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Let $\left(c_{n}\right)_{n},\,\left(d_{n}\right)_{n}$ two successions of complex numbers and let $N$ a large natural number.Is it true that $$\left|\underset{n=1}{\overset{N}{\sum}}c_{n}d_{n}\right|\leq\underset{n=1,\dots,N}{\max}\left|c_{n}\right|\left|\underset{n=1}{\overset{N}{\sum}}d_{n}\right|?$$It would be fine $$\left|\underset{n=1}{\overset{N}{\sum}}c_{n}d_{n}\right|=O\left(\underset{n=1,\dots,N}{\max}\left|c_{n}\right|\left|\underset{n=1}{\overset{N}{\sum}}d_{n}\right|\right).$$ It is important to me that the sum remains in the modulus. Thank you.

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  • $\begingroup$ What are your thoughts? What have you tried? $\endgroup$ – Calvin Lin Mar 3 '14 at 21:39
  • $\begingroup$ @CalvinLin I tried to use Holder inequality but I'm not able to keep the sum in the modulus. $\endgroup$ – user118929 Mar 3 '14 at 21:42
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    $\begingroup$ $c_n = d_n = (-1)^n$ explains why. $\endgroup$ – Daniel Fischer Mar 3 '14 at 22:13
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The answer is no. If the $d_n$ are such that $d_n\ne0$, $1\le n\le N$, and $\sum_{n=1}^Nd_n=0$, it would imply that $$ \sum_{n=1}^nc_n\,d_n=0 $$ for all possible choices of $c_n$, which is clearly imposible (choose $c_n=\overline{ d_n}$.)

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