Introduction to Fourier Series

Basic Knowledge

We focus on a class of functions

(1) Continuous functions $[0, L]$

(2) Piecewise continuous functions (only finitely many discontinuities)

(3) Riemann integrable functions

(4) Functions on the circle (correspondence with $2\pi$ periodic functions on $\mathbb{R}$ such that $f(0)=f(2\pi)$.

Definition 2.1

Given an integrable function $f:[a,b]\to\mathbb{C}$, we define the Fourier series of $f$ as

\[\begin{align*} \sum_{n=-\infty}^{\infty} \hat{f}(n)e^{\frac{2\pi inx}{b-a}}, \end{align*}\]

where

\[\begin{align*} \hat{f}(n) = \frac{1}{b-a}\int_a^b f(x) \exp\left(-\frac{2\pi inx }{b-a}\right)dx \end{align*}\]

denotes the $n$-th Fourier coefficient of $f$ for $n\in\mathbb{N}$.

Remark

One can define inner product $\langle f,g \rangle = \frac{1}{b-a}\int_a^b f(x) \bar{g}(x) dx$. Then

\[\begin{align*} \hat{f}(n) = \left\langle f, \exp\left(\frac{2\pi inx}{b-a}\right) \right\rangle . \end{align*}\]

Example

The Fourier series of the $2\pi$-periodic odd function defined on $[0,\pi]$ by $f(\theta) = \theta(\pi-\theta)$. For $\theta>0$ the given function is $f(\theta)=\theta(\pi-\theta)$ with its derivative $f^\prime(\theta)=\pi-2\theta$. On the other hand, $f(\theta)= \theta(\pi+\theta)$ with its derivative $f^\prime(\theta)=\pi+2\theta$ for $\theta<0$. Then using the integration by parts,

\[\begin{align*} \hat{f}(n) &= \frac{1}{2\pi} \int_{-\pi}^\pi f(\theta) \exp\left( \frac{-2\pi in\theta}{2\pi} \right) d\theta \\ &=\frac{1}{2\pi}\left(\left[\frac{-f(\theta)\exp(-in\theta)}{in} \right]_{-\pi}^\pi + \int_{-\pi}^\pi f^\prime(\theta) \frac{e^{-in\theta}}{in} d\theta \right) \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi f^\prime(\theta) \frac{e^{-in\theta}}{in} d\theta \quad (\because f(\pi)=f(-\pi)=0) \\ &=\frac{1}{2\pi}\left(\int_{-\pi}^0 (\pi+2\theta)\frac{e^{-in\theta}}{in}d\theta + \int_{0}^\pi (\pi-2\theta)\frac{e^{-in\theta}}{in}d\theta \right) \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi \pi \frac{e^{-in\theta}}{in}d\theta + \frac{1}{in\pi}\left(\int_{-\pi}^0 \theta e^{-in\theta} d\theta - \int_0^\pi \theta e^{-in\theta}d\theta \right) \end{align*}\]

By Euler formula $e^{i\theta}=\cos\theta+i\sin\theta$, for all non-zero integer $k\in\mathbb{Z}\setminus {0}$

\[\begin{align*} \int_{-\pi}^\pi e^{ikx} dx &= \left[\frac{1}{ik} e^{ikx} \right]_{-\pi}^\pi \\ &=\frac{1}{ik}(e^{ik\pi}-e^{-ik\pi}) \\ &= \frac{1}{ik}\left(\cos(k\pi)+i\sin(k\pi) - \cos(-k\pi)-i\sin(-k\pi) \right) \\ &= \frac{1}{ik}\left(\cos(k\pi) + i\sin(k\pi) -\cos(k\pi)+i\sin(k\pi) \right) \\ &=\frac{1}{ik}2i\sin(k\pi) \\ &=0. \end{align*}\]

Then

\[\begin{align*} \int_{-\pi}^\pi \pi \frac{e^{-in\theta}}{in}d\theta &= \frac{\pi}{in}\int_{-\pi}^\pi e^{-in\theta}d\theta \\ &=0. \end{align*}\]

