Fourier transform of translation in $L^2$. For a function $f : \mathbb{R} \longrightarrow \mathbb{R}$, let $(\tau_y f)(x) = f(x - y)$. If $f \in L^1(\mathbb{R})$, then it is straightforward to show that $\widehat{\tau_y f}(\xi) = e^{-2\pi j \xi y} \widehat{f}(\xi)$. If $f \in L^2(\mathbb{R})$, I am wondering if the same equality holds (where $\widehat{f}$ refers to the unique linear map from $L^2(\mathbb{R}) \longrightarrow L^2(\mathbb{R})$ which agrees with the Fourier transform on functions $f \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$). 
 A: $L^1$ is dense in $L^2$. For example, if $f \in L^2$, then $f_R=\chi_{[-R,R]}f \in L^1\cap L^2$ where $\chi_{[-R,R]}$ is the characteristic function of the finite interval $[-R,R]$; and $\lim_{R\rightarrow\infty}\|f-f_R\|_{L^2}=0$. The operators $\tau_y$, multiplication by $e^{-2\pi j\xi y}$ and the Fourier transform are isometric on $L^2$. Therefore, $L^2$ limits may be freely interchanged with any of these operations. Using your result for $L^1$ functions then gives you what you want. Explicitly, assume $\lim$ in the following refers to $L^2$ limits, and $\mathscr{F}$ is the Fourier transform:
\begin{align}
        \mathscr{F}(\tau_y f)&=\mathscr{F}(\tau_y\lim_{R\rightarrow\infty}f_R) \\
   & = \mathscr{F}(\lim_{R\rightarrow\infty}\tau_y f_R) \\
   & = \lim_{R\rightarrow\infty}\mathscr{F}(\tau_y f_R) \\
   & = \lim_{R\rightarrow\infty}e^{-2\pi j\xi y}\mathscr{F}(f_R) \\
   & = e^{-2\pi j\xi y}\lim_{R\rightarrow\infty}\mathscr{F}(f_R) \\
   & = e^{-2\pi j\xi y}\mathscr{F}(\lim_{R\rightarrow\infty}f_R) \\
   & = e^{-2\pi j\xi y}\mathscr{F}(f)
\end{align}
The above holds in the $L^2$ sense because all limits are in $L^2$.
