let ${ \phi_{n} } $ be sequence of orthogonal functions on $[a,b]$ If the series $ \sum_{n=1}^{\infty } a_{n}\phi_{n}(x) $ converges uniformly to a function $f(x)$ on $[a,b]$ prove that for each $n \in \mathbb{N} $, $a_{n}$ is the Fourier coefficient of $f$.
Above problem is an exercise of my introduction to real analysis class. I have no idea how to solve it. It would be very helpful if someone give me some hints or direction of proving. Thanks in advance. and sorry for bad english.
Assuming the $\phi_n$ are orthonormal, the $n$th Fourier coefficient of $f$ is given by $\hat{f_n} = \langle \phi_n, f \rangle$.
Let $S_m = \sum_{k=1}^m a_k \phi_k$, and suppose $m \ge n$. We see that $\langle \phi_n, S_m \rangle = a_n$.
Furthermore, we have (again for $n \ge m$) $|\hat{f_n}-a_n| = |\langle \phi_n, f-S_m \rangle| \le \|f-S_m\|_2 \le (b-a) \sup_{x \in [a,b]} |f(x)-S_m(x)|$.
We are given that $\lim_{m \to \infty} \sup_{x \in [a,b]} |f(x)-S_m(x)| = 0$, from which it follows that $|\hat{f_n}-a_n|=0$ or $\hat{f_n}=a_n$.