TITLE: Baby Rudin - Theorem 1.35 Cauchy Schwartz
QUESTION [5 upvotes]: I'm stumped on the following difficulty while reading baby Rudin (p.15).
Let $A=\sum|a_j|^2, B=\sum|b_j|^2, C=\sum a_j \overline{b}_j$:
$$\begin{align}
\sum|Ba_j-Cb_j|^2 &= \sum(Ba_j-Cb_j)(B\overline{a}_j-\overline{Cb_j}) \\
&= B^2\sum|a_j|^2{\bf -B\overline{C}\sum a_j\overline{b}_j-BC\sum \overline{a}_j b_j}+|C|^2\sum|b_j|^2 \\
&=B^2A-B|C|^2=B(AB-|C|^2)
\end{align}$$ 
Note that the second line of the equation, the book indicates that the middle 2 term together to be zero. Why is that?

REPLY [3 votes]: Not quite. It's the last three terms in the second line that reduce to $-B|C|^{2}$. To see this, note that $\sum \overline{a_{j}}b_{j}=\sum\overline{a_{j}\overline{b_{j}}}=\overline{\sum a_{j}\overline{b_{j}}}=\overline{C}$. Hence, 
$$
\begin{aligned}
&-B\overline{C}\sum a_{j}\overline{b_{j}}-BC\sum\overline{a_{j}}b_{j}+|C|^{2}\sum|b_{j}|^{2}\\
=&-B\overline{C}C-BC\overline{C}+C\overline{C}B\\
=&-BC\overline{C}\\
=&-B|C|^{2}
\end{aligned}
$$