Let the Fourier Transfor $F:L_2(\mathbb{R})\to L_2(\mathbb{R})$ defined by $\displaystyle Fg(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(t)e^{ixt}dt\,\,$ for $\,\,g\in L^2(\mathbb{R})$.
I need to show that $F^4=I$. Can you give me some suggestions to try this please
I know that $\displaystyle F^{-1}h(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}h(t)e^{-ixt}dt\,\,$ but but I'm not sure if that could be used to prove that $F^4=I$.
Given $g\in L_2(\mathbb R)$, you can prove that $(F \circ F)g(x)=g(-x)$ (as can be seen in this question: Fourier transform of the Fourier transform?. Here, they define Fourier Transform without $\frac{1}{\sqrt{2\pi}}$, but I'm sure you can deal with that).
Then, you just need to see that $F^4=F^2\circ F^2$.