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