My question comes from trying to understand a technical step in [this paper](http://65.54.113.26/Publication/4171765/ruzsa-s-problem-on-sets-of-recurrence) 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 $4RL$).



Inequality on Trigonometric polynomials
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". :)


