I'm writing a Bachelor-thesis in mathematics which is to be submitted in a couple of days, and would be thankfull if the following arguments concerning the extension of the Fourier transform from the Schwartz-class $\mathcal{S}(\mathbb{R})$ to $L^2(\mathbb{R})$ could be proof-read by someone knowledgeble:

"Since the compactly supported $C^{\infty}$-functions are dense in $L^2(\mathbb{R})$, as shown in [13], and these functions are contained in the Schwartz-class by definition, the same holds for $\mathcal{S}(\mathbb{R})$. As a consequence of this fact, all the result that will henceforth be stated for Schwartz-functions carries over (in some adequate sense, e.g. almost everywhere) to the whole space $L^2$ by an approximation procedure that will not be carried out explicitly. We can now state the Fourier inversion theorem, which proof can be found in the appendix.

Proposition. For every $f\in\mathcal{S}(\mathbb{R})$ the following integral formula holds

$$f(x) = \int\limits_{-\infty} ^\infty \widehat{f}(\omega) \mathrm{e}^{2\pi i \omega x} \mathrm{d}\omega.$$

Consequently, the Fourier transform acts bijectively on $\mathcal{S}(\mathbb{R})$. Moreover, the inversion theorem implies, and can be shown to be equivalent to, Plancherel's theorem which states that the Fourier transform preserves the norm of every $f\in \mathcal{S}(\mathbb{R})$, i.e. the energy of a signal equals the energy of its transform.


For $f\in \mathcal{S}(\mathbb{R})$ there holds

$$\int\limits_{-\infty} ^\infty \ |f(x)|^2 \mathrm{d}x=\int\limits_{-\infty} ^\infty \ |\hat{f}(\omega)|^2 \mathrm{d}\omega$$

In particular this means that the Fourier transform is also a unitary operator, and hence an isometry on $\mathcal{S}(\mathbb{R})$. Consequently, we can extend the Fourier transform to $L^2(\mathbb{R})$ as follows. Since $\mathcal{S}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$, for every $f\in L^2$ there exists a sequence $f_n\in \mathcal{S}(\mathbb{R})$ s.t. $\lVert f_n-f\rVert\to 0$ as $n\to\infty$. Hence, by Plancherel we have $$\lVert f_n-f_m\rVert=\lVert \widehat{f}_n-\widehat{f}_m\rVert$$ so $\widehat{f}_n$ is a Cauchy-sequence in $L^2$, and hence converges to a unique limit which we define as the Fourier transform of $f$. So $\mathcal{F}$ extends to a unitary operator on the whole space $L^2(\mathbb{R})$, and consequently the Fourier transform acts isometrically on this space."

Thank's a lot for your attention


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.