t} f(t)| dt{}\]
\[{}= \int_0^1 |f(t)| dt{}\]
because \(|e^{-2 \pi i n t}| = 1\). For completeness, we sketch a proof of the triangle inequality in the complex case:
Proposition
Suppose that \(f\) is a complex-valued Riemann integrable function on some interval \([a,b]\). Then \(|f(t)|\) is Riemann integrable and
\[ \left| \int_a^b f(t) dt \right| \leq \int_a^b |f(t)| dt. \]
Proof
Meta (Main Ideas)
First observe that
\[{}|f(x)| \leq |f(y)| + |f(x) - f(y)|{}\]
and
\[{}|f(y)| \leq |f(x)| + |f(x) - f(y)|.{}\]
Combining these inequalities gives that
\[ ||f(x)| - |f(y)|| \leq |f(x) - f(y)|. \]
Use this to show that when the real part of \(f\) takes values between \(A\) and \(B\) on some interval \(I\) and the imaginary part of \(f\) takes values between \(C\) and \(D\), then \(|f(x)|\) and \(|f(y)|\) differ by at most \((B-A) + (D-C)\) for all \(x,y \in I\). Then incorporate this into the definition of upper and lower sums to show that \(U(|f|,\mathcal P) - L(|f|,\mathcal P)\) is at most the sum of \(U(f_R,\mathcal P) - L(f_R,\mathcal P)\) and \(U(f_I,\mathcal P) - L(f_I,\mathcal P)\), which can be used to show that \(|f|\) is Riemann integrable.

Next, prove that \(\int_0^1 c f(t) dt = c \int_0^1 f(t) dt\) when \(f\) is complex-valued and \(c\) is complex. The trick is to write out what the real and imaginary parts of both sides must be and then show that real parts of both sides are equal and likewise for imaginary parts.

Now use the triangle inequality for real functions:
\[{}\left| \int_0^1 \operatorname{Re}(c f(t)) dt \right|{}\]
\[{}\leq \int_0^1 |\operatorname{Re}(c f(t))| dt.{}\]
\[{}\leq \int_0^1 |c f(t)| dt{}\]
\[{}= |c| \int_0^1 |f(t)| dt.{}\]
We can always find some \(c\) with \(|c| = 1\) such that \(c \int_0^1 f(t) dt\) is real, so for this value of \(c\),
\[ c \int_0^1 f(t) dt = \int_0^1 \operatorname{Re}(c f(t)) dt \]
and we conclude that
\[ \left| c \int_0^1 f(t) dt \right| \leq |c| \int_0^1 |f(t)| dt. \]
Since \(|c| = 1\), we can cancel it from both sides to deduce the triangle inequality.
Step 1: Simple/Step Functions Let us call a function \(f : [0,1] \rightarrow \mathbb C\) simple when there is a partition \(\mathcal P\) of \([0,1]\) such that \(f\) is constant on the interior of every interval \(I \in \mathcal P\). In particular, if \(\chi_I\) is the characteristic function of the interval \(I\) (meaning it is one on all points in \(I\) and zero on the complement of \(I\)), we must have that
\[ f = \sum_{I \in \mathcal P} c_I \chi_I \]
everywhere except possibly at endpoints of intervals (which is a finite subset of \([0,1]\)) for some constants \(c_I\).

Now for such a function \(f\),
\[{}\widehat{f}(n) = \int_0^1 e^{-2 \pi i n t} \sum_{I \in \mathcal P} c_I \chi_{I}(t) dt{}\]
\[{}= \sum_{I \in \mathcal P} c_I \int_{I} e^{-2 \pi i n t} dt{}\]
where \(\int_I\) denotes the integral over the interval \(I\). If \(I = [a_I,b_I]\) and \(n \neq 0\),
\[ \int_I e^{-2 \pi i n t} dt = \left. \frac{e^{-2 \pi i n t}}{-2 \pi i n} \right|_{a_I}^{b_I}. \]
Because \(|e^{-2 \pi i n t}| = 1\), the triangle inequality gives that
\[ \left| \int_I e^{-2 \pi i nt} dt \right| \leq \frac{1}{2 \pi |n|} + \frac{1}{2 \pi |n|}. \]
Importing this inequality back into the formula for \(\widehat{f}(n)\) gives that
\[ |\widehat{f}(n)| \leq \sum_{I \in \mathcal P} \frac{|c_I|}{\pi |n|}\]
for all \(n \rightarrow 0\). This clearly tends to zero as \(n \rightarrow \pm \infty\) because the sum on the right-hand side is a finite sum and its only dependence on \(n\) is via the denominator of each term.

Step 2: Approximation by Simple Functions Suppose now that \(f : [0,1] \rightarrow \mathbb{C}\) is Riemann integrable. If \(f_R\) and \(f_I\) are the real and imaginary parts of \(f\), respectively, we know by definition that both are Riemann integrable as well. Thus (by taking a common refinement), there must be a partition \(\mathcal P\) such that the upper sums of \(f_R\) and \(f_I\) with respect to \(\mathcal P\) differ from the respective lower sums of \(f_R\) and \(f_I\) by at most \(\epsilon/4\) for any fixed \(\epsilon\).

For the partition \(\mathcal P\) just constructed, let \(g\) be the simple function which for each \(I \in \mathcal P\) is constant on the interior of \(I\) and agrees with the value of \(f\) at the midpoint of \(I\). It follows that the real part of \(g\) is a simple function which on every \(I\) lies between the supremum and infimum of \(f_R\) on \(I\). Since \(f_R(t)\) also lies between the supremum and infimum of \(f_R\) on \(I\) when \(t \in I\), it follows that
\[ |\operatorname{Re}(g(t)) - f_R(t)| \leq \sup_{x \in I} f_R(x) - \inf_{x \in I} f_I(x) \]
for all \(t\) in the interior of \(I\), which implies that
\[{}\int_0^1 |\operatorname{Re}(g(t)) - f_R(t)| dt{}\]
\[{}\leq U(f_R,\mathcal P) - L(f_R,\mathcal P) < \frac{\epsilon}{4}. {}\]
Establishing the analogous inequality for imaginary parts and summing gives that
\[ \int_0^1 |g(t) - f(t)| dt \leq \frac{\epsilon}{2}. \]
By the triangle inequality for integrals, this means that
\[ |\widehat{g}(n) - \widehat{f}(n)| \leq \frac{\epsilon}{2}\]
for each \(n \in \mathbb{Z}\). We know that \(g\) is simple, and so in particular, \(|\widehat{g}(n)| < \frac{\epsilon}{2}\) when \(|n| > N_0\) for some \(N_0\) depending on \(\epsilon\). But now by the triangle inequality for complex numbers, this means that when \(|n| > N_0\),
\[ |\widehat{f}(n)| \leq |\widehat{g}(n)| + |\widehat{g}(n) - \widehat{f}(n)| < \epsilon\]
as desired.