Does Plancherel's Identity hold in $L^2$?

Question

We have the following result, known as Plancherel's Identity that states:

If $f,g \in \mathcal{S}(\mathbb{R})$, then $(f,g) = (\hat{f},\hat{g})$

where $\mathcal{S}(\mathbb{R})$ is the Schwartz space, $( \cdot,\cdot)$ denotes the $L^2$ product on $\mathcal{S}(\mathbb{R})$ and $\hat{}$ denotes the Fourier transform.

I was wondering if this result holds more generally. In particular, if we can use Plancherel's Identity when $f,g \in L^2$, such that $\hat{f},\hat{g} \in L^2$.

Thanks for your suggestions.

Answer 1

Yes, it is true, and in fact the $L^2$ version is also called Plancherel's theorem, and can be found in most textbooks.

Once you have the Schwartz version, then the $L^2$ version follows quickly. You now have that the Fourier transform is a linear isometry from the dense subspace $\mathcal{S}(\mathbb{R}) \subset L^2$ into $L^2$. In particular it is bounded, so it extends to a bounded linear map defined on all of $L^2$, which is how the Fourier transform of a general $L^2$ function is defined. Using continuity, one easily checks that this map is again an isometry, and that is equivalent to the desired identity.