The version of Heisenberg's uncertainty product I am interested in is
$$\left(\int_{-\infty}^{\infty} t^2|f(t)|^2\,dt\right)\left(\int_{-\infty}^{\infty}t^2|\mathcal{F}f(t)|^2\,dt\right),\tag{1}$$
where $\mathcal{F}$ denotes the Fourier transform and $f$ is assumed to be normalized to $1$. I am aware of multiple different approaches to showing that the Gaussian is the minimizer of the uncertainty product however I've always felt that a variational approach should exist (but I can't seem to find one).
Defining $\xi(t) = tf(t)$ and $\eta(t) = t\mathcal{F}f(t)$, the uncertainty product becomes $\langle \xi,\xi\rangle\langle \eta,\eta\rangle$. Making use of the Cauchy-Schwarz inequality, we get that $\langle\xi,\xi\rangle\langle\eta,\eta\rangle \ge |\langle\xi,\eta\rangle|^2$ with equality holding if and only if $\xi = \lambda\eta$, i.e. $tf(t) = \lambda t\mathcal{F}f(t)$.
This says that $f$ is an eigenfunction of the Fourier transform. Thus if we want to find the minimum of $(1)$, we need only to find the minimum amongst all eigenfunctions of the Fourier transform. Since the eigenvalues of the Fourier transform are fourth roots of unity, $\lambda = i^k$ for some $k$ and substituting $\mathcal{F}f(t) = \lambda^{-1} f(t)$ into $(1)$, we get
$$\left(\int_{-\infty}^{\infty} t^2|f(t)|^2\,dt\right)^2.$$
The minimizer of this would give us the minimizer of the Heisenberg uncertainty product. Of course this should come out to the Gaussian but this is where I get stuck. I'm not really sure how to show that the Gaussian is the minimizer of this expression. I tried using constrained calculus of variations techniques (via Euler-Lagrange equations) but it hasn't led anywhere fruitful. It seems like a variational approach should work here but I can't quite seem to figure it out. I'll sketch my thoughts below. Any insight is greatly appreciated!
Define the functional $\mathcal{J}$ by
$$\mathcal{J}[f] = \int_{-\infty}^{\infty} t^2|f(t)|^2\,dt.$$
Clearly the minimizer of $\mathcal{J}$ is the zero function so we need to apply a constraint, $\mathcal{I}$, defined by
$$\mathcal{I}[f] = \int_{-\infty}^{\infty}|f(t)|^2\,dt.$$
Assume (for now) that $f$ is real-valued and define $L(t,f,f') = t^2f(t)^2$ and $I(t,f,f') = f(t)^2$. The Euler-Lagrange equation for a constrained system (read: system with a Lagrange multiplier) is
$$\frac{d}{dt}\frac{\partial L}{\partial f'}-\frac{\partial L}{\partial f} = \lambda\left(\frac{d}{dt}\frac{\partial I}{\partial f'}-\frac{\partial I}{\partial f}\right).$$
Since neither $L$ nor $I$ are dependent upon $f'$, we're only left with
$$ 2t^2f(t) = 2\lambda f(t)$$
which clearly has only the zero solution. However from the outset, we assumed $f$ had norm one, particularly that it is normalizable. The zero function is of course not normalizable. I've clearly missed something important but I can't quite figure out what.
I've found the culprit in the midst here. The philosophy is fine but it cannot seemingly be implemented as variational problem, at least as I have set it up. The problem is that at no point in the above minimization procedure was it required that $f$ be an eigenfunction of the Fourier transform (which it was supposed to be). Without this constraint, there is no true minimum to be had. We can construct a sequence of $L^2$ normalized functions such that $\int_{-\infty}^{\infty} t^2|f(t)|^2\,dt$ goes to zero (particularly, consider bounded approximate identities, e.g. $f_{\lambda}(t) = \frac{1}{\sqrt{2\lambda}}\chi_{[-\lambda,\lambda]}(t))$. However such functions are not eigenfunctions of the Fourier transform and therein lies the problem. I'm not sure how to remedy this as of yet; perhaps it's a futile effort entirely.