I am new to Fourier series and here is what I understand (please correct me where I am wrong) :
The trigonometric series $a_0/2 + Σ(a_ncos(nx)+b_nsin(nx))$ will be called a Fourier series of $f$ if $a_n$ and $b_n$ are the Fourier coefficients of $f$. It may or may not converge or even if it does the sum may be different from $f$.
I want to show that if the trigonometric series does uniformly converge on $[-π,π]$ to $f$, then the trigonometric series is the Fourier series of $f$. My book does this by multiplying with $cos(nx)$ then integrating the new series term by term and then multiplying with $sin(nx)$ and then does the same. But how do we know that the new series will also be uniformly convergent? If it's not, then there is no reason for us to believe that we can integrate the series term by term.
Suppose the partial sums $S_N$ of the trigonometric series converge uniformly on $[-\pi,\pi]$ to $f$:
$$\tag{*}f(x) = \lim_{N \to \infty}S_N(x) =\frac{a_0}{2} + \sum_{k=1}^\infty (a_k \cos kx + b_k \sin kx).$$
By uniform convergence, since $S_N$ is continuous, the limit function $f$ is continuous and we can integrate the series termwise (after multiplying both sides of (*) by $\sin nx$ or $\cos nx$), to obtain
$$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos nx \,dx , \\ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin nx \,dx. $$
Obtaining this result uses the orthogonality of the trigonometric functions.
Thus, $a_n$ and $b_n$ are the Fourier coefficients.
Uniform convergence is sufficient here and, without uniform convergence, not every pointwise convergent trigonometric sum is a Fourier series.
For example, $$\sum_{n=2}^\infty \frac{ \sin nx}{\log n}$$
converges by the Dirichlet test but is not a Fourier series.