The complex exponential

Many of the computations involving Fourier series are simpler when we use the complex exponential instead of sine and cosine, so in this section we will consider complex valued solutions to the wave equation.

Euler’s Formulas

By definition

$$e^{i\theta} = \cos \theta + i\sin\theta\quad\text{for all }\theta\in\R. \tag{1}$$

This implies

$$e^{-i\theta} = \cos \theta - i\sin\theta\quad\text{for all }\theta\in\R$$

and hence

$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \qquad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} \tag{2}$$

Derivative of the complex exponential

Euler’s definition of $e^{i\theta}$ implies

$$\frac{de^{i\theta}}{d\theta} = i e^{i\theta}. \tag{3}$$

In the language of Linear Algebra, $e^{i\theta}$ is an eigenvector of the linear transformation $f\mapsto \frac{df}{d\theta}$ with eigenvalue $i$.

Trigonometric polynomials

A function of the form

$$f(x) = A_0 + \sum_{n=1}^N \bigl(A_n\cos nx + B_n\sin nx\bigr)$$

where $A_n, B_n\in \mathbb{C}$ are constants is called a trigonometric polynomial. Using Euler’s formulas we can rewrite a trigonometric polynomial in terms of complex exponentials:

$$\begin{aligned} f(x) &= A_0 + \sum_{n=1}^N \bigl(A_n\cos nx + B_n\sin nx\bigr) \\ &=A_0 +\sum_{n=1}^N \frac{A_n-iB_n}{2}e^{inx} + \frac{A_n+iB_n}{2} e^{-inx} \\ &=\sum_{n=-N}^N \hat f_n e^{inx}, \end{aligned}$$

provided we define

$$\hat f_n = \begin{cases} (A_n-iB_n)/2 & n\lt 0 \\ (A_n+iB_n)/2 & n\gt 0 \\ A_0 & n=0 \end{cases}$$

Finite string, Fourier solution

The wave equation is linear. If $u, v:\R^2\to\R$ are solutions to the wave equation,

then $w(x, t) = au(x, t)+bv(x, t)$ is also a solution for any choice of $a, b\in\R$. This is true both for classical solutions and for weak solutions; i.e. if $u,v$ are classical solutions, then $w=au+bv$ is a classical solution, and if $u,v$ are weak solutions, then $w=au+bv$ is a weak solution.

Solutions that are periodic in space. Instead of considering arbitrary solutions to the wave equation, we look for solutions for which

$$u(x+2\pi, t) = u(x, t)$$

holds for all $x, t$. By direct substitution in the equation we can verify that each of the functions

$$e^{int}e^{inx}, \qquad e^{-int}e^{inx}$$

are classical solutions of the wave equation, and that they are $2\pi$–periodic in the $x$ variable. It follows that for any choice of $\hat a_n, \hat b_n\in\mathbb{C}$ the linear combination

$$u(x, t) = \sum_{n=-N}^N \hat a_ne^{int}e^{inx} + \hat b_ne^{-int}e^{inx}\tag{4}$$

is again a solution of the wave equation. If we have infinitely many coefficients $\hat a_n, \hat b_n$ then we could try to show that the series converges and that its limit is a weak or classical solution of the wave equation.

Sometimes it is convenient to rewrite (4) by applying Euler’s formula $e^{\pm int} = \cos nt \pm i\sin nt$, with result

$$u(x, t) = \sum_{n=-N}^N \hat u_n\cos (nt) e^{inx} + \hat v_n\sin(nt)e^{inx} \tag{5}$$

where $\hat u_n = \hat a_n+\hat b_n$ and $\hat v_n = i(\hat a_n-\hat b_n)$.