With the change of variable $\gamma=-\theta$,

\[\begin{align*} \hat{f}(n) &= \frac{1}{in\pi}\left(\int_{-\pi}^0 \theta e^{-in\theta} d\theta - \int_0^\pi \theta e^{-in\theta}d\theta \right) \\ &= \frac{1}{in\pi}\left(-\int^\pi_0 \gamma e^{in\gamma}d\gamma -\int_{0}^\pi \gamma e^{-in\gamma}d\gamma\right) \\ &= -\frac{1}{in\pi}\left(\int_{0}^\pi \theta(e^{in\theta}+e^{-in\theta})d\theta \right). \end{align*}\]

Since

\[\begin{align*} \int e^{in\theta} + e^{-in\theta}d\theta &= \frac{1}{in}\left( e^{in\theta}-e^{-in\theta}\right) + C \\ &=\frac{1}{in}(cos(n\theta) + i\sin(n\theta) - \cos(-n\theta)-i\sin(-n\theta)) + C \\ &= \frac{1}{in}2i\sin(n\theta) +C \\ &= \frac{2}{n}\sin(n\theta)+C, \end{align*}\]

we get the following equation with the integration by parts:

\[\begin{align*} \hat{f}(n) &= -\frac{2}{in\pi}\left(\left[\frac{\theta}{n}\sin(n\theta) \right]_0^\pi - \int_0^\pi \frac{1}{n}\sin(n\theta)d\theta\right) \\ &= \frac{2}{in^2\pi}\int_0^\pi \sin(n\theta)d\theta \\ &= \frac{2}{in^2\pi}\left[-\frac{1}{n}\cos(n\theta) \right]_0^\pi \\ &= \frac{2}{in^2\pi}\left( \frac{1-\cos(n\pi)}{n}\right) \\ &= \frac{2}{in^3\pi}(1+(-1)^{n+1}) \\ &= \frac{4}{in^3\pi} \end{align*}\]

for odd $n$, and $0$ otherwise.

Thus,

\[\begin{align*} f(\theta) \sim \sum_{k \text{ odd}}\frac{4}{ik^3\pi}e^{ikx} \end{align*}\] \[\tag*{$\square$}\]

Definition 2.2

We define $N$-th partial sum of the Fourier series of $f, N\in\mathbb{N}$ by

\[\begin{align*} S_N(f)(x) := \sum_{n=-N}^N\hat{f}(n)\exp\left(\frac{2\pi inx}{L} \right). \end{align*}\]

Remark

“Convergence of the Fourier series to $f$” will always mean convergence of the partial sum $S_N(f)$ to $f$.

Definition 3.1

We define the Dirichlet Kernel $D_N$ by

\[\begin{align*} D_N(x) = \sum_{n=-N}^N e^{inx} \end{align*}\]

where $x\in [-\pi, \pi]$.

Lemma 3.2

\(\begin{align*} D_N(x) = \frac{\sin((N+\frac{1}{2})x)}{\sin (x/2)} \end{align*}\)

<Proof>

\[\begin{align*} D_N(x) &= \sum_{n=0}^N e^{inx} + \sum_{n=-1}^{-N}e^{inx} \\ &= \frac{1-(e^{ix})^{N+1}}{1-e^{ix}} + \frac{e^{-ix}(1-(e^{-ix})^N)}{1-e^{-ix}} \\ &= \frac{1-e^{(N+1)ix}}{1-e^{ix}} + \frac{1-e^{-Nix}}{e^{ix}-1} \\ &=\frac{1-e^{(N+1)ix}}{1-e^{ix}} + \frac{e^{-Nix}-1}{1-e^{ix}} \\ &= \frac{e^{-Nix}-e^{(N+1)ix}}{1-e^{ix}} \\ &= \frac{e^{-(N+\frac{1}{2})ix}-e^{(N+\frac{1}{2})ix}}{e^{-\frac{ix}{2}}-e^{\frac{ix}{2}}} \\ &=\frac{\sin((N+\frac{1}{2})x)}{\sin(x/2)} \end{align*}\] \[\tag*{$\square$}\]

