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Suppose $f:\mathbb R\to\mathbb R$ is a smooth $2\pi$-periodic function. Consider the following statements:

  1. There exist $a,\phi\in\mathbb R$ such that $f(\theta)\geq a\cos(\theta+\phi)$ for all $\theta\in\mathbb R$.
  2. $\sum_{k=1}^n f(\theta+2\pi k/n)\geq0$ for all $\theta\in\mathbb R$ and integers $n>1$.

Clearly (1) implies (2). Does (2) imply (1)?

More generally fix a finite set $N\subseteq\mathbb Z_{>0}$, and consider the statements

  1. There exists $a_n,\phi_n\in\mathbb R$ such that $f(\theta)\geq\sum_{n\in N}a_n\cos(n\theta+\phi_n)$ for all $\theta\in\mathbb R$.
  2. $\sum_{k=1}^n f(\theta+2\pi k/n)\geq0$ for all $\theta\in\mathbb R$ and integers $n$ not dividing any element of $N$.

Again (3) implies (4). Does (4) imply (3)?

My thoughts for the simpler version: Consider the Fourier series for $f$ (which converges nicely since $f$ is smooth): $$ f(\theta)=\sum_{k\geq0}a_k\cos(k\theta+\phi_k). $$ Then (2) implies $$ \sum_{n|k}a_k\cos(k\theta+\phi_k)\geq0 $$ for all $n>1$. It is sufficient to prove $$ \sum_{n\neq1}a_k\cos(k\theta+\phi_k)\geq0. $$ But no nonnegative linear combination of the former inequalities can produce the latter; just look at the coefficients of $a_2$, $a_3$ and $a_6$.

stewbasic
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1 Answers1

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After some graphing I found that my conjecture is false. Indeed $f(\theta)=1+\cos(2\theta)+\cos(3\theta)$ is a counterexample. Note that $$ \frac1n\sum_{k=1}^nf(\theta+2\pi k/n)=\begin{cases} 1+\cos(2\theta)&\text{if }n=2,\\ 1+\cos(3\theta)&\text{if }n=3,\\ 1&\text{if }n>3. \end{cases} $$ In each case it is $\geq0$ for all $\theta$. However, suppose $$ f(\theta)\geq a\cos(\theta+\phi) $$ for some $a,\phi\in\mathbb R$. We can rewrite this as $$ f(\theta)\geq p\cos(\theta)+q\sin(\theta) $$ where $p,q\in\mathbb R$. But $f$ is even, so substituting $\theta\mapsto-\theta$ and averaging, $$ f(\theta)=1+\cos(2\theta)+\cos(3\theta)\geq p\cos(\theta). $$ Setting $\theta=\pi$ we obtain $p\geq-1$. Setting $\theta=\pi/2-\epsilon$ where $\epsilon\in(0,\pi/2)$, we obtain $$ 1-\cos(2\epsilon)-\sin(3\epsilon)\geq p\sin(\epsilon)\geq-\sin(\epsilon). $$ To first order this gives $-3\epsilon\geq-\epsilon$, a contradiction.

stewbasic
  • 6,131