Is there an uncertainty principle for Fourier pairs everywhere dominated by $t^{-A}$? Hardy's uncertainty principle states that a real function $f$ and its Fourier transform $\widehat{f}$ may not both decay faster at infinity than the standard Gaussian $e^{-\pi t^2}$, unless $f = 0$. In a sense, $e^{- \pi t^2}$ is the closest a non-zero function can come to having both $f$ and $\widehat{f}$ almost compactly supported.
Restrict now to Schwartz functions normalized by $f(0) = \widehat{f}(0) = 1$. 
The standard Gaussian $e^{-\pi t^2}$ is the unique Gaussian function respecting this normalization, and it is self-dual with the optimal asymptotic decay, which is asymptotically much faster than $t^{-A}$ for any $A$. However, for $A > 10$ say, this Gaussian is not everywhere dominated by $t^{-A}$.
Question. Is there an upper bound on the values $A > 0$ for which there exists a Schwartz function $f$ having $f(0) = \widehat{f}(0) = 1$ and $|f(t)|,|\widehat{f}(t)| < t^{-A}$ for all $t \in \mathbb{R}$?
 A: Using compactness arguments one can show that such an $A$ exists, although this argument does not make it easy to extract an effective value for $A$ (but presumably one of the real variable proofs of the Hardy uncertainty principle, such as the one in this blog post of mine, could be modified for this purpose).
The key lemma is: 

Lemma  If $I,J$ are intervals, and $f$ is a Schwartz function with $\hat f$ supported on $I$, then $\| f \|_{L^\infty({\mathbb R})} \leq C_{I,J} \| f \|_{L^\infty({\mathbb R} \backslash J)}$ for some constant $C_{I,J}$ depending only on $J$.

Proof:  We can normalise $\| f \|_{L^\infty({\mathbb R})} = 1$.  If the claim failed, then we can find a sequence $f_n$ in the unit sphere of $L^\infty({\mathbb R})$ with Fourier transform supported in $I$ which goes to zero in $L^\infty({\mathbb R} \backslash J)$.  We can extract a weak (tempered distributional) limit $f$ which then vanishes almost everywhere on ${\mathbb R} \backslash J$; because all the $f_n$ have uniformly compactly supported Fourier transforms, one can upgrade weak convergence to local uniform convergence and so $f$ is not identically zero.  But then $f$ and its Fourier transform are both compactly supported, contradicting the uncertainty principle. $\Box$
(In fact this argument shows the stronger assertion $\|f\|_{L^\infty(J)} \leq C_{I,J,J'} \|f\|_{L^\infty(J')}$ for any compact intervals $I,J,J'$.)
Returning to your problem, if you truncate $f$ to $[-2,2]$ by a suitable smooth cutoff then the resulting truncated function $g$ has a Fourier transform that is exponentially small in $A$ outside of $[-2,2]$, hence by the above lemma is exponentially small in $A$ on all of ${\mathbb R}$.  By the Fourier inversion formula we conclude that $\hat f(0)$ is also exponentially small in $A$, giving a contradiction for $A$ large enough.
