Does any $L_1(-\pi, \pi)$ function whose Fourier coefficients are 0 equal 0 a.e.?

Question

Suppose $f \in L_1(-\pi,\pi)$, satisfying $\hat{f}(n) = 0, \forall n\in \mathbb{Z}$ (which means all Fourier coefficients are $0$). Does $f = 0$ almost everywhere in $(-\pi,\pi)$?

The Fourier coefficients of $f$ are defined by $\hat{f}(n) = \int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta$

This is a similar question, but it does not provide a clear answer: Functions whose Fourier coefficients are all zero

Answer 1

This follows immediately from Parseval's identity, $$ \int_{-\pi}^\pi |f(\theta)|^2d\theta = \sum_{n=0}^\infty |\hat{f}(n)|^2. $$ Since the coefficients are zero, so is the function's square integral, and thus so is the function.

Answer 2

We know that a $L^1(-\pi, \pi)$ function's continuous point is dense in $(-\pi,\pi)$.

Use the Theorem 2.1 in Stein's 'Fourier Analysis'(Page 39), we know that when $\hat{f}(n)=0$ for all $n\in \mathbb{Z}$, $f(x)=0$ at all continuous point, and hence $f=0$ a.e. $x\in (-\pi,\pi)$.