Let $F$ denote the Fourier transform over some group. What is known about the following quantity?


$$\gamma:=\inf\_{x\neq 0}\frac{\|Fx\|\_1}{\|F|x|\|\_1}$$


Here, $|x|$ denotes the pointwise absolute value of $x$. We know $\gamma\leq1$ since $x$ can be pointwise nonnegative. In the case where $F$ is the DFT, we also know $\gamma>0$ by a compactness argument, and computer simulations give $\gamma\leq0.72$ by taking $x$ to be some random bandlimited even function.


**EDIT:** Following Josep's comment, we actually know $\gamma\geq1/\sqrt{n}$ in the DFT case by passing to the 2-norm and applying Parseval.


I would mostly like to know if $\gamma\gg0$ (i.e., there exists a constant $\epsilon>0$ independent of $n$ such that $\gamma\geq\epsilon$), since this would imply that $\|Fx\|\_1$ is small (i.e., $Fx$ is concentrated) only if $\|F|x|\|\_1$ is also small. This would confirm my intuition that phase doesn't play much of a role in the concentration of a function's spectrum. Presumably, this question is natural enough to have been studied already.



Can phase significantly concentrate a function's spectrum?
Observe that if $n$ and $m$ are relatively prime, $\gamma\_{nm}\leq \gamma\_n \gamma\_m$ because you can combine functions on $\mathbb Z/n$ and $\mathbb Z/m$ using the Chinese remainder theorem.


So if we show $\gamma\_n \leq 1- \epsilon$ for some $\epsilon$ and all sufficiently large $n$, then we can make $\gamma\_n$ arbitrarily small.


Probably $x(t) = \sin( 2\pi t/n)$ works for this. I think it's clear that in this case the ratio is asymptotic to


$$ \frac{ \sum\_{k \in \mathbb Z}\left| \int\_{0}^{2 \pi} \sin (x) e^{ i k x} \right|}{ \sum\_{k \in \mathbb Z}\left| \int\_{0}^{2 \pi} |\sin (x)| e^{ i k x} \right|}$$


where the numerator is clearly $2\pi$ and the denominator is


$$ \sum\_{k \in 2 \mathbb Z} \left|\frac{-4}{k^2-1}\right|= 8 $$


This gives $\pi/4$, agreeing with Dustin's numerical experiments, which is bounded away from $1$.




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The asymptotic statement is from the Poisson summation-like formula. If $f$ is a function on the unit circle, and $f\_n$ is the function on $\mathbb Z/n$ you get by restricting it to the $n$ roots of unity, then


$$\hat{f\_n} ( a) =\sum\_{k \in \mathbb Z} \hat{f} ( a + nk)$$


so 


$$ \lim\_{n \to \infty} ||\hat{f}\_n||\_1 = || \hat{f} ||\_1$$


The integral I found the calculations too fiddly so I just did it on wolframalpha, but you can do it by breaking up $[0,2\pi]$ into $[0,\pi]$ and $[\pi, 2\pi]$. The sum is by telescoping from $\frac{1}{k-1} - \frac{1}{k+1} = \frac{2}{k^2-1}$.




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Here's a method for arbitrary $\mathbb Z/n$. Let $a\_1, \dots, a\_k$ be elements of $\mathbb Z/n$ and consider


$$x(t) = \prod\_{i=1}^k \sin (a\_k x/n) $$


Supposing that $\sum\_{i=1}^k \pm a\_i$ is nonvanishing, the $L^1$ norm of $Fx$ is $1 $ if you normalize it properly.


Now I claim for any group homomorphism $G \to H$, given a function on $H$, the Fourier transform of its pullback to $G$ can be computed on a character of $G$ by summing the Fourier transform of $H$ over all characters of $H$ that pull back to that character of $G$. This is the same as the Poisson summation formula and is proved the same way. Here I'm normalizing by the measure of the group.


Apply this to the map $\mathbb Z/n \to (\mathbb R/\mathbb Z)^n$ that sends $t$ to $(a\_1t/n, \dots, a\_kt/n)$


Let $g(m)$ be the Fourier transform of $|\sin x|$. Then the Fourier transform of $\left| \prod\_{i=1}^k sin x\_i \right| $ is $\prod\_{i=1}^k g(m\_i)$.


$$F |x| ( \xi) = \sum\_{(y\_1, \dots, y\_k) \in \mathbb Z^k } 1\_{\sum\_{i=1}^k a\_i y\_i \equiv \xi } \prod\_{i=1}^k g(y\_i)$$


This is the sum of $\prod\_{i=1}^k g(y\_i)$ over the lattice in $\mathbb Z^k$ satisfying the condition $\sum\_{i=1}^k a\_i y\_i \equiv \xi $. 


Now we know that $\prod\_{i=1}^k g(y\_i)$ has $L\_1$ norm $ (4/\pi)^k$, and at most $(2/3) (4/\pi)^k$ of its mass is in the box where $|y\_i| < Ck$, because $1-O(1/k)$ of the mass of $g(y)$ is in the box where $|Y| < k $.


So as long as no equation of the form $\sum\_{i=1}^k y\_i a\_i =0$ for $y\_i$ not all $0$ but $|y\_i| < 2 C k$ holds, then no two points in that box get sent to the same $\chi$, so the $L\_2$ norm goes down by at most $1-2/3$ (in the worst case where everything outside the box cancels something inside the box.)


When can we find $a\_1, \dots, a\_k$ such that no such equation holds? If $n=p$ is a prime then by selecting $a\_1, \dots, a\_k$ randomly, the probability of any given equation holds is $1/p$, so as long as $p > (2 C k)^k$ then sometimes none of the equations hold.


If we're not a prime then we can just do the Chinese Remainder Theorem thing, or do the probabilistic argument more carefully.


