How do they do this w.l.o.g. so freely (Fourier series).

Question

Theorem 2.1. Suppose that $f$ is an integrable function on the circle with $\hat{f}(n) = 0$ for all $n \in \mathbb{Z}$. Then $f(\theta_0) = 0$ whenever $f$ is continuous at the point $\theta_0$.

Proof. We suppose first that f is real-valued, and argue by contradiction. Assume, without loss of generality, that $f$ is defined on the interval $[-\pi, \pi]$.

I won't get why this w.l.o.g. works until I finish the proof and can check that it works, but are there any insights beforehand I could have regarding this?

Answer 1

We know that $f$ is defined on the circle. That means that $f=f(e^{i\theta})$, where $\theta$ is an angle ranging around the circle.

There are many ranges that you could use for $\theta$: $[0,2\pi],[-2\pi,0],[1,1+2\pi]$, but these are all essentially equivalent (note, for example that the theorem says nothing about the parametrization of $\theta$) and it's easier just to choose one - say, $[-\pi,\pi]$.

Answer 2

I know that if $f$ is defined instead on $[a,b]$, then there's an invertible line $\ell(x) = (b-a)x/(2\pi) + (b+a)/2$ such that $g(x) = f(\ell(x))$, $g$ on $[-\pi, \pi]$ and $f(x) = g(\ell^{-1}(x))$. Also, $\ell$ is a diffeomorphism and a homeomorphism.

Let $f$ be defined on $[a,b]$. Then $\hat{f}(n) = \frac{1}{b-a}\int_a^b f(x) e^{-2\pi i n x/(b-a)} dx$. And if $g(x) = f(\ell(x))$ is defined on $[-\pi, \pi]$, then $\hat{g}(n) = \frac{1}{\pi - (-\pi)} \int_{-\pi}^{\pi} g(x) e^{- 2 \pi i n x / (2 \pi)}dx = \\ \frac{1}{\ell^{-1}(b) - \ell^{-1}(a)} \int_{\ell^{-1}(a)}^{\ell^{-1}(b)} f(\ell(x)) e^{- 2 \pi i n x / (\ell^{-1}(b) - \ell^{-1}(a))} dx $.

Making the substitution $u = \ell(x)$ we have $du = d\ell/dx = (b-a)/(2\pi)$ so if in the integral we can perform your standard $u,du$ substitution, then we have that $\hat{g}(n) = \frac{1}{b-a}\int_a^b f(u) e^{-2 \pi i n \ell^{-1}(u)/(2\pi)} du$.

The exponent of $e$ in the last integrand is $- in \frac{(2\pi u - (b+a)\pi)}{b-a}$.

So $\hat{g}(n) = \frac{(-1)^{n \frac{-(b+a)}{b-a}}}{b-a}\int_a^b f(u) e^{-2 \pi i n u / (b-a)} du$.

In other words a reparametrization of the interval over which $f$ is defined yields coefficients that are constant multiples of the original coefficients of $f$ before the interval is reparameterized. So re-examining the theorem, there exists a reparameterization to $[-\pi, \pi]$, $g$, of $f$ such that $\hat{f}(n) = 0$ iff $\hat{g}(n) = 0$. So when we prove the thoerem over $[-\pi, \pi]$, then we've proved it over all intervals $[a,b]$.

Proof. The theorem goes "if $\hat{f}(n) = 0, \forall n \in \mathbb{Z}$, then $f(\theta_0)= 0$ whenever $f$ is continuous at the point $\theta_0$." If $g = f\circ\ell$ is a reparameterization and we've prove the theorem for $g$. Then we've proved that "if $\hat{f}(n) = 0, \forall n \in \mathbb{Z}$, then $f(\ell(\theta_0)) = 0$ at continuity points $\theta_0$." The last part is the same as "$f(\phi_0) = 0$ at continuity points $\phi_0 = \ell(\theta_0)$", since if $f\circ \ell$ is continuous at $\theta_0$, and any open set in $\mathbb{R}$ is the inverse image of some open set in $\mathbb{R}$ under homeomorphism $\ell$, $f$ is continuous at $\phi_0$. Conversely, if $f$ is continuous at some $\phi_0$ then there's $\theta_0$ such that $f\circ\ell$ is continuous at $\theta_0$. I think that proves it. QED.