3
$\begingroup$

This is another question on the possible shape of sets $A,B\subset \mathbb{R}^d,d\geq 2,$ where resp. a non-null Schwarz function $f$ and its Fourier transform can vanish.

A nice remark by Christian Remling is that choosing separable functions yields that both functions can vanish on a half-space. Can a function and its Fourier transform both vanish on a convex cone?? The theorem by Shapiro mentioned there says that $A$ and $B$ cannot be resp. a "major convex cone" (strictly larger than a half space) and a non-empty open set.

Hence a remaining question is what can happen if $A$ is a "minor cone", say the quadrant of points with non-negative coordinates. As we saw, $B$ can be a half-space, but not a major cone. Is it possible that $B$ has a "gap in each coordinate"?

Here is a one-sentence question: is it possible to find a non-null Schwartz function $f$ on $\mathbb R^2$ which vanishes on a quarter-space and which Fourier transform $\hat f(x,y)$ vanishes for $x\in O$ or $y\in O'$ where $O,O'$ are non-empty open sets of $\mathbb R$.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ I don't understand your question: Christian Remling's comment which you referred to points out that if $g$ and $\hat h$ both vanish on a half-line, then $f(x,y) := g(x)\,h(y)$ vanishes on a half-plane and $\hat f$ also does. But vanishing on a half-plane is stronger than vanishing on a quarter-plane, so it seems to me that this $f$ exactly answers the question of your last paragraph. $\endgroup$
    – Gro-Tsen
    Commented Nov 20 at 10:50
  • $\begingroup$ Yes but vanishing on OxO' is stronger (or at least different) than vanishing on a half plane $\endgroup$
    – kaleidoscop
    Commented Nov 20 at 10:55
  • 2
    $\begingroup$ Now I'm even more confused. Vanishing on $O\times O'$ with $O,O'$ nonempty open sets of $\mathbb{R}$ just means the same as vanishing on a nonempty open set of $\mathbb{R}^2$, since the former are a basis of the latter; and anyway the half-plane we're talking about is of the form $\mathbb{R} \times \mathbb{R}_{\gt 0}$. What more do you want? $\endgroup$
    – Gro-Tsen
    Commented Nov 20 at 11:07
  • $\begingroup$ Sorry, what I meant by "vanishes on a gap for each coordinate" came out completely wrong, I actually meant that it should vanish on infinite vertical and horizontal gaps, the post edited. $\endgroup$
    – kaleidoscop
    Commented Nov 20 at 11:44

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .