Cauchy-Schwartz Inequality:
I am trying to prove Question 2 of Problem 2.9:
Question
I have the following difficulties:
1) I am not able to understand how (1) in the figure is arrived at.
2) I am not able to understand how (2) in the figure is arrived at.
3) How is $ | \sum _{i=1}^{n} z_i | \le \sum _{i=1}^{n} |z_i| $ equivalent to $ | \sum _{i=1}^{n} x_iy_i^* | \le \sum _{i=1}^{n} |x_iy_i^*| $ (Equation 3) ?
Solution
(1) $$\left(\sum_{i=1}^nz_i\right)^2=\sum_{i=1}^n\sum_{j=1}^nz_iz_j$$
(2) \begin{align*} (ab+cd)^2&=a^2b^2+2abcd+c^2d^2\\ &=(a^2+c^2)b^2-c^2b^2+2abcd+(c^2+a^2)d^2-a^2d^2\\ &=(a^2+c^2)(b^2+d^2)-(c^2b^2-2abcd+a^2d^2)\\ &=(a^2+c^2)(b^2+d^2)-\underbrace{(cb-ad)^2}_{\geq 0 }\\ &\leq (a^2+c^2)(b^2+d^2) \end{align*}
(3)
The implication is obvious. For the converse, set $z_j=a_j+ib_j$. Set $x_j=a_j+ib_j$ and $y_j=1$. Then $z_j=x_jy_j^*$.
[This is mostly the same answer as Surb; I started writing one with a few more details before he posted his.]
Equation (1) follows from the distributive property of addition over multiplication: $$ \left(\sum_{i=1} z_{i,R}\right)^2 = \left(\sum_{i=1} z_{i,R}\right)\left(\sum_{m=1} z_{m,R}\right) = \sum_{i=1} \sum_{m=1} z_{i,R} z_{m,R} $$ That last step is a summation form (with more terms) of the “FOIL” identity $$ (a+b)(c+d) = ac + ad + dc + bd $$ If you do the same thing with $\left(\sum_{i=1} z_{i,I}\right)^2$ you get (1).
Equation (2) follows from the observation that the square of any real number is nonnegative. We have \begin{align*} (ac+bd)^2 &= a^2c^2 + 2abcd + b^2d^2 \\ 0 \leq (ad-bc)^2 &= a^2d^2 - 2abcd + b^2 c^2 \end{align*} So if you add both sides of the two together, you get $$ (ac+bd)^2 \leq a^2c^2 + a^2 d^2 + b^2 d^2 + b^2 c^2 = (a^2+b^2)(c^2+d^2) $$ So (2) now follows from setting $a=a_{i,R}$, $b=z_{i,I}$ $c=z_{m,R}$, $d=z_{m,I}$ over and over for each pair $(i,m)$.
For (3), this is nothing more than a substitution. If $z_i = x_i y_i^*$, then \begin{align*} \left|\sum_{i=1}^{m} z_i\right| &= \left|\sum_{i=1^m} x_i y_i^*\right| \\ \sum_{i=1}^m |z_i| &= \sum_{i=1}^m |x_i y^*_i| = \sum_{i=1}^m |x_i| |y^*_i| \end{align*}