The initial value problem. Suppose we want to find a solution to

$$\begin{aligned} &u_{tt}=u_{xx}, \text{ for all }x,t\\ &u(x+2\pi, t)=u(x, t), \text{ for all }x, t\\ &u(x, 0) = f(x),\quad u_t(x, 0) = g(x) \text{ for all }x \end{aligned}$$

by looking for a function $u(x, t)$ that is given by (5). The function already satisfies the PDE, and it is $2\pi$–periodic in the $x$ variable, so we only have to choose the coefficients $\hat u_n, \hat v_n$ so that $u$ satisfies the initial conditions. It follows from (5) that

$$f(x) = u(x, 0) = \sum_n \hat u_ne^{inx}, \qquad g(x) = u_t(x, 0) = \sum_n -n\hat v_n e^{inx}.$$

If $f$ and $g$ are trigonometric polynomials given by

$$f(x) = \sum_{n=-N}^N \hat f_ne^{inx},\qquad g(x) = \sum_{n=-N}^N \hat g_ne^{inx},\qquad$$

then we should choose $\hat u_n, \hat v_n$ so that

$$\hat u_n = \hat f_n,\qquad \hat v_n = \frac{\hat g_n}{n}.$$

Thus the solution is

$$u(x, t) = \sum_{n=-N}^N \left\{\hat f_n\cos(nt)+\frac{\hat g_n}{n}\sin(nt)\right\}\; e^{inx}. \tag{6}$$

How to find the coefficients $\hat f_n, \hat g_n$ if we know $f$ and $g$.

If $f(x)=\sum_{-N}^N \hat f_n e^{inx}$ then Fourier multiplied the equation with $e^{-ikx}$ and integrated from $0$ to $2\pi$:

$$\int_0^{2\pi} f(x)e^{-ikx}dx= \int_0^{2\pi}\sum_{n=-N}^N \hat f_n e^{i(n-k)x}dx= \sum_{n=-N}^N \hat f_n \int_0^{2\pi} e^{i(n-k)x}dx= 2\pi \hat f_k,$$

because when $k\neq n$

$$\int_0^{2\pi} e^{i(n-k)x}dx= \left[\frac{e^{i(n-k)x}}{i(n-k)}\right]_{x=0}^\pi =0$$

while when $k=n$ we get

$$\int_0^{2\pi} e^{i(n-k)x}dx= \int_0^{2\pi} 1\;dx=2\pi.$$

Thus, if we know the function $f$ then its coefficients are given by

$$\hat f_k = \frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-ikx}dx \tag{7}$$

Including infinitely many terms

If we completely ignore questions about convergence, then we could let $N\to\infty$ and claim that if the initial functions $f$ and $g$ are given by

$$f(x) = \sum_{-\infty}^\infty \hat f_n e^{inx}, \qquad g(x) = \sum_{-\infty}^\infty \hat g_n e^{inx},\tag{8}$$

then the $2\pi$–periodic solution to the wave equation with initial values $u(x, 0)=f(x)$ and $u_t(x, 0)=g(x)$ is given by

$$u(x, t) = \sum_{-\infty}^\infty \bigl\{\hat f_n \cos(nt) + \frac{\hat g_n}{n}\sin(nt)\bigr\} e^{inx} \tag{9}$$

So if you can write the initial function and time derivative as a Fourier series (8) then we have a solution given in (9). This leaves us with a few questions

• for which $2\pi$–periodic functions $f$ and $g$ can we find a Fourier expansion (8)?
• in what sense do the series (8) and (9) converge?

Fourier’s remarkable answer to the first question was: every $2\pi$–periodic function $f$ has a Fourier expansion and the coefficients can be found by computing the integrals in (7).

For example, the “sawtooth function” is given by

$$f(x) = \frac{\pi-x}{2} \text{ for }0\lt x\lt 2\pi$$

Fourier computed the coefficients in its expansion and claimed

$$f(x) = \sum_{n=1}^\infty \frac{\sin nx}{n} = \sin x + \tfrac12 \sin 2x + \tfrac13 \sin 3x + \cdots$$

Wikipedia offers a one paragraph history of Fourier’s claims.

Digression: convergence of Fourier series

There is not enough space in this aside on Fourier series to do the topic justice. If you want to read more, the book Fourier Analysis by Stein and Shakarchi (Princeton lectures in analysis) offers a good introduction. Alternatively, T.W.Körner’s book Fourier Analysis (yes, same title) from Cambridge University press is very readable.

Questions: was Fourier right?

