Suppose $f \in \mathcal S(\mathbb R)$ is a Schwartz function. Then the Fourier transform is invertible, with nice properties like $$\int f(x) \overline{\hat g(x)} \mathrm{d} x = \int \hat f(x) \overline{ g(x)} \mathrm{d} x.$$ In physics, if $$\hat f(\zeta) = \int f(x)e^{-2\pi i x \zeta}\mathrm{d} x,$$ then $x$ is often called "time" while $\zeta$ is called "frequency". This is very suggestive - I still don't have any intuition for Fourier transforms, but it hints that the variables are somehow inverted.
Suppose $\int|f(x)|^2\mathrm{d}x = 1$ and $\int|g(\zeta)|^2\mathrm{d}\zeta = 1$. In general it is certainly not true for arbitrary $\alpha$ that $$\Big(\int|x|^{\alpha}|f(x)|^2\mathrm{d}x\Big)\Big( \int|\zeta|^{\alpha}|g(\zeta)|^2\mathrm{d}\zeta\Big) = 1.$$
However, (noting that the first formula at the top implies $\int|\hat f(\zeta)|^2\mathrm{d}\zeta = 1$), if the two variables correspond to inverses in some sense (as in the Fourier transform), it raises the question of whether or not we can make conclusions about the value of
$$P_\alpha (f) = \Big(\int|x|^{\alpha}|f(x)|^2\mathrm{d}x\Big)\Big( \int|\zeta|^{\alpha}|\hat f(\zeta)|^2\mathrm{d}\zeta\Big).$$
By plugging in the Gaussian for $f(x)$, it is evident that $P_{\alpha}(f) \neq 1$ in general. This made me wonder if we can find bounds - is $P_\alpha (f)$ strictly positive? Is it bounded above? Perhaps we can only draw conclusions when $\alpha$ takes certain values (like 0!) or if our functions satisfy a few more properties.
I figured that $\alpha = 2$ would be the easiest to work with (because it's easy to play around with when using with the Cauchy-Schwarz inequality), but I couldn't prove anything non-trivial.