Why the coefficient $1/L$ in the fourier coefficients?

Question

In Stein's book on Fourier analysis, he defines the Fourier coefficients as $\hat{f}(n)=\frac{1}{L}\int_a^b f(x) e^{-2 \pi i n x /L}$ where the interval $[a,b]$ has length $L$. I understand that we need the $1/L$ in front of the integral in order for the basis to be orthonormal, but why do we need the $/L$ inside the exponential. Without it our basis $(e^{-2 \pi i nx})_{n\in \mathbb{}Z}$ would still be orthonormal, no?

Answer 1

Short answer is: no, your basis won't be orthonormal.

For longer answer, let us clarify some things. We are looking at Hilbert space $L^2([a,b])$ and we will fix a scalar product $$\langle f, g \rangle = \frac 1L \int_a^b f(x)\overline{g(x)}\, dx,$$ where $L = b - a$. As you say, $1/L$ is here so $1\colon [a,b]\to \mathbb C$ is normalized, i.e. $\langle 1,1\rangle = 1$.

Now, we want to find a coefficient $c$ such that $\{e_n\}_{n\in\mathbb Z}$ is orthonormal basis, where $e_n\colon [a,b]\to\mathbb C$, $e_n(x) = e^{incx}$.

Quick calculation of $\langle e_n,e_n\rangle$:

$$\frac 1 L \int_a^b e_n(x)\overline{e_n(x)}\, dx = \frac 1 L\int_a^b e^{incx}e^{-incx}\,dx = \frac 1 L \int_a^b 1\,dx = 1.$$

We need orthogonality as well, so for each $n\in\mathbb Z$, $n\neq 0$, we need $\langle e_n,1\rangle = 0.$ This is enough to claim that $\{e_n\}_{n\in\mathbb Z}$ is orthonormal since $\langle e_n, e_m\rangle = \langle e_{n-m},1\rangle$.

Now,

$$0 = \langle e_n,1\rangle = \frac 1L \int_a^b e^{incx}\,dx = \frac {e^{incb} - e^{inca}}{incL} \iff e^{incb}=e^{inca} \iff (\exists k\in\mathbb Z)\ incL = 2k\pi i.$$

Thus, we conclude that $\{e_n\}_n$ is orthonormal if and only if for each $n$ there is an integer $k$ such that $incL = 2k\pi i$, i.e. if and only if $\frac{cL}{2\pi}n \in \mathbb Z$, for all $n$, i.e. if and only if $\frac{cL}{2\pi}$ is an integer.

So, for any $k\in\mathbb Z$, $c = \frac{2\pi k}{L}$ will give an orthonormal set. If $L$ happens to be integer, just choose $k = L$ and $\{e^{2\pi i n x}\}_{n\in\mathbb Z}$ will be orthonormal, but this is not true if $L$ is not an integer. Warning, though, if $k \neq 1$, your set won't be a basis.

Intuitively, path $x\mapsto e^{incx}$ is a path along the unit circle in $\mathbb C$ and integral $\int_a^b e^{incx}\,dx$ will be $0$ if and only if the path $x\mapsto e^{incx}\colon [a,b]\to\mathbb C$ makes integral number of turns along the unit circle. Choosing $c = \frac{2\pi}L$ gives us that $e_n$ makes precisely $n$ turns.

Answer 2

"Without it our basis would still be orthonormal, no?" No it wouldn't. Did you try to compute the scalar products to verify your claim?