# l2 norms, rapidly decreasing functions and fourier transforms

Let $f\colon \mathbb{R} \to \mathbb{C}$ be a rapidly decreasing (rd) function. Let $\mathcal{F}(f)$ be the Fourier transform of $f$. It is known that

1) $\| \mathcal{F}(f) \|_2 = \| f \|_2$ (Plancherel theorem for rd functions) and

2) the space of rd functions from $\mathbb{R}$ to $\mathbb{C}$ is dense in $L^2$.

I have to prove the relation (1) in the case $f \in L^1 \cap L^2$.

Any ideas? My attempt was to note that there is a sequence of rd functions that converges to a $L^2$ function in $L^2$-norm, but how to get exactly the equality about norms?

• How is the Fourier transform for $L^2$ functions defined? The Fourier integral does in general not exist, so you need some form of approximation in the definition, which is used in your book/course? – Daniel Fischer Sep 20 '13 at 21:57
• Thank you. I fixed the question, hope it is ok now. – user91757 Sep 20 '13 at 22:09

Let $M \subset L^2(\mathbb{R})$ the space of rapidly decreasing functions, and $\mathcal{F}_0 = \mathcal{F}\lvert_M$ the restriction of the Fourier transform to $M$. Then $\mathcal{F}_0\colon M \to L^2(\mathbb{R})$ is a linear isometry, and since $L^2(\mathbb{R})$ is complete, it can be extended to a linear isometry $\Phi \colon L^2(\mathbb{R}) \to L^2(\mathbb{R})$.
Now let $f \in L^1 \cap L^2$. Then there is a sequence $(s_n)$ of rapidly decreasing functions (we can even choose smooth functions with compact support) that converges to $f$ simultaneously in the $\lVert\,\cdot\,\rVert_1$ and $\lVert\,\cdot\,\rVert_2$ norms. (If that's not known, it follows relatively easily by approximating $f$ first with a bounded function with compact support simultaneously in both norms, then Hölder's inequality with the characteristic function of the compact support yields the result.)
Then $\mathcal{F}_0(s_n)$ converges uniformly to $\mathcal{F}(f)$, since $\lVert s_n - f\rVert_1 \to 0$, and $\mathcal{F}_0(s_n)$ converges to $\Phi(f)$ in $L^2(\mathbb{R})$. All sequences that converge in $L^2$ contain subsequences that converge pointwise almost everywhere to the limit, so that means $\Phi(f) = \mathcal{F}(f)$. Hence, since $\Phi$ is an isometry,
$$\lVert \mathcal{F}(f)\rVert_2 = \lVert \Phi(f)\rVert_2 = \lVert f\rVert_2.$$