Let $p(x)=\sum_{k=1}^m [a_k\cos(n_kx)+b_k\sin(n_kx)]$ be a null average trigonometric polynomial (null average means that is $\int_\mathbb T p =0$ or, equivalently, there are no $a_0$ and $b_0$). Denote with $N$ the order of $p$, namely the number of times that $(a_k,b_k)\neq(0,0)$. We assume that $\int_\mathbb T p^2=1$.

I have done some numerical computations and I think the following may be true, there is an optimal constant $C(N)>0$ depending only on $N$, such that

$$\frac12\|p\|_{L^1(\mathbb T)}=\int\limits_{\{p>0\}\cap\mathbb T} p(s)\,ds=-\int\limits_{\{p<0\}\cap\mathbb T} p(s)\,ds\geq C(N).$$

The optimal $C(N)$ actually exists, but I am interested in an explicit expression or a way to find $C(N)$. When $N=2$ clearly the equality holds and $C(2)=\frac{2}{\sqrt{\pi}}$. Thank you.

  • $\begingroup$ Fejer kernel is one of the most basic examples of non-negative trigonometric polynomials. For those, it is clear that the middle equality cannot hold. $\endgroup$
    – Itay
    Mar 30 '18 at 12:01
  • $\begingroup$ @Itay Fejer kernel is not null average, $n_0$ is allowed to be zero or, in other words, $\int_\mathbb T F_n\neq 0$. $\endgroup$ Mar 30 '18 at 12:21
  • $\begingroup$ You can get some lower bound by using the normalized Dirichlet kernel. Take $p(x)=D(x)-1=\frac{1}{\sqrt{2n}}(\sum_{|k|<n}e^{ikx}-1)$. It is easy to see that this polynomial is null average and is positive on the interval $I=(-\frac{c}{n},\frac{c}{n})$, with some positive constant $c$. A direct computation gives $\intop_{\{p>0\}\cap \mathbb{T}} p(x)dx>\intop_{I} p(x)dx>\frac{C}{\sqrt{n}}$, where $C>0$ is some absolute constant. My feeling is that you cannot do much better. $\endgroup$
    – Itay
    Mar 30 '18 at 14:48
  • $\begingroup$ In that case, it is even better to use the Nikolskii inequalty $2\int_{\mathbb{\{p>0\}}\cap \mathbb T}p= \int_{\mathbb T}|p|\geq \frac1{2\sqrt n}$ where $n$ is de degree of $p$. But this estimate is far from being optimal. For example, when $N=2$ and $n$ is big, this lower bound is telling nothing. $\endgroup$ Mar 30 '18 at 19:37
  • $\begingroup$ You can easily get $\int|p|^4\le 2N$ whence, by Holder, you still have $C(N)\approx N^{-1/2}$. $\endgroup$
    – fedja
    Mar 31 '18 at 0:25

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