There are the definitions which we need for the proof:
Let {$\phi_n$} be orthonormal on $X$. Suppose $\sum |c_n|^2 $ converges, and put $s_n$ = $c_1\phi_1 + ... + c_n\phi_n $. Then there exists a function $f$ $\in$ $\mathscr L^2(\mu)$ such that {$s_n$} converges to $f$ in $\mathscr L^2(\mu)$, and such that
$f$ $\sim$ $\sum_{n=1}^\infty$ $c_n$$\phi_n$ .
There is the proof:
For $n$ $\gt$ $m$,
$||s_n - s_m||^2$ $=$ $|c_{m+1}|^2+...+|c_n|^2$,
so that {$s_n$} is a Cauchy sequence in $\mathscr L^2(\mu)$. By the previous theorem, there is a function $f$ $\in$ $\mathscr L^2(\mu)$ such that
$\lim_{n\to \infty}$ $||f - s_n||$ $=$ $0$.
Now, for $n$ $\gt$ $k$,
$\int_{X} f $$\bar \phi_k$ $d\mu$ $-$ $c_k$ $=$ $\int_{X} f $$\bar \phi_k$ $d\mu$ $-$ $\int_{X} s_n $$\bar \phi_k$ $d\mu$,
so that $|\int_{X} f \bar \phi_k d\mu - c_k|$ $\leq$ $||f-s_n||$ $\cdot$ $||\phi_k||$ $+$ $||f-s_n||$.
Letting $n\to \infty$, we see that
$c_k$ $=$ $\int_{X} f $$\bar \phi_k$ $d\mu$ ($k = 1,2,3,...$),
and the proof is complete.
I don't understand why is $c_k$ equal of $\int_{X} s_n $$\bar \phi_k$ $d\mu$ in this equation ( $\int_{X} f $$\bar \phi_k$ $d\mu$ $-$ $c_k$ $=$ $\int_{X} f $$\bar \phi_k$ $d\mu$ $-$ $\int_{X} s_n $$\bar \phi_k$ $d\mu$) and I don't understand how do we get the last inequality $|\int_{X} f \bar \phi_k d\mu - c_k|$ $\leq$ $||f-s_n||$ $\cdot$ $||\phi_k||$ $+$ $||f-s_n||$
Any help would be appreciated.
$c_k=\int s_n\overline {\phi_k}$ follows by writing down the definition of $s_n$ and splitting the integral into $n$ terms. (Recall that $\phi_k$'s are orthonormal).
$$|\int_{X} f \bar \phi_k d\mu - c_k|\leq ||f-s_n||\cdot ||\phi_k||+||f-s_n||$$ should be $$|\int_{X} f \bar \phi_k d\mu - c_k|\leq ||f-s_n||\cdot ||\phi_k||$$ (no need for the last term) and it follows from the fact that $\int_{X} f \bar \phi_k d\mu - c_k =\int (f-s_n)\overline {\phi_k}d\mu$ and $|\int (f-s_n)\overline {\phi_k}d\mu|\leq \|f-s_n\|\|\overline {\phi_k}\|=\|f-s_n\|\|\phi_k\|$.