# Bounding a Fourier coefficient of a non-negative periodic function in terms of its $L^2$-norm

This question is motivated by the earlier MO question: Show that $(\sum_{k=1}^{n}x_{k}\cos{k})^2+(\sum_{k=1}^{n}x_{k}\sin{k})^2\le (2+\frac{n}{4})\sum_{k=1}^{n}x^2_{k}$ . It is a cleaned up asymptotic version of that question.

Let $$f$$ be a non-negative function, periodic with period $$1$$, and square integrable on $${\Bbb R}/{\Bbb Z}$$. Is it true that $$|{\widehat f}(1)|^2 = \Big| \int_0^1 f(x) e^{-2\pi ix} dx \Big|^2 \le \frac 14 \int_0^1 f(x)^2 dx \ \ ?$$ Equality is attained for example when $$f(x) = \max(0, \cos(2\pi x))$$.

Note that $$|\widehat f(1)| =|\widehat f(-1)|$$ and, since $$f$$ is non-negative, $$|\widehat f(1)| \le \widehat f(0)$$. Therefore $$\int_0^1 f(x)^2 dx = \sum_n |\widehat f(n)|^2 \ge 3 |\widehat f(1)|^2,$$ so that the estimate holds with $$1/3$$ in place of $$1/4$$. There is a lot of scope to improve this argument, and with a more careful application of Bessel's inequality I could get the constant $$1/4+1/4\pi$$. But the claimed inequality looks very clean, and I wonder if (i) it is true!, (ii) is known in some context, and (iii) (hopefully) has an elegant proof?

## 1 Answer

Your conjecture is indeed correct. The proof is based on the following simple yet powerful trick I learned many years ago: $$|z| = \sup_{|v| = 1} \Re (zv)$$. Therefore it is enough to only bound from above $$\Re \left(\int_0^1 vf(x)e^{-2\pi ix}dx\right)$$ for all $$v\in \mathbb{T}$$. And now it is clear by Cauchy-Schwarz that $$f$$ should be proportional to $$\max(0, \Re(ve^{-2\pi ix}))$$, and all such functions gives us $$\frac{1}{4}$$ ($$f$$ from the OP corresponds to the choice $$v = 1$$).

• Cool! Missed that completely; nice proof! Mar 11, 2020 at 13:07
• And your proof works for the other problem too of course. Mar 11, 2020 at 13:16
• @Lucia hmm, are you sure?.. I can see it only in asymptotic sense, with $n(1/4 + \varepsilon)$ Mar 11, 2020 at 13:18
• Right, I did have in mind asymptotic version ... Mar 11, 2020 at 13:21