Baby rudin theorem 8.11

Question

Let {$\phi_n$} be orthonormal on [a,b]. Let $s_n(x) = \sum_{m=1}^n c_m\phi_m(x)$ be the nth partial sum of the Fourier series of f, and suppose $t_n(x) = \sum_{m=1}^n y_m\phi_m(x)$. Then $\int_a^b |f-s_n|^2dx\le\int_a^b|f-t_n|^2dx,$ and equality holds if and only if $y_m=c_m$, (m = 1,,,,,,,,n). Proof
let $\int$ denote the integral over [a,b], $\sum$ the sum from 1 to n. Then $\int f\bar t_n$ = $\int f\sum \bar y_m\bar\phi_m$ = $\sum c_m\bar y_m$ by the definition of {$c_m$}, $\int |t_n|^2 = \int t_n\bar t_n = \int \sum y_m\phi_m\sum\bar y_k \bar\phi_k = \sum |y_m|^2$ since {$\phi_m$} is orthonormal, and so $\int |f-t_n|^2 = \int|f|^2 - \int f\bar t_n - \int \bar ft_n + \int |t_n|^2 $
I don't understand the last equality. Any help would be appreciated.

Answer 1

For any two complex numbes $a$ and $b$ we have $|a-b|^{2}=(a-b)(\overline {a-b})=a\overline a+b\overline b -a\overline b-b\overline a=|a|^{2}+|b|^{2} -a\overline b-b\overline a$