I have a function $f$ such that $|f(x)|\leq e^{-x^2/2}$ hence in $\mathcal{L}^2(\mathbb{R})\cap\mathcal{L}^1(\mathbb{R})$ and thus we can compute the Fourier transform $$\hat{f} (\xi) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb R} f(x) e^{-i\xi x} dx.$$
By Plancherel's theorem we know that $$\int_{\mathbb R} |\hat{f}(\xi)|^2 d\xi = \int_{\mathbb R} |f(x)|^2 dx\leq \int_{\mathbb R} e^{-x^2}dx \leq \sqrt{\pi} $$ where the last follows from the fact that $|f(x)|\leq e^{-x^2/2}$.
But actually I want to prove more. I would like to see if: $$\int_{\mathbb R} |\frac{\hat{f}(\xi)}{\xi^2}|^2 d\xi \leq C$$ for a positive finite constant $C$. How do I proceed? Is there a way to prove it using Plancherel's theorem? Btw! If it makes things easier we can assume that $f$ has compact support :)
Thanks!
If you take $f\left(x\right)=\exp\left(-x^{2}/2\right)$ you have $$\hat{f}\left(\xi\right)=\exp\left(-\xi^{2}/2\right)$$hence $$\int_{\mathbb{R}}\left|\frac{\hat{f}\left(\xi\right)}{\xi^{2}}\right|^{2}d\xi=\int_{\mathbb{R}}\frac{\exp\left(-\xi^{2}\right)}{\xi^{4}}d\xi=\infty.$$