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Surjectivity of the Fourier Transform on Schwartz Space

Consider the Fourier transform on Schwartz space, given by \begin{equation} \mathcal{F}(f)(\xi)= \hat{f}(\xi) = (2\pi)^{-\frac{1}{2}}\int e^{-i\xi x} f(x) \, \mathrm dx \end{equation}

I understand a proof in my notes that shows that we have a left inverse \begin{equation} \mathcal{F}^{-1}(\hat{f})(x) = (2\pi)^{-\frac{1}{2}}\int e^{i\xi x} \hat{f}(\xi) \, \mathrm d\xi \end{equation}

but how do I know that this is also the right inverse?

Many thanks for hints!

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marked as duplicate by Hans Lundmark, Jennifer Dylan, William, Nate Eldredge, J. M. Aug 24 '12 at 13:18

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We have for a function $g\in\mathcal S(\mathbb R)$: $\mathcal F(h)(x)=\mathcal F^{-1}(\widetilde h)(x)$ where $\widetilde h(x)=h(-x)$ (it'sjust a substitution). Applying this result to $h(x)=\mathcal F^{-1}(f)(x)$, we have \begin{align*} \mathcal F(\mathcal F^{-1}(f))(x)&=\mathcal F\left(\frac 1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{i\xi x}f(\xi)d\xi\right)\\ &=\mathcal F^{-1}\left(\frac 1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{i\xi -x}f(\xi)d\xi\right)\\ &=\mathcal F^{-1}(\mathcal F(f))(x)\\ &=f(x), \end{align*} since $\mathcal F^{-1}$ is the left-inverse of $\mathcal F$.

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