Could anyone please point out if there is any mistakes in the following arguments for the extension of the Fourier transform from $\mathcal{S}(\mathbb{R})$ to $L^2(\mathbb{R})$:
"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.
Proposition.
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