= Solution
Use the angular-frequency <Fourier transform> convention
$$
\widehat f(\lambda)=\int_{\mathbb R}f(x)e^{-i\lambda x}\,dx.
$$
A precise sufficient interpretation of the regularity assumption is $f\ne0$, $f\in H^1(\mathbb R)$ in the <first-order Sobolev space> and $xf\in L^2(\mathbb R)$; in particular, nonzero <Schwartz functions> are admissible. We use the <Plancherel theorem>, in the normalization $\|\widehat f\|_2^2=2\pi\|f\|_2^2$, together with its derivative identity $\widehat{f'}(\lambda)=i\lambda\widehat f(\lambda)$. Both identities hold in $L^2$ for $f\in H^1$, by extending their identities for <Schwartz functions>. Thus the frequency factor is
$$
\frac{\int\lambda^2|\widehat f(\lambda)|^2\,d\lambda}{\int|\widehat f(\lambda)|^2\,d\lambda}=\frac{\|f'\|_2^2}{\|f\|_2^2}.
$$
For a <Schwartz function>, <integration by parts> gives $\|f\|_2^2=-2\operatorname{Re}\int x f'(x)\overline{f(x)}\,dx$. This remains true under the stated assumptions: insert a smooth cutoff $\chi_R(x)=\chi(x/R)$ into the derivative of $x|f|^2$, where $\chi$ is one near zero and compactly supported. Integration gives
$$
\int(\chi_R+x\chi_R')|f|^2\,dx=-2\operatorname{Re}\int\chi_R x f'\overline f\,dx.
$$
The cutoff-derivative term tends to zero, since $x\chi_R'$ is bounded independently of $R$ and supported in a tail. The other terms converge by <dominated convergence> and the integrability of $|f'||xf|$, furnished by the <Cauchy-Schwarz inequality>. Hence
$$
\|f\|_2^2=-2\operatorname{Re}\langle f',xf\rangle\leq2|\langle f',xf\rangle|\leq2\|f'\|_2\|xf\|_2.
$$
After squaring and using the <Plancherel theorem>, we obtain the <uncentred Fourier uncertainty principle>:
$$
\boxed{\frac{\int x^2|f(x)|^2\,dx}{\int|f(x)|^2\,dx}\frac{\int\lambda^2|\widehat f(\lambda)|^2\,d\lambda}{\int|\widehat f(\lambda)|^2\,d\lambda}\geq\frac14.}
$$
If equality holds, both displayed inequalities must be equalities. The equality condition in the <Cauchy-Schwarz inequality> gives $f'=cxf$ almost everywhere for a constant $c\in\mathbb C$; the real-part inequality and the identity for $\|f\|_2^2$ then force $c=-a$ with $a>0$ real. The <weak derivative> equation has the solution $f(x)=C e^{-ax^2/2}$: locally multiply by $e^{ax^2/2}$ to obtain a function whose weak derivative is zero, hence a constant. Conversely, for this <Gaussian function> the normalized second moments are $1/(2a)$ in position and $a/2$ in angular frequency, whose product is $1/4$. Therefore
$$
\boxed{\text{Equality holds exactly for }f(x)=C e^{-ax^2/2},\quad a>0,\ C\in\mathbb C\setminus\{0\}.}
$$
These are raw second moments, rather than variances about their respective means. Consequently translations and nonzero frequency modulations of the <Gaussian function> are not equality cases here; the <equality case of the Heisenberg uncertainty relation> for centred variances has those additional freedoms.
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