 Solution

For $\epsilon >0$, insert a Gaussian function into the proposed inverse Fourier transform:The Fourier bounds for Schwartz functions show that $\hat{\varphi }$ is an integrable function, so dominated convergence givesOn the other hand, Fubini's theorem applies because $\Vert \varphi {\Vert }_1\int e^{-\epsilon {\xi }^2/2}d\xi <\infty$. The given Fourier transform of a Gaussian, rescaled by $\sqrt{\epsilon }$, yieldsThe last integrand is bounded by $\Vert \varphi {\Vert }_{\infty }{e}^{-z^2/2}/\sqrt{2\pi }$. Since $\varphi$ is continuous, another application of dominated convergence makes this tend to $\varphi (x)$. Equating the limits proves the Fourier inversion theorem:The Gaussian functions above form an approximate identity; no Fourier inversion was used in evaluating their integrals.