Surjectivity of the Fourier Transform on Schwartz Space

I understand that, for $f \in S(\mathbb{R})$ (the Schwartz space) the transform $$\tag1 Tf(\xi) = (2\pi)^{-\frac{1}{2}} \int_\mathbb{R} e^{i\xi x}f(x) \,dx$$

defines a left inverse for the Fourier Transform $$f(x) \mapsto \mathcal{F}(\xi) = \hat{f}(\xi) = (2\pi)^{-\frac{1}{2}}\int_\mathbb{R} e^{-i\xi x} f(x) \, dx$$

Now I need to show that it is a right inverse, or (equivalently) that $\mathcal{F}$ is surjective.

Here is where I am struggeling to proceed. The proof that I am currently reading states that surjectivity follows from the fact that $(1)$ maps $f(x)$ to $\hat{f}(-\xi)$, but why does this mean that $\mathcal{F}$ is surjective?

Many thanks for your help!

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Your fact implies applying the transform 4 times returns the original function. So the inverse transform is the same as applying the transform 3 times. – Ragib Zaman Feb 4 '12 at 3:07

1 Answer

The Fourier transform converts differentiation into multiplication by polynomials. The Schwartz functions die at $\infty$ despite multiplication by polynomials and iterated differentation.

They are topologized as a locally convex spaces by norms that multiply by polynomials and differentiate. The result is that the Fourier transform sends them homeomorphically onto themselves.

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