Oscillating sums II

Question

Let $\left\{a_k\right\}_{0\leqslant k\leqslant n}$ and $\left\{b_k\right\}_{0\leqslant k\leqslant n}$ be two positive decreasing sequences such that $c_1a_k\leqslant b_k\leqslant c_2 a_k$ for all $k$ for some positive constants $c_1,c_2$. Let $\epsilon>0$ and $f>0$ a function such that $$\sum_{k=0}^na_k\sin(k\epsilon)\geqslant f(n).$$

Question. Is it true that $$\sum_{k=0}^nb_k\sin(k\epsilon)\geqslant cf(n)$$ for some constant $c$ (independent of $n$, $\epsilon$)?

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Without the monotonicity condition on the $a_k$'s and $b_k$'s, the answer is no, as shown in this answer.

Answer 1

$\newcommand\ep\varepsilon$The answer is still no (assuming you wanted $c>0$). Indeed, let $$s_n:=\sum_{k=0}^n a_k\sin(k\ep),\quad t_n:=\sum_{k=0}^n b_k\sin(k\ep),$$ where $\ep=3\pi/4$, $a_0=b_0=10$, $a_1=1$, $b_1=9$, and $a_k=b_k=2^{-k}$ for $k\ge2$.

Then $c_1a_k\le b_k\le c_2a_k$ for $c_1=1$, $c_2=9$, and all $k\ge0$. Moreover, $s_0>0$ and $s_1=-1 + 5\sqrt2>6>0$, whence $s_n>6-\sum_{k=2}^\infty 2^{-k}>0$ for $n\ge2$. So, $s_n>0$ for all $n\ge0$. So, letting $f(n):=s_n$ for all $n\ge0$, we get a positive function $f$ such that $s_n\le f(n)$ for all $n\ge0$. Thus, all your conditions on the $a_k$'s and $b_k$'s are satisfied.

However, $t_1=-9 + 5\sqrt2<-19/20<0$ and, moreover, $t_n<-19/20+\sum_{k=2}^\infty 2^{-k}<0$ for all $n\ge2$. Thus, the condition $t_n\le cf(n)$ fails to hold for any $c>0$ and any $n\ge1$.