Let $p$ be a prime. For $f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ let its Fourier transform be:
$$\hat f(n) = \frac{1}{\sqrt{p}}\sum_{x \in \mathbb{Z}/p \mathbb{Z}} f(x)\, e\left(\frac{-xn}{p}\right)$$
Terence Tao proves in "An uncertainty principle for cyclic groups of prime order" that for $f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ one has:
$$|S_1| + |S_2|\ge p+1$$
If $f$ is supported in $S_1$ and $\hat f$ is supported in $S_2$. This is much stronger than the traditional bound:
$$|S_1||S_2| \ge p$$
Can we prove an approximate version of this inequality, say, in which we lower bound
$$|S_1| + |S_2|$$
assuming only that $S_1$ and $S_2$ contain most of the mass of $f$ and $\hat f$ (rather than all of the mass)?
The proof of the lower bound to $|S_1||S_2|$ is "only analytical", so it caries over under this weaker assumption, but Tao's lower bound for $|S_1|+|S_2|$ relies strongly on algebra, in that it is important that $f$ and $\hat f$ are exactly $0$ outside the support, so the proof doesn't immediately carry over.
Is there such an approximate version of Tao's uncertainty principle?
For a concrete case, if the Fourier transform is normalized so that $\sum |f(x)|^2 = \sum |\hat f(x)|^2 = 1$, can we rule out the possibility that $\sum_{x \in S_1} |f(x)|^2 > 1-\epsilon$, and $\sum_{x\in S_2} |\hat f(x)|^2 > 1-\epsilon$ with $|S_1|, |S_2| < \epsilon p$?
For the applications that I am interested I need something quite weaker, but I couldn't really find a counter-example to the above possibility.
Thanks in advance
Best Answer
No. One just has to apply the standard example showing the classical uncertainty principle is sharp:
Let $f(a) = \sum_{n \in \mathbb Z} e^{- \pi ( a+pn)^2 / p}$. Then $\hat{f}$ is proportional to $f$.
But $1-\epsilon$ of its mass is contained in an interval of width something like $O ( \sqrt{ p \log (1/\epsilon)})$, which is much smaller than $\epsilon p$.