Calculate the Fourier transform of the following function: $$f(x)=\frac{1}{\sqrt{t}} g\left(\frac{x-u}{t}\right) e^{2 \pi i \xi x}$$ where $t>0$, $\xi,u\in\mathbb R$.
It seems to be quite cumbersome to me. We have three operators applied to the same function $g$. They are:
Translation / time shifting: For any real number $x_0$, if $h(x)=f\left(x-x_0\right)$, then $\hat{h}(\nu)=e^{-2 \pi i x_0 \nu} \hat{f}(\nu)$.
Modulation / frequency shifting: For any real number $\nu_0$, if $h(x)=e^{2 \pi i x \nu_0} f(x)$, then $\hat{h}(\nu)=\hat{f}\left(\nu-\nu_0\right)$.
Time scaling: For a non-zero real number $a$, if $h(x)=f(a x)$, then $$ \hat{h}(\nu)=\frac{1}{|a|} \hat{f}\left(\frac{\nu}{a}\right) . $$ How can I use togheter to obtain the Fourier transform of $f$?
Thank you!!
To make it easy to understand, we define the following operators: $$\left(\tau_{x_0}f\right)(x)=f(x-x_0),\ \ \ \left(M_{x_0}f\right)(x)=e^{2\pi ix \cdot x_0}f(x),\ \ \ \left(S_af\right)(x)=f(ax),\ \ x_0\in\mathbb R,a>0.$$ Then $$\left(\tau_{x_0}f\right)\hat{}=M_{-x_0}\hat f,\ \ \ \left(M_{x_0}f\right)\hat{}=\tau_{x_0}\hat f,\ \ \ \left(S_af\right)\hat{}(\nu)=\frac1a\hat f\left(\frac\nu a\right),\qquad \nu\in\mathbb R, a>0.$$ Now we come back to the original question, where $$f=\frac1{\sqrt t}\left(M_\xi\left(S_{1/t}\left(\tau_{u/t}g\right)\right)\right).$$ We have $$\left(\tau_{u/t}g\right)\hat{}(\nu)=e^{-2\pi i\frac ut\nu}\ \hat g(\nu),$$ $$\left(S_{1/t}\left(\tau_{u/t}g\right)\right)\hat{}(\nu)=t\left(\tau_{u/t}g\right)\hat{}(t\nu)=te^{-2\pi i u\nu}\ \hat g(t\nu), $$ $$\hat f(\nu)=\frac1{\sqrt{t}}\left(M_\xi\left(S_{1/t}\left(\tau_{u/t}g\right)\right)\right)\hat{}(\nu)=\frac1{\sqrt{t}}\left(S_{1/t}\left(\tau_{u/t}g\right)\right)\hat{}(\nu-\xi)=\sqrt te^{-2\pi i u(\nu-\xi)}\ \hat g(t(\nu-\xi)).$$ So we are done.
Indeed, I prefer to directly compute the intgral through the definition. Because these properties are really hard to remember for me. Here is the calculation: \begin{align*} \hat f(\nu)&=\int_\mathbb Rf(x)e^{-2\pi ix\nu}\,dx\\ &=\frac1{\sqrt t}\int_\mathbb Rg\left(\frac{x-u}{t}\right) e^{2 \pi i \xi x}e^{-2\pi ix\nu}\,dx\\ &=\frac1{\sqrt t}t\int_\mathbb Rg(y)e^{-2\pi i(u+ty)(\nu-\xi)}\,dy\\ &=\sqrt te^{-2\pi i u(\nu-\xi)}\ \hat g(t(\nu-\xi)) \end{align*}