I've seen stated several times without proof that the Fourier transform (giving the Fourier coefficients $(c_k(f))$ of a function $f$) defined on $L^1(\mathbb{T}^n)$ (and not just on $L^2$) is injective.
I am wondering how to prove that. I've see a proof that the Fourier transform is injective on $L^2$, but how to extend the injectivity to $L^1$ ?
A relevant lemma is: if $f\in L^1(\mathbb{T}^n)$ is such that $\int_{\mathbb{T}^n} fg=0$ for every $g\in C(\mathbb{T}^n)$, then $f=0$ a.e. One way to prove this is to assume $f$ isn't $0$ a.e., picking a Lebesgue point $x_0$ where $f(x_0)\ne 0$, and constructing a sequence of continuous functions $g_n$ that approximate the delta-function at $x_0$. Then $\int fg_n\to f(x_0)$ yields a contradiction. See also: Integral of the product of a bounded and any continuous function.
Given the above lemma, one appeals to Weierstrass approximation theorem (trigonometric form). If all Fourier coefficients of $f$ are zero, then $\int_{\mathbb{T}^n} fg=0$ for every trigonometric polynomial $g$. But such polynomials are dense in $C(\mathbb{T}^n)$, hence the integral vanishes for every continuous $g$.