3
$\begingroup$

Let S be a set of integers and denote the characteristic function of S as $\chi_{S}(n)$. Define an operator on the space of trig functions by the relation $\hat{Tf}(n) = \chi_{S}(n) \hat{f}(n)$. Here $\hat{f}(n)$ is the n-th Fourier coefficient of f.

For $p\geq 2$ we'll call S a $L^p$ multiplier set (or just $L^p$ multiplier) if there is an inequality of the form $\Vert Tf\Vert\_{p} \leq c \Vert f\Vert\_p$. If this inequality holds for some p but fails for $p+\epsilon$ for every $\epsilon>0$, we'll say that S is a strict $L^p$ multiplier.

Note that every set is a $L^2$ multiplier and if S is a $L^p$ multiplier for some p then it is a $L^q$ multiplier for $2 \leq q \leq p$. Moreover, it follows from a result of Zygmund that almost every (in the obvious sense) set is a strict $L^2$ multiplier. (I also think you can prove this via Khintchine's inequality, but I haven't checked this argument.)

Do strict $L^p$ multiplier sets exist for every $p>2$? Note that this is similar to the $\Lambda(p)$ problem, however, I don't see how to transform a strict $\Lambda(p)$ set into a strict $L^p$ multiplier set.

$\endgroup$

1 Answer 1

7
$\begingroup$

Let M_p be the class of L^p muliplier sets, as considered in this equation. It is known:

  1. This is an algebra of sets, but not a sigma-algebra. [Not sure of the reference.]

  2. The inclusion M_p \subsetneq M_q is strict for 2\le p

MR1728363 (2001g:42013) Mockenhaupt, Gerd(5-NSW-SM); Ricker, Werner J.(5-NSW-SM) Idempotent multipliers for $L^p(\bf R)$.

$\endgroup$
1
  • $\begingroup$ Thanks! It turns out that the proof follows easily from Bourgain's \Lambda(p) set construction. Let E be a strict \Lambda(p) set. Clearly E is a is a L^{p} multiplier, since ||Tf||_{p} << ||f||_{2} << ||f||_{p}. Conversely, if E was a L^{p+\epsilon} multiplier, we can show it must be a \Labmda(p+\epsilon') set by interpolating the estimates ||Tf||_{p} << ||f||_{2} with ||Tf||_{p+\epsilon} << ||f||_{p+\epsilon}. It was this last step that I was missing. $\endgroup$
    – Mark Lewko
    Commented Oct 31, 2009 at 3:25

You must log in to answer this question.

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