Injectivity of the Fourier transform on manifolds

Question

Let $\mathcal M$ be a smooth compact submanifold of $\mathbb R^n$ (equipped with the standard surface area measure), and define the Fourier transform of $f \in L^2(\mathcal M)$ to be $$\hat{f}(\xi) = \int_{\mathcal M} f(x) \cdot e^{2\pi i \cdot \langle x, \xi \rangle} \,\text{d} x$$ for $\xi \in \mathbb R^n$. Suppose that $\hat{f} = 0$; how would you show that $f = 0$?

Answer 1

It is the Fourier transform of a distribution $F$ compactly supported on $\mathcal{M}$. Let $F_k(x) = F \ast k^n e^{-\pi k^2|x|^2} = \int_\mathcal{M} k^n e^{-\pi k^2|x-y|^2}f(y)dy$ it is Schwartz and $\hat{F_k}(\xi)= \hat{F}(\xi) e^{-\pi |\xi|^2/k^2} = \hat{f}(\xi) e^{-\pi |\xi|^2/k^2}$. If $f$ is non-zero in $L^2(\mathcal{M})$ then $F_k$ is non-zero for $k$ large enough contradicting that $\hat{f} = 0 \implies \hat{F_k} = 0$