# 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?

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 $$v = 1$$).
• @Lucia hmm, are you sure?.. I can see it only in asymptotic sense, with $n(1/4 + \varepsilon)$ Mar 11, 2020 at 13:18