Remark

Dumb question. What are the Fourier coefficients of $D_N$?

\[\begin{align*} \hat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^\pi D_N(x) e^{-inx}dx \\ &=\sum_{m=-N}^N \frac{1}{2\pi} \int_{-\pi}^\pi e^{imx}e^{-inx}dx \\ &= \mathbf{1}\{n\leq N \} \end{align*}\]

Definition 3.3

We define Poisson kernel $P_r(\theta)$ by

\[\begin{align*} P_r(\theta) = \sum_{n=-\infty}^\infty r^{\lvert n \rvert} e^{in\theta} \end{align*}\]

where $\theta\in [-\pi, \pi]$ and $r\in [0,1)$.

Remark

Note that \(\sum_{n=-\infty}^\infty \left\lvert r^{\lvert n \rvert} e^{in\theta}\right\rvert = \sum_{n=-\infty}^\infty r^{\lvert n\rvert}\)

and $\sum_{n=-\infty}^\infty r^{\lvert n \rvert}$ converges. By Weierstrass M-Test, $P_r(\theta)$ converges uniformly. Then the Fourier coefficients of Poisson kernel are

\[\begin{align*} \hat{P}_r(n) & = \frac{1}{2\pi}\int_{-\pi}^\pi \sum_{m=-\infty}^\infty r^{\lvert m \rvert} e^{im\theta} \overline{e^{in\theta}}d\theta \\ &=\sum_{m=-\infty}^\infty \frac{1}{2\pi}r^{\lvert m\rvert} \int_{-\pi}^\pi e^{im\theta}e^{-in\theta}d\theta \quad (\because \text{ uniform convergent}) \\ &=r^{\lvert m\rvert}. \end{align*}\]

Lemma 3.4

$P_r(\theta) = \frac{1-r^2}{1-2r\cos\theta+r^2}$

<Proof>

Letting $\omega = re^{i\theta}$, we have a sum of geometric series:

\[\begin{align*} P_r(\theta) &= \sum_{n=0}^\infty \omega^n + \sum_{n=1}^\infty \bar{\omega}^n \\ &=\frac{1}{1-\omega} + \frac{\bar{\omega}}{1-\bar{\omega}} \\ &= \frac{1-\bar{\omega}+\bar{\omega}-\omega\bar{\omega}}{1-\omega-\bar{\omega} + \omega\bar{\omega}} \\ &=\frac{1 -r^2}{1-2r\cos\theta +r^2} \end{align*}\] \[\tag*{$\square$}\]

Theorem 4.1 (Uniqueness)

Let $f$ be integrable function on the circle. Suppose $\hat{f}(n)=0$ for all $n\in\mathbb{Z}$. If $f$ is continuous at $\theta_0$, then $f(\theta_0)=0$.

<Proof> Without loss of generality, assume $\theta_0=0$ and $f(\theta_0) >0$. By continuity of $f$ there is a $\delta>0$ such that

\[\begin{align*} x\in N_{\delta}(0) &\Rightarrow \lvert f(x) - f(0)\rvert <\frac{1}{2}f(0) \\ &\Rightarrow f(x) > \frac{1}{2}f(0) \end{align*}\]

Now, we create a sequence of trigometric polynomials (finite linear combination of $\{ e^{inx}: n\in\mathbb{Z}\}$ as follows: Let

\[\begin{align*} p(\theta) = \cos\theta + \epsilon \end{align*}\]

where $\epsilon>0$ is chosen so small that out side of $(-\delta, \delta)$, we still have

\[\begin{align*} \lvert p(\theta) \rvert < 1- \frac{\epsilon}{2}. \end{align*}\]

Since $\cos\theta$ is continuous, we can choose a small $\eta>0$ such that

\[\begin{align*} \theta \in (-\eta, \eta) &\Rightarrow \lvert p(\theta) - p(0)\rvert < \frac{\epsilon}{2} \\ &\Rightarrow p(\theta) > 1 + \frac{\epsilon}{2} \end{align*}\]

