4
$\begingroup$

Let $G$ be a locally compact group, let $A(G)$ be the Fourier algebra of $G$. We think of $A(G)$ as a subalgebra of $C_0(G)$.

Question 1: Let $f\in A(G)$ be a function that is pointwise positive. Does the function $\sqrt{f}$ belong to $A(G)$?

The motivation for this Question is the following:

Question 2: Given $f\in A(G)$, does the function of absolute values, $|f|$, belong to $A(G)$?

Since $A(G)$ is closed under passing to complex conjugation, a positive answer to Question 1 would imply a positve answer to Question 2.

Additionally, if $f,|f|\in A(G)$, is there a relation between the norms of $f$ and $|f|$ in $A(G)$?

Improve this question
$\endgroup$
6
  • $\begingroup$ So for example when $G = \bf R$ you're hoping that if $f$ is the Fourier transform of a continuous function $F$ (with $F(x) \rightarrow 0$ as $x \rightarrow \pm \infty$) then $|\,f\,|$ is also the Fourier transform of some continuous function $F\,$? $\endgroup$
    – Noam D. Elkies
    Commented Mar 4, 2015 at 14:56
  • $\begingroup$ That definition is what I'm not sure of. Are you saying then that the two $F$'s (sorry for using the same letter above; it's too late now to fix) are to be $L^1$? $\endgroup$
    – Noam D. Elkies
    Commented Mar 4, 2015 at 15:04
  • 1
    $\begingroup$ Yes, for every locally compact abelian group $G$, the Fourier algebra $A(G)$ is naturally isometrically isomorphic to $L^1(\widehat{G})$ via the Fourier transform. In particular, since $\widehat{\mathbb{R}}\cong\mathbb{R}$, we have $A(\mathbb{R})=\{\widehat{f} : f\in L^1(\mathbb{R})\}$, with the norm coming from $L^1(\mathbb{R})$. $\endgroup$
    – Hannes Thiel
    Commented Mar 4, 2015 at 15:34
  • $\begingroup$ @Noam: For $G=\mathbb{R}$, my question comes down to: If $f\in C_0(\mathbb{R})$ is the Fourier transform of some function in $L^1(\mathbb{R})$, is then $|f|$ the Fourier transform of some other function in $L^1(\mathbb{R})$? $\endgroup$
    – Hannes Thiel
    Commented Mar 4, 2015 at 15:36
  • $\begingroup$ Thanks for this eplanation/translation. So $C_0({\mathbb R})$ means continuous and vanishing at infinity? $\endgroup$
    – Noam D. Elkies
    Commented Mar 4, 2015 at 16:18

2 Answers 2

Reset to default
1
$\begingroup$

This won't work. I want to show that we can't take square roots in $A(\mathbb R)$. My function will be of the type $$ f(x) = \sum h_n \varphi\left( \frac{x-a_n}{L_n}\right) ,\quad\quad\quad\quad (1) $$ and here the individual summands will have disjoint supports. I will take $h_n\in\ell^2$, $h_n\notin\ell^1$. Since the $L^1$ norm of the Fourier transform of $\psi((x-a)/L)$ is independent of $a,L$, the first property will make sure that $\widehat{f^2}\in L^1$, that is $f^2\in A(\mathbb R)$.

So it is now enough to find $\varphi\ge 0$ and $h_n,a_n,L_n$ such that $f\notin A(\mathbb R)$. Fix a $\varphi\ge 0$ that is supported by $[0,1]$, with $\varphi'(0)>0$ and $\varphi$ is smooth otherwise. Then (after normalizing suitably) we will have that $|\widehat{\varphi}(t)|=1/t^2+O(t^{-3})$, and of course $\widehat{\varphi}$ is bounded. Thus the Fourier transform of the $n$th summand of (1) is of the order $$ \frac{h_n}{L_n x^2} + O(h_nL_n^{-2}x^{-3}) , $$ and it is bounded by $Ch_nL_n$. Thus for sufficiently large $B$, this summand on its own on the interval $B/L_n\le x\le 2B/L_n$ would make a contribution $\gtrsim h_n/B$ to the $L^1$ norm of $\widehat{f}$.

We will be done if we can make sure that the other summands cannot completely cancel out this contribution. The summands with $k>n$ cannot contribute more than $CBh_{n+1}L_{n+1}/L_n$, and the ones with $k<n$ can be treated as above, with $h_n/L_n$ replaced by $h_k/L_k$. So all competing contributions are much smaller if we take $L_n$'s that converge very rapidly to zero.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ The idea (now well hidden I'm afraid) was really that this shouldn't hold because taking powers of functions can improve smoothness (for something like $x^{1/2}$, say), so makes it easier for the Fourier transform to be in $L^1$. $\endgroup$
    – Christian Remling
    Commented Mar 4, 2015 at 20:29
5
$\begingroup$

To complement Christian Remling's nice concrete explanation, let me just add that more is known.

Your 2nd question can be rephrased as asking: does the function $x\mapsto |x|$ ``operate in'' the Fourier algebra? This kind of question used to be of interest to people working on Banach algebras of functions. For the Fourier algebra of a LCA group the answer is: a function which operates in A(G) for G abelian must be real-analytic. This is (apparently) proved in

MR0116185 (22 #6980) H. Helson, J.-P. Kahane, Y. Katznelson, W. Rudin The functions which operate on Fourier transforms. Acta Math. 102 1959 135–157.

In particular, this implies that your Q2, and hence your Q1, have negative answers. Of course for particular $f$ you may be able to do better by the holomorphic functional calculus for Banach algebras.

You can find some generalizations to Fourier algebras of nonabelian groups by looking at the citations as listed in MathSciNet. My impression is that in most cases one just lifts the result from the abelian case by using Herz's restriction theorem.

Improve this answer
$\endgroup$
1
  • $\begingroup$ @HannesThiel You're welcome. I should note that probably people have tried to do similar things for Figa-Talamanca--Herz algebras although I have no idea if they succeeded $\endgroup$
    – Yemon Choi
    Commented Mar 5, 2015 at 11:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .