Rudin 1.37 (d) is trivial to Rudin but hell for me

Hi everyone,

I am working on my own through Rudin's Principles of Mathematical Analysis and, after the demonstration of Cauchy - Schwarz Inequality, in Theorem 1.37, part (d), Rudin states:

$$|x \cdot y| \leqslant |x||y|$$

When he explains how to prove this, he simply states that this is an immediate consequence of Schwarz Inequality, which he defines as follows:

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

If someone can explain me how this two things are identical I would appreciate it a lot. My toughts so far:

$$|x \cdot y| \leqslant |x||y|$$ Take the square of this, which is:

$$(x \cdot y)(x \cdot y) \leqslant \sum x_i^2 \sum y_i^2$$ and hence,

$$(\sum x_iy_i)^2 \leqslant \sum x_i^2 \sum y_i^2$$ which is NOT the Schwarz Inequality!
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 Sure it is? He just conjugates the $b_j$ to make the formula compatible with complex spaces. It's the same reason he slaps absolute value bars everywhere.
 I understand the part of the conjugate, actually the whole part on the left is pretty clear to me. The problem is the right side of the inequality. I don't understand how $$\sum x_i^2 \sum y_i^2 = \sum_{j=1}^{n} |x|^2 \sum_{j=1}^{n} |y|^2$$

Rudin 1.37 (d) is trivial to Rudin but hell for me

You mean $|x_i|$ on the right, not $|x|$, right? Again, it's just slapping absolute value bars around to make the inequality compatible with complex spaces. For real numbers, there is no difference.

 Tags cauchy-schwarz, hell, rudin, schwarz inequality, trivial

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