Let $p_k(\theta) = [p(\theta)]^k$. We know that $\hat{f}(n) = \langle f, e^{inx}\rangle = 0$ for all $n\in\mathbb{Z}$ Thus, $\langle f, \sum_{n=1}^N c_ne^{inx}\rangle=0$ for all $N\in\mathbb{N}$. Since $\cos\theta = \frac{e^{i\theta}+ e^{-i\theta}}{2}$,

\[\begin{align*} (\cos\theta + \epsilon)^n &= \sum_{k=0}^n{n\choose k}\epsilon^{n-k}\cos^k\theta \\ &=\sum_{k=0}^n{n \choose k}\epsilon^{n-k} \sum_{l=0}^k \frac{e^{il\theta}+ e^{-i(k-l)\theta}}{2^k} \\ &\in \text{span}\{e^{inx}: n\in\mathbb{Z} \}. \end{align*}\]

Thus, $\langle f, p_k\rangle$ should be zero. But we will show that $\lim_{k\to\infty}\langle f, p_k\rangle =\infty$ which is a contradiction.

Now consider

\[\begin{align*} 2\pi\langle f, p_k(\theta) \rangle &= \int_{-\pi}^\pi f(\theta) p_k(\theta) d\theta \\ &=\int_{(-\eta, \eta)} f(\theta) p_k(\theta) d\theta+\int_{(-\delta, \delta)^c} f(\theta) p_k(\theta) d\theta + \int_{(-\delta, \delta)\setminus (-\eta, \eta)} f(\theta) p_k(\theta) d\theta \end{align*}\]

(1) In the $\eta$-neighborhood, we have

\[\begin{align*} \int_{-\eta}^\eta f(\theta) p_k(\theta) d\theta \geq \frac{f(0)}{2}\left(1+\frac{\epsilon}{2}\right)^k 2\eta \end{align*}\]

and it goes to infinity as $k\to\infty$.

(2) Outside of $\delta$-neighborhood, we have

\[\begin{align*} \int_{[-\pi, \pi]\setminus (-\delta, \delta)} f(\theta) p_k(\theta)d\theta &\leq \left\lvert\int_{[-\pi, \pi]\setminus (-\delta, \delta)} f(\theta) p_k(\theta)d\theta\right\rvert \\ &\leq \int_{[-\pi, \pi]\setminus (-\delta, \delta)} \lvert f(\theta) \rvert \lvert p_k(\theta)\rvert d\theta \\ &\leq 2\pi B\left( 1-\frac{\epsilon}{2}\right)^k \end{align*}\]

where $B$ is the bound on integrable function $f$. It converges to $0$ as $k\to \infty$.

(3) Between $(-\delta, \delta)$ and $(-\eta, \eta)$, $f(\theta) > \frac{f(0)}{2} >0$. Since we can take $\delta <\frac{\pi}{2}$, $p(\theta)>0$. Thus,

\[\begin{align*} \int_{(-\delta, \delta)\setminus (-\eta, \eta)} f(\theta) p_k(\theta) d\theta > 0. \end{align*}\]

Together, the above prove that

\[\begin{align*} \lim_{k\to\infty}\int_{-\pi}^\pi f(\theta)p_k(\theta) d\theta = \infty \end{align*}\]

, which contradicts to $\langle f, p_k \rangle = 0$.

\[\tag*{$\square$}\]

Corollary 2.1

Suppose $f$ is continuous on the circle. If $\hat{f}(n)=0$ for all $n\in\mathbb{Z}$, then $f\equiv 0$. If $f,g$ are $2\pi$ periodic function and $\hat{f}(n)=\hat{g}(n)$ for all $n\in\mathbb{Z}$, then $f= g$.

Corollary 2.2

Suppose $f$ is a continuous function on the circle, and $\sum_{n=-\infty}^\infty \lvert \hat{f}(n)\rvert <\infty$. Then

\[\begin{align*} \lim_{N\to\infty}S_N(f)(\theta) &=f(\theta) \end{align*}\]

uniformly on the circle.