Let $f$ be a $2\pi$–periodic function. Does the Fourier series for $f$ converge, and does it converge to $f$? To be more precise, define the partial sums of the series

$$S_Nf(x) = \sum_{-N}^N \hat f_n e^{ i n x},\qquad \hat f_n = \frac{1}{2\pi}\int_0^{2\pi} e^{-inx}f(x)\; dx.$$

Under what conditions can we guarantee that $\lim_{N\to\infty} S_Nf(x) = f(x)$? Is the converge uniform? What other notions of convergence might apply here?

The Dirichlet kernel. For any $2\pi$–periodic function $f$ the partial sums of the Fourier series of $f$ are given by

$$S_Nf(x) = \int_0^{2\pi} D_N(x-\xi) f(\xi)\; d\xi,\quad\text{where}\quad D_N(x) = \frac{1}{2\pi} \frac{\sin(N+\frac12)x}{\sin \frac{x}{2}}$$

The function $D_N$ is called the Dirichlet kernel. It satisfies

$$\int_0^{2\pi} D_N(x-\xi)\;d\xi = 1 \text{ for all }x\in\R.$$

Proof.

$$S_Nf(x) = \frac{1}{2\pi}\sum_{n=-N}^N \int_0^{2\pi} e^{in(x-\xi)}f(\xi)\;d\xi = \frac{1}{2\pi}\int_0^{2\pi} \sum_{n=-N}^N e^{in(x-\xi)}f(\xi)\;d\xi =\int_0^{2\pi} D_N(x-\xi) f(\xi)\; d\xi$$

where

$$D_N(x) = \frac{1}{2\pi}\sum_{-N}^N e^{inx}$$

We can compute the sum by using the formula for geometric sums:

$$\begin{aligned} \sum_{-N}^N e^{inx} &= e^{-iNx}+e^{-i(N-1)x}+\cdots+e^{i(N-1)x}+e^{iNx}\\ &= e^{-iNx} \bigl\{1+e^{ix} + e^{2ix} + \cdots + e^{2iNx}\bigr\} \\ &=e^{-iNx} \frac{e^{(2N+1)ix} - 1}{e^{ix}-1}\\ &=\frac{e^{i(N+\frac{1}{2})x} - e^{-i(N+\frac{1}{2})x}}{e^{ix/2}-e^{-ix/2}}\\ &=\frac{\sin(N+\frac{1}{2})x}{\sin \frac{x}{2}} \end{aligned}$$

Riemann–Lebesgue Lemma. If $f:[a,b]\to\R$ is Riemann integrable, then

$$\lim_{\lambda\to\infty} \int_a^b \sin(\lambda x) f(x)\; dx = \lim_{\lambda\to\infty} \int_a^b \cos(\lambda x) f(x)\; dx =0.$$

Proof. We show that $\int \sin(\lambda x)f(x)dx\to0$, the proof that $\int \cos(\lambda x)f(x)dx\to0$ being nearly identical.

Let $\epsilon\gt0$ be given. Then, by definition of Riemann integrability of the function $f$, there exists a partition $a=x_0\lt x_1\lt x_2\lt \cdots\lt x_{m-1}\lt x_m=b$, and numbers $m_i\lt M_i$ such that

• $m_k\leq f(x)\leq M_k$ for all $x\in [x_{k-1}, x_k]$, and all $k=1, \ldots, m$
• $\sum_{k=1}^m (M_k-m_k)(x_k-x_{k-1}) \lt \epsilon/\bigl(2(b-a)\bigr)$

Define the step function

$$s(x) = m_k \text{ for all $x\in (x_{k-1}, x_k]$ and for all $k=1, 2, \ldots, m$}$$

Then $s(x)\leq f(x)$ and $0\leq\int_a^b (f(x)-s(x))\;dx \lt \epsilon/\bigl(2(b-a)\bigr)$.

