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 $t \in \mathbb{R}$?


1 Answer 1


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.

  • $\begingroup$ Thanks much, this is extremely helpful! I realized after your answer that what I really wanted to know was a weaker variant: Under $f(0) = \hat{f}(0) = 1$, can it happen that both $f$ and $\hat{f}$ are almost supported on $[-1,1]$, in the sense that the $L^1(\mathbb{R} \setminus [-1,1])$-norms of $f, \hat{f}$ are arbitrarily small? A counterexample to the first question would have had this property, but the constraint of uniform decay is extremely restrictive, as the answer explains. Could it be said right away if there is an UP for Fourier pairs almost supported on $[-1,1]$ in this sense? $\endgroup$ Nov 16, 2016 at 11:12
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    $\begingroup$ I think basically the same compactness argument applies to exclude this scenario also. $\endgroup$
    – Terry Tao
    Nov 16, 2016 at 17:42
  • $\begingroup$ I see. Thank you, I'll reflect on this more carefully. $\endgroup$ Nov 16, 2016 at 18:51

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