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Simple question: do we really need the conjugate in the inequality?

$$ |\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2 $$

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The answer is No. – Tom Nov 17 '12 at 12:31
Of course no: replace all $b_j$ by their conjugates, the inequality is still true, and $|b_j|=|\bar{b}_j|$. – Jean-Claude Arbaut Nov 4 '14 at 8:29
up vote 6 down vote accepted

The Cauchy-Schwarz inequality says that $|\langle a,b \rangle| \leq \| a \| \|b\|$. In $\mathbb C^n$, the inner product is \begin{equation} \langle a, b \rangle = \sum_{j=1}^n a_j \bar{b_j}. \end{equation}

That's why the Cauchy-Schwarz inequality in $\mathbb C^n$ has conjugates in it.

While it is true that you could omit the conjugates in your inequality and still have a true statement, that would only take us further away from the nice statement that $|\langle a,b \rangle| \leq \| a \| \|b\|$.

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Ok thanks for the answer. Haven't read about the inner product for complex numbers yet – A. Napster Nov 17 '12 at 12:44

The answer is no, because $|b_j|^2=|\bar{b}_j|^2$.

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Simple answer, yes. Look back to the definition of the inner product (for complex numbers) and note the complex conjugate symmetry (as opposed to pure symmetry)

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If you replace $b_j$ by $\overline{b_j}$ on both sides, you have no more $\overline{.}$ on the left and on the right, you don't have any either since you have $|b_j| = |\overline{b_j}|$... – xavierm02 Nov 17 '12 at 12:33

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