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 $t \in \mathbb{R}$*?



Is there an uncertainty principle for Fourier pairs everywhere dominated by $t^{-A}$?
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](https://terrytao.wordpress.com/2009/02/18/hardys-uncertainty-principle/), could be modified for this purpose).


The key lemma is: 



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> *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 $J$.
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*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.


