= Solution
Let $R\phi(x)=\phi(-x)$, and define reflection of a <tempered distribution> by $\langle Ru,\phi\rangle=\langle u,R\phi\rangle$. The <Fourier inversion theorem> on the <Schwartz space> says $\widehat{\widehat\phi}=2\pi R\phi$. Thus
$$
\langle\widehat{\widehat u},\phi\rangle
=\langle u,\widehat{\widehat\phi}\rangle
=2\pi\langle u,R\phi\rangle
=2\pi\langle Ru,\phi\rangle.
$$
Reflection commutes with the <Fourier transform>, since $\widehat{R\phi}=R\widehat\phi$ and the same relation follows by duality for <tempered distributions>. Define the <inverse Fourier transform> on <tempered distributions> by
$$
\mathcal F^{-1}v=\frac1{2\pi}R\widehat v,
\qquad
\langle\mathcal F^{-1}v,\phi\rangle=\langle v,\mathcal F^{-1}\phi\rangle.
$$
It follows that
$$
\boxed{\mathcal F^{-1}\mathcal Fu=u=\mathcal F\mathcal F^{-1}u\quad(u\in\mathcal S'(\mathbb R)).}
$$
In particular $u=(2\pi)^{-1}R\widehat{\widehat u}$ is the general <Fourier inversion> formula. When $\widehat u$ is an <integrable function>, this is the <regular distribution> of $(2\pi)^{-1}\int\widehat u(\xi)e^{ix\xi}\,d\xi$; an arbitrary <tempered distribution> need not have an ordinary integral representation.
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