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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 $f \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$).

steve
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  • If $f \in L^2$, then $\tau_y f \in L^2$ also. If $\hat f$ exists, what's the worry? – Jon Warneke Apr 28 '16 at 04:06
  • for defining the Fourier transform on $L^2$ you'll need to consider for example $\displaystyle\hat{f}\epsilon(\xi) = \int{-\infty}^\infty f(x) e^{-2 i \pi \xi x} e^{-\epsilon^2 x^2} dx$. once you proved with the Parseval theorem that when $\epsilon \to 0$ : $\hat{f}_\epsilon$ converges (to $\hat{f}$) in $L^2$ and that $|\hat{f}| = |f|$, you proved that the Fourier transform $f \to \hat{f}$ is an unitary operator $L^2 \to L^2$ and that obviously the shift theorem holds – reuns Apr 28 '16 at 04:09

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$L^1\cap L^2$ is dense in $L^2$. (The intersection is necessary here because $L^1$ is not contained 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 $L^2$.

Disintegrating By Parts
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