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Suppose $a_{1},a_{2},a_{3}...a_{n}$ is a complex sequence satisfying $\bigl\lvert\left(\sum_{{k=1}}^{n}a_{k}b_{k}\right)\bigr\rvert \leq1$ for all $b_1,b_2,...,b_n$ such that $\left(\sum_{{k=1}}^{n}\mid b_{k}\mid^{2}\right)\leq1$. Show that $\left(\sum_{{k=1}}^{n}\mid a_{k}\mid^{2}\right)\leq1$

I'm considering to prove by contradiction of Cauchy-Schwarz inequality but don't know where to start. The conclusion seems obvious.

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If $a=0$ then the inequality holds. Now suppose $\lVert a \rVert=\left(\sum|a_j|^2\right)^{1/2} > 0$. Let $b_j = \overline{a_j}/\lVert a \rVert$. Then $\sum |b_j|^2=1$ and $\lVert a \rVert = \sum a_j b_j \leq 1 $.

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  • $\begingroup$ beat me to it!! $\endgroup$ – Greg Martin Feb 4 '15 at 18:26
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Suppose $\textbf{a,b}$ are complex vectors such that $\textbf{a}=(a_1,a_2,...,a_n) and \textbf{b}=(b_1,b_2,...,b_n)$.

Then $|\textbf{a}|^2=\sum_{k=1}^{n}|a_k|^2$ and $|\textbf{b}|^2=\sum_{k=1}^{n}|b_k|^2$.

Start proof by supposing $|\textbf{a}|^2 >1$.

Let $\textbf{b}=\frac{\textbf{a}}{|\textbf{a}|}$,so $|\textbf{b}|^2=\sum_{k=1}^{n}|b_k|^2 =1$,which satisfies the condition given in the statement.

According to the statement,we should have $|(\sum_{k=1}^{n}a_kb_k)|\leq1$.However,we actually have $$|(\sum_{k=1}^{n}a_kb_k)|=|\textbf{a $\cdot$ b}|=|\textbf{a} \cdot \frac{\textbf{a}} {|\textbf{a}|}|=|\textbf{a}|>1$$

Therefore,$|\textbf{a}|^2$(that is ,$\sum_{k=1}^{n}|a_k|^2$)cannot be greater than $1$.The proof is complete.

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    $\begingroup$ This misses absolute values and complex conjugations in several places. The "contradiction" part is a red herring. It plays no role in the argument. $\endgroup$ – WimC Feb 5 '15 at 9:46

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