# Inequality on Trigonometric polynomials

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 $4RL$).

• 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"? Feb 14, 2013 at 2:43
• Can you provide the links, Bill? Feb 14, 2013 at 6:27
• Here is one: mathoverflow.net/questions/101859/… Feb 14, 2013 at 6:46
• 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). Feb 14, 2013 at 11:23

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". :)
• 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... Feb 14, 2013 at 18:53
• @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 :)). Feb 14, 2013 at 19:00