To show that $\int_a^b \sin(\lambda x) f(x)\; dx\to0$ we rewrite the integral as follows

$$\int_a^b \sin(\lambda x) f(x)\; dx =\int_a^b \sin(\lambda x) s(x)\; dx + \int_a^b \sin(\lambda x) \bigl(f(x)-s(x)\bigr)\; dx = A+B.$$

We can compute the first term $A$ explicitly:

$$A= \int_a^b \sin(\lambda x) s(x)\; dx = \sum_{k=1}^m \int_{x_{k-1}}^{x_k} \sin(\lambda x) m_kdx = \sum_{k=1}^m m_k\left[\frac{-\cos(\lambda x)}{\lambda}\right]_{x_{k-1}}^{x_k} =\frac {1}{\lambda}\sum_{k=1}^m m_k\left[\cos\lambda x_{k-1} - \cos\lambda x_{k}\right]$$

which implies

$$\left\vert A\right\vert \leq \frac{2}{\lambda} \sum_{k=1}^m |m_k| .$$

Therefore, if we define $\lambda_\epsilon = \dfrac{\epsilon}{4\sum m_k}$ then $|A|\lt \epsilon/2$ holds for all $\lambda\gt\lambda_\epsilon$.

Next we show that $B$ is also small. No matter what $\lambda$ is, we always have

$$|B|=\left\vert \int_a^b \sin(\lambda x) \bigl(f(x)-s(x)\bigr)\; dx \right\vert \leq \int_a^b \left\vert \sin(\lambda x) \right\vert \big|f(x)-s(x)\big|\; dx \leq (b-a)\cdot \frac{\epsilon}{2(b-a)} = \frac\epsilon 2.$$

Therefore we get $|A+B| \leq |A|+|B| \lt \epsilon/2 + \epsilon/2 = \epsilon$ for all $\lambda \gt \lambda_\epsilon$.

Definition. A function $f:\R\to\R$ is called Lipschitz continuous at $x\in\R$ if there exists a $C\gt 0$ such that

$$|f(\xi)-f(x)|\leq C|x-\xi| \text{ for all }\xi\in\R.$$

Convergence of a Fourier series at a point of Lipschitz continuity

Assume that $f:\R\to\R$ is $2\pi$–periodic, Riemann integrable, and that $f$ is Lipschitz continuous at $a$. Then

$$f(a) = \sum_{k\in \Z} \hat f_k e^{ika}, \text{ i.e. } \lim_{N\to\infty} \sum_{k=-N}^N \hat f_k e^{ika} = f(a)$$

Proof. It follows from $\int_0^{2\pi} D_N(\xi-a)d\xi=1$ that $f(a) = \int_0^{2\pi} D_N(\xi-a)f(a)d\xi$. Therefore

$$S_Nf(a)- f(a) = \int_0^{2\pi} D_N(\xi - a) \bigl(f(\xi)-f(a)\bigr) \; d\xi.$$

Substitute $\xi=a+s$, and use the definition of the Dirichlet kernel:

$$\begin{aligned} S_Nf(a)- f(a) &= \int_{-\pi}^{\pi} D_N(s) \bigl(f(a+s)-f(a)\bigr) \; ds\\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \sin\bigl((N+\tfrac12)s\bigr)\; \frac{f(a+s)-f(a)}{\sin s/2} \; ds\\ &=\int_{-\pi}^\pi \sin\bigl((N+\tfrac12)s\bigr) \; g(s)\;ds \end{aligned}$$

where

$$g(s)\stackrel{\sf def}{=} \frac{f(a+s)-f(a)}{2\pi\sin s/2} .$$

This function is bounded because

$$|g(s)|\leq \frac{C|s|}{2\pi|\sin (s/2)|} \leq \frac C2 \text{ for all } s\in (-\pi, \pi), s\neq0.$$

Here we have used that $\sin x\geq \frac{\pi}{2}x$ for all $x\in[0, \frac\pi2]$.
using the boundedness of $g$ and the fact that $f$ is Riemann integrable, it is now a somewhat lengthy exercise in real analysis to show that $g$ is also Riemann integrable on $[-\pi, \pi]$. We can therefore invoke the Riemann–Lebesgue lemma and conclude that there is a $\lambda_\epsilon$ such that for all $\lambda\gt\lambda_\epsilon$ one has

$$\left\vert\int_{-\pi}^{\pi} \sin (\lambda s) g(s) ds\right\vert \lt \epsilon.$$

and therefore $|S_Nf(a)- f(a)| \lt \epsilon$ for all $\lambda\gt \lambda_\epsilon$, i.e. for all $N>\lambda_\epsilon-\frac12$.

