How do you prove translation invariance of Fourier transform?
Solution 1:
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
$$ \hat{g}(x) = \int f(s) e^{-2\pi i(s - h)x} \mathrm{d}s = e^{2\pi i h x}\int f(s) e^{-2\pi i s x} \mathrm{d}s = e^{2 \pi i h x} \hat{f}(x) $$