I'm not very knowledgeable in QM, and I know many physics books derive the uncertainty principle using commutators, but as an exercise in my PDE book (by Asmar), I should be able to derive it from one of the theorems in the section. I know that the Heisenberg uncertainty principle says that $\sigma_{x}\sigma_{p}\geq\frac{\hbar}{2}$. In my textbook, Asmar defines, for a normalized function $f$, the uncertainty about a point $a$ as the second moment about $a$ and writes $\Delta_af$. In this section, there is a theorem saying $\Delta_af\Delta_{\alpha}\hat{f}\geq\frac{1}{4}$ $\forall a$ on $x$-axis and $\forall\alpha$ on $\omega$ axis (where $\hat{f}$ is the Fourier transform of $f$).
So, I tried starting with the fact that if $f$ is the normalized wave function, the momentum distribution is given by $g(p)=\frac{1}{\sqrt{\hbar}}\hat{f}(\frac{p}{\hbar}), -\infty<p<\infty.$ I want to show that $\Delta_\alpha g=\hbar^2\Delta_{\frac{\alpha}{\hbar}}\hat{f}$, but I'm a little stuck on how to make the change of variables to show this. I see that if you're taking the second moment about $\alpha$ with $g$, you're taking the second moment about $\frac{\alpha}{\hbar}$ with $\hat{f}$, so I think I see why the $\Delta_{\frac{\alpha}{\hbar}}\hat{f}$ is there, but that's about it. I'm confused about the factor of $\hbar^2$, because I know that when you take the variance of a random variable $X$ scaled by $c$, you get var$(cX)=c^2$var$(X)$ and so I'm getting a factor of $\frac{1}{\hbar}$, instead of a factor of $\hbar^2$. I was thinking the correct change of variables would be $\alpha\rightarrow\hbar\alpha$, but i'm not sure. Can anyone fill in the gaps?