<Proof>

Since

\[\begin{align*} \sum_{n=-N}^N \lvert \hat{f}(n)\rvert \left\lvert e^{in\theta}\right\rvert = \sum_{n=-N}^N\lvert \hat{f}(n)\rvert \end{align*}\]

and it absolutely converges and is independent of $\theta$, $S_N(f)$ converges uniformly on the circle by Weierstrass M-test. Since $S_N(f)$ is continuous and uniform convergent, $S_N(f)$ uniformly converges to some continuous function.

Fourier coefficient of $\lim_{N\to\infty}\sum_{n=-N}^N \hat{f}(n)e^{in\theta}$ is

\[\begin{align*} \left\langle\lim_{N\to\infty}\sum_{n=-N}^N \hat{f}(n)e^{in\theta}, e^{im\theta} \right\rangle &= \frac{1}{2\pi}\int_{-\pi}^\pi \lim_{N\to\infty} \sum_{n=-N}^N \hat{f}(n)e^{in\theta} e^{-im\theta}d\theta \\ &=\frac{1}{2\pi}\lim_{N\to\infty}\sum_{n=-N}^N \int_{-\pi}^\pi \hat{f}(n)e^{in\theta} e^{-im\theta}d\theta \\ &=\frac{1}{2\pi}\lim_{N\to\infty} \sum_{n=-N}^N \hat{f}(n)\int_{-\pi}^\pi \delta_{n,m}d\theta \\ &=\hat{f}(m) \end{align*}\]

So two continuous functions have the identical Fourier coefficients. By previous corollary, they must be the same functions.

\[\tag*{$\square$}\]

Big O-Notation

We say $f(x) = O(g(x))$ as $x\to a$ if there exists a $C>0$ such that

\[\begin{align*} \lim_{x\to a} \frac{f(x)}{g(x)} \leq C. \end{align*}\]

Corollary 2.4

Let $f$ be a function on the circle. If $f$ is twice continuously differentiable (i.e. is of class $\mathscr{C}^2$), then $\hat{f}(n) = O(1/\lvert n \rvert^2)$ as $\lvert n\rvert\to \infty$.

<Proof>

\[\begin{align*} 2\pi \hat{f}(n) &= \int_0^{2\pi} f(\theta) e^{-in\theta}d\theta \\ &=\left[ f(\theta) \frac{-e^{-in\theta}}{in}\right]_0^{2\pi} + \frac{1}{in}\int_0^{2\pi} f^\prime(\theta) e^{-in\theta}d\theta \\ &=\frac{1}{in}\int_0^{2\pi}f^\prime (\theta)e^{-in\theta}d\theta \\ &=\frac{1}{in}\left[f^\prime(\theta)\frac{-e^{-in\theta}}{in} \right]^{2\pi}_0 + \frac{1}{(in)^2}\int_0^{2\pi}f^{\prime\prime}(\theta)e^{-in\theta}d\theta \\ &= -\frac{1}{n^2} \int_0^{2\pi}f^{\prime\prime}(\theta)e^{-in\theta}d\theta \end{align*}\]

So

\[\begin{align*} \lvert \hat{f}(n) \rvert n^2 &\leq \frac{1}{2\pi}\int_0^{2\pi} \lvert f^{\prime\prime}(\theta) e^{-in\theta}\rvert d\theta \\ &=\frac{1}{2\pi}\int_0^{2\pi}\lvert f^{\prime\prime}(\theta)\rvert d\theta \\ &\leq C \end{align*}\] \[\tag*{$\square$}\]

Definition 2.5

Let $f$ be a function for which there exists a constant $A\in\mathbb{R}$ such that for all $x,h$

\[\begin{align*} \lvert f(x+h) - f(x) \rvert \leq A \lvert h \rvert^\alpha, \end{align*}\]

Then we say that $f$ satisfies a Holder continuous of order $\alpha$.

Remark

If a function is Holder continuous order $\alpha> 1/2$, then the Fourier series converges absolutely and thus uniformly to $f$.

Reference

• Elias M. Stein and Rami Shakarchi 『Fourier Analysis: An Introduction』
• Math 139 Fourier Analysis Notes