Assume $f(x)$ is a smooth $2\pi$ periodic function and can be decomposed as $$f(x)=\sum_{n=-\infty}^\infty a_ne^{inx}$$ A function sequence $f_m(x)$ satisfies $$\int_0^{2\pi}f_m(x)e^{-inx}dx\to a_n,\ m\to\infty,\ \forall n$$ Does this implies $f_m$ converge to $f(x)$, and in what sense?(pointwise, $L_2$?) And what if $f(x)$ belongs to some other function class?
I'm not sure about this, like in a Hilbert space $H$ with orthonormal basis $\{e_i\}$, $$\left \langle x_n,e_i \right \rangle \to \left \langle x_0,e_i \right \rangle,\ n\to \infty, \forall i$$ doesn't imply $x_n\to x_0$ (also need $|x_n|$ is bounded, I only know the result but don't know any counterexample). So maybe we can't draw any conclusion from the above?
Pointwise convergence of Fourier coefficients is too weak for such conclusion. Let's take $f\equiv 0$ and $f_m = e^{m(1+ix)}$. Then the convergence of coefficients holds, but $|f_m|\to\infty$ pointwise.
If you also assume that the coefficients are uniformly bounded in $\ell^2$ norm, then weak convergence in $L^2$ holds. That is, $\int f_m g\to \int fg$ for every $g\in L^2$. This still does not imply that $f_m\to f$ in $L^2$, or pointwise.