Relating Fourier transform theory on two distinct subspaces In Fourier transform theory (on $\mathbb{R}$), three vector spaces play a very important role: $L^1(\Bbb R)$, $L^2(\Bbb R)$ and the Schwartz space $\mathcal{S}(\Bbb R)$. Arguably the nicer spaces of the three are the Schwartz space and $L^2(\Bbb R)$. To prove that the Fourier transform is unitary on $L^2(\Bbb R)$, it is sometimes shown that it is an $L^2$ isometry on $\mathcal{S}(\Bbb R)$ and that it is also invertible on $\mathcal{S}(\Bbb R)$. Since $\mathcal{S}(\Bbb R)$ is dense in $L^2(\Bbb R)$, the results extend nicely.
The Schwartz space is extremely nice but it is a little specialized for the Fourier transform. As such, for more general integral operators, we won't have such a nice space to work with. However, as we often do in measure theory, we do have characteristic functions to work with. Particularly, it's not hard to see that the set $V = \{\chi_{[a,b]}:a<b\in\Bbb R\}$ is linearly dense in $L^2(\Bbb R)$. For any reasonable (non-singular) kernel, we would expect that its corresponding integral transform is "nice" on $V$ (i.e. bounded).
Taking these cues, if one were to do Fourier theory on $V$, one would find that the Fourier transform is an $L^2$ isometry. $V$ here and $\mathcal{S}(\Bbb R)$ of course have trivial intersection. This leads me into my question: could we somehow conclude that because the Fourier transform is an $L^2$ isometry on $V$ that it is an $L^2$ isometry on $\mathcal{S}(\Bbb R)$ as an integral transform? Not as an extension of the integral transform on $V$ to $L^2(\Bbb R)$ and then restricted to $\mathcal{S}(\Bbb R)$ (which would trivially be an isometry) but as an integral transform.
I guess a further addendum would be that: since the Fourier transform (as an integral transform) is an $L^2$ isometry on $\mathcal{S}(\Bbb R)$, could we argue that it is also (as an integral transform) an $L^2$ isometry on $V$?
Of course this idea can be made much more general by considering an operator which is naturally defined (and has nice properties) on two distinct subspaces however I'm mostly curious about this particular example.
 A: In attacking a generalization of this problem for research, I found a solution that I think is somewhat cute. Suppose that we know that the Fourier transform is an $L^2$ isometry on the dense subspace $\mathcal{S}(\mathbb{R})$ of $L^2(\mathbb{R})$ and we wish to show it is an $L^2$ isometry on the dense subspace $V = \operatorname{span}\{\chi_{[a,b]}(x):a,b\in \mathbb{R}, a<b\}$ of $L^2(\mathbb{R})$. Here the Fourier transform, $\mathcal{F}$, of $f$ is defined to be
$$ \mathcal{F}f(y) = \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi}} e^{-ixy} f(x)\, dx.$$
The real problem at the heart of it is showing that the Fourier transform applied to a function in $V$ is in $L^2(\mathbb{R})$. Once we've done that, isometry follows quickly.
Let $f\in\mathcal{S}(\mathbb{R})$ and $g\in V$. We know that
$$ \int_{\mathbb{R}^2} \left|e^{-ixy} f(x)g(y)\right|\,dx\,dy < \infty. $$
allowing for integrals to be interchanged, i.e.
$$ \int_{\mathbb{R}}\overline{\left(\int_{\mathbb{R}} e^{ixy} g(x)\,dx\right)} f(y)\,dy = \int_{\mathbb{R}} \left(\int_{\mathbb{R}} e^{-ixy} f(y)\,dy \right) \overline{g(x)}\,dx.$$
We also have that the integrals $\displaystyle \int_{\mathbb{R}} e^{-ixy} f(x)\,dx$ and $\displaystyle \int_{\mathbb{R}} e^{-ixy} g(x)\,dx$ are well-defined (by a simple Schwartz space estimate and Lebesgue integrability).
Define the linear functional $\varphi:\mathcal{S}(\mathbb{R})\to\mathbb{C}$ by
$$\varphi(f) = \frac{1}{\sqrt{2}}\int_{\mathbb{R}}\overline{\left(\int_{\mathbb{R}} e^{ixy} g(x)\,dx\right)} f(y)\,dy.$$
This linear functional is bounded since from the above,
$$\varphi(f) = \langle \mathcal{F}f,g\rangle$$
and so by Cauchy-Schwarz
$$|\varphi(f)| \le \|g\| \|\mathcal{F}f\| = \|g\| \|f\|,$$
proving boundedness.
Since $\mathcal{S}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$, Hahn-Banach tells us that $\varphi$ may be extended to a \emph{unique} bounded linear functional on $L^2(\mathbb{R})$, call it $\widetilde{\varphi}$,, with equivalent norm. Thus $\widetilde{\varphi}|_{\mathcal{S}(\mathbb{R})} = \varphi$ and $\|\widetilde{\varphi}\| = \|\varphi\|$. However, the Riesz representation theorem tells us that a bounded linear functional, say $\widetilde{\varphi}$, on $L^2(\mathbb{R})$ can be identified with $h\in L^2(\mathbb{R})$ such that $\widetilde{\varphi}(f) = \langle f,h\rangle$ for all $f\in L^2(\mathbb{R})$.
But then by virtue of inner products on $L^2(\mathbb{R})$ being given by integrals, we see that
$$\int_{\mathbb{R}} f(y)\overline{h(y)}\,dy = \int_{\mathbb{R}} f(y) \overline{\left(\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi}} e^{ixy} g(x)\,dx\right)}\,dy.$$
Since these are equal on a dense subspace of $L^2(\mathbb{R})$ (and so their difference is the $0$ functional on $L^2(\mathbb{R})$), we conclude that $h(-y) = \displaystyle\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{-ixy} g(x)\,dx$ for y a.e. Particularly, $\displaystyle y\mapsto\int_{\mathbb{R}} e^{-ixy} g(x)\,dx$ defines a function in $L^2(\mathbb{R})$. From some simple analysis (e.g. picking an orthonormal basis in $\mathcal{S}(\mathbb{R})$, it is easy to see that $\|\mathcal{F}g\| = \|g\|$.
Edit: Note that this proof would actually work for any $g\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$.