Theorem. If $f$ is a $2\pi$–periodic $C^2$ function then its Fourier coefficients $\hat f_n$ satisfy

$$|\hat f_n| \leq \frac{\Vert f''\Vert_\infty}{n^2} \qquad \text{ for all }n\neq 0.$$

The Fourier series $f(x) = \sum_{-\infty}^\infty \hat f_n e^{ i nx}$ converges uniformly, and absolutely.

Proof. Integration by parts in the definition of $\hat f_n$ gives us

$$2\pi\hat f_n = \int_0^{2\pi} e^{-inx}f(x)\; dx = \left[\frac{e^{-inx}}{-in}f(x)\right]_0^{2\pi} - \frac{1}{-in}\int_0^{2\pi} e^{-inx}f'(x)\; dx = \frac{1}{in}\int_0^{2\pi} e^{-inx}f'(x)\; dx.$$

Integrating by parts again we get

$$2\pi\hat f_n= \frac{-1}{n^2}\int_0^{2\pi} e^{-inx}f''(x)\; dx.$$

This implies

$$2\pi|\hat f_n| =\left|\frac{-1}{n^2}\int_0^{2\pi} e^{-inx}f''(x)\; dx\right| \leq \frac{1}{n^2}\int_0^{2\pi} \left|e^{-inx}f''(x)\right|\; dx \leq \frac{1}{n^2}\int_0^{2\pi} \|f''\|_\infty dx = 2\pi\frac{\|f''\|_\infty}{n^2}.$$

The terms in the Fourier series for $f$ are bounded by

$$\left|\hat f_n e^{inx}\right| =|\hat f_n|\leq \frac{\|f''\|_\infty}{n^2}\qquad (n\neq0).$$

Therefore, by the Weierstrass M-test, the series $\sum_{-\infty}^\infty \hat f_n e^{ i nx}$ converges uniformly, and absolutely.

Since $f$ is $C^1$ it is also Hölder continuous at all $x\in\R$. Therefore Theorem 3.5 implies that $\sum_{-\infty}^\infty \hat f_n e^{ i nx} = f(x)$ for all $x\in\R$.

Fourier series and the $L^2$ Inner Product

An inner product

Let $\mathcal R_{\rm per}$ be the set of Riemann integrable $2\pi$–periodic functions, and let $\mathcal C_{\rm per}$ be the subset of continuous $2\pi$–periodic functions. For any two functions $f, g\in \mathcal R_{\rm per}$ we define

$$\langle f, g\rangle = \int_0^{2\pi} f(x)\overline{g(x)}\; dx\,.$$

Then $\langle \dots \rangle$ has the following properties:

• Symmetry: $\langle f, g\rangle = \overline{\langle g, f\rangle}$
• Bilinear: $\langle f+g, h\rangle = \langle f, h\rangle + \langle g, h\rangle$ for all $f, g, h\in \mathcal R_{\rm per}$
• Bilinear: $\langle cf, g\rangle = c\langle f, g\rangle$ and $\langle f, cg\rangle = \bar c\langle f, g\rangle$ for all $f, g\in \mathcal R_{\rm per}$ and $c\in\mathbb xC$.
• Nonnegative: $\langle f, f\rangle \geq 0$ for all $f\in \mathcal R_{\rm per}$

Theorem. If $f\in \mathcal C_{\rm per}$ and $\langle f, f\rangle = 0$ then $f=0$.

On the other hand there exist functions $f\in\mathcal R_{\rm per}$ with $f\neq 0$ and $\langle f, f\rangle = 0$.

Proof. Arguing by contradiction we suppose $\langle f, f\rangle = 0$ and $f(x_0)\neq 0$ for some $x_0\in [0, 2\pi)$. Since $f$ is continuous, there is an $\delta\gt0$ such that $|f(x)|\geq \frac12 |f(x_0)|$ for all $x\in (x_0-\delta, x_0+\delta)$. This implies

$$0=\langle f, f\rangle = \int_0^{2\pi} |f(x)|^2dx \geq \int_{x_0-\delta}^{x_0+\delta} |f(x)|^2dx \geq \bigl(\tfrac12 |f(x_0)|\bigr)^2 \cdot (2\delta) >0.$$

This is a contradiction, so we conclude that $f(x_0)=0$ for all $x_0$. $\qquad$

To prove the second part of the theorem, consider the function $f$ for which $f(x)=0$ at all $x\in\R$ except $x=n\pi$ ($n\in\Z$) where $f(n\pi)=1$. Then $f$ is Riemann integrable and $2\pi$–periodic, and $\langle f, f\rangle = 0$ even though $f\neq 0$.

