Injectivity of the Fourier transform on $L^1$ without inversion Is there a proof of the injectivity of the Fourier transform on $L^1({\bf R})$ that does not rely on an inversion formula?
The proofs I have seen in the literature ultimately rely either on the Fourier inversion formula for $L^2$ or Schwartz functions, on the Plancherel formula in $L^2$ which is more or less equivalent to the inversion formula, or on some inversion formula in $L^1$ using the Fejer kernel.
 A: Following the comment of Denis Chaperon de Lauzières, I had a look at the book of Rudin, which indeed leads to a proof of the injectivity based on the Stone-Weierstrass theorem and does not rely on the $L^2$ theory.
One needs an integrable function $k$ with compactly supported transform. W. Rudin uses the inversion formula here but this can be obtained by a direct explicit computation if we are on ${\bf R}$, e.g. with $k(x) = {\sin^2 x\over x^2}$. Then, using Fubini theorem, 
$$\langle e^{i\xi x}, \overline{\hat{k}} f \rangle  = \langle \hat{k}, e^{-i\xi x} f \rangle = \langle k, \hat{f}(.-\xi)\rangle = 0.$$
From the Stone-Weierstrass theorem (wrt the compact-open topology), we see that $\langle g, \overline{\hat{k}} f \rangle = 0$ for all continuous $g$ and by density, $\overline{\hat{k}} f = 0$. We get the result by dilating the support of $\hat{k}$.
Answering Alexandre Eremenko's comment, this gives a way to solve quickly the heat equation without needing to set up the whole $L^2$ theory. Indeed, denoting the heat kernel by $K_t$, and assuming we know how to compute its transform, getting to the heat equation in the phase space 
$$\hat{u} = \widehat{K_t}\widehat{u_0} = \widehat{K_t * u_0}$$
only uses the most basic properties of the Fourier transform on $L^1$ and injectivity gives the result. The inverse formulas are not needed here. 
