# A Fourier-analytic inequality used by Jean Bourgain

I am currently reading Jean Bourgain's 1986 paper A Szemerédi type theorem for sets of positive density in $R^k$ and would appreciate some help in understanding a Fourier-analytic estimate used in that article. I suspect that my question is relatively elementary, but my knowledge of Fourier analysis is not very strong and there are no experts in the topic at my current place of work, so I would very much appreciate a pointer.

In Bourgain's argument, $f \colon \mathbb{R}^d \to [0,1]$ is a nonzero measurable function supported in a fixed bounded measurable set $A$, and the $L^2$ norm of $f$ is fixed. For each $\lambda>0$ we define $P_\lambda \colon \mathbb{R}^d \to \mathbb{R}$ to be the function whose Fourier transform $\hat{P}_\lambda(\xi):=\int_{\mathbb{R}^d} e^{-2\pi i \langle x,\xi\rangle} P_\lambda(x)dx$ is given by $\hat{P}_\lambda(\xi)=e^{-\lambda\|\xi\|}$ for all $\xi \in \mathbb{R}^d$. Parameters $\delta, t>0$ are introduced, and the parameter $\delta$ is subsequently fixed at some small value which depends on $\|f\|_2$ (and possibly on $A$) but not on the precise choice of $f$. It is then claimed that by taking $t$ small enough, the quantity $$\|(f * P_{\delta t}) - (f * P_{\delta^{-1}t})\|_2$$ can be made arbitarily small in a manner which is uniform with respect to $f$. It is clear to me that this quantity must converge to zero as $t \to 0$ when $\delta$ and $f$ are fixed, but it is not clear to me why a single value $t$ can be chosen which works simultaneously for all $f$ (where $\|f\|_2$ is fixed and the support of $f$ lies in $A$). Bourgain's paper seems to use a quantitative bound which I infer to resemble $$\|(f * P_{\delta t}) - (f * P_{\delta^{-1}t})\|_2 \leq C\|f\|_2\frac{\log (1/\delta)}{\log (1/t)}.$$ Certainly it is stated that in order to make the above difference small (relative to $\delta^{1/4}$ and $\|f\|_2$) it is sufficient that $\log (1/t)$ should be a large multiple of $\log (1/\delta)$. Can anyone see more precisely what estimate is being used here, or at least how the above quantity can be bounded uniformly with respect to $f$?

Thanks!

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Just a quick guess: Convolution is linear, Fourier transform is an isometry which converts convolution into product, and then explicit calculation? – Stopple Jul 10 '12 at 15:53
Stopple: I've tried that without achieving anything conclusive. The square of the $L^2$ norm of the difference is $\int_{\mathbb{R}^2} |\hat{f}(\xi)|^2 |e^{-\delta t \|\xi\|}-e^{-\delta^{-1} t\|\xi\|}|^2 d\xi$. As $t \to 0$ the integrand converges to zero pointwise while being bounded by the integrable function $2|\hat{f}(\xi)|^2$ so this converges to zero for fixed $f$ by the Dominated Convergence Theorem, but it is not clear to me why this would be uniform. The supremum over $\xi$ of $|e^{-\delta t \|\xi\|}-e^{-\delta^{-1} t\|\xi\|}|^2$ doesn't depend on $t$ so bounding it also doesn't help. – Ian Morris Jul 10 '12 at 16:17
^^ By "uniform" I mean "uniform with respect to $f$ in the category being considered", and by "2|\hat{f}(\xi)^2|" I mean "4|\hat{f}(\xi)|^2". – Ian Morris Jul 10 '12 at 16:18

I assume you are referring to the argument in page 313 of Jean's paper http://www.springerlink.com/content/a343g53033872345/ . The point here is that the bound does not hold for all $t$, but for a single $t$ (out of $J$ possible choices $t_1,\dots,t_J$); note that Jean crucially refers in the paper to a "suitable" $t$ rather than an arbitrary $t$. This is a pigeonholing argument, based on the estimation of
$$\sum_{j=1}^J \| f * P_{\delta t_j} - f * P_{\delta^{-1} t_j} \|_{L^2}^2$$
which can be done by Plancherel's theorem and routine computations (if the $t_j$ are lacunary, as noted in Jean's paper).