Theorem. $\mathcal C_{\rm per}$ with $\langle f, g \rangle$ defined as above is a complex vector space with an inner product.

See the analysis appendix for a review of complex inner products.

We define the $L^2$ norm of $f\in\mathcal R_{\rm per}$ to be $\|f\|_2$ where

$$\|f\|_2^2 = \langle f, f\rangle = \int_0^{2\pi} |f(x)|^2\;dx.$$

The $L^2$ and the supremum norm. The $L^2$ norm and the supremum norm $\|f\|_\infty = \sup_x|f(x)|$ are generally different. They always satisfy

$$\|f\|_2 \leq \sqrt{2\pi} \|f\|_\infty \text{ for any }f\in\mathcal R_{\rm per}.$$

This follows from the following short computation

$$\|f\|_2^2 = \int_0^{2\pi} |f(x)|^2dx \leq \int_0^{2\pi}\|f\|_\infty^2dx=2\pi\|f\|_\infty^2.$$

Fourier coefficients as inner products. Let $e_k(x)= e^{ikx}$. Then for any $k\in\Z$ and any $l\neq k$ one has

$$\|e_k\|^2 = 2\pi, \text{ and } \langle e_k, e_l\rangle = 0, \text{ i.e. }e_k\perp e_l.$$

If $f=\hat f_{-N}e_{-N}+\cdots+\hat f_Ne_N$ then we have

$$\langle f, e_k\rangle = \hat f_k \|e_k\|^2 = 2\pi \hat f_k.$$

The $k^{\rm th}$ Fourier coefficient of $f\in\mathcal R_{\rm per}$ is given by

$$\hat f_k = \frac{1}{2\pi} \langle f, e_k\rangle$$

The $N^{\rm th}$ partial sum of the Fourier series of $f\in\mathcal R_{\rm per}$ is given by

$$S_Nf = \frac{1}{2\pi}\sum_{k=-N}^N \langle f, e_k\rangle e_k$$

Bessel’s inequality. For any $f\in\mathcal R_{\rm per}$ we have

$$\|S_Nf\|_2 ^2 + \|f-S_Nf\|_2^2 = \|f\|_2^2.$$

In particular $\|S_Nf\|_2\leq \|f\|_2$ always holds.

Proof. We begin by showing that $S_Nf \perp (f-S_Nf)$. For any integer $k$ with $|k|\leq N$ we have

$$\begin{aligned} \langle e_k, f-S_Nf\rangle &= \langle e_k, f\rangle - \frac{1}{2\pi}\sum_{l=-N}^n \langle e_k, \langle f, e_l\rangle e_l\rangle\\ &= \langle e_k, f\rangle - \frac{1}{2\pi}\sum_{l=-N}^n \overline{\langle f, e_l\rangle}\langle e_k, e_l\rangle\\ &= \langle e_k, f\rangle - \frac{1}{2\pi}\sum_{l=-N}^n \langle e_l, f\rangle\langle e_k, e_l\rangle &\qquad&\langle e_k, e_l\rangle=0 \text{ if }k\neq l \\ &= \langle e_k, f\rangle -\frac{1}{2\pi}\langle e_k, f\rangle\langle e_k, e_k\rangle &\qquad&\langle e_k, e_k\rangle=2\pi \\ &=0. \end{aligned}$$

Thus $e_k\perp f-S_Nf$ for all $k=-N, \dots, N$. Since $S_Nf$ is a linear combination of $e_{-N}, \dots, e_N$ it follows that $S_Nf\perp f-S_N f$.

