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The Fourier transform, is defined as an operator on $L^1(\mathbb{R})$, taking a function $f$ and sending it to $$\hat f(\omega) = \int_{\mathbb{R}}f(x)e^{-i\omega x}dx$$

It is also known that the Fourier transform can be extended in a single way to be a continuous operator on $span(L^2(\mathbb{R}) \cup L^1(\mathbb{R}))$ (the extension is started from the Schwartz space, which is dense in both $L^2(\mathbb{R})$ and $L^1(\mathbb{R})$). It is an extension, so it agrees with the original integral definition on $L^1(\mathbb{R})$.

Do the basic properties of shift, modulation etc. are also preserved in the extension?

For example, it is known that the Fourier transform of $f(x-t)$ is $e^{i\omega t} \hat f(\omega)$, where $f(\omega)$ is the Fourier transform of $f(x)$. Is this also true in the extended operator on $L^2(\mathbb{R})$?

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Yes, these properties are all preserved. To see this, note that the extension of the Fourier transform to $L^2$ functions can be written down explicitly in several ways and one of these is $$ \hat f(\omega) = \lim_{R\to\infty} \int_{-R}^R f(x)e^{-i\omega x}dx $$ (note that $f$ restricted to $[-R,R]$ is an $L^1$-function and hence, all integrals exist) and the limit is in the sense of $L^2$. Since the properties you mention hold for $L^1$ functions, they hold under the limit and hence, by continuity of the extension, are preserved by the limit.

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