# Fourier transform of a Schwartz space function and norm

The Schwartz space (of rapidly decreasing functions) is the set of all $C^{\infty}$ functions $f\colon\mathbb{R}\to\mathbb{C}$ such that $$x^jf^{(k)}(x) \to 0$$ for all integers $j,k\geq0$ as $x \to {\pm\infty}$.

Let $f$ be a rapidly decreasing function in Schwartz space and let $\mathcal{F}(f)$ its Fourier transform. If $$\| f \|_2 = \| \mathcal{F}(f) \|_2 \quad,$$ can we deduce that $\mathcal{F}(f)$ is a Schwartz function?

You don't need their norms to be equal - this is a consequence of the normalization factor chosen for the Fourier transform, and is called the Plancherel Theorem.

To show that the Fourier transform of a Schwartz function is Schwartz, it suffices to show the following:

• $k^\alpha\hat{f}(k)=i^{-\alpha}\mathcal{F}\left[f^{(\alpha)}\right]$
• $\frac{d^\alpha}{dk^\alpha}\hat{f}(k)=(-ik)^\alpha\hat{f}(k)$

Combine these two facts with the Riemann-Lebesgue lemma and you've got it.

Note: I'm using the following convention for the Fourier transform:

$$\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ikx}dx$$

• Why do those those two facts and the Riemann-Lebesgue lemma imply the result? – Robert Cardona Nov 3 '14 at 3:55
• Since $f$ is Schwartz, all its derivatives will be $L^1$, so Riemann-Lebesgue gives that $\mathcal{F}[f^{(\alpha)}]$ goes to zero for any $\alpha$. The two facts above relate this to decay of $\hat{f}$ and its derivatives, which is needed to prove that $\hat{f}$ is Schwartz. – icurays1 Nov 3 '14 at 19:18