Heisenberg uncertainty and Gaussian equality

Put $g(y)=e^{-2\pi ib\cdot (y+a)}f(y+a)$. By [F6] it is Schwartz, and [F3] gives $\widehat{g}(\eta )=e^{2\pi ia\cdot \eta }\widehat{f}(\eta +b)$. Translation substitution and unit modulus therefore identify $\mathrm{\parallel }g{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$, $\mathrm{\parallel }\mathrm{\mid }y\mathrm{\mid }g{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\mid }x-a\mathrm{\mid }f{\mathrm{\parallel }}_2$, and $\mathrm{\parallel }\mathrm{\mid }\eta \mathrm{\mid }\widehat{g}{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\mid }\xi -b\mathrm{\mid }\widehat{f}{\mathrm{\parallel }}_2$. It suffices to prove the zero-centre assertion for $g$.

For each coordinate $j$, apply [F5] along that line to $u=x_{j}g$ and $v=\bar{g}$. The endpoint product $x_j\mathrm{\mid }g{\mathrm{\mid }}^2$ tends to zero at both ends by rapid decay, and both differentiated products are line-integrable. Their full-space integrability follows from [F6] and [F4], so [F7] permits integrating the identity in the other coordinates. It yields $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=-2\mathrm{Re}\int x_{j}g\overline{{\mathrm{\partial }}_{j}g}$. Sum over $j$ and define tuples $A=(x_{j}g{)}_j$, $B=({\mathrm{\partial }}_{j}g{)}_j$. Then $n\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=-2\mathrm{Re}⟨A,B⟩\le 2\mathrm{\mid }⟨A,B⟩\mathrm{\mid }\le 2\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }\mathrm{.}$ By [F1] and [F2], $\mathrm{\parallel }B{\mathrm{\parallel }}^2=4{\pi }^2\int \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }\widehat{g}(\xi ){\mathrm{\mid }}^{2}d\xi$, while $\mathrm{\parallel }A\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\mid }x\mathrm{\mid }g{\mathrm{\parallel }}_2$. This proves the inequality with the claimed constant.

Suppose $g\mathrm{\ne }0$ and equality holds. Step 2.1 gives $n\mathrm{\parallel }g{\mathrm{\parallel }}_2^2>0$, so both tuple norms are nonzero. Equality in [F4] makes $B=dA$ for one complex scalar $d$. Equality in the real-part bound, together with $⟨A,B⟩=\bar{d}\mathrm{\parallel }A{\mathrm{\parallel }}^2$, forces $d$ to be a negative real number. Write $d=-\lambda$, $\lambda >0$. The identities ${\mathrm{\partial }}_{j}g=-\lambda x_{j}g$ hold a.e., hence everywhere by continuity and the positive measure of nondegenerate boxes. Thus every partial derivative of $G(x)=e^{\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2}g(x)$ is zero. Applying the interval FTC [F5] along successive coordinate segments shows $G(x)=G(0)=c$, so $g(x)=c{e}^{-\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2}$. Nonzero $g$ forces $c\mathrm{\ne }0$. Undoing step 1.1 gives exactly the displayed form for $f$.

Conversely let $g(x)=c{e}^{-\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2}$ with $\lambda >0$, $c\mathrm{\ne }0$. By [F8] it is Schwartz, and direct differentiation gives $B=-\lambda A$. Thus both inequalities in step 2.1 are equalities, giving equality in the uncertainty inequality. By step 1.1 the translated and modulated functions have the same equality property. If $f=0$, both sides are zero directly. The common scalar across all coordinates is essential to the radial equality assertion proved here.