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My question comes from trying to understand a technical step in this paper by Bourgain.

Let $R,L$ be positive integers and let $f(x)=\sum_{|n|\leq RL}a_ne^{2\pi inx}$ be a trigonometric polynomial. Assume $f(x)>0$. Let $F_L(x)=\sum_{|n|\leq L}\frac{L-|n|}Le^{2\pi inx}$ be the Fejer kernel. Define (as usual) the convolution $$(f*F_L)(x)=\int_0^1f(t)F_L(x-t)dt=\sum_{|n|\leq L}a_n\frac{L-|n|}Le^{2\pi inx}$$ Does it follow (and how) that $$f(x)\leq10R(f*F_L)(x)$$?

In the paper we have a concrete $f(x)$, so maybe this is not true in general. There we have $$a_n=1-\cos\left(2\pi\frac{RL-|n|}N\right)$$ where $N$ is a large positive integer, (at least bigger than $4RL$).

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    $\begingroup$ Don't we have a long list of MO questions that begin with "My question comes from trying to understand a technical step in this paper by Bourgain"? $\endgroup$
    – Bill Johnson
    Commented Feb 14, 2013 at 2:43
  • $\begingroup$ Can you provide the links, Bill? $\endgroup$
    – fedja
    Commented Feb 14, 2013 at 6:27
  • $\begingroup$ Here is one: mathoverflow.net/questions/101859/… $\endgroup$
    – user21162
    Commented Feb 14, 2013 at 6:46
  • $\begingroup$ Yeah, but that one was due to a misreading and the author had the privilege to get an explanation from Terry Tao himself, so I wouldn't complain too much if I were in his shoes (LOL). Of course, as stated in the post, the statement was just patently false. You can put almost all energy to any given sufficiently large annulus $R<|\xi|<2R$ you want (just take your favorite bounded compactly supported function $f$ and move the main bulk of $\widehat f$ anywhere you want by adjusting the phase). $\endgroup$
    – fedja
    Commented Feb 14, 2013 at 11:23

1 Answer 1

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That is true for all non-negative trigonometric polynomials, though not entirely obvious unless you are a Fourier analyst yourself. To see it, just note that the convolution with $K_{RL}=2F_{2RL}-F_{RL}$ recovers $f$ faithfully and $F_{RL}\le RF_L$. Of course, to Jean such things are as obvious as $2\times 2+1=5$ (he writes $10$ instead of $5$ just out of the traditional analyst's habit to have a 100% security margin in the constants), but I agree that it may be a bit perplexing for poor mortals like you and me. Joe Diestel just told me at a beer party tonight that the most common phrase in Bourgain's early papers was "By standard techniques we conclude from here that". :)

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    $\begingroup$ I was trying to be humorous with my other comment, fedja, but Joe is correct about Jean's early papers. You had an easier time reading his papers in harmonic analysis as by then he put in a few more details for us mortals. $\endgroup$
    – Bill Johnson
    Commented Feb 14, 2013 at 16:16
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    $\begingroup$ Thanks! It still took me some time to work with these hints but I got it now. Just to make sure I understood it, I think we actually get $$f=K_{RL}*f\leq2F_{2RL}*f\leq4RF_l*f$$ so with a $4$ and not $5$, right? It's good to know that I'm not alone having a hard time understanding Bourgain's papers... $\endgroup$
    – Joel Moreira
    Commented Feb 14, 2013 at 18:53
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    $\begingroup$ @Joel Yes, it is 4, but the other traditional analyst's habit is to make sure that an occasional wrong sign in an identity does not invalidate the argument :). @Bill Yeah, my first exposure to Jean's writing was reading his opus on the existence of $\Lambda(p)$ sets for $2<p<4$ in French (surprisingly, that one was written well and it took me just a couple of months to figure out what was going on there :)). $\endgroup$
    – fedja
    Commented Feb 14, 2013 at 19:00

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