Bessel’s inequality now follows from Pythagoras: since $S_Nf\perp f-S_N f$ we have

$$\|f\|_2^2 = \|f-S_Nf+S_Nf\|_2^2 = \|f-S_Nf\|_2^2 + \|S_Nf\|_2^2.$$

Partial sums as best approximations. If $g=\sum_{-N}^N \hat g_k e_k$ is a trigonometric polynomial of degree at most $N$ then

$$\|f-S_Nf\|_2 \leq \|f-g\|_2,$$

with strict inequality if $g\neq S_Nf$.

Proof. First note that $S_Nf-g$ is a linear combination of $e_{-N}, \dots, e_N$, and therefore that $S_Nf-g \perp S_N-f$. Then apply Pythagoras to $f-g = f-S_Nf+S_Nf-g$, to get

$$\|f-g\|_2^2 = \|f-S_Nf\|_2^2 + \|S_Nf-g\|_2^2.$$

This implies $\|f-S_Nf\|_2 \leq \|f-g\|_2$, and if $\|f-S_Nf\|_2 = \|f-g\|_2$ then $\|S_Nf-g\|_2=0$.
Since $S_Nf-g$ is a trigonometric polynomial it is a continuous function. Therefore $\|S_Nf-g\|_2=0$ implies that $S_f-g=0$, i.e. $S_Nf=g$.

Convergence in $L^2$ of Fourier series. Let $f\in\mathcal R_{\rm per}$. Then

$$\lim_{N\to\infty} \|S_Nf-f\|_2 = 0$$

and

$$\|f\|_2^2 = \frac{1}{2\pi} \sum_{n\in\Z} |\hat f_n|^2.$$

This statement is known as the Plancherel or Parseval identity.

Proof. Let $\epsilon\gt0$ be given. Since $f$ is Riemann integrable, a $C^2$ function $g$ exists that is $2\pi$–periodic, and that satisfies

$$\int_0^{2\pi} |f(x)-g(x)|^2dx\leq \frac{\epsilon^2}{4}, \text{ i.e. } \|f-g\|_2\lt \frac{\epsilon}{2}.$$

(details in lecture). Since $g$ is $C^2$, we know that the Fourier series of $g$ converges uniformly to $g$, i.e.

$$\|S_Ng-g\|_\infty \to 0.$$

For any given $\epsilon_2\gt0$ we can therefore find an $N_{\epsilon}$ such that $||S_Ng-g\|_\infty \lt \frac\epsilon{2\sqrt{2\pi}}$ for all $N\geq N_{\epsilon_2}$. This implies

$$\|S_Ng-g\|_2\leq \sqrt{2\pi} \; \|S_Ng-g\|_\infty \lt \frac{\epsilon}{2}.$$

We now have for all $N\geq N_\epsilon$

$$\begin{aligned} \|S_Nf-f\|_2 &=\|S_Nf-S_Ng+S_Ng-g\|_2 \\ &\leq\|S_Nf-S_Ng\|_2 + \|S_Ng-g\|_2 \\ &=\|S_N(f-g)\|_2 + \|S_Ng-g\|_2 \\ &\leq\|f-g\|_2 + \|S_Ng-g\|_2 \\ &\lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{aligned}$$

The Parseval—Plancherel identity now follows by letting $N\to\infty$ in $\|f\|_2^2=\|S_Nf\|_2^2+\|f-S_Nf\|_2^2$.

Problems

Consider the solutions $u:\R^2\to\R$ to the wave equation described in (4).

Find functions $F, G:\R\to\R$ such that $u(x, t)= F(x+t)+G(x-t)$ for all $x, t$. (Hint: don’t use the derivation of d’Alembert’s solution but instead use $e^{i\theta}e^{i\phi} = e^{i(\theta+\phi)}$ and take a good look at (4).

Various questions about the Dirichlet kernel and Fourier coefficients

(a) Show that $\int_0^{2\pi} D_N(s)\; ds =1$.
(b) The $k^{\rm th}$ Fourier coefficient $\hat f_k$ of a $2\pi$–periodic function $f:\R\to\mathbb{C}$ is defined as an integral from $x=0$ to $x=2\pi$. Show that $\hat f_k = \frac{1}{2\pi}\int_{-\pi}^\pi e^{-ikx} f(x)\;dx$.