Can phase significantly concentrate a function's spectrum?

Question

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.

Answer 1

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.

Answer 2

I believe no such constant exists. $\newcommand{\cF}{{\mathcal F}}\newcommand{\FA}{{\rm A}}$

Let $G$ be a LCA group with Pontryagin dual $\Gamma$, let $\cF_\Gamma: L^1(\Gamma)\to C_0(G)$ be the Fourier transform for $\Gamma$, and let $\FA(G)$ denote the image $\cF_\Gamma(L^1(\Gamma))$ equipped with the norm pushed forward from $L^1(\Gamma)$. Then if I understood your question correctly, it is equivalent — via Fourier inversion, etc — to asking if there exists some contant $\epsilon>0$ such that $\Vert\,|x|\,\Vert_{\FA(G)} \leq \epsilon^{-1}\Vert x \Vert_{\FA(G)}$ for all $x$.

In the older language of "functions operating in Banach algebras", this asks if the function $x\mapsto |x|$ operates in the Banach algebra $\FA(G)$, and as mentioned in my answer to someone else's MO question the answer is negative for $G={\bf T}$ or $G={\bf R}$.

My gut feeling is that if we take $f_n(t):= |\sin (2\pi n t)|$ for $t\in {\bf R}/{\bf Z} \cong {\bf T}$, then $\Vert f_n\Vert_{\FA({\bf R}/{\bf Z})} \to +\infty$ as $n\to + \infty$, but I haven't tried to do explicit estimates to prove this. Update: my closing paragraph was not thought through properly, as Will Sawin has pointed out in comments.