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Let $\left\{a_k\right\}_{0\leqslant k\leqslant n}$ and $\left\{b_k\right\}_{0\leqslant k\leqslant n}$ be two positive 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 $f$ 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 $n$, $\epsilon$)?

Edit: After Iosif Pinelis'answer, I should add that $a_k$ and $b_k$ are decreasing sequences.

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  • $\begingroup$ Do you mean: Assume that there is a function $f$ such that for all $n>0$ and all $\epsilon\in(0,1)$, $$\sum_{k=0}^n a_k \sin (k\epsilon) \geq f(n)$$ or do you mean that for a specific $\epsilon>0$, $$\sum_{k=0}^n a_k \sin (k\epsilon) \geq f(n)?$$ $\endgroup$
    – username
    Commented Apr 21, 2021 at 14:26
  • $\begingroup$ For a specific $\epsilon\in(0,1)$. $\endgroup$
    – coco
    Commented Apr 21, 2021 at 14:28
  • $\begingroup$ If the constant $c$ is independent of $\epsilon$, then it should be true for all $\epsilon$. Indeed, given $\epsilon>0$, just take $f(n)= \sum_{k=0}^n a_k \sin (k\epsilon)$. $\endgroup$
    – username
    Commented Apr 21, 2021 at 14:34

1 Answer 1

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No. E.g., let $a_k=2+\sin (k \epsilon )$ and $b_k=1$. Then for each real $\epsilon>0$ we have $$\sum_{k=0}^na_k\sin(k\epsilon)\asymp n,$$ whereas $$\sum_{k=0}^nb_k\sin(k\epsilon)=O(1).$$

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  • $\begingroup$ Thank you for your answer. And if we assume that $a_k$ is decreasing? $\endgroup$
    – coco
    Commented Apr 21, 2021 at 14:37
  • $\begingroup$ @coco : I think then the answer will change to yes. Since your original question has been answered, I suggest you post the additional question separately. $\endgroup$
    – Iosif Pinelis
    Commented Apr 21, 2021 at 15:10

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