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Let $f$ be a rapidly decreasing function in the sense that it lies in the Schwartz space $\mathcal{S}(\Bbb{R})$.

Then $\widehat{f(x+h)} = \hat{f}(\omega) e^{i 2 \pi h \omega}$, where $\hat{f}(\omega)$ is the Fourier transform of $f(x)$.

How do I prove that? This might not depend on the rapidly decreasing part.

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    $\begingroup$ literally just plug $g(x) = f(x+h)$ into the Fourier transform formula and you will see this comes out. $\endgroup$
    – Deven Ware
    Nov 8, 2013 at 23:43
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    $\begingroup$ There is a u substitution in the integral as well. $\endgroup$ Nov 8, 2013 at 23:45
  • $\begingroup$ @DevenWare Could you make this an answer and demonstrate? I tried that $\endgroup$ Nov 8, 2013 at 23:46
  • $\begingroup$ @EnjoysMath Done. $\endgroup$
    – Deven Ware
    Nov 8, 2013 at 23:54

1 Answer 1

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Let $g(x) = f(x+h)$. Then we calculate

$$\hat{g}(x) = \int g(t)e^{-2 \pi i x t} \mathrm{d}t = \int f(t + h) e^{-2 \pi i x t} \mathrm{d}t $$

Then make the $u$-sub $s = t + h$ and you get

$$ \hat{g}(x) = \int f(s) e^{-2\pi i(s - h)x} \mathrm{d}s = e^{2\pi i h x}\int f(s) e^{-2\pi i s x} \mathrm{d}s = e^{2 \pi i h x} \hat{f}(x) $$

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