A Schwartz function is identically zero on $\mathbb R^2$ if its integral on every line in the plane is zero

Question

If the integral of a Schwartz function is zero on every line in the plane then it is zero.

I think maybe Fourier transform in one variable is useful. But I can't success.

Is there any hint to prove this?

Thanks

Answer 1

To start, note that $$\hat f(0,t)=\int_{\Bbb R}\left(\int_{\Bbb R} f(x,y)e^{-ity}\,dx\right)\,dy=\int_{\Bbb R}0\,dy=0$$

There are various ways you could modify this to show $\hat f(s,t)=0$. The most fun is this: If $O$ is an orthogonal $2\times 2$ matrix (or rather the linear transformation defined by such a matrix) then $$\widehat{f\circ O}=\hat f\circ O.$$

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That last bit about the orthogonal transformation is a standard thing. But since I stated it incorrectly at first I might give a proof. We're going to shift notation, taking $x,\xi\in\Bbb R^2$, so that, after inserting the $\pi$'s as in your favorite normalization for the Fourier transform, $$\hat f(\xi)=\int_{\Bbb R^2}f(x)e^{-ix\cdot\xi}\,dx.$$ Since Lebesgue measure is rotation-invariant and $O^{-1}=O^T$ we see $$\int f(Ox)e^{-ix\cdot\xi}\,dx=\int f(x)e^{-i(O^Tx)\cdot\xi}\,dx=\int f(x)e^{-ix\cdot(O\xi)}\,dx.$$